Modern Physics (PHY 3305) · perspective of special relativity? Remember: and From those...

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Velocity, Energy and Matter (Ch. 2.6-2.7)SteveSekula, 29 January 2010 (created 13 December 2009)

CHAPTERS 2.6-2.7

Review of last lecture

We explored one of the popular so-called "paradoxes" of specialrelativity and realized that there is no paradox when Einstein'spostulates are applied thoroughly to the questionWe explored kinematics in special relativity - motion in a frame ofreference - beginning with the Doppler effect for light.

Today

Briefly discuss the transformation of velocitiesChallenge classical notions of momentum and energy and discuss theimplications of new realities for societyWe will begin a discussion of the wave nature of light.

Special Relativistic Transformation of Velocities

One of the most stunning revelations of special relativity proceeds from thevanilla-sounding "transformation of velocities." Recall that the Galileantransformation told us:

Special relativity seems to be a better description of the relationshipbetween space and time. What is the "correct" transformation from the

Modern Physics (PHY 3305) Lecture NotesModern Physics (PHY 3305) Lecture Notes

FRAME S : u = u0 + ·

FRAME S 0 : u0 = uÀ ·

Modern Physics (PHY 3305) - Lecture Notes file:///home/sekula/Documents/Notebooks/Modern...

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perspective of special relativity?

Remember:

and

From those relationships, we can derive how the motion of objects in oneframe are related to the motions in another frame. For instance, in frame :

u

In Frame :

u

The Mathematical Bag of Tricks: Binomial Expansion

We keep seeing things that look like , like:

or

or in the Doppler Shift

All of these are variations on

u x=dt = d

dx dx dt =@x

@x00 +

@x

@t00

S

0 =uÀ ·

1Àc2·u

S 0

=u0 + ·

1 +c2·u0

(a ) + x n

Í 1 ) ; · =1q

1À (·=c)2Ñ ( À Ì2 À1=2

1=Í 1 ) ; · = ( À Ì2 +1=2

(1 =c) : + · À1

(a ) : + x n


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When you see that, and your calculator fails you because , you applythe Binomial Expansion:

where

Here are some useful cases:

This will be helpful for you when, for instance, you are faced withcalculating relativistic effects for velocities much smaller than that of light,but where the relativistic effects, though tiny, can have a large impact onthe outcome.

Momentum and Energy

Remember the classical conservation of momentum:

What about when you apply the classical transformation of velocities - thatis, is momentum conserved in classical physics using the GalileanTransformation?

The answer is yes. If anyone would like to see this, consider a situationwhere two objects of mass and collide head-on. In frame , theirinitial momentum and final momentum is given by

m u u u u :

In another frame, , moving along the direction of the collision at speed ,the form of the conservation equation for the collision should be:

x < < 1

f (x) a ) a x a x a x :: = ( + x n = an + n nÀ1 +2!

n(n )À 1 nÀ2 2 +3!

n(n )(n )À 1 À 2 nÀ3 3 + :

(x ) 2 < a2

:: 1

(1 )À x2= 1 + x2 + x4 + x6 + :

x x x ::

s1

(1 )À x2= 1 +

2

1 2 +8

1 4 +1

166 + :

X

pinitial~ =X

pfinal~

m 1 m 2 S

1i1 +m2

i2 = m1

f1 +m2

f2

S 0 ·

m u u u u 10i1 +m2

0i2 = m1

0f1 +m2

0f2


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if the conservation of momentum is preserved under different states ofmotion.

So: are the forms of momentum conservation independent of the motion?We can apply the Galilean Transformation, , and rewrite thesecond equation in terms of quantities in frame . All the terms with cancel, and the find that the answer is yes: the statement that classicalmomentum is invariant under the classical relativistic transformation istrue.

But if you instead use the special-relativistic transformation of velocities,you will find the above no longer works. Momentum conservation takes adifferent form depending on the state of motion of the objects in questionrelative to an observer.

Discuss what to do next

So what do we do?

We can throw out momentum conservation as a principle of nature.We can change the definition of momentum and find one that isconserved.

Einstein chose to preserve momentum and energy conservation, ala hisFirst Postulate. This was again based on a lack of experimental evidence tothe contrary. Instead, he assumed that the form of momentum needed to beamended - that the classical form for momentum, , must beincomplete. So what's the correct form?

The Short Version: Momentum and Energy

Think through the problem of the definition of momentum in class

If is wrong, what is the right form?

What do we know about momentum as low speeds?

we know it experimentally takes the form

That means that whatever the form of momentum, , as a function ofvelocity it must reduce to for speeds much smaller than that of light.

u 0 = uÀ ·S ·

p u = m

p u = m

p u = m

p (u) =Mp u = m


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Do we have a relativistic quantity that behaves this way, which we couldmultiply times ?

What quantity has a minimum value of 1, and is very close to 1 when?

The answer is the gamma function, . So we can postulate that

momentum takes the more general form:

, which has the right behavior when .

