PH604
Special Relativity (8 lectures)
• Newtonian Mechanics and the Aether
• Einstein’s special relativity and Lorentz transformation
and its consequences
• Causality and the interval
• Relativistic Mechanics
• Optics and apparent effects
Books: “Special Relativity, a first encounter”, Domenico Giulini, Oxford
“Introduction to the Relativity Principle”, G.Barton, Wiley
+ many others in Section QC.6
1
Newtonian Mechanics and the Aether
1. Newtonian Mechanics and Newton’s law of Inertia
2. The relativity principle of Galileo and Newtonian
3. Questions with regard to Newtonian Mechanics
4. The “Aether” – does it exist?
5. Michelson – Morley Experiment
Books: “Special Relativity, a first encounter”, Domenico Giulini, Oxford
“Introduction to the Relativity Principle”, G.Barton, Wiley
+ many others in Section QC.6
2
1.Newtonian Mechanics and Newton’s law of Inertia
--Newton’s Law: m a = F: Predict the motions of the planets, moons,
comets, cannon balls, etc
--This law is actually not always correct! (surprised?)
http://www.phys.vt.edu/~takeuchi/relativity/notes/section02.html
--Inertial Frame: A frame in which the Newton’s law is correct.
--Any frame that is moving at a constant relative velocity to the first
inertial frame is also an inertial frame.
--The frames in which Newton’s law does NOT hold that are accelerating
with respect to inertial frames and are called non-inertial frames.
3
2. The relativity principle of Galileo and Newtonian
Y
S
Y’ S’
Two inertial reference frames S
and S’ moving with a constant
velocity u relative to each other
u
( u is // to x and x’)
O
Z’
O
’
X
X’
A moving object is described in
S: as (x,y,z,t) and in S’: as (x’, y’, z’, t’)
z Common sense shows the two measurements are related by:
x '  x  ut;
Or in vector form:
v x'  v x  u
y '  y;
(1.1) '
vy  vy
(1.2) v ’ = v - u
z '  z;
'
a’ = a
v

v
z
z
t'  t
--This is the Galilean transformation. Note the universal time, t=t’
--They would assert that Mechanics only deals with relative
motion and that ‘absolute’ motion can never be measured.
4
3. Questions with regard to Newtonian Mechanics
i) phenomena on a very small scale we need Quantum Mechanics;
ii) Phenomena where the speed of motion is near the speed of light “c”
we need relativity
We shall be concerned with case ii) in the 8 lectures in this course.
Modern experiment that shows the limitation of Newtonian mechanics:
Van de Graaf accelerator Experiment:
[American Journal of Physics, Volume 32,
Issue 7, pp. 551-555 (1964). ]
Target B
Accelerator
Pulsed electrons beam
Measure the rise in temperature
D
T  Kinetic Energy of electrons (between 0.5-15 MeV)
The V. of electrons can be determined by: V = D / time
A relation between V2 vs K.E of the electrons can be plotted.
5
Newtonian Mechanics:
K.E. = ½ mv2
v2
C2
O
--N-M prediction is valid at low energy (velocities).
K.E.
--Experiment: Vmax  3108(ms-1)C
--The Vmax of the electrons appear to equal the speed of light in Vacuum.
--Other ‘massless particles’ such as neutrinos appear only to move
at C as well
6
4. Speed of Light: existence of Aether ?
Maxwell’s electromagnetic theory predicted that light
should travel with a constant speed in vacuum,
irrespective of reference frames:
1
C
 0 0
How light propagates through a vacuum ?
--All other wave motions known, needed some form of ‘medium’
-- Wave velocity would be relative to the ‘medium’
Suggestion:
Perharps, even a vacuum contains a very tenuous ‘medium’ --- the ‘Aether’,
then the constant velocity of light is relative to this absolute frame, and the
speed of light in other ‘inertial’ systems would not be C.
if so, can we detect it?
Direct measurement of the relative motion to aether is difficult, but If it
existed in space, we should be able to measure the motion of the
Earth relative to aether -- Michelson-Morley (1887).
7
5. Michelson-Morley Experiment –Detect the Earth moving
through the Aether??
--In 1887 Michelson and Morley built an interferometer
To measure the movement of the
Earth through the Aether.
Mirror 1
Even though this instrument can be
a few meters in size, it can detect
changes in distance of hundreds of
nanometers
Mirror 2
Light source
beam splitter
Detector
8
Interferometer, stationary in
the Aether
Interferometer Moving
Through the Aether
9
The time for light to travel
along l2 arm (cross stream)
t 2 (c 2  v 2 )  l22
Ct
l2
The time for light to
travel along l1 arm and
back: (downstream)
Aether
wind speed
Vt
travel along l2
arm and back:
t 2  2t 
v
2l 2
v
c 1 2
c
l1
l
 1
cv cv
2l1

