relativity chapter 1. a brief overview of modern physics 20 th century revolution: - 1900 max planck...

RelativityRelativity

Chapter 1Chapter 1

A Brief Overview of Modern PhysicsA Brief Overview of Modern Physics

2020thth Century revolution: Century revolution:- - 1900 Max Planck1900 Max Planck

Basic ideas leading to Quantum theoryBasic ideas leading to Quantum theory

- - 1905 Einstein1905 Einstein Special Theory of RelativitySpecial Theory of Relativity

2121stst Century Century Story is still incompleteStory is still incomplete

Basic Problems

Newtonian mechanics fails to describe Newtonian mechanics fails to describe properly the motion of objects whose speeds properly the motion of objects whose speeds approach that of lightapproach that of light

Newtonian mechanics is a limited theoryNewtonian mechanics is a limited theory– It places no upper limit on speedIt places no upper limit on speed

– It is contrary to modern experimental resultsIt is contrary to modern experimental results

– Newtonian mechanics becomes a specialized Newtonian mechanics becomes a specialized case of Einstein’s special theory of relativity case of Einstein’s special theory of relativity when speeds are much less than the speed of when speeds are much less than the speed of lightlight

Galilean RelativityGalilean Relativity

To describe a physical event, a frame of To describe a physical event, a frame of reference must be establishedreference must be established

There is no absolute inertial frame of There is no absolute inertial frame of referencereference– This means that the results of an experiment This means that the results of an experiment

performed in a vehicle moving with uniform performed in a vehicle moving with uniform velocity will be identical to the results of the velocity will be identical to the results of the same experiment performed in a stationary same experiment performed in a stationary vehiclevehicle

Galilean RelativityGalilean Relativity Reminders about inertial framesReminders about inertial frames

– Objects subjected to no forces will experience no Objects subjected to no forces will experience no accelerationacceleration

– Any system moving at constant velocity with respect to Any system moving at constant velocity with respect to an inertial frame must also be in an inertial framean inertial frame must also be in an inertial frame

According to the According to the principle of Galilean relativityprinciple of Galilean relativity, , the laws of mechanics are the same in all inertial the laws of mechanics are the same in all inertial frames of referenceframes of reference


The observer in the The observer in the truck throws a ball truck throws a ball straight upstraight up– It appears to move in It appears to move in

a vertical patha vertical path– The law of gravity The law of gravity

and equations of and equations of motion under uniform motion under uniform acceleration are acceleration are obeyedobeyed


There is a stationary observer on the groundThere is a stationary observer on the ground– Views the path of the ball thrown to be a parabolaViews the path of the ball thrown to be a parabola– The ball has a velocity to the right equal to the The ball has a velocity to the right equal to the

velocity of the truckvelocity of the truck

Galilean Relativity – conclusionGalilean Relativity – conclusion

The two observers disagree on the shape of the ball’s path

Both agree that the motion obeys the law of gravity and Newton’s laws of motion

Both agree on how long the ball was in the air

Conclusion: There is no preferred frame of reference for describing the laws of mechanics

The cornerstone of the theory of special The cornerstone of the theory of special relativity isrelativity is the Principle of Relativity:

The Laws of Physics are the same in all inertial frames of reference.

We shall see that many surprising consequences follow We shall see that many surprising consequences follow from this innocuous looking statement. from this innocuous looking statement.

Frames of Reference and Newton's LawsFrames of Reference and Newton's Laws

Let us review Newton's mechanics in terms of frames of Let us review Newton's mechanics in terms of frames of

reference.reference.

A "frame of reference" is just a set of coordinates - something A "frame of reference" is just a set of coordinates - something you use to measure the things that matter in Newtonian you use to measure the things that matter in Newtonian mechanical problems - like positions and velocities, so we mechanical problems - like positions and velocities, so we also need a clock.also need a clock.

A point in space is specified A point in space is specified by its three coordinates by its three coordinates ((xx,,yy,,zz) and an "event" like, ) and an "event" like, say, a little explosion by a say, a little explosion by a place and time – (place and time – (xx,,yy,,zz,,tt). ).

The "laws of physics" we shall consider are those of The "laws of physics" we shall consider are those of Newtonian mechanics, as expressed by Newton's laws Newtonian mechanics, as expressed by Newton's laws of motion, with gravitational forces and also contact of motion, with gravitational forces and also contact

forces from objects pushing against each other.forces from objects pushing against each other. __________________________________________________________

For example, knowing the universal gravitational For example, knowing the universal gravitational constant from experiment (and the masses constant from experiment (and the masses involved), it is possible from Newton's second law, involved), it is possible from Newton's second law,

force = mass x accelerationforce = mass x acceleration, ,

to predict future planetary motions with great to predict future planetary motions with great accuracy. accuracy.

Suppose we know from experiment that Suppose we know from experiment that these laws of mechanics are true in one frame of these laws of mechanics are true in one frame of reference. How do they look in another frame, reference. How do they look in another frame, moving with respect to the first frame? To figure moving with respect to the first frame? To figure out, we have to find how to get from position, out, we have to find how to get from position, velocity and acceleration in one frame to the velocity and acceleration in one frame to the corresponding quantities in the second frame. corresponding quantities in the second frame.

Obviously, the two frames must have a Obviously, the two frames must have a constant constant relative velocity, otherwise the law of relative velocity, otherwise the law of inertia won't hold in both of them. inertia won't hold in both of them.

Let's choose the coordinates so that this velocity is along Let's choose the coordinates so that this velocity is along the the xx-axis of both of them. -axis of both of them.

Notice we also throw in a clock with each frame.Notice we also throw in a clock with each frame.

Now what are the coordinates of the event (Now what are the coordinates of the event (xx,,yy,,zz,,tt) in ) in S'S'? ? It's easy to seeIt's easy to see t't' = = tt - we synchronized the clocks when - we synchronized the clocks when O‘O‘passed passed OO. Also, evidently, . Also, evidently, y'y' = = yy and and z'z' = = zz, from the figure., from the figure.We can also see that We can also see that x x = = x'x' + +vtvt. Thus (. Thus (x,y,z,tx,y,z,t) in ) in SScorresponds to (corresponds to (x',y',z', t'x',y',z', t' ) in ) in S'S', where , where

That's how That's how positionspositions transform - these are known as the transform - these are known as the GalileanGalilean transformations transformations..

tt

zz

yy

vtxx

What about What about velocitiesvelocities ? The velocity in ? The velocity in S'S' in the in the x'x' direction direction

This is just the addition of velocities formula This is just the addition of velocities formula

vuvdt

dxvtx

dt

d

dt

xd

td

xdu xx

)(

vuu xx

How does How does accelerationacceleration transform? transform?

dt

duvu

dt

d

dt

ud

td

ud xx

xx

)(

the the acceleration is the sameacceleration is the same in both frames. This in both frames. This again is obvious - the acceleration is the rate of again is obvious - the acceleration is the rate of change of velocity, and the velocities of the same change of velocity, and the velocities of the same particle measured in the two frames differ by a particle measured in the two frames differ by a constantconstant factor - the relative velocity of the two factor - the relative velocity of the two frames.frames.

xx aa

Since Since vv is constant we have is constant we have

If we now look at the motion under gravitational forces, for If we now look at the motion under gravitational forces, for example, example,

we get the same law on going to another inertial frame we get the same law on going to another inertial frame because because every term in the above equation stays the sameevery term in the above equation stays the same..

rr

mGmam ˆ

221

1

Note that acceleration is the rate of change of momentum Note that acceleration is the rate of change of momentum - this is the same in both frames. So, in a collision, if total - this is the same in both frames. So, in a collision, if total momentum is conserved in one frame (the sum of momentum is conserved in one frame (the sum of individual rates of change of momentum is zero) the individual rates of change of momentum is zero) the same is true in same is true in allall inertial frames.inertial frames.