In fact, if you now write down the conservation of momentum equation forthis form of the momentum, and correctly transform the velocities from oneframe to another, you will find that momentum is conserved again.

Einstein sought then an expression for the total energy of a body in motion.Again, starting from the form of the conservation of Energy equation, hemade some arguments about what must happen as a consequence ofrelativity. In addition, he knew that the energy of motion of a body MUSTtake the form

E onstant mu

when , since that was what was known from experiments.

What he found was stunning. He found he could express the total energy-squared of a body in motion as

That is, the total energy of a body is related to the sum of its mass and itsmomentum.

The Long Version: Momentum and Energy

Let's apply Einstein's postulates, specifically:

The laws of physics are invariant for observers in relative uniform

mu

u < < c

1p1À(·=c)2

p mu = Íu

u < < c

= c +2

1 2

u < < c

E 2 = mcÀ

2Á2

+ (pc)2


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motion

Under that postulate, the form of energy and momentum conservation mustbe unchanged by motion; in addition, energy and momentum must beconserved in all inertial frames. The classical version of these equationsfor, say, an elastic collision (as above) is:

where (classically):

E (0) mu

(the second equation has an additional constant term allowed by Newton'slaws:

Let us generalize these ingredients in order to deduce the correctrelativistic form:

where and are unknown functions of the motion of the relativeframes. We DO know that in the limit that :

E 1b +E2b = E1a +E2a

p 1b + p2b = p1a + p2a

p u = m

= E +2

1 2

W dx (dp=dt)dx (du=dt)(udt) udu mu ) =

ZF =

Z=

Zm =

Zm = C +

2

1 2

p (u) =M

E (u) = E

M(u) E(u) u ! 0

lim M(u) u!0 = m

lim u!0 @u2

@E(u)=

2

m


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Let's start with momentum. There is so far only one relativistic functionthat we know which, in the limit , goes to 1: . We can then

guess that:

That leaves us to determine the form of relativistic energy. Energy istrickier because it has a non-linear dependence on velocity of the body in aframe, and it can have a leading constant term in addition to the term.To get us thinking about how to attack this, let's again think about . Thebinomial expansion for is:

We're looking for something that yields, for :

constant m

So it seems that we should just try out:

in the limit of low velocities:

which then leads us to conclude that:

u ! 0 Í u = 1q1À

c2u2

p (u) mu =M = Íu

u 2

Í uÍ u

Í (u ) u = 1 +2

1

c2u2

+O 4

u ! 0

+2

1

c2u2

E(u) E(0) = Íu

lim E(u) (0) E(0) u!0 Ù E +2

1

c2u2

E(0) c = m 2


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This is incredible. This says that the total energy of a body at rest can bedescribed by its mass. This also tells us that when energy is given off by abody (as in the form of radiation), its mass decreases correspondingly by

. This is not outside of reason.

problem: a lightbulb radiates photons, which themselves have no mass,with a power of 75 watts. In one year, how much mass is lost by thelightbulb? Can this be detected using any common measuringinstrument?

answer: 75 watts means 75 joules/s. A year is. Let's assume you

leave this light on ALL the time. In one year, the lightbulb thusradiates , where . What is the massequivalent lost? . This cannotbe measured with any common instrument (a scale, the Wii Fit,etc). Thus, you never would have noticed this mass disappearing,so this idea is not outside of reason because it is not within yourexperience.

The last equation for energy can be used to find the complete and generalexpression for the total energy of a body:

E(u) E(0) (mc ) m c

The Binomial Expansion of can be determined using:

where :

Plugging into our energy-squared equation:

Ám E=c = Á 2

3600s=hour 4hours=day 65days=year :2 0 s Â 2 Â 3 = 3 Â 1 7

2:4 0 J Â 1 9 1J kg =s = 1 Ám2 2

ÁE=c :4 0 =(3 0 ) :6 0 kg 2 = 2 Â 1 9 Â 1 8 2 = 2 Â 1 À8

2 = Íu2 2 = Íu

2 2 2 =1

1Àc2u2

2 4

1=(1 ) À x2

(1 ) ::: + y À1 = 1À y + y2 À y3

y x = À 2

(1 ) :: À x2 À1 = 1 + x2 + x4 + x6 + :


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So the total energy of a body is simply given by the sum of the rest-massand the momentum of the body. Simple, elegant, and ground-breaking!

At this poiint, we simple redefine and we have the famousequation:

which describes the total energy of any body in any inertial referenceframe.

Consequences of Einstein's Energy-Momentum-Mass Revelation

Let's discuss some of the consequences of this innocuous-looking equation:

What happens to the total energy of a body at rest in a given referenceframe?

In that case,

and all of the energy of the body is described by its total mass. Whyis this so odd? This equation tells you that if you then heat theobject, so that it gains internal energy through temperatureincrease but NOT kinetic energy through motion, the mass of theobject should increase! That is mind-blowing. If you know the massof something at rest, you know EVERYTHING about its total energyregardless of the form of that energy (light it emits, heat itcontains, etc.)

