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Rest Mass and Temperature

  1. May 17, 2014 #1
    I'm a high school chem and physics teacher, participating in a recreational online chem course hosted by MIT.

    One of the practice questions in this course asked how many atoms a certain mass would contain at 300K. This got me thinking about mass-energy equivalence, and whether the atomic mass units listed on the periodic table are defined according to a specific temperature, analogous to the way other values are defined at STP. This would mean that the number of atoms in a sample of mass X would vary inversely with the temperature of X.

    My research led me to these conclusions: (and I know this is an incomplete understanding)...temperature is kinetic energy. Kinetic energy can alter the apparent mass of a system because gravity couples to momentum and energy, not just mass. Thus the mass may appear to change, but this is simply the aggregate effect of the (unchanged) mass and the kinetic energy. I found several sources that support this view, or at the very least, the statement that non-relativistic changes in temperature/kinetic energy will alter mass, although the values may be too small for us to measure with existing tools.

    However, the response from the class's forum, including an MIT student administrating the forum, is that "mass is NOT temperature dependent" and will in no way affect it. I'm not sure if I'm completely misinterpreting mass-energy equivalence, but this doesn't make sense to me.
     
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  3. May 17, 2014 #2

    WannabeNewton

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    Energy certainly is temperature dependent but rest mass is not temperature dependent. Don't confuse rest mass with the energy of a particle as the two are only equal in the rest frame of the particle.
     
  4. May 17, 2014 #3

    PeterDonis

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    To follow up what WannabeNewton said, atomic mass units are units of rest mass, so they are independent of temperature. However, there is another wrinkle that's worth mentioning. You say:

    More precisely, the internal motions of the parts of a system can increase the *rest* mass of the system. For example, consider a solid object made up of lots of atoms in a crystal lattice. These atoms are not completely stationary; they are constantly undergoing small oscillations around their equilibrium positions in the lattice. If we increase the temperature of the object, the kinetic energy in these oscillations increases, and this shows up as a (very small) increase in the total rest mass of the object. This is probably what was being referred to by the sources you mentioned that seemed to support the view that changes in kinetic energy will change mass.

    Note that this change in the rest mass of the solid object does *not* change the number of atoms in the object, nor does it change the rest masses of the individual atoms. It only changes the kinetic energy of the atom's oscillations about their equilibrium positions in the crystal lattice. But those oscillations do not produce any net motion of the object as a whole, so their kinetic energy contributes to the object's rest mass.
     
  5. May 17, 2014 #4
    Thank you, it sounds like we share the same understanding; the atoms retain a fixed individual rest mass, while the rest mass of the solid object increases with energy. And, yes, this does not change the number of atoms in the object, but (and this was the crux of my original inquiry) it would change the number of atoms necessary to produce a solid object with mass X. Or at least, I'm hoping that's a sensible conclusion. It seems that others are arguing that the very small change in rest mass is, in fact, nonexistant.
     
  6. May 17, 2014 #5

    xox

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    The total energy of a system of particles is:

    [tex]E=c^2\Sigma {\gamma_i m_i}[/tex]

    The total momentum of a system of particles is:

    [tex]\vec{p}=\Sigma{\gamma_i m_i \vec{v_i}}[/tex]

    The total rest mass [itex]M[/itex] of a system of particles satisfies the relationship:

    [tex]E^2-(\vec{p}c)^2=M^2c^4[/tex]

    We can always find [itex]\vec{V}[/itex] such that:

    [tex]E=\gamma(V) M c^2[/tex]
    [tex]\vec{p}=\gamma(V) M \vec{V}[/tex]

    Obviously, from the above:

    [tex]\gamma(V)M=\Sigma {\gamma_i m_i}[/tex]

    [tex]\vec{V}=\frac{\Sigma {\gamma_i m_i \vec{v_i}}}{\Sigma {\gamma_i m_i }}[/tex]

    and

    [tex]M=\frac{\Sigma{\gamma_i m_i}}{\gamma(V)}[/tex]

    What Peter Donis has been telling you is that we can prove that:

    [tex]M \ge \Sigma{m_i}[/tex]

    You can try to prove the above for the case [itex]i=2[/itex], it is pretty easy.
     
    Last edited: May 17, 2014
  7. May 17, 2014 #6

    Astronuc

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    This interpretation is not quite sensible.
    Or rather, negligible.

    To put the matter into perspective, consider the energy equivalence of 300K with respect to electron binding energies and nuclear rest mass.

    1 eV is equivalent to a temperature of ~11604.52 K. (from NIST)
    http://physics.nist.gov/cuu/Constants/energy.html

    300 K is equivalent to 300 K/(11604.52 K/eV) = 0.02585 eV.

    Compare that to an electron binding energy, which is on the order to eVs. For example, the ionization energy of the electron in a hydrogen atom is ~13.6 eV.

    The rest energy of an electron is ~ 511 keV.

    The rest energy equivalence of an atomic mass unit is 931.494 MeV, so the contribution of 300 K to an atomic mass unit would be 1 part in 36 billion (3.6 E10), so it would be negligible.

    But associating kinetic energy of a mass with a change in it's rest mass is not correct.
     
  8. May 17, 2014 #7

    Bill_K

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    But since a typical sample contains ~1023 atoms, a change of one part in 36 billion means that ~1013 fewer atoms will be necessary. Not so negligible!
     
