Problems Understanding Mass Energy Equivalence So I've had a problem for a long time now understanding energy-mass equivalence, in particular  I've had a lot of trouble understanding how something like interatomic potentials can be seen as mass. I've heard other people on the site say it depends on how you probe the system, which doesn't make too much sense to me, I don't see why mass should change based on 'what level you view the system at'. Can someone help me understand mass-energy equivalency? 
 A: My answer is complementary to Rob's. When people talk about mass, there are three related concepts that are often bandied about, and this gives rise to confusion when one is not careful to tell or make the difference between them clear:


*

*Rest Mass;

*Gravitational Mass, i.e. the "strength" that something has in the Einstein Field Equations when it is thought of as a source of spacetime curvature. Otherwise put in the Newtonian picture: the source strength (either $M$ or $m$) of the gravitational force in Newton's inverse square law;

*Inertia, or what I like to call colloquially "shove stubborness": a measure of how hard you have to shove something to make it accelerate at one unit relative to an initially co-moving inertial frame.


Gravitational mass and intertia, for systems that behave as rigid bodies or points, are the same as verified by the Eötvös experiment, as taken as an axiom called the "Equivalence Principle" and generalised in the Einstein field equations. Gravitational mass and intertiaare the concepts that you should "equate" (subject to some provisos) with "energy": if a rigid point's total energy is $E$, then, to calculate the metric for the spacetime outside the source, you get the right answer if you write $E/c^2$ for the "effective mass" $M$ where the Schwarzschild radius $r_s = 2\,G\,M/c^2$. The energy "frozen" in your bond potential energies. (as with Gauss's law, you can also replace any spherically symmetric "matter" distribution with a point in the Schwarzschild solution as long as you are calculating the metric outside the distibution). Whatever energy you add to your "stuff", being stored potential energy, heat, whatever, the source strength $M$ in the Schwarzschild radius rises by the amount $E/c^2$.  The same goes for inertia: if you confine a system with total energy $E$ to a box, its inertia (i.e. force needed to make it accelerate at unit acceleration relative to an initially co-moving inertial frame) is $E/c^2$. You can ponder and play around with simple thought experiments like mine here  and here to explore this.
Rest mass is the distinct, but related concept. Early discussions talked about relativistic mass, which is what Bohan Lu's answer talks about. The faster the body goes, the harder it is to shove, thus leading to a divergence in the impulse you must impart to get the body relatively moving at the speed of light. I believe this is a useful concept to build intuition when one is beginning to understand these ideas, but one needs to move on from it: it is not Lorentz invariant. Some particles are always observed to have relative speed $c$ independent of one's inertial frame. These are massless particles. Their total energy $E$ is given by $p\,c$ where $p$ is the magnitude of their momentum. All others are particles which have an inertial frame wherein they are at rest. In this frame, their total energy is $E = m\,c^2$, and in this frame alone the rest mass $m$ equals the concepts 2. and 3. above. In all other frames, $E=m\,c^2$ is not the full picture and their total energy fulfills $E^2 = p^2\,c^2 + m^2\,c^4$. $m$ is Lorentz invariant, but witness how it is not generally the appropriate property to calculate "interia" and "gravitiation" from.
A: You say

in particular I've had a lot of trouble understanding how something like interatomic potentials can be seen as mass.

which I agree sounds confusing. There's a relationship between energy and mass, but there's a murkier relationship between potential and mass.
Interatomic interactions are a bad place to start because the energies involved are so small.  If you have a molecule that can be dissociated with visible light, its binding energy is a few electron volts. Its mass (in energy units) is probably a few billion electron volts.  It's just not a very big correction.
It's probably a little easier to think about in nuclear interactions, where the binding energies may be a percent-scale correction to the mass.  Suppose I have a carbon-12 atom, which has a mass of exactly 12 atomic mass units, or 12 u.  (This is exact, because carbon-12 is the definition of the atomic mass unit.)  Let's split this atom into helium-4 atoms. I know that the mass of each helium atom will be approximately 4 u.  But if I measure the masses, with a spectrometer that's calibrated so that carbon-12 gives 12.000 000 u, I find that each helium atom has a mass of 4.002 602 u. In order to make my three helium atoms, I had to generate roughly 0.0078 u of mass that didn't exist before!
Furthermore, you have probably noticed that carbon atoms do not spontaneously convert into helium atoms.  In your experiment where you are splitting the carbon to make helium, you will discover that you get zero production of helium while the energy that you put into your interaction is less than about 7.3 MeV. But when you "turn on" the helium production by putting in, say, 10.3 MeV, you find that the three helium nuclei that you produce have a kinetic energy of about 1 MeV each.
This is because energy and mass are equivalent.  Adding 1 MeV of energy to a nucleus has the same effect on its dynamics as increasing its mass by 0.001 703 u.  A carbon-12 nucleus can be thought of as three helium nuclei that are stuck together; to unstick them you have to add at least 7.3 MeV of energy, and that energy will continue to be stored in the system as additional mass in the helium nucleus.  Conversely, when three helium nuclei fused in a long-ago star to form that carbon nucleus, that new carbon nucleus released a 7.3 MeV photon which  wandered out into the heat bath of the core of the star and eventually made it to the surface as a few million quanta of visible light.
Similarly, if you have a uranium-235 atom, it'll weigh 235.043 908 u. Put it in with its friends and it may fission into a strontium-90 (mass 89.907 784 u), a xenon-143 (mass 142.935 191 u) and two neutrons (mass 1.008 660 u). Add these up and you have lost nearly 0.1 u!  This reaction also releases lots of energy, as you are probably aware.
A: Think of it this in terms of how you intuitively verify $E=mc^2$. Imagine a particle moving at a speed close to the speed of light. According to Lorentz transformation, the closer the speed of an object to the speed of light, the more work you have to do to increase the speed of the object by equal amount. The increase in kinetic energy will not change if you increase the speed of the object by equal amount. So one wonders where the extra work goes if the kinetic energy doesn't change. Again because of Lorentz transformation, the mass of the particle will increase. One then would naturally consider the extra work contributes to the accumulation of extra mass. So in relativistic scenarios, namely, $v \approx c$, energy can be turned into mass. Does that help?
