Comment author:SilasBarta
10 March 2010 04:10:00AM
*
7 points
[-]

Another pertinent example might be this: a metal shaft could spin so fast that its atoms' velocity distribution could be the same as that of the (hotter!) gaseous form of the same metal. Yet the spinning of the shaft does not evaporate the metal.

Why? Because, to a typical observer of the shaft, its degrees of freedom are significantly more constrained. So, since the observer knows more about the shaft (that its atoms are in a solid lattice that moves in relative unison), that makes the shaft colder -- and it allows you to extract more mechanical work from the shaft than if it were a hot gas with the same average particle velocities!

Comment author:wnoise
10 March 2010 06:01:44AM
0 points
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That is not in general possible. The speed at radius r is v = w r. Taking an arbitrary axisymmetric mass-distribution rho(r), we have a distribution of mass at speed v = w r that is U(r) = 2 pi h r^2 rho(r) and U(v) = 2 pi h v^2/w^2 rho(v/w) A monatomic gas at temperature T has a kinetic energy distribution of 2 Sqrt(E / Pi (kT)^3) exp(-E/kT) (dE), and a speed distribution of sqrt(2(m/kT)^3 / pi) v^2 exp(-m v^2 / 2 kT) (dv). By carefully tailoring rho(r) to an exponential, you can match this distribution (up to some finite cut-off, of course), at one specific match of angular speed w and temperature T.

This, of course, only matches speeds, not velocities, which will be a three-dimensional distribution. This spinning shaft of course, v_z = 0.

Comment author:SilasBarta
11 March 2010 03:55:08PM
1 point
[-]

For a gas in a mass-sealed container, average vx, vy, vz are also zero, just as in the shaft, so this trivially matches.

You're correct that I should have said speeds, and I should have mentioned that the shaft requires a special shape/density distribution, but the point stands: The molecular property distribution doesn't by itself tell you how cold something is, or how much energy can be extracted -- it's also relevant how its degrees of freedom are constrained, which shows how your knowledge of (mutual information with) the system matters.

Comment author:wnoise
11 March 2010 07:10:26PM
0 points
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For a gas in a mass-sealed container, average vx, vy, vz are also zero, just as in the shaft, so this trivially matches.

Yes, but the average tells you little that's meaningful: it's equivalent to the overall velocity of the gas as whole.

I agree with your overall point. Temperature is not directly a property of the system, but of how we can characterize it, including what constraints there are on it. I just think that this wasn't a great example for that point of view, precisely because you claimed agreement of distributions that doesn't exist.

A better way to say explain this can use the shaft. A shaft has the very strong constraint that position and velocity are perfectly correlated. This constraint lets us extract virtually all the kinetic energy out. A gas that had the same distribution of velocities, but lacked this constraint, would be very hard to extract useful energy from. No simple arrangement could do much better than treating it as a thermalized gas with the same average kinetic energy (and it would quickly evolve so that the velocity distribution would match this).

## Comments (64)

Old*7 points [-]Another pertinent example might be this: a metal shaft could spin so fast that its atoms' velocity distribution could be the same as that of the (hotter!) gaseous form of the same metal. Yet the spinning of the shaft does not evaporate the metal.

Why? Because, to a typical observer of the shaft, its degrees of freedom are significantly more constrained. So, since the observer knows more about the shaft (that its atoms are in a solid lattice that moves in relative unison), that

makesthe shaft colder -- and it allows you to extract more mechanical work from the shaft than if it were a hot gas with the same average particle velocities!That is not in general possible. The speed at radius r is v = w r. Taking an arbitrary axisymmetric mass-distribution rho(r), we have a distribution of mass at speed v = w r that is U(r) = 2 pi h r^2 rho(r) and U(v) = 2 pi h v^2/w^2 rho(v/w) A monatomic gas at temperature T has a kinetic energy distribution of 2 Sqrt(E / Pi (kT)^3) exp(-E/kT) (dE), and a speed distribution of sqrt(2(m/kT)^3 / pi) v^2 exp(-m v^2 / 2 kT) (dv). By carefully tailoring rho(r) to an exponential, you can match this distribution (up to some finite cut-off, of course), at one specific match of angular speed w and temperature T.

This, of course, only matches speeds, not velocities, which will be a three-dimensional distribution. This spinning shaft of course, v_z = 0.

For a gas in a mass-sealed container, average vx, vy, vz are also zero, just as in the shaft, so this trivially matches.

You're correct that I should have said speeds, and I should have mentioned that the shaft requires a special shape/density distribution, but the point stands: The molecular property distribution doesn't by itself tell you how cold something is, or how much energy can be extracted -- it's also relevant how its degrees of freedom are constrained, which shows how your knowledge of (mutual information with) the system matters.

Yes, but the average tells you little that's meaningful: it's equivalent to the overall velocity of the gas as whole.

I agree with your overall point. Temperature is not directly a property of the system, but of how we can characterize it, including what constraints there are on it. I just think that this wasn't a great example for that point of view, precisely because you claimed agreement of distributions that doesn't exist.

A better way to say explain this can use the shaft. A shaft has the very strong constraint that position and velocity are perfectly correlated. This constraint lets us extract virtually all the kinetic energy out. A gas that had the same distribution of velocities, but lacked this constraint, would be very hard to extract useful energy from. No simple arrangement could do much better than treating it as a thermalized gas with the same average kinetic energy (and it would quickly evolve so that the velocity distribution would match this).

Sure it does -- it just requires a special shaft shaping/material. Hence the "a metal shaft

could..."Otherwise, we're in agreement.

It's a really great point; thank you.