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I Questions about binding energy

  1. Oct 12, 2016 #1
    Two of them, actually:

    1) My understanding is that rest energy, as an invariant, is an "absolute" or "fixed" quantity. Everyone agrees on "how much" there is. But I'm also accustomed to thinking of potential energy as a quantity without a fixed value, so that we only care about the difference in potential energy between two different configurations. Since potential energy within a system contributes to the system's rest energy, something's got to give here, right? Does the mass defect allow us to assign "fixed" value to potential energy, too?

    2) If binding energy is negative,* then why is it said that some 99% of the rest energy of a proton comes from the kinetic and potential energies associated with its constituent quarks (with the rest energies of the quarks contributing only about 1%)? Wouldn't the strong nuclear potential energy associated with the quarks decrease the proton's rest energy?

    *Or if you prefer the convention that calls binding energy positive, read this as "If binding energy must be subtracted to calculate a bound system's rest energy ..."
     
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  3. Oct 12, 2016 #2

    Simon Bridge

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    The rest energy is not a "fixed" value - that is not a useful way to think of it - it is invariant in that all observers agree that if the object were at rest in their frame, it will have that energy. Since the mass deficit is a difference in rest masses, then it follows that all observers will agree about that too.

    ... for the force inside a nucleon as opposed to that between nucleons?
    http://hyperphysics.phy-astr.gsu.edu/hbase/forces/funfor.html

    I think your questions are centered around this: What is the definition of potential energy?
     
  4. Oct 12, 2016 #3
    By "fixed" I didn't mean unchangeable. Poor word choice. I understand that the mass of an open system can change. My point is that we can talk about an absolute "amount" of rest energy. We're not limited to talking about differences in energy here. But if we can't say the same for the potential energy within the system (that contributes to the system's "absolute" rest energy), then isn't there a contradiction? I'm sure I'm missing something. That's why I'm asking about it.

    And yes, in my second question I was talking about a single nucleon as a system. I've read that about 99% of the mass of a proton comes from the kinetic and potential energy "within" it. Is this potential energy not binding energy, and thus negative (or to be subtracted)?
     
  5. Oct 12, 2016 #4

    pervect

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    I believe the actual situation is a bit more complicated. In Newtonian theory, both rest energy and potential energy would be relative quantities. However, rest energy is rarely (perhaps never?) used as such in Newtonian physics - just because it is totally arbitrary, there's no need to give it a name.

    The idea of rest energy in Special Relativity is different as you note. One doesn't really have a choice anymore - "everyone agrees on how much", as you say. This arises from an important relationship between energy and momentum and mass that doesn't exist in Newtonian theory. The mathematical form of this relationship says that ##E^2 - p^2 = m^2##, where E is energy, p is momentum, and m is mass, and we've assumed that c=1. Chosing units where c=1 makes this relationship simpler, but it might be confusing. If one desires a more general relationship that does not require setting c=1, it becomes ##E^2 - (pc)^2 = (mc^2)^2##. Use whichever approach seems simpler to you, they should be equivalent.

    This relationship will be true only for one value of E, as can be seen by setting the momentum p, equal to zero. When we set p=0, we wind up with the iconic relationship E=mc^2, which defines rest energy. The point is that this concept is no longer Newtonian - the idea that E is fixed comes from relativity, and its origin is an idea that may seem odd at first glance, the idea that energy and momentum should be considered as two parts of something larger, in a manner similar to the way that we talk about "space-time" in special relativity as a unified concept, rather than separate unrelated concepts of space and time.
     
  6. Oct 12, 2016 #5

    PeterDonis

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    No, but there is a need to be very careful about describing the actual physics involved.

    First, the concept of "potential energy" only applies to a stationary system--roughly, one whose properties as seen from far away don't change with time. So we are not talking about a completely general concept; we're only talking about a particular class of systems.

    Second, for that class of systems, the rest energy is "absolute"--i.e., the rest energy itself is a meaningful number, not just differences in it--only in the sense that, as seen from far away, the system as a whole has a 4-momentum vector with an invariant length, which is the rest energy. But this rest energy is not, in general, the sum of rest energies of components of the system. It's not an absolute "amount of stuff" contained in the system. It's just an external property the system has.

