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Mass and thermodynamics in relativity

  1. Jun 21, 2014 #1
    I think I have come across something that invalidates something one of my Physics professors said back in college, though. Someone check me if I'm wrong.

    We talked about how increasing the potential energy of an object will increase its mass. This seems perfectly reasonable, especially if one takes mass to be equivalent to rest energy. Thus, if we develop a super storage battery and dump a few trillion Joules of electrical energy into it, we would find the mass of the charged battery to be higher than that of the empty battery.

    As I recall, he also mentioned something else.

    Let us take a pair of identical massive balls, connected by a massless, friction free spring. They sell them at imaginary hardware stores all over. If we either compress or stretch the spring, the mass of the system should rise, since we have increased its potential energy. Let's assume we've stretched it. If we release the balls, they will begin moving towards each other, converting the potential energy of the system into kinetic energy for each of the masses. The kinetic energy of the system will not change, of course, because the center of mass doesn't move. Now, taking mass to be Lorentz invariant, which all the smart boys and girls are insisting is the case, the mass of each ball does not change. As the balls move toward the point where the spring is relaxed, the mass of the system should return to its original value. The balls then start compressing the spring, until they eventually come to a halt WRT each other and start moving apart again. In short, we've created ourselves a nice, little perpetual clock - patent pending. The interesting thing is we see the mass of the clock regularly increase and decrease at twice the fundamental frequency of the clock. Am I on track so far?

    So now let's increase the number of masses and spring into the trillions. In other words, let's talk about a gas. Well, sort of. In a gas, the molecules are not really very influenced very much by the other molecules except as a pair of them get very close together, at which point the pair undertake some extremely high accelerations. Most of the time is spent by most of the molecules moving in a fairly straight, if rather short, line at a more or less constant velocity. It seems to me, then, that dumping heat into a perfectly rigid container filled with a gas would not increase its mass nearly as much as we would expect, despite the fact we have increased its potential energy a huge amount. I submit this is because we haven't really increased its potential energy. It only seems that way at the macroscopic level. Rather, we have actually increased its kinetic energy, which doesn't increase its mass. The reason it doesn't look like we have increased its kinetic energy is all the individual momentum vectors sun to zero (well, almost zero - we'll ignore Brownian motion, here). Have I run amok, here?

    It also seems to me the same would hold for a solid or a liquid, just not quite as strongly, since the molecules in non-gaseous object are much closer together, thus spending much more of their time accelerating. In any case, I think my prof was wrong.
    Last edited by a moderator: Jun 21, 2014
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  3. Jun 21, 2014 #2


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    More kinetic energy, in a bound state, will also increase the mass of the bound state. Why would it not?
  4. Jun 22, 2014 #3


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    Of course the kinetic energy changes if the balls start to accelerate towards each other.

    The energy of the system is always constant, therefore the mass of the whole system is constant. The energy stored in the spring contributes to the mass in the same way the kinetic energy contributes. The mass of the total system is not just the sum of the masses of the balls.
  5. Jun 22, 2014 #4


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    Sorry, you aren't on the right track. The system is isolated, so it has some total energy E that's constant over time. Potential energy is constantly being converted into kinetic energy, then back again, but the total value of E is constant. The total momentum is also constant. In relativistic units where c=1, the invariant mass is given by the formula E^2 - p^2 = m^2, so m is constant. By m here, I mean the invariant mass. If you want the factors of 'c' in the above, it becmoes E^2 - (pc)^2 = (mc^2)^2.

    Note that the system as a whole has more energy when you stretch the string than not, so a system with a stretched string has a higher mass m than a system without the string stretched. However, the total energy is a constant as a function of time, as is the mass.

    Note that this implies that the mass of the system is not just the sum of the mass of the two balls. Mass don[t, in general, add in special relativity. Energy does add, and so does momentum. When you calculate mass via the formula E^2 - p^2 (or the version with the extra factors of c if you're using non relativistic units), you add together E to get the energy, and you add togther p to get the momentum, but you calculate m from the new values of E and p.

