Http://www.nearingzero.net.

Presentation on theme: "Http://www.nearingzero.net."— Presentation transcript:

Relativistic Mechanics Mass, Energy, and Momentum
General Relativity “In science one tries to tell people, in such a way as to be understood by everyone, something that no one ever knew before. But in poetry, it's the exact opposite.”—P. A. M. Dirac

B throws ball in this direction
Relativistic Momentum We believe very strongly that momentum is conserved. Let’s see what effect a relativistic calculation has on momentum. Imagine this collision: person A on the ground throws a ball straight at the side of a railroad flatcar moving past him. Person B moving with the flatcar throws an identical ball straight out at the same speed (relative to the flatcar). The two balls meet head-on and rebound. v flatcar’s velocity B B throws ball in this direction B’s ball also has a velocity component in this direction… …so that B’s ball has a total velocity along this direction and follows this path A throws ball in this direction A collision! momentum is (??) conserved collision!

B throws ball in this direction
v flatcar’s velocity B throws ball in this direction B’s ball also has a velocity component in this direction… …so that B’s ball has a total velocity along this direction and follows this path A throws ball in this direction collision! collision! momentum is (??) conserved Read Beiser for a more detailed description of this thought experiment. Because of time dilation, A and B measure different travel times for ball thrown from the flatcar. They will calculate different initial and final momenta. They will not agree that momentum is conserved.

Non-conservation of momentum is an alarming idea
Non-conservation of momentum is an alarming idea. What can we do to fix this situation? It turns out that if the ball thrown by the moving* flatcar observer has a greater mass than the ball thrown by the stationary observer, momentum is conserved: where m is the mass of the ball at rest, and m(v) is the mass it needs to have when it is moving. *So we are taking the point of view of the observer on earth.

In the old days* we then said “OK, m is the mass of the object at rest and m(v) is its mass when it is moving. Let’s call m0 the rest mass and m the mass when it is moving (‘relativistic mass’).” This notation is consistent with our equations for time dilation and length contraction, so we have Mass depends on speed. This was Einstein’s original approach, but later he said it is “not good.” *I.E., back in the dinosaur ages, when I studied relativity in college. Also, in the previous edition of our modern physics text.

If we define momentum as where
The “new” approach is to say “Look, mass is mass. We believe it is something fundamental. If we believe in conservation of momentum, we had better change our definition of momentum.” If we define momentum as where then “mass is mass,” momentum is conserved in our thought experiment (and in real life), and relativistic momentum reduces to classical (Newtonian) momentum in the limit v0.* *More satisfying than saying “mass changes with velocity.”

In the old days, rest mass was relativistically invariant
In the old days, rest mass was relativistically invariant. Now mass is relativistically invariant. Same reality, just different use of words. Let’s make this new notation official. A consequence of this new definition of momentum:

You’ll need to translate unchanged solutions to the new notation.
So in this most recent version of his text, Beiser is always going to use the symbol m for mass. It’s what we called “proper mass” or “rest mass” in the old days. Many of my posted exam solutions use m0 for rest mass and m for m(v). This is now outdated! Am I going to change my exam solutions? No! They will slowly cycle out of circulation. By the school year, file exam solutions will be correct. You’ll need to translate unchanged solutions to the new notation. Will this “catch” some people? Yes, those* who haven’t paid attention. Remember: file exams before Fall 2002 use the “old” notation. *Don’t be one of those!

More consequences of this new definition of momentum:
relativistic momentum increases without bound as vc; c is the “speed limit” For finite F, a0 as vc.* No finite force can accelerate an object having nonzero mass up to the speed of light! classical momentum increases linearly with velocity; no limit to classical velocity *example 1.5

The correction factor for momentum is
“When can I use classical mechanics, and when do I have to use relativistic mechanics?” The correction factor for momentum is Here’s a table showing the correction factor for different objects: Object v v/c m(v)/m jogger 10 km/h ≈ 1 space shuttle 104 m/s electron 106 m/s 0.0033 1.001 108 m/s 0.333 1.061 subtract 1, multiply by 100 to get % error

Mass and Energy From your first-semester physics course: Use the definition of  and integrate by parts to get Assuming potential energy is zero (we can always choose coordinates to do this), we interpret mc2 as total energy. The “box” indicates an OSE.

