It is a postulate of Kinetic theory that the speed of molecules ranging from 0 to infinity. But we know that no object can exceed the speed of light. Then how can this be possible?
Asked
Active
Viewed 82 times
2
-
Can you please give a reference/quote of where you read that the speed could range from 0 - infinity? – LostCause Jul 22 '19 at 16:45
-
nothing is "approximately infinity". – JEB Jul 22 '19 at 18:21
-
@JohnRennie I do not consider this question a duplicate of the question indicated in the link. That question is about the correct relativistic velocity distribution function. The present one is asking how to justify the validity of the classical formula, taking into account that it allows for superluminal velocities and yet it works. – GiorgioP-DoomsdayClockIsAt-90 Jul 22 '19 at 19:00
-
The question has a perfectly reasonable answer within the non-relativistic treatment. – GiorgioP-DoomsdayClockIsAt-90 Jul 22 '19 at 19:02
-
1You might just as well ask why anybody bothers to talk about Newton's laws of motion after Einstein proved them wrong. The answer is, Newton's laws are good enough for describing many phenomena, and Newton's math is a whole lot easier to work with than Einstein's math. I am sure that somebody could formulate a relativistic, kinetic theory of gasses; but just the fact that nobody's done it yet is pretty good proof that nobody's needed it yet. – Solomon Slow Jul 22 '19 at 20:48
1 Answers
3
Under some reasonable hypotheses about general properties of the equilibrium velocity distribution function, kinetic theory is able to derive (it is not a postulate) the Maxwellian velocity distribution $$ f({\bf v}) = \left( \frac{m}{2\pi k_B T} \right)^{3/2} \exp \left( -\frac{m v^2}{2 k_B T} \right)~~~~~~~~~~~~~[1] $$ where $m$ is the mass of a single molecule, $k_B$ the Boltzmann's constant, $T$ the temperature in kelvins, $v^2 = v_x^2+v_y^2+v_z^2$, and each cartesian component of the velocity goes from $-\infty$ to $+\infty$. The product $f({\bf v}) d v_x d v_y d v_z$ represents the fraction of the particles with velocity components in the intervals $[v_x,v_x+d v_x ]$, $[v_y,v_y+d v_y ]$, and $[v_z,v_z+d v_z ]$. We can pass from the distribution of velocities to the distribution of the modulus of the velocity by multiplying $[1]$ by the volume element $4 \pi v^2 dv$. Of course $0<v<+\infty$.
Notice, that in the derivation of this result, the hypothesis has been done that classical mechanics has unlimited validity. Therefore it is not a surprise that no upper or lower limit for each component of velocity, and then no upper limit for the speed of each particle appears in the formulae.
The same result can also be obtained by using Statistical Mechanics in the canonical ensemble for a non-relativistic classical system. In that case, it is possible to interpret the unboundedness of the velocity as due to the possibility of unbounded fluctuations of energy, due to the energy exchanges with a thermostat. Once again, the underlying Classical Mechanics formalism does allow the possibility of arbitrarily large values for the speed of a particle.
At this point, the obvious question would be: how is that the Maxwell distribution provides a satisfactory description of usual laboratory experiments if it contains a condition (the possibility of arbitrarily high speed) which is known to be false within the more general Special Relativity (SR)?
The answer is that it is true, the Maxwell distribution is not coherent with SR. However, provided the temperature is not too large (and we can estimate what's the meaning of large in the present context, see later) it provides an extremely good quantitative approximation.
Even without knowing the precise form of the relativistic velocity distribution, we may judge the validity of the Maxwell result from a self-consistency argument: as long as the fraction of particles with speed comparable with the speed of light is negligible, we can safely use the classical result.
The fraction of particles with speed higher than a threshold $v_0$ is given by $$ 4 \pi \left( \frac{m}{2\pi k_B T} \right)^{3/2} \int_{v_0}^{+\infty} v^2 \exp \left( -\frac{m v^2}{2 k_B T} \right) dv $$ or, using the dimensionless variable $x=\left( \frac{m v^2}{2 k_B T} \right)^{1/2}$, $$ \frac{4}{\pi^{1/2}} \int_{x_0}^{+\infty} x^2 \exp \left( -x^2 \right) dx $$
Therefore, is clear that even with $v_0\approx c/1000$, and temperatures of the order of a few thousands kelvin degrees, $x_0\gg 1$ the fraction of relativistic particle is negligible even in the worst case of a gas of hydrogen atoms.