    A couple of advanced points:

    The quantity (E,p) is called the energy-momentum 4 vector. The quantity E^2-p^2 , in relativistic units, is just the "length" of this vector. It is necessary and sufficient for E^2 - p^2 to be constant for the 4-vector to be Lorentz invariant. When applied to physical systems, the energy-momentum 4-vector can be defined for point particles, and for closed systems. Most introductory textbooks don't go beyond this, so I won't either, but I will add that different techniques are needed to describe the distribution of mass-energy if you have a non-closed system.
  6. Jun 22, 2014 #5
    Because it isn't a bound state, at least not when the balls pass the zero crossing point of the force function. At that point the spring has no effect whatsoever on the motion of the balls. At that point the potential energy of the system is zero, of course, but the point is the mass of the system would seem to be the same as the mass of the system when it was at rest. The momentum of the system is at all times zero, so the rest energy due to the motion of the masses would also be zero at every point in time. Am I missing something, yet?

    Getting rid of the spring, the mass of the system is the same if the two balls are at rest, moving toward each other, or moving apart, is that not true? If not, then that would seem to be what I am missing. If so, then any additional mass in the system is due strictly to the energy imparted to the massless spring. It is seeming to me it must be true, since the compressed or stretched spring has potential energy associated with it whether the balls are present and none at all when it is relaxed.
  7. Jun 22, 2014 #6


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    Nope. Thanks to kinetic energy, a ball has more energy when it is moving than when it is sitting still. :smile:

    The first ball has energy E1 > m1 c2, and the second ball has energy E2 > m2 c2. Together, they have energy E = E1 + E2 > m1 c2 + m2 c2, and the mass of the combined system is M = E/c2 > m1 + m2.
  8. Jun 22, 2014 #7


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    It is still a bound state when the ball is located at the minimum of the potential. Whether or not it is a bound state has to do with whether or not additional energy input is required to separate them, not whether or not there is a force.

    The force function can be zero and the state may be bound, and conversely the force may be large and the state may be unbound.

    I am not certain, but I think that what you are missing is the fact that the mass of a system is not equal to the sum of the masses of the constituent particles.

    Energy, momentum, and mass are related by: ##m^2 c^2 = E^2/c^2 - p^2##. In units where c=1 for example, you could have a photon with E=1 and p=1 and a photon with E=1 and p=-1. Apply the formula for each photon and you get m=0 for each.

    Now, consider the system composed of the pair of photons. For this system E=2 and p=0. Apply the formula for the system and you get m=2 for the system. So the mass of the system is greater than the sum of the masses of the constituent particles.
    Last edited: Jun 22, 2014
  9. Jun 22, 2014 #8
    The kinetic energy of each ball changes, of course. How does the kinetic energy of the system WRT an external observer change? Whatever its mass, its position in space remains fixed relative to the external observer. The Lorentz transform for the center of mass of the system thus collapses to 1, and since the velocity is zero, [itex]\frac{1}{2}mv^2[/itex] is zero at every point in time. The same is true for the momentum. Whether the relative velocity of the two masses is high enough to make the value of the Lorentz transform deviate measurably from 1 or not, the total momentum is still zero. Now, of course the kinetic energy of the two balls changes, but that is not the question, at all. I am not talking about the kinetic energy in the frame of reference of either ball. I am talking about the KE (and PE) in the frame of reference of an external observer. The same is true of a vessel filled with gas.

    That is not a given, at all. Following that reasoning, when the binding energy of the nucleons in an atom is released during atomic fission, the total mass of the system would be the same after the event as before. That is not the case. The system's mass decreases as the nuclear forces diminish, and the system goes from a bound one to an unbound one over the course of a timescale so small it must be considered in QM terms. The same thing is essentially true here, except that the timescale is much, much longer, and of course the energies are vastly lower here, and the system is cyclic. It is still going from a bound state to an unbound state (or vice-versa) as the force exerted by the spring falls to zero.

