Theory of Special Relativity
Solved Problems
Michael Tsamparlis
e-mail: mtsampa@phys.uoa.gr
Contents
Mathematical Background: Chapter 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1
1 Mathematical Background: Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5
2 Classic Experiments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
7
3 The Position Four-Vector . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
9
Relativistic Kinematics: Theory for Chapters 4 and 5 . . . . . . . . . . . . . . . . . . 21
4 Four-Velocity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23
5 Four-Acceleration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31
6 Three-Momentum–Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35
Dynamics: Theory of Chapters 7–11 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41
7 Relativistic Optics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43
8 Four-Force . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49
9 Plane Waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53
10 Relativistic Reactions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55
Electromagnetic Field: Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71
11 Electromagnetic Field . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 75
iii
iv
Contents
Solutions
1 Mathematical Background: Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . 87
2 Classic Experiments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 97
3 The Position Four-Vector . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101
4 Four Velocity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 145
5 Four-Acceleration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 173
6 Three-Momentum – Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 189
7 Relativistic Optics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 211
8 Four-Force . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 229
9 Plane Waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 241
10 Relativistic Reactions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 243
11 Electromagnetic Field . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 315
Mathematical Background: Chapter 1
Spacetime or Minkowski space is a four-dimensional vector space endowed with
the Lorentz metric. The elements of the space are called four-vectors. A four-vector
is denoted by a kernel letter and a Latin index which takes the values 0, 1, 2, 3.
There are two types of four-vectors: the contravariant which we write as Ai and the
1
, x 2 ,⎞
x 3 ) the contravariant vectors are
covariant Ai . In a coordinate system Σ(x 0 , x ⎛
0
A
⎜
A1 ⎟
⎟
represented by column matrices, e.g., Ai = ⎜
⎝ A2 ⎠ and the covariant vectors by
A3 Σ
row matrices, e.g., Ai = (A0 , A1 , A2 , A3 ). The qualifying characteristic of a fourvector is that under the action of a Lorentz transformation it transforms linearly and
homogeneously. That is, if [L ii ] is the matrix representing a Lorentz transformation
between the coordinate frames Σ, Σ⎛ , and⎞if in the two frames
⎛ 0 ⎞the components of
A
A0
⎜ A1 ⎟
⎜ A1 ⎟
i
i
i
⎟
⎟
⎜
a contravariant vector A are A = ⎜
⎝ A2 ⎠ and A = ⎝ A2 ⎠ , respectively, it
A3 Σ
A3 Σ
holds
⎛
⎛ 0⎞
⎞
A0
A
1
⎜ A1 ⎟
⎜ A ⎟
⎜ ⎟ = [L i ] ⎜ 2 ⎟ .
i
2
⎝A ⎠
⎝A ⎠
A3 Σ
A3 Σ
The components of the covariant four-vectors are transformed as follows:
(A0 , A1 , A2 , A3 )Σ = (A0 , A1 , A2 , A3 )Σ [L ii ]−1 .
The length of a four-vector is defined by the invariant ηi j Ai A j , where η is the
Lorentz metric. The length of a four-vector can be > 0, < 0, or 0. We call the
four-vectors with negative length as timelike, with positive length spacelike, and
with zero length as null.
1
2
Mathematical Background: Chapter 1
In spacetime there are infinitely many coordinate systems. From all of them
we distinguish the ones in which the Lorentz metric has the reduced or canonical
form ηi j = diag(−1, 1, 1, 1). We call these coordinate frames Lorentz Cartesian
frames (LCF). In physics these frames are identified with the relativistic coordinate frames and will be called relativistic inertial frames (RIO). If Σ is an LCF,
i
then (and
⎞ then) the length of the four-vector A , which in Σ has components
⎛ 0only
A
⎜ A1 ⎟
i
⎟
A =⎜
⎝ A2 ⎠ , is given by the relation
A3 Σ
Ai Ai = −(A0 )2 + (A1 )2 + (A2 )2 + (A3 )2 .
⎞
⎛ 0⎞
A0
B
⎜ A1 ⎟
⎜ B1 ⎟
i
i
⎟
⎜ ⎟
More generally if A = ⎜
⎝ A2 ⎠ and B = ⎝ B 2 ⎠ are two four-vectors, then the
A3 Σ
B3 Σ
inner product (in Σ!)
⎛
Ai B i = −A0 B 0 + A1 B 1 + A2 B 2 + A3 B 3 .
In spacetime there are infinitely many types of coordinate transformations. Lorentz
transformations are special transformations which satisfy (among others) two basic
conditions:
(a) They are linear.
(b) They preserve the inner product of the four-vectors.
The geometric characteristic of Lorentz transformations is that they preserve the
canonical form diag(−1, 1, 1, 1) of the metric. Therefore, they relate an LCF with
an LCF or a RIO with a RIO. The same holds for the Galilean transformation in
Newtonian Physics.
Lorentz transformations constitute a group which we call the Poincaré group.
The Poincaré group has 10 parameters of which 4 correspond to translations in
spacetime and 6 to rotations. The rotations form a closed subgroup of the Poincaré
group which we call the Lorentz group. This latter has four components (parts) one
of which is a group because it is the only one which contains the identity. The other
three parts are not groups. We call the elements of the first part as the proper Lorentz
transformations. In the following we shall use these transformations only.
Consider two LCF Σ, Σ which are moving so that their spatial axes are parallel
and the relative velocity of Σ in Σ is u = βc. Then the proper Lorentz transformation relating Σ, Σ has the form
L(u) =
γ
−γ u μ
μ
γ u μ δνμ + γu−1
2 u uν
,
Mathematical Background: Chapter 1
3
⎞
ux
where u μ = (u x , u y , u z ) , u μ = ⎝ u y ⎠ , and δνμ +
uz
matrix:
⎛
⎛
⎜
⎜
⎜
⎜
⎝
1+
γ −1 2
ux
u2
γ −1
ux uz
u2
γ −1
ux u y
u2
1+
γ −1 μ
u uν
u2
γ −1 2
uy
u2
γ −1
u y uz
u2
1+
is the symmetric 3 × 3
⎞
⎟
⎟
⎟.
⎟
⎠
γ −1 2
uz
u2
In the special case, Σ, Σ are moving so that the axes y, y and z, z are kept
parallel whereas the x -axis slides along the x-axis we say that Σ, Σ are moving in
the standard configuration with relative speed u = βc. The Lorentz transformation
for this special motion is called boost and has the following representation:
⎡
γ
⎢ −βγ
L=⎢
⎣ 0
0
−βγ
γ
0
0
⎡
⎤
0
γ
⎢ βγ
0 ⎥
−1
⎥, L = ⎢
⎣ 0
0 ⎦
1
0
0
0
1
0
βγ
γ
0
0
0
0
1
0
⎤
0
0 ⎥
⎥.
0 ⎦
1
Consider a RIO Σ and another RIO Σ whose space axes are parallel to those of
Σ and moves with velocity u (wrt Σ). Suppose a four-vector Ai has components
⎞
A0
⎜ Ax ⎟
⎟
Ai = ⎜
⎝ Ay ⎠ ,
Az Σ
⎛
⎞
A0
⎜ Ax ⎟
⎟
Ai = ⎜
⎝ A y ⎠ ,
A z Σ
⎛
then the (proper) Lorentz transformation gives
A·u
,
A =γ A −
c
u
A·u
A = A−γ A0 + (γ − 1) 2 u.
c
u
0
0
In the special case of a boost the transformation equations read
A0 = γ (A0 − β A x ),
Ax = γ (A x − β A0 ),
Ay = A y ,
Az = A z .
These relations hold for an arbitrary four-vector!
4
Mathematical Background: Chapter 1
The basic four-vector in Special Relativity is the position four-vector x i of a relativistic mass point (ReMaP) say P. During the course of its existence the “motion”
of P traces a curve in Minkowski space which we call the world line of P. If in a
RIO Σ a ReMaP P at the time moment ct (of Σ!) has position vector r (in Σ!), then
the four-position vector of P in Σ has components:
xi =
ct
r
Σ
.
The length of the position four-vector x i x i = −τ 2 c2 defines the proper time of
P which is an invariant characteristic of P. Physically the proper time is identified
with the clock time in the proper frame Σ+P of P. This frame need not be a RIO.
The proper timeis related with the coordinate time t in a RIO Σ, in which P has
γ -factor γ = 1/ 1 − β 2 by the relation
t = γ τ > τ.
This relation is called time dilation formula.
Consider two points in the three-space of a ReMaP P which the proper moment
τ of P have spatial distance d P . Let dΣ be the difference of the spatial coordinate
distance of the two points in the RIO Σ in which P has factor γ . Then
dΣ =
dP
< dP .
γ
This relation is called the length contraction formula.
During the course of time there have been proposed many thought experiments
with the view to prove the inconsistency, hence the validity, of the Theory of Special Relativity. Most of them have to do with the length contraction and the time
dilation because the concepts of space and time are fundamental to our thinking.
These “experiments” have been called paradoxes and, contrary to their intension,
have established the validity and the consistency of the theory.
Chapter 1
Mathematical Background: Problems
⎛
(1)
(2)
(3)
(4)
⎞
e xc
⎜ xe x ⎟
⎟
In the LCF Σ the vector field Ai has components Ai = ⎜
⎝ 0 ⎠ . Compute the
x Σ
i
divergence A,i in the LCF Σ which moves wrt Σ in the standard configuration
with velocity u = kcx̂ (k < 1).
In the Euclidean Cartesian frame (ECF) {x μ , μ = 1, 2, 3} of E 3 a tensor of type
(0, 3) is defined as follows: Tμνρ = xμ2 + 2xν2 + xρ2 . Compute the div and the
curl of the vector Tμνμ .
Consider the Euclidean tensor Aμν = xμ2 + xν2 . Compute Aμν,ν and Aμν,μν .
Consider the two Lorentz Cartesian frames Σ(x, y, z) and Σ (x , y , z ) which
are related with the transformation
x 1 = − sinh φx 0 + cosh φx 1
x 2 = x 2,
x 3 = x 3,
x 0 = − sinh φx 1 + cosh φx 0
where φ is a real parameter.
(a) Show that the transformation is a Lorentz transformation.
(b) A four-vector V i in Σ(x, y, z) has components (0, 1, 0, 1)t . Compute the
components of V i in Σ (x , y , z ).
The only non-zero components in Σ (x , y , z ) of a tensor Ti j of type (0, 2)
are the T0 0 = T3 3 = 1. Compute the components of Ti j in Σ(x, y, z). Then
compute in Σ(x, y, z) the covariant vector Ti j V i and the invariant Ti j V i V j .
(c) Derive the results of (b) using matrix multiplication (the notation is obvious) as follows:
[V ]Σ = [L][V ]Σ , [Ti j ]Σ = [L −1 ]t [Ti j ]Σ [L −1 ], [a]Σ = ([T ]t [V ]Σ )t ,
ai V i = [a]Σ [V ]Σ .
5
6
1 Mathematical Background: Problems
(5) Consider the two-dimensional Lorentzian metric ds 2 , in which the coordinates
(t, x) has the form
ds 2 =
1
(−dt 2 + d x 2 ).
t2
Compute the geodesic equations and solve them. Comment on the type of
motion they describe.
(6) Determine the covariant Lorentz transformation using reasonable physical
assumptions.
Chapter 2
Classic Experiments
2.1. (a) The speed of a monochromatic beam of light propagating in a homogeneous and isotropic medium with refraction index n at rest equals v = nc < c.
When the medium is moving this formula is not valid. J. A. Fresnel (in 1817)
studied the speed of light in moving isotropic and homogeneous media and suggested that the speed of light in these media is given by the formula v = nc + κ V
where κ is a coefficient, which he called drag coefficient, and V is the projection
of the speed of the medium in the direction of propagation of light. The physical
explanation given by Fresnel was that the light was “dragged” from the moving medium. Years later (in 1891) he proposed the following experiment which
justified his explanation.
A monochromatic light source is producing a beam which by means of a
semi-transparent glass tile is split into two synchronous beams, each beam propagating in opposite directions. The two beams interfere in the telescope T due to
the difference in to the optical path. The displacement of the interference fringes
measures the number of the wavelengths. If N = Δλ/λ is the displacement
per wavelength, reduced, is the total distance of traveling of the beam in the
medium (water), v is the speed of the medium in the tube, and n the refraction
index (n = 1.33) of the medium, compute the drag coefficient.
(b) Using the relativistic composition rule for three-velocities justify the Fresnel
hypothesis. The explanation of the dragging of light by a moving medium without any extra hypothesis has been an additional experimental verification of
Special Relativity.
2.2. (a) A variation of the classical Fizeau experiment is the following. Two identical glass tubes are filled with water, sealed and are placed parallel to each other
on a horizontal circular table which can rotate about its center as in Fig. 2.1.
A source produces a monochromatic beam which is split by means of a halftransparent glass tile in two synchronous beams. One of the beams is directed
directly into the telescope and the other is reaching the telescope after it has
being propagated through the glass tubes by means of a system of reflectors.
The two beams interfere in the telescope creating interference fringes. When
the table is rotating with constant angular velocity about the perpendicular axis
7
8
2 Classic Experiments
Fig. 2.1 The Fizeau experiment
through its center the water in the tubes is moving with the result the difference
of the optical paths to change and consequently the interference fringes to be
displaced. Compute the change in the optical path when the angular speed of
the table is ω.
(b) If the dragging coefficient κ = 1 − n12 , = 20 cm, n = 1.33, R = 20 cm, and
λ = 5300 Å compute the angular speed ω so that the maximum difference of
the optical path is αλ (0 < α < 1).
Chapter 3
The Position Four-Vector
3.1 The Speed of Light
3.1.1. Consider two inertial Newtonian frames N , N and let v be the velocity of N relative to N . Let A be a Newtonian vector physical quantity which in N , N is given
by the expressions A, A , respectively. Then the action of the Galileo transformation
on the quantity A implies
A = A + vt,
where t is time in N (or N , since t = t ). Let B be another Newtonian vector physical quantity of the same nature as A and expression B, B in N , N , respectively.
(a) Show that under the action of the Galileo transformation
B − A = B − A.
Conclude that the Newtonian relative vector physical quantities (position, velocity, acceleration, momentum, etc.) are invariant (in the sense that they do not
change) under the action of the Galileo transformation.
(b) Based on (a) deduce that Newton’s law of the gravitational field between two
masses m 1 , m 2 gives
F1 2 =
Gm 1 m 2
(r2 − r1 ),
|r2 − r1 |3
and similarly the Coulomb’s law for charges q1 , q2 gives
F1 2 = −
K q1 q2
(r2 − r1 ),
|r2 − r1 |3
that is, F1 2 is covariant wrt the action of the Galileo transformation. Generalize to conclude that every law of Newtonian Physics, which is expressed in
terms of the relative position, relative velocity, etc., is covariant wrt the Galileo
transformation, hence compatible with the Galileo’s Principle of Relativity.
9
10
3 The Position Four-Vector
3.1.2. A strong laser beam which is rotating in the Earth about a perpendicular axis
with angular speed ω = 1 rotations/s reaches the Moon. If the Earth–Moon distance
equals 380×103 km, calculate the speed at which the light spot moves on the surface
of the Moon. Explain the result.
3.2 The Lorentz Transformation
3.2.1. A particle is moving around a circle on the x–y plane of a RIO Σ with constant angular speed. Describe the world line of the particle in spacetime.
3.2.2. The clocks 1, 2 which are synchronized at the event A (that is, t A = t A = 0)
start to move from the origin along the x-axis of the RIO Σ and are met again at the
event B. Clock 1 moves with constant speed υ0 so that
x = υ0 t, υ0 = constant > 0
and clock 2 moves with constant acceleration a0 so that
x=
1 2
a0 t , a0 = constant > 0.
2
Calculate the proper time measured by each clock during the transportation from
event A to event B. Compare the results and comment on the result.
3.2.3. It is given that the frames Σ (, x, y, z) and Σ ( , x , y , z ) are related by
the following linear transformation:
⎧ = a + bx + cy + dz
⎪
⎪
⎨ x = b + ax
,
⎪ y = y
⎪
⎩ z =z
where the coefficients are constants. Determine a, b, c, d so that the above transformation is the (general) Lorentz transformation relating Σ and Σ . Give a kinematic
interpretation of the transformation.
3.2.4. In the LCF Σ consider the coordinate transformation Σ (l, x, y, z) →
(u, υ, y, z) which is defined by the relations
1
u = √ (l + x),
2
1
υ = √ (l − x).
2
3.2
The Lorentz Transformation
11
(a) Compute the metric ds 2 = −dl 2 + d x 2 + dy 2 + dz 2 in the new coordinates
(u, υ, y, z).
∂2
∂2
(b) Write the box operator ≡ 2 − 2 2 in the new coordinates.
∂x
c ∂t
(c) Show that this coordinate transformation is not a Lorentz transformation. Comment on the result.
3.2.5. Consider two RIO Σ, Σ in Minkowski space whose space axes are parallel
and their relative velocity is u.
(a) Determine the Lorentz transformation between Σ, Σ assuming that
– Along the direction of the relative velocity Σ, Σ are related by a boost and
– The spatial directions perpendicular to the relative velocity u remain the
same.
(b) Discuss the limit of the transformation equations when β → 0 (Newtonian
limit).
(c) Determine the change of the temporal and the spatial differences under the
action of the Lorentz transformation for this general motion. Discuss the result.
3.2.6. Consider the Euclidean Cartesian frame Σ1 and a direction defined by the unit
vector n in Σ1 . Define a rotation R of the coordinates so that the x-axis becomes
parallel to the direction n. Use the result to derive the general Lorentz transportation.
3.2.7. The general Lorentz transformation between two RIO Σ, Σ with relative
velocity u for the position four-vector is given by the relations
l = γ l −
r·u
c
,
r = r − γ ut + (γ − 1)
(u · r)u
.
u2
Show that the second relation can be written as
r = γ (r − ut) + (γ − 1)
u × (u × r)
.
u2
Therefore, r consists of one part in the plane (r, u) and another part normal to the
plane (u, r).
3.2.8. Consider a transformation which in a Cartesian frame is represented by the
block matrix:
10
L=
,
0A
where A is a 3 × 3 matrix.
12
3 The Position Four-Vector
(a) Prove that L is a Lorentz transformation if and only if A is a Euclidean transformation.
(b) Show that these Lorentz transformations form a closed subgroup of the Lorentz
group. We call as these Lorentz transformations 3-Euclidean rotations. Consider a second-order tensor Ti j and show that under the action of these Lorentz
transformations its parts transform as follows:
T0 0 = T00 ,
Tμ 0 = ATμ0 ,
Tμ ν = ATμν At .
Deduce that under the action of three-Euclidean rotations a tensor transforms as
three independent quantities: An invariant (the T00 ), a Euclidean vector (the Tμ0 ),
and a Euclidean tensor of second order (the tensor Tμν ).
3.2.9. It can be shown that the proper Lorentz transformation L ∈ L+↑ leaves the
direction of a null four-vector invariant. That is, if α is a null four-vector (α α = 0)
then L +↑ (α ) = Aα , A > 0. A null tetrad consists of four linearly independent four vectors (α , m α , x α , y α ), such that the only non-vanishing inner products
among the various vectors are x α xα = y α yα = m α α = 1. It follows that a null
tetrad consists of two null vectors (the α , m α ) and two spacelike unit four-vectors
(the x α , y α ). Compute the action of the (proper) Lorentz transformation on the null
tetrad (α , m α , x α , y α ) assuming that the four-vector α is an eigenvector of the
(proper) Lorentz transformation.
3.2.10. The proper Lorentz transformation with beta factor β = (β μ ) is given by
the matrix expression
L +↑ (β) =
γ
−γβμ
μ
−γβμ δνμ + γβ−1
2 β βν
,
or equivalently by the relations
A0 = γ (1 − β · A),
A = A +
γ −1
(β
β2
· A)β − γ A0 β.
Define the dyadic φ(β) associated with the factor β by the relation
φ(β)μν = δνμ +
γ −1 μ
β βν .
β2
Obviously the quantity φ(β) is a Euclidean tensor (not a Lorentz tensor) of type
(1, 1) which in an arbitrary RIO Σ is represented by a 3 × 3 matrix. In terms of φ(β)
Lorentz transformation is written as follows:
3.2
The Lorentz Transformation
13
L +↑ (β) =
γ
−γβμ
−γβμ φ(β)μν
,
or equivalently
A0 = γ (1 − β · A)
A = φ(β)A − γ A0 β.
