How to compute a mean square average

[JD_PM]

Homework Statement

We know that

$$< (x_1 - x_2)^2 > = \frac{K_bT}{K}$$

$$< (x_2 - x_3)^2 > = \frac{K_bT}{\gamma}$$

What's $< (x_3 - x_1)^2 >$ equal to?

Homework Equations

The Attempt at a Solution

I have tried: $< (x_2 - x_3)^2 > - < (x_1 - x_2)^2 >$ but did not get $< (x_3 - x_1)^2 >$

How can I get it?

Thanks
[/B]

 

[RPinPA]

$<(x_1 - x_2)^2> = <x_1^2 - 2x_1x_2 + x_2^2 > = <x_1^2> - 2<x_1 x_2> + <x_2^2>$ and similarly for $<(x_3 - x_1)^2>$ and $<(x_2 - x_3)^2>$.

That should give you a clue on how to combine them to get your desired result, but you need to know something about the cross terms like $<x_1 x_2>$. You didn't tell us what these variables are, but information you have on them will probably tell you what those terms are (perhaps 0).

I'm just guessing here. I don't know what it is you're trying to get or what you've tried and why you didn't get the desired result. Perhaps show your work so far?

 

[JD_PM]

$<(x_1 - x_2)^2> = <x_1^2 - 2x_1x_2 + x_2^2 > = <x_1^2> - 2<x_1 x_2> + <x_2^2>$ and similarly for $<(x_3 - x_1)^2>$ and $<(x_2 - x_3)^2>$.

That should give you a clue on how to combine them to get your desired result, but you need to know something about the cross terms like $<x_1 x_2>$. You didn't tell us what these variables are, but information you have on them will probably tell you what those terms are (perhaps 0).

I'm just guessing here. I don't know what it is you're trying to get or what you've tried and why you didn't get the desired result. Perhaps show your work so far?

Yes thanks I think I got it. The key is that the cross products are zero. I expanded both and did $< (x_2 - x_3)^2 > - < (x_1 - x_2)^2 >$

$$<(x_1 - x_2)^2> = <x_1^2 - 2x_1x_2 + x_2^2 > = <x_1^2> - 2<x_1 x_2> + <x_2^2>$$

$$<(x_2 - x_3)^2> = <x_2^2 - 2x_2x_3 + x_3^2 > = <x_2^2> - 2<x_2 x_3> + <x_3^2>$$

$$< (x_2 - x_3)^2 > - < (x_1 - x_2)^2 > = <x_3^2> - 2<x_2 x_3> + 2<x_1 x_2> - <x_1^2> = <x_3^2> - <x_1^2> = < (x_3 - x_1)^2 >$$

I think it is OK but please let me know if you agree.