When an object is moving, its total energy is
When an object is at rest KE = 0, and any energy that remains is interpreted as the object’s rest energy E0. When an object is moving, its total energy is This is really just a variation of the OSE on the previous slide. This is the closest you’ll come to seeing E=mc2 in this class. In the “old days,” E=mc2 would have been written E=mc2.

These equations have a number of interesting implications.
Mass and energy are two different aspects of the same “thing.” Conservation of energy is actually conservation of mass-energy. The c2 in E0=mc2 means a little mass is “worth” a lot of energy. Your lunch: an example of relativity at work in “everyday life.” Total energy is conserved but not relativistically invariant. Rest (or proper) mass is relativistically invariant. Mass is not conserved! (But it is for the purposes of chemistry.)

Example: when 1 kg (how much is that
Example: when 1 kg (how much is that?) of dynamite explodes, it releases 5.4x106 joules of energy. How much mass disappears? This is actually a conservation of mass-energy problem. If this material were presented in Physics 23, I would make you start with your conservation of mass-energy OSE and derive the appropriate equations from there. For now, it is sufficient to realize that the problem is just asking “what is the mass equivalent of 5.4x106 joules of energy?” Conservation of mass is a very good approximation!

If we are to claim relativistic mechanics as a replacement theory for Newtonian mechanics, then relativistic mechanics had better reduce to Newtonian mechanics in the limit of small relative velocities. Our text shows that for v<<c,

When can I get away with using KE = mv2/2, and when do I have to use KE = mc2 - mc2?
Use Newtonian KE every time you can get away with it! Use relativistic KE only when you must! If v = 1x107 m/s (fast!) then mv2/2 is off by only 0.08%. Probably OK to use mv2/2. If v = 0.5 c, then mv2/2 is off by 19%. Better use relativity. I rarely* try to trap you into the “wrong” calculation. Often you will do that without my help. Sometimes I will ask you to make a judgment, but I will always give you the criteria. I might ask, if an error no greater than 5% is tolerable, is a relativistic calculation necessary? *Typically once or twice a semester, and it really bothers my conscience when I succeed.

Total energy and magnitude of momentum are given by
Energy and Momentum Total energy and magnitude of momentum are given by With a bit of algebra, you can show The quantities on the LHS and RHS of the above equation are relativistically invariant (same for all inertial observers). Rearranging: See p. 31 for comment on particle vs. system of particles.

Is it possible for a particle to have no mass
Is it possible for a particle to have no mass? If m = 0, what are E and p? For a particle with m = 0 and v < c, then E=0 and p=0. A “non-particle.” No such particle.` But if m = 0 and v = c, then the two equations above are indeterminate. We can’t use them to say anything about our proposed particle.

graviton is to gravity as photon is to E&M field
If m = 0 and v = c, we must use The energy of such a particle is E = pc. We could detect this particle! It could exist! Do you know of any massless particles?  photon  neutrino*  graviton** graviton is to gravity as photon is to E&M field *Doubtful. Nobel prize for you if you prove mneutrino = 0 or mneutrino ≠ 0. **Maybe. Nobel prize for its discoverer. Problem: gravitational fields are much, much weaker than E&M fields.

The momentum carried by massless particles is nonzero (E = pc).
Particles having KE >> E0 (or pc >> mc2) become more photon-like and behave more like waves. The momentum carried by massless particles is nonzero (E = pc). Quickly concealing his pocket laser pointer? Could you stop a freight train with a flashlight? Could you stop a beam of atoms with a laser beam?

A note on units. We will use the electron volt (eV) as an energy unit throughout this course. Variations on the eV: 1 meV = 10-3 eV (milli) 1 keV = 103 eV (kilo) 1 MeV = 106 eV (mega) 1 GeV = 109 eV (giga) There are a few rare cases where mixing eV units in calculations with SI units leads to numerical errors. I will warn you when we come to those cases.

Because mass and energy are convenient, we sometimes write masses in “energy units.”
An electron has a rest mass of 9.11x10-31 kg. If you plug that mass into E0 = mc2, you get an energy of 511,000 eV, or 511 keV, or MeV, or 0.511x10-3 GeV. We sometimes write the electron mass as MeV/c2. It is also possible to express momentum in “energy units.” An electron might have a momentum of 0.3 MeV/c. If you are making a calculation with an equation like and you want to use MeV/c2 for the electron mass, please do. It often simplifies the calculation. But watch out…

What is the total energy of an electron that has a momentum of 1
What is the total energy of an electron that has a momentum of 1.0 MeV/c? Notice the convenient cancellation of the c’s in the 2nd step. Avoid the common mistake: don’t divide by an extra c2 or multiply by an extra c2 in the 2nd step.