Consider two successive Lorentz transformations L +↑ (β 1 ), L +↑ (β 2 ) and show
that their composition L +↑ (β 2 )L +↑ (β 1 ) can be written in the form
L +↑ (β 2 )L +↑ (β 1 ) =
−γ12 β12μ
γ12
μ
−γ12 β21μ Γ12ν
,
where
γ12 = γ1 γ2 (1 + β 1 · β 2 ) = γ21 ,
β21μ =
1
1 + β1 · β2
β12μ =
1
1 + β1 · β2
μ
μ
β1μ
γ2 − 1
+ 1 + 2 (β 1 · β 2 ) β2μ ,
γ2
β2 γ2
β2μ
γ1 − 1
+ 1 + 2 (β 1 · β 2 ) β1μ ,
γ1
β1 γ1
ρ
μ
Γ12ν = φ2ρ φ1ν + γ1 γ2 β2 β1ν .
Show that
2
= β 12 · β 12 =
β12
(β 1 −β 2 )2 − (β 1 ×β 2 )2
2
= β21
.
(1+β 1 ·β 2 )2
Also prove that γ12 (= γ21 ) is the gamma factor for the β-factor β12 (= β21 ), that
1
is, γ12 = 1−β
2 .
12
From relation β12 = β21 deduce that β 12 , β 21 differ only in direction and show
that the angle ψ they form (in Euclidean three-space) is given by the relation
cos ψ = 1 −
2 sin2 φ
,
1 + 2A cos φ + A2
1/2
1 +1)(γ2 +1)
2
where A = (γ
(1 ≤ A ≤ ∞) and cos φ = ββ11·β
is the angle between
(γ1 −1)(γ2 −1)
β2
β 1 , β 2 . Considering the angle ψ as a rotation of β 12 → β 21 determine the direction
ê of the axis of rotation.
14
3 The Position Four-Vector
Using the fact that the composite Lorentz transformation can be decomposed in
the product of a proper Lorentz transformation and a Euclidean rotation R shows
μ
that the Euclidean tensor Γ12ν of the composite transformation can be written as
follows: Γ = φ(β 21 )R, that is,
r = φ(β 21 )ρ − γ A0 β 21 ,
A0 = γ12 (1 − ρ·β 21 ),
where ρ = RA and φ(β 21 ) = 1 + γ12β 2−1 β 21 ⊗ β 21 .
12
Prove that the rotation R is given by the following formula:
R(ê, ψ) = 1 + ω1 (β 1 ⊗ β 2 + β 2 ⊗ β 1 )
+ ω2 (β 1 ⊗ β 1 + β 2 ⊗ β 2 ) + ω3 (β 1 ⊗ β 2 − β 2 ⊗ β 1 ),
where ω1 = ω2 A + ω3 and
ω2 =
2 cos φ
,
1 + 2A cos φ + A2
ω3 =
−2
sin ψ
.
=
1 + 2A cos φ + A2
cos ψ
(3.1)
3.3 The Boost
3.3.1. In the RIO Σ(x) the four-vector Ai has components Ai = (0, 0, sin(5x 1 −
3x 0 ), sin(5x 1 − 3x 0 )). Calculate the components of Ai in the RIO Σ (x ) which is
moving wrt Σ in the standard configuration with speed u.
3.3.2. A point P leaves the origin of a RIO Σ at the time t = 0 of Σ. After proper
time Δτ of P a light signal is sent toward the origin of Σ and reaches the origin at
the moment t of Σ. Calculate the speed of P in Σ.
3.3.3. Consider two RIO Σ, Σ which are related in the standard configuration with
speed βc. Show that for every moment in Σ there exists only one plane in spacetime
on which the clocks ofΣ coincide with those of Σ . Show that this plane moves in
1/2
2
2
. Show that this result stays valid even if
Σ with speed u = cυ 1 − 1 − υc2
the origins of Σ, Σ change arbitrarily.
3.3.4. In order tostudy the boost
it is enough to study its action on the basis
1
0
vectors ê0 =
, ê1 =
(because it is a linear transformation). Consider
0
1
γ βγ
L=
and show that
βγ γ
3.4
Length Contraction
15
1
γ
=
0 βγ 0
βγ
=L
=
.
1
γ
ê0,L = L
ê1,L
Show that the Lorentz length of e0,L , e1,L equals ±1 whereas their Euclidean
length is ≥ 1. Find the locus of the vectors e0,L , e1,L considered as Euclidean vectors
on the Euclidean plane!
3.3.5. The RIO Σ, Σ are moving in the standard configuration with speed 3c/5.
Draw on the Euclidean plane (x, ) of Σ the axes x , of Σ .
3.3.6. (a) Show that if two events A, B are simultaneous and occur at different
three-space points in some RIO Σ, then there does not exist a limit in the time
difference of the events in another RIO Σ , whereas their spatial distance varies
from infinity to a minimum which is determined from their spatial distance in Σ.
(b) If two events A, B occur at the same point in the three-space of a RIO Σ show
that their temporal order (that is, the sequence at which they occur or, equivalently,
their causal relation) is the same for all RIO.
3.4 Length Contraction
3.4.1. A particle is produced in the laboratory whose lifetime is 10−6 s with constant
velocity and speed 2.7 × 108 m/s.
(a) Compute the lifetime of the particle in the laboratory.
(b) Compute the distance traveled by the particle in the laboratory before it disintegrates.
(c) Compare the results with the corresponding Newtonian results.
3.4.2. Two particles A, B are moving along the x-axis of the LCF Σ, each with
velocity factor β when they hit a standing wall (in Σ) with a time difference Δt.
Calculate the spatial distance of the particles in their proper frames.
3.4.3. The points A, B are at a distance d along the x-axis of the LCF Σ. At some
moment in Σ a light signal is emitted simultaneously from each point toward the
other. Show that in the LCF Σ which is moving in the standard way wrt Σ with
speed u, the emittance of the light signals is not simultaneous and more specifically
and when the second
the first emittance is done earlier at a time interval γ (u)ud
c2
c−u
emittance occurs the distance of the points in Σ is d c+u .
16
3 The Position Four-Vector
3.4.4. In the LCF Σ two light signals are moving along the x-axis so that their
distance d remains constant. Calculate the distance d of the signals in an LCF Σ
which is moving wrt Σ in the standard configuration with speed u.
3.4.5. A mass m of a gas of molecular weight k rests in a cylindrical container
of base radius R and height L. Calculate the average number of molecules per cm3
(this is called the number density) which is counted by an observer Σ who is moving
with speed v wrt the cylinder and in direction (a) parallel to the axis of the cylinder
and (b) normal to the axis of the cylinder.
3.5 Time Dilation
3.5.1. A photon is emitted from the origin and in the x y plane of the LCF Σ and
subsequently it is absorbed at the point (r, θ ) where r, θ are polar coordinates in
the x y plane of Σ. Another LCF Σ is moving in the standard configuration wrt Σ
with speed u. Calculate the time difference of the emittance and the absorption of
the photon in Σ as well as the spatial polar coordinates in Σ of the point where
absorption occurs.
3.5.2. A π -meson is moving in the outer atmosphere of the Earth when it disintegrates into a μ-meson of speed β = 0.985 and a neutrino. If before its disintegration
the μ-meson moves a distance d = 4 km wrt the observer in the Earth, calculate the
lifetime of the μ-meson.
3.5.3. The π -meson has a lifetime τ0 = 2.6 × 10−8 s. Consider a beam of N0
π -mesons of speed β in the LCF Σ and determine the distance l traveled by these
particles in Σ at which half of the particles have disintegrated. Plot l(β).
3.5.4. A beam of negative pions passes through a target of liquid hydrogen. With a
system of measuring devices, a current of charged particles along the initial direction of the beam of pions is found. One of the pions interacts with a proton of the
target at the point whose coordinates in the laboratory are x = −0.500 m, y =
0.500 m, z = 0.200 m, t = 0.000 s. At this event, which we call event 1, a
K -meson is produced. The K -meson is neutral and propagates with constant velocity until it disintegrates into two charged pions (K → π + π − ). This disintegration,
which we call event 2, in to the laboratory has coordinates x = 0.880 m, y =
0.480 m, z = 0.250 m, t = 5.31 × 10−9 .
(i) Calculate the lifetime of the K -meson.
(ii) Calculate the speed of K -meson in the laboratory.
3.5.5. An ideal clock is realized by fixing two perfect mirrors along a rod and let a
photon be reflected elastically between the mirrors. The unit of time (the “second”)
3.5
Time Dilation
17
is measured by a successive reflection of the photon on the same mirror. If the rod
slides along the x-axis of an LCF Σ with velocity factor β, calculate the unit of time
in Σ.
3.5.6. The Feynmann clock (see Fig. 3.1) consists of an electric bulb Λ which emits
a beam of photons which is elastically reflected on a plane mirror and is received by
the photocell Φ. The photocell activates with a special electronic device with zero
activation time again the bulb Λ when a new beam of photons is emitted and so on.
α
y
d
V
F
x
Fig. 3.1 The Feynmann clock
Let Σ be the “proper” observer of the clock and let d1 =
distance of the reflecting mirror from Φ, Λ in Σ.
d
cos α
be the spatial
(i) Consider an LCF Σ which is moving in the standard configuration along the
x-axis with speed u and show that in Σ the clock is slow (time dilation).
(ii) Repeat the calculations if Σ moves in the standard configuration along the
y-axis.
3.5.7. A beam of unstable particles is propagating with speed u = 35 c along the
x-axis of the LCF Σ. Along the path of the beam there have been placed two counters A, B at a distance of 9 m. The first counter measures a flow of 1000 particles per
second and the second counter 250 particles per second. Assuming that the reduction in the number of particles is due to decay only, calculate the half-life1 time
of the particles and their half life-time in an LCF Σ , which moves in the standard
configuration normal to the direction of the beam with speed v = 35 c.
1 The half-life time is the time required in the rest frame of the radioactive source for half the
number of particles to decay.
18
3 The Position Four-Vector
3.6 Problems with Rods
3.6.1. In the LCF Σ two light signals are moving along the x-axis so that their
distance d remains constant. Calculate the distance d of the signals in an LCF Σ
which is moving wrt Σ in the standard configuration with speed u.
3.6.2. A particle is produced in the laboratory whose lifetime is 10−6 s with constant
velocity and speed 2.7 × 108 m/s.
(a) Compute the lifetime of the particle in the laboratory.
(b) Compute the distance traveled by the particle in the laboratory before it disintegrates.
(c) Compare the results with the corresponding Newtonian results.
3.6.3. A rod is moving along the x-axis of an LCF Σ with constant speed u. It is
measured that in Σ the rod requires a time interval T to pass in front of a fixed point
P of the x-axis, whereas the corresponding time interval in the proper frame Σ−
of the rod is T . Calculate the proper length L 0 of the rod. This experiment shows
that one can compute relativistic spatial distances using Newtonian experimental
methods.
3.6.4. A rod of (proper) length l is moving with velocity factor β along the positive
x-axis of an LCF Σ. A particle moves in the opposite direction along the x-axis with
velocity factor β. Compute the time needed (in Σ) in order for the particle to travel
across the rod.
3.6.5. A spaceship 2 of length 2 moves antiparallel to another spaceship 1. The
observer at the end of spaceship 1 observes that it takes time T1 in order the spaceship 2 to pass in front of him.
(a) Determine the relative speed u of the spaceships.
(b) If the observer at the end of spaceship 2 measures that it takes time T2 for
spaceship 1 to pass in front of him, calculate the length 1 of spaceship 1.
3.6.6. Two rigid rods 1,2 of length L 1 = L 2 = L move along the x-axis of the LCF
Σ toward each other with common speed u. Determine the length of each rod in the
frame of the other.
3.6.7. From the end points A, B of a train of proper length l which is traveling with
constant speed β are emitted two light signals toward its midpoint M.
(a) For the observer Σ in the ground the light signals arrive simultaneously at M.
Compute the time difference of the emittance of the signals for the observer Σ
in the train (i) algebraically and (ii) geometrically.
3.6
Problems with Rods
19
(b) If the light signals are emitted simultaneously for the observer in the train what
is the time difference at which they reach the middle M of the train for the
observer in the ground?
3.6.8. In the x y plane of the LCF Σ a rod of proper length l slides with speed u so
that one of its end points is always on the x-axis while the rod is making a constant
angle θ with the x-axis in the proper frame of the rod. Find the length of the rod in Σ.
3.6.9. A rod AB is moving in the x y plane of the LCF Σ with velocity v =
(vx , v y , 0). Another LCF Σ is moving wrt Σ in the standard way with velocity
u = ui. If at the moment t of Σ the projection of the end points of the rod in Σ and
Σ are l x , l y and l x , l y , respectively, show that
l x = γ (u) l x 1 − vcx2u
,
v u
l y = l y − γ (u) cy2 l x
where
−1/2
u2
γ (u) = 1 − 2
.
c
3.6.10. A rod is resting along the x -axis of the LCF Σ when it starts sliding along
the x -axis with constant velocity u = uj . Show that in the LCF Σ in which Σ is
moving in the standard configuration with speed v, the rod appears to make an angle
with the x-axis.
θ = tan−1 γvcuv
2
3.6.11. A right circular cone of angle 2θ and contiguous surface area S0 is resting
with its axis along the x-axis of the LCF Σ (see Fig. 3.2). Another LCF Σ moves
wrt Σ in the standard way with speed β. Calculate the angle 2θ and the contiguous
surface S0 of the cone in Σ . Discuss the Newtonian and the ultra-relativistic limit
of the results.
yy
y
v
y’
x,x’ x x
Fig. 3.2 The right circular cone
20
3 The Position Four-Vector
3.6.12. A rod AB slides along the x-axis of the LCF Σ with speed β ( c = 1). At
the same moment in the proper frame of the rod two light signals are emitted toward
the origin of Σ. If the proper length of the rod is L 0 calculate the length of the rod
L̃ which sees (not measures!) the observer at the origin of Σ. How this length varies
with the speed of the rod in Σ? Compare L̃ with the measured length of the rod
in Σ.
Relativistic Kinematics:
Theory for Chapters 4 and 5
Consider a relativistic mass point (ReMaP) P with position vector x i and proper
time τ . The four-velocity and the four-acceleration of P are defined by the relations:
ui =
dxi
,
dτ
ai =
du i
.
dτ
The four-vectors u i , a i satisfy the conditions:
u i u i = −c2 , u i ai = 0, a i ai = (a+ )2 ,
where a+ is the proper acceleration of P.
ct
, three-velocity
Consider a LCF Σ in which P has position vector x =
r Σ
v, and three-acceleration a. Then the four-velocity and the four-acceleration of P
have components
i
ui =
γc
γv
Σ
,
ai =
ca0
a0 v + γ 2 a
Σ
,
(1)
where a0 = c12 γ 4 v · a = c12 γ 4 v · a = γ γ̇ .
We call the special one-dimensional motion with constant proper acceleration
a+ as hyperbolic motion. This motion is covariant, that is, if P executes hyperbolic
motion in one LCF Σ then it does so in all LCF.
Consider two LCF Σ, Σ with parallel axes and relative velocity u and suppose
that in Σ, Σ the ReMaP P has three-velocities v, v and three-accelerations a, a ,
respectively. Then under the proper Lorentz transformation which relates Σ and Σ
the four-vector of the four-velocity u i transforms as follows:
u · v
γv = γu γv c 1 − 2 ,
c
γu v · u
1
v =
v+
− 1 γu u ,
γu Q
γu + 1 c 2
where Q = 1 −
(2)
(3)
u·v
.
c2
21
22
Relativistic Kinematics: Theory for Chapters 4 and 5
The angle θ between the three-velocities v, u is
tan θ =
v sin θ γu (v cos θ + u)
(4)
and the speed is
v =
1
2
1+
uv cos θ 2
c2
!
u 2v2
2 (v ) + u + 2v u cos θ − 2 sin θ .
c
2
2
(5)
Concerning the four-acceleration of P we have the relation
u · v
u·a
u·a 1
a =
−1 u
3 1 − 2 a + 2 v + 2
c
c
u
γu
γu2 1 − u·v
c2
1
(6)
and for the magnitude
"
#
+ 2
= γ 6 a6 − (a × β)2 .
a
(7)
In case Σ, Σ are moving in the standard configuration with speed u = βc the boost
relating Σ, Σ gives for the four-velocity:
uυ γυ = γu γυ 1 − 2x ,
c
−
u
υ
x
υx =
,
uυ 1 − c2x
υy
,
υ y =
uυ γu 1 − c2x
υz
.
υz =
uυ γu 1 − c2x
(8)
The last three relations have been named Relativistic Law of composition of threevelocities. Concerning the four-acceleration the boost gives
1
1−
1
a y =
γu2 1 −
1
az =
γu2 1 −
ax =
γu3
ax ,
uvx 3
2
c
uv y uvx a
1
−
ax ,
+
y
uvx 2
c2
c2
c2
uvx uvz a
1
−
ax .
+
z
uvx 2
c2
c2
2
c
(9)
Chapter 4
Four-Velocity
4.1. The LCF Σ moves with velocity u wrt the LCF Σ. A particle P moves in Σ
with velocity v , which makes an angle θ with the direction of u (in Σ ). Calculate
(a) The velocity v of P in Σ.
(b) The angle θ between v and u in Σ.
(c) The speed v2 of P in Σ.
4.2. (a) Consider the vectors A and u and show the identity/decomposition
A=
A·u
1
u + 2 u × (A × u).
2
u
u
Define A = A·u
u and A⊥ = u12 u × (A × u) and write A = A + A⊥ .
u2
(b) A relativistic mass point has velocity v1 , v2 in the LCF Σ1 , Σ2 , respectively.
Let u the velocity of Σ2 wrt the LCF Σ1 . Prove the relations
#
1 u · v1
1 "
−
1
u=
(v1 ) − u ,
2
Q
u
Q
1 u · v1 1
v1 −
u =
(v2 )⊥ =
(v1 )⊥ ,
γu Q
u2
γu Q
(v2 ) =
where the and ⊥ refer to the direction of the velocity u and Q = 1 −
(4.1)
(4.2)
u·v
.
c2
4.3. The LCF Σ, Σ have parallel axes and Σ is moving wrt Σ with velocity u. A
relativistic particle P has three-velocities v, v in Σ and Σ , respectively.
(a) Show that
v + [
v=
γ u · v
+ 1] γ u
γ + 1 c2
.
u · v
γ (1 + 2 )
c
23
24
4 Four-Velocity
(b) If u is parallel to the x-axis, that is, we have a boost, and the velocity v has
components in the plane x − y only, show that
tan θ =
v sin θ ,
cos θ + u)
γ (v where u, v are the speeds of u, v and θ, θ are the angles made by v, v with
the axes x, x , respectively. Show that for photons this relation becomes
tan θ =
sin θ .
γ (β + cos θ )
Finally, show that the speeds are related as follows:
v2 =
1
γ 2 (1 +
1
[v 2 + u 2 + 2uv cos θ − β 2 v 2 sin2 θ ].
cos θ )2
uv c2
4.4. (a) Show that the rapidity α of a particle in the LCF Σ is related with the β
factor of the particle in Σ with the relation
1
1+β
.
φ = ln(γ + γβ) = ln
2
1−β
(b) Let (l, x)Σ , (l , x )Σ be the components of the position vector of a particle in the
LCF Σ, Σ which are related with a boost L(β). Show that
x + l = eφ (x + l).
(c) Show that the rapidity of successive boosts in the same direction equals the sum
of the rapidities of each boost.
4.5. In the LCF Σ the four-vector Ai has components Ai =
A0
A
.
Σ
(a) Calculate the β-factor of Ai in Σ if Ai is timelike.
(b) What happens when Ai is spacelike?
(c) A particle is moving in the LCF Σ so that its four-velocity is u =
i
Calculate the β-factor of the particle in Σ.
2
.
i+j
4.6. The spaceships A, B are moving in the same direction along the x-axis of the
LCF Σ with speeds u, υ, respectively. At some moment a light signal is emitted from
A, reflected at B, and then returns to A after a time period T. If the spatial distance
of the two spaceships at the moment of emittance of the light signal according to A
is l, determine the speed of A in terms of υ, l, T .
4 Four-Velocity
25
4.7. An LCF Σ is moving with velocity u wrt the LCF Σ . We consider a particle
with velocities v, v in Σ, Σ , respectively, and direction angles θ and θ with the
direction of the relative velocity u. Show that the rapidities ξ , ζ , ζ which correspond
to the velocities u, v, u satisfy the relations
tanh ζ sin θ
,
cosh ξ (tanh ζ cos θ − tanh ξ )
cosh ζ = cosh ξ cosh ζ − sinh ξ sinh ζ cosh θ.
tan θ =
4.8. The observer Γ is moving away from the observer A with speed 0.6 c along
a common direction x while observer B approaches A along the same direction
with speed 0.8 c (see Fig. 4.1). Γ is emitting continuously a beam of green light
(λ = 0, 54 μm) at an angle 170◦ with the direction of its motion. The beam is
reflected (elastically) at B and is deflected to A.
Calculate
(a) The relative speed of B, Γ.
(b) The angle with the x-direction and the wavelength of the radiation received
by A.
°
Fig. 4.1 The path of the light beam
4.9. In the laboratory two beams of particles with velocities v1 , v2 are produced.
(a) Calculate the relative speed of the beams.
