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Dr. Mayeen Uddin Khandaker [Bilash]

BSc (Hons.); MSc in Physics [1st class 1st]

University of Chittagong, Bangladesh

PhD in Nuclear Physics, South Korea

Training Fellowship: Michigan State University, USA

Post-Doctorate: Korea Atomic Energy Res. Inst., Korea

Guest Researcher: Tohoku University, Sendai, Japan

Visiting Researcher: France Atomic Energy, Saclay, Paris

Consultant: International Atomic Energy Agency, Austria

Special Theory of Relativity 1. Background of special theory of relativity

Introduction Frames of Reference Postulates of Relativity Galilean Transformation

2. Relativistic kinematics Lorentz transformation Time dilation (Relativity of time) Length contraction (Relativity of length) Relativistic velocity transformation Doppler effect

3. Relativistic dynamics Relativity of mass Relativistic momentum & Newton’s second law of motion Equivalence of mass and energy

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Before the Lecture

Let’s see the picture and gain concept about the Relativity

1. BACKGROUND OF SPECIAL THEORY OF RELATIVITY 1.1 INTRODUCTION: In 1905 a young physicist of 26 named Albert Einstein showed how measurements of time and space are affected by motion between an observer and what is being observed. Theoretical concept of the special relativity was introduced in Einstein's paper "On the Electrodynamics of Moving Bodies".

Although motivated by a desire to gain deeper insight into the nature of electromagnetism, Einstein, in his theory, extended and generalized Newtonian mechanics as well. He correctly predicted the results of mechanical experiments over the complete range of speeds from u/c = 0 to u/c → 1. Newtonian mechanics was revealed to be an important special case of a more general theory. In developing this theory of relativity, Einstein critically examined the procedures used to measure length and time intervals. These procedures require the use of light signals and, in fact, an assumption about the way light is propagated is one of the two central hypotheses upon which the theory is based. His theory resulted in a completely new view of the nature of space and time.

The connection between mechanics and electromagnetism is not surprising because light, which plays a basic role in making the fundamental space and time measurements that underlie mechanics, is an electromagnetic phenomenon. However, our low-speed Newtonian environment is so much a part of our daily life that almost everyone has some conceptual difficulty in understanding Einstein’s ideas of space-time when he first studies them. Einstein may have put his finger on the difficulty when he said “Common sense is that layer of prejudices laid down in the mind prior to the age of eighteen.”

1.2. FRAMES OF REFERENCE: The first step is to clarify what we mean by motion. When we say that something is moving, what we mean is that its position relative to something else is changing. A passenger moves relative to an airplane; the airplane moves relative of the earth; the earth moves relative to the sun; the sun moves relative to the galaxy of stars (the milky way) of which it is a member and so on. In each case a frame of reference is part of the description of the motion. To say that something is moving always implies a specific frame of reference.

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In order to describe the motion of moving bodies, we need to state where the object is at any given time. But to state where an object is, we need to measure its position relative to something else, right? So we need a reference point from which to define the position of objects. Once we have chosen such

a point, which is called the origin, we can specify the position of the object by saying, for instance, that the object is distance x to the east, distance y to the north, and distance z up from the origin. We also need a clock so that we can specify at what time t the object was at the given position.

When we have the origin and the directions in which to measure the distance from the origin set up, and a clock to measure the time, we say that we have a frame of reference or simply a frame.

1.2.1 Inertial Frames

Newton's first law, the law of inertia, states that

1. if an object is at rest it will stay at rest if no force is acting on it, and 2. if an object is moving it will keep on moving at constant velocity if no force is acting on it.

This law is actually not always correct! (surprised?) It depends on which frame you are using to describe the motion of the object. For instance, if you are using the moving object itself as the origin of your frame of reference, it is always at rest no matter what forces are acting on it.

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Another example is the motion of objects inside the space shuttle in orbit. They do not move relative to the space shuttle even though gravity is acting on them.

Or take the bowling ball you have in your car trunk. When you accelerate it will roll backward and when you brake it will roll forward even though there are no forces acting on it in those directions.

So when we talk about the law of inertia, we are assuming that a frame exists in which the law is correct. Such a frame is called an inertial frame. If one such inertial frame exists, then an infinite number of other inertial frames exist since any frame that is moving at a constant relative velocity to the first inertial frame is also an inertial frame.

The frames in which the law of inertia does NOT hold are those that are accelerating with respect to inertial frames. They are called non-inertial frames.

An inertial frame of reference is one in which Newton’s first law of motion holds. In such a frame, an object at rest remains at rest and an object in motion continues to move at constant velocity (constant speed and direction) if no force acts on it. Any frame of reference that moves at constant velocity relative to an inertial frame is itself an inertial frame.

Special theory of relativity deals with problems that involve inertial frames of reference. General relativity published by Einstein a decade later deals with problem that involve frames of reference accelerated with respect to one another.

1.3. POSTULATES OF SPECIAL THEORY OF RELATIVITY: Some History

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Newton discovered his laws in the 17th century. Using Newton's laws, physicists in the 18th and 19th century were able to predict the motions of the planets, moons, comets, cannon balls, etc.

In the later half of the 19th century, however, Maxwell discovered the equations, which now bear his name, that govern the behaviour of electricity and magnetism, the basis of all modern technology. In particular, his equations predicted that electromagnetic waves propagate at the speed of light c=3×108m/sec, establishing that light was in fact a form of electromagnetic radiation.

