JR0014 : DCP-02-024-23

Guru Vishnu

Apr 9, 2020 04:57

> Consider a collection of large number of particles each with a speed $v$. The direction of velocity is randomly distributed in the collection. The magnitude of relative velocity between a pair of particles averaged over all the pairs is:
> (a) zero
> (b) greater than $v$
> (c) less than $v$
> (d) $v$

@JohnRennie Sir, could you explain what is meant by - "magnitude of relative velocity between a pair of particles averaged over all the pairs"?

John Rennie

We have a group of particles, and they all have the same speed $v$ i.e. the magnitude of their velocities is always $v$.

Guru Vishnu

Yes sir.

John Rennie

So take two particles with velocities $v_1$ and $v_2$

We are told $|v_1| = |v_2|$ but the directions are different and that means the relative velocity given by:

$$ \mathbf v_r = \mathbf v_1 - \mathbf v_2 $$

Guru Vishnu

Ok sir.

John Rennie

Can have a magnitude ranging from zero if the particles are moving in the same direction to $2v$ if the particles are moving exactly opposite to each other.

Guru Vishnu

Apr 9, 2020 05:03

I see. The magnitude of $\vec{v_r}$ lies between $0$ and $2v$. So is the average $v$?

John Rennie

That's a good question, and I'll have to think about it ...

Guru Vishnu

And is this what meant by "averaged over all the pairs", sir?

@JohnRennie Ok sir. Could you ping me once you've completed thinking? :-)

John Rennie

@GuruVishnu Yes, for all possible pairs $ij$ calculate $v_{rel(ij)}$ and average all the relative speeds.

Guru Vishnu

@JohnRennie I considered this fact - the magnitude of relative velocity ranges from zero to $2v$ and since the velocity distribution is random, all possible magnitudes of relative velocity is equally favoured and hence the weighted average is nothing but the arithmetic mean of the extremes - $v$. Is this approach correct? If yes then the book might be incorrect.

Book says the answer is - greater than $v$.

John Rennie

At the moment I can't think how to approach the question.

It's not obvious to me that your approach is correct ...

For any two particles the relative velocity depends only on the angle between the two velocities. Then my questions are:
1. how is the relative angle distributed (I'd guess uniformly)
2. what is the functional dependence on the angle

For example if the angles are uniformly distributed than the relative speed will only be exactly $v$ if the dependence on the angle is linear.

Does this make sense so far?

Guru Vishnu

Apr 9, 2020 05:16

I also assumed the angle to be uniformly distributed - that is no particular angle is biased.

@JohnRennie But I didn't get what is meant by "functional dependence" of the angle?

Is that referring to expressing the angle as a function of something?

John Rennie

Suppose $|v_r| = f(\theta)$ where $\theta$ is the angle between the velocities and $f$ is some function.

Then $\langle |v_r| \rangle = \frac{1}{2\pi}\int_0^{2\pi} f(\theta) d\theta$

Yes?

Guru Vishnu

@JohnRennie Just for clarification. Is $<$ and $>$ meant for time averaging? I haven't used/seen that notation much.

John Rennie

Yes, that notation means the mean. So $\langle x \rangle$ is the mean value of $x$.

Guru Vishnu

Ok sir. So is it an alternative to $\bar x$?

John Rennie

Don't know. I just recall learning it as a student.

Guru Vishnu

Apr 9, 2020 05:24

Ok sir. No problem.

John Rennie

Anyhow, the point is that unless $|v_r| = k\theta$, i.e. a linear relationship, the average will not be $(v_{max} - v_{min})/2$

Guru Vishnu

Ok sir. As of now, we shall proceed. But I didn't completely understand this point. I need to think about it.

John Rennie

Let me draw a graph to illustrate the point:

(I'm going to omit the modulus symbols so $v$ is a speed here)

There is some graph $v = f(\theta)$

Guru Vishnu

Ok sir.

John Rennie

Suppose we want the average value of $v$.

Let's now draw the average value on the graph ...

Let's take a nice intuitive example by making it a velocity time graph, and we want the average velocity.

OK so far?

Guru Vishnu

Apr 9, 2020 05:36

@JohnRennie I understand $v_{av}$ lies between $0$ and $v_{max}$, could you tell why isn't it biased towards the down half as the length of the curve is more in the bottom part than the upper part, sir?

John Rennie

If we take the example of a velocity time graph then the average velocity is the total distance moved divided by the total time taken. Yes?

Guru Vishnu

@JohnRennie Yes if it's displacement over distance.

This definition loos like average speed.

John Rennie

And the total distance moved is the area under the graph i.e. the area under the wavy line. That's the area we get from $\int v(t) dt$. Yes?

Guru Vishnu

Yes sir.

John Rennie

So the average velocity is given by $v =\frac{1}{T}\int_0^T v(t) dt$

Guru Vishnu

Apr 9, 2020 05:40

Ok but shouldn't we have $d\theta$ in place of $dt$ as we are averaging over angle instead of time?

John Rennie

21 mins ago, by John Rennie

Then $\langle |v_r| \rangle = \frac{1}{2\pi}\int_0^{2\pi} f(\theta) d\theta$

Guru Vishnu

Ok sir. This seems to represent the area below the curve divided by the total angle in the final graph.

John Rennie

Yes. I used the example of a velocity time graph because that's nice and easy to understand, but it works for any function. To get the average value of the function over a range we integrate over that range then divide by the range.

Guru Vishnu

Ok sir. Understood so far.

John Rennie

If you look at my second picture, with the yellow rectangle, then what I'm saying is that the area of the yellow rectangle has to be equal to the area under the graph.

Guru Vishnu

Apr 9, 2020 05:46

@JohnRennie Ok sir. Got that point. Intuitive :-)

John Rennie

So in this case we have a relative speed that depends on the angle between the velocities, and assuming the angles are uniformly distributed between zero and $2\pi$ we want the average speed.

So we simply integrate the relative speed over the range zero to $2\pi$ then divide by $2\pi$.

Guru Vishnu

Ok sir.

John Rennie

And you can do that calculation.

Guru Vishnu

Ok sir. Let me try. Thank you for your help.

John Rennie

Hmm, actually, can you easily do that calculation? It's easy in 2D but this is a 3D problem ...

Guru Vishnu

Apr 9, 2020 05:54

@JohnRennie Why not? I don't think it's that complicated. $|\vec v|=2v\sin(\theta/2)$. Plugging this into:

13 mins ago, by John Rennie

21 mins ago, by John Rennie

Then $\langle |v_r| \rangle = \frac{1}{2\pi}\int_0^{2\pi} f(\theta) d\theta$

we get:

$\left<|\vec v|\right>=\frac{4v}{\pi}$

And successfully it's greater than $v$ thus matching with the correct answer!

@JohnRennie: Thank you sir :-)

John Rennie

:-)

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