v2
c (1  2 )
c
t1 
l1
If the light has frequency of f, the number of fringes that corresponds with
differences, t1-t2 of the light travel in the two arms is:




2
l
2
l
1
2

f  (t1  t 2 )  f  

2
2 
  v 
v
 c1  2  c 1  2 
c 
  c 
10
Since the test was to see if any fringes moved as the whole apparatus
was turned through 90o.Then the roles of l1 and l2 would be exchanged,
and the new number of fringes would be


2l1
2l2
f  (t '1 t '2 )  f  



v2
v2
c1  2
c 1 2
c
c






 

So the observed number of fringe shift on rotation through 90o should be:




2 f (l1  l2 )  1
1  f (l1  l2 ) v 2 l1  l2 v 2
'
'
f  (t1  t 2 )  f  (t1  t 2 ) 



2
2
2


c
c
c
 c2
 v 
v
1 2 
 1  2 
c
c 


11
Michelson & Morley made apparatus long enough to detect 1/3 of a
fringe, with =500nm, so that l1 + l2 =17m, Nfringe = 108v2 /(3c2)
--But they could detect no shift at all (at any time of year!)
--The only possible conclusion from this series of very difficult
experiments was that the whole concept of an all-pervading aether
was wrong from the start.
12
Einstein’s special relativity and Lorentz
transformation and its consequences
1.
2.
3.
4.
5.
6.
Einstein’s special relativity
Events and space-time in Relativity
Proper time and the invariant interval
Lorentz transformation
Consequences of the Lorentz transformation
Velocity transformation
13
1. Principle of Relativity by Einstein (1905)
It is based on the following two postulates:
1) The laws of physics are the same for all observers in uniform motion
relative to one another (principle of relativity),
- need a transformation of coordinates which preserves the laws of
physics
2) The speed of light in a vacuum is the same for all observers,
regardless of their relative motion or of the motion of the source of the
light. http://en.wikipedia.org/wiki/Theory_of_relativity
V
V
A
B
A
B
Observer in the car: the light pulse
reaches A and B at the same time
Observer to whom the car is
moving with relative V: the light
pulse reaches A before B
Simultaneity breaks down  time cannot be regarded as a universal entity
- need a different transformation from Galileo’s but will converge to it for
14
V<<C
2. Events and space-time in Relativity
When and where is the object under our interest An Event in Relativity.
--An event is a point defined by (t, x, y, z), which describes the
precise location of a “happening” which occurs at a precise point in
space and at a precise time.
--“Space-time” is often
depicted as a “Minkowski
diagram”.
decelerated
Time (ct)
constant
accelerated
Space r
15
3 Proper time and the invariant interval
3.1 invariant interval
In 3-dimensional EUCLIDIAN space: P1
P2
In coordinate system O:
P1 ( x1 , y1 , z1 ), P2 ( x2 , y2 , z2 )
In coordinate system O’:
P1 ( x'1 , y'1 , z'1 ), P2 ( x'2 , y'2 , z'2 )
r 2  x 2  y 2  z 2  x'2 y'2 z'2
In relativity, we would like to find a similar quantity for pairs of events,
that is frame-independent, or the same for all observers, that is
invariant interval
(s)2  (ct )2  (r )2  (ct )2  (x)2  (y)2  (z)2
∆t is the difference in time between the events
∆r is the difference between the places of occurrence of the events.
16
3.2 Events, INTERVAL AND THE METRIC
A metric specifies the interval between two events
17
3.3 Proper time (length) the invariant interval
The proper time between two events is the time experienced by an
observer in whose frame the events take place at the same point.
According to the definition of the interval between two events:
c 
( s ) 2
i) If (s)2  (ct )2  (r )2  0 , the interval is said to be “timelike”
--there always is such a frame since positive interval means: | c |  r
so a frame moving at vector v
= (∆r) /(∆t), in which the events take
place at the same point, is moving at a speed < c
ii) If (s)2  (ct )2  (r )2  0 , the interval is said to be “spacelike”
--It is still invariant even though there is no frame in which both events
take place at the same point. (or (c∆t)2 < 0).
--There is no such frame because necessarily it would have to move
faster than the speed of light.
18
Sometimes the proper distance is defined to be the distance separating
two events in the frame in which they occur at the same time. It only
makes sense if the interval is negative, and it is related to the interval by
 
