Is mass a representation of the amount of energy, or is it just another form of it?

Question

I've just started learning about Nuclear Physics, and here's something I'm not able to quite understand.

I have been taught the following:

1. During a nuclear reaction, mass-energy is conserved.
2. During nuclear fission, it is the mass defect that is converted to Kinetic Energy and radiation.

Here's what I don't understand.

1. If mass can be converted to energy (as in the case of the fission of a nucleus), then when total energy of the nucleus decreases, mass of the nucleus should increase in order to conserve the sum.
   Yet, the nucleus weighs lesser than the sum of the masses of the individual neutrons and protons (kept out of interacting distance with each other and at rest).
   
   My current understanding is that $E=mc^2$ represents how much energy a given amount of mass represents, so that it can be added to the other kinds(Kinetic, Potential, etc) of energies to get the total energy of the system.
2. When a nucleus splits during a fission, it splits in order to attain a lower energy (more stable) situation. This means, that potential energy of the nucleus reduces. According to my current understanding, the mass has to increase in order to conserve mass-energy. Here, it does indeed turn out that the sum of the masses of the split nucleii is greater than the mass of the un-split parent.
   I am also aware that energy leaves the system in the form of photons, but I currently of the understanding that some of the potential energy went into the motion of the particles (Kinetic Energy), some of it into radiation, and some of it got converted to mass.
   I understand $$E_{lost} \to E_{potential} + E_{kinetic} + E_{radiation} + E_{mass} + \dots$$ and the amount by which the mass increases can be found by $$m=\frac{E_{mass}}{c^2}$$

It appears to me that in some places, mass represents the total energy of the system, and that a system with higher energy has more mass. That mass isn't a form of energy, but just an effect of it. And one that is common to all kinds of energies.

However, in some places, energy can be converted to-and-from mass. That when energy decreases, mass increases to conserve the sum.
What gives? How is it supposed to be dealt with? What sort of mental picture should I have?

Answer 1

Unfortunately, this is one area where there are a lot of incorrect pop-sci presentations on this topic, largely because $E=mc^2$ is such a famous formula.

$E=mc^2$ does not mean that energy can be converted to mass. If a physicist wanted to write the idea that energy can be converted to mass then the expression would be $\Delta E = -\Delta m c^2$. Instead, $E=mc^2$ means that a system with mass $m$ has energy $E=mc^2$. The energy is not converted into mass the system has both the energy and the mass at all times.

However, despite its fame, $E=mc^2$ is not a generally applicable formula. The general formula is $m^2c^2=E^2/c^2-p^2$ where $p$ is the momentum of the system. For $p=0$ the general formula easily reduces to the famous formula. So, the famous formula is limited to systems with no momentum. For systems with momentum the more general formula is needed.

It appears to me that in some places, mass represents the total energy of the system, and that a system with higher energy has more mass. That mass isn't a form of energy, but just an effect of it. And one that is common to all kinds of energies.

This is essentially correct. However, I wouldn't say that one is an effect of the other as causes come before effects, but energy and mass co-exist.

However, in some places, energy can be converted to-and-from mass. That when energy decreases, mass increases to conserve the sum.

This is not correct. The primary issue is that the mass of a system is not equal to the sum of the masses of its constituents. Let's take a very simple example: annihilation of a positron and an electron.

Before the annihilation the positron and electron (assume both are at rest) each have a mass of $511\mathrm{\ keV/c^2}$, an energy of $511\mathrm{\ keV}$, and a momentum $0 \mathrm{\ keV/c}$. The momentum and energy of the system is the sum of the momenta and energies of the constituents so $0\mathrm{\ keV/c}$ and $1022 \mathrm{\ keV}$ respectively. So the mass of the system is calculated using either the general or the famous equation to get $1022\mathrm{\ keV/c^2}$.

After annihilation the momentum, energy, and mass of the system are unchanged. But now the resulting photons have energy of $511\mathrm{\ keV}$ and momentum of $\pm 511 \mathrm{\ keV/c}$. So using the general expression we can calculate the mass of each photon as $0\mathrm{\ keV/c^2}$. So the mass of the system of two photons is $1022\mathrm{\ keV/c^2}$ and the sum of the individual masses is 0.

Answer 2

At the nuclear scale, mass is a form of energy, that is how you should interpret it. That is why the heat $Q$ of a nuclear reaction/process (which is an amount of energy) is defined as the difference in masses between the initial and final masses of the system

$$Q = (m_{in}- m_{fin})c^2.$$

When the system loses mass, $Q>0$, we have an exothermic process. The energy (heat) produced is not lost, it will be present in the form of kinetic energy, including the energy of any releases photons, since all of their energy is associated to motion. Particularly, $Q-\Delta m c^2 =0$, so that the change in total energy is zero - as expected from the conservation law.

If for example a proton and neutron assemble to form a deuteron, a bound system with less mass than the sum of their parts, you have this reaction:

$$(m_p+m_n)c^2 - m_d c^2 = Q ≈ 2.2 \ \mathrm{MeV},$$

this energy being precisely the binding energy of the deuteron. Indeed, as the deuteron is a stable state, the system transforms part of its mass into energy - mainly as kinetic energy and a gamma. The new nucleus weighs less than the sum of its parts, but that is not a problem since global energy conservation stil holds.