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In my physical chemistry book, it says:

In the study of molecular speeds, we must consider a range of speeds. If we don’t, the probability would be zero. This probability is proportional to the range $\mathrm du_x$. Maxwell, based on probability theory, deduced that it is also proportional to $\mathrm e^{-\frac\beta3}$ or $\mathrm e^{-\frac{\beta u^2}2}$, where $\beta$ is a constant. Therefore, we can express the differential probability $\mathrm dP_x$ that a molecule has a speed component along the $x$-axis between $u_x$ and $u_x +\mathrm du_x$ as $\mathrm dP_x = B e^{-mu_x^2\beta/2}\mathrm du_x$

My question is trying to understand what $\mathrm dP_x$ is. I always like to think of the original functions, if we were to divide both sides by $\mathrm du_x$ and integrate, does that mean we are getting a cumulative distribution function or probability density function or what? if $\mathrm dP_x$ is a small change in $P_x$, what is $P_x$?

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  • $\begingroup$ Ah, if you want to scrutinise a textbook, that textbook must first be written in a rigorous enough way for that kind of scrutiny to make sense. The textbook you are reading is fudging enough to be correct, but not going to be rigorous enough for you to ask that question. You will need a lot more mathematics to actually be able to pose and answer that question. $\endgroup$
    – naturallyInconsistent
    Commented Mar 6 at 16:44
  • $\begingroup$ Should I give up on understanding it if I have not built enough mathematical background then? I have done till calc 2 and general statistics... $\endgroup$
    – Kintoke
    Commented Mar 6 at 16:50
  • $\begingroup$ No, it is very understandable, just that it is pretty much non-rigorous. It simply is saying "The probability density of this particular x slice in velocity". The issue is that this is a handwavy definition, but it will suffice for now. $\endgroup$
    – naturallyInconsistent
    Commented Mar 6 at 17:03
  • $\begingroup$ Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking. $\endgroup$
    – Community Bot
    Commented Mar 6 at 18:08

1 Answer 1

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$P_x$ is a probability distribution of speed in the $x$ direction, If you integrate between $u=0$ and $u=\infty$ the result of the integral is 1. Where $P_x$ is a probability density with units of inverse velocity. $P_x$ is a function of $u$, that is $P_x=f(u)$.

In this context, it is not true that $dP_x=(dP_x/du)du$, but rather, $dP_x=P_x du$, which is a very different thing.

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  • $\begingroup$ if Px is a probability distribution I would imagine that you mean a probability distribution function (PDF). And as far as I know, in a PDF, the y-axis (Px) would represent probability density and not probability and this will have the unit of the multiplicative inverse of whatever on the x-axis.I n the case of speed, the units of Px should be s/m. I thought that the distinction between probability and probability density is important. Am I getting something wrong? $\endgroup$
    – Kintoke
    Commented Mar 7 at 20:27
  • $\begingroup$ As for my question, I am trying to understand what dPx and Px are. So if Px is the probability density, shouldn't the probability that the velocity is between ux and ux+dux be the integral of Px from ux and ux + dux? I do not get the the role of getting dPx at all. $\endgroup$
    – Kintoke
    Commented Mar 8 at 0:09
  • $\begingroup$ You are absolutely right, I was sloppy in my explanation. Plus the probability is about speeds, not momentum, I read it too fast. You are right about the probability being the integral, but because the integration interval is small (a differential), then the integral becomes just the product $P_x du$, that you call $dP_x$. Is this last step that is causing you trouble? $\endgroup$
    – Wolphram jonny
    Commented Mar 9 at 14:29
  • $\begingroup$ I think it is an abuse of notation, as $dP_x$ should have units if you define it as the differential of $P_x$, but it is unitless because you define it as $P_x du$, which is a different thing. It is the same sloppiness I incurred, I believe most physicists do that because they understand in their mind what they are talking about and are less rigorous in terms of math/units/names. $\endgroup$
    – Wolphram jonny
    Commented Mar 9 at 14:35
  • $\begingroup$ So in this context, it is not true that $dP_x=(dP_x/du) du$, but rather, $dP_x=P_x du$ $\endgroup$
    – Wolphram jonny
    Commented Mar 9 at 14:41

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