What happens when the mass of an object is zero? Is that even allowed?

E(u) (0) (1 ) ) :: 2 = E 2 + (c

u 2 + (c

u 4 + :

E(u) c u c u u (1=c ) u (1=c ) :: 2 = m2 4 +m2 2 2 +m2 4 +m2 6 2 +m2 8 4 + :

E(u) mc ) u c (1 =c =c =c :: 2 = ( 2 2 +m2 2 2 + u2 2 + u4 4 + u6 6 + :

E(u) mc ) Í mu) c mc ) pc) 2 = ( 2 2 + ( u2 2 = ( 2 2 + ( 2

E(u) Ñ E

E c c 2 = m2 4 + p2 2

E c = m 2


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When mass is zero, . It seems that there is no prior reason tobelieve such a thing cannot exist; the equation allows it, but whatdoes the equation mean about such an object? If you write this as

, the ONLY way that the energy of such an object can beanything but zero is if so that . While relativity fails totell you about the properties of this object (e.g. exactly what itsenergy is!), it doesn't rule out that such objects can exist and ifthey do they must move at the speed of light. What moves at thespeed of light? Light. Light is a massless particle, but we won'tlearn much more about it from relativity since it's undefinable inrelativity.

Let's put this together into a question: if a light bulb is just sitting they,emitting light, what is happening to its mass?

The equation tells us not only that mass is energy, but thatwhen energy is given off by a body (as in the form of radiation), itsmass decreases correspondingly by . How could this behappening and us not be noticing?

Consider a 75W lightbulb. It radiates photons, which themselveshave no mass. In one year, how much mass is lost by the lightbulb?Can this be detected using any common measuring instrument?Well, 75 watts means 75 joules/s. A year is

. Let's assume youleave this light on ALL the time. In one year, the lightbulb thusradiates , where . What is the massequivalent lost?

This cannot be measured with any common instrument (a scale, theWii Fit, etc). Thus, you never would have noticed this massdisappearing, so this idea is not outside of reason because it is notwithin your experience.

Discuss the implications of changing a little mass into energy

Let's Talk a little bit about light

Relativity includes the speed of light, allows for things like like to exist (asmassless objects), but fails to tell us about the properties of light. What is

E c = p

E mu = Íuu = c Í u =1

E c = m 2

Ám E=c = Á 2

3600s=hour 4hours=day 65days=year :2 0 s Â 2 Â 3 = 3 Â 1 7

2:4 0 J Â 1 9 1J kg =s = 1 Ám2 2

ÁE=c :4 0 =(3 0 ) :6 0 kg: 2 = 2 Â 1 9 Â 1 8 2 = 2 Â 1 À8


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light? To answer that, we'll begin with a discussion about waves.

The Wave Nature of Radiation (Light)

Radiation is energy transmitted from one body to another; for instance,radio waves, light, and heat are all examples of common radiation. Theclassical understanding of radiation was that it was a wave by nature. Thatis, radiation (e.g. light) can be described as a phenomenon characterizedby a wavelength and a frequency and an amplitude, transmitted at somevelocity from one body to another. Waves are continuous phenomena; theydeliver arbitrary amounts of energy and that amount of energy is encodedin the intensity of the radiation - more intensity = more energy.

The evidence for the wave-nature of light was multi-fold, and is summarizedhere:

Key properties of waves are:diffraction

(from http://en.wikipedia.org/wiki/Diffraction)

interference


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(from http://en.wikipedia.org/wiki/Interference_%28wave_propagation%29)

PhET JAVA Demonstration of Wave Properties(from http://phet.colorado.edu/sims/wave-interference/wave-interference_en.jnlp)

Light was observed to diffract (that is, strike an obstacle and bend aroundit or, in the case of a small opening, spread out beyond the opening). Youcan see below an example of diffraction around the edges of a razor bladeusing blue laser light:

(from http://micro.magnet.fsu.edu/primer/lightandcolor/diffractionintro.html)

Interference of light with itself was also observed. An example below isobtained using a Michelson interferometer (just as in the Michelson-Morleyexperiment), where a single beam of green laser light is split and made totravel two slightly different path lengths, then recombined:


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In the mid-1800s, physicists were confident that light/radiation wastransmitted via electromagnetic waves.

In addition, classical physics had produced a theory of heat(thermodynamics) that allowed you to predict the spectrum of intensities ofheat emitted from a body. Together, they used these two pieces ofunderstanding to study the properties of objects called "blackbodies."

Next time:

We will explore a parallel set of developments that accompanied theSpecial Theory of Relativity: quantum physics. We will begin with themystery surrounding black-body radiation and its power spectrum(predicted vs. observed). We will then discuss Max Planck's attempt toresolve that mystery, and the consequences of his hypothesis. We willexplore Einstein's contribution to the development of this field, and beginto understand the nature of energy and matter. This will complement ourstudies of space and time, and will form the foundation upon which the restof the course will proceed.


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