Maxwell’s Equations of Electromagnetismin Vacuum

0q

AdE

0 AdB

dt

ddE B

dt

ddB E

00

Gauss’ Law for ElectrostaticsGauss’ Law for Electrostatics

Gauss’ Law for MagnetismGauss’ Law for Magnetism

Faraday’s Law of InductionFaraday’s Law of Induction

Ampere’s LawAmpere’s Law

1

2

The Equations of ElectromagnetismThe Equations of Electromagnetism

E dAq

0

B dA 0

..monopole..

?...there’s no magnetic monopole....!!

Gauss’s Laws

4

The Equations of ElectromagnetismThe Equations of Electromagnetism

dt

ddE B

dt

ddB E

00

3

.. if you change a magnetic field you induce an electricfield.........

.. if you change a magnetic field you induce an electricfield.........

Faraday’s LawFaraday’s Law

Ampere’s LawAmpere’s Law

.. if you change an electric field you induce a magneticfield.........

.. if you change an electric field you induce a magneticfield.........

B dl ddt

E 0 0 E dl d

dtB

Electromagnetic WavesElectromagnetic Waves

Faraday’s law:Faraday’s law: dB/dt electric fieldMaxwell’s modification of Ampere’s law Maxwell’s modification of Ampere’s law dE/dt magnetic field

These two equations can be solved simultaneously.These two equations can be solved simultaneously.

The result is:The result is:EE(x, t) = E(x, t) = EPP sin (kx- sin (kx-t)t)

BB(x, t) = B(x, t) = BPP sin (kx- sin (kx-t)t) z

j

Plane Electromagnetic WavesPlane Electromagnetic Waves

x

Ey

Bz

E(x, t) = EP sin (kx-t)

B(x, t) = BP sin (kx-t) z

j

cc

Plane Electromagnetic WavesPlane Electromagnetic Waves

x

Ey

Bz

E(x, t) = EP sin (kx-t)

B(x, t) = BP sin (kx-t) z

j

ccNotes: Waves are in Phase, Notes: Waves are in Phase, but fields oriented at but fields oriented at 909000.. k=2k=2ππ//λλ.. Speed of wave is Speed of wave is c=c=ωω/k (= /k (= ffλλ))

c m s 1 3 100 0

8/ / At all times

E=cB

It was recognized that the It was recognized that the Maxwell Maxwell equations equations did not obey the principles of did not obey the principles of NewtonianNewtonian relativity. i.e. the equations were not invariant relativity. i.e. the equations were not invariant when transformed between the inertial when transformed between the inertial reference frames using the reference frames using the Galilean Galilean transformation.transformation.

Lets consider an example of infinitely long Lets consider an example of infinitely long wire with a uniform negative charge density wire with a uniform negative charge density λλ per unit length and a point charge per unit length and a point charge qq located a distance located a distance yy11 above the wire. above the wire.

The observer inThe observer in SS andand SS’’ see identical electric field see identical electric field

at distanceat distance yy11=y=y11’’ from an infinity long wire from an infinity long wire

carrying uniform chargecarrying uniform charge λλ per unit length. per unit length. Observers in bothObservers in both SS andand S’S’ measure a force measure a force on chargeon charge qq due to the line of charge.due to the line of charge.

1

2

y

kE

1

2

y

kqF

However, the However, the S’S’ observer measured and additional force observer measured and additional force

due to the magnetic field at due to the magnetic field at yy11’’ arising from the motion arising from the motion

of the wire in the of the wire in the -x’-x’ direction. Thus, the electromagnetic force direction. Thus, the electromagnetic force does not have the same form in different inertial systems, does not have the same form in different inertial systems, implying that Maxwell’s equations are not invariant under a implying that Maxwell’s equations are not invariant under a Galilean transformation. Galilean transformation.

1

20

2 y

qv

Speed of the LightSpeed of the Light

It was postulated in the nineteenth century It was postulated in the nineteenth century that electromagnetic waves, like other waves, that electromagnetic waves, like other waves, propagated in a suitable material media, called the propagated in a suitable material media, called the ether. ether.

In according with this postulate the ether In according with this postulate the ether filed the entire universe including the interior of the filed the entire universe including the interior of the matter. matter.

It had the inconsistent properties of being It had the inconsistent properties of being extremely rigid (in order to support the stress of extremely rigid (in order to support the stress of the high electromagnetic wave speed), while the high electromagnetic wave speed), while offering no observable resistance to motion of the offering no observable resistance to motion of the planet, which was fully accounted for by Newton’s planet, which was fully accounted for by Newton’s law of gravitation. law of gravitation.

Speed of the LightSpeed of the Light

The implication of this postulate is that a The implication of this postulate is that a light wave, moving with velocity light wave, moving with velocity cc with respect with respect to the ether, would travel at velocity to the ether, would travel at velocity c’=c +vc’=c +v with respect to a frame of reference moving with respect to a frame of reference moving through the ether at through the ether at vv..

This would require that Maxwell’s This would require that Maxwell’s equations have a different form in the moving equations have a different form in the moving frame so as to predict the speed of light to be frame so as to predict the speed of light to be c’c’, instead of, instead of

00

1

c

Conflict Between Mechanics and E&MConflict Between Mechanics and E&M

A.A. MechanicsMechanics

Galilean relativity states that it is impossible Galilean relativity states that it is impossible for an observer to experimentally for an observer to experimentally distinguish between uniform motion in a distinguish between uniform motion in a straight line and absolute rest. Thus, all straight line and absolute rest. Thus, all states of uniform motion are equal.states of uniform motion are equal.

Conflict Between Mechanics and E&MConflict Between Mechanics and E&M

B.B. E&ME&M

Initially-Initially-

The initial interpretation of the speed of The initial interpretation of the speed of light in Maxwell's theory was this light in Maxwell's theory was this cc was the was the speed of light seen by observers in absolute rest speed of light seen by observers in absolute rest with respect to the ether. with respect to the ether.

In other reference frames, the speed of In other reference frames, the speed of light would be different from light would be different from cc and could be and could be obtained by the Galilean transformation. obtained by the Galilean transformation.

Problem-Problem-

It would now be possible for an It would now be possible for an observer to distinguish between different observer to distinguish between different states of uniform motion by measuring the states of uniform motion by measuring the speed of light or doing other electricity, speed of light or doing other electricity, magnetism, and optics experiments.magnetism, and optics experiments.

Possible SolutionsPossible Solutions

1. 1. Maxwell's theory of electricity and Maxwell's theory of electricity and magnetism was flawed. It was approximately 20 magnetism was flawed. It was approximately 20 years old while Newton's mechanics was years old while Newton's mechanics was approximately 200 years old. approximately 200 years old.

2. 2. Galilean relativity was incorrect. You can Galilean relativity was incorrect. You can detect absolute motion!detect absolute motion!

3. 3. Something else was wrong with Something else was wrong with mechanics (I.e Galilean transformation).mechanics (I.e Galilean transformation).

Experimental ResultsExperimental Results

Most physicists felt that Maxwell's equations Most physicists felt that Maxwell's equations were probably in error. were probably in error.

Numerous experiments were performed to Numerous experiments were performed to detect the motion of the earth through the ether detect the motion of the earth through the ether wind. wind.

The most famous of these experiments was The most famous of these experiments was thethe Michelson-MorleyMichelson-Morley experiment. Because of experiment. Because of the tremendous precision of their interferometer, it the tremendous precision of their interferometer, it was impossible for was impossible for MichelsonMichelson and and MorleyMorley to miss to miss detecting the effect of the earth's motion through detecting the effect of the earth's motion through the ether unless mechanics was flawed!the ether unless mechanics was flawed!

The Michelson-Morley experiment is a race between light beams. The Michelson-Morley experiment is a race between light beams. The incoming light beam is split into two beams by a half-silvered The incoming light beam is split into two beams by a half-silvered mirror. The beams follow perpendicular paths reflecting off full mirror. The beams follow perpendicular paths reflecting off full mirrors before recombining back at the half mirror. Time mirrors before recombining back at the half mirror. Time differences are seen in the interference pattern on the screen. differences are seen in the interference pattern on the screen.

