# The Equivalence of Mass and Energy

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1. Aug 2, 2015

Hello

I've been trying to get confirmation, or otherwise, that the total mass of an atom can be considered to be equal to the sum of the masses of the individual atomic particles it contains plus the mass equivelent of the internal energy, (PE+KE), of the particles. I have also been trying to find a reference to a work that refers to this question.

In my searches I came here to PF and read the FAQ ........"What is the mass-energy equivelence".

The FAQ pointed out that E and m are equivelent when p = 0, this being when internal energy is part of mass. So far so good but it gives the example of a box containing hot springs being more difficult to push than a box containing cold springs and that's where I get stuck.

I sort of understand why the hot box might seem to be more massive but I don't see how that can be justified by assuming that p=0. A main difference between the hot and cold springs is that in the latter case each particle has greater energy and this alternates between PE and KE. For each particle separately p = 0 only during those instances that it comes momentarily to rest.

A main problem I see is that the equation relating E, m, p and c relates to a single particle and to kinetic energy. How does it relate to an assembly of particle whose movements are somewhat random with respect to eachother and how does it relate to potential energy?

Thank you.

2. Aug 2, 2015

### Staff: Mentor

The mass of an atom is its total energy at rest divided by the speed of light squared, yes. It is possible to split this energy into mass of the nucleus, electron masses, kinetic and potential energy of the electrons. Inside the nucleus (for the nucleus mass) those categories are not meaningful any more.

p=0 is a condition for the box as a whole. The box has to stay at rest. Particles inside can move around as long as their total momentum (the sum over all particles) is zero.

3. Aug 3, 2015

Thank you very much mfb. I would be grateful if you or anyone else could provide a reference to confirm that the total mass of the atom is as I suggested and you confirmed. I have been searching for ages and haven't found anything. My main interest is in the experimental determination of atomic mass. I'm reasonably familiar with the experimental techniques used and how the results are processed but I would like to know a little bit more about the theoretical side. Thank you again.

4. Aug 3, 2015

### Staff: Mentor

Literally every textbook about special relativity.
The contribution from the electrons is well-understood, predicting the mass of the nucleus from theory alone is hard.

5. Aug 3, 2015

6. Aug 3, 2015

### PAllen

Along with dealing with several other issues, the following analyzes the contribution of the KE of electrons in atoms to the passive gravitational mass of an atom (which must equal inertial mass, to exceedingly high precision, per experiment and theory). Some of the references may more directly address your question:

http://arxiv.org/abs/gr-qc/9909014

Last edited: Aug 3, 2015
7. Aug 3, 2015

Thank you very much mfb, DaleSpam and PAllen,
I've scanned the references and they look great. Now I've got some reading to do.

8. Aug 3, 2015

### bcrowell

Staff Emeritus
The equation $m^2=E^2-p^2$ (in units with c=1) is the definition of the mass of a system, so it applies regardless of whether the system has internal structure.

9. Aug 3, 2015

### Staff: Mentor

For a very simple example of how what bcrowell says here will play out, consider a system containing two particles of equal mass and connected with a massless spring so that they're always moving in in opposite directions with equal speed as they fly apart and then back towards one another.

If you consider the system to be the two particles and the spring, you'll find that $p$ is zero (equal masses moving in opposite directions), $E$ is constant (energy is conserved), and you'll be able to calculate the rest mass of the system using $m^2=E^2-p^2$.

If you consider either particle to be the system, you will find that the kinetic energy and momentum are changing over time, but you'll be able to calculate the rest mass of the particle.

The thing you have to watch out for is that you cannot assume that the rest mass of the two-particle-plus-massless-spring will be equal to the sum of the rest masses of the two individual particles. It won't be, because you won't be accounting for the potential energy in the spring or the kinetic energy of the particles.

10. Aug 4, 2015

Thank you bcrowell and Nugatory. I think my question has been answered well by all people here.

11. Aug 4, 2015

### vanhees71

In Nugatory's example of #9, the total invariant mass (and that's the only notion of mass accepted by modern physicists) is the sum of the masses of the two particles plus the total intrinsic excitation energy (divided by $c^2$), which sum is constant and the total energy of the system as measured in the center-momentum frame, where the center of momentum is at rest. The excitation energy consists of the kinetic energy of the particle and the potential energy in the spring at any moment.

This example is, however, pretty tricky to formulate a concrete dynamical model for this thought experiment.

A simpler example for the meaning of the energy-mass-equivalence principle is a cavity with walls at a given temperature, taking into account the black-body radiation. A pretty straight-forward calculation shows that the invariant mass of the cavity is given by the mass of the cavity itself plus the thermal radiation energy (in natural units $\hbar=c=1$):
$$m_{\text{tot}}=m_{\text{cavity}}+U_{\text{rad}} \quad \text{with} \quad U=\frac{\pi^2 V_0}{15} T^4,$$
where $V_0$ is the proper volume of the cavity (i.e., the volume measured by an observer at rest with respect to the cavity). As one can see that's a manifestly covariant expression by definition since the proper volume as well as the temperature and the rest mass of the cavity are all Lorentz scalars!