30 Question 2013 Part 1

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Applied Physics [300218 (15)] Unit-1 Theory of Relativity

ALOKEVERMA, DEPARTMENTOFAPPLIEDPHYSICS, SRITECH, NEWRAIPUR Page 1

Question-1: Give Galileos Principle of relativityAnswer: A particle has different coordinates in different frames at the same instant. The equationswhich relate the coordinates of two frames of references is called the transformation equations. The

equation relating the coordinates of a particle in two inertial frames are called the Galilean

transformation.

Suppose we are in inertial frame of reference S and find the coordinates of some event that

occurs at the time t are x, y, z.

An observer located in a different inertial frame S which is moving with respect to S at the

constant velocityv

, will find that the same event occurs at time t and has the position coordinates x

y, and z.

Fig.1. Galilean Transformation

Assume that v

is in positive x (+x) directions. When origins of S and S coincide, measurements in the

x direction made in S is greater than those of S by tv

(distance).

Hence, vtxtvx'x t't&z'z;y'y No relative motion ------- (1)

The universal nature of this as assumed in classical physics is expressed by the equation t't . These set

of equations are are known as Galilean transformations.Similarly, the inverse Galilean transformation can written as

'vt'xx 'tt&'zz;'yy ------- (2)

Velocity and acceleration Transformation

From the first equation from the set eq. (1);

vtx'x Differentiating it with respect to t, one gets

vdt

dx

dt

'dx

Since t=t from eq. (1), then

'dt

'dx

dt

'dx v

dt

dx

'dt

'dx Analogously,

dt

dz

'dt

'dz;

dt

dy

'dt

'dy

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Second Method:-A particle has different coordinates in different frames at the same instant. The equations which relate

the coordinates of two frames of references is called the transformation equations. The equation

relating the coordinates of a particle in two inertial frames are called the Galilean transformation.

In fig.1. is an inertial frame in which an event occurs at any instant t at a point P. The coordinates of

the points P are (x,y,z,t). If another frame Sis moving with a velocity v relative to the frame S along its

X-axis and the coordinate axes of the frame S are parallel to that at the frame S, with their origins Oand O coincident at t=t=0, the coordinates of the same event at the point P in frame S will be

(x,y,z,t). The coordinates of the point P in frame S and S are related as:

t't;z'z;y'y;vtx'x ------ (1)'tt;'zz;'yy;'vt'xx ----- (2)

Fig.1. Galilean Transformation

If the position vectors of the point P at any instant in frames S and S are r and 'r respectively in, then

the vector triangle OOP,

P'O'OOOP vt'rr vtr'r ------ (3)

Differentiating above equation with respect time t, we get

vdt

rd

dt

'rd vV'V ------- (4)

The equation (4) is called the Galilean Law of Addition of Velocity.

Again differentiating above equation (4) with respect time t, we get

0dt

Vd

dt

'Vd

We Know that adt

Vdand'a

dt

'Vd ,thereforea'a ------ (5)

Thus the acceleration of the particle observed in frames S and S is the same. In other words, the frameS is also inertial like the frame S .

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2

2

2

2

2

2

2

c

v1c

lt

c

v1c

lt ------ (1)

Now the total time of travel of light from the plate P to the mirror M 1and the back from M1 to the

plate P is

2

221

2

2

2

21

c2

v1

c

l2

c

v1

c

l2

c

v1c

l2t2t ; {By binomial expansion}

------ (2

And the total time of travel of light from the plate P to the mirror M 2and the back from M2 to the plate

P is

c2

vc

lvcvc

vc

l

)vc(vc

vc

l

vc

l

vc

lttt

2222

22

bf2

2

2

2

2

2

222

c

v1c

l2

c

v1c

l c2

vc

l c2t

2

21

2

2

c

v1

c

l2

c

v1

c

l2; {By binomial expansion} ----- (3)

The time difference in the time of travel of the light in the two mutually perpendicular directions

2

2

2

2

2

2

2

2

12

c2

v1

c

v1

c

l2

c2

v1

c

l2

c

v1

c

l2ttt

2

2

2

2

2

2

2

2

c2

v

c

v

c

l2

c2

v1

c

v1

c

l2

3

2

2

2

2

22

c

l v

c2

v

c

l2

c2

vv2

c

l2t

----- (4)

Hence, the path difference for the light rays travelling in the two mutually perpendicular directions is

2

2

3

2

c

l v

c

l v.ctcx ------ (5)

Now the entire apparatus is rotated by 90. Therefore the path difference is

2

2

c

l vx ------ (6)

Thus, the change in path difference between the light waves coming from the two directions due to

rotation of the apparatus by 90 is

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2

2

2

2

2

2

2

2

2

2

c

l v2

c

l v

c

l v

c

l v

c

l vxxx

------- (7)

In the Michelson-Morley experiment .Sec/m103c.;Sec/m103v;m11l 84

m1022

103

10322

103

103112

c

l v2x

8

16

8

28

24

2

2

---- (8

For the yellow light (mean wavelength =5500=5.510-7m) the number of fringe shifted is

4.055

22

105.5

1022xn

7

8 ---- (9)

This result is called negative result of Michelson-Morley experiment because Michelson and Morley

observed no fringe shift when the experiment was repeated after six month.

