I am aware that the 1D MB Distribution is usually derived by counting the momentum states in various directions. However, I would like to know if the following method using the 3 dimensional speed distribution should also work for the derivation. This link states that the 3 dimensional MB Distribution in terms of speed $P(v)$ is. $$P(v)=\left(\frac{m}{2\pi k_B T}\right)^{\frac{3}{2}} \cdot \exp\left(-\frac{mv^2}{2k_BT}\right) \cdot 4\pi v^2 \cdot dv$$ This would mean that the probability for a speed $v$ in 1 unit of volume in speed space would be: $$\frac{P(v)}{4\pi v^2\cdot dv}=\left(\frac{m}{2\pi k_B T}\right)^{\frac{3}{2}} \cdot \exp\left(-\frac{mv^2}{2k_BT}\right)$$ If I want to compute the probability of speed $v$ but with a certain fixed value of a dimensional speed component, say $v_z$, I would predict that this would give the 1 dimensional speed distribution $v_z$ that belongs to $v$. I would deduce that this can be derived by calculating the following circular shell volume in speed space, shown in blue lines:

enter image description here

This circular shell volume $V$ basically contains a fixed value of $v_z$ while varying $v_y$ and $v_x$. I would therefore reason that if I formulate this volume and multiply it with the probability distribution for the speed per 1 unit of volume (previous formula), I'd get the 1 dimensional MB distribution for $v_z$. Since this circular shell volume can also be drawn at the below side of the sphere, a factor of $2$ must be added.

However, I have trouble formulating this volume in the first place. It would have a width of $v\cdot d\theta$, a thickness of $dv$ and a length of $\sin(\theta) \cdot v \cdot \int^{2\pi}_0 d\phi$ (the circumference at height $v_z$) which means that the volume is equal to: $$V = v^2 \cdot d\theta \cdot dv \cdot \sin(\theta) \cdot \int^{2\pi}_0 d\phi$$ From this link, I can see that $dv$ can be rewritten in terms of the 3 speed components and $\theta$ and $d\phi$: $$dv = \frac{dv_xdv_ydv_z}{v^2 \cdot \sin(\theta) \cdot d\theta \cdot d\phi}$$ Substituting $dv$ with this formula would give the following formula for the volume: $$V = \frac{dv_xdv_ydv_z}{d\phi}\cdot \int^{2\pi}_0 d\phi$$ I'm not sure how to rewrite this to be able to continue with the derivation.

  • $\begingroup$ @Greg.Paul Yeah, I noticed this afterwards but I don't know how to rewrite the integration otherwise. $\endgroup$ – JohnnyGui Aug 28 '19 at 14:30
  • $\begingroup$ Speed is proportional to momentum for nonrelativistic particles, but you had better stick to momentum when dealing with relativistic velocities. $\endgroup$ – Bert Barrois Aug 28 '19 at 14:59
  • $\begingroup$ @BertBarrois Am I dealing with relativistic velocities here? From what I can gather from the link I gave, it does not mention relativistic velocities. What I did is derive a concept based on that link that, at least in my mind, should give the 1D MB Distribution. But I want to verify this and know what the reason is if the concept is faulty. $\endgroup$ – JohnnyGui Aug 28 '19 at 19:42
  • $\begingroup$ If it’s relativistic, you had best use ${{d}^{D}}p\exp \left( -\beta \sqrt{{{p}^{2}}+{{m}^{2}}} \right)$ in any number of dimensions. $\endgroup$ – Bert Barrois Aug 29 '19 at 10:25

The derivation in the previous answer gives you $P(v_z)$, but my understanding is that you want $P(v_z |v)$, that is, the probability of finding $v_z$ at a fixed $v$.

You should integrate about the z-axis a ring of radius $R=v \sin\theta$ with fixed $v$ and $v_z$. The element of volume will be:

$dV=dR dv_z ds=v\sin\theta d\phi dv_z \sin\theta dv=v\sin^2\theta dv_z dv d\phi$

You can eliminate $\sin\theta$ using $v_z=v\cos\theta$ and get $\sin^2\theta= 1-(\frac{v_z}{v})^2$ and get:

$P(v,v_z)dv dv_z=2 \pi\left(\frac{m}{2\pi k_B T}\right)^{\frac{3}{2}} v(1-(\frac{v_z}{v})^2) \exp\left(-\frac{mv^2}{2k_BT}\right)dv dv_z$

