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The Maxwell speed distribution equation is given as

$$f(v) = 4\pi \biggl(\frac{m}{2\pi kT}\biggr)^{3/2}\exp\biggl(-\frac{mv^2}{2kT}\biggr)v^2.$$

The left hand side gives the fraction of molecules and is a rational number. But the right hand side is a product of several kinds of numbers, such as integers, fractions, irrational numbers, and transcendental numbers.

Since a rational number can only be equal to a rational number, the right hand side must also be a rational number. Thus, in order that an inconsistency/impossibility does not arise, how do we ascertain that we always get a rational number on the right hand side?

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  • $\begingroup$ @Omry If we don't get a rational number on the right hand side, the equality would not hold, since a rational number can only be equal to a rational number.Since a rational number cannot be equal to an irrational number or a transcendental number an impossibility arises. $\endgroup$ – user55356 Jun 6 '15 at 15:28
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    $\begingroup$ Let me present you with another problem: if we have a finite number of molecules, there will be a finite number of speeds, which means that the left hand side should be zero for almost every value of $v$. How come the right hand side isn't zero? $\endgroup$ – Javier Jun 6 '15 at 15:33
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A few issues with this argument. First of all, you're clearly right that the RHS can yield an irrational/transcendental number. It would require insane mathematical coincidence for this not to be so. (It's really hard to have $e^x$ be algebraic, for instance.) And there's nothing in the formula about how many particles there are; we could imagine $N =2$, which would imply that the LHS can only be $0, .5, 1$!

First of all, let's recognize that this is a probabilistic formula like many in statistical mechanics. It holds on average, and yields the probability that an atom has velocity $v$. In general, you can interpret this as "fraction of atoms with that velocity" but clearly there are mathematical difficulties with that interpretation.

Next, many formulas in statistical mechanics hold only in the thermodynamic limit, which means that they assume $N \to \infty$. This then allows for any real number to appear on the LHS, since we have a continuum of values rather than just something that is $m/N$.

Finally, remember that physics is emphatically not math, and really never can be. Any real experiment inevitably brings in complications that we neglect so as to make the problem analytically tractable, so the physical systems we experiment with are never the mathematical systems we calculate with. We can aim for a close congruence between the two, but let's not forget that they really are different things.

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  • $\begingroup$ I agree with all with what you say. 'It holds on average' , what do we mean by this? It holds some times and fails some times? Maxwell says about the nature of 2nd law that it is statistical in nature! Does it imply it holds but not always? $\endgroup$ – user55356 Jun 7 '15 at 15:15
  • $\begingroup$ Well, it really depends how you interpret the quantity $f(v)$. You're talking about it as the atomic fraction; I would usually call it a probability. It is always the case that the probability that a particular atom has speed $v$ is $f(v)$, but it might not be the case at a particular instant in time that $N f(v)$ of them have speed $v$ due to fluctuations. We assume that the system passes through many, many different configurations all the time, such that we can take an average over all the possibilities. This is called the ergodic hypothesis. $\endgroup$ – zeldredge Jun 7 '15 at 17:24

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