(b) In the ring of colliding beams (SR) at CERN a researcher wants to produce
beams which have the maximum relative speed. Assuming that the beams have
the same speed in the ring, calculate the relative orientation of the beams which
the researcher must employ.
4.10. In the laboratory frame two electrons have kinetic energy 1MeV each and are
moving in opposite directions along the x-axis. Calculate
(a) The speed of each electron in the lab system.
(b) Their relative velocity.
(Mass of electron m = 0.51 MeV/c2 ).
4.11. The relativistic observers 1, 2 are moving along the positive semi-axis O x, O y
of the LCF Σ with velocities v and kv, respectively. If 1 sees 2 to move in a direction
26
4 Four-Velocity
which makes an angle φ with the positive semi-axis O x, calculate v as a function
of k. Discuss the result for various values of the parameter k.
4.12. (a) A particle P of energy E and mass m is moving along the positive semiaxis y of the LCF Σ. An observer O who moves along the negative semi-axis x of Σ
sees P to slide in a direction which makes an angle φ with the positive semi-axis x.
(a) Calculate the speed of the observer in terms of the angle φ. Also calculate the
energy of P measured by the observer.
(b) Solve the problem in case m = 0, that is, P is a photon.
4.13. The LCF Σ2 moves in the standard configuration wrt the LCF Σ1 with velocity v21 = v1 i. Consider a third LCF Σ3 which moves in the standard way wrt Σ2
along the common y-axis with velocity v32 = v2 j2 . Suppose that the velocity u13 of
Σ3 wrt Σ1 makes an angle θ1 with the axis x1 of Σ1 and θ3 with the x3 -axis of Σ3 .
(a) Show that
tan θ1 =
v2
,
v1 γv1
tan θ3 =
v2 γv2
.
v1
(b) Compute the deviation δθ = θ3 − θ1 in terms of the velocity factors β1 , β2 .
Show that when β1 , β2 1 (Newtonian limit), δθ = 12 β1 β2 .
Application
In the LCF Σ1 a particle moves in a circular orbit with constant speed β 1.
Consider two adjacent positions of the particle along its orbit, the 2, 3 say, and let
Σ2 , Σ3 be the instantaneous LCF at the points 2, 3, respectively. Choose Σ2 , Σ3
so that their space axes are parallel (in the Euclidean sense!) and show that after a
complete rotation of the particle in Σ1 the proper frame of the particle has deviation
πβ 2 in a direction opposite to the three-velocity of the particle. (The phenomenon
of the braking of Euclidean parallelism after a complete rotation is called Thomas
effect.)
4.14. Consider the linear three-dimensional space R3 whose elements are the threevelocities of the relativistic particles, i.e., the position vector of each point of the
space is determined/represents a three-velocity. Let us write the four-velocity v i of
a relativistic particle in the LCF Σ, Σ as follows:
γc
γc
vi =
,
.
γ v Σ
γv Σ
If the relative velocity of Σ, Σ is u show that v and v are related as follows (this is
the effect of the Lorentz transformation in the three-velocity space):
v =
v+u
v·u .
1+ 2
c
(4.3)
4 Four-Velocity
27
Introduce in the three-velocity space E3 spherical coordinates (v, χ , ψ), where v is
the length of v (the speed) and the angles χ , ψ are defined by the relations (Fig. 4.2).
vx = v cos χ , v y = v sin χ cos ψ, vz = v sin χ sin ψ,
(4.4)
so that
v = v cos χ i + v sin χ cos ψj + v sin χ sin ψk .
Fig. 4.2 Spherical coordinates in the velocity 3-space
(a) Introduce a new coordinate system (v , χ , ψ ) using the transformed velocity v
and relations (4.4) and write (4.3) as a coordinate transformation (v, χ , ψ) →
(v , χ , ψ ) assuming that Σ , Σ are moving in the standard configuration with
velocity u = ui.
(b) Compute the Newtonian result by considering u/c → 0. Discuss the result in
the three-velocity space E3 considering the v, v , u as position vectors.
(c) Show that if v c then v c (assuming always that u c) and when v = c
then v = c for all u < c.
(d) Show that for each value of the angle χ the angle χ diminishes as the relative
speed of Σ, Σ increases. Discuss the case of the photons (v = v = c) and
show that in this case the following relation holds:
tan χ =
1 sin χ .
γ β + cos χ Show that this relation can be written equivalently:
χ
tan =
2
$
χ
1−β
tan .
1+β
2
Deduce that for each value of χ the angle χ diminishes as the relative speed u
(β → 1) of Σ, Σ increases.
28
4 Four-Velocity
ct1
4.15. Two particles 1, 2 in the LCF Σ have position vectors x1i =
, xi =
r1 Σ 2
ct2
cγ1
cγ2
i
i
and four-velocities v1 =
, v2 =
, respectively. Let
r2 Σ
γ1 v1 Σ
γ2 v2 Σ
Σ2 the proper frame of particle 2 and
that the position four-vector and the
suppose
ct12
i
12
, v = dr
.
three-velocity of 1 in Σ2 are x1 =
dt12
r12 Σ 12
2
(a) Calculate v12 in terms of v1 , v2 .
(b) Show that v12 = −v21 , while |v12 | = |v21 |. Calculate the angle between the
three-vectors v12 , v21 .
4.16. Show that the γ -factor of an observer with four-velocity u i wrt another
observer with four-velocity v i is given by γ (u, v) = − c12 u i vi . Consider a third
observer with four-velocity wi and compute the γ -factor γ (u, w) of u i wrt wi in
terms of the γ -factors γ (u, v) and γ (v, w).
4.17. (a) Show that two successive boosts in the same direction with speeds u 1 , u 2
are equivalent to a single boost in the same direction and with speed
u=
u1 + u2
u1u2 .
1+ 2
c
(b) Consider the boost along the x-direction with speed u 1 and the boost in the
y-direction with speed u 2 . Show that the result of composition of the two boosts
depends on the order they are composed (the composition is not commutative).
Also show that the composition of these boosts is not a boost in either case.
4.18. A relativistic particle in the LCF Σ has four-velocity u =
i
the rapidity a(∈ R) of the particle in Σ with the relation
γc
γv
. Define
Σ
tanh a = β.
(a) Show that
γ = cosh a,
βγ = sinh a
i
i
and conclude
that the four-velocity u in terms of the rapidity is written as u =
c cosh a
, where ê is the unit in the direction of the three-velocity. Let θ, φ
c sinh a ê Σ
be the angles made by ê with the x-axis and the z-axis of Σ. Then show that
4 Four-Velocity
29
ê = sin θ cos φi + sin θ sin φj + cos θ k
and use this to write u i in the form
⎛
⎞
c cosh a
⎜ c sinh a sin θ cos φ ⎟
⎟
ui = ⎜
⎝ c sinh a sin θ cos φ ⎠ .
c sinh a cos θ
Σ
(b) Consider two LCF Σ1 , Σ2 with parallel axes which are moving with relative
velocity u and a relativistic particle which in Σ1 , Σ2 has four-velocity
ui =
c cosh a1
c sinh a1 ê1
Σ1
,
ui =
c cosh a2
c sinh a2 ê2
Σ2
.
Calculate the relation between a1 , a2 . Examine the special case when Σ1 , Σ2
are moving in the standard configuration.
% cos,
(c) Define the trigonometric functions cos,
% sin,
% c%
ot with the relations
cosi
% x = cosh x
% x = sinh x
sini
cosi
% x = tanh x
c%
oti x = coth x .
Show that these functions satisfy the same relations as the real trigonometric
functions sin, cos, tan, cot. Show that if in the complex plane we consider a
triangle with sides ia0 , ia1 , ia2 and corresponding angles x0 , x1 , x2 as in Fig.
4.3, then the relations which have been found in the (b) part express the sine
and the cosine law for this triangle.
Such triangles and angles are the subject of spherical Euclidean geometry.
The space of this geometry is the Euclidean sphere (= space of two dimensions
with constant curvature equal to 1) whereas the classical Euclidean geometry is
based on the Euclidean flat space (space of constant curvature equal to 0).1
1 The space whose points are the three-velocities (the three-velocity space) is a space of constant
curvature equal to −1. Such a space is called hyperbolic. The difference between spherical, flat,
and hyperbolic geometry is measured with the defect ε, which is defined by the relation ε =
π − (x0 + x1 + x2 ). In Euclidean geometry ε = 0 while in spherical geometry ε is negative and its
absolute value equals the area of the triangle (012). In hyperbolic geometry ε is positive and can
be shown that it is given by the relation
1
(cosh a2 − 1)(cosh a0 − 1) 2
ε
sin =
sin x2
2
2(1 + cosh a1 )
(and cyclically for the three tips of the triangle).
The non-vanishing of ε is due to the curvature of the space and it is a measure of deviation from
the Euclidean geometry. Note that ε = 0 implies that a spherical or hyperbolic triangle does not
“close” in flat Euclidean geometry.
30
4 Four-Velocity
Fig. 4.3 The triangle in spherical Euclidean geometry
4.19. Consider in spacetime the timelike straight line2 z 1 i (τ ) = a1 i + u 1 i τ , where
a1 i , u 1 i are constant four-vectors, u 1 i u 1i = −1, and τ is an affine parameter along
the line. Let P be a spacetime point outside the straight line. As it is well known at
every point of spacetime there exists a unique light cone with vertex at that point.
Consider the light cone with vertex at P and let R and A be the intersection points
of this light cone with the straight line z 1 i (τ A > τ R ).
(a) Calculate the values τ R , τ A .
(b) Consider the point Q of z 1 i defined for the parameter value τaver. = 12 (τ R +
τ A ) and show that Q P i is perpendicular to z 1 i ; therefore, its length defines the
minimum distance of P from the straight line z 1 i .
(c) Let z 2 i (τ2 ) = a1 i + u 2 i τ2 (u 2 i u 2i = −1) be another timelike straight line which
intersects z 1 i (τ ) at the point a1 i . Assume that P is the point of z 2 i defined by
the parameter value τ2,P , that is,
z P i = a1 i + u 2 i τ2,P .
Compute the normal P Q i and its length in terms of the four-vectors u 1 i , u 2 i
and the value τ P . Use these results to the case that z 1 i , z 2 i are the world lines of
two relativistic inertial observers (RIO) with four-velocities u 1 i , u 2 i , respectively,
to show that the measure of the relative velocity of these RIO is defined covariantly
and in an intrinsic geometric manner.3
2
The results are general; therefore, we write gi j instead of ηi j .
3
See A. Aurilia and F. Rohrlich Am. J. Phys. (1975) 43, 261–264.
Chapter 5
Four-Acceleration
5.1. A relativistic mass point (ReMaP) P is moving along the x-axis of the LCF
Σ.
(a) Assuming that P moves in an arbitrary manner compute the four-velocity and
the four-acceleration of P.
(b) Consider the initial conditions v(0) = 0, x(0) = x0 and calculate the motion for
the case of hyperbolic motion (i.e., a+ = constant).
(c) Assume a + = 2ckτ (k positive constant) and compute the motion of P.
5.2. A relativistic mass point P moves so that its position vector x i is related with
its four-acceleration a i with the relation a i = k 2 x i (k > 0).
(a) Show that during the motion x i is at all times normal to the four-velocity vector
u i (this “motion” in spacetime is the analogue of uniform circular motion).
(b) Calculate the equation of motion of P.
(c) Compute the motion of P in an LCF Σ if P starts from rest from the point x0
of x-axis.
5.3. The world line of a relativistic mass point P is described by the equations
x 2 − c2 t 2 = a 2 , y = z = 0.
(a) Show that a parametric form of the world line is
ct = a sinh φ,
x = a cosh φ,
y = z = 0.
What is the kinematic meaning of the parameter φ and how it is related with the
proper time of P?
(b) Compute the four-velocity and the four-acceleration of P and find how the later
is related to the position four-vector of P. Comment on the result.
31
32
5 Four-Acceleration
(c) If the mass of P is m show that the necessary (inertial) four-force to sustain
the motion of P is F i = am2 x i . Does the energy of P varies during its motion
(c = 1)?
5.4. The relativistic point P moves in spacetime so that its four-velocity u i and the
i
= a 2 u i (aε R).
four-acceleration a i are related by da
dτ
(a) Show that a 2 is the length of the four-acceleration.
(b) Consider an arbitrary LCF Σ in which P has three-velocity u and threeacceleration a and show that in Σ the covariant equation of motion is equivalent
to the following two equations:
(γ γ̇ ) = a 2 ,
(γ 3 a)· = 0.
Solve the equations of motion in case the motion is taking place along the x-axis.
5.5. A spaceship departs from the origin of the LCF Σ and moves along the x-axis
with constant proper acceleration a + = 10 m/s2 until it attains the speed v1 = 0.5 c.
Calculate the time the√spaceship moved in Σ and in its proper frame. Repeat the
calculations for v1 = 0.9999 c and a + = 20 m/s2 .
Assume c = 3 × 108 m/s and that the year has 365 days.
5.6. A relativistic mass particle P moves along the x-axis of the LCF Σ with constant proper acceleration a + .
(a) Show that the acceleration of P in Σ is not constant and it is given by the
relation γu3 ax = a + , where u is the instantaneous speed of the particle in Σ.
(b) An elastic string brakes when its proper frame is elongated to a length equal
twice its natural length. Assume that the string is resting along the x-axis of the
LCF Σ along its natural length when all its points start accelerating with the
same proper acceleration a + . Calculate (in Σ) the time required for the string to
brake.
5.7. In the LCF Σ a particle moves uniformly along a circular trajectory of radius
R with period T. Calculate the period and the distance covered for one complete
rotation in the proper frame of the particle.
Application
In a space platform an astronaut starts a journey along a circular path of radius R
with uniform speed v. When the astronaut returns back to the basis will he be older
or younger from another astronaut who stayed in the basis during the trip?
5.8. (a) The four-acceleration of a particle P whose three-velocity
and three
a0 c
i
acceleration in the LCF Σ are v, a, respectively, is given by a =
a0 v + γ 2 a
5 Four-Acceleration
33
where a0 = γ 2 + v·a
. Let a+ be the proper acceleration of P. Show the relations.
c2
a =
1 +
a ,
γ3 a⊥ =
1 +
a ,
γ2 ⊥
where the parallel and the perpendicular parts are wrt the direction of the threevelocity v.
(b) A particle P of charge q and mass m is accelerating in the laboratory along
the x-axis under the action of the constant electric field E = Ei. Calculate
the motion of P in the lab frame assuming the initial conditions v(0) = 0,
x(0) = qcmE .
(c) The same particle P enters in the laboratory with speed u normal to a constant
magnetic field B. Calculate the motion of the P in the lab frame.
It is given that the electromagnetic field in two LCF Σ, Σ whose relative
velocity is v transforms as follows:
E = E
1
E⊥ = γ E + v × B
c
⊥
B = B
1
E⊥ = γ B − v × E .
c
⊥
5.9. The LCF Σ, Σ have their spatial axes parallel and the velocity of Σ wrt Σ
is u.
(a) Show that the transformation of the three-acceleration of a particle P in Σ and
Σ is given by the relation
a =
1
γu2 1 −
u·v 3
c2
u · v
u·a
u·a
1− 2 a+ 2 v+ 2
c
c
u
1
−1 u .
γu
Write this relation in case Σ, Σ are related with a boost along the x-axis with
speed u.
(b) Show that the parallel and the normal projections a and a⊥ of the threeacceleration of P relative to the direction of the velocity u are transformed as
follows:
1
3 a ,
1 − u·v
2
c
1
1
a
a⊥ =
+
u
×
(a
×
v)
.
3 ⊥
c2
γu2 1 − u·v
2
c
a =
γu2
5.10. The four-acceleration of a particle P in an LCF Σ in which the threevelocity
of the particle are v and a, respectively, is
and the three-acceleration
ca0
dγ
i
where a0 = γ dt .
a =
a0 v + γ 2 a Σ
34
5 Four-Acceleration
(a) Prove that a0 = c12 γ 4 v·a. Show
& & that if&the& speed of P is constant then the proper
&. Also show that if the velocity of P in
acceleration a+ has length &a+ & = γ 2 & dv
dt
& &
& &2
& .
Σ does not change direction and changes only the speed, then &a+ & = γ 6 & dv
dt
(b) Show that in general
+ 2
"
#
a
= γ 6 a6 − (a × β)2 .
5.11. Two rockets are accelerating along the x-axis of the LCF Σ in opposite directions with constant acceleration a. The equation of their world lines are
x1 (τ1 ) = −
c2
c2
cosh ψ1 , ct1 =
sinh ψ1
a
a
and
x2 (τ2 ) = −
where ψi =
aτi
c
,
c2
c2
cosh ψ2 , ct2 =
sinh ψ2 ,
a
a
i = 1, 2, is the rapidity of each rocket.
(a) Draw a spacetime diagram of the world lines of the two rockets.
(b) Define the four-vector Z i (τ ) = x2i (τ ) − x1i (−τ ), where xii (τ ) is the position
vector of the i = 1, 2 rocket at its proper time τ. The four-vector Z i defines a
synchronization of the proper times of the rockets (recall that a synchronization
between two observers is a 1 to 1 correspondence of their proper time). Prove
that
– The (Lorentz) length Z i (τ )Z i (τ ) is independent of τ. Explain the geometric significance of this result.
– Z i (τ ) is normal to the four-velocity of rocket 1 at the event x1i (−τ ) and to the
four-velocity of rocket 2 at the event x2i (τ ).
Chapter 6
Three-Momentum–Energy
6.1 Problems on Energy and Three-Momentum
6.1.1. In the SI system the charge of the electron equals 1.6021 × 10−19 Cb and its
mass 9.1 × 10−28 g. Calculate the mass of the electron in natural units.
6.1.2. Calculate the speed of a particle in an LCF Σ in which
(a) The kinetic energy of the particle equals its internal energy. What happens when
the particle is a photon?
(b) The kinetic energy of the particle equals its mass.
6.1.3. A particle P of mass m in the LCF Σ has energy E and three-momentum p.
(a) Show that the velocity of the particle both in Special Relativity and in Newtonian Physics is given by the formula u = ddpE . Is this relation valid for photons?
(b) Show that
dp = vγ dm + md(γ v),
d E = mc2 dγ + γ c2 dm,
and conclude that the change in energy and in three-momentum may be due
to either a change in γ (equivalently the speed of P) or a change in the mass
independent of the change in velocity, e.g., heating of the particle.
Note: This implies that there are two types of forces in Special Relativity:
– The inertial forces which cause change in velocity and
– The (so-called) pure forces, which cause change of the internal energy of the
particles only.
It is emphasized that change in the mass does not mean that the mass is not
invariant. It is another matter to be constant and another to be invariant!
35
36
6 Three-Momentum–Energy
6.1.4. Show that if in a given motion the three-momentum of a particle in an LCF Σ
changes only in direction, then the energy is conserved in Σ. Examine the converse.
6.1.5. In the LCF Σ a particle of mass m is displaced by dr under the action of a
3-force f.
(a) Show that the work done by f in Σ is dW = d(mc2 γ ).
(b) Show that if the change in the kinetic energy
T1→2 of a particle in a LCF Σ
'γ
between the states 1 and 2 is T1→2 = γ12 dW , then T1→2 = mc2 (γ2 − γ1 ).
Conclude that it is required infinite energy for a particle to reach speed c.
6.1.6. (a) Show that if the temporal components of two four-vectors are equal in
all LCF, then the four-vectors are equal. (We have shown in another exercise
that the same holds for the spatial components.)
(b) Apply (a) to the four-momentum vector and show that in Special Relativity
the conservation of energy implies the conservation of three-momentum and
the conservation of three-momentum implies the conservation of energy. This
result means that
– The two conservation laws of Newtonian Physics are contained in the law of
conservation of four-momentum of Special Relativity
– In Special Relativity the conservation energy is bound by the conservation of
three-momentum and conversely.
6.1.7. Prove that
(a) If the spatial parts of two four-vectors are equal in all LCF, then the four-vectors
are equal.
(b) If the three-velocities of two particles of different masses are equal in one LCF,
then they are equal in all LCF.
(c) If the four-momentum of two particles of different masses are equal in one LCF,
then they are not equal in any other LCF.
6.1.8. The world line of a particle of energy E in the LCF Σ is described by the
equations
(
x=
(
r cos θ cos φdλ,
y=
(
r cos θ sin φdλ,
z=
(
r sin θ dλ,
t=
r dλ,
where r, θ, φ are (smooth) functions of the parameter λ. Determine the type of the
particle and compute its four-momentum.
6.2
Relative Three-Momentum – Colliding Beams
37
6.1.9. Assuming the boost along the x-axis show that the quantity
that is,
d3p
E
is invariant,
d 3p
d 3 p
.
=
E
E
[The quantity
d3p
E
is the invariant volume element in the space of three-momenta].
