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Several Questions Concerning Mass-Energy

  1. Jun 17, 2012 #1
    I have a few questions concerning mass that I'm hoping some kind soul more knowledgeable than myself can answer for me. Most of these concern mass energy equivalence, although two, I think, are classical questions that I have not even had to consider for several years (although they just passed me through modern physics with an A about a week ago, scary). If I should break this post in two and post the more classical questions in the other forum, please let me know.

    1. Say I have some object, for example lets say a baseball (or an atom w/e). Then lets say, from my frame of reference the ball is stationary. Then from what I've learned I would say that the energy of the ball is related to its mass by E = mc^2. Now if I think about it, the ball is composed of a lot of electrons, protons, and neutrons, and none of these are at rest. Does this mean that the sum of the kinetic and potential energies of all of these particles would be equal to mc^2? How do values measured in different 'levels of scale' relate to one another when looking at an object/system?

    2. This is related to question 1. If you say an object is 'at rest' with respect to some observer, does that just mean its center of mass has fixed space coordinates with respect to that observer or does it mean something else (since the individual components of the objects are not at rest)?

    3. In regards to "mass-energy equivalence", if I take an object and heat it up, does its mass increase? Lets say I have a sealed box full of gas. If I heat the gas up will the mass of the system be greater? Would it weigh more on a hypothetical perfect accuracy scale? On that note, if I trapped light in a box, would the box with light trapped in it have more mass than an identical box without light trapped in it? I've read the FAQ, and i'm still a little fuzzy on what 'effective mass' is for light, as light has 0 rest mass.

    4. This is related to question 3. If a system having more energy implies that it has more mass, does this mean that if I had an isolated system of two particles each with nonzero mass and zero charge, then if the objects were further away the system would have more mass than if the objects were closer together since there would be more gravitational potential energy? If so, how is that mass to be interpreted? If not, why not?

    5. If light has some sort of "effective mass", since it has momentum, can you apply a force to light? I'm thinking the answer is 'no' as light, I think always has 0 acceleration meaning that no force is ever applied to it. However I'm not sure about this, I mean can light rotate? On that note if a force cannot be applied to light, then wouldn't that mean it's momentum would have to be an unchanging constant as F = dp/dt and F = 0 (in this case)?

    Any help, information, or even redirection to where it can be obtained, would be extremely appreciated.

    Thank you for your time and consideration.


  2. jcsd
  3. Jun 17, 2012 #2
    Yes, it does. But, regarding the mass of the atom, its biggest contribution comes from the mass of the nucleus. The total rest mass of the electrons is about [itex]Z/(1850 \, A)[/itex] of the mass of the nucleons. Then, there is the binding energy of the electrons in the atom. It is the sum of the kinetic and the potential energy in the electric field of each electron. It is supposed to be negative (meaning the potential energy due to attraction to the nucleus dominates the kinetic energy of the electrons and the repulsion between them), making the system bound.

    When you go down to the nucleus, the mass of the nucleus is equal the sum of the rest masses of the protons and neutrons (nucleons), plus the binding energy of the nucleus, which is also negative. Again, the rest mass of the nucleons has a dominant contribution.

    Going back to the proton, or the neutron, they are composed of three quarks, which interchange self-interacting gluons. At this point, the binding energy of the system is much bigger than the sum of the rest masses of the quarks.

    Yes. Although, in Relativity it is better to use the term "Center of Inertia", which is a reference frame in which the total linear momentum of the system is zero.

    Yes, for all questions. In fact, Einstein proved the [itex]E = m c^{2}[/itex] by considering the absorption and emission of photons.

    Yes, the increase in mass is due to the work you employed in moving the attracting masses apart.

    Well, you can have constant speed, but accelerated motion, if the direction of the velocity (and, momentum) changes.

    However, photons do not feel forces simply because they are not charged, and because gravitation is considered a pseudo-force according to GR instead of a real force that requires a coupling constant.
  4. Jun 17, 2012 #3


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    I think the usual term is "center of momentum" frame which happens to have the same acronym "COM" as "center of mass". I've also seen "zero-momentum frame" which more explicitly relates to the definition.
  5. Jun 17, 2012 #4


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    Hi, StillNihilist (interesting handle that), and welcome to PF! These all look like questions that are appropriate for this forum.

    Yes. One terminology point: "mass" here means "rest mass", also called "invariant mass".

    Basically, yes. See further comments under question 2 below.

