The Special Theory of
Relativity
Chapter III
1. Relativistic dynamics
2. Momentum and energy
3. E=mc2
4. Relativistic particle scattering
Excercise
Relativistic Momentum
The formula for
relativistic momentum
can be derived by
requiring that the
conservation of
momentum during
collisions remain
valid in all inertial
reference frames.
Note: that does NOT
mean that the
momentum is equal in
different reference
frames
Go over this and derive !
Result

p=

mv
1 − v2 / c2

= γm v
Relativistic Force
Newtons second law remains valid (without proof)

 dp d
 d 
mv
F=
= γmv = 
dt dt
dt  1 − v 2 / c 2




1) For every physical law it has to be established how they
transform in relativity (under Lorentz transformations)
2) Quantities (like F) not the same in reference frames
Relativistic acceleration

 dp d

 dγ
dv
= γmv = γm
+ mv
F=
dt dt
dt
dt

 dγ
= γm a + m v
dt
The force vector does not point in the same direction as the
acceleration vector
dγ d  v 
= 1 − 2 
dt dt  c 
2
−1 / 2
1 v 
= − 1 − 2 
2 c 
2
−3 / 2
v
 − 2v 
⋅ 2  = γ 3 2
c
 c 
Note: in case of acceleration γ is not constant
Relativistic Mass
From the momentum:

p=

mv
1 − v2 / c2

= γm v
Gamma and the rest mass are combined to form
the relativistic mass:
mrel =
m
1 − v2 / c2
Relativistic Energy
Work done to increase the speed of a particle from v=0 (i-state) to v=v (f state):
f
f
f
f
dp
dp
W = ∫ Fdx = ∫ dx = ∫ vdt = ∫ vdp = ∫ d ( pv ) − ∫ pdv
i
i dt
i dt
i
i
i
f
f
∫ d ( pv ) =
i
f
f
pv i
= (γmv )v
f
v
mv
i
0
1 − v2 / c2
− ∫ pdv = − ∫
dv = mc
vdp = d ( pv) − pdv
because
2
1− v / c
2
2
v
= mc 2 1 − v 2 / c 2 − mc 2
0
use
So:
W = γmv +
2
mc 2
γ
− mc
2
(
γ 2v 2 + c 2 )
=
m − mc 2 = (γ − 1)mc 2
(
)
d
 v 
1 − v 2 / c 2 = − 2  / 1 − v 2 / c 2
dv
c 
γ
Kinetic energy of the particle is
K = (γ − 1)mc 2
1) Amount of kinetic
energy depends on
inertial frame
2) Amount of kinetic
energy reduces to
classical value at low v
3) Note
1
K ≠ mv 2
2
Mass and Energy
The kinetic energy
K = (γ − 1)mc 2
Can be written as the total energy:
Where the difference is the rest energy:
E = γmc 2 =
mc 2
1 − v2 / c2
E = mc 2
The last equation is Einstein famous equation
implying that mass is equivalent to energy
The energy of a particle at rest.
Note that mc2 is the same as seen from all reference frames;
It is an invariant upon frame transformation
Mass, Energy, Momentum
Energy
E = γmc
2
Momentum
p = γmv
Combining these relations gives
E = p c +m c
2
2 2
Hence also the following
Is an invariant under Lorentz
transformations
2 4
E −p c
2
2 2
Mass, Energy, Momentum
for light particles
Light particles have no “rest” mass (m=0), but they have energy
E = p c +m c
2
2 2
2 4
hence
Energy in the quantum picture
E h
Hence momentum p =
=
c λ
E
p=
c
E = hν =
hc
λ
Compton scattering
A photon (is a light particle) collides with an electron and its energy (so its wavelength) must change !!
Before collision
hc
photon
E = hf =
electron
Ee = me c 2
λ
p=
h
λ
After collision
hc
λ'
photon
E'=
electron
e
Etot
= γme c 2
p' =
h
λ'
pe = γme v
e
Ekin
= (γ − 1)me c 2
Write the momentum conservation equations along the x-coordinate and along the y-coordinate.
Write the energy conservation equation.
Then solve the equations and determine the wavelength λ’ for angle φ.
Compton scattering
Conservation of energy
hc
λ
=
hc
+ (γ − 1)me c 2
λ'
Conservation of momentum
h
Along x:
Along y:
Three equations with 3 unknowns,
eliminate v and θ
Compton scattering:
λ
=
0=
h
cos φ + γme v cosθ
λ'
h
sin φ − γme v sin θ
λ'
h
(1 − cosφ )
λ'= λ +
me c
Compton scattering
Arthur Compton
∆λ =
h
(1 − cosφ ) = λC (1 − cosφ )
me c
The Nobel Prize
in Physics 1927
"for his discovery of
the effect named after him"
Note that λC ~ 0.00243 nm
So the effects is not so well visible with visible light
Compton performed his experiment with x-rays
Doppler Shift for Light
The Doppler shift for light for c=constant in all inertial frames.
λ0 = c∆t0 = c / f 0
Two effects in relativistic Doppler
1) Moving waves + 2) Time dilation!!
1)
λ = c∆t − v∆t
2) ∆t
= ∆t0 / 1 − v 2 / c 2
λ = (c − v )∆t =
(c − v )∆t0
1 − v2 / c2
= λ0
c−v
c+v
When source moving toward observer
Doppler Shift for Light
Hence, one can derive the observed frequency and
wavelength:
If the source and observer are moving
away from each other, v changes sign.
Remember: higher pitch, blue shift when moving toward each other
Doppler Shift for Light
Speeding through a red light.
A driver claims that he did not go through a red light because
the light was Doppler shifted and appeared green. Calculate the
speed of a driver in order for a red light to appear green.
λ = 500nm ; λ0 =650 nm
v = 0.26 c