    Well, of course! That is the fundamental precept of the thought experiment. When the spring is compressed or expanded, there is a rest energy, and thus a mass, associated with the varying length of the spring, irrespective of whether it is attached to any masses or what the magnitude of those masses might be. The only mass associated with the spring at any point in time is given by m=[itex]\frac{1}{c^2}\int_0^a \vec{f}(s)\cdot\,d\vec{s}[/itex], where a is the absolute value of the difference in length of the unstressed spring and the length of the spring at the moment in question. [itex]\vec{f}(s)[/itex] is of course the vector function describing the force exerted by the spring at every value of the length of the spring. The magnitude of [itex]\vec{s}[/itex] is the length of the spring. This is completely independent of anything but the displacement of the balls. It remains true if we get rid of the balls altogether and replace them with a massless vice or pair of hands. In such a case, the mass of the system is constantly going up and down as the hands alternately stretch and compress the spring. (Yes, it has to be an open system. We can't consider what is happening to the energy as it is delivered to or from the hands.)
  10. Jun 22, 2014 #9


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    Have you ever had a course in physics? This is not rocket science. It is not even Special Relativity, it is basic Newtonian Mechanics. There's a limit to what we can explain to you in this format, and if you seriously maintain that energy is not conserved, what you need to do is pick up a physics book.
  11. Jun 22, 2014 #10
    Absolutely. No force is required to separate them, and consequently no additional energy. Detach the spring at that instant, and they will continue to move apart. Now, whether we decide it is inappropriate to consider a bound state to have a magnitude (it is certainly not valid for QM, but this isn't QM), which is to say the state becomes more bound as the force increases, or equally invalid to say the system becomes instantaneously un-bound and re-bound at the zero-crossing point is moving astray of the issue. We can detach the spring at any instance in the experiment, but prevent it from relaxing. At that point, the masses stop accelerating, continuing to go on to either collide or fly apart. There is at that point some mass associated with the spring due to its potential energy. I see no way there can be any mass associated with the motion of the two masses, other than their rest mass.

    Surely, but I see no way in this instance to associate a fixed value to the binding energy. Unless the binding energy is constant, the rest energy is not constant and thus the mass is not constant. Even taking the system to be eternally bound, when it is at rest before the experiment starts, the mass of the system consists solely of the sum of the masses of the two balls. If we stretch the spring, but do not yet release it, I believe we agree the mass of the system has increased due solely to the increase in potential energy of the spring. (We are ignoring the gravitational attraction of the two masses, of course.) When the balls are released, at the moment the balls reach the zero crossing point, all the potential energy of the spring has been converted to kinetic energy. The value of that kinetic energy is precisely equal to the potential energy of the stretched spring, but I do not see how increasing the kinetic energy of the two balls increases the rest energy of any component of the system, nor of the system as a whole. To the contrary, the only rest energy I can see changing at any point in time is that of the spring.

    Not at all. I may be missing some component of the calculation. It is going to be the sum of all the rest energies of the masses, plus any rest energy associated with any force interaction within the system.

    Of course. Photons are massless.

    Yes, of course. If one of the photons is absorbed by a small mass in the system, however, while the momentum and the total energy of the system do not change, part of the momentum of the photon is changed into kinetic energy, and the rest is changed into heat. There is no way to get around the fact the mass of the system has changed, however. The question is, has it gone down only by the amount of mass represented by [itex]\frac{KE}{c^2}[/itex], where KE is the change in kinetic energy of the mass when it absorbs the poton, or is it [itex]\frac{pc}{c^2}[/itex]? If my suppositions are correct, then it is the latter, or perhaps some value between the two.
    Last edited: Jun 22, 2014
  12. Jun 22, 2014 #11


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    This is incorrect. The mass of the SYSTEM is indeed the same before atomic fission as after. What changes is the sum of the masses of the constituent particles. See my previous post.

    Consider a nucleus of mass 1 in some units (where c=1). It could, hypothetically, fission into two products each of mass 0.3 with v = ±0.8 c. It is easy to show that the mass of the system of the two particles is still 1, although the mass of each particle is 0.3
  13. Jun 22, 2014 #12
    Apparently you missed the very first sentence in the thread:

    The fact is the number of physics classes I took was more than 20. My major was physics.

