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Homework Help: Proving the relation using multivariable calculus

  1. Oct 18, 2014 #1
    1. The problem


    I am trying to prove the following relation in cartesian coordinates. We were given a hint to use integration by parts, as well as the fact that we know $d \vec r = dx\,dy\,dz$ (volume integral).

    $$\int f(\vec r)\ \nabla \cdot \vec A(\vec r) \, d \vec r = -\int \vec A(\vec r)\ \cdot \nabla f(\vec r)\,d\vec r$$


    2. Relevant equations


    3. The attempt at a solution

    If I use integration by parts, we have:

    $$uv - \int(v\ du)$$

    Does anyone see the way in which to prove this relation?

    I can start by saying:

    $$u=f(r)$$
    $$du = \nabla fdr$$
    $$dv = \nabla \cdot A(r)dr$$
    $$v = A(r)$$

    Does this look correct?

    I am not sure if I am doing integration by parts correctly, because I am not sure how to handle gradient/divergence in this manner. Is there anyone out there who can provide some insight?
    m statement, all variables and given/known data

     
  2. jcsd
  3. Oct 18, 2014 #2

    Orodruin

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    Hint: In cartesian coordinates, ##\vec\nabla = \sum_i \vec e_i \frac{\partial}{\partial x_i}##
     
  4. Oct 18, 2014 #3
    I am sorry, but I am not entirely sure what do with your hint? Why would I need to introduce summations?
     
  5. Oct 18, 2014 #4
    I am sorry, but I am not entirely sure what do with your hint? Why would I need to introduce summations?
     
  6. Oct 18, 2014 #5

    Orodruin

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    I suggest you use it and tackle one term at a time. That the ##\vec\nabla## operator can be written in this way in cartesian coordinates should be familiar.
     
  7. Oct 19, 2014 #6
    You are referring to the del symbol, but in the context of this problem, the integral on the left side is the dot product. Are you aware of this?
     
  8. Oct 19, 2014 #7

    Orodruin

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    Perfectly aware, this changes nothing. Are you aware of the following?
    ∇f = grad(f)
    ∇⋅A = div(A)
     
  9. Oct 19, 2014 #8
    I am aware, so are you saying to integrate the right side instead of the left? And have a triple integral?
     
  10. Oct 19, 2014 #9
    Do you agree with the integration by parts that I did? Does that work for the left side, and you are suggesting a way to solve the right side?
     
  11. Oct 19, 2014 #10

    Orodruin

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    You can do it in any direction you like. Integration by parts would work in both directions. Why don't you try writing out one of the sides using the expansion of ∇ and show us what you get? (If you know the Einstein summation convention I would suggest using it for brevity.)
     
  12. Oct 19, 2014 #11
    Can you explain to me why the integration by parts as I show in my original post is not the corrrect way to do it? I believe you, I just don't understand why that method is wrong. I am not sure what the Einstein summation method is.

    If I were to right out the right side using the convention you said, it would just be the integral over 3 summations wouldn't it? The d//dx*e_x component, and similarly for y and z. then dr = dxdydz so i guess i could split into 3 separate integrals? One for dx, dy and dz?
     
  13. Oct 19, 2014 #12

    vela

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    The suggestion is to write the integrals in terms of stuff you should already know instead of guessing at how integration by part works with divergence and gradient. For example,
    $$\int f(\vec{r})(\nabla \cdot \vec{A})\,d\vec{r} = \int f(\vec{r})\left[\frac{\partial}{\partial x} A_x + \frac{\partial}{\partial y} A_y + \frac{\partial}{\partial z} A_z \right]\,d\vec{r}.$$ You should know how to apply integration by parts to
    $$\iiint f(\vec{r})\frac{\partial A_x}{\partial x}dx\,dy\,dz$$ as well as the other terms. What do you get when you do that?
     
  14. Oct 19, 2014 #13
    Let's see here:

    $$
     
  15. Oct 19, 2014 #14
    $$\int(f(r)\frac{dAx}{dx}dx)$$

    So $$dv =\frac{dA_x}{dx}$$
    $$v = A_x$$

    $$u = f(r)dx$$

    Am I starting right?
     
  16. Oct 19, 2014 #15
    Not sure if that is the right start
     
  17. Oct 19, 2014 #16

    vela

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    No, it's not correct because of sloppy notation. Writing ##dv = A_x'## doesn't make sense. It should be ##dv = A_x' \,dx##. Similarly, writing ##u=f(r)\,dx## doesn't make sense either.

    Do you know how to do integration by parts? The fact that you seem to be guessing here suggests you don't really understand how to do it. If you're rusty, reviewing it would be a good idea.
     
  18. Oct 19, 2014 #17
    I do generally know how to integration by parts. I think I am confused by gradient and divergence function in here.

    So, in that case, $$v=A_x$$

    and

    $$u=f(r)$$

    And then

    $$du=f'(r)dr$$

    And then $$uv-\int(vdu)$$

    And multiply by y and z components?
     
  19. Oct 19, 2014 #18

    vela

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    What is f'(r) supposed to mean? What are you differentiating with respect to? What is dr?
     
  20. Oct 19, 2014 #19
    f(r) is in the original problem.
    You even wrote it in the problem before saying I needed to do integration by parts. I was differentiating wrt r. This is why I am confused, I would think it would be with respect to x.

    Since this is r_vector, do i just do x-component, where f(r)_x = xi?
    So f(x) =x

    so u =x
    du =dx

    Am I on the right track?
     
  21. Oct 19, 2014 #20
    Well, I think I made a mistake. r_vector = xi+yj+zk, though I have f(r), so I have f(xi+yj+zk)
     
  22. Oct 19, 2014 #21
    And I know dr = dxdydz
     
  23. Oct 19, 2014 #22

    vela

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    Integration by parts comes from the product rule: (fg)' = f'g + fg', which you rearrange to get f'g = (fg)' - fg'. When you integrate both sides, you end up with
    $$\int f'g = fg - \int fg'.$$ Note that the primes all denote differentiation with respect to the same variable.

    Now in the integral
    $$\iiint f(\vec{r})\frac{\partial A_x}{\partial x}dx\,dy\,dz$$
    the differentiation is with respect to ##x##, right?

    So you start with ##u = f(\vec{r}) = f(x,y,z)## and ##dv = \frac{\partial A_x}{\partial x}dx##. This gives you ##v=A_x##. Now what should ##du## be? Don't just say f'. You have to explicitly say what variable you're differentiating with respect to.
     
  24. Oct 19, 2014 #23
    I think

    $$u=f(x,y,z)$$

    So

    $$du=f(x',y,z)dx$$
     
  25. Oct 19, 2014 #24

    So:

    I think

    $$u=f(x,y,z)$$

    So

    $$du=f(x',y,z)dx$$

    Therefore,

    $$uv-\int(vdu)$$

    $$f(x,y,z)A_x-\int(A_x*f(x',y,z)dx)$$
     
  26. Oct 19, 2014 #25
    How will the uv terms disappear to show the relation that we want? When I tried this without substituting r for x,y, and z, I could not figure out how to get the uv term to disappear.

    I know that integration by parts is $$uv-\int(vdu)$$

    and if only uv disappears, we have proved the relation. How will doing it for each component change anything? Have you figured that part out? I am stumped
     
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