special theory of relativity

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:


Aim: To determine the speed of earth (experimentally) w.r.t. ether.


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


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


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



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

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 s

kmscvtand

scc

cv

tt

4.108.34998.0

8.34/)998.0(1

102.2

122

6

2

2

0

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

cv

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

xxcv

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 =

[1 + (v/c2)ux´ ]

[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 2Y 2 VB 'T0 Y

momentum conservation: mAVA mBVB

Relativistic mass (cont.)

momentum conservation in S :

mAVA 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

cv

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

cv

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

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

special theory of relativity

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