Special theory of relativity
Mechanics (Study of motion of particles)
Classical Mechanics (macro)
Quantum Mechanics (micro)
Non-relativistic (v<<c)
Relativistic (v~c)
Special theory of relativity
Frame of Reference
x
y
z
O
Coordinate system
clock
Coordinate system + clock = Frame of reference
Frames of Reference
FRAME OF REFERENCE
Inertial frame Non-inertial frame
Obeys Newton Ist law of motion or law of inertia
Does not obey Newton Ist law of motion
or law of inertia
Rest frame
Moving with uniform speed
Accelerating frame
Rotating frame
Michelson-Morley Experiment:
Michelson-Morley Experiment:
Aim: To determine the speed of earth (experimentally) w.r.t. ether.
Michelson-Morley Experiment:
2
2
0
2
2
3
2
2
2
21
2
22/1
2
2
222
2
21
2
2
221
2)(
90
2
2
21
21
22
12
122
c
vxSo
throughrotatedisupsetwholewhen
c
vtcxor
c
v
c
v
cttttherefore
c
v
cc
v
cvct
c
v
cc
v
cvc
c
vcvct
total
=∆
−
=∆×=∆
=
=−=∆
+=
−=
−=
+=
−=
−=
++
−=
−
−
Michelson-Morley Experiment:
Now we have to calculate the number of interference fringes shifted from the centre of cross wires
For this we have to use one logic as one wavelength shifts one fringe, so if Δx contains Δn wavelengths then the number of interference fringes shifted from the centre of cross-wires is
totalx)(∆
4.037.02)(
2
2
≈==∆=∆c
vxn total
λλ
λ
Michelson-Morley Experiment:
Actual view of the experimental set-up
The Postulates of Special theory of Relativity
On June 30, 1905 Einstein gave two postulates of special theory of relativity on the results of Michelson-Morley experiment:
1. The Principle of Relativity:
The laws of physics are the same in all inertial frames of reference.
2. The Constancy of Speed of Light in Vacuum:
The speed of light in vacuum has the same value c in all inertial frames of reference.
The speed of light in vacuum is actually the only speed that is absolute and the same for all observers as was stated in the second postulate.
Transformation Equations:
Equations relating the position vectors of the same particle with reference to the two frames of reference are called as transformation equations.
s
O
r
S’
O’
r’v
Transformation Equations:
Transformation Equations:
Galilean Transformation
(v<<c) NRM
Lorentz Transformation
(V~c) RM
x x´
y´y
v
x´ = x – vt
y´ = y
z´ = z
t´ = tTime is absolute
K K´
O´O
vt x´x
Galilean Transformation:
Lorentz transformations:
where
In reverse transformation
Numericals based on Lorentz transformation:
Q1. Show that is invariant under Lorentz transformation.
Hint: we have to prove that
On the basis of Lorentz transformations
22222 tczyx −++
22222 tczyx −++ 22222 '''' tczyx −++=
Length Contraction:
Numerical problem related to Length contraction
Q A meter stick moving with respect to an observer appears only 500 mm long to her. What is its relative speed?
Hint: On the basis of length contraction.
2
2
0 1C
V−= mmm
m
5.0500
10
===
Numerical problem related to Length contraction
Q A spacecraft antenna is at an angle10o relative to an axis of the spacecraft. If the spacecraft moves away from the earth at a speed 0.7c, what is the angle of the antenna as seen from the earth?
Hint: tan (10o) = P/ 0
this gives P in terms of 0 =…..
Also
theta comes equal to 13.840 as is small in comparison to 0.
100
Spacecraft antenna
V=0.7cearth
0
θ PP
2
2
0 1C
V−=
Length Contraction:
L = L0 1- v2/c2
The length of an object is measured to be shorter when it is moving relative to the observer than when it is at rest.
Length were observer is moving relative to the length being measured.
Length were observer is at rest relative to the length being measured.
Time Dilation:
Time Dilation:
Time dilation:
t/
v
c (C2-v2)1/2
0
−
=
−
=−
=
=
2
2
/
2/1
2
2
0
22
0
0/
1
1
22
2
cv
tt
cv
cvc
t
ct
Numerical Problems related to Time Dilation:
Q. An observer on a spacecraft moving at 0.700 c relative to the earth finds that a car takes 40.0 minutes to make a trip. How long does the trip take to the driver of the car?
Hint: on the basis of time dilation
car =0.7 c
Numerical Problems related to Time Dilation:
Q Two observers, A on earth and B in a spacecraft whose speed is 2.00 x 10 8 m/s, both set their watches to the same time when the ship is at rest on earth. How much time must elapse by A’s reckoning before the watches differ by 1.00 s ?