Unlike Newton's equation, however, Maxwell's equations were not invariant under the Galilei transformation. This is evident from the fact that Maxwell's equations predicted the speed of light. But the speed of something depends on which inertial frame you are observing it from. So Maxwell's equations could only be correct in one particular inertial frame!

This did not go well with the belief that the laws of physics should be the same regardless of which inertial frame you were making the observation from.

To overcome this difficulty, physicists decided that light had to travel through a medium which they called ether, just like sound had to travel through air. The speed of light predicted by Maxwell's equations was interpreted as that relative to ether. So if the speed of light you observe is different from that predicted by Maxwell, it would NOT mean that the laws of physics is different in your inertial frame from others. Your inertial frame is simply moving relative to ether!

About a hundred years ago, a famous experiment was conducted called the Michelson-Morley experiment which measured the dependence of the speed of light on the speed of the observer. To everyones surprise, it turned out that the speed of light was always the same no matter what the motion of the observer was!

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This was a very peculiar situation indeed. Physicist were trying to come up with a theory that explained why the speed of light could be different for different inertial frames but the Michelson-Morley experiment showed that that was not what they should have been worried about. The puzzle now was: how could the speed of light be the same regardless of which inertial frame you were in?

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It was Einstein who, in 1905, pointed out that the only way to understand this was to change our notion of simultaneity. This was his famous special theory of relativity.

Special relativity is based on two postulates which are contradictory in classical mechanics:

1. The laws of physics are the same in all inertial frames of reference (observers in uniform motion relative to one another)

2. The speed of light in vacuum is same in all inertial frames of reference (all observers, regardless of their relative motion or of the motion of the source of the light).

Consequences of Einstein’s postulates:

These innocent sounding propositions have far reaching implications. Some of these are: Time dilation: Moving clocks tick slower than an observer's "stationary" clock.

Length contraction: Objects are observed to be shortened in the direction that they are moving with respect to the observer.

Relativity of simultaneity: Two events that appear simultaneous to an observer A will not be simultaneous to an observer B if B is moving with respect to A.

Mass-energy equivalence: According to the relationship E = mc², energy and mass are equivalent and transmutable.

The defining feature of special relativity is the replacement of the Galilean transformations of classical mechanics by the Lorentz transformations. (See Maxwell's equations of electromagnetism and introduction to special relativity).

1.4. GALILEAN TRANSFORMATION: Suppose we are in an inertial frame of reference S and find the coordinates of some event that occurs at the time t are x, y, z. Another observer located in a different inertial frame S’ which is moving with respect to S at constant velocity u, will find that the same event occurs at the time t’ and has the coordinates x’, y’, z’. How are the measurements x, y, z, t related to x’, y’, z’, t’?

If the clocks in both systems are started when the origins of S and S’ coincide, measurements in the x-direction made in S will be greater than those made in S’ by the amount of ut, which is the distance S’ has moved in the x-direction. That is,

,utxx ….. …… ….. [1a]

There is no relative motion in the y and z directions, and so

yy ….. …… ….. [1b]

zz ….. …… ….. [1c]

It is assumed that time can be defined independently of any particular frame of reference. This is an implicit assumption of classical physics, which is expressed in the transformation equations by the

O’ p

y y’

x X’

z z’

S S’ u

y

O

y’

x’

ut

x

Fig.6: Frame S’ moves in the X-direction with speed u relative to frame S.

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absence of a transformation for t. We can make this assumption of the universal nature of time explicit by adding to the Galilean transformations the equation

tt ….. …… ….. [1d]

The set of equations [1a] to [1d] is known as Galilean transformation.

To convert velocity components measured in the S frame to their equivalents in the S’ frame according to the Galilean transformation, we get

uvdtdxv xx

'' ….. …… ….. [2a]

yy vdtdyv

'' ….. …… ….. [2b]

zz vdtdzv

'' ….. …… ….. [2c]

We can see that different velocities are assigned to a particle by different observers when the observers in relative motion.

Although the Galilean transformation and the corresponding velocity transformation seem straightforward enough, they violate both the postulates of special relativity. The first postulate calls for the same equations of physics in both the S and S’ inertial frames, but the equations of electricity and magnetism become very different when the Galilean transformation is used to convert quantities measured in one frame into their equivalents in the other. The second postulate calls for the same value of the speed of light c when determined in S or S’. If we measure the speed of light in the x-direction in the S system to be c, however, in the S’ system it will be

ucc according to Eq.[2a].

Clearly a different transformation is required if the postulates of special relativity are to be satisfied.

2. RELATIVISTIC KINEMATICS LORENTZ TRANSFORMATION

What Einstein's special theory of relativity says is that to understand why the speed of light is constant, we have to modify the way in which we translate the observation in one inertial frame to that of another. The Galilei transformation

is wrong. The correct relation is

This is called the Lorentz transformation.

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You can see that if the relative velocity v between the two frames are much smaller than the speed of light c, then the ratio v/c can be neglected in this relation and we recover the Galilei transformation. So the reason why we did not have any problems with the Galilei transformation up to now is that v was small enough for it to be a good approximation of the Lorentz transformation.

2.1. Derivation of LORENTZ TRANSFORMATION EQUATIONS

A reasonable guess about the nature of the correct relationship between x and x’ is

)( utxkx ….. ….. ….. [3]

Here k is a factor that does not depend on either x or t but may be a function of u.