    That's an odd statement. I don't recall any Newtonian references to a mass associated with a compressed spring, or one associated with heating up a gas-filed vessel. It's true none of the velocities are high enough to be considered Relativistic, but that is beside the point. It's also true the masses involved in any real-world experiment of this nature would be immeasurably small, but that is also beside the point.

    I NEVER suggested energy is not conserved, or at least not in a closed system. I am very well aware that Conversation of Energy and Conservation of Momentum are two of the most fundamental laws of Physics. I learned that well over 40 years ago.

    Please do not treat me like a child nor an ignoramous. Just because I may be mistaken does not mean I am stupid or uneducated. I still have all the physics books from my classes in the 1970s - all the ones but a few which were ruined by a plumbing accident.
  14. Jun 22, 2014 #13


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    No, this is incorrect. Please see the formula posted above.

    This is a self contradiction. Look at the formula. If the system's momentum and energy are unchanged then so is its mass. Conservation of energy and momentum automatically imply conservation of mass since mass is a function of energy and momentum only.
    Last edited: Jun 22, 2014
  15. Jun 22, 2014 #14
    Check me if I am wrong, but that is only true if the photons produced by the the fission have momenta that all add to zero.

    Doesn't that assume no gamma photon is emitted? Nonetheless, your point is taken. Any net momentum imparted to the two particles will have to be offset by the momentum of any photons, so we are back at p=0 and m=[itex]\frac{E}{c^2}[/itex].

    Perhaps I am getting a glimmer of where the truth lies, or maybe not.

    In corresponding with Dave Knapp and other physicists on FIDOnet a couple of decades ago and based upon my earlier studies in physics, I got the impression a photon does not curve space. I believe this is still correct, since the mass of the photon is zero. If we take some number of photons whose total momentum is smaller than the sum of the magnitudes of all the momenta, then unlike the single photon, pc is much smaller than the total energy, and the residual is associated with a momentum of zero, so Einstein's equation collapses to e=m[itex]c^2[/itex]. Am I correct in concluding the observed space is curved by gravity?

    Interesting. Where would the center of mass be (assuming a non-uniform wavelength and position of the photons), and how could a gravitational field affect it? I suppose the field would red-shift some of the photons and blue-shift others, thus changing the momentum of the center of mass?

    This would seem to relate to a question I asked one of the professors in a discussion in college. (Not one of my professors, just one of the Physicists I knew.) The question in effect was, "where does the rest mass reside?" I don't mean geometrically, I mean in terms of the fabric of space. If we take a QED approach, an exchange of virtual photons applied to the components of the system obviously acts on the "solid" parts of the system (those possessing charged particles), yet the acceleration of the system includes a mass not associated with any of those parts directly.
  16. Jun 22, 2014 #15
    There is something else? What? We have point particles, we have binding energies between them, and we have massless partciles whose momenta sum to zero. What else is there?

    No, because mass is not fundamentally a function of energy. It is only a function of rest energy. That said, the assertion the rest energy of a closed system must remain constant does present an extremely strong argument. It relies on the notion that any conversion of rest energy to some other form must have products with zero momentum. I admit I am having a great deal of trouble imagining how it could ever not be the case, but I am also having trouble seeing how the motion of a pair of masses whose momentum is zero produces rest energy.
  17. Jun 22, 2014 #16


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    It is a function of the total energy in the center of momentum frame (and the factor is just c2). Kinetic energy of the balls is a part of this energy.

    In the original rest frame. Yes, this is just momentum conservation.
  18. Jun 22, 2014 #17


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    Because the energies of the two masses don't cancel each other the way their momenta do.
  19. Jun 22, 2014 #18


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    The invariant mass of a point particle IS fundamentally a function of energy and momentum. Your forty-year old physics books might not mention this point, but if you find a modern textbook on relativity it most likely will.

    A couple of references: Goldstein, "Classical Mechanics". A graduate level mechanics textbook, this has both relataivistic and non-relativistic treatments of mechanics.

    Taylor & Wheeler, "Space-time Physics". An undergraduate textbook on relataivity.

