# Can we find particles about zero speed in Maxwell-boltzmann distribution?

The distribution is:

$$f_\mathbf{v} (v_x, v_y, v_z) = \left(\frac{m}{2 \pi kT} \right)^{3/2} \exp \left[- \frac{m(v_x^2 + v_y^2 + v_z^2)}{2kT} \right]$$

So, if I look at each dimension, it goes like $e^{-x}$. We will most likely to find particles with speed close to zero.

But if the distribution is shown in spherical coordinates:

$$f(v) = \sqrt{\left(\frac{m}{2 \pi kT}\right)^3}\, 4\pi v^2 e^{- \frac{mv^2}{2kT}}$$ And a rough graph:

Then I find it quite unlikely to find particles with speed around zero.

Why is the discrepancy? Can someone explain physically? Or is it just a math problem (I know that spherical coordinate is singular at $v=0$. So I keep saying "around zero". I think at least we should find many particles with speed close to zero.).

Given any small volume in velocity space, the patch near the origin will have the highest probability of having particles. But there aren't many patches near 0.

Not far from the origin, the probability is still pretty high, and there is more volume to integrate over.

Farther away, there is lots of volume to integrate over, but the probability is very low.

• I see your point. But I have another doubt. Looking at each dimension, the average value for speed is zero. But over all, the average value for speed is nonzero. How to account for this? Oct 30, 2016 at 3:53
• The average velocity is 0 because there is just as much negative X (leftward) as positive (rightward). The average speed is not 0. Oct 30, 2016 at 4:02

This question bothered me for a long time also, and I do not find statisfying answers elsewhere. I just leave here an explanation that I think is correct.

The reason for that the Maxwell-Boltzmann distribution $$g(v)=C_1 e^{(-C_2 v^2)} 4\pi v^2$$ seems nonintuitive comes from two aspects, 1. we missed some details in the derivation of $$g(v)$$, 2. we can not extend the intuitive relation between propability density and propability in the 1-d case to the 3-d case simply.

First of all we have to recall the derivation process of $$g(v)$$.

We believe that in the 1-d case, the probability that particles is considered near $$v_x$$ is: $$P(v_x)=C_1 e^{-C_2 v_x^2}dv_x$$ where $$C_1$$ and $$C_2$$ have some physical meanings but we do not stress them here.

In the 3-d case, we start to consider the velocity vector $$\vec{v}=[v_x,v_y,v_z]^T$$, and the probability that particles is considered near $$\vec{v}$$ is: $$P(\vec{v})=P(v_x)P(v_y)P(v_z)$$

Under the assumption of isotropy, $$P(\vec{v})$$ has the form: $$P(\vec{v})=P(v_x)P(v_y)P(v_z)=C_1^3 e^{-C_2 (v_x^2+v_y^2 + v_z^2)} dv_x dv_y dv_z \tag{1}$$

At this time, a good discovery comes out: propability of velocity vector has no relation with its direction, but is only related to its length $$v_x^2+v_y^2 + v_z^2$$. Then based on eqn(1) and the new discovery, we can write the propability of speed which is an extension of eqn(1) (I believe you understand why there is $$4\pi v^2$$): $$P(v)=P(v_x)P(v_y)P(v_z)=C_1^3 e^{-C_2 (v_x^2+v_y^2 + v_z^2)} 4\pi v^2 dv \tag{2}$$ Here comes some pitfalls. eqn(2) looks like eqn(1) but they are totally different. eqn(2) is a equation about the probabilty of speed $$v$$, and eqn(1) is about the probablity of velocity $$\vec{v}$$. Moreover, the volume element also changes. The only common point here is that eqn(1) and eqn(2) shares the same probablity density distribution of velocity $$\hat{f} (\vec{v})=C_1^3 e^{-C_2 (v_x^2+v_y^2 + v_z^2)}$$. We write the eqn(2) more clearly: $$P(v) = \hat{f} (\vec{v}) dV$$

In the 1-d case, we pick up an interval $$(v_{x1}, v_{x2})$$ with constant length $$L=v_{x2}-v_{x1}$$, we believe that the greater probability density $$\hat{f} (v_x)$$ means the greater probability, $$\hat{f}_1 \cdot L > \hat{f}_2 \cdot L$$ when $$\hat{f}_1 > \hat{f}_2$$.

However, in the 3-d velocity space, we still pick up an speed interval $$(v_1,v_2)$$ where $$\Delta v = const$$ -- here the volume of integral element is no longer proportional to $$\Delta v$$ ! With constant $$\Delta v$$, the situation changes : $$\hat{f}_1 \cdot dV_1 \quad ? \quad \hat{f}_2 dV_2 \quad when \quad \hat{f}_1 > \hat{f}_2$$.

Here the volume increses with the cubic power of speed $$v$$. Where the probability density is large enough, the corresponding $$dV$$ is not necessarily large. We can no longer intuitively think that the probability density and probability have obvious positive correlation or linear relationship.

At the final step, we go further on the eqn(2): $$P(v)=g(v)dv, \quad where \quad g(v)=C_1 e^{(-C_2 v^2)}\cdot 4\pi v^2$$ Here $$g(v)$$ is the Maxwell-Boltzmann distribution, i.e., the probability density distribution of speed.

Let's have a brief overview. 1. The $$v$$ in $$g(v)$$ is the speed, which comes from the velocity $$\vec{v}$$ in the derivation. 2. During the derivation from the 1-d case to the 3-d case, the increase of dimensions leads to the fact that probability and probability density are no longer positively correlated. 3. We have to have a clear understanding of what changes in the derivation process from velocity probability density to speed velocity probabitity density ($$\vec{v} \rightarrow v$$, and $$dv_x dv_y dv_z \rightarrow$$ spherical shell), and what does not change in the derivation process (velocity probability density $$\hat{f}(\vec{v})$$). 4. From $$\hat{f}(\vec{v})$$ to $$g(v)$$, the volume element changes from 3-d to 1-d. $$g(v)$$ contains the information of $$\hat{f}(\vec{v})$$. If we understand this four aspects and the derivation process of $$g(v)$$ clearly, we may not think the Maxwell-Boltzmann distribution is nonintuitive again.