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Because the equations must have the same form in both S and S’, we need only change the sign of u (in order to take into account the difference in the direction of relative motion) to write the corresponding equation for x in terms of x’ and t’:

)( tuxkx ….. ….. ….. [4]

The factor k must be the same in both frames of reference since there is no difference between S and S’ other than in the sign of v.

Let us consider that a light pulse that starts at the origin of S at t=0. Since we have assumed that the origins are coincident at t’ = t = 0, the pulse also starts at the origin of S’ at t’=0. Einstein’s postulates require that the equation for the x-component of the wave front of the light pulse is

ctx in frame S

and tcx in frame S’.

Substituting ct for x and ct’ for x’ in Eq.[3] and [4], we get

tucktutckct )()( … … [5]

and tuckutctktc )()( … … [6]

We can eliminate either t’ or t from [5] and [6] and determine k. We get

22

2

1

1

cu

k

or

22

1

1

cu

k

… …[7]

Substituting

)( '' utxkx for x in )( utxkx , we get

])([ uttuxkkx …. …. [8]

which can be solved for t in terms of x and t .

Now, simplification of eq. [8] gives,

]9[.........)(])([

2 kuttuxkuttuxkkx

From [9], we can get

22

22

21

22

12

22

/1/

]1[

)(]1[)(cucuxtu

cu

xtuxcu

kxtuxkut

…. …. [10]

From [10], we can write

)/(/1/ 2

22

2

cuxtkcucuxtt

…. … [11]

The complete relativistic transformation is

),( tuxkx ….. …… ….. [12a]

yy , ….. …… ….. [12b]

zz , ….. …… ….. [12c]

),( 2cxutkt

….. …… ….. [12d]

The inverse transformation is

)( utxkx ….. …… ….. [13a]

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yy ….. …… ….. [13b]

zz ….. …… ….. [13c]

)( 2cuxtkt ….. …… ….. [13d]

From Eq. [12d] & [13d], we can see that the time coordinate of one inertial system depends on both the time and the space coordinates of another inertial system. Hence, instead of treating space and time separately, as is quite properly done in classical theory, it is natural in relativity to treat them together.

2.2. TIME DILATION (THE RELATIVITY OF TIME) Let us consider a clock at the point x in the moving frameS . When an observer in S finds that the time is 1t an observer in S will find it to be t1, where from Eq.[12d]

)/( 211 cxutkt

After a time interval of t0 (to observer inS ), the observer in the moving system S finds that the time is now 2t according to his clock. That is, 120 ttt .

The observer in S, however, measures the same time to be

)/( 222 cxutkt .

So, to observer in S the duration of time interval t is

22

001212

/1)(

cutktttkttt

….. ….. ….. [14]

Eq.[14] summarizes the effect known as time dilation. According to Eq. [14], observer in S measures a longer time interval than the observer in S’ measures. This is a general result within special relativity, which we can summarize as follows:

Consider an occurrence that has a duration Δto. An observer S’ fixed with respect to that occurrence (i.e its beginning and end take place at the same point in space, according to S’) measures the interval Δt0, which is known as the proper time. An observer S moving with respect to S’ measures a longer time interval Δt for the same occurrence. The interval Δt is always longer than Δt0, no matter what the magnitude or direction of u is.

The time dilation effect has been verified experimentally with decaying elementary particles as well as with precise atomic clocks carried aboard aircraft.

Problem solving strategy: First make sure that you understand the difference between the proper time t0 in a particular reference frame and the dilated time t. The proper time t0 in a particular reference frame is the time interval between two events that occur at the same point in that reference frame. Then the longer time t is the time interval between the same two events as measured in a second reference frame that has a speed u with relative to the first frame. The two events will be at different points in this reference frame.

Check that your answers make sense; t is never smaller than t0, and u is never larger than c.

EXAMPLE 1: Two events occur at the same place xo at times t1 and t2 in S. What is the time interval between them inS ?

)/(' 2011 cuxtkt , )/(' 2

022 cuxtkt

t’2 – t’1 = k (t2 – t1).

“A moving clock ticks slower than a clock at rest”

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This is our familiar result for time dilation since (t2 - t1) is the proper time interval.

EXAMPLE 2: Muons are elementary particles with a (proper) lifetime of 2.2 µs. They are produced with very high speeds in the upper atmosphere when cosmic rays collide with air molecules. Take the height Lo of the atmosphere to be 100km in the reference frame of the earth. Find the minimum speed that enables the muons to survive the journey to the surface of the earth.

Solution: The birth and decay of the muon can be considered as the ticks of a clock. In the frame of reference of the Earth (Observer O) this clock is moving and therefore its ticks are slowed by the time dilation effect. If the muon is moving at a speed that is close to c, the time necessary for the muon to travel from the top of the atmosphere to the surface of the Earth is:

sxkm

cLt 333

1000.3100

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If the muon is to be observed at the surface of the earth, it must live for at least 333 µs in the earths’ frame of reference. In the muon’s frame of reference, the interval between its birth and decay is a proper time interval of 2.2 µs.

Now

221

22333

cu

ss

/

.

Solving we get u=0.999978c.

If it were not for the time dilation effect, muons would not survive to reach the earths’ surface. The observation of these muons is a direct verification of the time dilation effect of special relativity.

EXAMPLE 3: A spacecraft is moving relative to earth. An observer on the earth finds that according to her clock, 3601s elapse between 1 PM and 2 PM on the spacecraft’s clock. What is the spacecrafts’ speed relative to the earth?

Solution: Here to = 3600s is the proper time interval on the earth and t = 3601s in the time interval in the moving frame as measured from the earth.

