Prev. lecture: Velocity as a 4-vector: 4-velocity U
U = (cγ(u), γ(u)u)
Since numerator and denominator are invariant, U is also invariant.
If we now consider the mass of a particle to be invariant, we arrive
at the 4-momentum of that particle:
P = mU = (γ(u)mc, γ(u)mu)
defining the relativistic 3-momentum p = γ(u)mu
Assignment 0: what is U 2 and what does that mean?
Lambert van Eijck
TN2612 - intro special relativity - lecture 5
1
4-acceleration?
If U2 = c 2 , what is left over of 4-acceleration?
Lambert van Eijck
TN2612 - intro special relativity - lecture 5
2
U 2 = c 2 graphically (for illustration only)
Figure: hypersurfaces (a) U = c, (b) U = −c and (c) U = ic.
Lambert van Eijck
TN2612 - intro special relativity - lecture 5
3
The zeroth element of P
P0 is the first term in the vector, so the ct term
P 0 = γ(u)mc
for u << c the truncated Taylor expansion yields:
γ(u) = q
leading to:
Lambert van Eijck
1
1−
u2
c2
≈1+
u2
2c 2
u2
)
2c 2
mcu 2
P 0 ≈ mc +
2c 2
2
mc + 12 mu 2
P0 ≈
c
E
0
⇒P =
c
P 0 ≈ mc(1 +
so this term includes the resting mass
(mc^2)
+ the kinetic coeffienct 1/2mu^2
TN2612 - intro special relativity - lecture 5
4
The zeroth element of P
combining:
P 0 = γ(u)mc
P0 =
E
c
equate these two expression and you get that
leads to:
E = γ(u)mc 2
which is contained in the 4-vector P:
E
P = ( , p)
c
Lambert van Eijck
TN2612 - intro special relativity - lecture 5
5
Borsele radioactive storage building COVRA
Figure: did they forget something?
Lambert van Eijck
TN2612 - intro special relativity - lecture 5
6
Overview of the 4-vectors
a) 4-vectors comply to the Lorentz transformation:
 0 
  0
A0
A
γ
−βγ 0 0
 10 


γ
0 0 A1 
A  −βγ

 20 = 
0
1 0 A2 
A 
0
A3
0
0 1
A3
b) The dot-product of 4-vectors is different from what you know:
A · A = (A0 )2 − (A1 )2 − (A2 )2 − (A3 )2
Lambert van Eijck
TN2612 - intro special relativity - lecture 5
7
a) Lorentz transformation of E when S 0 moves along x
basically like a Lorentz transformation for coordinates, but now for energy and momentum
E 0 = γ(V )(E − VP 1 )
0
P 1 = γ(V )(P 1 −
V
E)
c2
0
P2 = P2
0
P3 = P3
but remember that γ(u 0 ) 6= γ(u).
Lambert van Eijck
TN2612 - intro special relativity - lecture 5
8
b) from X to E in Minkovski space
We have defined an event:
X = X µ = (ct, x, y , z),
the velocity of an object moving 4-dimensional space:
U=
dX µ
= γ(u)(c, ux , uy , uz ) = γ(u)(c, u),
dτ
and its corresponding momentum (where mass m is invariant):
E
P = mU = γ(u)(mc, m u) = γ(u)(mc, p) = ( , γ(u)mu)
c
leading to the energy definition:
P·P=
Lambert van Eijck
dot product: multiply all elements in corresponding position of
each vector by the same position in the other, then add all the
products and the result is the dot product
E2
− p2 = m2 c 2 ⇒ E 2 = m2 c 4 + p2 c 2
c2
TN2612 - intro special relativity - lecture 5
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conservation laws in Special Relativity
Initial and final 4-momentum are conserved in a collision:
Pi = Pf
Which already contains energy conservation through P 0 .
Instead of the Newtonian conservation laws for mass, momentum and energy, we are left with only the conservation of
4-momentum in Special Relativity.
Lambert van Eijck
TN2612 - intro special relativity - lecture 5
10
Example of conservation of P
particle 1 with mass m is moving at velocity V towards particle 2
with equal mass, but at rest. After the head-on collision of the
particles, they move as one particle with mass M and speed VM .
a) What is the relativistic momentum of the particle M after the
collision?
b) What is its velocity?
c) What is the mass M of the new particle?
solution on the next page, so please try yourself first!
Lambert van Eijck
TN2612 - intro special relativity - lecture 5
11
Example of conservation of P
particle 1 with mass m is moving at velocity V towards particle 2
with equal mass, but at rest. After the head-on collision of the
particles, they move as one particle with mass M and speed VM .
a) PM = (γ(VM )Mc, γ(VM )MVM )
b)
VM =
cf. ½V in the classical case.
γ(V )
V
γ(V ) + 1
when the speed V is very small compared to c, gamma goes to 1
and equation becomes 1/2V
c) P2 before and after yields M = m
classical case.
Lambert van Eijck
p
2 + 2γ(V ) cf 2m in the
TN2612 - intro special relativity - lecture 5
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photons
For a massive particle:
classical momentum is m*u
E
P = mU = γ(u)(mc, m u) = γ(u)(mc, p) = ( , γ(u)mu)
c
but for a photon m = 0 and its energy is defined by its frequency f :
E = hf = h
c
= pc ⇒ p = E /c
λ
and so, the 4-momentum of a photon becomes:
E E
P = ( , n̂)
c c
where n̂ is the propagation direction of the photon.
Lambert van Eijck
TN2612 - intro special relativity - lecture 5
13
photons
E E
P = ( , n̂)
c c
We previously found that for massive particles:
P · P = m2 c 2
Which leads to the requirement for photons:
E2 E2
− 2 = m2 c 2
c2
c
Since the speed of light c is a prerequisite of Special Relativity, the
mass m of a photon must be zero for this equation to hold.
mphoton = 0
Lambert van Eijck
TN2612 - intro special relativity - lecture 5
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