Vector calculus eq. needs translation

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Goddar
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Homework Statement


Hi, this is not part of a problem but just an equation I'm having a hard time to decipher (for the reference the original one is in "Statistical Mechanics" by Pathria, eq. 3.7.16)
We define: r = |r2r1|,
Where bold letters are vectors, and we basically integrate a function of 2 vectors over a volume V. So here's the beast:
[tex]\int[/tex][tex]\int[/tex]g(r2r1)dr1dr2=V[tex]\int[/tex]g(r)(4[tex]\pi[/tex]r2dr

(the integral on the right side runs now from 0 to [tex]\pi[/tex]. Sorry the pi's shouldn't appear like raised powers)
I understand the idea, roughly, but can't make the math rigorous: mainly, i don't know what to do with dr1dr2 to obtain the right side...
Thanks for helping
 
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He's made a change of variables to [tex]\mathbf{R} = \mathbf{r}_1+\mathbf{r}_2[/tex] and [tex]\mathbf{r} = \mathbf{r}_2-\mathbf{r}_1[/tex]. I don't have the text, but presumably [tex]g[/tex] is a function of [tex]| \mathbf{r}_2-\mathbf{r}_1|[/tex] only. However I believe that the integral on the RHS must still vary from [tex]0[/tex] to [tex]\infty[/tex].
 
The RHS integral is indeed from 0 to infinity, my mistake.. But g is a function of r2r1 so while a scalar, it's a function of a vector (the vector-difference of the r's).
As defined here however, r is not a vector but the scalar |r2r1|.
If i define a vector R = r2+r1,
for instance, i can express the integral as:
[tex]\int[/tex][tex]\int[/tex]g(r2r1)(dRdr1)dr1
But i still can't see how to fill the gap...
 
Goddar said:
The RHS integral is indeed from 0 to infinity, my mistake.. But g is a function of r2r1 so while a scalar, it's a function of a vector (the vector-difference of the r's).
As defined here however, r is not a vector but the scalar |r2r1|.
If i define a vector R = r2+r1,
for instance, i can express the integral as:
[tex]\int[/tex][tex]\int[/tex]g(r2r1)(dRdr1)dr1
But i still can't see how to fill the gap...

If we change variables to

[tex] \mathbf{R} = \mathbf{r}_1+\mathbf{r}_2, \mathbf{r} = \mathbf{r}_2-\mathbf{r}_1,[/tex]

then

[tex]d\mathbf{r}_1d\mathbf{r}_2 = d\mathbf{R} d\mathbf{r}.[/tex]

I don't have the book handy, so you might want to define [tex]g[/tex] if you need help getting further along.
 
g is not given explicitly, it's kept general as a function of r. But i think your answer is precisely where I'm lost:
it seems like dRdr should be the equivalent of (dr1)2–(dr2)2 to me...
component-wise, i can't seem to make sense of these expressions. Then if you're right, the integration over dR should yield a factor of V and switching from dr to dr would give the integrand a factor of 4πr and send the limit of integration to infinity?
 
Goddar said:
g is not given explicitly, it's kept general as a function of r. But i think your answer is precisely where I'm lost:
it seems like dRdr should be the equivalent of (dr1)2–(dr2)2 to me...
component-wise, i can't seem to make sense of these expressions.

You might want to read up on the Jacobian matrix, since that's the correct way to compute the new measure after a change of variables. http://en.wikipedia.org/wiki/Jacobian_matrix_and_determinant

Then if you're right, the integration over dR should yield a factor of V and switching from dr to dr would give the integrand a factor of 4πr and send the limit of integration to infinity?

Yes.