Relativistic mechanics
--Scalars
-- 4-vectors
-- 4-D velocity
-- 4-momentum, rest mass
-- conservation laws
-- Collisions
-- Photons and Compton scattering
-- Velocity addition (revisited) and the Doppler shift
-- 4-force
1. Scalars
A scalar is a quantity that is the same in all reference frames, or for all observers.
It is an invariant number.
E.g., ( s )
2
,  (t rest , proper time ), l (lrest , proper length )
But the time interval ∆t, or the distance ∆x between two events, or the length l
separating two worldlines are not scalars: they do not have frame-independent
values.
2.

4-vectors x  (ct , x, y, z )
This 4-vector defined above is actually a frame-independent object, although the
components of it are not frame-independent, because they transform by the
Lorentz transformation.
E.g., in 3-space, the Different observers set up different coordinate systems and
assign different coordinates to two points C and L, say Canterbury and London.
--They may assign different coordinates to the point of the two cities
--They agree on the 3-displacement r separating C and L., the distance
between the two points, etc.
With each 4-displacement we can associate a scalar the interval (s)2 along the
vector. The interval associated with the above defined 4-vector is
(s) 2  (ct ) 2  (r ) 2  (ct ) 2  (x) 2  (y ) 2  (z ) 2
Because of the similarity of this expression to that of the dot product between
3-vectors in three dimensions, we also denote this interval by a dot product and
also by  2
 
2
2
2
2
x
 x  x  (ct )  (x)  ( y )  ( z )
and we will sometimes refer to this as the magnitude or length of the 4-vector.
--We can generalize this dot product to a dot product between any two 4-vectors


a  (at , a x , a y , a z ) and b  (bt , bx , by , bz ) :
 
a  b  at bt  a x bx  a y by  a z bz
--When frames are changed, 4-displacement transform according to the Lorentz
transformation, and obeys associativity over addition and commutativity :
  
   
i ) a  (b  c )  a  b  a  c ;
   
a b  b  a
ii) A 4-vector multiplied or divided by a scalar is another 4-vector
3. 4-velocity
In 3-dimensional space, 3-velocity is defined by
r̂ dr̂
vˆ  lim

t 0 t
dt
where ∆t is the time it takes the object in question to go the 3-displacement ∆ r.

x in place of
Can we put the 4-displacement
the 3-displacement r so that we have


dx
u 
dt
??
However, this in itself won't do, because we are dividing a 4-vector by a
non-scalar (time intervals are not scalars); the quotient will not transform
according to the Lorentz transformation.
The fix is to replace ∆t by the proper time ∆ corresponding to the interval of
the 4-displacement; the 4-velocity is then

x

u  lim
 0 
where :
  t

dt dx dy dz
dt dx dy dz
 dx
u
 (c , , , )  (c , 
,
,  )  ( c,  vx ,  v y ,  vz )
d
d d d d
dt dt dt dt
where
(v x , v y , v z )
are the components of the 3-velocity
vˆ 
dr̂
dt
Although it is unpleasant to do so, we often write 4-vectors as two-component
objects with the rest component a single number and the second a 3-vector. In

this notation
u  ( c,  vˆ)

u
--What is the magnitude of
?
2

u
u
The magnitude
must be the same in all frames because
is a 4-vector.
Let us change into the frame in which the object in question is at rest.
In this frame

u  ( c, 0,0,0) for vˆ  (0,0,0) and
2
u  c 2 or

u c
 1
It is a scalar so it must have this value
You can also show this
 2 in all

 frames.
2
by calculating the dot product of u  u  u  c
A little strange? Some particles move quickly, some slowly, but for all particles,
the magnitude of the 4-velocity is c. But this is not strange, because we need
the magnitude to be a scalar, the same in all frames.
If you change frames, some of the particles that were moving quickly before
now move slowly, and some of them are stopped altogether. Speeds (magnitudes
of 3-velocities) are relative; the magnitude of the 4-velocity has to be invariant.
4. 4-momentum, rest mass and conservation laws


In spacetime 4-momentum p is mass m times 4-velocity u
--Under this definition, the mass must be a scalar if the 4-momentum is going
to be a 4-vector.


