Sample Questions
0
0
Q.1 : Consider two inertial reference frames S and S , where S is moving with constant speed
v along the positive x-direction relative to S. Now imagine an observer in S sees a (electrically
charged) particle of rest mass m and charge q, start from rest at the origin of S and accelerate
uniformly with acceleration a in the positive y-direction. Find the velocity vector and acceleration
0
vector of the same particle as seen from S . Find the rate of change of momentum (vector) of the
0
particle in both S and S .
1
0
A. In reference frame S we have ~u ≡ (ux , uy , uz ) = (0, at, 0). In S if γ = (1 − v 2 /c2 )− 2 ,
0
0
0
0
dx = γ(dx − v dt) ; dt = γ(dt − v/c2 dx) ; dy = dy ; dz = dz
Hence,
0
dx
ux − v
= −v
ux = 0 =
dt
1 − ux v/c2
0
0
0
uy =
uy
dy
at
=
0 =
dt
γ(1 − ux v/c2 )
γ
0
uz = uz = 0
0
0
0
Hence ~u = (−v, at
, 0). Now d~u = (0, γa dt, 0), hence d~u = (0, adt, 0). Hence ~a =
γ
0
and ~a = (0, a, 0). Hence ~a = (0, γa2 , 0). Now,
m~u
m(0, at, 0)
= q
2 2
1 − ~u2 /c2
1 − ac2t
p~ = q
Thus,
d~p
m(0, a, 0)
=
2 2 3
dt
(1 − ac2t ) 2
0
m(−v, at
, 0)
γ
p~ = q
1−
0
d~p =
m(1 −
v2
c2
(1 −
v2
c2
a2 t2
)(0, adt
, 0)
γ 2 c2
γ
3
2
2
2
v
− γa2 ct 2 ) 2
c2
−
−
+
a2 t2
γ 2 c2
, 0)
m(−v, at
γ
(1 −
2
0
d~p =
v2
c2
, m adt
, 0)
(−mv aγ 2tdt
c2
γ3
(1 −
v2
c2
−
a2 t2 32
)
γ 2 c2
−
a2 t2
γ 2 c2
)
3
2
(
a2 tdt
)
γ 2 c2
0
d~
u
dt0
= (0, γa dtdt0 , 0)
2
2
2
0
(−mv γa3 ct2 , m γa4 , 0)
(−mv ac2t , m γa , 0)
d~p
=
=
2
2 2 3
2 2 3
dt0
(1 − vc2 − γa2 ct 2 ) 2
(1 − ac2t ) 2
(1)
Q.2 It is given that the particle in reference frame S in Q.1 is accelerating because there is a
uniform electric field in the positive y direction (and no magnetic field). Find the magnitude of that
0
electric field. What fields do an observer in S see ? Calculate all the components of both electric
0
and magnetic fields as seen from S . Explicitly verify both the relations,
d
~ + ~u × B)
~
p~ = q(E
dt
c
in reference frame S and
0
d 0
~ 0 + ~u × B
~ 0)
~ = q(E
0p
dt
c
0
in reference frame S .
