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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

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B Relativistic Momentum We believe very strongly that momentum is conserved. Lets 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. A v flatcars velocity B throws ball in this direction Bs ball also has a velocity component in this direction… …so that Bs ball has a total velocity along this direction and follows this path A throws ball in this direction collision! collision! momentum is (??) conserved

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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. Read Beiser for a more detailed description of this thought experiment. B A v flatcars velocity B throws ball in this direction Bs ball also has a velocity component in this direction… …so that Bs ball has a total velocity along this direction and follows this path A throws ball in this direction collision! collision! momentum is (??) conserved

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where m is the mass of the ball at rest, and m(v) is the mass it needs to have when it is moving. 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: *So we are taking the point of view of the observer on earth.

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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. Lets call m 0 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 *I.E., back in the dinosaur ages, when I studied relativity in college. Also, in the previous edition of our modern physics text. Mass depends on speed. This was Einsteins original approach, but later he said it is not good.

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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.

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

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So in this most recent version of his text, Beiser is always going to use the symbol m for mass. Its what we called proper mass or rest mass in the old days. Many of my posted exam solutions use m 0 for rest mass and m for m(v). This is now outdated! Am I going to change my exam solutions? Youll need to translate unchanged solutions to the new notation. No! They will slowly cycle out of circulation. By the school year, file exam solutions will be correct. Will this catch some people? Yes, those* who havent paid attention. Remember: file exams before Fall 2002 use the old notation. *Dont be one of those!

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More consequences of this new definition of momentum: 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 relativistic momentum increases without bound as v c; c is the speed limit *example 1.5

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When can I use classical mechanics, and when do I have to use relativistic mechanics? Objectvv/cm(v)/m jogger10 km/h space shuttle 10 4 m/s electron10 6 m/s electron10 8 m/s The correction factor for momentum is Heres a table showing the correction factor for different objects: subtract 1, multiply by 100 to get % error

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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 mc 2 as total energy. The box indicates an OSE.

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When an object is at rest KE = 0, and any energy that remains is interpreted as the objects rest energy E 0. When an object is moving, its total energy is This is the closest youll come to seeing E=mc 2 in this class. In the old days, E= mc 2 would have been written E=mc 2. This is really just a variation of the OSE on the previous slide.

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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 c 2 in E 0 =mc 2 means a little mass is worth a lot of energy. 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.) Your lunch: an example of relativity at work in everyday life.

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Example: when 1 kg (how much is that?) of dynamite explodes, it releases 5.4x10 6 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 5.4x10 6 joules of energy? Conservation of mass is a very good approximation!

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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<

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Use Newtonian KE every time you can get away with it! Use relativistic KE only when you must! If v = 1x10 7 m/s (fast!) then mv 2 /2 is off by only 0.08%. Probably OK to use mv 2 /2. If v = 0.5 c, then mv 2 /2 is off by 19%. Better use relativity. When can I get away with using KE = mv 2 /2, and when do I have to use KE = mc 2 - mc 2 ? 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.

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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: Energy and Momentum See p. 31 for comment on particle vs. system of particles.

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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 cant use them to say anything about our proposed particle.

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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* *Doubtful. Nobel prize for you if you prove m neutrino = 0 or m neutrino 0. graviton** **Maybe. Nobel prize for its discoverer. Problem: gravitational fields are much, much weaker than E&M fields. graviton is to gravity as photon is to E&M field

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Particles having KE >> E 0 (or pc >> mc 2 ) become more photon-like and behave more like waves. The momentum carried by massless particles is nonzero (E = pc). Could you stop a freight train with a flashlight? Could you stop a beam of atoms with a laser beam?laser beam Quickly concealing his pocket laser pointer?

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A note on units. We will use the electron volt (eV) as an energy unit throughout this course. Variations on the eV: 1 meV = eV (milli) 1 keV = 10 3 eV (kilo) 1 MeV = 10 6 eV (mega) 1 GeV = 10 9 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.

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An electron has a rest mass of 9.11x kg. If you plug that mass into E 0 = mc 2, 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/c 2. 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/c 2 for the electron mass, please do. It often simplifies the calculation. But watch out… Because mass and energy are convenient, we sometimes write masses in energy units.

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What is the total energy of an electron that has a momentum of 1.0 MeV/c? Avoid the common mistake: dont divide by an extra c 2 or multiply by an extra c 2 in the 2 nd step. Notice the convenient cancellation of the cs in the 2 nd step.

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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 = Gm 1 m 2 /r 2 ? Not necessarily! Experimentally, the two kinds of mass are the same to within better than one part in 10 12, 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.

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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 Einsteins great triumphs. We will hear more about light and gravity in the next chapter. "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 Before I give the first quiz, I would like to work a few problems in class. Ive already done time dilation and length contraction problems. I should work some problems with energy and momentum.

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Problem At what speed does the kinetic energy of a particle equal its rest energy? Kinetic energy: Rest energy: When the two are equal E 0 =K so that: Only if you can guarantee that you will never make a mistake. Is it a good idea to combine several algebraic steps in one line?

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cross-multiply square both sides solve for v 2 /c 2 simplify solve for v 2

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Problem A particle has a kinetic energy 20 times its rest energy. Find the speed of the particle in terms of c. Im showing this problem to demonstrate the use of a really handy derived equation (so it might appear on your quiz, but probably wont appear on your OSE sheet). rearrange plug in expression for algebra OSE I cant do this problem. You didnt give me the mass.

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rewriting equation, so you can see the next step ÷ LHS top and bottom by mc 2 square both sides solve for v 2 /c 2

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solve for v; not official, but often useful By now youve 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 mc 2, and the problem says use K=20mc 2. K MUST be the relativistic kinetic energy. K=½mv 2 is wrong here! take of both sides for the exam, you should have this equation on your 3x5 card

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use K=20mc 2 algebra math Important conceptual result: if an objects 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.

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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/c 2 for mass units. OSE plug in expression for

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substitute numerical values The electron mass in energy units is MeV/c 2. notice how conveniently the units lead to cancellation dont forget to square the 0.6

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My Mathcad solution is rather klunky. For the quiz or exam, work the problem the way I did here. I dont 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?

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Is this answer correct? 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.)

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1 1 1

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roundoff error crept into the 3 rd 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.

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