# Potential energy and mass/energy equivalence

1. Jan 8, 2014

### VantagePoint72

How does potential energy fit into mass-energy equivalence in SR? As with all forms of energy, a potential energy of $E$ added to static system ought to increase the system's mass by $E/c^2$. This is often illustrated by saying that a compressed spring has slightly more mass than an uncompressed spring. But how does this square with the fact that potential energy is only specified up to an additive constant?

For instance, suppose we have a system of charged particles separated by some fixed distance. Each particle, with charge $q_i$ (say, all positive charges) and produces a potential of $\phi_i$. Then the potential energy of the system is $U = \frac{1}{2} \sum_i q_i \phi_i$ (with the 1/2 factor to avoid double counting). If each particle has a rest mass of $m$ and say there are $N$ particles, what is the mass of this system? That's an unambiguous, measurable quantity—we could, for instance, use the equivalence principle to do this by measuring the strength of the gravitational field induced by the system—and it must be greater than just $Nm$ since work was done to assemble the system. Naïvely (at least, I assume it's naïvely since it seems to be wrong), the rest energy seems like it ought to be $Nm + U/c^2$. This doesn't make sense, though, because the $\phi_i$'s are not uniquely specified. I can shift each of them by some constant amount, corresponding to picking a different reference point for the potential. So how is the effect of potential energy on the system's mass unambiguously determined?

2. Jan 8, 2014

### Staff: Mentor

Yes, but it's important to understand why this is. It's because work was done on the spring to compress it; the amount of work done (divided by $c^2$ if you insist on using funny units ) is equal to the increase in mass of the system.

In the spring example, you're not measuring the "absolute" amount of potential energy; you're only measuring the *difference* in potential energy before and after you did the work on the spring to compress it. The additive constant drops out because it was there both before and after, so it's not part of what changed as the work was being done.

But the more comprehensive answer is that, if you're going to count potential energy as part of the mass of a system, then you no longer have the freedom to pick the additive constant arbitrarily. In other words, you have to pick out which state, physically, has zero potential energy. For the spring, for example, the zero potential energy state is the unstressed state, i.e., the state in which the spring is in equilibrium with no force being applied.

It's also worth noting that "potential energy" can't be modeled just with SR when gravity is involved. See below.

Here you have to be more careful picking the zero potential energy state. It is the state with zero interaction between the particles; i.e., the state where all the particles are infinitely separated from each other, so the Coulomb force between each pair of particles goes to zero. Only this choice of the "zero" for the potential energy will give the right answer for the mass, which, as you note, can be measured directly.

Why is that the right choice? Suppose we have two particles with the same charge at infinite separation. What is the total mass of the system? Obviously it's just the sum of the rest masses of the particles, i.e., the potential energy is zero.

Now we bring the particles closer together. To do that, we must do work on the particles, just as we did work to compress the spring in the example above. That adds energy to the system, increasing its mass--and sure enough, the potential $\phi$ between two particles with like charges increases as their separation $r$ decreases. If we measure the mass of the system after we've brought the particles closer, it will increase by $\phi$ (or $\phi / c^2$ if, once again, you insist on funny units ).

Conversely, if we have two oppositely charged particles at infinite separation, we can *extract* work from the system by letting them come closer together. The work we extract is just equal to the (negative) potential $\phi$ between the particles, and sure enough, if we measure the mass of the system when the particles are closer together, it *decreases* by $\phi$.

As I noted above, if you do this, you're no longer using SR, you're using GR. There is no gravity in SR. But in GR, potential energy due to gravity works similarly to what I outlined above.

If you had to do work to assemble the system, as in the spring and like charged particle cases above, yes. But in other cases, such as the oppositely charged particle case above, you can *extract* work in assembling the system, and the potential energy is then negative, so the final mass is less than the initial mass before assembly. Gravity works like this: if I have two masses at infinite separation, and I bring them closer together, I can extract work from the process, purely due to the mutual gravity of the masses. The final mass of the bound system will be less than the sum of the two masses; the difference is the negative potential energy due to gravity.

3. Jan 9, 2014

### Meir Achuz

$$(p^\mu-A^\mu)(p_\mu-A_\mu)=(m+S)^2$$
where A is a 4-vector potential and S is a scalar potential.

Last edited: Jan 9, 2014
4. Jan 10, 2014

### VantagePoint72

Thanks, PeterDonis! Meir Achuz, it's not clear to me what your post explains; but, I think I've got it from Peter's reply.

5. Jan 10, 2014

### Meir Achuz

In non-covariant form, the relation $E^2={\bf p}^2+m^2$ with no potential becomes
$$(E-V)^2=({\bf p-A})^2+(m+S)^2$$ in the presence of a Lorentz vector potential
$(V,{\bf A})$ and a Lorentz scalar potential S.

Last edited: Jan 10, 2014