p  mu  (mc, mvx , mv y , mvz )  (mc, mvˆ)
--The mass m of an object as far as we are concerned is its rest mass, or
the mass we would measure if we were at rest with respect to the object.
--Again, by switching into the rest frame of the particle, or by calculating
the magnitude, we can show:

p  mc
As with 4-velocity, it is strange but true that the magnitude of the 4-momentum
does not depend on speed.
Why introduce all these 4-vectors, and in particular the 4-momentum?
--all the laws of physics must be same in all uniformly moving reference frames
--only scalars and 4-vectors are truly frame-independent, relativistically
invariant conservation of momentum must take a slightly different form.
--In all interactions, collisions and decays of objects, the total 4-momentum is
conserved (of course we don’t consider any external force here).


p  mu  (mc, mvˆ)
--Furthermore,
p̂
E
c
We are actually re-defining E and
p̂
to be: E  mc2
pˆ  mvˆ
and
You better forget any other expressions you learned for E or p in non-relativistic
mechanics.
pˆ c 2
vˆ 
E
A very useful equation suggested by the
new, correct expressions for E and p̂
Taking the magnitude-squared of

p
We get a relation between m, E and
p  pˆ
2
2
  E
p  m 2c 2  p  p     p 2
c
which, after multiplication by c2 and rearrangement becomes
This is the famous equation of Einstein's, which becomes
E 2  m2c 4
when the particle is at rest
pˆ  0
E 2  m2c 4  p 2c 2
In the low-speed limit  
1
2
v
 1
c
1
v2
ˆ  mvˆ(1   )
p
 mvˆ  m 2 vˆ  mvˆ
2
c
1

1
2
2
E  mc (1   ) 2  mc 2  mv 2
2
2

i.e., the momentum has the classical form, and the energy is just Einstein's
famous mc2 plus the classical kinetic energy mv2/2. But remember, these
formulae only apply when v << c.
5. Conservation laws


pq
For a single particle: 4-momenum
before an action = that after:
For a multi-particle system:

i
Summed over All the 4-momenta of
all the components of the whole
system before interaction


pi   qi
i
Summed over all the 4-momenta
of all the components of the whole
system after interaction
6. Collisions
In non-relativistic mechanics collisions divide into two classes:
elastic
inelastic
energy and 3-momentum
only 3-momentum is conserved
are conserved.
In relativistic mechanics 4-momentum, and in particular the time component
or energy, is conserved in all collisions;
No distinction is made between elastic and inelastic collisions.
Before the collision
m
v̂
After the collision
M
’
m
v̂ '
Non-relativistic theory gives: M’=2m, v’ =v/2


pm  (mc, mv,0,0); ps  (mc,0,0,0)
 
p  pm  pˆ s  [(  1)mc, mv,0,0)]

q  ( ' M ' c,  ' M ' v' ,0,0)
By conservation of 4-momentum before and after collision, which means
that the two 4-vectors are equal, component by component,
g ' M'c = [g +1]mc
and g ' M'v' = gmv
The ratio of these two components should provide v’/c;
v' 
v
v

 1 2

q
The magnitude of
should be M‘c; we use

q  M ' c;
2
2
q  p
v2
v2
2
M '  [  1] m   m 2  [1  2   (1  2 )]m 2  2(  1)m 2
c
c
2
2
2
2
2
M '  2(  1) m  2m
--the mass M’ of the final product is greater than the sum of the masses of
its progenitors, 2m.
--So the non-relativistic answers are incorrect,
Q: Where does the extra rest mass come from?
A: The answer is energy.
In this classically inelastic collision, some of the kinetic energy is lost.
But total energy is conserved. Even in classical mechanics the energy is not
actually lost, it is just converted into other forms, like heat in the ball, or
rotational energy of the final product, or in vibrational waves or sound travelling
through the material of the ball.
Strange as it may sound, this internal energy actually increases the mass
of the product of the collision in relativistic mechanics.
The consequences of this are strange. For example, a brick becomes more
massive when one heats it up. Or, a tourist becomes less massive as he or she
burns calories climbing the steps of the Effiel Tower.
All these statements are true, but it is important to remember that the effect
is very very small unless the internal energy of the object in question
is on the same order as mc2.
For a brick of 1 kg, mc2 is 1020 Joules, or 3 *1013 kWh, a household energy
consumption over about ten billion years (roughly the age of the Universe!)
For this reason, macroscopic objects (like bricks or balls of putty) cannot
possibly be put into states of relativistic motion in Earth-bound experiments.
Only subatomic and atomic particles can be accelerated to relativistic speeds,
and even these require huge machines (accelerators) with huge power
supplies.
6. Photons and Compton scattering
6.1 properties of photon
i) Can something have zero rest mass?
2
2 4
2 2
From E  m c  p c  E = p c
(p is the magnitude of the 3-momentum)
Substitute E=pc into v = p c2/E = c
So massless particles would always have to travel at v = c, the speed of light.
Strange??
Photons, or particles of light, have zero rest mass, and this is why they always
travel at the speed of light.
ii) The magnitude of a photon's 4-momentum