6.2 Relative Three-Momentum – Colliding Beams
6.2.1. Two particles 1, 2 of masses m 1 and m 2 , respectively, are connected with a
spring of constant k and negligible mass. The two particles are resting on a smooth
plane when a photon is emitted from particle 1 toward particle 2 where it is absorbed.
Due to this the system of the two masses starts to oscillate with simple harmonic
motion. Assuming that the emittance of the photon causes a reduction Δm in the
mass of 1 and that the time of traveling of the photon from one particle to the other
is negligible compared to the
period of oscillation, show that the maximal extension
1
√
+ m12 .
of the spring is Δx = cΔm
m1
k
Examine Δx for the limiting values of the constant k.
6.2.2. The linear accelerator of electrons in Stanford has length 3 km and accelerates electrons to a final energy of E = 20 GeV. Given that the energy of the
electrons in the accelerator is a linear function of the distance they covered, find the
total length of the accelerator “seen” by the electrons (m e = 0.511 MeV).
6.2.3. (a) Show that in Special Relativity the three-velocity v of a relativistic particle P with three-momentum p and energy E (all quantities in the same LCF Σ)
2
is given by the relation v = pcE .
(b) Two relativistic mass points 1, 2 have four-velocities u i1 , u i2 and four-momentum
p1i , p1i , respectively. Define the relative energy E 12 and the relative threemomentum p12 of 1 wrt 2 with the components of the four-vector p1i in the
proper frame of the four-vector p2i , that is,
p1i
=
E 12 /c
p12
Σ+
2
and show that the relative velocity of 1 wrt 2 is given by the relation v12 = Ep1212
in accordance with part (a) of the problem. Calculate the velocity v21 of 2 wrt 1
and show that |v12 | = |v21 | and, in general, v21 = Ep2121 = v12 .
38
6 Three-Momentum–Energy
6.2.4. In DESY (Hamburg) operates the accelerator HERA, which produces two
head-on colliding beams of electrons and protons of energy 30 GeV/c2 and
800 GeV/c2 , respectively. Calculate
(a) The available energy in the CM of the beams.
(b) The speed of the CM in the laboratory frame.
(c) The three-momentum of the beams in the CM frame.
It is given that m p = 0.94 Gev/c2 , m e = 0.5 Mev/c2 .
6.2.5. In the ring of colliding beams I.S.R. at CERN two beams of protons each of
three-momentum p = 25 GeV/c, collide at an angle 15◦ as in Fig. 6.1. Calculate
the energy of one beam in the rest frame of the other, assuming that the mass of the
proton is 1 Gev/c2 .
Fig. 6.1 Colliding beams
6.2.6. A particle of mass m 1 = 1 Gev and kinetic energy T1 = 25 GeV collides with
another particle of mass m 2 which rests in the laboratory.
(a) Calculate the available energy for the production of particles if the mass m 2
takes the values 1.5 and 100 GeV.
(b) If the value of the available energies is smaller than T1 , explain the deficit in the
energy balance.
(c) Compute the maximal value of the available energy for this configuration.
6.2.7. Consider two photons 1, 2 which in the LCF Σ are propagating along the
directions ê1 , ê2 with energies E 1 , E 2 , respectively.
(a) Prove that there are infinite LCF Σ in which the two photons have the same
energy.
(b) Assume that Σ has parallel axes with Σ and calculate the common energy E of the photons in Σ as well as the β-factor of Σ wrt Σ in the following cases:
(b1) The photons are propagating
– In Σ in directions which coincide with those in Σ.
– In Σ in opposite directions.
6.3
Scattering Densities
39
(b2)
– The relative direction of propagation of the photons in Σ is along the positive
direction of the bisector in the first quarter of the x–y plane of Σ.
– The photons propagate along the positive directions along the x, y axes of Σ.
6.2.8. (a) Show that if two photons are moving along parallel orbits in one LCF
they are moving in parallel orbits in all LCF. Based on this result conclude that
if two photons are moving along parallel orbits in one LCF their four-momenta
are parallel.
(b) A beam of similar photons is observed in two LCF Σ, Σ which are moving
in the standard configuration along the common x-axis with speed factor β. If
the densities of the beam (= number of photons per unit spatial volume) and the
frequencies of the photons in Σ, Σ are ρ, ρ and ν, ν , respectively, show that
if the beam propagates along the y-axis of Σ the following relation holds:
ρ : ρ = ν : ν.
Compute the relation between the densities when the beam propagates along
the common x-axis.
6.3 Scattering Densities
6.3.1. A mass dm of proper volume d V0 is moving with velocities v and v in the
the mass densities and ρ E = dd VE ,
LCF Σ and Σ , respectively. Let ρ = ddmV , ρ = ddm
V
d E
ρ E = d V the energy densities of the mass dm in Σ and Σ , respectively. Calculate
the relation between the quantities
(a) d V and d V ,
(b) ρ and ρ ,
(c) ρ E and ρ E .
6.3.2. A beam of atoms of radius R1 is moving along a fixed direction in the laboratory with speed u when enters a region in which there is a gas consisting of atoms
of radius R2 .
(a) If in the gas there are k atoms per unit of volume and the gas is assumed to be at
rest in the laboratory, calculate the percentage of atoms of the beam which are
scattered in time t in the laboratory frame.
(b) If the atoms of the gas are moving in the laboratory in the opposite direction
of the beam with speed V, compute again the percentage of atoms of the beam
which are scattered in time t in the laboratory frame assuming that the proper
density of the gas is k.
(Hint: Reduce all quantities in the lab frame.)
40
6 Three-Momentum–Energy
6.4 Reflection of Light on Mirror
6.4.1. A plane mirror K is moving in the LCF Σ with constant speed β and in a
direction which is
(a) Parallel to its plane.
(b) Normal to its plane.
A photon of frequency ν is reflected on the mirror with angle of incidence θ (see
Fig. 6.2). Compute the angle of reflection and the frequency of the reflected photon
in both cases of motion of the mirror. Examine if there is a speed of the mirror
for which the frequency of the reflected photon remains unchanged in Σ. Assume
c = 1.
Fig. 6.2 Reflection on a moving mirror
6.4.2. A particle is moving in the x–y plane of an LCF Σ so that its velocity makes
an angle θ with the positive semi-axis x. The particle is reflected elastically on a
smooth wall which moves perpendicularly to the x-axis with speed u. Calculate the
reflection angle of the particle in Σ.
6.4.3. In the LCF Σ(x, y, z) a mirror moves so that its plane remains always parallel
to the y–z plane and its velocity is u = u x̂. A photon of frequency ν0 in Σ moves
along the direction √12 (x̂ + ŷ) when it is reflected on the mirror. Determine the
frequency and the direction of motion of the reflected photon in Σ (a) in Newtonian
Physics and (b) in Special Relativity.
Dynamics: Theory of Chapters 7–11
Consider a relativistic mass particle (ReMaP) P of mass m which has a four-velocity
u i . The four-momentum of P is the four-vector pi = mu i . It has length pi pi =
γc
and the four-momentum
−mc2 . In an LCF Σ the four-velocity of P is u i =
γv Σ
E/c
, where E = mγ c2 is the total energy (i.e., kinetic plus “potential”)
pi =
p Σ
of P and p = mv is the three-momentum of P, both in the LCF Σ. The length of
pi implies the relation
E=
p2 c2 + m 2 c4 .
i
i
i
, p(2)
, . . . , p(n)
be the four-momenta of n ReMaPs. The center of momenLet p(1)
tum (CM) particle is the (virtual) ReMaP whose four-momentum is
P =
i
n
)
i
p(A)
=
A=1
* n
)
E (A) ,
A=1
n
)
+
p(A)
A=1
and its mass is M = − c12 P i Pi . The B and Γ factors of the CM particle are given by
the relations
,n
,n
A=1 p(A)
A=1 E (A)
,
Γ=
.
B = ,n
Mc2
A=1 E (A)
The four-force F i on a ReMaP P with proper time τ and four-momentum pi is
defined by the relation
Fi =
dpi
= ma i ,
dτ
where the second equality applies whenthe mass m of P is constant. In an LCF Σ
ca0
i
in which P has four-acceleration a =
, the four - force F i = ma i
a0 v + γ 2 a Σ
41
42
Dynamics: Theory of Chapters 7–11
has decomposition
Fi = γ
1
c
f·v
f
Σ
,
where f is the three-force applied to P according to Σ.
In an LCF Σ the equations of motion of a ReMaP on which a three-force f (in Σ!)
is applied are
dE
mc2 3 dβ 2
= mcγ 3 β · a =
γ
,
dt
2
dt
dp
= mγ 3 (β · a)β + mγ ca.
dt
In the collision or interaction of a set of ReMaP, the total four-momentum
remains the same (is conserved). The number of the particles is not necessarily
conserved and when this is the case we say that the collision is elastic.
We say that a reaction occurs at the threshold if in the proper frame of the CM
particle (the CM frame) the sum of three-momenta of all the interacting particles
equals zero. If three particles with masses m 1 , m 2 , m 3 are involved, the condition
for threshold is that the triangle function
λ2 (m 1 , m 2 , m 3 ) = (m 21 + m 22 − m 23 )2 − 4m 21 m 22 = 0.
Chapter 7
Relativistic Optics
7.1. Two photons with energies E 1 and E 2 = k E 1 , k < 1, propagate along different
directions along the x-axis of an LCF Σ.
(a) Calculate the β-factor of another LCF Σ which is moving in the standard
configuration wrt Σ along the x-axis in which the two photons have the same
frequency. Calculate the common energy of the photons in Σ .
(b) An observer is moving with constant speed along the x-axis receding from a
fixed light source which emits blue light of frequency ν = 7 × 5 × 1013 Hz
and approaching another fixed source which emits red light of frequency ν =
5 × 1013 Hz. Calculate the speed for which the observer sees the light from the
two sources to have the same color. Calculate the frequency of the common
color.
As an application of the above, discuss the following methodology for the determination of the color of a galaxy. Suppose we know the color of the galaxy A which
recedes (reference galaxy) and we wish to determine the color of a galaxy B which
approaches. According to the above we speed up along the line joining the two
galaxies until the colors of the two galaxies appear to be the same. Then we measure
our speed with a speedometer and determine the color of the approaching galaxy.
Is it possible to develop with electronic methods an artificial Doppler effect which
will replace the galaxy A?
7.2. Photons 1, 2 of frequency ν1 , ν2 , respectively, are propagating in the LCF Σ in
directions which make an angle θ .
(a) Calculate the velocity β of the CM of the photons in Σ.
(b) Calculate the velocity β of the CM when the photons propagate along the positive and negative directions of the axes x, y, respectively. What happens in this
case if ν1 = ν2 ?
(c) Repeat the same if the photons move along the x-axis in opposite directions.
43
44
7 Relativistic Optics
7.3. In the LCF Σ the ReMaP P has four-momentum pi = mc(20, 8, 6, 0)t where
m is a constant.
(a) Determine the mass of P.
(b) Consider the LCF Σ which moves wrt Σ in the standard configuration with
speed u = 0.6c. Determine the energy and the direction of motion of P in Σ .
7.4. (a) Show that two null four-vectors are parallel if and only if their space parts
are parallel.
(b) In the LCF Σ two photons 1, 2 whose frequencies are ν1 , ν2 (in Σ!), respectively, are emitted from the points A, B along the x-axis of Σ parallel to the
y-axis. Compute the relative velocity and the four-momenta of the photons in
an LCF Σ which moves wrt Σ in the standard way along the common x, x axis
with speed u.
7.5. The observers 1, 2 with four-velocities u 1 , u 2 are observing a photon with fourmomentum P i .
(a) Show that the ratio ν1 :ν2 of the frequencies observed by the observers in their
proper frame equals u 1 P:u 2 P.
(b) If u is the velocity of 1 wrt 2 use (a) to compute the Doppler formula assuming
that in the proper frame of 2 the angle between u and the spatial propagation of
the photon is a.
7.6. During the study of the spectrum of the light from a distant galaxy, it is found
that the spectral line with wavelength 7300 Å corresponds to the line of the spectrum
of hydrogen, which in the laboratory has wavelength 4870 Å. If we assume that the
galaxy is receding along the radial direction and furthermore that the observed shift
in the spectrum is only due to the relativistic Doppler effect, calculate the speed of
the receding galaxy from the Earth.
From other methods we know that the distance of the receding galaxy from our
galaxy (the Earth) is 5 million light years. Assuming that at the Big Bang the two
galaxies were created simultaneously and since then the foreign galaxy recedes with
the same speed, calculate the time of the Big Bang in the Earth and in the proper
frame of the receding galaxy.
7.7. In order to compute the receding velocity of a galaxy one measures the wavelength of a line in the spectrum of the light emitted by the galaxy and compares it
with the wavelength of the same line in the spectrum of the light in the laboratory.
It has been found that the spectral line of Na with wavelength 5890 Å on the Earth
has wavelength 6100 Å in the spectrum of the galaxy. Determine the speed at which
the galaxy is receding from the Earth.
7.8. A light source is placed at some point along the positive semi-axis x of one
LCF Σ and emits monochromatic radiation of frequency ν0 toward the origin of Σ.
7 Relativistic Optics
45
This radiation is observed by another observer Σ who passes through the origin of
Σ with a velocity in the plane x-y at an angle θ with the positive direction of the
axis x.
(a) Determine the frequency ν and the direction of the radiation measured by Σ .
(b) Determine the angle θ so that the frequency measured by Σ is ν0 . Is this angle
uniquely determined?
7.9. A rocket moves with speed u along the x-axis of an LCF Σ. In order the
observer in the rocket to measure the speed of the rocket in Σ emits monochromatic
radiation of frequency ν0 which is reflected at the two ends of the rocket and returns
to the observer. If the returned radiation in Σ has frequency ν̄ and arrives at the
observer with a time difference ΔT, compute the speed and the length L of the
rocket. The rocket is assumed to be a solid, that is, all its points have the same speed
in Σ.
7.10. (a) Consider a photon with four-momentum f i and an observer with fourvelocity u i and show that the frequency ν measured by the observer is given by
the relation
ν=−
1 i
f ui .
h
(b) An emitter E and an receiver R of photons are fixed in a centrifugal machine at
a distance r from the axis of rotation and at an angle a, both r and a measured
when the machine is resting in the laboratory. The machine starts rotating with
constant angular velocity ω as measured in the laboratory. Calculate the redshift
z of the photon, which is defined by the relation
z=
λR − λE
,
λE
where λ E is the wavelength of the emitted photon and λ R is the wavelength of
the absorbed photon.
7.11. Consider a photon with four-momentum pi and define the frequency fourvector f i of the photon as follows: pi = hc f i ( f i is a four-vector because both h, c
are universal constants, hence invariants). Calculate the components of f i in an LCF
in which the photon has energy E and it is propagating along the three-direction with
unit vector ê:
(a) In terms of the frequency of the photon.
(b) In terms of the wave vector k which is defined with the relation k = ν ê, where
ν is the frequency of the photon in Σ, when the photon is considered as an
electromagnetic wave.
46
7 Relativistic Optics
Consider a mass particle with four-momentum pi and define with the same formula the four-vector of frequency f i of the particle. Compute f i in an LCF Σ
in terms of the energy and the three-momentum of the the particle in Σ. Finally,
compute the three-velocity of the particle in terms of the wave vector (de Broglie
wave vector) of the particle. Comment on the results.
7.12. The usual geometric representation of the boost is in terms of the hyperbolic
angles in the Euclidean plane via the introduction of the rapidity ψε R with the
relations β = tanh ψ. In order to define a geometric representation of the boost in
terms of the standard trigonometric functions, we define β = sin θ (−1 ≤ β ≤ 1,
−π/2 ≤ θ ≤ π ).
(a) Write the boost in terms of the angle θ instead of the parameter β.
(b) Express the components of the four-velocity in terms of θ and making use of
a trigonometric identity show that u i u i = −c2 . Calculate the components of
four-momentum and making use of the same trigonometric identity show that
E 2 = p 2 c2 + m 2 c4 .
(c) Consider a particle with energy E and three-momentum p which moves along
a straight line with constant velocity in the LCF Σ. Define for the particle a
frequency and one wavelength with the de Broglie relations E = hν, p = h/λ.
2
h
, ν0 = mch
Subsequently show that λ = λ0 cot θ, ν = ν0 / cos θ where λ0 = mc
is the wavelength and the Compton frequency of the particle.
Show that the phase speed of the particle u ph = λν equals u ph = sinc θ which
dν
implies u ph > c. Define the group velocity of the particle u gr = d(1/λ)
and show
that it equals the velocity of the particle in Σ.
Also show that u gr u ph = c2 and explain the result. Finally, making use of the
trigonometric identity sec2 θ − tan2 θ = 1 shows that ν 2 = c2 (k02 + k 2 ) (relativistic
dispersion formula of de Broglie waves), where k = 1/λ is the wave number of the
de Broglie waves (see V. Majernik (1986), Am. J. Phys. 54, 536–538).
7.13. (a) Prove that necessary and sufficient condition for the quotient of the components of two tensorial four-quantities to be (Lorentz) invariant is that the
four-quantities are four-vectors.
(b) Application
The wave nature of matter is expressed with the frequency four-vector, whereas
the particle nature of matter is expressed with the four-momentum.
The dual perspective of the wave and the particle nature of matter is expressed
by the statement:
The frequency four-vector and the four-momentum are parallel.
The zeroth component of the four-momentum is the energy E of the particle
while the corresponding component of the frequency four-vector is the fre-
7 Relativistic Optics
47
quency ν of the wave nature. These physical quantities are related by Plank
relation E = hν, where h is a universal constant (therefore, invariant in all theories of physics, whatever the spacetime symmetry group). Use these data and
part (a) to produce the de Broglie relation which relates the three-momentum
p with the wave vector k of a particle. Conclude that the dual nature of relativistic matter is inherent in Special Relativity and does not hold in Newtonian
Physics (the electromagnetic field is not invariant under the Galileo group) (see
R. Newburgh (1956), Lett. Nuovo Cimento. 29, 195–196; P. Dirac (1924) Proc.
Cambridge Philos. Soc. 22, 432).
Chapter 8
Four-Force
8.1. In the LCF Σ two mass points 1, 2 are in contact exerting forces of action–
reaction f12 and f21 , respectively, to one another. Prove the third Newton’s law f12 =
−f21 (Law of action–reaction) assuming that the masses of the two particles remain
constant.
8.2. Write the equations of motion of a particle of mass m in the LCF Σ in which
the three-force f is applied on the particle. Write these equations in the following
special cases:
(a) The three-velocity of the particle is collinear with the three-force (⇔ the direction of the velocity is constant).
(b) The three-force is always normal to the three-velocity (⇔ the speed is constant).
(c) The speed is c.
8.3. A particle of mass m is moving along the x-axis of the LCF Σ under the action
2
of the attractive force 2mc
. If at t = 0, x = 2 show that the motion of the particle in
x2
.
Σ is simple harmonic motion with period 4π
c
8.4. A particle of mass m is moving along the x-axis of the LCF Σ under the action
of the force:
f=−
mc3 ω2 x
i,
(c2 − ω2 a 2 + ω2 x 2 )3/2
where ω, a are positive constants. If at t = 0 the particle is at the origin of Σ and
has velocity v = ωai, show that the speed satisfies the relation v 2 = ω2 (a 2 − x 2 ).
What this implies for the motion of the particle?
8.5. A particle of mass m is resting at the position x = a (a > 0) in the LCF Σ
when the three-force f = −mω2 xi is applied to
the particle. Show that the speed on
√
2 +ω2 a 2
.
the particle when passes the origin equals ωac2c24c
+ω2 a 2
49
50
8 Four-Force
8.6. A particle of mass m moves with speed υ along the x-axis of the LCF Σ when
. Show that the time interval required in Σ in order that the
it suffers the reaction mv
k
#
" 5
speed of the particle will be reduced from 45 c to 35 c is ln 32 + 12
k.
8.7. Assuming Special Relativity holds for the gravitational field near the Earth
(which is not true!), calculate the “corrections” in the flight time, the range, the
maximal height, and the angle of firing of a projectile as it is seen by an observer
moving with velocity u = ui.
8.8. A ReMaP P moves in the LCF Σ with velocity v under the action of the threeforce f.
(a) Show that the analysis of the four-force on P in Σ are the following:
F i = γ (v)
f · v/c
f
Σ
.
Give the physical significance of the zeroth component.
(b) Prove that F i Vi = 0 where V i is the four-velocity of P.
(c) Consider another LCF Σ which moves wrt Σ in the standard configuration with
speed u and compute the components of the four-force in Σ .
(d) Show that a central three-force f = −kr in Σ is not, in general, transformed to
a central three-force in Σ .
8.9. (a) Using the identity
A×(B × C) = (A · C)B − (A · B)C,
show that
A=
1
B×(B × A)
(A · B)B −
.
B2
B2
(8.1)
This relation decomposes a vector A parallel and normal to a vector B.