TheoryTheory

We will simplify the calculations by We will simplify the calculations by assuming that assuming that LL11 = L = L22 = L = L..

The time required to complete path 1 The time required to complete path 1 (horizontal path) is given by(horizontal path) is given by

where we have used the Galilean where we have used the Galilean transformation and velocity = distance/time.transformation and velocity = distance/time.

uc

L

uc

LT1

22221

uc

2Lcucuc

uc

LT

21

c

u1

1

c

2LT

Since Since u<<cu<<c, we can use the binomial , we can use the binomial approximation:approximation:

2

1 c

u1

c

2LT

We can determine the time required to complete We can determine the time required to complete path path 2 2 (vertical path) using the distance diagram (vertical path) using the distance diagram below:below:

Using the Pythagorean theorem, we have:Using the Pythagorean theorem, we have:

2

222

2

2

TuL

2

Tc

u(T2/2) u(T2/2)

4

TuL

4

Tc 22

22

22

2

222

22 L4Tuc

2

2

2

22

22

2

cu

1c

4L

uc

4LT

22

c

u1c

L2T

Again, using the binomial approximation:Again, using the binomial approximation:

2

2 c

u

2

11

c

L2T

Thus, the time difference for the two paths Thus, the time difference for the two paths is approximatelyis approximately

We can now calculate the phase shift in We can now calculate the phase shift in terms of wavelengths as follows:terms of wavelengths as follows:

22

21 c

u

c

L

c

u

2

12T-TΔT

c

L

cλf cT

λ cTλ

Thus, the phase shift in terms of a fraction Thus, the phase shift in terms of a fraction of a wavelength is given byof a wavelength is given by

2

c

uLTcλ

2

c

u

λ

L

λ

λ

Using a sodium light, Using a sodium light, = 590 nm = 590 nm, and a , and a interferometer with interferometer with L = 11 mL = 11 m, we have, we have

This was a very large shift (20%) and This was a very large shift (20%) and couldn't have been over looked.couldn't have been over looked.

0.210m10x5.90

m11

λ

λΔ 24

7

Result - Result - No shift was ever observed No shift was ever observed regardless of when the experiment was regardless of when the experiment was performed or how the interferometer was performed or how the interferometer was orientated!orientated!

Einstein’s PostulatesEinstein’s Postulates• In 1905 In 1905 Albert EinsteinAlbert Einstein published a paper on the published a paper on the

electrodynamics of moving bodieselectrodynamics of moving bodies. In this paper, he . In this paper, he postulated that absolute motion can not be detected postulated that absolute motion can not be detected by any experiment. That is, there is no ether. The by any experiment. That is, there is no ether. The reference frame connected with earth is considered reference frame connected with earth is considered to be at rest and the velocity of the light will be the to be at rest and the velocity of the light will be the same in any direction. His theory of same in any direction. His theory of special relativityspecial relativity can be derived from two postulates:can be derived from two postulates:

• Postulate 1Postulate 1:: Absolute uniform motion can not be detected.

• Postulate 2Postulate 2:: The speed of light is independent of the motion of the source.

The Lorentz TransformationThe Lorentz TransformationGalilean transformations:Galilean transformations:

x = x’ + vt’, y = y’, z = z’, t = t’x = x’ + vt’, y = y’, z = z’, t = t’

The inverse transformations areThe inverse transformations are

x’ = x – vt, y’ = y, z’ = z, t’ = tx’ = x – vt, y’ = y, z’ = z, t’ = t

These equations are consistent with experimental These equations are consistent with experimental observations as long as observations as long as vv is much less than is much less than cc. They lead to . They lead to the familiar classical addition law for velocities. If a particle the familiar classical addition law for velocities. If a particle has velocity has velocity uuxx = dx/dt = dx/dt in frame in frame SS, its velocity in frame , its velocity in frame S’S’ is is

from here we havefrom here we have

vuvdt

dxvtx

dt

d

dt

xd

td

xdu xx

)(

xxx

x atd

ud

dt

dua

The Lorentz Transformation

It should be clear that the Galilean transformation is not consistent with Einstein’s postulates of special relativity.

If light moved along the x axis with speed ux’=c in S’, these equations imply that the speed in S is ux=c+v rather than ux=c, which is not consistent with Einstein’s postulates and experiment.

The classical transformation equations must therefore be modified .

The Lorentz Transformation

We assume that the relativistic transformation equation for x is the same as the classical equation except for a constant multiplier on the right side:

where γ is a constant that can depend on v and c but not on coordinates. The inverse transformation in this case

)''( vtxx

)(' vtxx

Lorentz Transformation

The transformation described by these equations is called the Lorentz transformation.

Lorentz’s equations replace the flawed Galileo transformation equations in relating the measurements of two different observers in uniform motion relative to each other.

Lorentz Transformation

2

2

1

1γ

c

uβ

β11γ

cu

2cxutγt'

zz'yy'

utxγx'

Addition of Velocities

We start by taking the derivative with respect to t' of the first Lorentz transformation equation:

From Calculus, we have that:

dt'

dtu

dt'

dxγ

dt'

dx'

dt'

dtu

dt'

dt

dt

dxγ

dt'

dx'

dt'

dtu

dt'

dt

dt

dxγ

dt'

dx'

dt'

dtu

dt

dxγ

dt'

dx'

Using the definition of velocity, we have:

dt'

dtuvγ'v

Differentiating the fourth Lorentz equation with respect to t, we obtain:

dt

dx

cu

dt

dtγ

dt

dt'2

vucc

γv

cu1γ

dt

dt' 2

22

2

2

2

c

vu1γ

1

v)uγ(c

c

dt'

dt

We now substitute this result into our velocity equation to obtain:

where v is velocity of object seen by unprimed observer, v ' is velocity of object seen by primed observer, u is velocity of the primed observer as seen by unprimed observer.

cβvc

uv

c

vu1

uv'v

2

EXERSICE: Show that both observers agree on the speed of light.

SOLUTION: We determine the speed v' as seen by the primed observer for a beam of light v = c seen by the unprimed observer. Using the velocity addition formula we have:

ccuc

uc

c

u1

uc

c

uc1

uc'v

2

Lorentz-Fitgerald (Length) Contraction Fitzgerald's length contraction is now a

direct consequence of the Lorentz transformation. Consider two different observers in relative motion who measure the length of a box as shown below:

observer sees stationary observer sees box moving box at u

SS S’S’

x1

At a single instant of time At a single instant of time tt, the unprimed observer , the unprimed observer measures the box's length in measures the box's length in SS as as

while the unprimed observer measures the box's length in while the unprimed observer measures the box's length in SS’’ as as

Using the Lorentz transformation equation for Using the Lorentz transformation equation for x'x', we have that, we have that

120 xxL

'1

'2' xxL

ut

xut

xL

12'

1212'xxututxx

L

Therefore, we have that

0'

LL

Lorentz-Fitzgerald (Length) ContractionLorentz-Fitzgerald (Length) Contraction

where where LL is the improper lengthis the improper length

LLoo is the proper length is the proper length

uu is the relative speed between the is the relative speed between the observers observers

2

cu1oLL

Length Contraction

Warning: Warning: Your everyday use of English terms can Your everyday use of English terms can get you in trouble in this material. The term get you in trouble in this material. The term "proper""proper" is not meant to indicate that its the "correct" or "right" is not meant to indicate that its the "correct" or "right" answer. Two observers might measure the length of answer. Two observers might measure the length of a train as a train as 10 m10 m when they see the train as when they see the train as stationary. If one observer rides in the train as it stationary. If one observer rides in the train as it moves down the track, he will measure the train's moves down the track, he will measure the train's length as length as 10 m10 m while an observer standing by the while an observer standing by the track sees the train as shorter than track sees the train as shorter than 10 m10 m. . Both Both Observers Are CORRECT!!! Observers Are CORRECT!!! TheThe question question "What "What is the length of the train?" is meaningless!!is the length of the train?" is meaningless!! You You must specify the frame. must specify the frame. There is no such thing as There is no such thing as absolute distance anymore!absolute distance anymore!

