The amount of energy a system has depends on the reference frame you view it from. If you pick a reference frame that is moving at high speed relative to a system, then you’ll judge the system as having fairly high energy, with much of that energy being kinetic energy (since you and the system have a high relative velocity). There is one reference frame in which the total energy of the system is the smallest possible from your point of view. You can’t make the energy you measure for the system any smaller by "transforming"sup[/sup] to another referencing frame. And, in that special reference frame, the energy you measure for the system is the mass of the system (E=mc[sup]2[/sup]). And, that special frame is the center of momentum frame of the system, where the (vector) sum of all the momenta of all the constituent parts is zero.
This is all true, but it is a consequence of a more general set of statements.
sup[/sup] “Transforming” here means “converting measured (or measurable) quantities (like energy) into what they would be if they were measured by an observer in a different reference frame (i.e., moving with a different velocity)”.
First, an analogy. Consider a long, thin rod. Let’s say one end of the rod is at location (0,0,0) in 3D space and the other end is at some (x,y,z) coordinate. If we calculate sqrt(x[sup]2[/sup]+y[sup]2[/sup]+z[sup]2[/sup]), we’ll get the length of the rod, thanks to Pythagoras.
If we rotate the rod around in space, keeping one end pinned at (0,0,0), we’ll end up with a different set of values for the other end’s (x,y,z) position. We can even tag the variables with “prime” symbols – i.e., (x’, y’, z’) – to remind us that we’ve rotated the stick around so we shouldn’t expect x and x’ (or y and y’…) to be the same numerical values. But we can still take sqrt(x’[sup]2[/sup]+y’[sup]2[/sup]+z’[sup]2[/sup]) to get the length of the rod.
What we have is a vector – the (x,y,z) position of the end of the rod – where the individual numerical values in the vector depend on how we’ve rotated the rod in space, yet regardless of the rotation we can still calculate the length of the rod through a fixed calculation.
Instead of actually rotating the rod we can rotate ourselves, rotating our chosen coordinate system with us. The change from (x,y,z) to (x’,y’,z’) is a transformation due to our picking a different reference frame to view the rod from. But the length of the rod is the same regardless of our choice of reference frame, and we can calculate it using the same calculation in any reference frame, always getting the same answer. The jargon of all this: the length of the rod is an “invariant” with respect to this particular transformation.
In relativity the transformations of interest are not rotations but rather changes of velocity. And the special vectors of interest have four components, not three. One such special vector is the “energy-momentum” four-vector of an object (we’ll generalize to multiple objects shortly). This four-vector’s components are (E, p[sub]x[/sub], p[sub]y[/sub], p[sub]z[/sub]). The first component is the energy of the object, and the next three components are the three components of momentum.
These are all measurable quantities, and you can write them down into this four-vector notation. The invariant quantity you can calculate with such four-vectors is analogous to a length. For the energy-momentum four-vector the quantity sqrt(E[sup]2[/sup] - p[sub]x[/sub][sup]2[/sup] - p[sub]y[/sub][sup]2[/sup] - p[sub]z[/sub][sup]2[/sup]) is an invariant. Note the minus signs, making this thing not actually a Euclidean length.
That invariant quantity has the same value regardless of the reference frame you measure (E, p[sub]x[/sub], p[sub]y[/sub], p[sub]z[/sub]) in. Changing reference frames means changing velocity (by definition), so the energy and the momenta all change when you change reference frames. But you will always get the same answer for sqrt(E[sup]2[/sup] - p[sub]x[/sub][sup]2[/sup] - p[sub]y[/sub][sup]2[/sup] - p[sub]z[/sub][sup]2[/sup]).
And the answer you always get is the mass of the object.
This is a good time to remind you that energy in relativity counts both rest mass and kinetic energy. If we pick a reference frame where our object is at rest, the the momentum parts of our four-vector are zero and the energy is just the rest mass. In any other reference frame, the energy would be higher.
So, the mass of an object is indeed the energy it has when viewed by a (relatively) stationary observer. But more generally, the mass of an object is also the invariant quantity calculable from the energy-momentum four-vector of an object as measured in any reference frame at all.
If our object splits into two or more objects, the total energy and total momentum of the system doesn’t change. Both energy and momentum are conserved in all processes, so no matter the process that busted the object apart, the total energy and total momentum will be unchanged. Thus, you can talk about the mass of the whole system as being the same invariant sqrt() quantity, where E=(total E of all the pieces), p[sub]x[/sub]=(total p[sub]x[/sub] of all the pieces), etc. And this mass will be the same as the mass of the original object. And it will be the mass of the system forever, as long as you don’t lose track of part of it.
I like to point out in these threads that “mass” has multiple practical definitions used in science and even in physics specifically. If anyone tries to slap you down for saying “mass is converted into energy” or “the mass changed”, send them my way. Like with most English words, the meaning of “mass” is context dependent and it isn’t always the strict relativistic version given above.