Conclusions from the negative result-

1. The velocity of earth is zero relative to the ether.

2. The speed of light does not depend on the motion of source or on the motion of observer.

3. The concept of ether to be stationary is found to be wrong.Question-3: Derive inverse Lorentz transformation equation for space and time co-ordinate. Showthat when V

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Fig.1. Two inertial frame in relative motion

22222

2

222

2

222

zyxctc

zyxt

c

zyx

c

OPt

0ctzyx 22222 ------- (1)

And the time taken by the light to reach from O to P as observed in the frame S is

c

zyx

c

POt

222

22222

2

222

2

zyxct

c

zyxt

0ctzyx 22222 ------- (2)

From eq.(1) and (2), we get2222222222

ctzyxctzyx ------- (3)Since the frame S is moving relative to the frame S along S-axis, the lengths in direction perpendicular

to the direction of motion are un affected i.e.,

zzandyy , ------- (4)Therefore from eq.(3),

222222ctxctx ------- (5)

The transformation between x and x coordinates is given by simple relationvtxkx ------- (6)

Where k is constant, independent of x and t.

And tvxkx ------- (7)Where k is constant, independent of x and t.

Substituting the value x from eq. (6) in eq. (7), we get

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2

2

2

2

c

v1

kk

1

kk

11

c

v

22

2

2

2

2

2

2

c

v1

c

v1

c

v1

c

v1k

k

1

2

2

c

v1

1k -------- (11)

Substituting the value of k and k in eq. (6) and (8), we get

2

2

2

2

2

c

v

1

cvxt

'tand

c

v

1

vtxx

--------- (12)

Thus the Lorentz transformations are

2

2

2

2

2

c

v1

cxvt

'tandzz;yy;

c

v1

vtxx

----- (13)

Question-4: Derive the formula for the variation of mass with velocity according to special theory ofrelativity.

OrDeduce the variation of mass with velocity. Show that where the symbols have theirusual meaning.Answer: According to the Newtonian mechanics, mass is an invariant quantity. If a stationary body ofmass m is acted upon by a finite and constant force F for a time t due to which it acquires a velocity v

then the gain in momentum of the body is

tFmvP ----- (1)The maximum momentum that can be gained by a body is

mcPmax

----- (2)Where c is the velocity of light.

According to the special theory of relativity the mass of body is not constant, but it depends on

the velocity of the body. When the body is at rest ( v=0 ), the mass of the body is m0and when the

velocity of the body becomes equal to the velocity of light ( v=c ), the mass of the body becomes infinite

Consider a frame S relative to which a particle is moving with the velocity v along the X-axis

The moving mass of the particle in frame S is m. Consider another frame S in which the particle is

stationary. The rest mass of the particle in frame S is m 0. If the displacement of the particle relative to

the frame S in time t is y along Y-axis, then the velocity of particle along Y-axis in the frame S is

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t

ymPmomentumi tsand

t

yv

0'yy

------ (3)Similarly,

t

ymPand

t

yv 0vy y ------ (4)

From the Lorentz transformations yy , hence yy and according to time dilation

2

2

c

v1

tt

t

y.

c

v1m

cv1

t

ym

t

ymP

2

2

22

y ----- (5)

Since the momentum of the particle is invariant, therefore Py=Py. From eq. (3) and eq. (4), we get

t

y.

c

v1m

t

ym

2

2

0

2

2

0 c

v1mm

2

2

0

c

v1

mm

------- (6)

The above expression gives the variation of mass with velocity.

Second Method:According to Newtonian Mechanics, the mass of an object is independent of its motion. But according

to Einstein, the mass of an object in motion is different from its mass at rest. We shall now verify this

by considering the hypothetical experiment of Tolman and Lews.Consider two systems of coordinates S and S, the later moving with velocity v relative to S

along positive direction of their common X-axis as shown is fig.1. We consider the collision of two

objects in system Sand observer it from the systemS. Let two objects masses m1and m2are travelling

parallel to X-axis with uand u, respectively. Let objects collide and after collision coalesce into a

single object.Here it should be noted that the definition of momentum = mass velocity)and the principles

of conservation of mass and momentum also hold well in relativity same as in classical mechanics. So

according to the conservation of mass, the mass of the coalesced objects after collision should be 2m

Since the two objects were moving with same velocity in opposite directions, therefore, after collision

they are at rest for an observer in system S . Using the law of addition of velocities, the velocities u

and u2in system Scorresponding to uanduare given by

2

1

c

uv1

vuu ------ (1)

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And

2

2

c

uv1

vuu ------ (2)

Further, we assume that the mass of the object moving with velocity u1 is m1and that of the object

moving with velocity u2be m2. According to the principle of conservation of momentum, we have

Momentum before collision= Momentum after collision

i.e. vmmumum 212211 ------- (3)where vis the velocity of single object after collision.