Now 22

0

/1 cutt

Therefore

smxsmxttcv /.)()/.()/( 62820 1017

3601360011099821

To-days spacecrafts are much slower than this. For instance, the highest speed of the Apollo 11 spacecraft that went to the moon was only 10,840 m/s and its clocks differed from those on the earth by less than one part in 109.

Most of the experiments that have confirmed time dilation made use of unstable nuclei and elementary particles which readily attain speeds not far from that of light.

The Concept of Simultaneity

As you can guess from the equations for the Lorentz transformation, the concept of time becomes a relative concept since we no longer have the simple relation t=t'. What we actually mean here is that the concept of simultaneity will depend on which frame you are in. This is a simple consequence of the fact that the speed of light is the same no matter which inertial frame you are in.

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Consider a moving train with a light bulb in the middle. If you turn the light bulb on, light will travel both toward the front of the train and also toward the back of the train with speed c=3×108m/sec.

From the point of view of the observer riding on the train, the distances from the light bulb to the front and back ends of the train are the same so the light will reach both ends at the same time .

However, from the point of view of the person on the ground, the front of the train is moving away from the light coming toward it while the back of the train is moving closer to the light coming toward it. This means that the distance covered by light going forward will be longer than the light going backwards. And since the speed of light is c in both directions for the observer on the ground also, the light will reach the back of the train before it reaches the front of the train.

What I am saying is that 2 events that seem to have happened at the same time for the observer on train do not happen at the same time for the observer on the ground. In fact, from the point of view of another observer moving faster than the train, like the observer in the sports car above, the light reaches the back of the train after it reaches the front of the train since in that observer's frame, it is the back of the train that is moving away from the light and the front of the train that is moving toward it.

So the concept of before and after actually depends on the observer.

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2.3. LENGTH CONTRACTION (RELATIVITY OF LENGTH) Let us consider a rod lying along the x axis in the moving frameS . An observer in this frame determines the coordinates of its ends to 1x and 2x , and so the proper length of the rod is

120 xxL .

In order to find 12 xxL , the length of the rod as measured in the stationary frame S at time t, we make use of Eq.[13a] to give

),( 11 utxkx and )( 22 utxkx .

Hence,

220

0121212 /1 cuL

kL

kxxut

kxut

kxxxL

,/1 220 cuLL …. ….. ….. [15]

Fq. [15] summarizes the effect known as length contraction. Observer S’ who is at rest with respect to the object, measures the rest length L0 (also known as proper Length, in analogy with the proper time). All observes in motion relative to S’ measure a shorter length, but only along the direction of motion; length measurements transverse to the direction of the motion are unaffected.

For ordinary speeds (u<<c), the effects of length contraction are too small to be compared.

Length contraction suggests that objects in motion are measured to have a shorter length than they do at rest. The objects do not actually shrink; there is merely a difference in the length measured by different observers. For example, to observers on earth a high speed rocket ship would appear to be contracted along its direction of motion, but to an observer on the ship it is the passing earth that appears to be contracted.

Problem solving strategy: First make sure that you understand the difference between the proper length l0 and the contracted length l. The proper length l0 in a particular reference frame is length measured on an object at rest in that frame. Then the length l is the length between the same two points in the object as measured in a second reference frame that has speed u relative to the first frame.

Both the lengths l0 and l are measured parallel to the direction of motion. Any length perpendicular to the direction of motion is unchanged.

Check that your answers make sense; L is never larger than L0, and u is never larger than c.

Faster means shorter

L0

L0 L0

L0

L L0

Fig.4: Some length-contracted objects. Notice that the shortening occurs only in the direction of motion.

X’

Y’

Z’

X1’ X2’

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EXAMPLE 4: In the reference frame of muon of Example 2 what is the apparent thickness of the earth’s atmosphere?

Solution:.,., In the reference frame of the muon, the earth as we found in example 1 is rushing towards it at a speed of u = 0.999978C. To an observer on the Earth the height of the atmosphere is its rest length Lo of 100 Km. To the muon, the moving earth has an atmosphere of height given by

mkmkmcuLL 660660999978011001 2220 .).()(/

Note that what appears as a time dilation in one frame of reference, can be regarded as a length contraction in another frame of reference.

EXAMPLE 5: A crew member on a space ship that flies past at a speed of 0.99c relative to the earth, measures its length obtaining 400 m. what length do observers measure on earth?

Solution:

The 400 m length of the space ship is the proper length l0 because it is measured in the frame in which it is at rest. We want to find the length L measured by the observers on earth which is given by

mmcuLL 4.56)99.0(1)400(/1 2220

2.4. RELATIVISTIC VELOCITY TRANSFORMATION: By differentiating the Lorentz transformation equations [12] for x, y, z and t, we get

),( tvdxdkdx yddy , zddz , )/( 2cxvdtdkdt .

Therefore,

x

xx

vcu

uv

tdxd

cu

utdxd

cxudtdktudxdk

dtdxv

22

2

11)/()(

…… ….. … [16a]

and ,

x

yy

vcu

cuv

tdxd

cu

cutdyd

xdcutdk

yd

xdcutdk

dydtdyv

2

22

2

22

22 1

1

1

1 //

)()( ….. [16b]

and similarly ,

x

zz

vcu

cuvv

2

22

1

1 / ….. …… ….. [16c]

If ,cv x that is, if light is emitted in the moving frame S in its direction of motion relative to S, an observer in frame S will measure the speed

cc

cuuc

vcu

uvvx

xx

22 11

Thus the observers in the car and on the road both find the same value for the speed of light, as they must.