(b) Let Σ, Σ be two LCF with parallel axes andrelative
0 β. A four-vector
velocity
0
A
A
,
. Show that
Ai in Σ, Σ is decomposed as follows: Ai =
A Σ
A Σ
A = γ (A − A0 β) + (γ − 1)
β×(β × A)
.
β2
This relation decomposes A parallel and normal to the relative velocity β of the
LCF Σ, Σ .
(c) Apply the above decomposition to the four-vector of four-force and show that
the following relations hold:
8 Four-Force
51
1
(v − u) · f,
Q
1
u×(v × u)
f⊥ + f −
· f e ,
f =
γu Q
Quc
f · v =
where Q = 1− u·v
. Comment on the physical meaning of the time component of
c2
the four-force. Concerning the spatial part of the four-force, show that provided
that f ⊥ = 0 this can always be written in the form of a Lorentz force and
calculate the “charge,” the “electric field,” and the “magnetic field” due to this
force.
8.10. In the LCF Σ the mass point P with mass m moves along a circular orbit of
radius R with constant speed u. Calculate the force exerted on m.
8.11. A particle P of mass m and charge e is moving in the field of another particle
of charge q of opposite sign (so that we have attraction) fixed at the origin of the
LCF Σ.
(a) Show that the equation of motion of P in Σ is
kr
d
(γ u) = − 3 ,
dt
r
d
k
2
(γ c ) = − 3 (u − r),
dt
r
where k = − eq
> 0.
m
(b) Prove that motion is planar.
(c) Use polar coordinates (r, θ ) on the plane of motion to show that γ r 2 θ = A =
const. and γ c2 rk = w =const.
(d) Introduce the coordinate u = r1 > 0 and eliminating the time, show that the
equation of motion becomes
wk
u2
d 2u
+ 1 − 2 2 u = 2 2.
dθ 2
A c
A c
2
Subsequently define the quantity p = 1 − Ak2 c2 and show that the orbit of P
in Σ can be written
2
1
w
c
kw
2 1
,
=u=
cos
+
c)
+
−
p
(ρθ
r
Ap 2
c4
2
Ac2
where C is a constant. Deduce that the condition for a bound trajectory is
−c2 < w < c2 and that the bound trajectory can be described with an ellipse
whose one focus is the origin of Σ and precession of the perihelion equals
52
8 Four-Force
2π 1p − 1 per rotation. Finally, show that for small velocities this quantity
can be approximated as follows:
π k2
4π 2 α 2
=
,
A2 c2
τ 2 e2 (1 − a 2 )
where α is the length of the small major semi-axis, τ is the period of the orbit,
and e is the eccentricity of the orbit.
8.12 In the LCF Σ the Lagrangian of a free particle of speed u is defined by L 0,Σ =
−αc where α is a quantity which is computed from the requirement that the action
t2',Σ
L 0,Σ dtΣ is invariant. Show that α = − mγ .
integral S =
t1 ,Σ
Chapter 9
Plane Waves
9.1. The equation of an electromagnetic wave which is propagating along the
direction ê = cos αi + sin αj of the x–y plane of the LCF Σ is
Ωi = Ai exp
2πiν
(x cos α + y sin α − ct) ,
c
where ν is the frequency of the wave in Σ. Compute the frequency and the angle of
propagation of the wave with the x -axis in the LCF Σ which moves wrt Σ in the
standard configuration with speed u. Consider c = 1.
53
Chapter 10
Relativistic Reactions
10.1 Center of Momentum Energy
10.1.1. (a) Calculate the available energy in the CM frame in the following experiments of proton to proton collisions. In the first experiment a beam of protons is
accelerated to energy E = 30 GeV and hits a resting target (e.g., liquid hydrogen). In the second experiment two beams of protons are accelerated to energy
E = 15 GeV and then they come onto head-on collision.
(b) At which energy in the first experiment the bullet beam must be accelerated in
order the available energy to equal that of the second experiment?
The mass m p = 0.94 GeV/c2 .
10.1.2. In Newtonian Physics the kinetic energy TΣ of a system of particles in the
frame Σ is related to the kinetic energy T ∗ of the system in the CM frame as follows:
T = T∗ +
1
MV2 ,
2
where M is the sum of the masses of the particles and V is the velocity of the CM
frame in Σ. This result is the Theorem of Kinetic Energy in Newtonian Physics.
Calculate the corresponding relation in Special Relativity and show that for small
speeds the relativistic formula reduces to the Newtonian one.
10.1.3. A particle of mass m and kinetic energy E collides with a similar particle
which rests in the laboratory.
√ Show that if E m, then the maximum available
energy in the CM frame is 2Em.
10.2 Elastic Collisions
10.2.1. A smooth elastic sphere 1 which moves with velocity u in the laboratory
collides centrally with another similar sphere 2, which rests in the laboratory. After
the collision the two spheres 1, 2 move in directions which make angles θ, φ with
the line of collision, respectively. Calculate the product tan θ tan φ in terms of the
55
56
10 Relativistic Reactions
velocity of the CM of the system of spheres in the laboratory. Find the Newtonian
limit of the result.
10.2.2. A beam of π -mesons is scattered elastically on a target of protons which
are at rest in the laboratory. Calculate the energy of a π -meson in order the corresponding scattered π -meson to have momentum 0.8 GeV /c in a direction normal
to the direction of the falling π -meson. Calculate the maximum and the minimum
energy of the scattered π -meson for scattering angle 90◦ in terms of the energy of
the falling beam in the laboratory.
The masses: m π = 0.14 GeV/c2 , m p = 1 GeV/c2 .
10.2.3. A photon of energy E γ collides head on with an electron with momentum pe . As a result of the collision the electron is scattered at an angle θ to its
direction of motion prior to collision while the photon attains energy E γ (all quantities refer in the laboratory frame L). Calculate E γ in terms of the other quantities.
Derive the expression of E γ when the electron is at rest in the laboratory.
10.2.4. A photon of frequency ν1 is scattered on a free particle of mass m which is
at rest in the laboratory (Compton scattering). If the scattering angle of the photon
in the laboratory is θ, show that the new frequency ν2 of the photon in the laboratory
is given by the relation
1
θ
1
2h
−
=
sin2 .
2
ν2
ν1
mc
2
Calculate the kinetic energy of the particle after the scattering of the photon. If the
photon is scattered backwards (θ = 180◦ ), calculate the speed of the particle in
terms of its mass.
10.2.5. A photon is absorbed by an atom of mass m which is at rest in the laboratory.
Due to the absorption, the atom is excited with one of its electrons changing to a
higher energy level by an energy difference E (excitation energy). Calculate the
frequency of the photon in terms of the excitation energy E and the mass m.
10.2.6. Compute the energy of the scattered electrons in the elastic electron – proton
scattering in the proper frame of the proton, in terms of the energy of the initial
electron and the scattering angle. Assume that the initial electrons have very high
energies and the masses of the electron and the proton are m, M and m M.
10.3 Reactions of the Form 1 → 2 + 3 (Decays)
10.3.1. An exited nucleus of mass m which rests in the laboratory frame (L) emits a
photon of energy hν. Due to the radiative transition, the mass of the excited nucleus
is reduced by Δm. Calculate the energy of the emitted photon taking into consideration the rebound of the nucleus.
10.3
Reactions of the Form 1 → 2 + 3 (Decays)
57
10.3.2. An unstable particle of mass M moves along the x-axis of the LCF Σ which
disintegrates into two particles 2, 3 of mass m 2 , m 3 , respectively, so that one of the
particles moves in a direction perpendicular to the direction of the mother particle.
Calculate the speed β1 of the mother particle in Σ in terms of the quantities mM2 , mM3
and the angle φ between the directions of motion of particles 1, 2 in Σ. Apply the
result to the special case m 2 = m 3 = m, φ = 90◦ . Consider c = 1.
10.3.3. The particle 1 of mass m 1 and energy E in the laboratory frame L disintegrates giving the particles 2, 3 whose masses are m 2 , m 3 , respectively. Assuming
that the direction of propagation of particle 2 with that of particle 1 in L is θ , calculate the energy and the three-momentum of the daughter particles in L in terms of
the quantities E, m 1 , m 2 , cos θ .
Calculate the energy and the three-momentum of particle 2 when it is emitted at
right angle to the direction of motion of particle 1 (c = 1).
10.3.4. A particle 1 of mass m 1 decays in two other particles 2,3 of mass m 2 , m 3 ,
respectively. Determine how the decay energy ΔE = (m 1 − m 2 − m 3 )c2 is distributed (as kinetic energy) among the daughter particles 2, 3 in the proper frame of
particle 1.
10.3.5. Among the products of a high-energy reaction is an unstable neutral particle which disintegrates into κ + and π − at a distance 0.9 mm from the point of
creation. From relevant measurements it is found that κ + and π − have momentum
10.122 GeV/c and 1.047 GeV/c, respectively. Furthermore the two particles have
velocities, which are in the same plane with the direction of motion of the neutral
particle at angles 2.79◦ and 28.13◦ , respectively, above and below that direction.
Calculate
(a) The mass of the neutral particle.
(b) The angle θ ∗ of κ + with the direction of motion of the neutral particle in the
CM.
(c) The lifetime of the neutral particle.
It is given m κ + = 490 MeV/c2 ,
m π − = 140 MeV/c2 .
10.3.6. One method (obsolete by now) to study the reactions of elementary particles
is to study their orbits in the bubble chamber in which the particles leave a visible
trace of bubbles which is taken on a photographic plate. The neutral particles do not
create visible orbits.
(a) Justify the following statement: The bigger the energy (equivalently the threemomentum) of a charged particle in the laboratory, the bigger its orbit on the
photographic plate.
(b) The particles K + and μ+ are at rest in the laboratory when they disintegrate
according to the reactions:
K + −→ μ+ + ν
,
π + −→ μ+ + ν.
58
10 Relativistic Reactions
The images of the orbits of the particles obtained on the photographic plate
for these reactions are the (a) and (b) as shown in Fig. 10.1.
Fig. 10.1 Disintegration in the bubble chamber
Determine which orbit corresponds to which reaction. The masses m K =
0.494 GeV/c2 , m π = 0.140 GeV/c2 , m μ = 0.1057 GeV/c2 are assumed to be
known.
(c) Do you trust your results and would you consider the method as a reliable criterion for the distinction of similar reactions?
10.3.7. A particle of mass M moves along the x-axis of the laboratory frame with
speed factor β when disintegrates in two particles 1, 2 of masses m 1 , m 2 , respectively.
(a) Decompose the three-momenta of the daughter particles 1, 2 perpendicular (⊥)
p −p
and parallel (||) to the x-axis and define the quantity α = p11 + p22 . Subsequently
show that this quantity satisfies the relation
α−B
A
2
+
p⊥
p∗
2
= 1,
where p⊥ is the common length of the perpendicular three-momenta in the
laboratory and p ∗ is the common length of the three-momentum in the CM.
Calculate the quantities A, B and show their geometric significance by a diagram.
(b) Calculate the maximum value of p ∗ and comment on the result.
Assume c = 1.
10.3.8. A π 0 moves in the laboratory with kinetic energy T when disintegrates in
two photons.
(a) If the photons are emitted in opposite directions in the laboratory, compute the
energies of the photons in the laboratory.
(b) If the photons are emitted at equal angles to the direction of motion of the pion
(all three trajectories are on the same plane), calculate the angle between the
directions of propagation of the photons (c = 1). Also compute the direction of
motion of the photons in the CM.
Application: Assume m π = 135 MeV/c2 , T = 1 GeV/c2 .
10.3
Reactions of the Form 1 → 2 + 3 (Decays)
59
10.3.9. Consider the fission A → B + C where the particles A, B, C have masses
m A , m B , m C , respectively.
(a) In the proper frame of A show that B has energy:
EB =
m 2A + m 2B − m C2 2
c .
2m A
(b) An atom of mass M which rests in the laboratory disintegrates emitting a photon
and goes to a new state with mass M − δ. Show that the energy hν of the photon
is hν < δc2 . Why in the Mösbauer effect hν = δc2 ?
(c) If A moves during its disintegration, find a relation between the angle of emission of B and the energies of the particles A, B.
Application
The K 0 is moving in the lab frame with kinetic energy T while disintegrates into
K 0 → π + + π − . Calculate the maximum energy of the K 0 in order to have production of π + at 90◦ .
Assume m K 0 = 0.5 GeV, m π = 0.140 GeV.
10.3.10. Three identical particles of mass m are placed on the vertices of a canonical
triangle which rest on the plane x − y of the LCF Σ. At some moment (in Σ!) each
of the particles decays in two other particles of mass am and bm, respectively, where
a, b are properly defined constants in the interval (0, 1) so that the particle with mass
am propagates along the height of the triangle from the corresponding vertex.
(a) Show that the daughter particles arrive simultaneously at the mass center of the
triangle.
(b) Assume that the three daughter particles interact producing a new particle of
mass M. Compute M assuming that the interaction occurs at the threshold.
10.3.11. A particle 1 of mass m 1 decays in two other particles 2,3 of mass m 2 =
am 1 and m 3 = bm 1 , respectively, where a, b are properly defined constants in the
interval (0, 1).
(a) Show that the energy and the length of the three-momentum of particle 2 in the
proper frame of particle 1 are
X
m 1 c2 ,
2
YZ
2
m 1 c,
1p =
2
2
1E
=
60
10 Relativistic Reactions
where
X = 1 + a 2 − b2 , Y 2 = (1 − a)2 − b2 , Z 2 = (1 + a)2 − b2 .
(b) Show that the β and the γ factors of particle 2 in the proper frame of particle 1
are
2
1β
=
YZ 2
X
, γ =
.
X 1
2a
(c) After its creation particle 2 decays into two particles 3, 4 of mass m 3 = am 2
and m 4 = bm 2 , respectively. Subsequently the particle 3 decays in particles
5, 6 and so on. Assuming that all decays occur along the same line and that all
particles 2n for n = 1, 2, . . . move in the same direction compute the energy of
the particle 2n in the proper frame of particle 1.
10.3.12. The most accurate method for the determination of the mass of neutrino
(which we consider to be zero) is the study of the reaction π 0 → μ + ν.
Consider this reaction and show that the mass of neutrino is given by the formula:
m 2ν = (m π − m μ )2 −
1
2m π μπ T,
c2
where μπ T is the kinetic energy of the muon in the proper frame of the pion.
Recent methods of spectroscopy of X-rays (see R. Shafer (1967) Phys. Rev.
163, 1451) have improved the accuracy of the value of mass of the pion to m π =
139.577 ± 0.013 MeV/c2 . Assuming m μ = 105.659 ± 0.002 MeV/c2 and μπ T =
4.122 ± 0.016 MeV/c2 , calculate the mass of the neutrino and determine the accuracy of the result.
10.3.13. (a) A π -meson disintegrates according to the reaction π → μ + ν. Show
that the kinetic energy of the emitted μ meson in the proper frame of the
π -meson (= CM frame) is
T =
(m π − m μ )2 2
c .
2m π
Numerical application: m π = 140 MeV/c2 , m μ = 105 MeV/c2 , m ν = 0 MeV/c2 .
(b) Consider three-particle (that is timelike or null) four-vectors Ai , B i , C i such
that Ai = B i + C i . Prove the identity:
B i Ci =
1
[−A2 + B 2 + C 2 ],
2
where −A2 = Ai Ai , −B 2 = B i Bi , and −C 2 = C i Ci (A, B, C ≥ 0). Assuming
B i to be timelike show that in the proper frame Σ B of B i the zeroth component
10.3
Reactions of the Form 1 → 2 + 3 (Decays)
61
of the four-vector C i is given from the relation
BC
0
=
1
[A2 − B 2 − C 2 ].
2A
Is there a relation between the parts (b) and (a) of the problem?
10.3.14. The excited nucleus N returns to its ground state by emitting a photon
(radiative transition) as follows:
N →N +γ
.
1
2
3
(a) In some LCF Σ (say the lab frame) define the quantities (i = 1, 2)
ui k
φi = γi 1 −
,
c
where γi = 1 −
u i2
c2
−1/2
ψi = γi
|ui × k|
,
c
(i = 1, 2 ) and show that the following relations hold:
M1 φ1 = M2 φ2 = α,
M1 ψ1 = M2 ψ2 = δ,
for all three-momenta and energies of the nucleus.
(b) Show the validity of the following identity (i = 1, 2):
γi =
1
(1 + φi2 + ψi2 ),
2φi
and from this show that the energy of the produced photon in Σ equals
hν =
c2
(M 2 − M22 ).
2α 1
(c) Let ν0 be the frequency of the photon produced from the radiative transition of
nucleus and E 0 the average energy of the states of the atom. Show that
ν=
E0
ν0 .
ac2
(d) Show that if ν1 (respectively ν2 ) is the frequency of the photon in the proper
frame of the nucleus before (respectively after) the radiative transition, then
ν
ν1
= 2,
φ1
φ2
that is, the quantity
νi
φi
is an invariant.
62
10 Relativistic Reactions
10.3.15. A nucleus X in excited state is resting in the laboratory when, by means
of a radiative transition, it emits a photon of energy E γ . If M is the mass of the
nucleus at the ground state, calculate the mass of the excited nucleus. Is it possible
to use the produced photons to excite other nuclei similar to X which are resting in
the laboratory? Justify your answer with calculations.
10.3.16. For every non-null four-vector Ai we introduce the symmetric tensor
h i j (A) = ηi j − Ak1Ak Ai A j which projects normal to Ai , that is, h i j (A)A j = 0.
(a) Let pi the four-momentum of a particle and p1i the four-momentum of another
particle. Show the identity/relation
j
p1 p j i
j
p + h ij ( p) p1 .
p k pk
p1i =
This identity defines the 1+3 decomposition of the four-vector p1i wrt the fourvector pi . In this decomposition the first part corresponds to the parallel part
i
i
and the second the perpendicular part p1⊥
.
p1
j
Show that the inner product p1 p j = 12 {( pi + p1i )2 − pi pi − p1i p1i }, that
is, it is expressed in terms of the lengths of the four-vectors. In order to give
physical meaning to this relation, we introduce the masses of the particles by
the relations pi pi = −m 2 c2 , p1i p1i = −m 21 c2 and then the inner product equals
−m E 1∗ , where E 1∗ is the energy of the particle with four-momentum p1i in the
proper frame Σ∗ of the particle with four-momentum pi . Set p2i = pi − p1i and
show that
E 1∗ =
m 1 + m 22 − m 21 2
c ,
2m
i
where p2i p2i = −m 22 c2 . Also show that the length of the normal part p1⊥
p1⊥i is
given by the relation
[m 2 − (m 1 + m 2 )2 ][m 2 − (m 1 − m 2 )2 ]
2m
2
c
λ(m, m 1 , m 2 ).
c2 =
2m
i
p1⊥
p1⊥i ≡ p1 2 c2 = E 1∗2 − m 21 c4 =
Finally, collect the results as the components of the four-vector p1i in Σ∗ as
follows:
⎞
⎛
m 1 +m 22 −m 21
∗ c
2m
E 1 /c
⎟
⎜
=
,
p1i =
p1 Σ∗ ⎝ [m 2 −(m +m )2 ][m 2 −(m −m )2 ] ∗ ⎠
1
2
1
2
cê
2M
Σ∗
where ê∗ is the unit along the three-momentum in the proper frame of pi .
10.5
Reactions of the Form 1 + 2 → 3 + 4
63
Deduce that reactions of the form 1 + 2 → 3 and 1 → 2 + 3 are determined completely in terms of five parameters. One set of such parameters is the
three masses of the particles and the angles of propagation of one particle in the
proper frame of the other particle. Needless to say that there are many other sets
of appropriate five parameters.
10.4 Reactions of the Form 1 + 2 → 3
10.4.1. Particles 1, 2 with masses m 1 , m 2 have velocities v1 , v2 , respectively, in the
LCF Σ when they collide creating a new particle of mass m 3 . Calculate the mass m 3
and show that m 3 ≥ m 1 + m 2 . Comment on the result (c = 1).
10.4.2. The particle Y (a bound state of b − b̄ quarks) has mass 9.46 GeV/c2 and it
is produced during the head on collision of e+ , e− beams each of energy 4.73 GeV.
If instead of head on colliding beams one uses e− as a standing target, calculate the
energy of the beam of e+ in the rest frame of e− for the Y particle to be produced
(m e± = 0.5 MeV/c2 , c = 1).
10.5 Reactions of the Form 1 + 2 → 3 + 4
10.5.1. In an experiment for the production of kaons and Λ hyperons according to
the reaction π − + p + → k 0 + Λ0 , it has been found that the π − in the laboratory
have three-momentum 2.50 GeV/c and the Λ0 have three-momentum 0.60 GeV/c
and are produced in a direction making an angle of 45◦ with the direction of the π − .
Calculate the three-momentum of k 0 in the laboratory and in the CM frame.
It is given that m π = 140 MeV/c2 , m p = 938 MeV/c2 , m k = 498 MeV/c2 ,
m Λ = 1116 MeV/c2 .