S 2
iii) if (s)2  (ct )2  (r )2  0 ,the interval is said to be “light-like” or
null.
This is the case in which
i.e., (ct )2  (r )2
Or, in which the two events lie on the worldline of a photon.
Because the speed of light is the same in all frames…….
…. an interval equal to zero in one frame must equal zero in all frames.
The three cases have different causal properties, which will be
discussed later.
19
4. A transformation formula – Lorentz Transformation
4.1 The formula fits into the Einstein’s two postulates
We assume that relative transformation equation for x is the same as
the Galileo Trans. except for a constant multiplier on the right side, i.e,
x'   ( x  ut )
x   ( x'ut ' )
(1)
(2)
where  is a constant which can depend on u and c but not on the
coordinates. (based on Postulate 1)
To find the  factor ?
By tracing the propagation of a light wave front in two different
reference frames, one of which is moving with a velocity of V
along x-axis w.r.t. the other.
20
Assume a light pulse that
starts at the origins of S
and S’ at t =t’=0
Y’
After a time interval the
front of the wave moves
Y
S
Y
u
Y’
It is recorded as:
S’
(X, t) in S
and
(X’, t’) in S’
O’
O
O
O’
X
XX’
z
Z’
z
X’
Z’
By Einstein’s postulates 2:
x = ct
x’=ct’
21
Substituting ct for x and ct’ for x’ in eqs. (1) and (2)
x'   ( x  ut) (1)
x   ( x'ut' ) (2)
ct '   (ct  ut)   (c  u)t
ct   (ct 'ut' )   (c  u)t ' (4)
t '  (c  u )
(3)  
(3' )
t
c
t'
c
(4)  
(4' )
t  (c  u )
Let (3’) = (4’)
2
c
1
1
  2 2
 
2
2 2
c u
u
1

u
/
c
1 2
c
2
(3)
u < c so  is
always > 1
If u~c,  
When u << c
~1
22
The relativistic transformation for x and and x’ is
x'   ( x  ut) (1)
x   ( x'ut' ) (2)
Lorentz transf.
If u << c
 ~1
x'  x  ut;
x  x'ut
Galileo transf.
23
The transformation between t and t’ can be derived:
For the wave front of light, x=ct, x’=ct’
Divide c into Eq.(1) x'  ct '   ( x  ut)   (c  u)t
t' 

c
(ct  ut )   (t  uct
)   [t 
2
Divide c into Eq(2)
t
c
(1)
ux
]
2
c
x  ct   (c  u)t '
(2)

'
(ct 'ut ' )   (t ' uct ' )   [t ' ux
]
2
2
c
c
c
The complete relativistic transformation (L.T.) is
ux'
x   ( x'ut ' ), y  y' , z  z ' , t   [t ' 2 ]
(5)
c
ux
x'   ( x  ut ), y'  y, z '  z, t '   [t  2 ]
(6)
c
24
4. 2 The interval of two events under Lorentz transformation.
For two events, (t1, x1,y1,z1) and (t2,x2,y2,z2), we define:
(T, X, Y, Z) = (t1-t2,x1-x2,y1-y2,z1-z2)
cT '   (cT 
then Lorentz transformation becomes
X '  (X 
2
c T
'2
v2
 X '   [c T  2 X 2  2vXT
c
 X 2  v 2T 2  2vXT ]
2
2

