Maxwell velocity distribution, in 1D or otherwise

Question

I learned from my textbook that Maxwell's velocity distribution gives: $$v_{rms} =\sqrt{\frac{3kT}{m}}$$ $$v_{avg} = \sqrt{\frac{8kT}{\pi m}}$$ Presumably this is for a three dimensions. This confuses me because for one-dimensional distributions the values are $$v_{x,rms} =\sqrt{\frac{kT}{m}}$$ $$v_{x,avg} = \sqrt{\frac{2kT}{\pi m}}$$

The fact that the 1D rms is $\frac{1}{\sqrt{3}}$ times the 3D rms is intuitive (and the site explains why.) I don't see why the 1D average, though, is $\frac{1}{2}$ the 3D average. Is the site correct? And what is the reason?

I think you are right. If there is 3D average, its value will not be different from 1D average. — user115350, May 17 '16 at 20:07
I have tangentially addressed this in several answers at http://physics.stackexchange.com/q/216819/59023, http://physics.stackexchange.com/a/218643/59023, and http://physics.stackexchange.com/a/179057/59023. — honeste_vivere, May 22 '16 at 18:00
You may also be interested in http://physics.stackexchange.com/a/257592/59023. — honeste_vivere, May 26 '16 at 15:01

score 1 · Accepted Answer · answered May 18 '16 at 06:56

The one dimensional Maxwell distribution for the $i-$component of the velocity vector is

$$f_{1D }(v_i) = \left(\frac{m}{2 \pi k T}\right)^{1/2} \exp\left(-\frac{m v_i^2}{2kT} \right)$$

Let's drop the $i$ and $1D$ subscripts for simplicity. You are looking for the average of the absolute value of $v$, $|v|$. To find $\langle |v| \rangle $, we have to perform the integration

$$\int_{-\infty}^{\infty} |v| \ f(v) \ d v$$

Now, since $f(v)$ depends only on the square of $v$ and we are in one dimension, $f(|v|) d|v| = f(v) d v$ and we have

$$\left(\frac{m}{2 \pi k T}\right)^{1/2} \int_{-\infty}^{\infty} |v| \exp\left(-\frac{m |v|^2}{2kT} \right) \ d |v|$$

Since the absolute value is present, this is equal to (let's change notation for clarity, $|v|=u$)

$$ 2 \left(\frac{m}{2 \pi k T}\right)^{1/2} \int_{0}^{\infty} u \exp\left(-\frac{m u^2}{2kT} \right) \ d u$$

The integral can be solved easily integrating by parts, and yields $\frac{kT}{m}$, so that:

$$ \left(\frac{m}{2 \pi k T}\right)^{1/2} \left(\frac{2 kT}{m}\right)$$

Simplifying, we obtain:

$$\langle |v| \rangle = \left( \frac{2kT}{\pi m} \right)^{1/2}$$

Q.E.D. :-)

P.S. Notice that the average of $v_i$ (with no absolute value) would be $0$ for symmetry. For clarity:

$$\langle v \rangle = \int_{-\infty}^{\infty} v \ f(v) \ d v = 0$$

$$\sqrt{\langle v \rangle^2} = v_{rms} = \sqrt{\int_{-\infty}^{\infty} v^2 \ f(v) \ d v }$$

$$\langle {|v|} \rangle = \int_{-\infty}^{\infty} |v| \ f(v) \ d v $$

score 0 · Answer 2 · answered May 19 '16 at 05:02

The definition of $v_{avg}$ for 3 dimension is $$v_{avg}=\int_0^\infty |v|f(v)dv$$ However, it needs to be clear that $$v^2=v_x^2+v_y^2+v_z^2$$

For 3D, the distribution can be derived from 1D distribution but a little different. $$f_{3D}(v) = \left(\frac{m}{2 \pi k T}\right)^{3/2} \exp\left(-\frac{m v^2}{2kT} \right)$$

After similar integration as above, you can get the same result as in your post.

The 3D average is not straightforward but you have to go through definition, which in general consistent with what we do in daily life.

Maxwell velocity distribution, in 1D or otherwise

2 Answers2