# 0 Kelvin body moving

As many books say:

Temperature is (proportional, almost, etc...) average kinetic energy of particles.

My question is this.

"Suppose there is a body somewhere in empty space which moves at velocity v thus possessing KE not equal to 0 Joules, but has temperature equal to 0K. Can this be possible in a theoretical sense?"

In my view it is possible since I think T is a measure of KE only for non-coherently moving particles. And if all particles of the body move coherently in one direction KE will not be equal to 0 Joules, but T will be 0 Kelvin.

P.S. Since in the case of classical Thermodynamics, the engine (e.g. Carnot engine) does not move as a whole; it does not possess any "coherent KE", only "random KE" and thus the heuristic "temperature is average KE" works just fine. But this "rule of thumb" breaks down if the body moves as a whole. Am I right?

• physics.stackexchange.com/q/83488 – user83548 Nov 20 '15 at 14:37
• If you can shift to a reference frame where the body is not moving, then you can discount its movement. – OrangeDog Nov 20 '15 at 16:02
• This thought experiment entails a steady-state current of particles; it is not a thermodynamic equilibrium situation. However, the temperature is defined only in a thermodynamic equilibrium setting; so, here, the temperature is not defined. – AlQuemist Nov 27 '15 at 13:34

## 5 Answers

I think you are right. A perhaps more precise relation between temperature and velocity is the Maxwell–Boltzmann distribution: \begin{equation*} P(\textbf{v}) = \left( \frac{m}{2\pi k_B T} \right)^{3/2} \text{exp} \left[-\frac{m ( \textbf{v} - \textbf{v}_0)^2}{2 k_B T} \right]. \end{equation*} where you see that the mean velocity $\textbf{v}_0$ and the temperature are independent. Only the variance of the velocity is related to temperature.

• Only variance is related to Temperature. Nice wording. Need to think about it. – coobit Nov 20 '15 at 18:38
• But if T goes to zero, P(v) is not well-defined. – aeismail Nov 22 '15 at 12:23
• @aeismail if you treat it as a distribution, it is well-defined to be Dirac delta. – Ruslan Nov 22 '15 at 12:45
• The Maxwell distribution of velocities is valid in a thermodynamic equilibrium situation, but the above thought experiment is not in thermodynamic equilibrium; it describes a steady-state current of particles in one direction. Note that the Wikipedia page states explicitly: The distribution is used for particle speeds in idealized gases where the particles move freely inside a stationary container... [with] collisions [to] exchange energy or momentum with each other or with their thermal environment;... the system of particles... have reached thermodynamic equilibrium.'' – AlQuemist Nov 27 '15 at 13:44
• Well, it is just a matter of defining a temperature. You can stick to the temperature defined only for system at thermodynamic equilibrium indeed. However, it is sometimes convenient to define a temperature even if not at thermodynamic equilibrium. In this sense, I believe the above "generalization" of Maxwell distribution is not completely meaningless. – Tony Nov 27 '15 at 14:24

I think your view is correct, and you can think about the following real word example. In labs here on earth, we can use laser cooling techniques to cool atoms to $\mu$K scales in the lab frame. But the lab is on earth, and the earth is moving very fast around the sun, and the sun is moving very fast around the galactic center and so on. We don't take into account this additional kinetic energy when considering the temperature.

If your object is moving at a constant velocity, you can always put yourself in a frame where the mean velocity of that object is zero, and what you are left with is the distribution of velocities that will define the temperature of the object.