Hint: initially t=t/=0
given that t- t/= 1 s
Our aim is to find ‘t’
For that change t/ in terms of t as
2 x 10 8 m/s
.93.3
/1000.2
1)11(1
8
2
2/
2
2/
stoequalcomestand
smvknowweascalculatedbecantNow
sc
vtttand
c
vtt
×=
=−−=−−=
Muon Paradox:
On the basis of Time Dilation:
muon can travel
vt0=(2.2 x10-6 s)(0.998c)=0.66 km
For earth frame the life time is
Extended
on the basis of time dilation6 Km
Earth
Muon frame
Life time is 2.2 µ s
kmscvtand
scc
cv
tt
4.108.34998.0
8.34/)998.0(1
102.2
122
6
2
2
0
=×=
=−
×=−
=−
µ
µ
0.998c
Numerical problem related to time dilation:
Q Find of the velocity of a spacecraft so that every day on it will correspond to 2 days on earth?
Hint:
2
2
0
0
1
2
1
c
v
tt
dayst
dayt
−=
==
Numerical problem related to Length contraction:
Q How fast should a spaceship move for its length to be contracted at 99% of its length?
Hint: on the basis of length contraction
.
1100
99
%99
1
2
2
00
0
2
2
0
calculatedbecanvthisFrom
c
v
ofthatgivenc
v
−==×
=
−=
Relativity of Simultaneity
• Events which are simultaneous in one frame may not be in another!
• Each observer is correct in their own frame of reference
Relativity of simultaneity:
How relativity affect simultaneity of events:
0/
//2
/1
=∆
==
timpliesThat
ttt
x1, x/
1
x2, x/
2
?0
0
21 =−=∆=∆
tttor
notortiffindtoaimOur
Relativity of simultaneity:
On the basis of Inverse Lorentz transformations:
( )Strweoussimulnotareeventstherefore
cv
xxc
v
t
cvcvx
tt
cvcvx
tt
..tan0
1
11
2
2
/1
/22
2
2
2
/2/
2
2
2
2
2
/1/
1
1
≠−
−=∆
−
+=
−
+=
Numerical Problem related to Relativity of simultaneity:
Q. An observer detects two explosions, one that occurs near her at a certain time and another that occurs 2.00 ms later 100 km away. Another observer finds that the two explosions occur at the same place. What time interval separates the explosions to the second observer?
Hint: v= 100/2 =50 km/ms=5 x 107 m/s
mstand
cv
cv
xxtt
tfindtoisaimour
xx
mst
97.1
1
)'(
0
2
/
2
2
212'/
/
/1
/2
=∆
−
−+∆=∆
∆
=−
=∆100 Km
Appears to be simultaneous
LORENTZ VELOCITY TRANSFORMATIONS:
OR Velocity addition theorem:
s s’ v~c
o o’
u, u’ x, y, z, t and
x’, y’, z’, t’
Our aim is to find the relation between u and u’ in terms of x, y, z components as u and u’ are along arbitrary directions.
ux = 1 + (v/c2)ux´
uy =
uz =
γ [1 + (v/c2)ux´ ]
γ [1 + (v/c2)ux´ ]
uy´
uz´
Addition of Velocities:
ux´ + v
Numerical problem related to Lorentz velocity transformation:
Q. Two spacecraft A and B are moving in opposite directions, An observer on the earth measures the speed of craft A to be 0.750 c and the speed of craft B to be 0.850 c. Find the velocity of craft B as observed by the crew on craft A.
Hint:
earthAB
.977.01
850.0
750.0
2
/
/
cu
cv
vuu
ufindtoisaimour
cu
cv
x
xx
x
x
−=−
−=
−==
Relativistic Inertia (“relativistic mass”)
z
x
y
x’
y’
z’v
S
S’
Y
V’B
VA
Ball A moves vertically only in frame S with speed VA , Ball B moves vertically only in frame S’ with speed VB ’= VA . Ball A rebounds in S with speed VA , Ball B rebounds in S’ with speed VB’ .
Y/2
Y/2
Collision in S Collision in S’
VA T0 = 2× Y 2 VB 'T0 = Y
momentum conservation : mAVA = mBVB
Relativistic mass (cont.)
Relativistic mass (cont.)
momentum conservation in S :
mAV A = mBVB
VB =Y T
T =T0 1− v c( )2
mA Y T0 = mB Y T0 1− v c( )2
mB = mA 1− v c( )2
( )massrest
1
Inertia) tic(Relativis Mass" icRelativist"
0
20
=−=
m
cvmm
.
Relativistic mass (cont.)