    Third, you are correct that, for this special class of systems, the potential energy contributes to the rest energy--but only if we define the "zero point" of potential energy properly. The proper definition is that potential energy goes to zero at spatial infinity, i.e., very far away from the system, which must, on this view, be confined to a finite region of space. Only on this definition does potential energy make a (negative, classically--but see below for some complications when quantum effects are brought in) contribution to the system's rest energy as seen from far away.

    It might help to consider a concrete example. Suppose we have a huge cloud of matter that collapses gravitationally into a small spherical object. The cloud starts out spread out over a very large region of space, with each individual piece of matter in the cloud at rest relative to all the others. The spherical object that is the endpoint of the collapse also has all individual pieces of matter at rest, and is at zero temperature, i.e., all possible energy that can be radiated away has been radiated away to infinity. However, no matter escapes during this process; all of the matter particles that were in the initial cloud are in the final spherical object. We can then define a "rest frame" for this entire process (the frame in which both the initial cloud and the final spherical object are at rest), and in this frame, the initial mass ##M_i## of the cloud will be larger than the final mass ##M_f## of the spherical object. The difference ##M_i - M_f## is the "binding energy" of the object, and is also the energy that must have been released as radiation to infinity during the collapse process. (Whether we define the sign of the binding energy as positive or negative is just a convention; either way the physical fact that ##M_i > M_f## is the same.)

    This case is a bit more complicated than the example I gave above, because now we are allowing for individual constituents of the system to move (as seen in the rest frame of the system as a whole). Also, there is a terminology issue here (not yours but in the pop science descriptions) regarding "potential energy".

    Allowing for the motion of individual constituents is easy to add to the model, because, as above, we have a rest frame defined for the entire system, so all we have to do is calculate the relativistic energy of each constituent in that frame, i.e., we calculate ##\gamma m## for each constituent, where ##m## is the constituent's rest mass and ##\gamma## is its relativistic gamma factor as evaluated in the system rest frame.

    Allowing for "potential energy" in the sense I defined it above--where it goes to zero at infinity--for each constituent is also easy to add to the model, because we can treat each constituent as being in a bound orbit in the system's rest frame, and any such bound orbit has a well-defined sum of kinetic and potential energy. In the simplest case the orbit is circular and the kinetic and potential energies are both constant (the first positive and the second negative), so we can consider them separately; but even for elliptical orbits we can consider their sum, which is a constant of the motion (and will be positive if we include the constituent's rest mass as above).

    (Note that in the case of a nucleon, the "orbits" are not classical orbits but ground state quantum energy levels; but the underlying principle is the same.)

    However, the term "potential energy" in the case of nucleons does not refer to a simple attractive potential as we have been discussing so far (where potential energy is zero at infinity and gets more and more negative as you get closer and closer to the center of mass of the system). The exact potential function for the strong interaction is not known, but it is known that it becomes repulsive at very short range, which means the "potential energy" of a quark inside a nucleon can actually be positive, not negative, for certain portions of its "orbit", which is actually, as above, a ground state quantum energy level--a more precise way of saying what I just said is that the quark has an amplitude to have a positive potential energy as well as a negative one. Even this is still a highly heuristic description; but the key point is that, when we are dealing with quantum systems of this sort, terms like "potential energy" and "binding energy" become fuzzy, and there can be contributions to the total rest energy of the system, as seen from far away, that have no simple classical interpretation such as we have been discussing.
     
  7. Oct 12, 2016 #6
    So if our system consists of two objects, we can speak of an "absolute amount" of potential energy within the system that contributes to the system's rest energy, as long as we define the "zero point" of that potential energy as the configuration in which our two objects are infinitely far apart? (Assuming "normal" inverse-square-law fields, as with Newtonian gravity or electromagnetism.) And this potential energy will indeed always be a negative contribution to the system's rest energy?

    And if I'm catching your drift here, hadronic quark-and-gluon soup is different because the strong nuclear force is weird and gets repulsive at close range, and that's why "potential energy" in a nucleon is positive?

    The electrostatic force can also be repulsive (for like charges). Does potential energy in that case likewise have a positive contribution to a system's rest energy? (Sorry if I'm going way off-course here.)

    I'm not sure I quite understand what you mean. Maybe the word "properties" is confusing me. If I took, say, the moon and Earth as my system, would I not say that my system has gravitational potential energy?

    Thanks for taking the time here, PD (and everyone else).
     
  8. Oct 12, 2016 #7

    PeterDonis

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    No. If the objects are infinitely far apart, then the system as a whole is not confined to a finite region of space.

