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I am reading a document on cosmology and particle physics (this s the first time I look into something like this). It frequently states the word "non-relativistic" but I do not understand what is meant by this. When I try to Google it, I don't get a straight answer. The passage I'm referring to is:

"This means that away from temperatures where particles become non-relativistic we find that the factor of proportionality, i.e. the slope of the decrease of the temperature, is constant and depends on the relativistic degrees of freedom. If one species drops out of equilibrium because it becomes non-relativistic, then its entropy density (like its energy density) decays exponentially. However, the net entropy has to stay constant so the particle that becomes non-relativistic has to transfer its entropy to the particle species that are still in thermal equilibrium. For example, when electrons and positions are in thermal equilibrium we have the reaction:..."

My question is: what does the author mean when he says the particles become non-relativistic?

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In general, taking the non-relativistic limit for a moving particle is to assume that its velocity is much smaller than the speed of light, i.e. that $$\frac{v}{c}\ll 1.$$

In this limit, the laws of Special Relativity coincide with the laws of Newtonian physics, and (most) relativistic effects can be ignored. For example, in Special Relativity the equations that govern the transformations from one inertial frame $S$ to another $S'$ that's moving with a velocity $v$ with respect to $S$ are the Lorentz Transformations:

\begin{aligned} x' &= \gamma \left( x - v t \right),\\ t' &= \gamma \left( t - \frac{v}{c^2} x \right), \end{aligned}

where $$\gamma = \frac{1}{\sqrt{1 - \left(\frac{v}{c}\right)^2}}.$$

In the limit $v/c \ll 1$, $\gamma = 1$, and you can see that these just reduce to the Galilean Transformations of Newtonian physics:

\begin{aligned} x' &= x - vt\\ t' &= t \end{aligned}

In the document you've linked, the author speaks about temperature since at thermal equilibrium the "temperature" of a substance is proportional to the average kinetic energy of the molecules that compose it. Therefore, as the "soup" that the document speaks of cools, the average kinetic energy of its constituents decreases. As a result, the velocities of these constituents start to decrease, meaning they become "less" relativistic.

There is, however, no strict threshold beyond which a particle is "non-relativistic". That being said, there is another way to roughly quantify this: the energy of a (massive) particle in Special Relativity is given by: $$E = \gamma m c^2 \approx m c^2 + \frac{1}{2} m v^2 + ...$$

The first term in the expansion above is the rest mass (energy) and the second is the "kinetic" energy. If a particle's total kinetic energy is much larger than its rest mass energy, you should be able to see that this means the particle is "relativistic".

Given that the average kinetic energy of a system of particles is simply related to the temperature, this basically means that (in units where $c = 1$ and the Boltzmann constant $k_B = 1$) when $T \gg m$, a massive particle can be considered to be relativistic, and when $T \ll m$ a massive particle can be considered to be non-relativistic. In other words, as a system of particles cools, $T \sim m$ is a rough threshold for the particle to be "non-relativistic".

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    $\begingroup$ Nice answer. It may also be worth noting that for non-relativistic particles the kinetic energy is proportional to $p^2$ while for relativistic particles it scales like $p$; because the scaling between kinetic energy and momentum enters the partition function through the Boltzmann factor, the two different regimes have very different statistical and thermodynamical properties (e.g. specific heat capacity, thermal conductivity, thermal equilibriation times, etc). $\endgroup$
    – J. Murray
    Jul 25 at 18:14
  • $\begingroup$ By calling the second term $(½)mv^2$ in the expansion of $E$ the "kinetic energy", you've built "non-relativistic" into your terminology. In particular, the next sentence about "[i]f a particle's kinetic energy is much larger than its rest energy" doesn't make sense with that notion of kinetic energy. $\endgroup$ Jul 25 at 18:28
  • $\begingroup$ @J.Murray Very good point, I didn't actually think of that. I should certainly add that to my answer at the end. I will, once I've worked it out myself :) Andreas, you're right of course that the "definition" I've given is quite tautological, but as far as I understand it, the boundaries aren't really that rigid. My way to deal with this was to put "kinetic" energy in quotes, implying that it isn't quite the correct definition, since it's "classical". However, I do not know of a better way to obtain the $T\sim m$ condition, I'd be very interested if you have a more rigorous way to obtain it! $\endgroup$
    – Philip
    Jul 25 at 19:25
  • $\begingroup$ But some phenomena, such as attraction of charges by currents flowing through conductors, are explained by special relativity at any speed. $\endgroup$ Jul 25 at 23:58
  • $\begingroup$ @RomanOdaisky Yep! I agree, which is why I said "...(most) relativistic effects can be ignored" :) I was actually thinking of such effects in which the sheer number of constituent particles can add the tiny relativistic corrections up to make the effect observable when I wrote that :) $\endgroup$
    – Philip
    Jul 26 at 6:05

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