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As far as my understanding goes, thermodynamics only allows us to calculate changes in internal energy, not the absolute quantity itself. But according to Einstein's mass energy equivalence, the mass of an object is due to energy itself. So if we were to use the rest mass in $E=mc^2$, wouldn't that give us the absolute internal energy?

One potential problem with this could be that internal energy only includes kinetic energies of and potential energies betweenatoms/molecules, whereas the rest mass energy would include all energies; kinetic and potential energies of the electrons, nuclear potential energies of the nucleons etc.. But a simple work around could be to just include all of these in the definition of internal energy; it makes more sense to include ALL of the "internal" energies instead only those associated with atoms, and .thermodynamics wouldn't be affected either since it only cared about changes.

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  • $\begingroup$ Kind of related (and certainly helpful when thinking about these issues): math.ucr.edu/home/baez/torsors.html $\endgroup$
    – PM 2Ring
    Dec 6 '20 at 16:28
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    $\begingroup$ I was going to suggest you read about “renormalization,” which is what happens when you try this. But the link by @PM2Ring is also relevant, and much more accessible. $\endgroup$
    – rob
    Dec 6 '20 at 21:15
  • $\begingroup$ en.wikipedia.org/wiki/… $\endgroup$
    – user65081
    Dec 9 '20 at 17:26
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$E = mc^2$ where m is the relativistic mass. $m_0$ is the classical or rest mass.

Consider a closed system at rest with no heat added and no work done, but with an internal chemical or nuclear reaction. At rest means no change in kinetic or gravitational potential energies of the overall system. $\Delta U = 0$ where $U$ is the total internal energy.

Classical thermodynamics considers the energy of the reaction as the "internal energy of formation", and defines $\Delta U$ as $U_{classical \enspace products} - U_{classical \enspace reactants} - U_{formation}. \Delta U = 0$ for this example. $U_{classical}$ is the heat capacity at constant volume for all the moles or nuclei in the system. (See one of the thermodynamics textbooks by Sonntag and van Wylen.) In an engineering thermodynamics context the (classical) internal energy is just the heat capacity.

The energy balance for classical thermodynamics was developed before $E=mc^2$ was understood, hence the need for considering the energy of formation. Using $E=mc^2$ we can express this energy of formation as a change in rest mass.

We can express $U_{formation}$ in terms of the rest masses of the reacting constituents. Consider the constituents in the system; atoms/molecules for a chemical reaction, nuclei for a nuclear reaction. For each constituent $U = U_{classical} + nm_0c^2$ where $m_0$ is the rest mass energy- of an atom/molecule or nucleus- and n the total number of moles or nuclei of the constituent. (See note (a) below.) So for the reaction $a + X ->b + Y$, we have $U_{classical \enspace a} + U_{classical \enspace X} + n_am_{0a}c^2 +n_Xm_{0X}c^2 = U_{classical \enspace b} + U_{classical \enspace Y} + n_bm_{0b}c^2 + n_Ym_{0Y}c^2$

So $U_{formation} = n_am_{0a}c^2 + n_Xm_{0X}c^2 - (n_bm_{0b}c^2 + n_Ym_{0Y}c^2)$; the internal energy of formation is equal to the change in the rest masses. If the product- b and Y- rest masses are less than the reactant- a and X- rest masses, $U_{formation}$ is positive and $U_{classical \enspace products}$ is greater than $U_{classical \enspace reactants}$, due a reduction in rest mass causing an increase in $U_{classical}$.

The change in rest mass is very small for a chemical reaction as contrasted with a nuclear reaction, but the same concept holds. In a chemical reaction the change in rest mass is dictated by the binding energy of the electrons in an atom/molecule. In a nuclear reaction the change in rest mass is dictated by the binding energy of the nucleons in a nucleus.

So far, we have considered the energies of all the reactant and product atoms/molecules in the system. We can also view this entire system, which is at rest, externally as having total internal energy $U_{total} = m_{0 \enspace system} c^2$ that is constant, where $m_0$ considers the overall classical internal energies and rest masses of the constituents in the system.
$m_{0 \enspace system} c^2$ = $U_{classical \enspace a} + U_{classical \enspace X} +n_a m_{0a}c^2 + n_Xm_{0X}c^2 = U_{classical \enspace b} + U_{classical \enspace Y} +n_bm_{0b}c^2 + n_Ym_{0Y}c^2$

So, viewing the system externally, the absolute internal energy is the the total rest mass energy for an isolated system (no heat, work, or mass transfer) that is at rest (no change in overall kinetic or potential energy).

Note (a). For a discussion of the reaction energetics on an atom to atom or nucleus to nucleus basis see my answer at Why is mass defect calculated by the rest mass (energy)? on this exchange.

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  • $\begingroup$ wait. Why do you consider the "total" energy as the sum of "classical" and "rest mass" energy? I thought that mass is caused by energy, not a contribution to it. Isn't rest mass energy a way of saying how much energy is contained for a specific mass, as opposed to the amount of energy due to mass. I $\endgroup$ Dec 9 '20 at 5:01
  • $\begingroup$ In an engineering thermodynamics context the (classical) internal energy is just the heat capacity. The energy balance for classical thermodynamics was developed before $E = mc^2$ was understood, hence the need for considering the energy of formation. Using $E = mc^2$ we can express this energy of formation as a change in rest mass. In $E = mc^2$ energy is mass and mass is energy, but mass is the relativistic mass. I have revised my answer to hopefully clarify this. $\endgroup$
    – John Darby
    Dec 9 '20 at 12:34
  • $\begingroup$ relativistic mass is an outdated concept $\endgroup$
    – fqq
    Dec 18 '20 at 20:22
  • $\begingroup$ @fqq Can you please provide a reference where I can update myself on why relativistic mass is outdated? In particular how is $E = mc^2$ interpreted without using the relativistic mass? Thanks. $\endgroup$
    – John Darby
    Dec 18 '20 at 20:41
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    $\begingroup$ @fqq Thanks. My last course on modern physics was in 1972. $\endgroup$
    – John Darby
    Dec 18 '20 at 20:57
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It is determined by convention, say, the gravitational potential of a $g$-mass at infinity is zero, thus the difference in $g$-potentials for a test mass to move it from infinity to a distance $r$ away from the $g$-mass is consequentially equal to the $g$-potential at that specific point a distance $r$ away from the $g$-mass.

Therefore, in Einstein's formula, if we consider the energy in the absence of a mass as zero at a specific point in space-time (similar to the zero potential at infinity), the difference in energies when the mass appears at that very point equals the absolute value of energy at that point, which can be defined by Einstein's formula.

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