The Energy-momentum formula considering internal energy

[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!

 

[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.

 

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

 

[Bill_K]

are the contributions to $m_0$ additive, e.g.
$m_0=\frac{E_0}{c^2}$, just the rest mass $m_0$
add heat $Q$:
$m_0^{\prime}=\frac{E_0}{c^2} + \frac{Q}{c^2} = m_0 + \Delta m_0$ ?

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.

 

[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?

 

[Dale]

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.

 

[samalkhaiat]

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

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

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

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

 

[bcrowell]

It holds for an isolated system. I have a discussion of this in my SR book , http://www.lightandmatter.com/sr/ , in section 9.3.4. Einstein's original proof was not valid for a system with internal structure; see Ohanian, "Einstein's E = mc^2 mistakes," http://arxiv.org/abs/0805.1400 .

 

[stevendaryl]

It holds for an isolated system. I have a discussion of this in my SR book , http://www.lightandmatter.com/sr/ , in section 9.3.4. Einstein's original proof was not valid for a system with internal structure; see Ohanian, "Einstein's E = mc^2 mistakes," http://arxiv.org/abs/0805.1400 .

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

 

[bcrowell]

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

His argument.

 

[Sunfire]

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.

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 $2m$?

 

[Dale]

Yes. And the mass is no longer m0

 

[Sunfire]

Yes. And the mass is no longer m0

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

 

[Dale]

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
$m_0^2 c^2=E^2/c^2-p^2$

 

[Dale]

initially, there is rest mass $m_0$ in its rest frame.

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

Then the mass $m_0$ starts moving so that its energy in the initial frame is $m$.

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 $m_0$.

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 $2m$?

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 $m_0 \sqrt{1+3\gamma^2}$.

 

[Sunfire]

Everything is clear, only this

That gives a four-momentum of $(2\gamma m_0 c, \gamma m_0 \mathbf{v})$

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

 

[Sunfire]

That gives a four-momentum of $(2\gamma m_0 c, \gamma m_0 \mathbf{v})$

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?

 

[Dale]

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.

 

[PeterDonis]

You added internal energy only, not momentum.

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.

 

[Dale]

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.

 

[Sunfire]

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

I was considering these 2 processes:

(A) volume of gas with mass $m_0$ is stationary in a lab. frame, moving with velocity $v$. From a stationary frame, the gas appears to have energy $m=\gamma m_0$. The gas is then pressurized (say, pistoned in from all directions) in the lab as the lab keeps moving. The pressure adds internal energy equal to $m$.

(B) volume of gas with mass $m_0$ is stationary in a stationary lab. frame. The gas is then pressurized, so that its total energy becomes $m$. The lab into which the gas is found, starts moving with velocity $v$, so that any mass $m_0$ at rest in the lab appears to have energy $m=\gamma m_0$ in the stationary frame.

 

[PeterDonis]

I was considering these 2 processes

Which, now that you've given some details, raise other subtleties. Don't feel bad, that often happens with relativity problems ; but that's why it's good to look at a concrete example. See further comments below.

(A) volume of gas with mass $m_0$ is stationary in a lab. frame, moving with velocity $v$. From a stationary frame, the gas appears to have energy $m=\gamma m_0$. The gas is then pressurized (say, pistoned in from all directions) in the lab as the lab keeps moving. The pressure adds internal energy equal to $m$.

So we have a system with total energy $m$ in the "observer" frame, i.e., the frame in which the "lab" (not a good choice of terminology, IMO, since in most relativity problems the "lab" frame is at rest, but we'll go with it ) is moving with velocity $v$. What happens when we pressurize the gas? Consider first the process as viewed from the lab frame: we have a set of pistons that push on the gas in all directions, so that no net momentum in the lab frame is imparted to the gas. So in the lab frame, we have a gas that starts with energy $m_0$ and ends with energy $m_0 + m$, and its momentum in the lab frame remains zero throughout.

Now transform to the observer frame. We have a system moving with velocity $v$ whose total energy starts out as $m = \gamma m_0$, and ends as $\gamma \left( m_0 + m \right) = m + \gamma m = m \left( 1 + \gamma \right)$.

But we've left out something: where did the energy come from to drive the pistons? Energy is conserved: if we applied pressure to the gas, we must have had some energy source in the lab frame, say a battery driving electric motors that pushed the pistons, that contained energy $m$ in that frame. So the *total* energy in the lab frame, counting the gas plus the battery, was $m_0 + m$ before, and is still $m_0 + m$ after; all we've done is move energy $m$ from the battery to the gas.

Now look at things in the observer frame in the light of what I've just said. We started out with a system with total energy $\gamma \left( m_0 + m \right) = m \left( 1 + \gamma \right)$ in this frame, and we ended with a system with the *same* total energy. The energy just got shifted around internally in the system; we started with energy $m$ in the gas and $\gamma m$ in the battery, and we ended with all the energy in the gas.

For completeness, if we want to calculate the momentum in the observer frame, the total momentum also is constant at $\gamma \left( m_0 + m \right) v = m v \left( 1 + \gamma \right)$. Before the experiment, the momentum is split between the gas and the battery; the gas has momentum $\gamma m_0 v = m v$ and the battery has momentum $\gamma m v$. After the experiment, all the momentum is in the gas. (Of course we're idealizing the battery as having zero rest mass when it's discharged; but let's not open more cans of worms than we have to. ) So if we just look at the gas by itself, yes, adding internal energy adds momentum as well, if the gas is moving. But if we look at the whole system, we didn't "add" any energy or momentum; we just shifted it around.