    The key equation E^2 - (pc)^2 = m^2 c^4 has already been mentioned several times, if you do a bit of reading you'll find out that we are not just making this stuff up , regardless of whether or not it was mentioned in your fourty-year-old physics texts.
  20. Jun 22, 2014 #19
    Yes, of course, or rather it is isolated once the hands release the balls. 'Same with the vessel filled with gas. It is not a closed system while the heat is being added. OF course that is not really relevant to the discussion at hand. The point is, it is not the total energy of a system, closed or not, that determines the mass. A system consisting of a single gamma photon has quite a significant energy, but no mass.

    Again, yes, of course. At any scale above the Heisenberg limit, momentum and energy are always conserved. It doesn't mean they can't change forms. Mass is but one of the forms of energy.

    I know you and the other senior members of this forum know these things all too well. Just please don't assume I don't. I am far from an expert in Relativity, which is why I'm posting these questions - along with wanting to participate in a lively discussion, of course. I know a number of professional Physicists who know comparatively little Relativity.

    I think you might want to re-state that. The first sentence is not consistent with the second, because the system specified in the first sentence is not closed while the system specified in the second is. It's not germane to the discussion at hand, however.

    Thank you.

    I see all of that, and with more consideration I may be able to figure out exactly where my conclusions have gone awry. Please don't you or anyone else feel that I am presenting myself to be more savvy than everyone else here. I do, however, demand that anyone who claims to understand something better than I convince me of where, exactly my misapprehension lies. Failure to do so means no one has taught anyone anything in the exchange. They have merely told the other to go memorize something. That's not science, it's training by authority, and it is distressingly close to being analogous to instilling religious dogma.

    To be sure, mathematics is often an excellent means of illustrating the point, but not always. After all, epicycles are a perfectly valid mathematical treatment of the observed planetary motions provided one applies an infinite number of diminishingly smaller ones. It provides absolutely no insight into the physical reality behind the observations, for all that it comes up with the correct numerical answers. For several reasons having nothing to do with the study of physics, I abandoned it as a career over 30 years ago. Since then I became an Engineer. In my field, it is the actual numerical answers that are important from a professional standpoint, not any particular understanding of the underlying paradigms. Science is different. It only cares about the actual physical values in that the theories must make accurate predictions in order to be valid. Well, sort of. Sometimes the theorist is happy if their predictions are within an order of magnitude. My boss calls me to the carpet if I am off by 1% :eek:

    In any case, I am struggling not to just come up with consistent calculations, but to be able to reconcile my understanding of the mechanisms of the universe with those equations. I'm not anywhere close to being on a par with Einstein, but I submit he clearly did the very same thing, time and time again, except that he started with his understanding and created the math to fit it. I therefore don't think it is inappropriate for me to do so, just in the reverse direction.

    By the way, I love your handle. I have some in-laws I swear are secretly from Perv.
  21. Jun 22, 2014 #20


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    The mass of an isolated system of particles is always conserved. Any particles produced, including photons, are considered part of the system.

    For example, a particle of m=1 can annihilate with its antiparticle at low velocity and produce a pair of photons with E=1 and p=±1. This conservation principle explains why annihilation reactions always produce at least two photons. It is not possible for a single photon to conserve the system mass.

    I think that for now you should probably stick with special relativity and flat spacetime. Until you understand the basics, this type of inquiry will be more confusing than valuable. When you are ready you should research p-p wave spacetimes.

    In particular you seem unaware of four-vectors. I have tried to avoid them thus far, but you will need to learn about those before you will be ready to tackle mass and energy concepts in GR.
    Last edited: Jun 22, 2014
  22. Jun 22, 2014 #21


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    The kinetic energy of the massive particles, and the energy of the massless particles.

    For a system with a well-defined rest frame then in the rest frame p=0 and the equation above reduces to ##mc=E/c##. However, not all systems have a well-defined rest frame, such as a system consisting of a single photon, or the rest frame of the system may be non-inertial, such as a rotating disk. The formula I provided above is the general formula*. It works for all systems regardless of whether or not they have an inertial rest frame.

