soothsayer said:
So if we have two identical boxes, and we fill one of them with light, even though the light has no rest mass, the energy from the light will contribute to the rest mass of the box? I mean I get that it would increase the weight, like, on a scale...
Yes, that is exactly what happens. And if you were to open a hole in the box and let out some of the light inside the box, the mass of the box+contents will decrease, just as it would if you opened a hole and let some of the gas inside the box out. Conversely, if more light to were enter through the hole than left, or more gas, the mass of the box would increase.
How does the box know it is filled with light, and how does that change its inherent properties? Is it because the light is being absorbed by the box? What if we have two identical boxes and poke a hole in one to let light in?
Suppose I put a lead shot in the box... How does the box know that it contains a lead shot so has a greater mass?
Here's a thought experiment. The box, when empty, has a rest mass of 1 kilogram. It contains one gram (rest mass) of matter and one gram (rest mass) of antimatter, so its total rest mass is 1.002 kg as long as the matter and antimatter are separated. But eventually the matter and antimatter will come into contact with each other, annihilate, and turn into (an amazing amount of amazingly energetic) photons that have zero rest mass. If the box is strong enough to contain the resulting explosion so that we on the outside stilll see the same closed system, does the mass of the box plus its contents change?
No. The total rest mass is still 1.002 kilograms.
Backing up from here, the easiest way of thinking about this is to say that energy has mass, given by ##E=mc^2##. The trick and pitfall is that
kinetic energy depends on speed, hence is frame-dependent. Thus, if you're going to use this equation, you can either:
1) Choose a frame in which the speed is zero so there's no kinetic energy and you are dealing with rest mass. This makes the math especially simple, and feels right for this thread where we're talking about a battery sitting on a table in front of us - no movement, no kinetic energy to mess with. The kinetic energy in the system is captured in the ##(pc)^2## term of ##E^2=(m_{0}c^2)^2+(pc)^2##, conveniently zero in this frame.
2) Go with the concept of relativistic mass given by ##m_r=\frac{m_0}{\sqrt{1-v^2}}## and use ##m_r## in ##E=mc^2## so that it includes the non-zero kinetic energy, so is not frame-independent.
3) Use ##E^2=(m_{0}c^2)^2+(pc)^2## (and note that if ##v## and hence ##p## is zero this option reduces to #1 above). As in #1, we're only dealing with rest mass and rest energy here; the frame-dependent kinetic energy is captured in the ##(pc)^2## term.