In my course on mathematical fluid dynamics, the temperature is defined by the relation


where $\vec{v_k}$ is the velocity of the $k$-th particle and $\vec{v}$ is the average velocity of the particles in the fluid element.

The sum divided by two is the kinetic energy of the particles caused by their random motion. So, does this mean that if, say, all particles would have the same mass ($\forall k \leq N: m_k=1)$ and same velocity, say for all $k\leq N: \vec{v_k}=(1,0,0)$, the temperature of the fluid element is zero?

Intuitively, I'd say both yes and no. Yes, because temperature is caused by the random motions of the particles, and we've eliminated that by making them all follow the same path. And no, because they're still moving, therefore they still have energy?

I've forgotten how temperature was defined in my thermodynamics class (I do maths), but I'd think both should be compatible. (something with entropy)


1 Answer 1


At zero temperature, strictly you have to use quantum theory not classical theory to describe a fluid. However, we can answer your question without knowing the details of the calculation. It is that a fluid at zero temperature can still flow.

Here are some examples.

  1. Liquid helium. This remains fluid at 1 atmosphere pressure in the limit $T \rightarrow 0$ and it can flow. For example, if it is all still in one reference frame, then in another reference frame moving with respect to the first it is in motion, all at the same velocity, but its entropy is still zero and its temperature is zero.

  2. Similar statements apply to cold gases when the density and pressure is low enough that they don't solidify.

  3. But what about more complicated flow, say flow down a tube with varying cross-section? I think that if the flow is completely deterministic, corresponding to a single quantum state, then there is no entropy and zero temperature.

  • $\begingroup$ Your answer refers to the one condition that matter is at zero temperature. I am starting from another condition: all particles move along one direction. This corresponds to zero temperature, now, if I take your comment into consideration, I'd suppose I have to take quantum effects into account in the "all along one direction"-setup. The only effect I can imagine at the moment is the uncertainty on momentum... Or is the point simply that there will always be random motions at random times due to quantum effects, and thus (1,0,0) is simply not physical? $\endgroup$ Sep 30, 2019 at 19:18
  • $\begingroup$ @PaleBlueDot One statement of the third law of thermodynamics is that it's impossible to actually achieve a zero-temperature state in a real, physical system. $\endgroup$
    – rob
    Oct 1, 2019 at 13:20
  • $\begingroup$ @rob, I know, but I'm not talking about why T = 0 is a problem, I'm talking about what problem arises when making all the particles flow along the same direction. I'm not satisfied with "this leads to T=0, which is a problem". There should be a more dynamical explanation. $\endgroup$ Oct 1, 2019 at 13:40

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