Why Is Average Relative Speed 4/3v in a Uniform Gas Speed Scenario?

[Oojee]

Homework Statement

For a gas in which all molecules travel with the same speed v, show that average relative speed = 4/3v (rather than sqrt(2) v which is the result obtained when we consider the actual distribution of molecular speeds.)

Homework Equations

The Attempt at a Solution

http://hyperphysics.phy-astr.gsu.edu/hbase/kinetic/menfre.html#c5

 

[hikaru1221]

How do you approach this problem? Show us your work so far.
By the way, I doubt the correctness of the method provided in the link.

 

[Oojee]

How do you approach this problem? Show us your work so far.
By the way, I doubt the correctness of the method provided in the link.

I used the method given in the link & for the case of same speed for each particle this method gives zero average relative speed. I cannot think of any other approach to the problem that is why I posted the question here.

 

[hikaru1221]

IMHO, the method in the link is a fallacy, or at least, it's a source of misunderstanding. The wrong step is that from this equation:
$$v_{rel}=\sqrt{\vec{v_1}^2-2\vec{v_1}\vec{v_2}+\vec{v_2}^2$$
we CANNOT go straight to this equation:
$$<v_{rel}>=\sqrt{<\vec{v_1}^2>-<2\vec{v_1}\vec{v_2}>+<\vec{v_2}^2>$$
simply because these two are different:
$$<\sqrt{\vec{v_1}^2-2\vec{v_1}\vec{v_2}+\vec{v_2}^2}>\neq \sqrt{<\vec{v_1}^2>-<2\vec{v_1}\vec{v_2}>+<\vec{v_2}^2>$$
That this method yields the correct result, I think, is just a mathematical coincidence.

We have the relative speed between 2 particles: $$v_{rel}=|\vec{v_1}-\vec{v_2}| = 2v|cos\phi |$$ where $$\phi$$ is one-half of the angle occupied by 2 vectors $$\vec{v_1}$$ and $$-\vec{v_2}$$. Now some questions:
1. Because of the randomness of the gas, in the viewpoint of each molecule, the average speed of all other molecules relative to it (let V denote this average speed) is the same for every molecule; that is, the average speed in the viewpoint of each molecule doesn't depend on which molecule considered. Do you agree?
2. From that, what can you deduce about the relation between the average relative speed of the whole system and V? Do we only need to compute V in order to calculate the needed average relative speed?
3. Calculating V: Pick an arbitrary molecule with velocity $$\vec{u}$$. You can use $$\vec{u}$$ as a fixed axis and calculate $$<|cos\phi |>$$. Notice the uniform distribution of velocities in every direction, which leads to spherical symmetry.

P.S.: The < > sign means average.

 

[rwisz]

The average relative speed of a gas can be calculated using the root-mean-square speed, which takes into account the distribution of molecular speeds. However, in the case where all molecules have the same speed, the root-mean-square speed will be the same as the average relative speed. This is because the root-mean-square speed is calculated by taking the square root of the sum of the squares of all the speeds, divided by the total number of molecules. In the case where all speeds are the same, this will result in the speed being multiplied by the square root of the number of molecules, which is equal to the total number of molecules. Therefore, the average relative speed in this case will be equal to the speed multiplied by the square root of the total number of molecules, which is equal to 4/3v. This is a simplified result compared to the root-mean-square speed, which takes into account the actual distribution of molecular speeds.