Is it right interpretation to interpret $E=mc^2$ using potential energy?

Question

I am wondering that it is the right interpretation to interpret $E=mc^2$ with potential energy. What I mean is this: When I studied nuclear fusion, there was missing mass. The hydrogen's nuclear fusion happens when four hydrogen nucleus fuse into one helium nucleus.

The hydrogen is one proton and the helium is two protons and two neutrons. Neutron is a little heavier than the proton. But there is something strange. Where is the missing mass? As a result, the helium should be heavier than the sum of four protons, but it is not.

So I interpreted this situation as using $E=mc^2$. The potential energy has negative energy, so as the perspective of $m=E/c^2$, it is possible to have negative mass.

But I can't be sure of this interpretation. How can I explain the missing mass?

Plus: Is it possible to interpret binding energy as using $E=mc^2$?

P.s. I'm a Korean student, so I am not used to writing in English. Please give me a comment, if you are find it hard to understand.

Answer 1

First of all

The potential energy has negative energy, so as the perspective of m=E/c2, it is possible to have negative mass

The potential energy has meaning only as a difference between two states not as absolute value. So the negative mass would be not really a mass but difference between masses of the two states. In this case, we could say that four protons have more mass than one hellium neuclei, because they contain more energy. Or put differently, if I wish to separate helium nuclei into its constituent parts I need put some energy in, which will manifest itself as raise in mass.

Second:

Old special theory of relativity defined several kind of masses to deal with the issues like the one you have. But because of a lot of confusion, physicists abandoned these and sticked with just one kind of mass - the rest mass. For the rest mass, it no longer is true that sum of the constituent masses is equal to the mass of the composite system. The mass of composite system is given by the norm of 4-momentum. In 2 particle system this is: $$-m^2c^2=p^2=\eta_{\mu\nu}\left(p^\mu_1+p^\mu_2\right)\left(p^\nu_1+p^\nu_2\right)=-m_1^2c^2-m_2^2c^2+2p_1\cdot p_2.$$ The last term depends on inner energy of the system.

I think, it is much easier to just accept that mass is not extensive quantity, then to try to push the interpretation of $E=mc^2$. The mass is supposed to be measure of "resistance to acceleration" of the object and it will be nice, if it was property of the object itself. Then we could use this as geometric quantity and work with it in abstract language, where no confusion between switching the frames can arise. This is true for the rest mass, not so for the other masses defined through the formula $E=mc^2$.

Answer 2

@Umaxo is absolutely correct, but I think your problem comes from the fact that you assume that four hydrogen nuclei fuse into a helium nucleus. But, in fact what actually happens in stars like the sun is:

A deuterium nuclei (which is hydrogen with a neutron) and a hydrogen nuclei (just a proton), fuse into Helium-3 (which is Helium with 1 neutron only). Then two such Helium-3 fuse into Helium-4 (Helium with 2 protons and 2 neutrons). To put it in a more concise way: $$(^2_1D) + (^1_1H) \rightarrow (^3_2He)$$ $$(^3_2He) + (^3_2He) \rightarrow (^4_2He) + p^{+} + p^{+}$$

So that is why there is no increase of mass in our Sun. Such processes are called proton-proton chain reactions.

Of course, there is some binding energy in the Helium nucleus. When you compare the expected mass of a helium nucleus and the experimental value of the same, you find that the experimental value is lower than the expected one. This difference of mass $\Delta m$, is known as the mass defect and it is exactly equal to: $$\Delta m = \frac{E_b}{c^2}$$ where $E_b$ is the binding energy. So that is where the $E = mc^2$ actually shows up.