General Relativity Something to think about. Is the mass that goes in F = ma (or the relativistic version) the same “thing” as the mass that goes in F = Gm1m2/r2? Not necessarily! Experimentally, the two “kinds” of mass are the same to within better than one part in 1012, and most of us believe they are the same anyway, so… “An observer in a closed laboratory cannot distinguish between the effects of a gravitational field or an acceleration of the lab.” The principle of equivalence.

We will hear more about light and gravity in the next chapter.
The principle of equivalence leads one to conclude that light must be deflected by a gravitational field. Experimental observation of this effect in 1919 was one of Einstein’s great triumphs. We will hear more about light and gravity in the next chapter. Before I give the first quiz, I would like to work a few problems in class. I’ve already done time dilation and length contraction problems. I should work some problems with energy and momentum. "If A equals success, then the formula is: A=X+Y+Z. X is work. Y is play. Z is keep your mouth shut.“—A. Einstein

When the two are equal E0=K so that:
Problem At what speed does the kinetic energy of a particle equal its rest energy? Kinetic energy: Rest energy: When the two are equal E0=K so that: Is it a good idea to combine several algebraic steps in one “line?” Only if you can guarantee that you will never make a mistake.

cross-multiply square both sides solve for v2/c2 simplify solve for v2

Problem A particle has a kinetic energy 20 times its rest energy. Find the speed of the particle in terms of c. “I can’t do this problem. You didn’t give me the mass.” I’m showing this problem to demonstrate the use of a really handy derived equation (so it might appear on your quiz, but probably won’t appear on your OSE sheet). OSE rearrange plug in expression for  algebra

rewriting equation, so you can see the next step
÷ LHS top and bottom by mc2 square both sides solve for v2/c2

But the rest energy is just mc2, and the problem says use K=20mc2.
take  of both sides solve for v; not “official,” but often useful for the exam, you should have this equation on your 3x5 card K MUST be the relativistic kinetic energy. K=½mv2 is wrong here! By now you’ve forgotten what the problem asked, so here it is again: “A particle has a kinetic energy 20 times its rest energy. Find the speed of the particle in terms of c.” But the rest energy is just mc2, and the problem says use K=20mc2.

use K=20mc2 algebra math Important conceptual result: if an object’s kinetic energy (or total energy) is much greater (“much” meaning a factor of 5 or 10 or so) than its rest energy, the object is going almost the speed of light.

“I thought momentum had units of [mass]·[velocity]?”
Problem Find the momentum (in MeV/c) of an electron whose speed is 0.6c. “I thought momentum had units of [mass]·[velocity]?” It does! And MeV/c works out to [mass]·[velocity], but is a much more convenient unit to use when working with relativistic particles. Physicists who deal with high-energy particles use MeV/c for momentum units, and MeV/c2 for mass units. OSE plug in expression for 

The electron mass in “energy units” is 0.511 MeV/c2.
substitute numerical values notice how conveniently the units lead to cancellation don’t forget to square the 0.6

My Mathcad solution is rather klunky
My Mathcad solution is rather klunky. For the quiz or exam, work the problem the way I did here. “I don’t trust these funny “energy units” for mass and momentum. Would I get the correct result if I work the problem in SI units?” How many different approaches are there to solving any physics problem? How many right answers are there to any physics problem?

If this were a multiple choice quiz problem and I asked for an answer in MeV/c, you would not see the above answer as a choice. If this were a “work-it-out-for-partial-credit” problem, I would take off a few points (one point for “math,” plus a few more points, depending on how annoyed I happened to be when I was grading). “Can you show me how to convert to energy units?” Sure. Keep in mind that in algebra, you can always multiply or divide by 1, or add or subtract 0, without changing your answer. (You may change the form of the answer, but not the value.)

1 1 1

Which way would you rather work this problem?
roundoff error crept into the 3rd decimal place Which way would you rather work this problem? Keep in mind that every time you punch a button on your calculator, and every time you write a symbol on paper, you have increased the chance that you made a mistake.* *Also true, I suppose, when you show all your steps, but if each step is simple, the chance for error goes way down, plus I can give more partial credit if I see where you made your mistake.