    If the system consisting of two objects is stationary and of finite size (for example, two objects orbiting at finite distance about their mutual center of mass), then a potential energy can be defined for the system as a whole, with a "zero point" infinitely far away from the two objects.

    For the case I just described (two objects orbiting at finite distance about their mutual center of mass), yes. In general, this will be true for classical bound systems. But it might not be for quantum systems; see below.

    Heuristically, sort of. A more accurate statement would be that this is a quantum system and doesn't have a classical potential energy; it has contributions to its total rest energy that have no simple classical interpretation.

    Yes, but in this classical case, there would have to be some other interaction present that constrained the charges; otherwise the system would not be stationary--the charges would just fly apart. The potential energy associated with that other interaction would be negative, and the overall contribution of both potential energies combined to the system's rest energy would be negative. (If it weren't, the system would again not be stationary--it would fly apart.)

    Once again, all this is assuming a classical system. Quantum systems are weird and don't really work this way. See above.

    Yes, because the moon and Earth together constitute a stationary system--two objects orbiting their mutual center of mass. (We are idealizing away all perturbations from other objects like the Sun and other planets--essentially we are assuming the Earth and moon are alone in the universe.) "Stationary" does not mean nothing can change with time; it just means that whatever changes are happening are periodic, so the system returns to the same state repeatedly.

    An example of a non-stationary system would be the two like charges flying apart (no other forces acting). This system is changing and the change is not periodic; it does not return to the same state repeatedly.
     
  9. Oct 12, 2016 #8
    So if we have a system of two electrons flying apart, we can't speak of the electrostatic potential energy of that system?

    But surely we can still speak of the rest energy (mass) of that system, which is just the total energy of the system in its center-of-momentum frame. Right? So how would we find the total energy of the system in that frame? We'd account for the rest energy of each individual electron, and then we'd add their kinetic energies (increasing as they accelerate away from each other, but at any given moment there's a value), and finally we'd have to add some energy contribution associated with their relative positions and like charge that decreases as the electrons fly apart so as to counterbalance the increase in kinetic energy (otherwise energy ain't conserved).

    If that last energy contribution isn't "potential" energy, what is it?

    I can't tell whether my misunderstanding is semantic or physical.
     
  10. Oct 12, 2016 #9

    PeterDonis

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    Not "electrostatic" because the system isn't static.

    Technically, yes, you can define a (positive) potential energy for the system as a whole, which gradually decreases as the electrons fly apart (and their kinetic energy correspondingly increases); so technically we can view the system as a whole as having a rest energy which is greater than the sum of the rest energies of the individual electrons. But this system is not a bound system and the usefulness of such a viewpoint is limited.
     
  11. Oct 12, 2016 #10
    Thank you very much, I think I'm getting most of this.

    Usefulness and quantum weirdness aside, is it fair to say that "potential energy" associated with a repulsive force is a positive contribution to a system's rest energy, whereas "potential energy" associated with an attractive force is a negative contribution to a system's rest energy? (And this is only accurate heuristically where QCD or QM in general is concerned?)

    I'm using scare quotes because I'm invoking a broader definition of potential energy than you prefer.

    I'm still unsure about this "zero point" bit, though:

    Sure, the objects couldn't actually be infinitely far apart, but to check my understanding: would it be accurate to say that, for an attractive force, the potential-energy contribution to the rest energy of a two-body system approaches zero in the limit that the distance between the two objects approaches infinity? (I'm regarding the distance as an initial condition.) And could we say the opposite for a repulsive force?
     
  12. Oct 12, 2016 #11

    PeterDonis

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    Yes. One way of thinking of this is, again, to consider a simple example of two objects starting at rest at infinite (very large) separation, and ending at rest at some finite separation. If the force between them is attractive, this process will emit energy; whereas if the force between them is repulsive, it will consume energy (i.e., it will take energy from some other source to push the objects closer together).

    Heuristically, yes. (There are complications if we include repulsive forces, since the system can't be bound in this case, as I mentioned before.)

    No. You would say the same for a repulsive force. More precisely, you would say the same assuming that in both cases the force weakens with increasing distance (e.g., the usual inverse square force), ultimately vanishing at infinite separation. The difference between an attractive and a repulsive force is that for an attractive force, the potential energy approaches zero from below as the distance increases, whereas for a repulsive force the potential energy approaches zero from above as the distance increases.
     
  13. Oct 13, 2016 #12
    You're a mensch.
     
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