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I know this has been asked before here and I understand how the formula changes when the Maxwell-Boltzmann Distribution (MBD) is written in terms of speed. But I am trying to understand this more intrinsically.

I don't understand why the probability for a particle with energy $E \geq E+dE$ is different from the probability of the speed that belongs to that energy range.

The answer that's being said (I think) is because when the MBD is written in terms of energy, a $dE$ covers a different number of particles than a $dv$ does when it is written in terms of speed.

However, I have a struggle with that answer because of the following:

Starting from a certain speed $v_0$, a $dv$ covers the number of particles that have speeds between $v_0 \geq v_0 + dv$. Those particles would have energies between $\frac{1}{2}mv_0^2 \geq \frac{1}{2}m(v_0 + dv)^2$ which is exactly what $dE$ is. Thus, $dE$ covers those same number of particles that have those speeds $v_0 \geq v_0 + dv$.

How would the number of particles differ?

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  • $\begingroup$ The way you wrote the question makes me think that you have not unterstood the answer given to the linked question. The probabilities are equal. The probability densities are not. $\endgroup$
    – paleonix
    Commented Mar 28, 2020 at 15:57
  • $\begingroup$ @Paul You're saying the probability within $E_0 \geq E_0 + dE$ is the same as the probability within $v_0 \geq v_0 + dv$ where $v_0$ corresponds to $E_0$? It gives me different answers. The densities however are the same; the density at $v_0$ is the same as the density at $\frac{1}{2}mv_0^2$. $\endgroup$
    – Phy
    Commented Mar 28, 2020 at 17:23
  • $\begingroup$ I'm not sure what you mean by $E_0 \geq E_0 + dE$. I guess you mean the interval between $E_0$ and $E_0 + dE$? $\endgroup$
    – paleonix
    Commented Mar 28, 2020 at 17:29
  • $\begingroup$ Either way, maybe take a look at Wikipedia. The distribution of energies is actually found by imposing that the probability (integral over probability density) is the same. You must have some kind of missunderstanding/error. $\endgroup$
    – paleonix
    Commented Mar 28, 2020 at 17:33
  • $\begingroup$ @Paul You're right, I got it mixed up. The densities are different but the probabilities within $dE$ and $dv$ are the same. It does surprise me though why then the most probable speed is different from the most probable energy since the number of particles should be the same whether it be in terms of speed $v_0$ or energy $E_0$. $\endgroup$
    – Phy
    Commented Mar 28, 2020 at 18:23

2 Answers 2

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It is confusing whats the difference between probabilities and probability densities. If you write the the probability distribution expressed with velocities you get $$ 1 = \int_0^\infty f(v)dv.$$ If you want to calculate the probability of a particle having velocity in some finite interval $a < v <b$ you get $$ P(a<v<b) = \int_a^b f(v)dv.$$ If you make a change of variable to energy $E = \frac{1}{2}mv^2$ you get $$ P(a<v<b) = \int_{ma^2/2}^{mb^2/2} f(v(E)) \frac{dv}{dE}dE.$$ Now you can in particular choose $a=v_0$ and $b=v_0+dv$ and the $\textit{probabilities}$ of being in interval $v_0<v<v_0 +dv$ is equal to $mv_0^2/2< E < m(v_0+dv)^2/2$. This just follows from making a variable change in an integral. However, the probability densities of having velocity $v$ and energy $E=\frac{1}{2}mv^2$ are not equal. This is because the probability density of having energy $E$ is not just $f(v(E))$ but you get a contribution from the derivative $\frac{dv}{dE}$.

So, in general the probability densities are not equal $$f(v) \neq f(v(E)) \frac{dv}{dE}.$$ But when you integrate the probability density you get equal probabilities for corresponding velocity and energy intervals.

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  • $\begingroup$ You're right, I got it all mixed up. It is the densities that are different. Is there a way to extrapolate this to explain why the most probable speed is different from the most probable energy? What is exactly wrong with the following statement: If the largest number of particles has an energy between $E_0 \geq E_0 + dE$, then that same large number of particles have the speed $v_0 \geq dv$ that corresponds to that energy $E_0 \geq E_0 + dE$ $\endgroup$
    – Phy
    Commented Mar 28, 2020 at 18:22
  • $\begingroup$ The most probable speed density is $v_{max}=\sqrt{\frac{2k_BT}{m}}$ and the most probable energy density is $E_{max} = \frac{k_BT}{2}$. They're not equal as you said but if you say that integrating them gives the same probability then that means that $$P(\frac{k_BT}{2}) \cdot dE = P(\sqrt{\frac{2k_BT}{m}})\cdot dv$$ But they do not give the same probabiity. Notice that they are not related to each other: $E_{max}\neq\frac{1}{2}mv_{max}^2$ $\endgroup$
    – Phy
    Commented Mar 28, 2020 at 23:15
  • $\begingroup$ The most probable speed is different from the most probable energy just because you get the derivative factor $\frac{dv}{dE}$. Why the probability densities are different is simply a consequence of that the probabilities has to be equal. The probabilities is what is ''real'' in some sense, what you can measure. The probability densities is defined as ''what you integrate over to get the probability''. So that the probability densities have different maxima can never be observed because you always measure the probabilities. $\endgroup$ Commented Mar 29, 2020 at 7:18
  • $\begingroup$ Thank you, this helped me a lot. One last question if you don't mind. The answer in this link physics.stackexchange.com/questions/329395/… explains that the discrepancy between the probability densities being a factor of $\frac{1}{2}$ is because $$\mathrm{d}v \propto \frac{1}{2} \mathrm{d}E$$ I can't derive this proportionality. How is this derived? $\endgroup$
    – Phy
    Commented Mar 30, 2020 at 20:34
  • $\begingroup$ I would say that this proportionality does not say much because you have a bunch of other factors as well. To derive the correct, and general expression take $$E = \frac{1}{2}mv^2 $$ and differentiate w.r.t. $v$ $$\frac{dE}{dv} = mv \implies dE = mv dv.$$ Then you can rearrange this in lots of ways if you want. $\endgroup$ Commented Mar 31, 2020 at 6:14
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$$ E = \frac 1 2 m v^2$$

$$ dE = mv dv$$

so:

$$ \frac{dE}{E} = \frac{mv dv}{\frac 1 2 m v^2}$$

$$\frac{dE}{E} = 2 \frac{dv}{v} $$

Or in the terms you presented:

$$ \frac 1 2 m(v_0 + dv)^2 = \frac 1 2 m(v_0 + 2mv_0dv)$$

(where $dv^2 \rightarrow 0$, has been used):

$$ = E_0 + mdv = E_0 + 2dE \ne E_0 + dE$$

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  • $\begingroup$ Wait, isn't $\frac{1}{2}m(v_0 + dv)^2 = \frac{1}{2}m(v_0^2 + 2v_0dv)$ (where $dv^2 = 0$ as you proposed)? Thus, meaning $$\frac{1}{2}m(v_0^2 + 2v_0dv) = \frac{1}{2}mv_0^2 + mv_0dv = E_0 + dE$$ $\endgroup$
    – Phy
    Commented Mar 28, 2020 at 17:38

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