10.5.2. In the study of the reaction 1+2 → 3+4 in the laboratory it has been found
that particle 1 has kinetic energy T1 , particle 2 is at rest, while particle 3 is produced
at 90◦ to the direction of motion of the bullet particle 1. Assuming that the masses of
the particles are known, calculate the energy E 3 of particle 3 in the laboratory. What
is the expression of E 3 when m 1 = m 3 = m k , m 2 = m 4 = m p ? Show that in this
case when T1 increases without a limit, the energy E 3 remains finite and compute
its maximum value. It is given that m k < m p .
10.5.3. An antiproton with kinetic energy 23 GeV in the laboratory falls on a proton
at rest. As a result of the collision two photons are produced. If the photons move
in the laboratory along the direction of the antiproton, calculate the energy of each
photon and the direction of its motion:
64
10 Relativistic Reactions
(a) In the laboratory frame L
(b) In the proper frame of the antiproton.
The mass of the proton is m p = 1 GeV/c2 , c = 1.
10.5.4. A particle X 1 of mass m which rests in the laboratory frame L absorbs a
photon of energy E. Subsequently the particle disintegrates into a particle X 2 and
a photon of energy E and in a direction normal to the direction of the reacting
photon. Compute the mass of the daughter particle X 2 and its three-momentum in
the laboratory frame. Comment on the result.
10.5.5. Consider the reaction
a + b → c + X.
By definition the active mass of particle X is given from the relation
m 2X = ( p1i + p2i )2 ,
where pai , pbi are the four-momenta of particles 1, 2. If√we denote the total energy
s, show that the maximum
of the reacting particles a, b in the CM frame Σ∗ with
√
value for the mass of the particle X is m X,max = c12 s − m c .
10.5.6. In an experiment concerning the reaction of protons at rest in the laboratory
with antiprotons, it has been found that p + p → π + C. The study of the spectrum
of the three-momenta of the charged π -mesons showed the production of π -mesons
of a single energy and three-momentum p = 190 MeV/c in the laboratory along
the direction of the motion of the antiprotons. This result suggests the existence of a
bound state C. If the energy of the antiprotons is E 1 = 10 GeV, calculate the mass
of the bound state C.
It is given that m p = m p = 0.938 GeV/c2 , m π = 0.139 GeV/c2 .
10.5.7. In a bubble chamber a K − meson reacts with a proton at rest,1 producing
a π + meson and an unknown particle X . The photograph of the reaction gives the
orbits of the particles as shown in Fig. 10.2.
From the observation of the figure we infer the following:
– The orbits of the particles are in a plane perpendicular to the magnetic field of
the chamber.
– The orbits of the daughter particles are on the same curve, C say, at the point of
reaction O.
1 Interact at rest means that the total three-momentum in the CM of the system of particles vanishes
and not that the particles are at rest in the CM. Therefore, in the current problem both the proton
and the K − meson have non-vanishing three-momenta in CM which, in this case, coincides with
the laboratory.
10.5
Reactions of the Form 1 + 2 → 3 + 4
65
Fig. 10.2 Orbits of particles in the bubble chamber
– The orbits of p and K − are along the principal normal of of orbit.
– The radius of curvature of the common orbit C at the point O is R = 0.340 m.
Given that the magnetic field in the bubble chamber is normal to the plane of the
orbit and has intensity B = 1.70 ± 0.07 T .
(a) Calculate the three-momentum of K − at the point O given that the threemomentum of the proton is 10 GeV/c .
(b) Calculate the mass and the charge of the unknown particle X.
It is given:
m π = 139.6 MeV/c2 , m p = 938.3 MeV/c2 , m K = 493.8 MeV/c2 ,
q = 1.6 × 10−19 Cb, 1 Joule = 6.24 × 1013 MeV.
10.5.8. In the reaction e+ +e− → γ +γ , the target particle e− rests in the laboratory
frame (L) and the photons in the CM frame are emitted normal to the direction of
motion of the bullet particle e+ .
(a) If the energy of the e+ in L is E, calculate the four-momenta of the photons
in L .
(b) Calculate the energy of the e+ in the laboratory for which the photons are emitted at angle π4 with the direction of motion of the e+ in the laboratory.
The mass m of the electron is known.
10.5.9. (a) Show that a photon is impossible to disintegrate in empty space to the
pair e− , e+ . Also show that this is possible if a nucleus is present. If m N is the
mass of the nucleus, calculate in the proper frame of the nucleus the minimum
energy of the photon in order to achieve the production of the pair e− , e+ . Discuss the cases m N m e and m N = m e .
(b) Show that a free electron cannot absorb or emit a photon.
10.5.10. A beam of particles of mass m is scattered elastically on a target of particles of mass M, the latter being at rest in the laboratory. If the energy of the beam
in the laboratory is E 1 , calculate the scattering angle in the laboratory as a function
of the scattering angle in the CM frame.
66
10 Relativistic Reactions
10.5.11. A particle of mass m is scattered elastically on another particle of mass
M which rests in the laboratory frame L . Calculate the angle of scattering of the
particles after the collision in terms of the scattering angle θ ∗ in the CM.
Application
Consider the scattered particle to be a neutrino (m = 0) and the target particle to be
an electron (M = m e ). More specifically consider the case E 1 = m e c2 and show
that θ ∗ cannot equal 0.
10.5.12. A particle 1 of mass m and energy E collides elastically with a similar
particle 2 which rests in the laboratory (lab). Show that in the two-dimensional
Euclidean plane the tip of three-momentum of each of the particles moves on an
ellipse whose axes depend on the energy and the mass of the particles.
10.6 Other Reactions
10.6.1. An electron of mass m and three-momentum pe interacts electromagnetically with a nucleus of mass M which is at rest in the laboratory. As a result of this
interaction the electron loses energy by emitting a photon [this radiation is called
bremsstrahlung radiation (brake radiation)]. Calculate kinematically the range of
energy of the emitted photon assuming that the nucleus is very heavy and the electron very fast so that it does not loose energy during the collision.
10.6.2. A particle of mass M is possible to disintegrate either to two particles of
mass m or to three particles of mass μ.
(a) Calculate the maximum speed β2 which is possible for one of the particles of
mass μ.
μ
m
,y = M
so
(b) Derive the condition which must be satisfied by the ratios x = M
that for the disintegration in two masses, the velocity factor β of the produced
particles of mass m is equal to the maximum velocity factor β2 determined in
part (a).
10.6.3. A k-meson which rests in the laboratory disintegrates to three π -mesons.
(a) Show that the trajectories of the three π -mesons are coplanar.
(b) If the three π -mesons are moving in directions which make an angle 120◦ one
with another, find their speed in the laboratory. If the lifetime of the pion is
2.6 × 10−8 s how far the pions travel in the laboratory before they disintegrate?
It is given that the mass of the k-meson is 494 MeV/c2 and the mass of the
π -meson is 140 MeV/c2 .
10.7
Threshold
67
10.6.4. Consider the reaction k̄ + p → k̄ + p +nπ where in the laboratory frame the
three-momentum of the beam of k-meson is 10 GeV/c. Find the maximum number
n of π -mesons which can be produced.
It is given that m k = 0.494 GeV, m π = 0.135 GeV, m p = 0.938 GeV.
10.6.5. A typical fission reaction is caused by slow neutrons according to the reaction
n 1 + 235
92 U → x + y + 2n 2 .
Calculate the velocity factor β of n 2 in the laboratory assuming that the reacting
particles n 1 , 235 U, and the products x, y are at rest in the laboratory. If the reaction
takes place with a certain quantity of 235 U, the products slow down due to collisions
with other nuclei until all their kinetic energy is transformed to heat. Assuming that
5 kg 235 U are involved in the process, calculate the thermal energy produced when
141
x = 93
38 Sr55 , y = 54 Xe87 .
The masses of the various nuclei involved are:
m x = 92.342327 amu, m y = 140.969100 amu, m U = 235.117496 amu, m N =
1.00896 amu.
Note: 1 amu = 1.66 × 10−27 kg, the Avogadro number N A = 6.022 × 1023 atoms
per mole, and the atomic weight of 235 U = 238.03.
10.7 Threshold
10.7.1. (a) A particle of mass m (bullet particle) reacts with another particle of
mass m (target particle) producing a number of various particles of total mass M. If
m + m < M, the reaction occurs only when the kinetic energy of the bullet particle
in the proper frame of the target particle is higher than a certain critical value (which
is called the threshold energy). Calculate that energy.
(b) Calculate the threshold energy of the reaction pp → ppp p̄. It is given that
m p = m p̄ = 1 GeV/c2 .
10.7.2. From the reaction of a π -meson with a proton which rests in the laboratory,
a k-meson and a Λ-hyperon are produced. Calculate
(a) The threshold energy of the reaction.
(b) The minimum kinetic energy of the π -meson in order the k-meson to be produced at right angles to the direction of the π -meson in the laboratory.
(c) The kinetic energy of the k-meson produced in (b) when the kinetic energy of
the π -meson in the laboratory is 1009 MeV.
The masses of the particles are m π = 140 MeV/c2 , m p = 940 MeV /c2 , m k =
494 MeV/c2 , m Λ = 1115 MeV/c2 . Take c = 1.
68
10 Relativistic Reactions
10.7.3. A target 1 of mass N is hit by a bullet beam of particles 2 of mass M and
the particles 1, 2 and the particle 3 of mass m are produced.
(a) Show that the γ -factor for the threshold energy of particle 2 is
γ =1+
m2
m(N + M)
+
.
NM
2N M
(b) Define the efficiency of the reaction k by the ratio
k=
Energy of produced particle 3 in CM
.
Required kinetic energy of bullet particle 2
Calculate the efficiency k in the above case and show that for large masses N of
the target the efficiency of the reaction tends to 1.
Application
Show that a photon cannot disintegrate spontaneously into a pair of positron–
electron. However, this is possible in the presence of a nucleus which acts as a
catalyst. If the mass of the nucleus is N and the mass of the electron m, calculate
the threshold frequency of the photon. Show that for large N the efficiency of the
reaction is 100%; hence, the nucleus tends to act as a pure catalyst.
10.7.4. In the laboratory frame L a photon collides with a nucleon N of kinetic
energy 1600 MeV according to the reaction γ + N → N +π . Calculate the threshold
energy of the photon in the laboratory.
masses m N = 1000 MeV/c2 and m π = 150 MeV/c2 .
10.8 General Problems on Reactions
10.8.1. Consider a linear space V n and let {x i } a coordinate system. The volume
element in the coordinates {x i } is defined by d V{x i } = d x 1 d x 2 . . . d x n . Let {x i }
be a second coordinate system in V n . Then the element of volume d V{x i } in the
i i coordinates {x i } is given from the relation d V{x i } = J xx i d V{x i } , where J xx i
i is the Jacobian of the transformation {x i } → {x i } given by the relation J xx i =
i ∂x
.
∂ xi
(a) In the Euclidean space E 3 consider the Euclidean coordinate system {x, y, z}
and the spherical coordinate system {r, φ, θ } defined by the coordinate
transformation.
x = r cos φ sin θ
y = r sin φ sin θ
z = r cos θ.
10.8
General Problems on Reactions
69
Calculate the volume element in spherical coordinates.
(b) Let Σ be an LCF in which we consider spherical coordinates and compute the
volume element as in (a). From the volume element separate the part which
contains only the angles and define the solid angle dΩ in the direction θ with
the relation dΩ = sin θ dθ dφ. Let Σ be a second LCF in which the solid angle
is dΩ = sin θ dθ dφ . The quantities dΩ , dΩ are related with the Lorentz
transformation defined by Σ, Σ . The exact relation depends on the space we
consider as the solid angle. For example, the solid angle transforms differently
in spacetime and in the momentum space. The same argument applies to the
element of volume.
In the LCF Σ a beam of photons propagates within the solid angle dΩ. Let
Σ be a second LCF which moves in the standard way wrt Σ with speed u = ux
(and has parallel axes). Compute the transformation formula dΩ → dΩ .
Application
Assuming that the stars in the night sky are distributed isotropically and uniformly for the observers on the Earth, calculate their distribution for observers
who move with large velocities wrt the Earth.
Electromagnetic Field: Theory
In an LCF Σ the electromagnetic field is expressed by the electric field E and the
magnetic field H and satisfies Maxwell equations. In an isotropic and homogeneous
medium the electric field E = D where D is the field of electric inductance and the dielectric constant of the medium. Similarly in this medium the magnetic field
is related to the magnetic inductance B with the relation B = μH where μ is the
magnetic permeability of the medium.
For empty space 0 μ0 = c12 .
In empty space the fields E, B in terms of the scalar (φ) and the vector (A)
potential are given by the expressions
E = −∇
∂A
φ
−
,
c
∂t
B = ∇ × A.
The potentials φ and A define the four-potential Ωi = (−φ/c, A). In terms of the
four-potential, Maxwell equations are written as follows:
Ωi,i = 0,
ij
,j
Ω
= −μ0 J i ,
where J i is the four-vector of current density and the comma indicates derivation
wrt the index that follows.
If a charge of density ρ0 has four-velocity u i , the four-current J i = ρ0 u i + j i
where u i ji = 0. ρ0 u i is the conduction current and j i is the convection current.
The electromagnetic field tensor is defined in terms of the four-potential as follows:
Fi j = Ω j,i − Ωi, j .
If in an LCF Σ the electric field and the magnetic field have components E =
(E x , E y , E z ), B = (Bx , B y , Bz ), the electromagnetic field tensor in Σ (in the SI
system of units) has components
71
72
Electromagnectic Field: Theory
⎛
⎜
⎜
Fi j = ⎜
⎜
⎝
0 − Ecx −
Ex
c
Ey
c
Ez
c
Ey
c
− Ecz
⎞
⎟
Bz −B y ⎟
⎟.
⎟
−Bz 0 Bx ⎠
B y −Bx 0
0
In terms of the electromagnetic field tensor Maxwell equations are written as
Fi j,k + F jk,i + Fki, j = 0,
F
ij
,j
= μ0 J i .
The four-force due to the action of the electromagnetic field Fi j on the charge q
whose four-velocity is u i is given by the relation
F i = q F μj u j .
The covariant form of the equation of motion of a charge q and mass m under the
action of the electromagnetic field Fi j is
q F ij u j =
d
(mu i ) .
dτ
In an LCF Σ in which the charge q has three-velocity v, the four-force gives the
three-force (Lorentz force)
F = q (E + v × B) .
In an LCF Σ the equations of motion of a charge can be written in various forms.
The zeroth component of the four-force gives
dE
= q(β · E),
dt
where E is the work done by the force F on the charge. From this relation we infer
that the magnetic field does not produce work (i.e. under the action of magnetic
field only the speed and consequently the β, γ -factors are constants of motion). The
spatial components of the four-force give
d
(mγ vx ) =
dt
d
mγ 2 a y = (mγ v y ) =
dt
d
2
mγ az = (mγ vz ) =
dt
mγ 3 ax =
dpx
,
dt
dp y
,
dt
dpz
,
dt
Electromagnectic Field: Theory
73
or in vector form
q(E + 1c v × B) = mγ 3 a|| ,
q(E + 1c v × B)⊥ = mγ 2 a⊥ .
Consider the LCF Σ which moves wrt the LCF Σ in the standard configuration
along the x-axis with velocity factor β. Let an electromagnetic field which in Σ
and Σ is described by the fields E = (E x , E y , E z ), B = (Bx , B y , Bz ) and E =
(E x , E y , E z ), B = (Bx , B y , Bz ), respectively. The boost relating Σ, Σ relates
the components of the fields as follows:
⎧
⎧
⎨ Bx = Bx
⎨ Ex = Ex
E y /c = γ (E y /c − β Bz ) B y = γ (B y + β E z /c)
⎩ ⎩ E z /c = γ (E z /c + β B y )
Bz = γ (Bz − β E y /c).
The vector form of these relations is
E = E
E⊥ = γ (E⊥ + βc × B),
B = B
1
B⊥ = γ (B⊥ − β × E)
c
where the parallel and the normal projections are understood wrt the relative velocity
u of Σ, Σ .
The electromagnetic field has two invariants under the Lorentz transformation
defined by
1
X ≡ − F i j Fi j
2
1
Y ≡ − ηi jkl F i j F kl .
8
These invariants in an LCF Σ are expressed in terms of the electric and the magnetic
field in Σ as follows:
X≡
1 2
E − B2
c2
,
Y ≡
1
E · B.
c
The equations of motion under the action of the electromagnetic field Fi j can be
obtained from the Lagrangian
.
L S R (r,r) = −mc
2
1−
u2
− q(φ − A · u)
c2
or the Hamiltonian
H = c (p−qA)2 + mc2 + qφ.
Chapter 11
Electromagnetic Field
11.1. (a) Show that the speed of a charge which moves under the action of a magnetic field only is a constant of motion.
(b) In the LCF Σ a charge moves under the action of the constant magnetic field
B = Bi. Assuming the initial conditions r(0) = y0 j + z 0 k, v(0) = υ0y j + υ0z k,
show that the orbit of the charge is the periphery of a circle. Compute the radius
(Larmor radius) of the circle and discuss the differences with the corresponding
classical orbit.
11.2. In the LCF Σ the observer the time moment t = 1 measures at the position
(x = 2, y = 3, z = 4) an electric field E = (0, E, 0) while the magnetic field
vanishes. In the LCF Σ , which moves in the standard configuration wrt Σ, the
event of measurement of the electromagnetic field has coordinates t = −1/4, x =
7/4, y = 3, z = 4.
(a) Calculate the speed of Σ wrt Σ.
(b) Calculate the electric and the magnetic field measured by the observer in Σ at
the same event.
11.3. In the LCF Σ there is an electromagnetic field whose electric and magnetic
fields are B = (0, B y , Bz ) and E = (0, E y , E z ). Given that the invariant Y = 1c B ·
E = 0, find the velocities of all LCF Σ for which the vector fields E , B are parallel.
11.4. Consider the four-vector Ωi which in the LCF Σ has components Ωi =
where Φ, A are scalar and vector fields, respectively.
Φ
c
A
(a) Define the vector fields E, B with the relations
E = −∇Φ +
∂
A,
∂t
B = ∇ × A,
75
76
11 Electromagnetic Field
and show that the components of the antisymmetric tensor1 Fi j = Ω j,i − Ωi, j
in Σ are the elements of the matrix:
⎛
⎞
E
0 − Ecx − cy − Ecz
⎜ Ex
⎟
⎜ c 0 Bz −B y ⎟
⎟,
Fi j = ⎜
⎜ Ey
⎟
⎝ c −Bz 0 Bx ⎠
Ez
B y −Bx 0
c
where we follow the convention that the first index counts columns and the
second index counts rows.
(b) Let Σ be another LCF which moves wrt Σ in the standard way along the x-axis
with speed factor β. Compute the components Fi j of the tensor F in Σ .
11.5. A charge q is moving along the z-axis of the LCF Σ with constant speed u.
Given that the moment t = 0 of Σ the particle passes through the origin of Σ
(a) Calculate the electromagnetic field (E, B) due to the particle in Σ. Find the
Newtonian limit of the fields E, B.
(b) Show that the electric field is isotropic (i.e., spherically symmetric) in Σ the
moment t = 0 only, while for t > 0 the field is anisotropic. Consider spherical
coordinates in Σ and write the fields in these coordinates. Show that in Σ the
field lines of the electric field are denser normal to the direction of the velocity
u of the particle and less dense parallel to the direction of u. Compute explicitly
the field lines of the magnetic field.
11.6. Along the x-axis of the LCF Σ the charges +q and −q are at rest at the
points x = ±l , respectively (electric dipole). Calculate the electromagnetic field in
the LCF Σ in which Σ moves in the standard way along the x-axis with speed u.
Calculate the asymptotic values of the field for distances l.
11.7. A conductor is resting along the x-axis of the LCF Σ when at its ends is
applied a constant potential difference and the subsequent development of constant
current I along the conductor.
(a) Show that the conductor remains electrically neutral in Σ after the application
of the potential difference.
(b) Let Σ be a second LCF which moves wrt Σ in the standard configuration along
the x-axis with speed β. Show that in Σ the conductor appears to be charged
and compute the charge density across the conductor. Explain the result in geometric terms.
1
The “,” between two indices indicates derivation wrt the index that follows.
11 Electromagnetic Field
77
11.8. Along the x-axis of the LCF Σ a charge is distributed with a uniform constant
density ρ0 , while a second point charge q of opposite sign rests at a distance r0 from
the x-axis.
(a) Calclulate in Σ the force exerted on the charge q.
(b) Calculate the same force on another LCF Σ which moves in the standard way
along the x-axis with speed u.
11.9. In the LCF Σ rests a straight conductor of infinite length which is charged
with a static charge of linear density λ0 > 0.
(a) Calculate the electric field in Σ created by the conductor at a point P in space
which is at a distance r0 from the conductor.
