The Energy-momentum formula considering internal energy

[PeterDonis]

If we have 2 relativistic masses moving with the same velocity but having different momenta; Once brought to rest, if it turns out that they both end up as identical rest masses

This is not possible. If they have the same velocity but different momenta, they must have different rest masses.

I wanted to understand what is the role of internal energy in relativistic mass and momentum.

As has already been said several times in this thread, internal energy is part of rest mass. So it plays the same role as anything else that is part of rest mass.

 

[Sunfire]

Then how about this simple example -

Gas with rest mass $m_0$ is at rest in a stationary frame.
The gas is pressurized, adding internal energy $m_0c^2$
The total energy becomes $2m_0c^2$, the rest mass is $2m_0$ ($m_0$ comes from its actual mass, the other $m_0$ comes from adding internal energy), the 3-momentum is zero.
This same gas moving with velocity $v$ means its total energy now is $mc^2=\gamma \times \mbox{the rest energy of the gas} \times c^2=2\gamma m_0c^2$; the 3-momentum is $2\gamma m_0\mathbf{v}$, the 4-momentum is $(2\gamma m_0 c, 2 \gamma m_0 \mathbf{v})$.

Correct?

 

[PeterDonis]

Correct?

Yes.

 

[Dale]

I think the OP needs to clarify exactly what physical process he is thinking of.

I think that specifying the exact physical process is certainly sufficient, but I think it is not necessary. Many different physical processes could result in the same changes.

What frame is the "internal energy" being added in? Adding energy in one frame may add both energy and momentum in another frame.

I think this is the key. What is added is not energy, it is four-momentum. The change in four-momentum must be specified, e.g. By specifying the frame where only energy is added.

For example, in my analysis above I assumed that only energy was added in the frame where the gas was moving at v. This could be accomplished by having it absorb a pair of opposed light pulses orthogonal to the motion. The result of that is a change in energy and not momentum.

However, even though momentum is not changing, the speed is. In the gas' frame the pulses are not orthogonal due to aberration. If you want to maintain speed then the energy must be added in the gas' frame, which will result in a change of momentum in the other frame.