  9. May 17, 2014 #8
    Yes, it seems like some of the answers I've been reading elsewhere do not distinguish between "there is a negligible effect" and "there is NO effect". The class's original question referred to 13g of gold, so the result, according to these values, would have been trivial on the macroscopic scale but nontrivial on the atomic.

    Ok, this is where I assumed my understanding was breaking down. From my understanding of the equations that xox posted (thank you, xox), and what PeterDonis wrote, momentum and velocity are included in the consideration of total rest mass M, which is an aggregate of mass and energy and does not alter the original rest mass mi of the constituent particles. This appears to be the divergent point: is there something that forbids kinetic energy from contributing to M?
     
  10. May 17, 2014 #9

    xox

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    Your understanding is correct. The mass of a system of particles is NOT equal to the sum of the masses of the particles composing the system. It is always larger than that sum. By how much it is larger is a function of the energy of the particles. You can try to prove the above for the case [itex]i=2[/itex], it is pretty easy.
    In general , I shy away from "literary" descriptions of physics and I ted to mathematical descriptions. "Literature" can be very imprecise, math, on the other hand, is very precise. (see my answer to your question).
     
    Last edited: May 17, 2014
  11. May 17, 2014 #10

    jtbell

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    If the particles are all at rest with respect to each other, that is, if there is an inertial reference frame in which all the particles are at rest; and the system does not have any binding energy; then the mass of the system does equal the sum of the masses of the particles.
     
  12. May 17, 2014 #11

    xox

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    Yes, this is the ONLY case when [itex]M=\Sigma{m_i}[/itex]. It is very improbable for such a frame to exist since it would require that all particles have the same velocity.
    In ALL other cases [itex]M>\Sigma{m_i}[/itex].
     
  13. May 17, 2014 #12

    PeterDonis

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    For a single object, no. But associating kinetic energy of *parts* of a system with the rest mass of the *system*, which is what I was describing, *is* correct; the rest mass of the system as a whole is *not* the sum of the rest masses of all the parts, if the parts are in relative motion (i.e., if the parts are moving in the center of mass frame of the system as a whole).
     
  14. May 17, 2014 #13

    xox

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    In the improbable (but not totally impossible) case when all particle velocities are equal:


    [tex]\vec{V}=\frac{\Sigma {\gamma_i m_i \vec{v_i}}}{\Sigma {\gamma_i m_i }}=\vec{v}[/tex]

    resulting into:

    [tex]M=\frac{\Sigma{\gamma_i m_i}}{\gamma(V)}=\Sigma{m_i}[/tex]
     
  15. May 17, 2014 #14

    jtbell

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    After I wrote my last post I went out for a walk, during which I realized that I hadn't fully appreciated the impact of my comment about binding energy, because I was focusing on the ∑mi = M case.

    Binding energy actually subtracts from the sum of the masses of the individual particles in a bound system! The most prominent example is an atomic nucleus, whose mass is less than the sum of the masses of the protons and neutrons that comprise it. In order to "rip" a nucleus apart completely into protons and neutrons, you have to supply at least enough energy to make up that difference in mass.
     
  16. May 17, 2014 #15

    xox

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    While it is true that binding energy is the difference between the sum of the energies of the particles composing a nucleus and the total energy of the nucleus, the OP is about a system of atoms, binding energy operates only at the level of the nucleus. Besides, the part that you posted about all components having the same velocity indicated that you knew that the discussion was about a system of atoms, you added the part about the binding energy later.
     
    Last edited: May 17, 2014
  17. May 17, 2014 #16

    PeterDonis

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    No, it doesn't. The electrons in the atom have a negative binding energy when bound to the nucleus, compared to free electrons. So the rest mass of the atom is smaller than the sum of the rest masses of the nucleus and the same number of free electrons. It's true that the difference is much smaller than the difference in the case of a nucleus (nuclear binding energies are in the millions of eV, whereas atomic binding energies are a few eV, or about a million times smaller); but it's still not zero.

    Also, atoms in a bound object like a solid have negative binding energy relative to identical atoms that are free. So the rest mass of the solid object is smaller than the sum of the rest masses of its constituent atoms. These binding energies (i.e., chemical bond energies) are typically an eV or less, so somewhat smaller than atomic binding energies; but again, that's still not zero.
     
    Last edited: May 17, 2014
  18. May 17, 2014 #17

    xox

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    I stand corrected, nevertheless the point is that the OP is about the system of atoms, not about the energies inside the atoms.

    I would take exception from the above language, since a better way is to express the above in terms of energy, not "rest mass".
     
  19. May 17, 2014 #18

    PeterDonis

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    Binding energy applies there too; see the edit I just made to my post.
     
  20. May 17, 2014 #19

    PeterDonis

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    Rest mass *is* energy; which term is more appropriate is a matter of taste and convention, not physics. I would say that if we are making measurements of mass, such as weighing something on a balance, the term "mass" is more appropriate. Given that convention, the statement I made is correct as it stands; it can be confirmed directly by sufficiently accurate measurements of the respective rest masses, in the same way we normally measure mass (not the way we measure energy).
     
  21. May 17, 2014 #20

    xox

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    "rest mass" is rest energy ([itex]m_0c^2[/itex]). Energy is [itex]\gamma m_0 c^2[/itex]. The particles constituting an atom are never at rest.
     
    Last edited: May 17, 2014
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