~ = 0. Hence,
A. In S , B
m(0, a, 0)
d
~
p~ =
2 2 3 = qE
dt
(1 − ac2t ) 2
Hence,
~
E(t)
=
m
(0, a, 0)
q
2 2 3
(1 − ac2t ) 2
~
; B(t)
=0
According to the transformation law of electric and magnetic fields,
0
0
Bx (t ) = Bx (t) = 0
v
0
0
By (t ) = γ(By (t) + Ez (t)) = 0
c
v
mv
a
0
0
Bz (t ) = γ(Bz (t) − Ey (t)) = −γ
c
q c (1 − ac22t2 ) 23
and
0
0
Ex (t ) = Ex (t) = 0
m
a
v
q
Ey (t ) = γ(Ey (t) − Bz (t)) = γ
2 2 3
c
(1 − ac2t ) 2
0
0
v
0
0
Ez (t ) = γ(Ez (t) + By (t)) = 0
c
3
0
Now we calculate the Lorentz force in S ,
0
0
0
~ 0 + ~u × B
~ 0)
F~ (t ) = q(E
c
Thus,
0
0
Fx (t ) = q(0 +
0
0
Fy (t ) = γ
at
mv
mv
a
a2 t
(−γ)
)
=
−
cγ
qc (1 − ac22t2 ) 32
c2 (1 − ac22t2 ) 23
ma
(1 −
a2 t2 32
)
c2
−γ
0
v2
ma
ma
2 t2 3 =
2 2 3
2
a
c (1 − c2 ) 2
γ(1 − ac2t ) 2
0
Fz (t ) = 0
Thus,
0
0
F~ (t ) =
a2 t, ma
, 0)
(− mv
c2
γ
(1 −
a2 t2 23
)
c2
But this is the same as Eq.(1). Hence,
0
0
~ 0 + ~u × B
~ 0 ) = d~p0
q(E
c
dt
Q.3 Withdrawn. Even though both the equations ∂µ T µ0 = 0 and ∂µ j µ = 0 are valid, it does
**not** mean that T µ0 is a four-vector just because j µ is a four-vector. In order to ascertain the
transformation we have to use,
T
0 µν
= Λµρ Λνσ T ρσ
0
where x µ = Λµν xν . Now if we put ν = 0 we get ( i = 1, 2, 3 ),
T
0 µ0
= Λµρ Λ0σ T ρσ = Λµ0 Λ0σ T 0σ + Λµρ Λ00 T ρ0 + Λµi Λ0σ T iσ + Λµρ Λ0i T ρi
0
Thus the transformed T µ0 will involve objects such as T iσ , this is not present in a four-vector
transformation. A four-vector transformation should simply be,
T
0 µ0
= Λµν T ν0
but it is not.
Q.4 Find ds2 in terms of appropriate parametric variables (such as θ, φ) on the following two
dimensional surfaces.
(a) Torus of smaller radius a and bigger radius b.
2
2
(b) Ellipsoid of revolution obtained by rotating the ellipse xa2 + yb2 = 1 about the x-axis.
A.
4
(a) In terms of cartesian coordinate unit vectors the position on the torus with respect to the
origin at the center of the big circle is ~r where,
~r = ρ~ + ~r
0
where,
0
~r = a cos(θ)(−ρ̂) + a sin(θ) k̂
where ρ~ is the vector that joins from the center of the big circle of radius b to the center of the small
0
circle of radius a and ~r is the vector joining the point in question (on the torus) from the center of
the small circle. Thus,
ρ~ = b cos(φ) î + b sin(φ) ĵ
Thus the position of a point on the surface of a torus from the center of the big circle is given by,
~r = (b − a cos(θ))(cos(φ) î + sin(φ) ĵ) + a sin(θ) k̂
where φ is the angle swept along the big circle and θ is the angle swept along the small circle. Now
~ · dr.
~
ds2 = dr
d~r = (a sin(θ)dθ cos(φ) − (b − a cos(θ))sin(φ) dφ) î
+(a sin(θ) sin(φ) dθ + (b − a cos(θ))cos(φ)dφ) ĵ + a cos(θ)dθ k̂
Therefore,
ds2 = a2 dθ2 + (b − a cos(θ))2 dφ2
Here there are no cross terms meaning that φ̂ and θ̂ are perpendicular to each other. This is not
surprising since φ̂ is perpendicular to the plane of the small circle whereas θ̂ is in the plane of the
small circle.
(b) A general ellipsoid symmetrically situated at the origin is described by the equation,
x2 y 2 z 2
+ 2 + 2 =1
a2
b
c
2
2
If I rotate the ellipse xa2 + yb2 = 1 about the x-axis, I get an ellipsoid of revolution with c = b. We
may parametrize by the following equations.
x = a sin(θ)cos(φ) ; y = b sin(θ)sin(φ) ; z = b cos(θ)
dx = a cos(θ)cos(φ)dθ − a sin(θ)sin(φ)dφ
dy = b cos(θ)sin(φ)dθ + b sin(θ)cos(φ)dφ
dz = −b sin(θ) dθ
5
Now ds2 = dx2 + dy 2 + dz 2 . Hence,
ds2 = ((a2 cos2 (φ) + b2 sin2 (φ))cos2 (θ) + b2 sin2 (θ))dθ2
+(b2 cos2 (φ) + a2 sin2 (φ)) sin2 (θ) dφ2 + (b2 − a2 )cos(φ)sin(φ)sin(2θ) dθdφ
The limiting case a = b comes out as expected.