Another interesting aspect of relativity Another interesting aspect of relativity is that moving clocks always run slower. is that moving clocks always run slower.

This is not an artifact of the clock! It is This is not an artifact of the clock! It is a consequence of the nature of time itself a consequence of the nature of time itself according to Einstein. according to Einstein.

All clocks will behave this way All clocks will behave this way include your biological processes!include your biological processes!

Time DilationTime Dilation

Let us consider a "Light Clock" in which Let us consider a "Light Clock" in which one unit of time corresponds to the time it takes one unit of time corresponds to the time it takes a light pulse to travel between two meters a a light pulse to travel between two meters a distancedistance LL apart. We will place the clock on a apart. We will place the clock on a train traveling at speedtrain traveling at speed uu with the unprimed with the unprimed observer.observer.


L

Light Path As Seen By Train ObserverLight Path As Seen By Train Observer

Mirror


For the unprimed observer on the train, For the unprimed observer on the train, the time it takes the light pulse to make a the time it takes the light pulse to make a single tick is given bysingle tick is given by

c

Lt


The primed observer standing by the railroad The primed observer standing by the railroad track see the train and clock pass track see the train and clock pass with a speed with a speed ofof uu. . Thus, the path of the light beam appears to Thus, the path of the light beam appears to follow the diagonal path shown below:follow the diagonal path shown below:

Lu

u t'

ct'

Time DilationTime DilationFrom the Pythagorean theorem,From the Pythagorean theorem, t't' is given byis given by

222 Lt'ut'c

22222 Lt'ut'c

2222 Lt'uc

2

2

2

22

22

c

u1c

L

uc

Lt'


We now plug in our results for the unprimed We now plug in our results for the unprimed observer and we find thatobserver and we find that

c

Lγ

c

u1c

Lt'

2

tγt'

Thus, our observers do not agree upon time!Thus, our observers do not agree upon time!

wherewhere TToo is the proper timeis the proper time

T’T’ is the improper timeis the improper time

Time Dilation EquationTime Dilation Equation

2

o

cu

1

TT

Time Dilation as a Function of SpeedTime Dilation as a Function of Speed

An astronomer on Earth observes a meteoroid in the southern An astronomer on Earth observes a meteoroid in the southern sky approaching the Earth at a speed of sky approaching the Earth at a speed of 0.8000.800cc. At the time of . At the time of its discovery the meteoroid is its discovery the meteoroid is 20.0 ly20.0 ly from the Earth. from the Earth. Calculate (a) the time interval required for the meteoroid to Calculate (a) the time interval required for the meteoroid to reach the Earth as measured by the Earthbound astronomer, reach the Earth as measured by the Earthbound astronomer, (b) this time interval as measured by a tourist on the (b) this time interval as measured by a tourist on the meteoroid, and (c) the distance to the Earth as measured by meteoroid, and (c) the distance to the Earth as measured by the tourist. the tourist.


20 ly 20 ly 125.0 yr

0.8 0.8 1 ly yrx c

tv c c

(a) The 0.8c and the 20 ly are measured in the Earth frame, so in this frame,

.

(b) We see a clock on the meteoroid moving, so we do not measure proper time; that clock measures proper time.

pt t 2

2 2

25.0 yr25.0 yr 1 0.8 25.0 yr 0.6 15.0 yr

1 1-p

tt

v c


(c) Method one: We measure the 20 ly on a stick stationary in our frame, so it is proper length. The tourist measures it to be contracted to

2

20 ly 20 ly12.0 ly

1.6671 1 0.8

pLL

Method two: The tourist sees the Earth approaching at

0.8 ly yr 15 yr 12.0 ly

Classical Doppler ShiftClassical Doppler Shift Anyone who has watched auto racing on TV Anyone who has watched auto racing on TV

is aware of the Doppler shift. is aware of the Doppler shift. As a race car approaches the camera, the As a race car approaches the camera, the

sound of its engine increases in pitch (frequency). sound of its engine increases in pitch (frequency). After the car passes the camera, the pitch of After the car passes the camera, the pitch of

its engine decreases. its engine decreases. We could use this pitch to determine the We could use this pitch to determine the

relative speed of the car. This technique is used relative speed of the car. This technique is used in many real world applications including in many real world applications including ultrasound imaging, Doppler radar, and to ultrasound imaging, Doppler radar, and to determine the motion of stars. determine the motion of stars.

I. Moving ObserverI. Moving Observer

Assume that we have a stationary audio Assume that we have a stationary audio source that produces sound waves of source that produces sound waves of frequencyfrequency f f and speedand speed v v. .

SourceSource

v

ObserverObserver x

y

A stationary observer sees the time between each A stationary observer sees the time between each wave aswave as

Speed

DistanceTime


v

λ

f

1T

If the observer is now moving at a velocity If the observer is now moving at a velocity uu relative to the relative to the source then the speed of the waves as seen by the observer source then the speed of the waves as seen by the observer isis

uvSpeed

where the positive sign is when the observer is moving where the positive sign is when the observer is moving toward the source. The time between waves is now toward the source. The time between waves is now

uv

λ

'f

1T'


Taking the ratio of our two results we get thatTaking the ratio of our two results we get that

v

uv

f

'f

T'

T

v

uvf'f

II. Moving SourceII. Moving SourceWe now consider the case in which the source is moving We now consider the case in which the source is moving

toward the observer. In this case, the wave's speed is toward the observer. In this case, the wave's speed is unchanged but the distance between wave fronts (wavelength) unchanged but the distance between wave fronts (wavelength) is reduced. For the source moving away from the observer the is reduced. For the source moving away from the observer the wavelength increased.wavelength increased.

l

SourceSource

u

ObserverObserver x

y

uT v

Tuλ'λ

II. Moving SourceII. Moving Source

λ

Tu1

λ

Tuλ

λ

'λ

λf

u1

λ

λ

'

v

uv

v

u1

λ

λ'

v

uv

c

λ'

λ

c

II. Moving SourceII. Moving Source

v

uv

'f

f

Thus, the frequency seen by the observer for a moving source Thus, the frequency seen by the observer for a moving source is given byis given by

uv

vf'f

Moving ObserverMoving Observer

Moving SourceMoving Source

v

uvf'f

uv

vf'f

Note: Note: The motion of the observer and source create different The motion of the observer and source create different effects. For sound, this difference is explained due to motion effects. For sound, this difference is explained due to motion relative to the preferred reference frame! This preferred frame relative to the preferred reference frame! This preferred frame is the reference frame stationary to the medium propagating is the reference frame stationary to the medium propagating the sound (air)!the sound (air)!

Relativistic Doppler EffectRelativistic Doppler Effect

For light in vacuum the distinction between For light in vacuum the distinction between motion the source and detector can not be done. motion the source and detector can not be done. Therefore the expression we derived for sound can Therefore the expression we derived for sound can not be correct for light.not be correct for light.

Consider a source of light moving toward the Consider a source of light moving toward the detector with velocitydetector with velocity vv, relative to the detector. If , relative to the detector. If source emitssource emits NN electromagnetic waves in timeelectromagnetic waves in time ΔΔttDD

(measured in the frame of the detector), the first (measured in the frame of the detector), the first wave will travel a distancewave will travel a distance ccΔΔttDD and the source will and the source will

travel the distancetravel the distance v v ΔΔttDD measures in the frame of measures in the frame of

detector.detector.