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We may write the complete set of relativistic velocity transformation as

x

zz

x

yy

x

xx

vcu

cuvvandv

cu

cuvv

vcu

uvv

2

22

2

22

2 1

/1,1

/1,

1

The inverse velocity transformation equations are

x

zz

x

yy

x

xx

vcu

cuvvandv

cu

cuvv

vcu

uvv2

22

2

22

2 1

/1,1

/1,

1

Table: The relativistic velocity transformation equations

x

xx

vcu

uvv21

x

xx

vcu

uvv

21

x

yy

vcu

cuvv

2

22

1

/1

x

yy

vcu

cuvv

2

22

1

/1

x

zz

vcu

cuvv2

22

1

/1

x

zz

vcu

cuvv

2

22

1

/1

EXAMPLE - 6

An observer in S frame is moving to the right at a speed u = 0.50c away from the stationary

observer in frame S. The observer in S measures the v of a particle moving to the right away

from her. What speed v does the observer in S measure for the particle if a) v = 0.60c? b) v =

0.99c? Solution : The relative speed between the frames of reference u = 0.5 c

The speed of the particle in the S frame v = 0.6 c

Speed of the particle in the S frame v =?

[a] We know,

x

xx

vcu

uvv

21= cc

cc

cc 846.03.11.1

6.05.01

5.06.0

2

[b] The speed of the particle in S frame v = 0.99 c

Hence v = ccc

cc

cc 997.0495.149.1

99.05.01

5.099.0

2

EXAMPLE - 7

A spaceship moving relative to the earth at a large speed fires a rocket toward the earth with a speed of 0.84c relative to the spaceship. An earth-based observer measures that the rocket is

'xv

u

S S

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approaching with a speed of 0.29c. What is the speed of the spaceship relative to the earth? Is the spaceship moving toward or away from the earth?

Assuming that the space ship is moving away at speed u

Speed of the rocket with respect to the spaceship is 'xv = -0.84 c

Speed of the rocket with respect to the earth is xv = -0.29 c

We know,

x

xx

vcu

uvv

21

or, cc

ccc

cc

cvvvvu

xx

xx 442.02436.155.0

)84.0)((29.0(1

)84.0(29.0

1 22

Since u is positive the spaceship is receding away from the earth.

2.5. DOPPLER EFFECT: Let us suppose that a source of light is moving with constant speed u toward an observer who is stationary in an inertial frame. As measured in its rest frame, the source emits light waves with frequency o and time period To= 1/o. What is the frequency of these waves as received by the observer?

Let T be time interval between emission of successive wave crests as observed in observer’s reference frame. Note that this is not the interval between the arrival of successive crests at his position, because the crests are emitted at different positions in observer’s frame.

During a time T the crest ahead of the source move a distance cT, and the source moves a shorter distance uT in the same direction. The distance between successive crests- that is, the wavelength – is thus = (c-u)T , as measured in observer’s frame. The frequency that the observer measures is c/.

Therefore

Tucc

)( …. ….. ….. [17]

Now the time interval T measured in observer’s frame is related to the time interval T0 between emissions of successive wave crest by the source [measured in the rest frame of the source] by

222

21 uc

cT

cu

TT oo

…. ….. ….. [18]

Or, since To = 1/o,

xv = -0.29 c u

S S

'xv = -0.84 c

uT cT

S

18

0

22

0

221

cuc

cTuc

T

To find , let us substitute the expression of 1/T into Eq.[17] and get

ucuc

cuc

ucc

00

22

, …. …. …. [19]

The above equation represents Doppler Effect of electromagnetic wave. This shows that, when the source moves toward the observer, the apparent frequency is greater than the emitted frequency o. the change = - o is called the Doppler frequency shift. When source moves away from the observer, we change the sign of u to get

ucuc

o

…. …. …. [20]

EXAMPLE - 8

A galaxy in the Constellation Ursa Major is receding from earth at a speed of 15,000 km/s. If one of the characteristic wavelength of the light the galaxy emits is 550 nm, what is the corresponding wavelength measured by astronomers on the earth?

Solution : Speed of light c = 3 x 108 m/s

Speed of recession of the constellation u = 15000 km/s = 1.5 x 107 m/s = 0.05 c

We know,

When the source moves away from the observer the observed frequency is given by

ucuc

or

or, ucuccc

or

or, nmnmccccnm

ucuc

or 23.57895.005.1)550(

5.005.550

(Ans)

EXAMPLE - 9

A highway patrolman measures the speed of cars approaching him with a device that sends

out electromagnetic waves with frequency fo and then measures the shift in frequency f of the

waves reflected from the moving car. What fractional frequency shift f/fo is produced by a car speeding at 35.4 m/s?

Solution : The patrolman sends out electromagnetic signal of frequency fo

The moving car reflects signal of frequency f

The receiver device receives signal of frequency f

Now f f

f f

19

ucucff oa

, and

ucucff aa

Or, ucucff oa

Or, uc

uffor

ucuc

ff

oo

a

2,,11

Or, 71036.24.35300000000

4.3522

xx

ucu

ff

o

(Ans)

3. RELATIVISTIC DYNAMICS 3.1. RELATIVITY OF MASS: To investigate what happens to the mass of an object as its speed increases, let us consider an elastic collision ( that is a collision in which kinetic energy is conserved ) between two particles A and B as witnessed by observers in the reference frames S and S’ [Fig. 5] which are in uniform relative motion .