2
2
c 2T '
2
2
(c T
2
 X
2
 X
'2
2
v
X)
c
v
cT )
c
v2
)(1  2 ), i.e.,
c
 c 2T 2  X
2
i.e., (S ' )2  (S )2
The interval of two events is an invariant under Lorentz Transformation.
25
For short: the interval is a Lorentz scalar.
4.3 Lorentz transfermation in 4-dimensional formula
The L-T could be formally defined as a genernal linear transformation
that leaves all intervals between any pair of events unaltered.
Introduce 4-D vector
 ct1

x
x   1
 y1


 z1







x  ct1 , x1 , y1 , z1 
S 212  (x1  x 2 )g (x1  x 2 )
1 0 0 0 
0  1 0 0 
Here we have introduced:

g
0 0  1 0 


0 0 0  1
26
L-T can be expressed as
ct '
 x' 
 
 y' 
 
z' 
X'  LX
v



c

v


  
c

0
0
0
0

0 0 

0 0


1 0
0 1
ct 
x 
 
y 
 
z 
27
5. Consequences of the Lorentz transformation
5.1 Time dilation
A light source in S’ a flash of light lasting t’ seconds
Y’
Y
The flash starts at time t’1, and goes off at time t’2,
as measured by a clock in S’
O
Z
O
’
t’ = t’2 – t’1
X
X’
Also: x’ = x’2 – x’1 = 0
Z’
How does an observer in S views the light on and off events?
Light on time:
ux1 '
t1   [t1 ' 2 ]
c
Light off time:
t 2   [t 2 '
ux2 '
]
2
c
28
x’ = 0
Light duration time measured by a clock in S:
t  t 2  t1 
(t 2'  t1' )  ( x2'  x1' )u / c 2
u2
1 2
c
t  t '
--S is the moving frame w.r.t the
strobe light
-- t is called tmoving
we have
t moving 

t '
u2
1 2
c
--S’ is the rest frame w.r.t the
strobeflash
-- t’ is called trest(the proper time)
t rest
u2
1 2
c
tmoving > trest, this effect is called time dilation on a moving clock,
or moving clock runs slower, i.e., Observer in S see’s the clock in S’
runs slower because the clock in S’ is moving w.r.t to him.
trest is also called the proper time (the shortest) of the two events.
29
Experimental demonstration of time dilation effect:
Cosmic Ray evidence for ‘time dilation’
 Mesons are formed at heights >
10 km in atmosphere.
Observations found that most of
them manage to survive down to
sea level –despite their half-life
being only
6
 1/ 2  2 10 s
Even moving in C, half should
have decayed in a distance of:
2 106  (3108 )  600m 10km
But: as they move so fast their
clocks (proper time) run slower
due to “ time dilation”.
If v=0.999c,  =22.37
http://www.jlab.org/~cecire/cosmic.jpg
  2 106  (3108 )  13.4km  10km30
The Twin Paradox
Right after their 20th birthday, L blasts off in a rocket ship for a space trip,
travelling at a speed 0.99c to a nearby star at 30 light year away, then
come back with same speed, while M stay on Earth..
-- In the view of M:
The journey will take time T = 2*30*c*year/0.99c = 60 year, so L will
return when M is 20+60=80 yr. How much will L aged over the same
period?
L was travelling at a high speed and L’s clock, including her internal
biological clock, were running slowly compared to M’s, therefore when
L reunite with M, L will have aged by T’ = T/  = 60/7 < 9 yr. So L is
younger than M.
-- In the view of L: M was travelling away at a high speed and M’s
clock, including her internal biological clock, were running slowly
compared to L’s, therefore when M reunite with L, M is younger than L.
Conclusion: In one frame of reference, L is younger while from
the other frame of reference, M is younger. This is the paradox.
http://www.phys.vt.edu/~takeuchi/relativity/notes/section15.html
31
Worldlines in the (M’s or Earth’s frame)
The invariant between two events D and R:
ct
S 2  (ct ) 2  x 2
R
L
M
M: x=0 the proper time for her is tDR
L: moves quickly, so (x’) 0, so her
proper time out to event T and back
again will be much smaller by factor of
 than tDR.
T
D
x
Let's draw this now in L's frame:
A problem: just what frame do we choose?
Frame S’ that is L's rest frame
on her way out to the space?
OR
Frame S’ that is L's rest frame
on her way back?
L changes frames at event T
This breaks the symmetry and resolves the paradox: M travels from event
D to event R in a single frame with no changes, while L changes frames.
L's worldline is crooked (non-inertial) while M's is straight (inertial)!
Therefore: M’s point of view is right, L will be younger than M
32
Causality and prohibition of motion faster than light.
33
5.2 Length contraction - A rod lies in x’ axis in S’, at rest relative to S’
y’
S’
-its two ends measured as x’1 and x’2
S
u
y
The length of the rod in S’ is
O’
L '  x2'  x1'
x’
O
z’
zx
x
What is the length of the rod measured in S?
Because the rod is moving relative to S, we should measure the
x-coordinates x1 and x2 of the ends of the rod at the same time,
i.e., t=t2-t1=0, L = x2 –x1
Using Eq 6
x1 '   ( x1  ut)
x2 '   ( x2  ut)
x'   ( x  ut) 
( x2  x1 )  u (t 2  t1 )
L
L'  x2  x1 