Using equation (1), (2) and (3), we get

vmm

c

uv1

vum

c

uv1

vum

21

2

1

2

1

2

2

2

1

cuv1

vuvmv

cuv1

vum

2

2

2

1

2

2

2

1

c

uv1

vuc

uvv

m

c

uv1

c

uvvvu

m

2

2

2

1

c

uv1

c

uv1

m

m ------- (4)

Now using equation (1), we get

2

2

2

2

2

1

c

1.

c

uv1

vu

c

u1

2

2 ccuv1

vu1

2

2c

uv1

c

vu

1

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2

2

2

22

c

uv1

c

vuv2u

1

2

2

2

2

22

22

2

2

2

1

c

uv1

c

v

c

uv2

c

u

c

uv1

c

u1

2

2

2

2

22

2

4

22

2

2

2

1

c

uv1

c

v

c

uv2

c

u

c

vu

c

uv21

c

u1

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

1

c

uv

1

c

u1

c

v1

c

uv

1

c

v1

c

u

c

v1

c

u1

------ (5

Similarly2

2

2

2

2

2

2

2

2

c

uv1

c

u1

c

v1

c

u1

------ (6)

From equation (5) and (6), we get

2

2

2

2

2

2

2

2

2

1

c

uv1

c

uv1

c

u1

c

u1

------ (7)

Therefore 0

2

2

1

2

2

2

2

1m

c

u1

c

u1

m

m

2

2

2

0

2

2

2

1

0

1

c

u1

mmand

c

u1

mm

------ (8)

Finally, we get in general form

2

2

0

c

v1

mm

The above expression gives the variation of mass with velocity.

Question-5:Deduce Einsteins mass Energy relation E=mc2 considering the variation of mass withvelocity.

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Answer:- According to the Einstein, the mass and velocity are equivalent to each other. If a mass m isconverted into energy, the energy produced is E=mc2, where c is the speed of light. This is called the

mass energy equivalence relation.

Let a particle of mass m moving with a velocity v is displaced by a distance dx by applying a

force F on it. The work done by the force is

dx.dt

dmvdx

dt

dvm

dx.dt

dmvdt

dvmdx.mvdt

ddk

dx.Fdwdk

dmvmvdvdk 2 ; vdt

dx

------ (1

But 2202

2

2

2

2

0cm

c

v1m

c

v

1

mm

2

0

2222mvmcm ------ (2)

On partial differentiation of eq. (2), we get

vdvm2dmmv2dmmc2

0vdvm2dmmv2dmmc2

222

222

mvdvdmvdmc 22 ------ (3)

From eq. (1) and (3), we get

dmcdk 2 ------- (4)

Hence gain in kinetic energy in acquiring the velocity v from rest is

2

0

2

0

2

m

m

2cmmcmmcdmck

0

------ (5)The eq. (5) gives the expression for the kinetic energy of the particle at relativistic velocity.

But at v=0, the rest energy of the particle = 20cm ------ (6)

Total energy of the particle E= kinetic energy + rest energy2

0

2

0

2cmcmmcE

2mcE ------ (7)

The above equation is called the Einstein equation of mass-energy equivalence.

Question-6. Calculate the percentage contraction of rod moving with a velocity 0.8c in a directioninclined at 60 to its own length.Answer: the component of the length of the rod along its direction of motion 00 L

2

160cosL

And the component of its length, perpendicular to its direction of motion 00 L2

360si nL

Where L0= length of the rod, placed along the x-axis, in frame S(at rest).

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In this case cos component undergoes change in length.

Hence c3.0c

c8.01L

2

1L

2

2

0x Other component

0y L2

3L

Total length of the rod in frame S (moving frame) is

00

2

0

2

0

2

0

2

0

L9165.0L84.0

L4

3L09.0L

2

3L3.0L

The percentage contraction produced in length of the rod

%34.8100L

L9165.0L

0

00 Question-7: What will be the apparent length of a meter stick measured by an observer at rest whenthe stick is moving along its length with a velocity equal to c2

3

?Answer:As

2

2

0 1

c

vLL

4

343

2

2

2

2

c

c

c

v

24

34

4

311 0

002

2

0

LLL

c

vLL

mL 10

.5.0

2

1mL

Question-8: Calculate the velocity of a particle at which its mass will be 5 times the mass at rest.Answer:

2

2

0

1c

v

mm

According to question 05mm , therefore we have

2

2

0

0

c

v1

mm5

25

11

c

v

25

1

c

v1

5

1

c

v1

c

v1

15

2

2

2

2

2

2

2

2

.sec/m1094.2

10398.0v

c98.0v

c25

24vc

25

24v

25

24

c

v

8

8

22

2

2

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Question-9: Show that a particle which travels with the speed of light must have a zero rest mass.Answer: The relativistic mass of a particle with speed v is

2

2

0

1c

v

mm

Therefore rest mass is2

2

0 1

c

vmm

when ,cv then 00 m Thus, a particle traveling with the speed of light must have a zero rest mass.

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30 Question 2013 Part 1