p

p
2
2
m c
2
2
E
2
2

  p  0  c  0;
 c 
E  pc
but this does not mean that the components are all zero.
--The time component squared, E2/ c2, is exactly cancelled
out by the sum
2
2
2
2
of the space components squared, px  p y  pz  p̂
--Thus the photon may be massless, but it carries momentum and energy,
and it should obey the law of conservation of 4-momentum.
6.2 Compton scattering.
The idea of the experiment is to beam photons of known momentum Q at a target
of stationary electrons,and measure the momenta Q’of the scattered photons as
a function of scattering angle.
We therefore want to derive an expression for Q’ as a function of .
Before the collision the 4-momenta of the photon and electron are:


p  (Q, Q,0,0); pe  (mc,0,0,0)


after they are: q  (Q' , Q' cos  , Q' sin  ,0); qe  (mc, mv cos  , sin  ,0)








2
2
The conservation law is p  pe  q  qe  ( p  q )  (qe  pe )
 
 
 
 
 
 
p  p  q  q  2 p  q  pe  pe  qe  qe  2 pe  qe
For all photons
 
p  p  0;
Also, in this case
And:
 
p  q  QQ'QQ' cos
 
pe  qe  m 2 c 2 ;
Equation (a) becomes:
and for all electrons
( a)
 
p  p  (mc ) 2
 2QQ ' (1  cos  )  2(1   )m2c 2
But by conservation of energy, ( −1)mc is just Q−Q’, and (a − b)/ab is just
1/b−1/a, so we have what we are looking for:
1
1
1


(1  cos  )
Q' Q
mc
This prediction of special relativity was confirmed in a beautiful experiment by
Compton (1923) and has been reconfirmed many times since by undergraduates
in physics lab courses.
In addition to providing quantitative confirmation of relativistic mechanics, this
experimental result is a demonstration of the fact that photons, though
massless, carry momentum and energy.
Quantum mechanics tells
The energy E of a photon is related to its wave frequency by E = h
Then
1 c
c 
 
 ;
Q E hv h
1 '

Q' h
so we can rewrite the Compton scattering
equation in its traditional form:
 ' 
h
(1  cos  )
mc
7. Particle decay and pair production
7.1 Particle decay:
An elementary particle of rest mass M decays from rest into a photon and
a new particle of rest mass M/2. Find its velocity.
û
M
/2
M
For 3-momentum conservation, the
particle moves in x direction, and
the photon moves in –x direction.
h

pM  (Mc,0,0,0)
hv hv

p ph  ( , ,0,0);
c
c
 hv  

 Mc   c  
0   hv  
  
  

0   c  
 

 0
0

 0


0
 0
By momentum
conservation:



pM  p M  p ph
2
(2) Into (1):
Mc  
Mc 
2 

Mu 
;
2 



Mc Mu

p M  (
,
,0,0)
2
2
2
Mc 
hv Mc

c
2
hv
Mu

c
2
(1)
( 2)
Mu Mc
M (u  c) M (u  c) Mc 1  u / c




2
2
2
2
2 1 u / c
u
2 1 2
c
Solve for u: u  0.6c
7.2 Pair production - gamma photon can not be converted to e- and e+
Show that the following pair production cannot occur without involvement of other
particles.
eLet m be the rest mass of electrons
and u, v the 3-velocities of electron
and positron.
e+
 hv 
 c   1m c   2 m c 
hv  mc 2 ( 1   2 ) (1)
  

 

mv

mu
hv

hv
2
x
1
x


p ph     
 m( 1u x   2 v x ) ( 2)
 1mu y    2 m v y 
c
c

0
0




0



0


 1u y   2 v y
(3)
Sub. (3) into (2):
uy
hv
 m 1 (u x  v x
)
c
vy
u
hv 1  ( ) 2
uy
c

