Causality in special relativity
--The ladder and barn paradox
-- Lorentz transformation in Spacetime diagram:
--Causality
-- Nothing can travel faster than C
1. The pole and barn paradox
In frame of barn, the pole should fit in the barn, If v=0.866c,  = 2, l(pole) =10m
v
l (pole)  lrest (pole) 1  ( ) 2  20 1  0.8662  10m
c
In frame of pole, the barn is shorter,  = 2, l(barn) =5 m, the situation is even
worse, the pole can’t fit into the barn at all
l =20m
v
l (barn)  lrest (barn) 1  ( ) 2  10 1  (0.866) 2  5m
c
They cannot be both right!!
--Let’s assume the doors of the barn are kept open in usual state, and are
designed in such a way that it can be triggered by passing of the pole to close
and then open again immediately after, so that the pole can keep a constant
motion passing through the barn.
--The back door of the barn: will close when the front of pole just approaches the
back door, then open again immediately and the front door will close just when
the rear end of pole passes it and open again immediately.
Barn’s frame:
t0 = 0
10 m
Back door closes then
opens
t1 =38.49ns, x1=0
(10m/0.886c)
Front door closes and
then open
t2= 38.49ns, x2 =10
pole moves out
t3 = 2*38.49=76.98ns
Pole’ frame
10 m
t’0 = 0
20 m
1010
m
Back door closes
then opens
t’ 1= 19.25ns
(5m/0.866c)
20 m
m
two doors close
Simultaneously.
10 m
10 m
By Lorentz Transformation:
The back door closes:
The front door closes:
5m
5m
two doors close
at different times.
Front door closes and then
open:
5m
20 m
t’2 =76.98ns,
(20m/0.886c)
5m
20 m
pole moves out:
t’3 = 19.25+76.98=96.23ns
vx
10109
t '1   (t1  2 )  2.(
 0.866c 10109 / c 2 )  19.25ns
c
0.886c
t '2   (t 2 
vx 2
)  2(38 .49  0)  76 .98 ns
c2
--The surprising result is that the back gate is seen to close earlier, before
the front of the pole reaches it.
--The door closings are not simultaneous in Pole’s frame, and they
permit the pole to pass through without hitting either doors.
2. Lorentz transformation in Spacetime diagram:
Ct’
Ct
a) How to set the second RF in the space
time diagram of the first frame?
S’
--The Ct’ and x’ cannot point in just any odd
x ’ direction because they must be oriented
such that the second observer (S’) also
measures speed c for the light pulse.
S
B
--They have to point in this way shown in
blue lines:
A
x
S’
b) How to measure with respect to
axes which are not right angles to one
another ?
--The diagram in the right shows the set of
all points (in purple) with some particular
value of x´.
Ct
’
x’
c) How is relativity of simultaneity described in space time diagram?
C t ’ Ct’’
Ct
Ct
S’
S’
x’’
x’
x’
x
x
Increasing speed
Ct’
d) what happens if the relative speed of the observers is larger?
´The point with x´´=0 covers even more ground in a given time interval than
did the point with x´=0. Thus, the ct´´ axis (the set of all points with x´´=0) is
inclined even more towards the light cone than the ct´ axis is, as shown in
the above right figure.
3. Causality
Causality means that cause precedes effect : an ordering in time
which every observer agrees upon.
Q: Whether it is possible to change the order of cause and effect just
by viewing two events from a different frame.
A: two events can only be cause and effect if they can be connected to
one another by something moving at speed less than or equal to the
speed of light.
Two such events are said to be causally connected.
Diagrammatically, event B is causally connected to event A if B lies
within or on the light cone centered at A:
tB > tA, B occurs after A
Ct
B
Can we perform a Lorentz
transformation such that
tB < tA, B occurs before A?
A
x
Here’s a representative case:
Ct’
Ct
t'B > t’A, B occurs after A
B
X’
x
A
In contrast, let’s suppose that it were possible to go into a frame moving faster
than light.
X‘
Then the ct´ axis would tilt
Ct
past the light cone, and the
B
Ct‘
order of events could be
reversed (B could occur
at a negative value of t´):
t'B < t’A, B occur before A
A
x
But it is not true for v >c.
4. Nothing can travel faster than C
It’s also easy to show that if a body were able to travel faster than the speed of
light, some observers would observe causality violationsEffects would precede
their causes.
Assume a person is born at event A and travels faster than the speed of light to
event B, where he dies. His world line is shown in red on the left
tB  t A
Another observer In S’:
The death occurs after the birth.
tB  t A
Ct
Ct
S
Ct‘
S’
X’
B
B
X
A
A
Such causality violations are not observed experimentally. This is evidence
that bodies cannot travel faster than the speed of light
X
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
,  (trest , propertime), l (lrest , properlength)
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 cites
--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 , ax , a y , az ) and b  (bt , bx , by , bz ) :
 
a  b  at bt  axbx  a y by  az 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
rˆ drˆ

t  0  t
dt
vˆ  lim
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
(vx , vy ,vz ) are the components of the 3-velocity
drˆ
vˆ 
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)
--What is the magnitude of
The magnitude

u

u

u
must be the same in all frames because
2
is a 4-vector.
Let us change into the frame in which the object in question is at rest.