(b) In order to compute the magnetic field created by a current of intensity I, consider the conductor to move along its length in Σ with constant speed u and
show that the magnetic field which is created in Σ at a distance r from the
conductor is given by the expression
B=
μ0
(I × r0 ).
2πr02
(c) A charge q is moving in Σ parallel to the conductor with the same speed u.
Calculate the force on the charge.
11.10. A particle of mass m and charge q is moving in the LCF Σ in which there is
an electromagnetic field (E, B).
(a) Write the equations of motion of the particle in Σ in component form and in
vector form parallel and normal to the velocity.
(b) Prove that the equations of motion of the particle in Σ can be written in the form
d
(mγ vx ) =
dt
d
mγ 2 a y = (mγ v y ) =
dt
d
mγ 2 az = (mγ vz ) =
dt
mγ 3 ax =
dpx
dt
dp y
,
dt
dpz
dt
where a = dv
and v = dr
is the three-acceleration and the three-velocity of the
dt
dt
particle in Σ.
(c) Show that the rate of change of the energy E of the particle in Σ is given by the
expression
78
11 Electromagnetic Field
dE
= q(β · E)
dt
and deduce that the magnetic field does no work.
(d) Write the equations of motion of the particle in Σ in terms of the proper time of
the particle.
11.11. The purpose of this problem is to show that using the Lorentz force one
can derive the generalization of Newton’s second law in Special Relativity and the
fundamental formula E 2 = c2 p2 + m 2 c4 .
(a) Show that if the spatial parts of two four-vectors are equal in all LCF, then the
four-vectors are equal.
(b) A particle of mass m and charge q is moving in the LCF Σ under the action of
the electromagnetic field which in Σ is described by the pair of fields (E, B). We
assume that the equation of motion of the particle in Σ is given by the equation
d
(mγ v),
dt
q (E + v × B) =
where t is time in Σ and v is the velocity of the particle in Σ. The LHS of this
equation is the Lorentz force.
(b1) Show that the RHS of this equation can be written in covariant form as
follows:
1 d
(mu μ )
γ dτ
μ = 1, 2, 3,
where τ is the proper time of the particle and u i is the four-velocity of the
particle.
(b2) The electromagnetic field tensor Fi j in Σ has components
⎛
0 − 1c E x − 1c E y − 1c E z
⎜ 1E
⎜ x
Fi j = ⎜ 1c
⎝ c Ey
−Bz
1
E
c z
By
0
⎞
Bz −B y ⎟
⎟
⎟ .
0
Bx ⎠
−Bx
0
Show that the LHS of the equation of motion can be written in covariant
form as follows:
q
Fμj u j
γ
μ = 1, 2, 3.
From (b1) and (b2) conclude that the covariant form of the equation of motion
of the charged particle under the action of the electromagnetic field Fi j has the
following form:
11 Electromagnetic Field
79
μ
qF ju j =
d
(mu μ ) .
dτ
This equation contains the spatial components of the four-vectors qc F ji u j and
d
(mu j ) and it is valid independently of the particular LCF Σ we have employed
dτ
to derive it. Making use of (a) deduce that the covariant equation of motion of a
charge q under the action of the electromagnetic field Fi j is
q F ij u j =
dpi
,
dτ
where pi = mu i is the four-momentum of the charge.
(c) Define the four-vector of the conduction current J i = qu i and the four-vector
F i = F ji J j and show that the covariant equation of motion is written as
Fi =
dpi
,
dτ
which is similar to the Second Law of Newton. This result suggests that we
i
generalize the force law in Special Relativity with the relation F i = dp
where
dτ
F i is the four-force (specified uniquely for every three-force F). In addition
in the case of a charged particle, we consider the four-vector F i = F i j J j to
be the four-force on the conduction four-current J j due to the electromagnetic
field Fi j .
(d) Use the zeroth component of the equation of motion to prove the relation E 2 =
p2 + m 2 c4 which expresses the total energy of the charged particle in terms of
its three-momentum and its mass.
11.12. In the LCF Σ propagates a cylindrical (parallel) beam of electrons of radius
R whose charge density in the proper frame of the beam is axially symmetric about
the axis of the beam. If the total charge of the beam is I calculate in Σ
(a) The force and the velocity of an electron at a distance r from the symmetry axis
of the beam.
(b) If the length of the beam is L calculate the widening ΔA of the initial diameter
of the beam assuming ΔA L.
11.13. Two electrons A, B are projected from a vertical plane with common velocity v = vi from the points A, B, respectively, which are at a distance d apart and
are moving toward a vertical screen which is placed at a distance L from the plane
of emission (see Fig. 11.1).
Due to mutual repulsion the electrons divert during their motion from their initial direction at distances Δy A , Δy B , respectively. This phenomenon is known as
space charge defocusing. To compute the degree of defocusing of the two beams,
80
11 Electromagnetic Field
Fig. 11.1 Space charge defocusing
we assume (the same assumption is made in the Newtonian solution to the problem)
that the electrons are emitted simultaneously in the laboratory frame and subsequently they are moving always on the same vertical plane so that the x-component
of their velocities is equal and constant. As a result the two electrons hit the screen
simultaneously in the laboratory frame.
(a) Calculate the deviation Δy A in terms of the distance L traveled by the electrons
and the velocity of the electrons when they hit the screen. What you expect the
Newtonian result to be? Discuss the importance of the result for the beams of
charged particles.
(b) In the laboratory frame calculate the electromagnetic field and the acceleration
of the charge A. Compare the result with (a).
(c) Transform the four-acceleration from the proper frame of the charges and discuss if the result agrees with those of (a) and (b).
11.14. The electromagnetic field in the LCF Σ is given by the fields E, B. Define in
the complex space C3 the complex vector K = E + icB and show that the Lorentz
transformation of the fields corresponds to a rotation of the vector K in C3 for the
imaginary angle iφ = cos−1 γ .
Subsequently show that the only invariants of the electromagnetic field are
1 2
E
− B2 and 1c E · B.
c2
11.15. In the LCF Σ there is the uniform electromagnetic field (E, B). Show that
(a) If E·B = 0, then there are infinitely many LCF Σ in which the electromagnetic
field has
– Only magnetic field (i.e., E = 0) when E2 < B2 c2 .
– Only electric field (i.e., B = 0) when E2 > B2 c2 .
(b) If E · B = 0, then there are infinitely many LCF Σ in which the fields E , B
are parallel.
Note: If the electromagnetic field is not uniform, then the above holds at every
point of Σ separately.
11 Electromagnetic Field
81
11.16. In the LCF Σ a particle of mass m and charge q is moving under the action
of the constant electric field E = E x̂. If at the moment t = 0, the particle starts
from the origin of Σ with velocity v(0) = v0 ŷ show that the motion takes place on
the plane 'x y and compute the equation of the orbit.
x
= sinh−1 ax , a = 0.]
[Hint: √ad2 +x
2
11.17. A particle of mass m and charge q moves freely in the magnetic field B =
(0, 0, B).
(a) Show that the total energy and the length of the three-momentum are constants
of motion.
(b) Assume the initial conditions.
x(1) = y(1) = 0, z(1) = k,
ẋ(1) = λ (λ > 0), ẏ(1) = 0, ż(1) = k,
where k, λ are real constants and show that the orbit of the particle in space is
spiral:
x(t) =
λt
cos (a ln t − θ ) ,
1 + a2
y(t) = −
λt
λ
sin (a ln t − θ ) −
sin θ,
1 + a2
1 + a2
z(t) = kt,
where a =
qB
and tan θ = a.
mγ0
Finally, prove that the particle moves on the surface
λ2
x + y +
sin θ
1 + a2
2
2
=
λ2
z2.
k 2 (1 + a 2 )
11.18. In the region 0 ≤ x ≤ a of the LCF Σ there is a constant magnetic field
B = (0, 0, B). A particle of mass m and charge q is moving with constant speed u
along the x-axis of Σ when enters the region 0 ≤ x ≤ a from the side x < 0.
(a) Show that the energy of the particle remains constant during its motion in the
field B.
(b) Show that the orbit of the particle is the circle:
x 2 + y 2 + 2ky = 0,
z = 0,
82
11 Electromagnetic Field
where k = muγ
. Finally, show that if k < a the charge will be reflected from
qB
the magnetic field.
11.19. In the LCF Σ there is a uniform electromagnetic field with E = acj and
B = 35 ak (a = constant). A particle of mass m and charge q which rests at the point
x0 i of the x-axis of Σ is left to move freely. Calculate the time interval required in
Σ in order the particle to return to the x-axis.
11.20. In the LCF Σ there is the homogeneous magnetic field B = Bz. A particle
of mass m and charge q moves on the plane x − y of Σ.
(a) Show that the speed v of the particle is a constant of motion and that with
proper choice of initial conditions the motion is a uniform rotational motion
B
.
with angular speed ω = γ q(v)m
(b) Consider a second LCF Σ which moves wrt Σ in the standard configuration
along the x-axis with speed u. Calculate u and B so that in Σ the electromagnetic field is
E = (0, −E 0 , 0) ,
B = (0, 0, B0 ) .
Describe the motion of a charged particle of (a) in Σ and show that the average
speed is EB00 and is parallel to the axis x .
11.21. In the LCF Σ the uniform electromagnetic field is decomposed in the magnetic field B = B ẑ and the electric field with intensity E = 4Bc lying in the plane
x − z and making an angle θ = 0 with x-axis.
(a) Calculate the invariant Y = 1c E · B and draw your conclusions.
(b) Calculate the velocities of all LCF Σ for which the fields E , B of the given
electromagnetic field are parallel.
11.22. A plane electromagnetic wave of frequency ν which propagates in the x − y
plane of the LCF Σ at a direction which makes an angle θ with the the x-axis
consists of the fields
E = (−AX sin θ, AX cos θ, 0), B = (0, 0, AX/c),
sin θ
). Calculate the frequency and the direction of
where X = sin 2π ν(t − x cos θ+y
c
propagation of the electromagnetic wave in another LCF Σ which moves wrt Σ in
the standard configuration with velocity u = c cos θ i.
11.23. Let (E, B) and (E , B ) be the electromagnetic field in two LCF Σ and Σ
whose axes are parallel and their relative velocity factor β. Prove the transformation
equations
11 Electromagnetic Field
83
E = E ,
E⊥ = γ (E⊥ + βc × B),
B = B ,
1
B⊥ = γ (B⊥ − β × E).
c
Application
An observer Σ finds that in his proper frame there exists only electric field E.
Show that the electromagnetic field (E , B ) in another LCF Σ satisfies the relation
B + 1c β × E = 0. Similarly show that if in Σ the magnetic field B = 0, then in Σ
the following condition is satisfied E − 1c β × B = 0.
Note: From the above we conclude that if one of the fields E, B vanishes in a LCF,
then in all other LCF Σ the corresponding fields (of the same electromagnetic field!)
are perpendicular. The inverse is also true, that is, if in one LCF the inner product
E · B = 0, then there exists an LCF in which there is only electric or only magnetic
field.
11.24. (a) Describe the propagation of an electromagnetic wave in a conducting
medium with electric permeability , magnetic permeability μ, and conductivity σ .
(b) Show that in a conducting medium of electric permeability , magnetic permeability μ in which there are no free charges and conduction currents, an
electromagnetic wave propagates with speed √1μ and the fields H and E are
given by the relations
H(t, r) = H0 ei(ωt−k1 ·r) ,
E(t, r) = E0 ei(ωt−k1 ·r) ,
√
where the wave vector k has length k = ω μ.
11.25. Maxwell equations for empty space in terms of the electromagnetic potential
Ωi are
Ωi,i = 0,
i, j
j
Ω
= −μ0 J i ,
where J i is the four-current of charge density.
(a) Show that if two four-vectors Ai , ki are such that Ai k i = 0, ki k i = 0, that is,
k i is null and Ai is normal to k i , then Ai is not in general a null four-vector.
(b) Show that in a region of space in which there are no free charges and electric
r
currents (that is, J i = 0) one solution of Maxwell equations is Ωi = Ai eikr x
where the four-quantities Ai , ki are constant four-vectors normal to each other,
84
11 Electromagnetic Field
i.e., Ai k i = 0 and k i is a null four-vector (i.e., ki k i = 0). We call this solution
of Maxwell equations as plane electromagnetic wave.
11.26. Calculate the Lagrangian and the Hamiltonian of a charged particle of mass
m and charge e which moves in the electromagnetic field φ, A. Compare the results
with the corresponding Newtonian.
Solutions
Chapter 1
Mathematical Background: Problems
(1) The quantity Ai,i is invariant; therefore, we compute in Σ and then we find it in
Σ by using the Lorentz transformation relating Σ, Σ . We have
x
Ai,i = A0,0 + A1,1 + A2,2 + A3,3 = A,x
= e x + xe x = (1 + x)e x .
Σ, Σ are related with a boost along the x-axis with speed kc, hence γ =
(k < 1). It follows that
x = γ (x + βl ) = √
1
1 − k2
√ 1
1−k 2
(x + kl ),
which gives that the divergence of Ai in Σ is given by the expression
√ 1 (x +kl )
1
Ai,i = 1 + √
(x + kl ) e 1−k 2
.
1 − k2
[Note: Compute the same result by transforming first Ai in Σ and then compute the divergence in Σ .]
(2) The components of the vector Bν = Tμνμ are
Bν = T1ν1 + T2ν2 + T3ν3 .
We compute
B1 = T111 + T212 + T313 ⇒ B1 = 8x12 + 2(x22 + x32 )
B2 = T121 + T222 + T323 ⇒ B2 = 8x22 + 2(x32 + x12 )
B3 = T131 + T232 + T333 ⇒ B3 = 8x32 + 2(x12 + x22 ).
The div
divBν =
∂Bν
∂B1
∂B2
∂B3
=
+
+
= 16(x1 + x2 + x3 )
∂ xν
∂ x1
∂ x2
∂ x3
87
88
1 Mathematical Background: Problems
and the curl
∂B3
∂B2 ∂B1
∂B3 ∂B2
∂B1
−
,
−
,
−
∂ x2
∂ x3 ∂ x3
∂ x1 ∂ x1
∂ x2
= 4(x2 − x3 , x3 − x1 , x1 − x2 ).
curlBν =
(3) (i)
Aμν,ν =
∂
(x 2
∂ x1 μ
=
∂ Aμν
∂ xν
=
∂ Aμ1
∂ x1
+ x12 ) +
+
∂ Aμ2
∂ x2
∂
(x 2
∂ x2 μ
+
∂ Aμ3
∂ x3
+ x22 ) +
=
∂
(x 2
∂ x3 μ
+ x32 ) =
= 2x1 (1 + δμ1 ) + 2x2 (1 + δμ2 ) + 2x3 (1 + δμ3 ) ⇒
Aμν,ν = 2
,3
ν=1
xν (1 + δνν ) .
(ii)
Aμν,μν = 4
,3
μ=1 δμμ
= 12 .
(4) (a) The matrix of the transformation is
⎡
cosh φ
⎢ − sinh φ
L=⎢
⎣
0
0
− sinh φ
cosh φ
0
0
0
0
1
0
⎤
0
0 ⎥
⎥.
0 ⎦
1
It is easy to show that η = L t ηL where η = diag(−1, 1, 1, 1). It follows
that L is a Lorentz transformation
(b) The transformation of the contravariant vector V i is given by the relation
∂xi i
V =
V = L ii V i ,
∂xi
i
where L = [L ii ]. We compute
V 0 = L i0 V i = L 00 V 0 + L 01 V 1 + L 02 V 2 + L 03 V 3 = − sinh φ,
V 1 = L i1 V i = L 10 V 0 + L 11 V 1 + L 12 V 2 + L 13 V 3 = cosh φ,
V 2 = L i2 V i = L 20 V 0 + L 21 V 1 + L 22 V 2 + L 23 V 3 = 0,
V 3 = L i3 V i = L 30 V 0 + L 31 V 1 + L 32 V 2 + L 33 V 3 = 1.
Hence [V i ] = (− sinh φ, cosh φ, 0, 1)t .
1 Mathematical Background: Problems
89
For the tensor Ti j we have
Ti j =
∂xi ∂x j
j
Ti j = L ii L j Ti j = L i1 L 1j T1 1 + L i3 L 3j T3 3 ,
i
j
∂x ∂x
= L i1 L 1j + L i3 L 3j
where we have used that the only non-zero components of Ti j in Σ (x , y , z )
are the T0 0 = T3 3 = 1. From this relation we compute the components of
the tensor Ti j in Σ(x, y, z) . For example, for the component T00 we have
T00 = L 10 L 10 + L 30 L 30 = sinh2 φ.
The result of the calculation is given in the following symmetric matrix
⎡
sinh2 φ
⎢ − sinh φ cosh φ
[Ti j ] = ⎢
⎣
0
0
− sinh φ cosh φ
cosh2 φ
0
0
0
0
0
0
⎤
0
0 ⎥
⎥.
0 ⎦
1
Concerning the vector ai = Ti j V j , we have in Σ(x, y, z):
a0 = T00 V 0 + T01 V 1 + T02 V 2 + T03 V 3 = − cosh φ sinh φ
a1 = T10 V 0 + T11 V 1 + T12 V 2 + T13 V 3 = cosh2 φ
a2 = T20 V 0 + T21 V 1 + T22 V 2 + T23 V 3 = 0
a3 = T30 V 0 + T31 V 1 + T32 V 2 + T33 V 3 = 1.
Finally, the invariant ai V i = Ti j V i V j is
ai V i = a0 V 0 + a1 V 1 + a2 V 2 + a3 V 3 = cosh2 φ + 1.
(c) It is left as an exercise for the reader.
(5) To find the geodesics we compute the Lagrangian of the metric and then
the geodesic equations are the Lagrange equations for each coordinate. The
Lagrangian of the metric ds 2 is
L=
1
1
gi j x i x j = 2 (−ṫ 2 + ẋ 2 ).
2
2t
The t-coordinate is
∂L
ṫ
=− 2,
∂ ṫ
t
∂L
1
= − 3 (−ṫ 2 + ẋ 2 ).
∂t
t
90
1 Mathematical Background: Problems
The t-geodesic equation is
d
ds
∂L
∂ ṫ
−
∂L
d
=0⇒
∂t
ds
1
ṫ
− 2 + 3 (−ṫ 2 + ẋ 2 ) = 0 ⇒
t
t
ẗ
2ṫ 2
ṫ 2
ẋ 2
ẗ
ṫ 2
ẋ 2
−
+
−
=
0
⇒
−
−
=0⇒
t2
t3
t3
t3
t2
t3
t3
t ẗ − ṫ 2 − ẋ 2 = 0.
The x- coordinate is
ẋ
∂L
= 2,
∂ ẋ
t
∂L
=0
∂x
Hence1 :
ẋ
t2
= 0 ⇒ ẋ = at 2 ⇒ a = const.
Next we integrate the geodesic equations.
Dividing the t-geodesic with t 2 we find
ṫ 2
ẗ
ẋ 2
− 2 − 2 =0⇒
t
t
t
.
.
ṫ
ṫ
− ax = 0.
− a ẋ = 0 ⇒
t
t
Hence,
ṫ
− ax = β,
t
The ẋ =
dx
ds
=
d x dt
dt ds
= x ṫ, where ≡
d
.
dt
β = const.
Therefore,
at 2 = x ṫ ⇒ ṫ =
at 2
.
x
1 From the equation of the geodesics, it is possible to compute the connection coefficients. It turns
out that the non-vanishing connection coefficients Γijk are the following:
Γ000 =
where t ←→ 0, x ←→ 1.
1
,
t3
Γ110 = Γ101 = −Γ011 = −
1
,
t3
1 Mathematical Background: Problems
91
Replacing we find
at
1
1
− ax = β ⇒ (ax + β)x = at ⇒ ax 2 + βx + γ = at 2
x
2
2
x2 +
(γ = const.).
2
2γ
2β
β 2
β
2γ
x+
= t2 ⇒ x +
.
= t2 +
−
a
a
a
a
a
We set
β
= −x0 ,
a
2
β
2γ
=A
−
a
a
and obtain the equation
(x − x0 )2 = t 2 + A,
where x0 , a are constants. We conclude that the geodesics are hyperbolae with
asymptotes the null lines on the light cone and origin the point x0 . Indeed the
last equation gives the solution
√
τ
x − x0 = |A| cosh √
|A|
√
τ
t = |A| sinh √
.
|A|
These equations represent the world line of a point accelerating along the x-axis
with constant proper acceleration √1|A| . τ is the proper time of the accelerating
point.
Note: From the result follows that the coordinates (t, x) which define the metric
ds 2 = t12 (−dt 2 + d x 2 ) are “inertial” wrt the accelerating point, in the sense that
the later moves along geodesics of the metric. This result justifies the Equivalence Principle which is fundamental in the Theory of General Relativity and in
a simplified form can be stated as follows:
Free fall in a gravitational field is along the geodesics of a non-flat metric
with Lorentzian character in a space free of gravity.
