# Temperature of a gas (assumptions about the particle speeds)

Temperature is related to average of particles kinetic energy.

I would like to ask about a singular state of a particle system of a little time interval.

The question is:

If all particles were going same direction, same speed, same mass, for that while they were avoiding external forces, internal colision, interaction, etc, then even within this interval they will have an average kinetic energy, so a definite temperature.

But, in this case, I think it's better to describe as a "movement of whole the system", instead of a "temperature of the system", because there is no relative shaking, then I see there is a kind of restriction about particle speeds, they have to be diverse to have a temperature.

I know in practice that's a very unusual process or even a ephemeral one, but I would like to think about theoretical meaning, How is this assumption of speed diversity defined within a gas model ?

The answer to your question is part of the large framework known as statistical mechanics, which provides very definite predictions for the distribution of velocities of gas particles in thermal equilibrium at some temperature $T$.

I think the simplest response is to say that temperature is not a measure of the average kinetic energy of the gas particles, but rather to the spread of velocities from the mean value. If this spread is zero - as in your example of all the gas particles moving in the same direction - then the temperature is also zero.

In general, for a gas at temperature $T$, the difference of particle velocities from the average velocity of the gas will follow a Maxwell-Boltzmann distribution. Thus, if the mean velocity is $\mathbf{v}_m$, then a particle will have velocity $\mathbf{v}_p=\mathbf{v}_m+\mathbf{v}$, where $v=|\mathbf{v}|$, with probability $$p(v)=\sqrt{\frac{2}{\pi}\left(\frac{m}{kT}\right)^3}v^2 \exp\left(-\frac{\frac{1}{2}mv^2}{kT}\right).$$

EDIT in response to comment:

To be more precise, the temperature of a system is proportional to the mean square of each degree of freedom that contributes a square to the energy and is in thermal equilibrium at that temperature. In this case (a gas at temperature $T$ moving at mean velovity $v$, however, not all particle velocities can be considered to be in thermal equilibrium, since the macroscopic degree of freedom (the mean velocity) has zero temperature. Thus one must eliminate it, and this leaves $N-1$ degrees of freedom (per spatial dimension, for $N$ particles) which are formally the difference between each particle's velocity and the mean velocity. It is these degrees of freedom that obey a Maxwell-Boltzmann distribution and the equipartition theorem.

• It seems a little strange to say that temperature is xxx to the spread of velocities from the mean, since temperature is a measure of the kinetic energy of the molecules, not the velocities. Then again, the grammar you used makes your claim technically ambiguous. Perhaps temperature can be a measure of statistical variance to get some means of a $\sum{v^2}$ metric. – Alan Rominger Oct 24 '12 at 19:48
• Temperature is really not a measure of either but, fundamentally, of how many new states are available with a given increase in energy. Usually the energy is a quadratic form in some way - like $E=\frac{1}{2}mv^2$ - and this abstract temperature comes down to the mean of $v^2$, which can be interpreted both as an energy and as a statistical quantity. – Emilio Pisanty Oct 25 '12 at 12:32
• @EmilioPisanty about Temperature being "fundamentally, of how many new states are available with a given increase in energy.", I think it's quite the opposite $T=\frac{\partial U}{\partial S}$, the more "new available states" are for a given increase in energy will let to a lower temperature – HDE Nov 6 '12 at 13:50
• The two are equivalent and depend on what representation you use. You can say $T=\frac{\partial U}{\partial S}$, meaning temperature is how much more (less) energy you'll need to (get from) increasing (decreasing) your state space by a unit volume, or you can say $\frac{1}{T}=\frac{\partial S}{\partial U}$, meaning inverse temperature is how much phase-space volume you'll get (lose) for a unit increase (decrease) in energy. It's the same thing. – Emilio Pisanty Nov 6 '12 at 13:54

In thermodynamics, all systems are considered in a frame where the center of mass is stationary.

This precludes the situation where are particles head in the same direction, since then the center of mass heads in the same direction.

The relation between temperature and velocity you describe is associated with a system approximately in thermal equilibrium. This means the particle energies have approximately a boltzmann distribution. So temprature doesn't apply to the case all the particles are going in the same direction.