The properties of A and B are identical when determined in reference frames in which they are at rest. The frames S and S’ are oriented as in Fig.5 with S’ moving in the +x direction with respect to S at the velocity v .

Before the collision, particle A had been at rest in frame S and particle B in frame S’. Then, at the same instant, A was thrown in the +y direction at the speed VA while B was thrown in the -Y direction at the speed BV

, where

BA VV …. …. …. [21]

Hence the behavior of A as seen from S is exactly the same as the behavior of B as seen from S’.

When the two particles collide, A rebounds in the -Y direction at the speed VA while B rebounds in the +Y direction at the speed BV . If the particles are thrown from positions Y apart, an observer in S finds that the collision occurs at y=Y/2 and one in S’ finds that it occurs at y’ = y = Y/2. The round-trip time t0 for A as measured in S is therefore,

AVYt 0 …. …. …. [22]

and it is the same for B in S’ : BV

Yt

0 …. …. …. [23]

If the linear momentum is conserved in the S frame, it must be true that

BBAA VmVm …. …. …. [24]

where mA and mB are the masses of A and B and VA and VB are their velocities as measured in the S frame. In S, speed VB is found from

tYVB …. …. …. [25]

where t is the time required for B to make its round trip as measured in S. In S’, however, the trip of B requires the time t0 , where

22

0

/1 cutt

…. …. …. [26]

From [25],

20

22

0

22

/1/1 cuVt

cuYV AB

…. …. …. [27]

since from [22] we know that 0t

YVA .

Inserting these expressions for VA and VB in [24], we see that the momentum is conserved provided that

22 /1 cumm BA …. …. …. [28]

Because A and B are identical when at rest with respect to an observer, the difference between mA and mB means that measurements of mass like those of space and time depend upon the relative speed between an observer and whatever he or she is observing.

In this example, both A and B are moving in S. In order to obtain a formula that gives the mass m of a body measured while in motion in terms of its mass mo when measured at rest, we need only consider a similar example in which VA and VB are very small compared with v. In this case an observer in S will see B approach A with the velocity v and making a glancing collision (since VB << v) and then continue on.

In S mA= m0 and mB = m

and so

Relativistic mass 22

0

/1 cumm

…. …. …. [29]

The mass of a body moving at the speed v relative to an observer is larger than its mass when at rest relative to the observer by the factor

22 /11 cu .

This mass increase is reciprocal, to an observer in S’, mA= m and mB=m0. Measured from the earth, a spacecraft in flight is shorter than its twin still on the ground and its mass is greater. To somebody on the spacecraft in flight, the ship on the ground also appears to be shorter and to have a greater mass.

Relativistic mass increases are significant only at speeds approaching that of light. At a speed one tenth that of light the mass increase amounts to 0.5 percent but this increase is over 100% at a speed nine-tenths that of light. Only atomic particles such as electrons, protons, mesons and so on have sufficiently highly speeds for relativistic effects to be measurable and in dealing with these particles, the ordinary laws of physics cannot be used.

Fig.5: An elastic collision as observed in two different frames of reference. The balls are initially Y apart, which is the same distance in both frames since S’ moves only in the X direction.

VA

V’B

Y

y y’

x x’

z z’

S S’ u

21

EXAMPLE 10: Find the mass of an electron (mo =9.1x10-31 kg) whose velocity is 0.99c.

Solution : Given u/c=0.99, so

.1064)99.0(1

101.9/1

31

2

31

220 kgxx

cumm

which is 7 times greater than the electron’s rest mass.

If v =c , m = , from which we can conclude that v can never equal c:

3.2. EQUIVALENCE OF MASS AND ENERGY :

Let now see how this relationship can be derived from what we have already done. The work W done on an object by a constant force of magnitude F that acts through the distance s where F

is in the

same direction as s

is given by

W = Fs

If no other forces act on the object and the object starts from rest, all the work done on it becomes K.E., so

K.E. = Fs

In the general case where F

need not be constant, the formula for K..E becomes

S u

oo

S mu

cuumduumdusd

dtumdsFdEK

0 0 022

0

0

]/1

[][][

.

Integrating by parts

ydxxyxdy ,

we get

uu

cuudum

cuuumEK

022

0

022

0

/1/1.

u

cuudum

cuum

022022

20

/1/1

Taking 1- u2/c2 = k2, we get

u du = - c2 k dk

and when u = 0 ; k = 1

and when u = u ; k = 221 cu

Changing he variable the above equation can be written as

22 /1

0

2

022

20

/1..

cu

kdkkcm

cuumEK

No material object can travel as fast as light

Where E = mc2 comes from

22

]32[........................../1

/1

20

2

2022

20

/1

1

2022

20

22

cmcm

cmcu

cm

dkcmcu

um cu

From Eq. [32], we see that the K.E of an object is equal to the increase in its mass due to its relativistic motion multiplied by the square of the speed of light.

Eq [32] may be written as

EKcmmc . Energy Total 20

2 …. …. …. [33]

If we interpret mc2 as the total energy E of the object, we see that when it is at rest and its K.E =0 , it nevertheless possesses the energy m0c2 . Accordingly m0c2 is called the rest energy Eo.