Q.5 : Consider a two dimensional surface described by the metric :
ds2 = (du2 + u2 dv 2 ) Cosh(2v) + 2u Sinh(2v) dudv
here Cosh(x) = (ex + e−x )/2 and Sinh(x) = (ex − e−x )/2.
Find the metric tensor gµν (u, v) and g µν (u, v) (choose x1 = u and x2 = v).
Find the connection coefficients Γµνσ (u, v).
From this find the equation of the geodesic.
Find the components of the Riemann curvature tensor.
What can you say about the nature of this surface ?
A.
Metric and Christoffel Symbols
g11 (u, v) = Cosh(2v) ; g22 (u, v) = u2 Cosh(2v) ; g12 (u, v) = g21 (u, v) = u Sinh(2v)
gµν =
Cosh(2v)
u Sinh(2v)
u Sinh(2v) u2 Cosh(2v)
!
g µν is the inverse of gµν , hence,
g
µν
=
Cosh(2v)
− u1 Sinh(2v)
− u1 Sinh(2v) u12 Cosh(2v)
!
From the notes,
1
Γµνσ = g µρ (gρν,σ + gνρ,σ − gσν,ρ )
2
First, it is better to symmetrize them. The geodesic equation has a term Γµσρ ẋσ ẋρ . But since
ẋσ ẋρ = ẋρ ẋσ we may write, Γµσρ ẋσ ẋρ = 12 (Γµσρ +Γµρσ )ẋσ ẋρ . Thus it is sufficient if I find the symmetrical
part,
1
1
Γµνσ → g µρ (gρν,σ + gνρ,σ − gσν,ρ ) + g µρ (gρσ,ν + gσρ,ν − gνσ,ρ )
4
4
or since gµν = gνµ we have,
1
Γµνσ → (g µρ gρν,σ + g µρ gρσ,ν − g µρ gνσ,ρ )
2
This new Γ has the property that Γµνσ = Γµσν .
6
1
Γµ11 = (g µρ gρ1,1 + g µρ gρ1,1 − g µρ g11,ρ )
2
1
Γµ22 = (g µρ gρ2,2 + g µρ gρ2,2 − g µρ g22,ρ )
2
1
Γµ12 = (g µρ gρ1,2 + g µρ gρ2,1 − g µρ g12,ρ )
2
gµν,1 =
gµν,2 =
g
µν
=
0
Sinh(2v)
Sinh(2v) 2u Cosh(2v)
!
2Sinh(2v)
2u Cosh(2v)
2u Cosh(2v) 2u2 Sinh(2v)
Cosh(2v)
− u1 Sinh(2v)
− u1 Sinh(2v) u12 Cosh(2v)
!
!
Thus,
Γ111 = −
1
1
Sinh(2v)Sinh(2v) + Sinh(2v)Sinh(2v) = 0
u
u
Γ122 = u
Γ112 = Γ121 = Cosh(2v)Sinh(2v) − Sinh(2v)Cosh(2v) = 0
Γ211 =
1
1
Cosh(2v)Sinh(2v)
−
Cosh(2v)Sinh(2v) = 0
u2
u2
Γ222 = 0
Γ212 = Γ221 =
1
u
The equations of the geodesic are,
d2 v
d2 u
1 a b
ẋ
ẋ
=
0
;
+
Γ
+ Γ2ab ẋa ẋb = 0
ab
ds2
ds2
Or the equations of the geodesic are,
ü + uv̇ 2 = 0
2
v̈ + u̇v̇ = 0
u
7
B.