(6) Let L be a homogeneous2 linear transformation in Minkowski space:
L:
x0
xμ
Σ
x0
→
xμ
,
(1.1)
Σ
Homogeneous means that we correspond Σ and Σ so that when the origins coincide the Lorentz
transformation is the identity. This is usually stated as when t = t = 0, x μ = x μ = 0.
2
92
1 Mathematical Background: Problems
where
0 the small case Greek indices is assumed to take the values 1, 2, 3 and
x
is a four-vector. In terms of components the transformation is written
xμ Σ
as follows:
x 0 = L 00 x 0 + L 0μ x μ
x
μ
=
μ
L0 x0
+
(1.2)
L μμ x μ .
(1.3)
The Lorentz transformation L is a particular linear transformation which is
defined by means of certain physical assumptions. The first assumption is the
so-called reciprocity principle which is the demand that the inverse transformation is the inverse of the matrix (L aa ) in Σ. This is expressed by the relation
x 0 = L 00 x 0 + L 0μ x μ
μ
μ
x μ = L 0 x 0 + L μ x μ .
Applying twice the transformation we find
x0
xμ
*
=
*
=
*
=
+
+
L 00 L 0μ
μ
L 0 L μμ
L 00 L 0μ
μ
L 0 L μμ
x0
xμ
L 00 L 0ν μ
μ
L 0 L ν μ
x0
xν
μ
L 00 L 00 + L 0μ L 0 L 00 L 0ν + L 0μ L ν L μ0 L 00 + L μμ L μ0 L μ0 L 0ν + L μμ L μν
+
x0
,
xν
from which the following conditions on the components of L follow:
L 00 L 0ν + L 0μ L μν = 0
μ
+
μ
L 0μ L 0
+
μ
L μμ L ν μ
L 0 L 00 + L μμ L 0 = 0
L 00
L 00
μ
L 0 L 0ν =1
=
μ
δν .
(1.4)
(1.5)
(1.6)
(1.7)
order to state the second requirement we assume that the four-vector
In
x0
is the position four-vector of a point P. Then the velocity of P in Σ is
xμ Σ
uμ =
dx μ
.
dx 0
1 Mathematical Background: Problems
93
Replacing dx μ , dx 0 from the transformation we find
μ
u
μ
L 0 dx 0 + L μμ dx μ
L 0 + L μμ u μ
dx μ
=
=
=
,
dx 0
L 00 dx 0 + L 0μ dx μ
L 00 + L 0μ u μ
μ
(1.8)
μ
where u μ = dx
the velocity of the point in Σ.
dx 0
μ
The speed of the origin of Σ wrt Σ is u μ |0 = L 0 /L 00 because the velocity
μ
u of Σ wrt itself vanishes. Similarly the velocity of the origin O of Σ wrt Σ
is
u μ |0 =
μ
L 0
.
L 00
Now we set the second requirement which is the “relativity of motion,” that
is, the relative velocity of the origins of Σ, Σ are opposite. This is expressed
mathematically by the condition:
Lμ
μ
u |0 = −u |0 ⇔ 00 = −u μ ⇒ L 0 = −L 00 u μ .
L0
μ
μ
(1.9)
We need one more requirement. This is the “synchronization” of the clocks,
which is expressed by the following condition:
L 0 = ±L 0 0 = ∓γ ⇒ L 00 = −L 0 0 = ±γ ,
L 0μ = L μ 0 ,
(1.10)
(1.11)
where3 γ is a parameter which has to be determined in terms of the relative
velocity of the frames Σ, Σ .
Requirement (1.10) gives for the requirement (1.9):
L μ0 = ∓γ u μ .
(1.12)
Requirement (1.11) simplifies requirement (4.17) as follows:
μ
L μμ L ν =
μ
δν +
μ
L 0 L 0ν =
μ
δν L μ L ν0 μ
+ 00 0 L 00 L 00 = δν + u μ u ν γ 2 .
L0 L0
(1.13)
From the above requirements we determine the Lorentz transformation L aa
relating Σ, Σ .
3
We raise and lower the indices with η = diag(−1, 1, 1, 1).
94
1 Mathematical Background: Problems
Determination of the element (00)
From (1.10) we have L 00 = ±γ . Therefore, all we have to do is to determine
the factor γ in terms of the relative speed of Σ, Σ . To do that we note that
condition (10.10) gives
L 00 L 00
L 0μ L μ
−1/2
+ 0 00 = 1 ⇒ γ 2 + γ 2 u μ (−u μ ) = 1 ⇒ γ 2 = 1 − u 2
. (1.14)
L 0 L 0
Determination of the elements (0μ), (μ0)
μ
This has been done because from (1.12) we have L 0 = ∓γ u μ and condition
t
μ
(1.11) gives4 L 0μ = L 0 .
Determination of the element L μμ
First we note that condition (4.15) gives
L μμ
μ
μ
L
L 0
μ
μ
+ L 0 = 0 ⇒ L μμ u μ = −L 0 = − 00 L 00 = ∓γ u μ .
L 00
L0
(1.15)
To determine the spatial part L μμ we solve the system of equations (1.13), (4.20).
We look for a solution of the form
L μμ
=
aδμμ
uμ uμ
+b 2 ,
u
(1.16)
where u 2 = u ρ u ρ . Substituting in (1.13) we get
*
aδμμ
uμ uμ
+b 2
u
+
μ
aδν + b
uμuν
u2
μ
= a 2 δν + 2ab
= δνμ + γ 2 u μ u ν .
μ
uμ uν
2 u uν
+
b
=
u2
u2
(1.17)
From the last equation we conclude that
a 2 = 1, 2ab + b2 = γ 2 u 2 .
(1.18)
Adding these equations and using (1.14), we find
(a + b)2 = 1 + γ 2 u 2 = γ 2 ,
4
Where t stands for transpose.
(1.19)
1 Mathematical Background: Problems
95
from which follows
b = ±γ ∓ 1, a = ±1.
(1.20)
We examine now (4.20). Substitution of (1.16) in (4.20) gives
*
aδμμ
uμ uμ
+b 2
u
+
u μ = (a + b) u μ = ±γ u μ ,
(1.21)
from which we find that
a + b = ±γ .
(1.22)
This condition gives us nothing new and we have the final solution:
L μμ
=
±δμμ
!
uμ uμ
uμ uμ
μ
+ ±(γ − 1) 2 = ± δμ + (γ − 1) 2
.
u
u
(1.23)
We conclude that the Lorentz transformation relating Σ, Σ is the following:
*
L Σ,Σ =
∓γu μ
uμ u
∓γ u μ ± δμμ + (γ − 1) u 2 μ
±γ
+
.
(1.24)
We note that we have determined four Lorentz transformations which satisfy
the assumed requirements. To distinguish between them we compute the determinant of the transformation. It is an easy exercise to show that
detL Σ,Σ = 1 + γ (ε − 1),
(1.25)
where ε = ±1. We call the Lorentz transformation with determinant equal to
+1 as the proper Lorentz transformation. Two of the four Lorentz transformations are proper Lorentz transformations. The rest two are improper Lorentz
transformations and their determinant is = 1.
Another selective rule is defined by means of the sign of the term L 00 .
We call the proper Lorentz transformations defined by the value +γ as the
orthochronous Lorentz transformations. These transformations preserve the
direction of time between Σ, Σ (that is, the dt dt > 0). The orthochronous
Lorentz transformations form a subgroup of the totality of Lorentz transformations (because they contain the identity transformation).
96
1 Mathematical Background: Problems
The various Lorentz transformations have different names as follows:
The orthochronous Lorentz transformation:
*
L Σ,Σ =
γ
−γ u μ
−γ u μ δμμ + (γ − 1)
+
uμ uμ
u2
.
(1.26)
.
(1.27)
The time inversion:
*
L Σ,Σ =
μ
−γ γ
u uμ u
γ u μ δμμ − (γ + 1) u 2 μ
+
The space inversion:
*
L Σ,Σ =
μ
−γ u
uμ u
−γ u μ δμμ − (γ + 1) u 2 μ
γ
+
.
(1.28)
.
(1.29)
The space and time inversion:
*
L Σ,Σ =
μ
−γ γ
u uμ u
γ u μ −δμμ + (γ + 1) u 2 μ
+
In the literature by proper Lorentz transformation it is understood the orthochronous
Lorentz transformation. We shall follow this abuse of terminology in this book.
Chapter 2
Classic Experiments
2.1. (a) The time required for a collinear passing of the light through the medium,
according to Fresnel hypothesis, is
t1 =
c
n
,
+ κV
where κ is the drag coefficient. Similarly for the opposite course we have
t2 =
c
n
.
− kV
The time difference between the two courses equals:
2κ V
2n 2 κ V
ΔT = t2 − t1 = 2
≈
.
c
2V 2
c2
−
κ
n
Using the relation T = λ/c relating the wavelength with the period of the wave
we have
N=
2kn 2 V
Δλ
ΔT
2n 2 κ V
N λc
=
=
≈
⇒κ=
.
λ
T
T c2
λc
2n 2 V
The experimental measurements agree with the above result thus justifying
the Fresnel hypothesis.
(b) In order to show that the Fresnel hypothesis results directly from the Theory
of Special Relativity without any further (i.e., the dragging) assumption, we
consider two RIO Σ, Σ which are related with a boost and are such that
r
r
The common x, x is along the direction of the propagation of light.
Their relative speed is V , that is, Σ is comoving with the medium.
Then in Σ the refraction index is given by n = vc where v is the speed of
light in the comoving (isotropic and homogeneous) medium. The speed of the
beam in Σ is given by v = nc and it is computed in terms of v , V by means of
97
98
2 Classic Experiments
the relativistic three-velocity composition rule
v=
v + V
,
1 + vc2V
where = 1 when the speed of the beam is collinear with the speed of Σ in Σ ,
and = −1 differently. Replacing we find
2 !
V
εV
+ εV
1−
+O
v=
=
n
nc
cn
1 + εV
nc
2 !
V
V
εV
1
V2
c
c
+O
.
= − 2 + εV −
= +ε 1− 2 V +O
n
n
cn
cn
n
n
cn
c
n
+ εV
c
We see that the speed of light in the moving medium to the first approximation
equals the speed ( = c/n) in the resting medium augmented by the term (1 −
1/n 2 )V = κ V , where κ = (1 − 1/n 2 ) is the drag coefficient introduced by
Fresnel.
2.2. During the rotation of the circular table the speed of each glass tube is v = ωr
where r is the distance of the center of the tube from the rotation center. Obviously
r=
1 2
4R − 2 .
2
Hence
v = ωr =
ω 2
4R − 2 .
2
The speed of light along the direction of rotation is nc + κv, where κ is the dragging
coefficient and in the opposite direction nc − κv. Therefore, the time of traveling
time is, respectively,
t1 =
c
n
2
+ κωr
,
t2 =
c
n
2
.
− κωr
2 Classic Experiments
99
The phase difference of the light signals in the telescope equals
t2 − t1 = 2
c
n
1
−
− κωr
2κωr
= 2 2
c
n
c
n
1
+ κωr
− κ 2 ω2 r 2
1
4κωr n 2
κ 2 ω2 r 2 n 2
c2
1 − c2
√
2
4κωr n
κωn 2 4R 2 − 2
≈
=
.
c2
c2
=
This time difference corresponds to a difference in the optical path equal to
√
κωn 2 4R 2 − 2
.
c(t2 − t1 ) =
c
1
(b) For n = 1.33, we have for the drag coefficient κ = 1 − n12 ⇒ κ1 − 1.33
2 = 0.25.
From the general formula, we have.
αλ =
2 1 −
1
n2
√
n 2 ω 4R 2 − 2
c
Replacing we calculate the result.
⇒ω=
2
n2
αλc
.
√
− 1 4R 2 − 2
Chapter 3
The Position Four-Vector
3.1 The Speed of Light
3.1.2. The speed of the light spot on the surface of the Moon is
v = ω · R = 1 s−1 × 380 × 103 km = 380 × 103 kms−1 .
We note that v > c! This does not contradict the assumption that the maximal
speed is the speed of light because the light spot on the surface of the Moon is not
the light beam itself but the image of the light source on the surface of the Moon.
Furthermore, the light spot itself does not transport information (the beam does!).
A similar case we have in the Galaxy Grab Nebula which is rotating and the ray
which emits reaches our telescopes 30 times per second (hence ω = 30 rotations/s)
whereas its distance from our Galaxy is many light years. If we consider a spherical
surface centered at our Galaxy and radius equal the distance of the Grab Nebula
from our Galaxy, then we see that the image of the light it emits moves on the
surface with huge speed compared to the speed of light c.
3.2 The Lorentz Transformation
3.2.1. Assume that the center of the circle is at the origin of the coordinates. Then
the motion of the particle in Σ is described by the equations
x = R cos ωt,
y = R sin ωt,
z = 0.
We infer that the world line of the particle is a helix on a cylinder of radius R and
axis along the time axis of Σ.
101
102
3 The Position Four-Vector
3.2.2. The proper time of clock 1 at the event B is
1
(τ B − τ A )1 =
c
(B
1
dτ1 =
c
A
(B (t B υ 2
1
0
2
2
2
tB .
1−
c dt1 − d x1 =
dt =
c
γ (υ0 )
A
0
Similarly the proper time of clock 2 equals
(τ B − τ A )2 =
=
=
a0
c
tB
2
1
c
't B c 2
a0
0
'B
dτ2 =
A
−
=
t 2 dt
'B 1
c
c2 dt22 − d x22 =
't B A
a0
2c
1−
0
- 2
c
a0
t
−
t2
+
2
c
a0
a0 t 2
c
dt
!t B
sin−1 tac0
0
1−
t B a0 2
c
+
c
a0
sin−1
t B a0 c
.
In order to compute the time t B , we note that at the event B the two clocks meet;
therefore, x B = υ0 t B , x B = 12 a0 t B2 . It follows:
υ0 t B =
1 2
2υ0
a0 t B ⇒ t B =
.
2
a0
Replacing the value of t B in the previous formulae we compute
(τ B − τ A )1 =
(τ B − τ A )2 =
υ 2 2υ
0
0
c
a0
υ 2
c
0
−1 2υ0
.
1−4
+
sin
c
a0
c
1−
-
υ0
a0
The argument of the function sin−1 must be ≤ 1. This means that υ0 must satisfy
the constraint 2υc 0 ≤ 1 or υ0 ≤ 2c . The kinematic interpretation of this condition is
that if υ0 > 2c , then the clocks 1, 2 do not meet.
In order to compare the periods (τ B −τ A )1 and (τ B −τ A )2 , we consider the limiting
case υ0 = 2c . Then
1 −1/2
2
γ (υ0 ) = 1 −
=√
4
3
from which follows
√
3 c
(τ B − τ A )1 =
2 a0
and
(τ B − τ A )2 =
π c
.
2 a0
We note that for this speed (τ B −τ A )2 > (τ B −τ A )1 which was to be expected because
clock 1 moves inertially along a straight line (geodesic) in spacetime and this is the
3.2
The Lorentz Transformation
103
(timelike) line with the minimal length between the events A, B. Obviously this
conclusion holds for all υ0 ≤ 2c .
Note: It is interesting to study the condition υ0 ≤ 2c along physical grounds. This is
done if we look at (a) spacetime intervals and (b) relative velocities.
(a) Spacetime intervals
It must hold dτ12 < 0 , dτ22 < 0 because the world lines of the clocks are timelike.
Therefore, we have the conditions:
Clock 1:
d x12 − c2 dt12 < 0 ⇒ 0 < υ0 < c.
Clock 2:
d x22 − c2 dt22 < 0 ⇒ 0 < a0 t2 < c.
During the course of motion, t2 increases and its maximal value is
tB =
2υ0
.
a0
Replacing in the last relation we find
υ0 <
c
.
2
This condition is necessary in order the two clocks to meet at the event B. If
υ0 = 2c , the two clocks never meet.
(b) Relative velocities
We compute the relative speed of clock 2 wrt clock 1. To do that we apply the boost
relating the RIO Σ and the proper frame of clock 1. The relativistic formula for the
transformation of three-velocities is
υx − υ0
,
1 − υcx 2υ0
υy
υy =
γ (υ0 ) 1 −
υz
υz =
γ (υ0 ) 1 −
υx =
υx υ0
c2
υx υ0
c2
,
.
Because the clocks move along the x-axis of Σ and noting that the speed of clock
2 in Σ is
υ2 =
dx
= a0 t
dt
104
3 The Position Four-Vector
and that of clock 1
υ0 > 0
We calculate for the speed of 2 relative to 1:
υ21 =
a0 t − υ0
.
1 − υ0cta2 0
It must hold υ21 < c. Hence we have the condition
&
&
&a t −υ &
0&
& 0
&
& < c.
& 1 − υ0cta2 0 &
At the event B we have
&
&
&
&
&
&
&a t − υ &
&
2υ
−
υ
0&
0
0 &&
& 0B
&
< c ⇒ 2υ02 + cυ0 − c2 < 0.
&
& < c ⇒&
2 &
& 1 − υ0ct B2 a0 &
& 1 − 2υ20 &
c
The roots of this equation are
c
υ21 = −c, .
2
The first root −c is rejected and we end up again with the condition 0 < υ0 < 2c .
3.2.3. The Lorentz transformation is defined by two (main) requirements:
(a) It is linear.
(b) Preserves the Lorentz inner product.
The given transformation is linear; therefore, it remains to examine only condition
(b). We consider the invariance of the fundamental form
x 2 + y 2 + z 2 − 2 = x + y + z − .
2
2
2
2
Replacing the primed coordinates by means of the transformation, we find
− 2 + x 2 + y 2 + z 2
= −(a + bx + cy + dz)2 + (b + ax)2 + y 2 + z 2
= (−a 2 + b2 )2 + (−b2 + a 2 )x 2 + (−c2 + 1)y 2 + (−d 2 + 1)z 2 −
−2acy − 2adz + 2(ba − ab)x − 2bcx y − 2bd x z − 2cdyz
= −2 + x 2 + y 2 + z 2
3.2
The Lorentz Transformation
⇒
105
√
−b2 + a 2 = 1 ⇒ a = ± b2 + 1
.
c=d=0
Therefore, the Lorentz transformation relating Σ (, x, y, z) and Σ ( , x , y , z ) is
given by the expression
⎧ ⎪
⎪
⎨ x
⎪
y
⎪
⎩ z
√
⎫
= ε1 b2 +
⎪
√1 + bx ⎪
⎬
= b + ε1 b2 + 1x
, where ε1 = ±1.
⎪
=y
⎪
⎭
=z
We define two new parameters β ∈ (−1, 1) , γ ∈ (0, +∞) with the relations
1 + b2 =
1
= γ 2 ⇒ b = ε2 γβ , where ε2 = ±1.
1 − β2
Then the transformation is written as follows:
⎧ ⎪
⎪
⎨ x
⎪ y
⎪
⎩ z
⎫
ε1 ε2 γ + ε2 γβx ⎪
⎪
⎬
ε2 βγ + ε1 ε2 γ x
.
y
⎪
⎪
⎭
z
=
=
=
=
These relations contain four families of Lorentz transformations depending on the
value of the constants ε1 , ε2 . We have
ε1 = 1, ε2 = 1
⎧ ⎪
⎪
⎨ x
⎪ y
⎪
⎩ z
⎫
= γ ( + βx) ⎪
⎪
⎬
= γ (x + β)
.
=y
⎪
⎪
⎭
=z
This type of Lorentz transformation is called time reversal.
ε1 = 1, ε2 = −1
⎧ ⎪
⎪
⎨ x
y
⎪
⎪
⎩ z
=
=
=
=
⎫
γ (− − βx) ⎪
⎪
⎬
γ (−x − β)
.
y
⎪
⎪
⎭
z
This type of Lorentz transformation is called spacetime reversal.
106
3 The Position Four-Vector
ε1 = −1, ε2 = 1
⎧ ⎪
⎪
⎨ x
y
⎪
⎪
⎩ z
=
=
=
=
⎫
γ (− + βx) ⎪
⎪
⎬
γ (−x + β)
.
y
⎪
⎪
⎭
z
This type of Lorentz transformation is called space reversal.
ε1 = −1, ε2 = −1
⎧ ⎪
⎪
⎨ x
y
⎪
⎪
⎩ z
⎫
= γ ( − βx) ⎪
⎪
⎬
= γ (x − β)
.
=y
⎪
⎪
⎭
=z
The final transformation is called the proper Lorentz transformation and it is the
one we apply, as a rule. The reason for this preference is (among others) that; (a)
preserves the causal relation of the events and (b) contains the identity; therefore,
the set of all proper Lorentz transformations forms a group.
Note: An additional important element of the Lorentz transformations has to do
with the orientation (left-handed and right-handed) of the coordinate frames. The
Lorentz transformation preserves the orientation of the frame only if its determinant
equals +1, otherwise the orientation changes (a fact we do not want in general). The
proper Lorentz trans