We, therefore, have

EKEE .0 …. …. …. [34]

Where, 200 cmE …. …. …. [35]

If the object is moving, its total energy is

22

202

/1 cvcmmcE

…. …. …. [36]

For low speeds, v/c <<1,

Then we can write

K.E=mc2-m0c2

= m0c2 2

1

22

1

cv - m0c2

= m0c2

2

2

211

cv - m0c2

[ neglecting the higher order terms]

= 21

m0v2

EXAMPLE 11: A stationary body explodes into two fragments each of rest mass 1.0 Kg that move apart at speeds 0.6C relative to the original body. Find the rest mass of the original body.

Solution: The total energy of the original body must equal the sum of the total energies of the fragments.

Hence

2

2

2

2

222

202

221

2012

0

000

)6.0(1

)0.1(

)6.0(1

)0.1(

/1/1

0.

ckgckg

cu

cm

cu

cmcm

EEEKEE

Therefore,

kgm 520 .

23

Mass can be created or destroyed but when this happens an equivalent amount of energy simultaneously vanishes or comes into being and vice versa.

The conversion factor between the unit of mass (the kg) and the unit of energy (the joule, J) is c2. So, 1 kg of matter has an energy content of

Jxsmxkgcm 162820 1091031 )/(*)(

This is enough to send a payload of a million tons to the moon.

Mass-less particles:

We know, ,/1

energy Total22

20

cu

cmE

and relativistic momentum 22

0

/1 cu

ump

When m0 = 0 and u<c, i.e. E = p = 0 i.e.

However, when mo =o and v =c, E=o/o and p=o/o which are indeterminate: E and p can have any values. The equations [b] & [c] are consistent with the existence of mass-less particles that possess energy and momentum provided that they travel with the speed of light.

There is another restriction on mass-less particles.

From Eq. [36], we can write

22

4202

/1 cucmE

and from [30], we can write

,/1 22

2202

cuump

22

222022

/1 or,

cucumcp

Subtracting 22cp from E2 yields

420

222 cmcpE

Therefore, for all particles, we have

E = 2222242 cpEcpcm oo … … … [37]

According to this formula, if a particle exists with m0 = 0, the relationship between its energy and momentum must be given by

pcE particles less-mass … … … [38]

Mass and energy are different aspects of the same thing

They can exist only if they move with the speed of light

A mass less particle with a speed less than that of light can have neither energy nor momentum

24

In fact, mass-less particles of two different kinds - the photon and the neutrino have indeed been discovered and their behavior is as expected.

Electronvolts: In atomic physics the usual unit of energy is electronvolt (eV) where 1 eV is the energy gained by an electron accelerated through a potential difference of 1 volt.

Since

W =QV

1 eV =(1.602x10-19 C)(1.00V)=1.602x10-19 J

The rest energies of elementary particles are often expressed in MeV and GeV and the corresponding rest masses in MeV/c2 and GeV/c2. In a similar way the MeV/c and GeV/c are sometimes convenient units of linear momentum.

EXAMPLE 12: An electron (m0 = 0.511 MeV/c2) and a photon (m0 =0) both have momenta of 2.00 MeV/c. Find the total energy of each. Solution:

The electron’s total energy is

MeVccMeVccMeVcpcmE 06420025110 2242222420 .)/.()/.(

Conclusion I hope I have succeeded in giving you a flavor of Einstein's special theory of relativity. Key points I would like you to take home with you are:

1. The speed of light is constant regardless of the inertial frame in which it is measured. All of the theory's conclusions are derived from this simple experimental fact.

2. Simultaneity is a relative concept. The chronological order in which two events occurred may depend on the frame of the observer.

3. Faster than light speed travel or communication is impossible. (Otherwise, causality will be broken.)

4. Interesting phenomena like time dilation and Lorentz contraction have been predicted and observed.

5. Equivalence of mass and energy 6. Physics is fun!

Famous People The following is a list of famous people whose names I have mentioned in my lecture notes.

Galileo Galilei (1564-1642) Sir Isaac Newton (1643-1727) James Clerk Maxwell (1831-1879) Hendrik Antoon Lorentz (1854-1928) Albert Einstein (1879-1955)

25

PROBLEMS ON SPECIAL THEORY OF RELATIVITY 1 Relativity of time intervals 1. A spaceship flies past Mars with a speed of 0.964c relative to the surface of the planet. When

the spaceship is directly overhead at an altitude of 1500 km, a very bright signal light on the Martian surface blinks on and then off. An observer on the Mars measures the signal was on for 82.4 s. What is the duration of the light pulse measured by the pilot of the spaceship?

2. Astronauts in a spaceship traveling at speed v=0.8 c relative to earth sign off from the space control, saying that they are going to have a nap for 1 hour and then call back. How long does their nap last as measured on earth?

3. The positive muon (+) is an unstable particle with and average lifetime of 2.2 s (measured in the rest frame of the muon). If the muon is made to travel at a very high speed relative to a laboratory, its average lifetime is measured in the laboratory to be 19 s. (i) Calculate the speed of the muon. (ii) What distance, measured in the laboratory, does the particle travel during its lifetime?

2 Relativity of lengths

1. As measured by an observer on earth, a spacecraft runway on earth has a length of 1200 m. a) What is the length of the runway as measured by a pilot of a spacecraft flying past at a speed of 8.0 107 m/s relative to the earth? b) An observer on the earth measured the time interval from when the spacecraft is directly over one end of the runway until it is directly over the other end. What result does she get? c) The pilot of the spacecraft measures the time it takes him to travel from one end of the runway to the other end. What value does he get?

2. Two spaceships, each 50 m long when measured at rest, travel towards each other with speeds of 0.85 c relative to the earth. (a) How long is each ship as measured by someone on earth? (b) At time t=0 on earth, the fronts of the ships are together as they just begin to pass each other. At what time on earth are their ends together?

