# In special relativity is mass just a measure of all other energy than kinetic?

Yes, in special relativity, the mass of a system is synonymous with the energy of the system in a frame where its momentum is zero. This, as you observe, would directly follow from the relation $E^2=p^2+m^2$. I will drop the factors of $c$ for convenience (or, in other words, I will use natural units and set $c=1$). Thus, in spirit, saying that the mass is a measure of all energy except the kinetic energy is correct with a couple of caveats:

- As you already notice in the updated version of your question, a many-particle system can have motion of its constituents which do not contribute to the overall momentum of the system but do contribute to the overall energy. And thus, they contribute to the mass of the system as a whole. Thus, the mass of the system as a whole does include contributions from kinetic energy but only from the kinetic energy that doesn't contribute to the overall momentum of the system.
- Due to the quadratic nature of the relation $E^2=p^2+m^2$, it is a bit problematic to directly identify $p^2$ with the kinetic energy of the system. Rather, the kinetic energy would be $\sqrt{p^2+m^2}-m$ which can be approximated to be $\frac{p^2}{2m}$ as usual for values of $p$ that are very small as compared to $m$. If you naively identify the $p^2$ as the kinetic energy, you wouldn't recover the correct non-relativistic limit.

Now, all your claims such as a hot cup of coffee having more mass than an otherwise identical but cold cup of coffee are true. However, this doesn't mean that the notion of mass doesn't anymore correspond to the property of inertia. Relativity doesn't change the notion of mass completely--it rather corrects it while unifying it with the notion of energy. In particular, it *would* be more difficult to accelerate a hot cup of coffee than a cold one if you can measure all the minuscule effects. So, the notion of mass in relativity is yet very much representative of the quality of inertia. The way to see this is to write down the expressions for momentum and energy in relativity. As you probably know, in relativity, $$p=\frac{mv}{\sqrt{1-v^2}}$$$$E=\frac{m}{\sqrt{1-v^2}}$$As you can see, it is the same $m$ that enters the formula for the energy also enters the formula for momentum. Thus, the same $m$ that represents both the measure of the energy in the rest frame (via entering the formula for energy) and the property of inertia (via entering the formula for momentum). This is the most basic conceptual unification represented in relativity and the genius of Einstein--the (rest) energy of a system is not independent of its inertia but two are the very same thing.

Now, finally, all of this doesn't mean that a cold object at rest shouldn't have any energy at all. It can very well have all sorts of reasons to have rest energy (and thus, mass). For example, even if all the constituents of a system are at rest and there is no interaction potential energy among them, the system as a whole would still have mass but it would simply be the sum of the masses of all its constituents. So, an object whose constituents are at rest simply means that its mass would not have contributions from the kinetic energy of its constituents. More importantly, there cannot be a massless system with no momentum (i.e., a system known to be rest must have mass). If something has no mass then having no momentum implies that its energy is also zero and this simply means that there is nothing.