Relativistic Doppler EffectRelativistic Doppler Effect

The wavelength of the light will be:The wavelength of the light will be:

N

tvtc DD '

The frequencyThe frequency f ’f ’ observed by the detector observed by the detector will therefore be:will therefore be:

DD t

N

cvvct

cNcf

1

1

)(''

Relativistic Doppler EffectRelativistic Doppler Effect If the frequency of the source isIf the frequency of the source is ff00 it will emitit will emit N=fN=f00ΔΔttSS

waves in the timewaves in the time ΔΔttSS measured by the source. Thenmeasured by the source. Then

D

S

D

S

D t

t

cv

f

t

tf

cvt

N

cv

f

11

1

1

1' 00

HereHere ΔΔttSS is the proper time interval ( the first wave and is the proper time interval ( the first wave and

thethe NNthth wave are emitted at the same place in the source’s wave are emitted at the same place in the source’s

reference frame). Therefore timesreference frame). Therefore times ΔΔttSS and and ΔΔttDD are related as:are related as:

2

2

1cv

ttt SSD

Relativistic Doppler EffectRelativistic Doppler Effect When the source and detector are moving When the source and detector are moving

towards one another:towards one another:

-- approachingapproaching

-- receding receding

0

2

2

0

1

11

1' f

cvcv

cv

ff

0

1

1' f

cvcv

f

0

1

1' f

cvcv

f

Relativistic Doppler ShiftRelativistic Doppler Shift Since light has no medium, there should be no Since light has no medium, there should be no

difference between moving the source and the observer. difference between moving the source and the observer. The problem with our previous derivation when dealing The problem with our previous derivation when dealing

with light is that we haven't considered that space and time with light is that we haven't considered that space and time coordinates are different for the source and observer. coordinates are different for the source and observer.

Thus, we must account for the contraction of space and Thus, we must account for the contraction of space and dilation of time due to motion. After accounting for differences dilation of time due to motion. After accounting for differences in time and space, we get the following result for both moving in time and space, we get the following result for both moving source and moving source and moving observer:observer:

0fuc

uc'f

Police radar detects the speed of a car as follows: Microwaves Police radar detects the speed of a car as follows: Microwaves of a precisely known frequency are broadcast toward the car. of a precisely known frequency are broadcast toward the car. The moving car reflects the microwaves with a The moving car reflects the microwaves with a Doppler shiftDoppler shift. . The reflected waves are received and combined with an The reflected waves are received and combined with an attenuated version of the transmitted wave. Beats occur attenuated version of the transmitted wave. Beats occur between the two microwave signals. The beat frequency is between the two microwave signals. The beat frequency is measured. (a) For an electromagnetic wave reflected back to measured. (a) For an electromagnetic wave reflected back to its source from a mirror approaching at speed its source from a mirror approaching at speed vv, show that the , show that the reflected wave has frequency reflected wave has frequency

where where ffsourcesource is the source frequency. (b) When is the source frequency. (b) When vv is much less is much less than than cc, the beat frequency is much smaller than the , the beat frequency is much smaller than the transmitted frequency. In this case use the approximation transmitted frequency. In this case use the approximation f f + + ffsourcesource ≈ 2 ≈ 2 ffsourcesource and show that the beat frequency can be and show that the beat frequency can be written as written as ffbeatbeat = 2 = 2vv/λ/λ. (c) What beat frequency is measured for . (c) What beat frequency is measured for a car speed of a car speed of 30.0 m/s30.0 m/s if the microwaves have frequency if the microwaves have frequency 10.0 GHz10.0 GHz? (d) If the beat frequency measurement is accurate ? (d) If the beat frequency measurement is accurate to to ±5 Hz±5 Hz, how accurate is the velocity measurement? , how accurate is the velocity measurement?

vc

vcff

source

(a) Let Let ffcc be the frequency as seen by the car: be the frequency as seen by the car: sourcecc v

ffc v

and, if and, if ff is the frequency of the reflected wave: is the frequency of the reflected wave: cc v

ffc v

Combining gives: Combining gives: sourcec v

ffc v

(b) Using the above result:(b) Using the above result: sourcef c v f c v

source source source2ff c ff v f v

The beat frequency is then:The beat frequency is then:

sourcebeat source

2 2f v vff f

c

(c) (c)

9

beat 8

2 30.0 m s 10.0 10 Hz 2 30.0 m s2000 Hz= 2.00 kHz

0.0300 m3.00 10 m sf

8

9source

3.00 10 m s3.00 cm

10.0 10 Hz

cf

(d) (d) beat

2

fv

beat 5 Hz 0.0300 m0.0750 m s 0.2 mi h

2 2

fv

A ball is thrown at A ball is thrown at 20.0 m/s20.0 m/s inside a boxcar moving inside a boxcar moving along the tracks at along the tracks at 40.0 m/s40.0 m/s. What is the speed of . What is the speed of the ball relative to the ground if the ball is thrown the ball relative to the ground if the ball is thrown (a) forward (b) backward (c) out the side door? (a) forward (b) backward (c) out the side door?

A ball is thrown at A ball is thrown at 20.0 m/s20.0 m/s inside a boxcar moving inside a boxcar moving along the tracks at along the tracks at 40.0 m/s40.0 m/s. What is the speed of . What is the speed of the ball relative to the ground if the ball is thrown the ball relative to the ground if the ball is thrown (a) forward (b) backward (c) out the side door? (a) forward (b) backward (c) out the side door?

(a)(a)

(b)(b)

(c)(c)

60.0 m sT Bv v v

20.0 m sT Bv v v

2 2 2 220 40 44.7 m sT Bv v v

A muon formed high in the Earth’s atmosphere A muon formed high in the Earth’s atmosphere travels at speed travels at speed v v = 0.990= 0.990cc for a distance of for a distance of 4.60 km4.60 km before it decays into an electron, a neutrino, and an before it decays into an electron, a neutrino, and an antineutrino . (a) How long does antineutrino . (a) How long does the muon live, as measured in its reference frame? the muon live, as measured in its reference frame? (b) How far does the muon travel, as measured in its (b) How far does the muon travel, as measured in its frame? frame?

vv e

A muon formed high in the Earth’s atmosphere travels at speed A muon formed high in the Earth’s atmosphere travels at speed v v = 0.990= 0.990cc for a distance of for a distance of 4.60 km4.60 km before it decays into an before it decays into an electron, a neutrino, and an antineutrino . electron, a neutrino, and an antineutrino . (a) How long does the muon live, as measured in its reference (a) How long does the muon live, as measured in its reference frame? (b) How far does the muon travel, as measured in its frame? (b) How far does the muon travel, as measured in its frame? frame?

vv e

0.990vc

For For

,

09.7990.1

1

1

12

2

2

cv

4.60 km0.990

tc

3

8

1 4.60 10 m2.18 s

7.09 0.990 3.00 10 m sp

tt

(a)(a) The muon’s lifetime as measured in the Earth’s The muon’s lifetime as measured in the Earth’s rest frame isrest frame is

and the lifetime measured in the muon’s rest frame isand the lifetime measured in the muon’s rest frame is

.

A muon formed high in the Earth’s atmosphere A muon formed high in the Earth’s atmosphere travels at speed travels at speed v v = 0.990= 0.990cc for a distance of for a distance of 4.60 km4.60 km before it decays into an electron, a neutrino, and an before it decays into an electron, a neutrino, and an antineutrino . (a) How long does antineutrino . (a) How long does the muon live, as measured in its reference frame? the muon live, as measured in its reference frame? (b) How far does the muon travel, as measured in its (b) How far does the muon travel, as measured in its frame? frame?

vv e

(b)(b) 2 34.60 10 m1 649 m

7.09p

p

LvL L

c

The identical twins Speedo and Goslo join a migration from The identical twins Speedo and Goslo join a migration from the Earth to Planet X. It is the Earth to Planet X. It is 20.0 ly20.0 ly away in a reference frame away in a reference frame in which both planets are at rest. The twins, of the same age, in which both planets are at rest. The twins, of the same age, depart at the same time on different spacecraft. Speedo’s depart at the same time on different spacecraft. Speedo’s craft travels steadily at craft travels steadily at 0.9500.950cc, and Goslo’s at , and Goslo’s at 0.7500.750cc. . Calculate the age difference between the twins after Goslo’s Calculate the age difference between the twins after Goslo’s spacecraft lands on Planet X. Which twin is the older? spacecraft lands on Planet X. Which twin is the older?

