Vector calculus eq. needs translation

In summary, the homework statement is saying that r is not a vector, but the scalar |r2–r1|. To calculate the right side of the equation, he has to change variables to \mathbf{R} = \mathbf{r}_1+\mathbf{r}_2, \mathbf{r} = \mathbf{r}_2-\mathbf{r}_1, and then use the Jacobian matrix to calculate the new measure after a change of variables.
  • #1
Goddar
205
16

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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  • #2
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].
 
  • #3
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...
 
  • #4
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.
 
  • #5
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?
 
  • #6
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.
 

FAQ: Vector calculus eq. needs translation

What is vector calculus?

Vector calculus is a branch of mathematics that deals with multivariable calculus and the study of vector fields and their derivatives. It is used to analyze and solve problems in physics, engineering, and other scientific fields.

Why does vector calculus need translation?

Vector calculus equations often involve vector operations and derivatives that are specific to vector fields. Translation is necessary to convert these equations into a language that can be understood and used by computers for numerical calculations and simulations.

What is the importance of vector calculus in scientific research?

Vector calculus is essential in scientific research as it provides a powerful tool for analyzing and solving complex problems involving vector fields. It is widely used in fields such as physics, engineering, and computer graphics to model and understand real-world phenomena.

Can anyone learn vector calculus?

Yes, anyone can learn vector calculus with proper instruction and practice. It is a fundamental subject in many scientific disciplines, and there are many resources available online and in textbooks to help individuals learn and understand it.

What are some common applications of vector calculus?

Vector calculus has numerous applications in various fields, including fluid dynamics, electromagnetism, computer graphics, and optimization problems. It is also used in the development of mathematical models and simulations for predicting and understanding natural phenomena.

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