Does rest-energy include kinetic energy in relativity?

[Elu314]

I just read this passage, which is really all I've read about relativity so excuse me if this is basic. I don't understand how rest-energy is defined. I thought it was the energy content of the mass only, not including the energy due to motion (hence, rest, right?). But this passage says it is the sum of both. So why is it called rest-energy, then?
(Here, S refers to the system consisting of a contained ideal gas)

"Einstein's equation also says, of course, that if there is a change to the inertial mass of S, then there is a concurrent change to the rest-energy of S. Thus, if we remove one molecule from the gas sample, the rest-energy of the gas sample will diminish by an amount equal to the sum of the molecule's kinetic energy and the mass of the molecule times c2. We can modify our example to make it a bit more telling if we consider the gas sample to be at a temperature of absolute zero, i.e., if we consider the gas sample when all of its molecules are in a state of relative rest. In this case, the rest-energy of S is simply the sum of the masses of the molecules times c2. Let us suppose for simplicity that there are n molecules each of rest-mass m. The rest-energy of S is then simply E = n·mc2. If we remove one of the molecules from the gas, then the rest-energy decreases by an amount ΔE=mc2 and the new rest-energy of S becomes E′ = (n − 1)mc2."

 

[HallsofIvy]

The gas is made up of a large number of molecules. The rest energy of the gas is its energy when its velocity, the average velocity of the molecules, is 0. The gas can be at rest while the individual molecules are not.

 

[tiny-tim]

Welcome to PF!

Hi Elu314! Welcome to PF!

"Einstein's equation also says, of course, that if there is a change to the inertial mass of S, then there is a concurrent change to the rest-energy of S.

Thus, if we remove one molecule from the gas sample, the rest-energy of the gas sample will diminish by an amount equal to the sum of the molecule's kinetic energy and the mass of the molecule times c2.

Yes … the kinetic energy referred to is the KE within the sample, not hte overall KE of the sample as a whole.

It's clearer with a solid rather than a gas …

a solid has a uniform motion, with an associated kinetic energy m[1/√(1 - v²/c²) - 1], but inside the solid the molecules are vibrating slightly (dependent on temperature) …

those vibrations also have kinetic energy, which goes to make up m …

if you increase the speed of the solid, without affecting the vibrations, then the kinetic energy will increase according to that formula with the same m … that m (including the kinetic energy of the vibrations) is the rest-mass of the solid at that temperature.

 

[yuiop]

Imagine you have 2 bricks each with a rest mass of 1kg. One brick (m1) is moving to the left at +v/c and the other (m2) is moving to the right at -v/c both relative to you. The total momentum of the system (both bricks) is m1*v/sqrt(1-(v/c)^2) - m2/sqrt(1-(v/c)^2) = 0. The total rest energy of the two bricks is m1*c^2/sqrt(1-(v/c)^2)+m2*c^2/sqrt(1-(v/c)^2) = 2kg*c^2/sqrt(1-(v/c)^2). This rest energy of the two bricks (considered as a system) is greater than 2kg*c^2 because it includes their energy due relative motion. It is valid to have a rest mass for the two bricks as a system, because the two bricks have a total momentum of zero in your rest frame. This is the definition of the rest mass of a system, i.e. the total energy of the system measured in the reference frame where the total momentum of the system is zero. For a gas, just imagine millions of small bricks going in random directions with a combined total momentum of zero.