The identical twins Speedo and Goslo join a migration from The identical twins Speedo and Goslo join a migration from the Earth to Planet X. It is the Earth to Planet X. It is 20.0 ly20.0 ly away in a reference frame away in a reference frame in which both planets are at rest. The twins, of the same age, in which both planets are at rest. The twins, of the same age, depart at the same time on different spacecraft. Speedo’s depart at the same time on different spacecraft. Speedo’s craft travels steadily at craft travels steadily at 0.9500.950cc, and Goslo’s at , and Goslo’s at 0.7500.750cc. . Calculate the age difference between the twins after Goslo’s Calculate the age difference between the twins after Goslo’s spacecraft lands on Planet X. Which twin is the older? spacecraft lands on Planet X. Which twin is the older?

In the Earth frame, Speedo’s trip lasts for a time:In the Earth frame, Speedo’s trip lasts for a time:

20.0 ly21.05 yr

0.950 ly yrx

tv

Speedo’s age advances only by the proper time intervalSpeedo’s age advances only by the proper time interval

221.05 yr 1 0.95 6.574 yrpt

t

during his trip.during his trip.

The identical twins Speedo and Goslo join a migration from the Earth to Planet X. The identical twins Speedo and Goslo join a migration from the Earth to Planet X. It is It is 20.0 ly20.0 ly away in a reference frame in which both planets are at rest. The away in a reference frame in which both planets are at rest. The twins, of the same age, depart at the same time on different spacecraft. Speedo’s twins, of the same age, depart at the same time on different spacecraft. Speedo’s craft travels steadily at craft travels steadily at 0.9500.950cc, and Goslo’s at , and Goslo’s at 0.7500.750cc. Calculate the age . Calculate the age difference between the twins after Goslo’s spacecraft lands on Planet X. Which difference between the twins after Goslo’s spacecraft lands on Planet X. Which twin is the older? twin is the older?

Similarly for Goslo,Similarly for Goslo,2

22

20.0 ly1 1 0.75 17.64 yr

0.750 ly yrpx v

tv c

While Speedo has landed on Planet X and is waiting for his While Speedo has landed on Planet X and is waiting for his brother, he ages by:brother, he ages by:

20.0 ly 20.0 ly5.614 yr

0.750 ly yr 0.950 ly yr

Goslo

17.64 yr 6.574 yr 5.614 yr 5.45 yr ThenThen

ends up older ends up older byby

.

An atomic clock moves at An atomic clock moves at 1 000 km/h1 000 km/h for for 1.00 h1.00 h as as measured by an identical clock on the Earth. How measured by an identical clock on the Earth. How many nanoseconds slow will the moving clock be many nanoseconds slow will the moving clock be compared with the Earth clock, at the end of the compared with the Earth clock, at the end of the 1.00-h1.00-h interval? interval?

An atomic clock moves at An atomic clock moves at 1 000 km/h1 000 km/h for for 1.00 h1.00 h as as measured by an identical clock on the Earth. How many measured by an identical clock on the Earth. How many nanoseconds slow will the moving clock be compared with nanoseconds slow will the moving clock be compared with the Earth clock, at the end of the the Earth clock, at the end of the 1.00-h1.00-h interval? interval?

Solution:Solution:

2 21

pp

tt t

v c

2 2

2 21 12

pv v

t t tc c

2

2

2

2

2

2

2211

21

c

vt

c

vt

c

vtttt p

An atomic clock moves at An atomic clock moves at 1 000 km/h1 000 km/h for for 1.00 h1.00 h as as measured by an identical clock on the Earth. How many measured by an identical clock on the Earth. How many nanoseconds slow will the moving clock be compared with nanoseconds slow will the moving clock be compared with the Earth clock, at the end of the the Earth clock, at the end of the 1.00-h1.00-h interval? interval?

Solution:Solution:

61.00 10 m1000 km h 277.8 m s

3 600 sv

nsss

c

vttt

sm

sm

c

v

p 54.11054.1)1027.9(2

3600

2

1027.9/103

/8.277

9272

78

A spacecraft is launched from the surface of the Earth with a A spacecraft is launched from the surface of the Earth with a velocity of velocity of 0.6000.600cc at an angle of at an angle of 50.0°50.0° above the horizontal above the horizontal positive positive x x axis. Another spacecraft is moving past, with a axis. Another spacecraft is moving past, with a velocity of velocity of 0.7000.700cc in the negative in the negative x x direction. Determine the direction. Determine the magnitude and direction of the velocity of the first spacecraft magnitude and direction of the velocity of the first spacecraft as measured by the pilot of the second spacecraft. as measured by the pilot of the second spacecraft.

A spacecraft is launched from the surface of the Earth with a A spacecraft is launched from the surface of the Earth with a velocity of velocity of 0.6000.600cc at an angle of at an angle of 50.0°50.0° above the horizontal above the horizontal positive positive x x axis. Another spacecraft is moving past, with a axis. Another spacecraft is moving past, with a velocity of velocity of 0.7000.700cc in the negative in the negative x x direction. Determine the direction. Determine the magnitude and direction of the velocity of the first spacecraft magnitude and direction of the velocity of the first spacecraft as measured by the pilot of the second spacecraft. as measured by the pilot of the second spacecraft. Solution:Solution:

0.7v cLet frame S be the Earth frame of reference. Then

0.6 cos50 0.386xu c c

0.6 sin50 0.459yu c c As measured from the S’ frame of the second spacecraft:

S

A spacecraft is launched from the surface of the Earth with a A spacecraft is launched from the surface of the Earth with a velocity of velocity of 0.6000.600cc at an angle of at an angle of 50.0°50.0° above the horizontal above the horizontal positive positive x x axis. Another spacecraft is moving past, with a axis. Another spacecraft is moving past, with a velocity of velocity of 0.7000.700cc in the negative in the negative x x direction. Determine the direction. Determine the magnitude and direction of the velocity of the first spacecraft magnitude and direction of the velocity of the first spacecraft as measured by the pilot of the second spacecraft. as measured by the pilot of the second spacecraft. Solution:Solution:

As measured from the S’ frame of the second spacecraft:

S

2 2

2

2

0.386 0.7 1.0860.855

1.271 1 0.386 0.7

0.459 1 0.7 0.459 0.7140.258

1 0.386 0.7 1.271

xx

x

yy

x

c cu v cu c

u v c c c c

u c cu c

u v c

2 20.855 0.285 0.893c c c Iu’I =

1 0.258tan 16.8 above the -axis

0.855c

xc

A moving rod is observed to have a length of A moving rod is observed to have a length of 2.00 m2.00 m and to and to be oriented at an angle of be oriented at an angle of 30.0°30.0° with respect to the direction with respect to the direction of motion, as shown in Figure. The rod has a speed of of motion, as shown in Figure. The rod has a speed of 0.9950.995cc. . (a) What is the proper length of the rod? (b) What is (a) What is the proper length of the rod? (b) What is the orientation angle in the proper frame?the orientation angle in the proper frame?

A moving rod is observed to have a length of A moving rod is observed to have a length of 2.00 m2.00 m and to and to be oriented at an angle of be oriented at an angle of 30.0°30.0° with respect to the direction with respect to the direction of motion, as shown in Figure. The rod has a speed of of motion, as shown in Figure. The rod has a speed of 0.9950.995cc. . (a) What is the proper length of the rod? (b) What is (a) What is the proper length of the rod? (b) What is the orientation angle in the proper frame?the orientation angle in the proper frame?

2 2 2

1 110.0

1 1 0.995v c

1 1 1cos 2.00 m 0.867 1.73 mxL L

1 1 1sin 2.00 m 0.500 1.00 myL L

2xL is a proper length, related to 1xL by2

1x

xL

L

.

2 110.0 17.3 mx xL L 2 1 1.00 my yL L

We have seen that Einstein’s postulates We have seen that Einstein’s postulates require important modifications in our ideas of require important modifications in our ideas of simultaneitysimultaneity and our measurements of and our measurements of time time

and and lengthlength. Einstein’s postulates also require . Einstein’s postulates also require modification of our concepts of modification of our concepts of mass, mass,

momentum, momentum, and and energyenergy.

