Is hot gas more massive than cold gas?

[rugerts]

Homework Statement

In principle, does a hot gas have more mass than the same gas when it is cold? In practice, would this be a measurable effect?

Homework Equations

E = γmc²

The Attempt at a Solution

Since there's more energy, there's more mass. But I don't think these effects would be measurable since the speed of the particle is much less than the speed of light.
Is this correct? This video () suggests that there's only one mass (called the rest mass) and that relativistic mass is not actually a thing. Can someone please help clear up the confusion as it pertains to this specific problem? The contents of the video make sense to me and now seem to contradict what I've suggested as a solution above. Thanks for your time.

 

[mfb]

Yes, the gas as a whole thing has more mass. The energy in its rest frame increases. The mass of the individual atoms, one by one, doesn't increase.
Currently the mass change is still below what we can measure - but there is just a factor 10-100 or so missing, it might become measurable in the near future.

 

[jbriggs444]

This video suggests that there's only one mass (called the rest mass) and that relativistic mass is not actually a thing. Can someone please help clear up the confusion as it pertains to this specific problem? The contents of the video make sense to me and now seem to contradict what I've suggested as a solution above.

The video is correct. The modern convention is that "mass" always means rest mass. The notion of "relativistic mass" is not particularly useful. In particular, relativistic mass ($\gamma m$) is just a synonym for total energy ($E=\gamma m c^2$) with a scaling factor of $c^2$. In units where c=1, energy and relativistic mass are numerically identical.

But it is important to realize that rest mass (or "invariant mass" or just "mass") is not an additive quantity. The mass of a system containing two or more objects is not always equal to the sum of the masses of the component objects.

Rest mass can be defined in two ways -- in terms of the total energy of a system in a frame of reference where its total momentum is zero ($m=\frac{E}{c^2}$ where p=0). Or as the norm of the energy-momentum four-vector. $m^2c^4 = E^2 - p^2c^2$. The energy-momentum four-vector is formed by putting energy in as the first element and the three components of momentum as the other three elements. The "norm" or "magnitude" of the four-vector is defined with a modified Pythagoras type formula. You take the sum of the squared components -- except that the squared momentum components count as negatives. Then you take the square root of the sum.

The two definitions are equivalent except that the latter still works for photons whose momentum can never be zero in any frame. [For photons, E=pc and m=0]

If you take two objects that each have a mass of 1 kg and are moving toward each other, the combined system has zero momentum in the combined center of mass frame. In that frame, total energy is 1 kg (times c squared) plus 1 kg (times c squared) plus the kinetic energy of each mass in the center of mass frame. That is, of course, just a tiny bit larger than 2 kg (times c squared). The rest mass of that system is larger than the sum of the rest masses of its pieces.

Hope that clarified more than it confused.

 

[haruspex]

If you take two objects that each have a mass of 1 kg and are moving toward each other, the combined system has zero momentum in the combined center of mass frame. In that frame, total energy is 1 kg (times c squared) plus 1 kg (times c squared) plus the kinetic energy of each mass in the center of mass frame. That is, of course, just a tiny bit larger than 2 kg (times c squared). The rest mass of that system is larger than the sum of the rest masses of its pieces.

If we take a black box view of the two object system and apply a force, does the apparent inertia depend on the relationship between the direction of the force and the velocity direction of the objects?

 

[rugerts]

The video is correct. The modern convention is that "mass" always means rest mass. The notion of "relativistic mass" is not particularly useful. In particular, relativistic mass ($\gamma m$) is just a synonym for total energy ($E=\gamma m c^2$) with a scaling factor of $c^2$. In units where c=1, energy and relativistic mass are numerically identical.

But it is important to realize that rest mass (or "invariant mass" or just "mass") is not an additive quantity. The mass of a system containing two or more objects is not always equal to the sum of the masses of the component objects.

Rest mass can be defined in two ways -- in terms of the total energy of a system in a frame of reference where its total momentum is zero ($m=\frac{E}{c^2}$ where p=0). Or as the norm of the energy-momentum four-vector. $m^2c^4 = E^2 - p^2c^2$. The energy-momentum four-vector is formed by putting energy in as the first element and the three components of momentum as the other three elements. The "norm" or "magnitude" of the four-vector is defined with a modified Pythagoras type formula. You take the sum of the squared components -- except that the squared momentum components count as negatives. Then you take the square root of the sum.

The two definitions are equivalent except that the latter still works for photons whose momentum can never be zero in any frame. [For photons, E=pc and m=0]

If you take two objects that each have a mass of 1 kg and are moving toward each other, the combined system has zero momentum in the combined center of mass frame. In that frame, total energy is 1 kg (times c squared) plus 1 kg (times c squared) plus the kinetic energy of each mass in the center of mass frame. That is, of course, just a tiny bit larger than 2 kg (times c squared). The rest mass of that system is larger than the sum of the rest masses of its pieces.

Hope that clarified more than it confused.

Quite a bit to take in but that doesn't imply your explanation was off, just that I have some more learning to do. Can I essentially take away that since, at rest, $m=\frac{E}{c^2}$, since the energy of the gas increases with increase in temperature, the mass increases? It only confuses because it doesn't seem as though we actually have more "stuff", it's just the same stuff with more energy. Also, since if the temperature increased, it actually seems as though $m=\frac{E}{c^2}$ can't be used, since that's for something at rest.

 

[PeroK]

Quite a bit to take in but that doesn't imply your explanation was off, just that I have some more learning to do. Can I essentially take away that since, at rest, $m=\frac{E}{c^2}$, since the energy of the gas increases with increase in temperature, the mass increases? It only confuses because it doesn't seem as though we actually have more "stuff", it's just the same stuff with more energy. Also, since if the temperature increased, it actually seems as though $m=\frac{E}{c^2}$ can't be used, since that's for something at rest.

You might also like to watch:

 

[jbriggs444]

Quite a bit to take in but that doesn't imply your explanation was off, just that I have some more learning to do. Can I essentially take away that since, at rest, $m=\frac{E}{c^2}$, since the energy of the gas increases with increase in temperature, the mass increases? It only confuses because it doesn't seem as though we actually have more "stuff", it's just the same stuff with more energy. Also, since if the temperature increased, it actually seems as though $m=\frac{E}{c^2}$ can't be used, since that's for something at rest.

It is for something whose momentum is zero. Which is not quite the same thing.

 

[mfb]

Internal motion is fine. $m=E/c^2$ works as long as the total momentum is zero (we are in the rest frame of the object we consider).

 

[rugerts]

You might also like to watch:

I actually don't like this one since the concept of relativistic mass is used and it's actually more confusing. I prefer Dr. Don Lincoln's (of the Fermilab video) explanation over his.

 

[rugerts]

It is for something whose momentum is zero. Which is not quite the same thing.

So, just to conclude this, was my final statement, that the mass "increases" since the energy of the system increases, valid? Still confused about the mass thing since there's no change in the amount of stuff, hence my quotes around increases. But, I suppose if there's a mass energy equivalence for something at rest, then it sort of transcends the idea of mass just being = amount of stuff.

 

[mfb]

So, just to conclude this, was my final statement, that the mass "increases" since the energy of the system increases, valid?

The mass increases since the energy of the system increases but its overall momentum does not (the object does't fly away in some direction).

Still confused about the mass thing since there's no change in the amount of stuff, hence my quotes around increases. But, I suppose if there's a mass energy equivalence for something at rest, then it sort of transcends the idea of mass just being = amount of stuff.

Right. You have this in more pronounced form e.g. in fission and some other nuclear reactions. The amount of stuff stays the same but the binding energy and therefore the mass changes.