    You may want to read the following two lectures:
    http://cas.umt.edu/physics/physics141/ch9_slides.pdf [Broken]
    http://cas.umt.edu/physics/physics141/ch10_slides.pdf [Broken]

    This is precisely how particle colliders produce massive particles. If the equation I posted above does not satisfy you then I don't know if I can give you an answer that would satisfy you about "how", but it is experimentally clear that the universe works this way to within the best experimental precision available.

    For me the "how" is quite obvious, but it relies on the concept of the four-momentum and the Minkowski norm from whence the equation for mass is derived. Please read the lectures I posted and ask follow up questions if needed.

    *EDIT: actually, the general formula is ##m^2 c^2 = g_{\mu\nu}p^{\mu}p^{\nu}##. The one I gave in post 7 works only for inertial frames in flat spacetime, this truly general one works in non-inertial frames and in curved spacetime.
    Last edited by a moderator: May 6, 2017
  23. Jun 23, 2014 #22
    How? The total energy of a single particle is given by

    [itex]E^2=p^2c^2 + m^4c^4[/itex]

    ...Woah, hold on a minute. Are you trying to claim that because E is a function of m, that m is a function of E? If so, that is just simply wrong.

    The fact a set of numbers A can be expressed as function of a second set of numbers B does NOT imply B is a function of A. Take the simple formula


    Y is a function of X, but X is not a function of Y. There is an implicit function composed of a pair of functions that relate X to Y, but no explicit function of Y in X. In Einstein's equation there isn't even any implicit function of E in m, unless the momentum is zero. Otherwise, the function for E is not separable for p and m. There is an implicit function for m composed of an infinite number of explicit functions of E as long as there is no massless element of p. In the general case, however, we cannot assert this, because there are cases where p is not strictly related only to m, and even ones where it is not related to m, at all. It's highly misleading to assert it without qualification even in the special case.

    Take a system comprised of a single point mass at rest wrt the observer and a single photon. The energy of the photon is completely independent of the mass of the stationary particle, agreed? Let the wavelength of the photon approach 0, and the Energy approaches infinity, but unless I am still totally in the woods, the mass does not change. There is a relationship between E and m, but m is not functional in E. For this case, it's perfectly constant, irrespective of the value of E.

    I'm not trying to be mathematically pedantic or intellectually obtuse, here. This genuinely sent me on an hours-long reflection trying to come up with a function of E that predicted a specific value of m for any given E. Unless I am missing some sort of variable underlying p for the general case which explicitly relates it to m, then there isn't one.

    Even given all that, I'm still not certain. Is there actually some function of E that explicitly predicts m in the general case that I am not seeing?

    Are you saying the invariant mass is not Lorentz-invariant? If so, then I'm afraid you lost me again.

    Of course you aren't! Where in any of my posts did I intimate in any way this is the case? I'm not asking anyone to justify themselves. I'm asking someone to explain it, or at least point me in the right direction. The suggested reading might do it, but so far I haven't gotten all the way there.

    It most certainly was, and I have never forgotten it, certainly not during this discussion. The fact one can calculate the value of m given the values of p and E does not mean m is a function of E. It's not even true if we limit it to the special case where


    In the spring system I proposed, I do see a glimmer of something in the formulation:

    [itex]E = \sqrt{\frac{m_1^2v_1^2}{\sqrt{1-\frac{v_1^2}{c^2}}}+m_1^2c^2} + \sqrt{\frac{m_2^2v_2^2}{\sqrt{1-\frac{v_2^2}{c^2}}}+m_2^2c^2} + \int_0^a \vec{f}(s)\cdot\,d\vec{s} [/itex]

    I'm too tired to expand it now, so I will get back to it later in the week or next weekend. ( I hope I got the formulation right. I can barely see or think at this point.)
    Last edited: Jun 23, 2014
  24. Jun 23, 2014 #23


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    That is the same equation that I pointed out in post 7 (except for your typo on the exponent for m).

    Frankly, it doesn't matter to me which terms you call functions of the other terms, or if you just want to say that they are all related by the equation or use some other English description of the math. As long as you consistently use that equation then your question in the OP is resolved.