Relativistic MomentumRelativistic Momentum

In classical mechanics, the momentum of a In classical mechanics, the momentum of a particle is defined as a product of its mass and its particle is defined as a product of its mass and its velocity, velocity, mumu. .

In an isolated system of particles, with no net In an isolated system of particles, with no net force acting on the system, the total momentum of force acting on the system, the total momentum of the system remains the same. However , we can see the system remains the same. However , we can see from a simple though experiment that the quantity from a simple though experiment that the quantity ΣΣmmiivvii is not conserved in isolated system.is not conserved in isolated system.

We consider two observers: observer We consider two observers: observer AA in reference frame in reference frame SS and observer and observer BB in frame in frame S’S’,, which is moving to the right in the which is moving to the right in the xx direction with speeddirection with speed vv with respect to the frame with respect to the frame SS. Each . Each has a ball of the masshas a ball of the mass mm. .

One observer throws his ball up with a speed One observer throws his ball up with a speed uu00 relative to relative to

him and the other throws his ball down with a speed him and the other throws his ball down with a speed uu00

relative to him, so that each ball travel a distance relative to him, so that each ball travel a distance LL, makes an , makes an

elastic collision with the other ball, and returns.elastic collision with the other ball, and returns.

Classically, each ball has vertical momentum of magnitude Classically, each ball has vertical momentum of magnitude mumu00. Since the vertical components of the momenta are . Since the vertical components of the momenta are

equal and opposite, the total vertical momentum are zero equal and opposite, the total vertical momentum are zero before the collision. The collision merely reverses the before the collision. The collision merely reverses the momentum of each ball, so the total vertical momentum is momentum of each ball, so the total vertical momentum is zero after the collision. zero after the collision.

Relativistically, however, the vertical components of the Relativistically, however, the vertical components of the velocities of the two balls as seen by the observers are not velocities of the two balls as seen by the observers are not equal and opposite. Thus, when they are reversed by the equal and opposite. Thus, when they are reversed by the collision, classical momentum is not conserved. As seen by collision, classical momentum is not conserved. As seen by AA in in frame frame SS,, the velocity of his ball is the velocity of his ball is uuAyAy=+u=+u00. Since the velocity of . Since the velocity of ball ball BB in frame in frame S’S’ is is uu’’

BxBx=0=0, , uuByBy’=-u’=-u00,, the the yy component of the component of the velocity of ball velocity of ball BB in frame in frame SS is is uuByBy=-u=-u00//γγ. So, the . So, the ΣΣ m miiuuii is is conserved only in the approximation that conserved only in the approximation that u<<cu<<c. .

We will define the relativistic momentum We will define the relativistic momentum pp of a particle to have of a particle to have the following properties: the following properties: 1. In collisions , 1. In collisions , pp is conserved. is conserved.2.As 2.As u/cu/c approaches approaches 00, , pp approaches approaches mumu. . We will show that quantityWe will show that quantity

is conserved in the elastic collision and we take this equation is conserved in the elastic collision and we take this equation for the definition of the relativistic momentum of a particle.for the definition of the relativistic momentum of a particle.

2

2

1cu

mup

Conservation of the Relativistic MomentumConservation of the Relativistic Momentum

One interpretation of this equation is that the mass of One interpretation of this equation is that the mass of an object increases with the speed. Then the an object increases with the speed. Then the quantity quantity

is called the relativistic mass. The mass of a particle is called the relativistic mass. The mass of a particle when it is at rest in some reference frame is then when it is at rest in some reference frame is then called its rest masscalled its rest mass ..

mmrelrel= = γγmmrestrest

2

2

1cu

mup

2

2

1cu

mmrel

Conservation of the Relativistic Conservation of the Relativistic MomentumMomentum

The speed of ball The speed of ball AA in in SS is is uu00, so the, so the yy component of its component of its

relativistic momentum is:relativistic momentum is:

The speed of The speed of BB in in SS is more complicated. Its is more complicated. Its xx component is component is vv and its and its yy component is component is –u–u00//γγ. .

2

20

0

1cu

mup

yA

2

2202

022

2

2

2

02222 1

c

vuuvu

c

vuvuuu

B

ByBxB

Using this result to compute Using this result to compute √1-(u√1-(u22BB/c/c22)),, we obtain we obtain

andand

2

20

2

2

4

2202

20

2

2

2

2

11

11

c

u

c

v

c

vu

c

u

c

v

c

uB

2

20

2

20

2

2

2

2

11

111c

u

c

u

c

v

c

uB


The The yy component of the relativistic momentum of component of the relativistic momentum of ball ball BB as seen in as seen in SS is therefore is therefore

2

20

0

2

2

11

1c

u

mu

cu

mup

B

ByBy

2

20

0

1c

u

mupBy


Since Since ppByBy=-p=-pAyAy’’ the the yy component of the total component of the total

momentum of the two balls is zero. If the speed momentum of the two balls is zero. If the speed of each ball is reversed by the collision, the total of each ball is reversed by the collision, the total momentum will remain zero and momentum will momentum will remain zero and momentum will be conserved.be conserved.

withwithmup 2

2

1

1

cu

Relativistic momentum as given by equationRelativistic momentum as given by equation

versus versus u/cu/c, where , where u u is speed of the object relative to is speed of the object relative to an observer. The magnitude of momentuman observer. The magnitude of momentum pp is plotted is plotted in units of in units of mcmc. The fainter dashed line shows the . The fainter dashed line shows the classical momentum classical momentum mumu..

2

2

1cu

mup

Relativistic EnergyRelativistic Energy

In classic mechanicIn classic mechanic: The work done by the net : The work done by the net force acting on a particle equals the change in the force acting on a particle equals the change in the kinetic energy of the particle.kinetic energy of the particle.

In relativistic mechanicIn relativistic mechanic: The net force acting on a : The net force acting on a particle to accelerate it from rest to some final particle to accelerate it from rest to some final velocity is equal to the rate of change of the velocity is equal to the rate of change of the relativistic momentum. The work done by the net relativistic momentum. The work done by the net force can then be calculated and set equal to the force can then be calculated and set equal to the change of kinetic energy.change of kinetic energy.


As in relativistic mechanic, we will define the kinetic As in relativistic mechanic, we will define the kinetic energy as the work done by the net force in energy as the work done by the net force in accelerating a particle from rest to some final accelerating a particle from rest to some final velocity velocity uuff..

where we have used where we have used u=ds/dtu=ds/dt..

2

2

1cu

muududpds

dt

dpdsFK net

du

cu

m

du

cuc

u

cuc

um

du

cu

cuc

uum

du

cu

u

cu

dm

cu

mud

3

2

2

3

2

22

2

3

2

22

2

2

23

2

22

2

2

2

2

2

2

11

111

11

1

111

2

2

1

1

1

1

1

1

Relativistic EnergyRelativistic EnergySubstituting the last result in the equation for kinetic Substituting the last result in the equation for kinetic energy we obtain:energy we obtain:

The second part of this expression, The second part of this expression, mcmc22, is independent of , is independent of the speed and called the the speed and called the rest energyrest energy EEoo of the particle.of the particle.

EE00=mc=mc22

The total relativistic energy:The total relativistic energy:

2

2

2

2

1

mc

cu

mcK

2

2

22

1c

u

mcmcKE


The work done by unbalanced force increases the The work done by unbalanced force increases the energy from the rest energy to the final energy energy from the rest energy to the final energy

where where

is the relativistic mass.is the relativistic mass.