• since the Earth isn't actually moving at constant velocity (otherwise it would fly on a tangent out of the Solar System), how does this factor in to how close to absolute zero we can cool an atom? – Michael Nov 20 '15 at 17:40
• @Michael I don't think you would see an effect. Think about dropping an object in a gravitational field. It is accelerating and thus in a non-inertial frame (if your frame is centered on the object), but the relative velocity distribution of the particles is unaffected because they all accelerate the same way. In fact, the temperature of ultracold atoms can be measured by releasing the atoms from the trap and measuring the free expansion of the gas, even when it is falling due to gravity. – tmwilson26 Nov 20 '15 at 18:10
• @tmwilson26: well, in general relativity it is in any case when you release them from the trap that they cease to be on average motionless relative to an accelerating frame. Freefall is the neutral state. – Steve Jessop Nov 20 '15 at 23:38
• isnt temperature relative? – bubakazouba Nov 22 '15 at 20:46
• @bubakazouba it's more related to the distribution of velocities. You do have to consider things in the reference frame where the mean velocity is zero. Otherwise defining the temperature in terms of the mean kinetic energy doesn't make sense. – tmwilson26 Nov 22 '15 at 21:37

It's simple. You can think of temperature as being the standard deviation of KE among all components (atoms) of a mass. This is significant because KE is a relative quantity, but temperature is absolute, and this relationship makes that possible. If all atoms are moving uniformly in the same direction, then the temperature would be 0.

• Simple indeed. I wonder why books don't use this simple heuristical definition: " the standard deviation of KE among all components "?! I would have saved me much time. – coobit Nov 21 '15 at 11:55
• In this case it's convenient to think of it that way, but it's actually more complicated because waves don't always act like billiard balls. – James Watkins Nov 23 '15 at 16:25

You are correct. The temperature is usually defined in terms of random motion of the atoms. This announcement from NIST talks of an atomic beam cooled to $30 \mu K$, so there is very little random energy. It doesn't give the speed of the beam, but that could be very high if the atoms are all moving in the same direction.

One way to define temperature is via statistical physics. It is a quantity used to describe an equilibrium distribution of possible states via Boltzmann factor. Given a total energy of a classical system

$$E(x_1,\ldots,x_N, p_1, \ldots p_N)$$,

where $x_1 \ldots x_N$ are the position coordinates, and $p_1 \ldots p_N$ are the momentum coordinates. The probability of the system being in a particular state in now given proportional to

$$P(x_1,\ldots,x_N, p_1, \ldots p_N) = Z^{-1} e^{-E(x_1,\ldots,x_N, p_1, \ldots p_N) / (k_B T)}$$

In classical system of particles, it almost always is the case, that the total energy is a sum of kinetic energy and velocity independent potential energy.

$$E(x_1,\ldots,x_N, p_1, \ldots p_N) = V(x_1,\ldots,x_N) + \sum_i^{N} \frac{p_i^2}{2m}$$

We could play with the idea of dividing the total kinetic energy of the system to three components, translation kinetic energy, rotation kinetic energy and vibrational kinetic energy with help of suitable coordinate system

$$E_{tot} = E_{trans} + E_{rot} + E_{vib}$$,

where the translation kinetic energy is given by

$$E_{trans} = \frac{P^2}{2M}$$,

where $P=\sum_i p_i$ and $M=\sum_i m_i$. Now, we can speculate, that the probability for the system center of mass having a certain velocity V is roughly proportional to

$$P(V) \approx e^{-\frac{1}{2}MV^2 / (k_BT)}.$$

Given that, the total kinetic energy is huge compared to $k_BT$ (which is just fractions of an eV in room temperaure), we can conclude that if we find a system with large V, this is a non-equilibrium system and temperature is defined only on equilibrium systems. Therefore, the definition of temperature is ill-defined here.

Ok, this was just purely formal. In practice, one does not consern one-self with such formalities, but just moves to center-of-momentum coordinates or something similar. For example, it is standard practice in molecular dynamics simulations to first initialize the velocities of atoms according to Maxwell-Boltzmann distribution, but then to remove total momentum and total angular momentum to prevent excess drift and rotation. The number of particles is usually so large, that one does not even consider the fact that there are few degrees of freedom less, than one would get just by based on particle number. An exception is few particle systems, where one needs to switch to reduced mass coordinates.

## protected by Qmechanic♦Nov 21 '15 at 1:46

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?