# I Feynman's explanation of atomic collisions in gas

1. Dec 26, 2017

### themagiciant95

In the first volume of his lectures' books, Feynman tries to mathematically describe that in a particular gas ( particular because in this case is the sum of 2 monoatomic gasses with different masses) as a result of the collisions beetween atoms it will be equally likely to find any pair moving in any direction in space , and to mathematically explain this, says:

The idea is that any area on a sphere centered at a collision point will have just
as many molecules going through it as go through any other equal area on the
sphere. So the result of the collisions will be to distribute the directions so that
equal areas on a sphere will have equal probabilities.

The paragraph in question is the 39-13 of the first volume. It's about the relation beetween kinetic energy and temperature.
I have 2 questions:

1) Could you explain me better what this sentence means?
2) If you know, why he use this example to explain the motion of the piston beetween 2 monoatomic gasses ?

2. Dec 26, 2017

### dRic2

I will try to answer your first question, but I didn't read the book, so I'll try my best. Imagine to "draw" a sphere centered in point of collision of two molecules like this:

(I found this online just to give you an example, but it will work with different particles and with a sphere of arbitrary radius).

You don't know nothing about the particles: you don't know the direction where they come from and so you can't predict the direction they will take after the impact.

Consider the area on the surface of the sphere that one particle will "hit" (go through) as it enters the sphere. Than en equal area should be "created" when the particle exits the sphere (after the collision). It is obvious that the two areas have to be the same because the are "created" by te same particle.

How many of these areas can you draw on the surface of the sphere? An infinite number (!) because the particles can arrive from any direction (note that you don't know anything about the particles so you can't choose a direction: you have to consider all the infinite possibility).

And agin since you don't know anything about te direction of the molecules you have to consider that any of areas that we discussed previously (on the surface of the sphere) has the same probability to be "hit". So, since those areas are connected to the direction of the molecules, every direction has the same probability to be taken.

That's how I figured it out, let me know. Sorry but I didn't read the book, so I don't know the example you are referring to in the second question.

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3. Dec 26, 2017

### jartsa

First Feynman explains that an elastic collision of two particles will cause a random change of direction of the colliding particles. Seems that many random changes of direction are needed to make the direction random, probably because the random changes of direction are most probably small random changes of direction.

Then Feynman tells us how "the direction has now become random" can be said in mathematics.

4. Dec 27, 2017

### themagiciant95

Suppose we have 2 particles colliding ,a particle has a velocity v1 and the other v2.
Is it true that i can find infinite combinations of v1-v2 that gives a certain Center of Mass Velocity ?

This because, Feynman goes on : (always about the collision of 2 particles)

"Now then, what is the distribution resulting from this? From our previous
argument we conclude this: that at equilibrium, all directions for w are equally
likely, relative to the direction of the motion of the CM.
There will be no
particular correlation, in the end, between the direction of the motion of the
relative velocity and that of the motion of the CM.

So the cosine of the
angle between w and vCM is zero on the average. That is:
$$<\bar{w}\cdot \bar{V}_{cm}>=0$$

Where w is v1-v2 and Vcm is the Center of mass' velocity"

In addition, what's the intuitive and practical meaning of
"So the cosine of the angle between w and vCM is zero on the average."

5. Dec 28, 2017

### themagiciant95

$$<\bar{w}\cdot \bar{v}_{cm}> = 0$$ for the collision of 2 particles of a gas composed of 2 monoatomic gasses with different masses (situated in a box), i think that he implicity extend this formula to all the collisions in the gas.