3. An unstable particle is created in the upper atmosphere from a cosmic ray and travels straight down toward the surface of the earth with a speed of 0.99860c relative to earth. A scientist in a laboratory on the earth’s surface measures that the particle is created at an altitude of 95 km. (i) As measured by the scientist , how much time does it take the particle to travel the 95 km to the surface of the earth? (ii) Use the length contraction formula to calculate the distance from where the particle is created to the surface of the earth as measured in the particle’s frame. (iii) In the particle’s frame, how much time does it take the particle to travel the 95 km to the surface of the earth? Calculate this by both the time dilation formula and from the distance calculated in part (ii). Do these two results agree?

3. Relativistic Momentum and Energy

1. a) At what speed is the momentum of a particle twice as great as the result obtained from the non-relativistic expression mou? b) At what speed is the momentum of a particle differs by 1% from the value obtained from the non-relativistic expression mou? Is it greater than that obtained from non-relativistic expression?

2. How much work must be done on a particle with mass mo to accelerate it a) from rest to a speed of 0.08c? b) from a speed of 0.9c to 0.98c? [Express your answers in terms of moc2 ]. C) How do your answers in part (a) and (b) compare?

26

3. What is the speed of a particle if its kinetic energy is 1% larger than 2

21 umo ?

4 Relativistic Doppler effect

1. A highway patrolman measures the speed of cars approaching him with a device that sends

out electromagnetic waves with frequency fo and then measures the shift in frequency f of the

waves reflected from the moving car. What fractional frequency shift f/fo is produced by a car

speeding at 35.4 m/s?

2. Certain wavelengths in the light from a galaxy in the Constellation Virgo are observed to be

0.4% longer than the corresponding light from earth sources. What is the radial speed of this

galaxy with respect to earth? Is it approaching or receding?

3. A spaceship, moving away from earth at a speed of 0.90 c, reports back by transmitting a

frequency (measured in the spaceship frame) of 100 MHz. To what frequency must earth

receivers be tuned to receive the report?

5. Velocity transformation 3-1 Consider the case of two spaceships approaching each other in the vicinity of a star. Relative

to the star, the spaceship A is approaching at 0.95c, while spaceship B is approaching from

the opposite direction at a velocity of 0.85c with respect to the star. What is the velocity of A

relative to B?

3-2 Spacecraft Alpha is moving at a speed 0.90c with respect to the earth. If the spacecraft Beta is

to pass Alpha at a relative speed of 0.50c in the same direction, what speed must Beta have

with respect to the earth?

3-3 Two twins of rest mass 60.00 kg are headed towards each other in spacecraft whose speeds

relative to the earth are 0.800c. What mass does each twin find for the other?

PROBLEMS ON SPECIAL THEORY OF RELATIVITY

4. An airplane 40.0 m in length in its rest system is moving at a uniform velocity with respect to earth at a speed of 630 m/s. (a) By what fraction of its rest length will it appear to be shortened to an observer on earth? (b) How long would it take by earth clocks for the airplane’s clock to fall behind by one microsecond?

5. We could define the length of a moving rod as the product of its velocity by the time interval between the instant that one end point of the rod passes a fixed marker and the instant that the other end point passes the same marker. Show that definition also leads to result of length contraction?

6. A meter stick moves past you at a great speed. Its motion relative to you is parallel to its long axis. If you measure the length of the moving meter stick to be 1.0 yd (1 yd = 0.9144 m), for example, by comparing it to a yardstick that is at rest relative to you, what is the speed with which the meter stick is moving relative to you?

7. An unstable high-energy particle enters a detector and leaves a track 1.05 mm long before it decays. Its speed relative to the detector was 0.992 c. What is its proper lifetime? That is, how

27

long would the particle have lasted before decay had it been at rest with respect to the detector?

8. Suppose a spaceship moves away from you along the x-direction with speed v= 0.80c relative to you. A second spaceship moves away from you and away from the first ship with the same speed 0.80c relative to the first ship along the x-direction. How fast is the second ship moving relative to you?

9. Spacecraft Alpha is moving at a speed 0.90c with respect to the earth. If the spacecraft Beta is to pass Alpha at a relative speed of 0.50c in the same direction, what speed must Beta have with respect to the earth?

10. Two twins of rest mass 60.00 kg are headed towards each other in spacecraft whose speeds relative to the earth are 0.800c. What mass does each twin find for the other?

11. In terms of c, what relative velocity u between a source of electromagnetic waves and an observer produces a) 1% decrease in frequency? b) a decrease by a factor of three of the frequency of the observed light?

12. A spaceship is moving away from earth at a speed of 0.20 c. A light source on the rear of the ship appears blue ( = 450 nm) to the passengers on the ship. What colour would that source appear to an observer on earth monitoring the receding spaceship?

13. A distant galaxy is moving away from earth at a speed of 1.85x107 m/s. Calculate the fractional red-shift (’ - o)/ o in the light from this galaxy.

14. Certain wavelengths in the light from a galaxy in the Constellation Virgo are observed to be 0.4% longer than the corresponding light from earth sources. What is the radial speed of this galaxy with respect to earth? Is it approaching or receding?

15. What is the percentage increase in the mass of an electron accelerated to a K.E of 500 MeV? Use rest mass of electron = 0.511 MeV/c2.

16. What is the speed of a particle if its kinetic energy is 1% larger than 2

21 umo ?

28