 (u x  v x
)
mc
vy
u 2
Sub. (3) into (1):
hv
1

(
)
u
hv
y
 mc 2 1 (1 
)
c
vy
c

mc
 c (1 
Compare (4) and (5):(u x  v x u y )  c(1  u y ) (6)
vy
vy
uy
vy
)
( 4)
(5)
For ux and vx < c
(6) can not be satisfied
Pair production needs an additional particle to carry off some momentum.
8 Velocity addition (revisited) and the Doppler shift
8 .1 Velocity addition revisited
In S, a particle of mass m moves in the x-direction at speed vx, so its 4-momentum
1

is
p  ( mc,  mv ,0,0) where  
1
1
x
1
In S’ moving at speed v, the 4-momentum of the particle:
ég ' mc ù
ég 2
- g 2b
ê
ú
g ' mv' x ú ê
-g 2 b
g2
ê
p' =
=ê
êg ' mv' y ú
ê0
0
ê
ú
ê
0
ë0
ëg ' mv' z û
ég 2g1mc - g 2g1mvx b ù
ê
ú
-g 2g1mcb + g1g 2 mvx ú
ê
=
ê
ú
0
ê
ú
0
ë
û
2
v
1  x2
c
0 0 ù ég1mc ù
úê
ú
0 0 ú êg1mvx ú
úê 0
ú
1 0
úê
ú
0 1û ë 0
û
with  
v
and  2 
c
1
v2
1 2
c
 1 2vx   1 2v
vx  v
 ' mv' x  1 2 mvx   1 2 mcv / c
v' x
'



 vx 
v vx
 ' mc  1 2 mc   1 2 mvx v / c
c  1 2c   1 2vx v / c
1 2
c
This is a much simpler derivation than that found before.
8.2 Photon makes an angle from x axis
S’
v
y’
Q’(frest )
S
y
Q(fobs )

’
x
x’
z
z

q' 
’ (Q' , Q' cos  ' , Q' sin  ' ,0)

q  (Q, Q cos  , Q sin  ,0);
 0 0  Q'
Q



Q cos  

 Q ' cos  '

0
0

 

;
q  
Q sin  
0
0
1 0 Q ' sin  ' 





0
0
0
0
1
0





Equate each component on both side:
Q  Q 'Q ' cos  '
 Q  Q' (1   cos  ' )  f obs  f em
v

c
1   cos  em
1  2
Q'
 (   cos  ' )
Q cos   Q 'Q ' cos  '  cos  
 (   cos  ' ) 
Q
 (1   cos  ' )
  cos  em
  cos  '
Q sin   Q ' sin  '

 cos  obs 
1   cos  '
1   cos  em
f obs  f em
Doppler effect from:
i) If q em = 0; fobs = fem
1+ b
1- b 2
1   cos  em
1  2
1+ b
1- b
= fem
the light is blue - shifted, when the light source is moving toward you.
When v  c : 1 
ii) If  em   ; f obs  f em
v

c
1
v
1
c
1 
1  2
v
f ob  f em /(1  )
c
;
 f em
classic D  E
1 
1 
the light is red  shited , when the light source is moving away from you.
When v  c : 1 
v

c
1
1
v
c
;
ii) If  em   / 2; f obs  f em
v
f ob  f em /(1  ) classic D  E
c
1
1  2
This predicts a very small transvers e Doppler effect.
cos  obs 
Aberration of light from:
if  em  0;
if  em   ;
if  em   / 2;
1
cos  obs  1;
cos  obs  1;
cos  obs   ;
  cos  em
1   cos  em
 obs  0
 obs  
0   obs   / 2
3
1
3
2
Light rays emitted by source in S’
2
Light rays observed in S
When v is very large so that =0.9, and cosobs =0.9, obs =26
http://www.anu.edu.au/Physics/Savage/TEE/site/tee/learning/aberration/aberration.html
9. 4-force
We recall the 4-velocity and 4-momentum are defined in terms of derivatives
with respect to proper time rather than coordinate time t . The definitions are

 dx
u
d
and


p  mu

Where x is spacetime position and m is rest mass

a
if we want to define a 4-vector form of acceleration
we will need to use
E

ˆ)
Because p  ( , p
c

du

a
d
and

dE dpˆ
K (
, )
cd d


dp
K 
d

, or a 4-vector force K ,
ˆ
dp
ˆ
 F
d
Also, if the rest mass m of the object in question is a constant (not true if the
object in question is doing work, because then it must be using up some of its rest


energy!),
i.e., if the rest mass is not changing then p and K
 
2 2
p p  m c
are orthogonal. In 3+1-dimensional spacetime,
 
d ( p  p)
orthogonality
0
d
is something quite different from orthogonality in 3-space:


dp   dp
it has nothing to do with 90 angles.
 p  p
0
d
 
pK  0
d