2
u
u2
1 2
1 2
c
c
'
'
34
Call L’ as the Lrest, since the rod is at rest to S’
L the Lmoving since it moves with velocity u relative to S,
Lmoving  Lrest 1  u / c
2
2
which shows the effect of length contraction on a moving rod.
-The length or the distance is measured differently by two observers
in relative motion
- One observer will measure a shorter length when the object is
moving relative to him/her
-The longest length is measured when the rod is at rest relative to
the observer---proper length
-Only lengths or distances parallel to the direction of the relative
motion are affected by length contraction
35
6. Velocity addition
A) Velocity Transformation
How velocities are transformed from one Ref. frame to another by
differentiating L.T. equations?
--In order to avoid confusion, we now use
reference frame S’ w.r.t. S in x direction.
Suppose a particle has a velocity in S’
in S:
Differentiate Eq. (5)
 for the speed of the
dx'
ux 
dt '
dx
ux 
dt
'
vdx'
dx   (dx'vdt ' ), dt   [dt ' 2 ]
c
36
dx  (dx'vdt ' )
ux  
vdx'
dt
 [dt' 2 ]
c
dx'
v
dt '

v dx'
1 2
c dt '
dy
uy  
dt
dy'
dt '

v dx '
 (1  2
)
'
c dt
dy'
vdx'
 [dt ' 2 ]
c
uz '
dz
dz'

uz  
'
vu
vdx
'
dt
 [dt' 2 ]  (1  2x )
c
c
L5
u x ' v

v ux'
1
c2

uy '
'
x
2
vu
 (1 
)
c
dz'
dt '

v dx '
 (1  2
)
'
c dt
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The velocity transformation
equation from S’ to S is
u x ' v
ux 
vu x'
1 2
c
uy '
uy 
(7 )
'
vu
 (1  2x )
c
uz '
uz 
vu x'
 (1  2 )
c
From Eq(7) and (8) we have:
The inverse velocity
transformation equation is
u v
ux '  x
vu
1  2x
c
uy
uy '
(8)
vu
 (1  2x )
c
uz
uz ' 
vu
 (1  2x )
c
i). When v and ux, ux’ << C,
ux = ux’ +v
ii). When ux‘=C,
and when ux=Cux’=C
ux=C,
, the L.T G.T
L.T. includes the constancy of the speed of light, as well as G.T. for the
low speed world.
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