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  • $\begingroup$ Yes, this is indeed what I want! I have a little trouble understanding your element of volume. Shouldn't the horizontal length be $v \sin(\theta) \cdot d\phi$, the height being $dv_z$ and the thickness $dv$? $\endgroup$ – JohnnyGui Oct 23 '19 at 21:50
  • $\begingroup$ The height is $dv_z$, the thickness is in the radial direction so is dR, and the "lenght", which I think you mean the differential of circumference is $v \sin(\theta) \cdot d\phi$, you are right, I missed a factor, so the results is wrong but the idea is the same $\endgroup$ – Wolphram jonny Oct 24 '19 at 0:34
  • $\begingroup$ Actually I might had make more that one algebraic error, I need to check all again. I guess it is right now. $\endgroup$ – Wolphram jonny Oct 24 '19 at 0:37
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    $\begingroup$ The thickness is not $dv$ because $dv$ points radially from the origin, it is dR because it points radially from the center of the ring $\endgroup$ – Wolphram jonny Oct 24 '19 at 0:41
  • $\begingroup$ Thanks for clarifying. Sorry, but I don't get how you deduced that $dR=\cos(\theta)dv$. If $R =\sin(\theta)\cdot v = \sin(\theta) \cdot \frac{v_z}{\cos(\theta)}$, then I get $dR=d\theta \cdot v_z \cdot (\frac{\sin^2(\theta)}{\cos^2(\theta)}+1)$ $\endgroup$ – JohnnyGui Oct 25 '19 at 17:30

I don't fully uderstand your reasoning, but here I show a simple derivation for $P(v_i)$. First you have \begin{gather} P(v_x)P(v_y)P(v_z) ~\mathrm{d}v_x\mathrm{d}v_y\mathrm{d}v_z=P(v)P(\theta,\phi) ~\mathrm{d}v\mathrm{d}\theta\mathrm{d}\phi \end{gather} And since the distribution is isotropic we have \begin{gather} P(\theta,\phi) ~\mathrm{d}\theta\mathrm{d}\phi = \frac{\mathrm{d}\Omega}{4\pi}= \frac{\sin \theta ~ \mathrm{d}\theta\mathrm{d}\phi}{4\pi} \end{gather} Hence \begin{align} P(v_x)P(v_y)P(v_z) ~\mathrm{d}v_x\mathrm{d}v_y\mathrm{d}v_z&=\left(\frac{m}{2\pi k_BT}\right)^\frac{3}{2} \exp\left(-\frac{mv^2}{2m k_BT}\right) 4\pi v^2 \mathrm{d}v\times \frac{\sin \theta ~\mathrm{d}\theta\mathrm{d}\phi}{4\pi}\\ P(v_x)P(v_y)P(v_z) ~\mathrm{d}v_x\mathrm{d}v_y\mathrm{d}v_z&=\left(\frac{m}{2\pi k_BT}\right)^\frac{3}{2} \exp\left(-\frac{mv^2}{2m k_BT}\right) v^2 \sin \theta ~\mathrm{d}v\mathrm{d}\theta\mathrm{d}\phi\\ P(v_x)P(v_y)P(v_z) ~\mathrm{d}v_x\mathrm{d}v_y\mathrm{d}v_z&=\left(\frac{m}{2\pi k_BT}\right)^\frac{3}{2} \exp\left(-\frac{m(v_x^2+v_y^2+v_z^2)}{2m k_BT}\right)~\mathrm{d}v_x\mathrm{d}v_y\mathrm{d}v_z \end{align} Finally you get \begin{gather} P(v_i)=\left(\frac{m}{2\pi k_BT}\right)^\frac{1}{2} \exp\left(-\frac{mv_i^2}{2m k_BT}\right) \end{gather}

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  • $\begingroup$ Thanks for your reply. What exact part of my reasoning don't you understand? I'm curious to know that because I want to know if my concept should give the correct 1D MB Distribution as well. Another question; does $P(\theta,\phi) d\theta d\phi$ represent an infinitesimally small surface piece on the velocity sphere? $\endgroup$ – JohnnyGui Sep 2 '19 at 13:27
  • $\begingroup$ I have improved my question, perhaps it makes it more clear. Another question, the link hyperphysics.phy-astr.gsu.edu/hbase/Kinetic/maxspe.html shows that $\int^\pi_0 \int^{2\pi}_0 v^2 \sin(\theta) d\theta d\phi = 4\pi v^2$, but how can this be integrated over $d\phi$ if the function itself does not contain $\phi$ at all? $\endgroup$ – JohnnyGui Sep 6 '19 at 22:00
  • $\begingroup$ you integrate it as a constant: $\phi v^2 sin \theta |_0^{2\pi}$ $\endgroup$ – Wolphram jonny Sep 7 '19 at 3:07
  • $\begingroup$ @Wolphramjonny Thanks for the info. Could you perhaps also help me with my question in the opening post? $\endgroup$ – JohnnyGui Sep 9 '19 at 22:14
  • $\begingroup$ Your mistake is that the element of volume that you chose does not have a constant $dv_z$, but a constant $d\theta$ $\endgroup$ – Wolphram jonny Sep 10 '19 at 17:46

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