2

2

2

2

1

cm

cu

mcrel

2

2

1cu

mmrel


We can obtain a useful expression for the We can obtain a useful expression for the velocity of a particle by multiplying equation for velocity of a particle by multiplying equation for relativistic momentum by relativistic momentum by cc22::

Eu

cu

umcpc

cu

mup

2

2

22

2

2

1

1

E

pc

c

u

Energies in atomic and nuclear physics are Energies in atomic and nuclear physics are usually expressed in units of electron volts usually expressed in units of electron volts ((eVeV)) or mega-electron volts or mega-electron volts ((MeVMeV):):

1eV = 1.602 x 101eV = 1.602 x 10-19-19JJ

A convenient unit for the masses of atomic A convenient unit for the masses of atomic particles is particles is eV/ceV/c22 or or MeV/cMeV/c22, which is the rest , which is the rest energy of the particle divided by energy of the particle divided by cc22..

The expression for kinetic energyThe expression for kinetic energy

does not look much like the classical expression does not look much like the classical expression

½(mu½(mu22)).. However when However when v<<cv<<c, we can approximate , we can approximate

1/√1-(u1/√1-(u22 /c /c22)) using the binomial expansion:using the binomial expansion:

(1+x)(1+x)nn =1 + nx =1 + nx + + n(n-1)xn(n-1)x22/2 + ……./2 + …….≈ 1 + nx≈ 1 + nx

2

2

2

2

1

mc

cu

mcK

ThenThen

and when and when v<<cv<<c

2

22

1

2

2

2

2 2

111

1

1

c

u

c

u

cu

22

22

2

2

2

2

11

2

111

1

1mu

c

umc

cu

mcK

The equation for relativistic momentum:The equation for relativistic momentum:

and equation for the total energy:and equation for the total energy:

can be combined to eliminate the speed can be combined to eliminate the speed uu::

EE2 2 = p= p22cc22 +(mc +(mc22))22

2

2

1cu

mup

2

2

22

1c

u

mcmcKE

EE2 2 = p= p22cc22 +(mc +(mc22))22

We received the relation for total energy, We received the relation for total energy, momentum, and rest energy.momentum, and rest energy.

If the energy of a particle is much grater than its If the energy of a particle is much grater than its rest energy rest energy mcmc22, the second term on the right , the second term on the right side of the last equation can be neglected, side of the last equation can be neglected, giving the useful approximation:giving the useful approximation:

E ≈ pcE ≈ pc (for(for E>>mcE>>mc22))

An unstable particle at rest breaks into two An unstable particle at rest breaks into two fragments of unequal mass. The mass of the first fragments of unequal mass. The mass of the first fragment is fragment is 2.50 × 102.50 × 10–28–28 kg kg, and that of the other is , and that of the other is 1.67 × 101.67 × 10–27–27 kg. If the lighter fragment has a speed kg. If the lighter fragment has a speed of of 0.8930.893cc after the breakup, what is the speed of after the breakup, what is the speed of the heavier fragment? the heavier fragment?

An unstable particle at rest breaks into two fragments of An unstable particle at rest breaks into two fragments of unequal mass. The mass of the first fragment is unequal mass. The mass of the first fragment is 2.50 × 102.50 × 10–28–28 kgkg, and that of the other is , and that of the other is 1.67 × 101.67 × 10–27–27 kg. If the lighter kg. If the lighter fragment has a speed of fragment has a speed of 0.8930.893cc after the breakup, what is the after the breakup, what is the speed of the heavier fragment? speed of the heavier fragment?

Relativistic momentum of the system of fragments Relativistic momentum of the system of fragments must be conservedmust be conserved. For . For total momentum to be zero after as it was before, we must have, with total momentum to be zero after as it was before, we must have, with subscript 2 referring to the heavier fragment, and subscript 1 to the lighter, subscript 2 referring to the heavier fragment, and subscript 1 to the lighter,

2 1p p

28

2 2 2 1 1 1 2

2.50 10 kg0.893

1 0.893mu mu c

27

2 28

22

1.67 10 kg4.960 10 kg

1

uc

u c

An unstable particle at rest breaks into two fragments of An unstable particle at rest breaks into two fragments of unequal mass. The mass of the first fragment is unequal mass. The mass of the first fragment is 2.50 × 102.50 × 10–28–28 kgkg, and that of the other is , and that of the other is 1.67 × 101.67 × 10–27–27 kg. If the lighter kg. If the lighter fragment has a speed of fragment has a speed of 0.8930.893cc after the breakup, what is the after the breakup, what is the speed of the heavier fragment? speed of the heavier fragment?

2 2272 2

28 2

1.67 101

4.960 10

u uc c

22212.3 1u

c

2 0.285u c

An unstable particle with a mass of An unstable particle with a mass of 3.34 × 103.34 × 10–27–27 kg kg is is initially at rest. The particle decays into two fragments initially at rest. The particle decays into two fragments that fly off along the that fly off along the x x axis with velocity components axis with velocity components 0.9870.987cc and and –0.868–0.868cc. . Find the masses of the Find the masses of the fragments. (fragments. (Suggestion: Suggestion: Conserve both energy and Conserve both energy and momentum.) momentum.)


1 2

12.01

1 0.868

2 2

16.22

1 0.987

1 2 totalE E E 2 2 2

1 1 2 2 totalmc m c m c

271 22.01 6.22 3.34 10 kgm m


27

1 22.01 6.22 3.34 10 kgm m This reduces to:This reduces to:

271 23.09 1.66 10 kgm m

1 2p p 1 1 1 2 2 2mu mu

1 22.01 0.868 6.22 0.987c m c m

1 23.52m m

281 8.84 10 kgm

282 2.51 10 kgm

A pion at rest (A pion at rest (mmππ = 273 = 273mmee) decays to a muon () decays to a muon (mmμμ = =

207207mmee) and an antineutrino ( ). The reaction is ) and an antineutrino ( ). The reaction is

written . Find the kinetic energy of the written . Find the kinetic energy of the muon and the energy of the antineutrino in electron volts. muon and the energy of the antineutrino in electron volts. ((Suggestion: Suggestion: Conserve both energy and momentum.) Conserve both energy and momentum.)

0vm

v




0vm

v

before decay after decay 0p p 207v ep p m u m u

vE E E 2 2vm c p c m c

207 273ve e

pm m

c

207 207 273e e eu

m m mc

2731 1.32

207uc

1

1.741

u c

u c

0.270uc




0vm

v

2 21 1 207 eK m c mc

2

11 207 0.511 MeV

1 0.270K

4.08 MeVK

vE E E 2 2 2273 207v eE m c m c mc

2

207273 0.511 MeV

1 0.270

29.6 MeV

v

v

E

E

SUMMARYSUMMARY

1. Einstein’s Postulates1. Einstein’s Postulates::

• Postulate 1Postulate 1: Absolute uniform motion : Absolute uniform motion can not be detected.can not be detected.

• Postulate 2Postulate 2: The speed of light is : The speed of light is independent of the motion of the independent of the motion of the source.source.

SUMMARYSUMMARY

• 2. The Lorentz Transformation:2. The Lorentz Transformation:

2cxutγt'

zz'yy'

utxγx'

SUMMARYSUMMARY

6. 6. The Velocity Transformation:The Velocity Transformation:

2cxvu

1

uxv'xv

2c

'xuv

1

'xv

xv

u

SUMMARYSUMMARY

• 3. Time Dilation:3. Time Dilation:

• 4. Length Contraction:4. Length Contraction:

• 5. 5. The Relativistic The Relativistic Doppler Effect:Doppler Effect: 0f

uc

uc'f

approaching

0fuc

uc'f

receding

2

o

cu

1

TT

0'

LL

SUMMARYSUMMARY

2

2

1cu

mup

)1(

1

2

2

22

2

mc

c

u

mcmcK

7.7. Relativistic Momentum: Relativistic Momentum:

8. Relativistic Energy:8. Relativistic Energy:

9. Rest Energy:9. Rest Energy: 20 mcE

10. Kinetic Energy10. Kinetic Energy

2

2

222

1c

u

mcmcKmcE

SUMMARYSUMMARY

10.10. Useful Formulas for Speed, Energy, and Momentum:Useful Formulas for Speed, Energy, and Momentum:

22222 mccpE

E

pc

c

u

2mcEforpcE

relativity chapter 1. a brief overview of modern physics 20 th century revolution: - 1900 max planck...

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