Solution of the Equations [Optional, not asked in the question]
To solve these equations we first note that the natural parametrization is in terms of the arc length
of the geodesic : u(s), v(s). However, we now choose to invert the second correspondence and make v
d
d
the independent parameter. This means the curve is u(v) and v̇ is also now a function of v, ds
= v̇ dv
.
u̇ =
du
du
= v̇
ds
dv
2
du̇
d du
dv̇ du
2 d u
ü =
= v̇ (v̇ ) = v̇
+ v̇
ds
dv dv
dv dv
dv 2
dv̇
d
= v̇ v̇
ds
dv
v̈ =
Substituting into the geodesic equations we get,
2
dv̇ du
2 d u
v̇
+ v̇
+ uv̇ 2 = 0
2
dv dv
dv
v̇
d
2 du
v̇ + v̇ v̇ = 0
dv
u dv
Or,
dv̇ du
d2 u
+ v̇
+ uv̇ = 0
dv dv
dv 2
dv̇ 2 du
+
v̇ = 0
dv u dv
Now multiply the second equation by
du
dv
and subtract from the first we get,
d2 u
2 du 2
+
u
−
( ) =0
dv 2
u dv
Now make the transformation w = u1 , or u =
we get,
1
.
w
0
0
00
00
Hence u = − ww2 , u = − ww2 +
0
2w 2
.
w3
Substituting
00
w −w =0
Hence, w = c1 Sinh(v) + c2 Cosh(v) and u =
u Sinh(v) = y. Then we get,
1
w
=
1
.
c1 Sinh(v)+c2 Cosh(v)
c1 x + c2 y = 1
which is nothing but the equation of a straight line.
Call u Cosh(v) = x and
8
C.
Riemann Curvature Tensor
The components of the Riemann curvature tensor are given by,
ρ
Rσµν
= ∂µ Γρνσ − ∂ν Γρµσ + Γρµλ Γλνσ − Γρνλ Γλµσ
We can write,
1
1
Γµσν = δµ,1 δσ,2 δν,2 u + δµ,2 ( δσ,1 δν,2 + δσ,2 δν,1 )
u
u
After simplifying,
ρ
Rσµν
= ∂µ Γρνσ − ∂ν Γρµσ + Γρµ1 δν,2 δσ,2 u − Γρν1 δµ,2 δσ,2 u
1
1
+ Γρµ2 (δν,2 δσ,1 + δν,1 δσ,2 ) − Γρν2 (δµ,1 δσ,2 + δµ,2 δσ,1 )
u
u
Thus,
1
Rσµν
= (δν,2 δµ,1 − δµ,2 δν,1 + δµ,2 δν,1 − δν,2 δµ,1 )δσ,2 = 0
and,
2
Rσµν
= (δν,1 δσ,2 + δν,2 δσ,1 )δµ,1 ∂1 Γ212 − (δµ,1 δσ,2 + δµ,2 δσ,1 )δν,1 ∂1 Γ212 + δµ,2 δν,2 δσ,2 − δν,2 δµ,2 δσ,2
+
1
1
δµ,1 (δν,2 δσ,1 + δν,1 δσ,2 ) − 2 δν,1 (δµ,1 δσ,2 + δµ,2 δσ,1 )
2
u
u
which means,
2
Rσµν
=0
ρ
This means Rσµν
= 0. Therefore the space is really flat.
D.
Transformation to Cartesian Coordinates
The space has been shown to be flat since all components of the Riemann curvature tensor are
zero. Thus there must be some transformation that will make the metric equal to ds2 = dx2 + dy 2
for some x(u, v), y(u, v). It is not easy to guess this but the answer is,
x = u Cosh(v) ; y = u Sinh(v)
Now if we calculate,
dx = du Cosh(v) + udv Sinh(v)
dy = du Sinh(v) + udv Cosh(v)
ds2 = dx2 + dy 2 = (du Cosh(v) + udvSinh(v))2 + (du Sinh(v) + udv Cosh(v))2
or,
ds2 = (du2 + u2 dv 2 ) Cosh(2v) + 2u Sinh(2v) dudv
The equation of the geodesic is therefore,
y = mx + c
or,
u Sinh(v) = m u Cosh(v) + c