u
 ( c, 0,0,0)
In this frame
2 2

u  c or u  c
for vˆ  (0,0,0) and
 1
It is a scalar so it must have this value
 2in all frames.
 You2can also show this
by calculating the dot product of u
 u u  c
You may find this 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 , mvy , 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 calculationg
the magnitude we find that 4-momentum, 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  (m c, mvˆ)
--Furthermore,
E
c
We are actually re-defining E and
pˆ
pˆ
to be:
E  mc2
and
pˆ  mvˆ
You better forget any other expressions you learned for E or p in non-relativistic
mechanics.
ˆpc 2
vˆ 
E
A very useful equation suggested by the
ˆ
new, correct expressions for E and p
Taking the magnitude-squared of

p
We get a relation between m, E and
p  pˆ
2
2
  E
p  m2c 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  m2 c 4  p 2 c 2
In the low-speed limit  
v
 1
c
1
2
1
v2
ˆ  mv
ˆ(1   )
ˆ
ˆ  mv
ˆ
p
 mv
m 2 v
2
c
1

1
E  m c2 (1   2 ) 2  m c2 
m v2
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.


pq
5. Conservation laws
For a single particle: 4-momenum
before an action = that after
For a multi-particle system:


p

q
 i  i
i
Summed over All the 4-momenta of
all the components of the whole
system before interaction
i
Summed over all the 4-momenta
of all the components of the whole
system after interaction
5. 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
vˆ
After the collision
M’
m
vˆ '
Non-relativiistic 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 collsion, which means
that the two 4-vectors are equal, component by component,
 ' M'c  [ 1]mc
and
 ' M'v'  mv
The ratio of these two components should provide v’/c;
v
v
v' 

  1 2
The magnitude of
qshould 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
2
2
E
2
2
p  m2c 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  p
ˆ
--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, mvcos , 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  m2c2 ;
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.
M/
2
M
h

pM  (Mc,0,0,0)
By momentum
conservation:



pM  p M  p ph 
2
ˆ
u
For 3-momentum conservation, the
particle moves in x direction, and
the photon moves in –x direction.
hv hv
Mc Mu


p ph  ( , ,0,0); p M  (
,
,0,0)
c
c
2
2
2
 hv  

 Mc  c  
0   hv  
  

  
0   c  
 

 0
0
 0

 
0
 0
Mc 
2 

Mu 
;
2 



Mc 
hv Mc

c
2
hv
Mu

c
2
(1)
( 2)
Mu Mc
M (u  c) M (u  c) Mc 1  u / c
Mc  




2
2
2
2
2 1 u / c
u
2 1 2
c
Solve for u: u  0.6c
(2) Into (1):
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
 
hv  m c2 ( 1   2 ) (1)
 c   1m c   2 m c 
  

 
hv

m
v

m
u
hv

2
x
1
x





 m( 1u x   2 vx ) (2)
p ph    


c
 1m uy    2 m v y 
c
  

 
 1u y   2v y
(3)
0
  0

 0
0 
Sub. (3) into (2):
u 2
hv 1  ( )
uy
c

 (u x  v x
) (4)
mc
vy
Sub. (3) into (1):
u 2
hv
1

(
)
uy
uy
hv
c
2
 m c  1 (1 
) 
 c(1 
) (5)
c
vy
mc
vy
uy
uy
Compare (4) and (5):
For ux and vx < c
(u x  vx
)  c(1 
) (6) (6) can
not be satisfied
vy
vy
uy
hv
 m 1 (u x  vx
)
c
vy
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  ( m c,  m v ,0,0) where  
1
1
x
1
In S’ moving at speed v, the 4-momentum of the particle:
2
v
1  x2
c
 ' m c   2
  2
0 0 1m c 

 



'
m
v'




0
0

m
v
x
v
2
2
1
x
 


p'  
 ' m v' y  0
 0

0
1 0

 


0
0
0
1
0

'
m
v'





z 
 21m c   21m vx  

 with   v and  
2
 21m c  1 2 m vx 
c
 


0


0


1
v2
1 2
c
 1 2vx   1 2v
vx  v
 ' m v' x  1 2 m vx   1 2 m cv / c
v' x
'



 vx 
v vx
 ' m c  1 2 m c   1 2 m vx v / c
c  1 2 c   1 2vx v / c
1 2
c
This is a much simpler derivation than that found before.
8.2 Photon makes a angle from x axis)
S’
y’
Q’(frest )
v
y
S
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 co s 

 Q ' co s '

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  em  0; f obs  f em
1 
 f em
1  2
1   cos em
1 
1 
1  2
t he light is blue  shit ed, when t he light source is movingt oward you.
When v  c : 1 
v

c
ii) If  em   ; f obs  f em
1
v
1
c
f ob  f em /(1 
;
1 
 f em
1  2
v
)
c
classic D  E
1 
1 
t he light is red  shit ed, when t he light source is movingaway 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  )
c
classic D  E
1
1  2
T hispredict sa verysmall t ransvers
e Dopplereffect .
cos obs
Aberration of light from:
if em  0;
cosobs
if em   ;
cosobs  1;
if em   / 2;
1
  cos em

1   cos em
 1; obs  0
cosobs   ;
3
obs  
0  obs   / 2
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
, or a 4-vector force K ,
we will need to use

du

a
d
E

ˆ)
Because p  ( , p
c
and

dE dpˆ
K (
, )
cd  d


dp
K 
d
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!),
p
i.e., if the rest mass is not changing then and K
 
p  p  m 2c 2
are orthogonal. In 3+1-dimensional spacetime,
 
d ( p  p)
orthogonality
0
d
is something quite different from orthogonality in 3-space:


dp 
 dp
 p  p
 0it has nothing to do with 90 angles.
d


pK  0
d