    Also, IMO your objection (though correct in general) is not relevant to the case at hand. E and m are both always non-negative for all known particles so you can indeed express each as a function of the other and momentum.
    Last edited: Jun 23, 2014
  25. Jun 23, 2014 #24


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    The invariant mass of each individual ball in your "two balls and a spring" scenario is Lorentz invariant. However, the invariant mass of the *total system* is *not* just the sum of the invariant masses of the two balls. It is not even the sum of the invariant masses of the two balls, plus the potential energy of the system. You have to count the kinetic energies of the balls, with respect to the system's center of mass frame (the frame in which the net momentum of the balls is zero). The reason is that, even though the invariant masses of the two balls are constant, the "spacetime direction" in which their 4-momentum vectors point is not; so when you add their 4-momentum vectors to get the balls' contribution to the system's total 4-momentum, the addition works differently depending on the system's internal state.

    To amplify on the above, I'll walk through how the addition goes for two cases: (1) spring fully extended, balls momentarily at rest relative to each other; (2) spring fully relaxed, balls moving at maximum velocity towards each other. (Case 3, spring fully compressed and balls momentarily at rest, works similarly to case 1, and case 4, spring fully relaxed, balls moving at maximum velocity away from each other, works similarly to case 2.) Everything is written relative to the frame in which the balls' net momentum is zero, i.e., the system center of mass frame.

    The general formula is ##P^{\mu} = p_1^{\mu} + p_2^{\mu} + V^{\mu}##, where ##P^{\mu}## is the total 4-momentum of the system, ##p_1^{\mu}## is the 4-momentum of ball #1, ##p_2^{\mu}## is the total 4-momentum of ball #2, and ##V^{\mu}## is the potential energy due to the spring, written as a 4-vector. In the system center of mass frame, we have ##P^{\mu} = \left( E, 0, 0, 0 \right)##, i.e., all spatial components zero, and ##E## is the total energy of the system, which is conserved; and ##V^{\mu} = \left( V, 0, 0, 0 \right)##, where ##V## is the potential energy due to the spring as we usually think of it, i.e., just a function of the spring's displacement. So in this frame, the spatial components of ##p_1^{\mu}## and ##p_2^{\mu}## must cancel, and we are left with their time components only. But those time components are just the *total* energy of ball #1 and ball #2 in the system center of mass frame, i.e., *including* their kinetic energies as well as their rest energies.

    So for the two cases given above, this gives:

    (1) Both balls are at rest, so no kinetic energy; and the spring is fully extended, so its potential energy is the maximum possible, which we'll just call ##V_{max}##:

    E = m_1 + m_2 + V_{max}

    (2) Both balls are in motion at maximum possible velocity, ##v_{max}##, so each one's total energy includes the maximum possible relativistic gamma factor; and the spring is fully relaxed, so zero potential energy:

    E = \frac{1}{\sqrt{1 - v_{max}^2}} \left( m_1 + m_2 \right)

    These two values of ##E## must be the same, since ##E## is conserved. But they can be, because of the gamma factor on the RHS of the second equation. From the system point of view, the extra "internal energy" over and above the rest energies of the two balls is being converted back and forth between potential energy in the spring and kinetic energies of the balls. Each individual ball still satisfies ##E^2 = p^2 + m^2## (note that I'm using units where ##c = 1##); but the 4-momentum vectors of the two balls "point in different directions" in spacetime in case #2 above, where the balls are in relative motion, so their kinetic energies can add even though their momenta cancel out.
  26. Jun 25, 2014 #25
    Is WRT a standard physics expression?

    Of what?

    And isn't the actual input of energy the initial squeeze or stretch, and not the cycling of balls? Isn't that that the actual source of mass under discussion? And if it is so that the Balls change mass as each is shuffled to and fro in harmonic reciprocity, is it not also true that the mass of the spring is also changing equally and oppositely?

    AH! But the spring is MASSLESS! So that cannot be! (WINK!)

    Where, pray tell, can I get one of those massless springs?

    This is a game of Three Card Monte, for physics students.
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