(B) volume of gas with mass $m_0$ is stationary in a stationary lab. frame. The gas is then pressurized, so that its total energy becomes $m$. The lab into which the gas is found, starts moving with velocity $v$, so that any mass $m_0$ at rest in the lab appears to have energy $m=\gamma m_0$ in the stationary frame.

Here we start with a system with energy $m_0$ in the observer frame. We add energy to it so its total energy is $m$. Of course that means, once again, that we must have had a battery or something storing the energy to push the pistons that compress the gas and increase its internal energy; here the battery must have stored energy $m - m_0$. But that's less of an issue here since nothing is moving until after the pistons do their work.

Then we start the gas moving with velocity $v$, which gives it a total energy of $\gamma m$ in the observer frame. So we must have had a *second* energy source, containing energy $\left( \gamma - 1 \right) m$, that we used to get the gas moving. The simplest way to do that would be to attach a small rocket to the "lab", containing that energy, and let it fire until its fuel is exhausted. This obviously imparts a momentum $\gamma m v$ to the gas, in the observer frame (and the rocket exhaust has a corresponding momentum of $- \gamma m v$, so total momentum is conserved).

(Note: technically, the rocket fuel would have to contain more energy than just $\left( \gamma - 1 \right) m$, because it has to supply the energy that goes into the rocket exhaust as well. Since that won't affect the final energy of the gas in the lab, I won't open that can of worms either. )

If we look at this in the "lab" frame after the lab is moving, the total energy is just $m$, as before; nothing has changed (assuming the rocket we attach is an idealized one that has zero energy once its fuel is exhausted, similar to what we assumed for the battery in the previous case above). We could, if we wanted, split this into an "original" energy $m_0$ of the gas, and an "added" energy $m - m_0$ that was supplied by the pistons; but again, that's all unchanged from before we set the gas moving. The process of setting the gas moving did not change its internal energy or its rest mass at all.

Obviously these two cases are different; I assume you intended that.

 

[pervect]

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

If the energy is being added in the rest frame and the momentum in the original rest frame is held constant, the final rest frame is equivalent to the original rest frame and both momentum and energy are being added in the lab frame.

If you add energy in the lab frame and allow the velocity to vary so that the momentum in the lab frame remains constant, then the change in velocity of the particle in the lab frame implies that the final rest frame isn't the same as the original rest frame.

I'm not too clear of the intent of the question, perhaps pointing out the ambiguity of the term "adding energy" without specifiying a frame will clarify the issues.

 

[PeterDonis]

What frame is the "internal energy" being added in?

See post #21 (and my response in #22).

 

[pervect]

See post #21 (and my response in #22).

I think you're making the same point at greater length - your post snuch in front of mine though.

 

[PeterDonis]

If you add energy in the lab frame and allow the velocity to vary so that the momentum in the lab frame remains constant, then the change in velocity of the particle in the lab frame implies that the final rest frame isn't the same as the original rest frame.

If I've interpreted post #21 correctly, this would be a different scenario from both of those described in that post. (I was interpreting part A of that post as saying that the lab's velocity remains constant in the observer frame as the gas is pressurized, though it doesn't say that explicitly.)

Also, "adding energy in the lab frame" is more complicated than it seems; as I pointed out in my post #22, the process the OP was describing doesn't, if I've interpreted it correctly, actually "add" any energy; it just transfers energy from the battery to the gas, both of which are at rest in the lab frame. Of course, there are other ways of supplying the energy to the pistons (for example, beaming it with a laser that is at rest in the observer frame), which would change the scenario.

Finally, a quick comment: allowing the velocity to vary as you've described it makes the "lab frame" non-inertial, correct?

 

[Sunfire]

allowing the velocity to vary as you've described it makes the "lab frame" non-inertial, correct?

Let's disregard the transition from 0 to $v$. We look at two different snapshots in time and are trying to make sense of them Let's consider the moving frame inertial.

Also, thank you for your detailed reply!

 

[Dale]

I have to correct myself. At first, I could not reconcile this statement by PeterDonis

So if we just look at the gas by itself, yes, adding internal energy adds momentum as well, if the gas is moving.

With my understanding of the four-force.

The spacelike components of the four-force are usually described as $d\mathbf{p}/d\tau=\gamma \mathbf{f}$, so the only way to change the momentum appears to be via a net force. Which seems to contradict his statement. This is the reason that I said earlier that:

For momentum to change you would need an unbalanced net force, not merely the addition of energy.

However, that description of the four-force assumes that the particle has a fixed mass, i.e. $dm/d\tau=0$. We are deliberately violating that assumption here, so my comment does not hold.

 

[Sunfire]

Then -

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, then one of them must have higher internal energy?

I am unsure if I am formulating this right. If anyone can reformulate it correctly, it will be helpful. Still to keep in mind - the best way to learn something is to consider the simplest scenario possible. In the end I wanted to understand what is the role of internal energy in relativistic mass and momentum. What are the main principles in this.

 

[Dale]

if it turns out that they both end up as identical rest masses, then one of them must have higher internal energy?

Internal energy is included in the rest mass.

 

[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.