Numerical problem related to relativistic mass:
Q: Find the mass of an electron (m0 = 9.1E-31 kg)whose speed is .99c
Hint:
2
31
99.01
101.9
−
×=−
cc
m
Numerical problem related to relativistic mass:
Q. An electron has a kinetic energy 0.100 MeV. Find its speed according to classical mechanics.
Hint: KE of electron =1/2 mv2 =0.100 MeV. 1/2 mv2 =0.100 x 1.602 x10-13 J1/2 mv2 =0.100 x 1.602 x10-13 Kg m2 s-2
1/2 (9.1 x 10 -31 ) v2 = 0.100 x 1.602 x10-13 Kg m2 s-2
v2 = 2 x 0.100 x 1.602 x10-13 Kg m2 s-2 / 9.1 x 10 -31
v2 = 0.0352 x 1018
v = 1.876 X 108 m/s.
F
dx
mo m
Momentum-energy relations:
E =m0c 2
1− v c( )2p =
m0v
1− v c( )2
E2 − p2c2 =m0
2c4
1− v c( )2−
m0
2v 2c2
1− v c( )2
=m0
2c4
1− v c( )21− v 2
c2
= m0
2c4
E2 = p2c2 +m0
2c4
for a massless particle (photon, neutrino, ...)
E = pc
(and v = c always)
Numerical question related to Relativistic mass and Momentum:
Q1 Find the momentum of an electron (in MeV/c) whose speed is 0.600 c.
Hint:
cMeVp
cMeVKgascMeVin
cc
cp
c
v
vmmvp
/381.0
/1079.1
11/
600.0179.1
600.010101.9
1
230
2
3031
2
2
0
=
×=
−
×××=
−==
−
Numerical problem related to relativistic mass
Q At what speed does the kinetic energy of a particle equal its rest energy?
Hint:
cv
cv
m
c
v
m
mm
cmmc
cmcmmc
energymassstKE
2
3
2
1
1
2
1
2
2
Re
2
2
0
2
2
0
0
20
2
20
20
2
=
=−
=−
==
=−=
Numerical question related to Relativistic mass and Momentum:
Q Find the momentum of an electron whose kinetic energy equals its rest mass energy of 511 keV.
Hint:
./885
/51133
33
43
12
3
1
2
32
1
2
20
000
2
2
0
2
2
0
0
20
20
2
ckeVp
ckeVc
cm
c
ccmcm
cm
cv
vmp
ascalculatedbecanmomentumvthisofbasistheon
cvm
cv
m
mm
cmcmmc
o
=
×==×==
−
=−
=
=⇒=−
==−
Relativistic Mass
It follows from the Lorentz transformation when collisions are described from a fixed and moving reference frame, where it arises as a result of
conservation of momentum.
For v = c, m =m0
The increase in relativistic effective mass makes the speed of light c the speed limit of the universe.
Mass–energy equivalence
E=MC2
3-meter-tall sculpture of Einstein's 1905 E = mc2 formula at the 2006 Walk of Ideas, Germany
In physics, mass–energy equivalence is the concept that any mass has an associated energy and vice versa. In special relativity this relationship is expressed using the mass–energy equivalence formulawhere
•E = energy, •m = mass, •c = the speed of light in a vacuum (celeritas),
Momentum and Energy Transformations
Consider a frame of reference S’ moving with a speed v along positive direction of X-axis w.r.t. a frame S. Let the origins of the two frames coincide when t=t’=0 and then a signal of light is sent. We may write
22222
22222
tczyxand
tczyx
′=′+′+′=++
These equations have the solutions given by Lorentz transformation equations i.e.
yy =′2
2
1c
v
vtxx
−
−=′ zz =′
2
2
2
1c
v
c
vxt
t
−
−=′
If p is the momentum of photon of light having components px, py, pz in frame S. then
2222zyx pppp ++=
But c
E
c
hhp === ν
λ
2
22222
2
22222
22
22
2222
2222
2
''''
=++
=++
==++
++=∴
c
EcpppSimilarly
c
Ecpppor
cc
E
c
Epppor
pppc
E
zyx
zyx
zyx
zyx
Comparing, we observe x, y, z and t correspond to px , py, pz and &
x’, y’, z’ and t’ correspond to p’x , p’y, p’z and respectively.
2c
E
2c
E′
2
2
2
1
'
cv
Ecv
pp
x
x
−
−=
yy pp ='
2
2
2
2
22
2
1
1
cv
vpEEor
cvc
vp
cE
c
E
x
x
−
−=′
−
−=
′
Therefore, the solutions can be evaluated as
zz pp ='
These are the transformation equations of momentum and energy.
The inverse transformations can be written as
2
2
2
1
''
cv
Ecv
pp
x
x
−
+=
'
'
zz
yy
pp
pp
=
=
and
2
2
1
'
cv
vpEE x
−
+′=
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