    If you are considering the object as a single system, then yes, its center of mass is what counts when you are determining whether the object, as a single system, is "at rest". That means that, when you are determining the rest mass of the system, as in #1 above, you do it in the center of mass (CoM) frame; all the energies of the individual parts, which may be moving in the CoM frame, are evaluated in that frame. This is important because energy is frame-dependent; it changes depending on your state of motion relative to the object whose energy you are measuring. (There is, actually, a frame-independent way to describe what I just described above, but I won't go into that now.)

    If you mean "rest mass", yes. This goes for both of the specific examples you gave (a box of gas and a "box of light").

    In situations where it makes sense to view light as having an "effective mass" (such as when you have light confined in a box and you want to know the total rest mass of the box), its effective mass is just its energy divided by c^2. (Note that the energy has to be measured in the CoM frame of the system as a whole--for example, of the box that contains the light.) This might seem confusing because you have a bunch of light, which has zero rest mass, somehow adding up to something that has nonzero rest mass. That's because rest mass is not additive in relativity; the thing that is additive is total energy (in a given frame). So you have to be careful to distinguish them.

    If you looked at it from the CoM frame of the total system (the two objects combined), yes, part of the contribution of the total invariant mass of the system as a whole would be the gravitational potential energy. Moving the two objects further apart would increase the gravitational potential energy, and therefore the total mass. (The added mass would come from the energy that was expended to move the objects further apart.)

    However, you have to be careful about interpreting such scenarios. Two objects with nonzero mass and zero charge will fall towards each other due to their mutual gravity, and as they do so, the total mass of the system, seen from far away, won't change: the gravitational potential energy decreases, but the total kinetic energy of the two objects (in the CoM frame) increases by the same amount. So the contribution to the total mass from gravitational potential energy won't stay the same for very long, if the objects are allowed to move freely.

    Yes. (Technically, it's the light's energy that can be considered "effective mass" in certain situations--see above for comments on that--but the light's energy is just its momentum times c, so they're basically the same thing.) Applying a force just means changing the light's momentum, and there are ways of doing that, though they are not commonly encountered. For example, check out Compton scattering:


    In the sense that photons, particles of light, have spin in the quantum mechanical sense, yes. I don't know if that's the sense of "rotate" you meant, though. The spin of a photon can't be changed; it always has the same magnitude, so it can't really be viewed as the result of applying a torque to the photon.
  6. Jun 18, 2012 #5
    Thanks for the replies. They were all much quicker than I thought, and all very helpful. I guess I have one follow up question to what Peter was saying "Note that the energy has to be measured in the CoM frame of the system as a whole". I thought that the useful part of rest mass was that it was invariant under the Lorentz Transformations? So using the relation:

    m^2c^4 = (E/c)^2 - (pc)^2

    The value m^2c^4, would always be the same regardless of what frame of reference you measured E and p in? Which would of course mean that m (obtained from that relation) would be the same regardless of which frames you obtained you E,p pair from? Or did you mean that if you use E = mc^2 then it had better be in the CoM frame, because you are saying p = 0 when you write E = mc^2?

    Sometimes I seem to get mixed up on 'proper' quantities, 'relativistic' quantities, and "invariant" quantities. I mean, I think proper just means something directly measured by an observer with respect to his own frame, in which his own velocity is 0 (or maybe those are 'ordinary' quantities). Relativistic quantities I think were modifications to classical definitions of energy, momentum, etc.. to make sure they are conserved in special relativity, or at least make putting them through Lorentz transforms easier. Where as invariant just means that it has the same value in all inertial reference frames.

    All of that said, when I'm looking at formulas on an irregular basis I think I sometimes forget if that E/t/m/etc... is ordinary or proper. So like if I see something like E = mc^2, I stop and go, "well do they mean 'm' as a direct measurement with some instrument in the frame of reference in which the object is at rest, or do they mean m as in m/(1 - v^2/c^2)^(1/2).

    Anyway, I really appreciate all the help. It's very... helpful.
  7. Jun 18, 2012 #6


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    Yes, m is invariant, but unless you are in the rest frame of the object (the CoM frame if the object has internal parts), m is not equal to the object's energy (divided by c^2). As you can see from the formula, if you are working in a frame in which the object is moving, you have to know both the object's energy and its momentum to calculate m.


    The terminology is not always used consistently, so it's easy to get confused. When in doubt, my advice is to look at the math, which forces you to be unambiguous about what you are talking about.
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