# The Energy-momentum formula considering internal energy

1. Mar 6, 2014

### Sunfire

Hello,

$E_{tot}^2=(pc)^2+(m_0 c^2)^2$ works fine for mass $m_0$ moving with relativistic speeds. What if the moving mass has internal energy also (say, heat). Does the energy-momentum relation still apply? What is the expression for the momentum $p$ then?

Because $p=\gamma m_0 v$ is okay, but perhaps only if we consider the case of $m_0$ not having any other energy, but kinetic- and rest- only.

Thanks!

2. Mar 6, 2014

### Bill_K

The exact same formula holds. Any internal energy makes a contribution to m0, and this contribution therefore affects both the total energy E and the momentum p.

3. Mar 6, 2014

### Sunfire

are the contributions to $m_0$ additive, e.g.

$m_0=\frac{E_0}{c^2}$, just the rest mass $m_0$

add heat $Q$:
$m_0^{\prime}=\frac{E_0}{c^2} + \frac{Q}{c^2} = m_0 + \Delta m_0$ ?

4. Mar 6, 2014

### Bill_K

Yes, provided you're careful what you mean by Q. The quantity m0 is the total energy in the rest frame of the system. If it's a system made up of particles, that includes the rest masses of the particles plus their kinetic and potential energies.

5. Mar 7, 2014

### Sunfire

If I am to rephrase the question:

rest mass $m_0$ is moving. Its energy becomes $m$.
If this mass had also internal energy $m$ (for instance, pressure energy), then the moving mass has energy $2m$, correct?

6. Mar 7, 2014

### Staff: Mentor

Your rephrasing is very unclear and possibly self-contradictory. A system with internal energy m cannot possibly have a (invariant) mass of m0 where m>m0. The internal energy contributes to the mass of the system.

Perhaps you meant that a system with mass m0, in a reference frame where its energy is m, then has an additional quantity of internal energy equal to m added to it? If so, then yes, the system then would have energy 2m, and its mass would be greater than m0.

7. Mar 7, 2014

### samalkhaiat

There is no mass other than the rest mass $m_{ 0 }$. The notion of "relativistic mass" has long gone. And, when you add heat you do not add anything to $m_{ 0 }$, you only make atoms move faster, i.e., you increase the kinetic-energy, that is all.

Sam

8. Mar 7, 2014

### bcrowell

Staff Emeritus
9. Mar 7, 2014

### stevendaryl

Staff Emeritus
Are you saying Einstein's conclusion was wrong, or just that his argument was wrong?

10. Mar 7, 2014

### bcrowell

Staff Emeritus
His argument.

11. Mar 8, 2014

### Sunfire

initially, there is rest mass $m_0$ in its rest frame. Then the mass $m_0$ starts moving so that its energy in the initial frame is $m$.
If at that moment, internal energy $m$ (say, pressure) is added to this mass, then is it true that the total energy of the mass becomes $2m$?

12. Mar 8, 2014

### Staff: Mentor

Yes. And the mass is no longer m0

13. Mar 8, 2014

### Sunfire

It is $m=\gamma m_0$; add the pressure energy and it becomes $2\gamma m_0$

14. Mar 8, 2014

### Staff: Mentor

No, the invariant mass will be less than 2m but more than m0. I can calculate it tomorrow, but need to sleep.

If you want to work it out the relationship is
$m_0^2 c^2=E^2/c^2-p^2$

15. Mar 9, 2014

### Staff: Mentor

That gives a four-momentum $(E/c,\mathbf{p})$ of $(m_0 c,\mathbf{0})$.

That gives a four-momentum of $(\gamma m_0 c, \gamma m_0 \mathbf{v})$, so the energy is $\gamma m_0 c^2$ and the mass is still $m_0$.

That gives a four-momentum of $(2\gamma m_0 c, \gamma m_0 \mathbf{v})$, so the energy is $2\gamma m_0 c^2$ and the mass is $m_0 \sqrt{1+3\gamma^2}$.

16. Mar 11, 2014

### Sunfire

Everything is clear, only this
is harder to grasp because one would think that the 3-momentum is $2\gamma m_0 \mathbf{v} = 2m\mathbf{v}$

Last edited: Mar 11, 2014
17. Mar 11, 2014

### Sunfire

I am afraid this statement may need clarification...
(1) We know that $E_{tot}=2\gamma m_0c^2$, thus the 1st component of the 4-momentum is indeed $P_1=E_{tot}/c=2\gamma m_0 c$.

(2) By definition, the 4-momentum $P$ is $P=\gamma(c,\mathbf{v})$x the rest mass. But the rest mass is not known yet; the equation $m_0^2 c^2=E^2/c^2-p^2$ is an equation from which we can find the rest mass, but $p$ also contains the rest mass.

We should probably write

$m_?^2 c^2=E_{tot}^2/c^2-p^2=E_{tot}^2/c^2-(\gamma m_? \mathbf{v})^2$, where $E_{tot}=2\gamma m_0c^2$. This leads to the conclusion that the rest mass $m_?$ is $m_?=2m_0$.

Then the 4-momentum is $(2\gamma m_0 c, 2\gamma m_0 \mathbf{v})$ Does this sound right?

18. Mar 11, 2014

### Staff: Mentor

No. You added internal energy only, not momentum. For momentum to change you would need an unbalanced net force, not merely the addition of energy.

In general, momentum does not "contain the rest mass". Consider a photon. It has momentum, but no rest mass.

19. Mar 11, 2014

### Staff: Mentor

But the "internal energy" is added in a frame in which the system is not at rest. So it's not entirely clear to me what "adding internal energy $m$" is supposed to mean, and I'm not sure it's possible to just "add internal energy" to a moving object without also adding momentum. (For example, if I add heat to a moving object, how do I add it? By shining a heat lamp at it? That adds momentum, not just energy, because the infrared radiation emitted by the lamp carries momentum.) I think the OP needs to clarify exactly what physical process he is thinking of.

20. Mar 11, 2014

### Staff: Mentor

Hmm, good point. I wasn't thinking of any specific process, just following what the OP wrote as closely as possible. But you could very well be right that what the OP wrote is not possible.

In any case, regardless of the mechanism, the formula for determining the mass given the energy and the momentum is as given above. If the mechanism adds momentum as well as energy then that would be accounted for in the expression.