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Integral limit trouble

  1. Feb 18, 2009 #1
    we have [itex]\mathbf{A(r)}=\frac{\mu_0 I}{4 \pi} \int dV' \mathbf{\frac{dl'}{|r-r'|}}[/itex]

    i need to show that [itex]A_x=-\frac{\mu_0 I}{4 \pi} \int_{-x}^x \frac{d \xi}{(\xi^2 + y^2 + z^2)^{-\frac{1}{2}}}[/itex]

    i said [itex]|\mathbf{r-r'}|=\sqrt{(x-x')^2+(y-y')^2+(z-z')^2}[/itex]

    then if we pick [itex]\mathbf{r'}[/itex] on the x axis, y'=z'=0

    then we let [itex]\xi=x-x' \Rightarrow dl'=-d \xi[/itex]

    so everything's looking good up till now but i can't get the limits on the integration to come out right.

    x' goes between [itex]-\infty[/itex] and [itex]+\infty[/itex] btw

    any ideas?
     
  2. jcsd
  3. Feb 19, 2009 #2

    gabbagabbahey

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    The way this is writen, it makes absolutely no sense. The LHS of the equation is a vector, and yet every term on the LHS is a scalar. How can a vector equal a scalar? And you seem to be integrating both over a volume dV' and along a curve dl'....how is that possible?

    Surely you meant to write:

    [tex]\mathbf{A}(\mathbf{r})=\frac{\mu_0 I}{4 \pi} \int \frac{\mathbf{dl'}}{|\mathbf{r}-\mathbf{r'}|}[/tex]

    right?

    How do you justify picking [itex]\mathbf{r'}[/itex] to be on the x axis? What physical problem is this in regards to?

    Why is that? What is the entire problem?
     
    Last edited: Feb 19, 2009
  4. Feb 23, 2009 #3
  5. Feb 23, 2009 #4

    gabbagabbahey

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    The figure isn't very clear, but I think you are supposed to assume that the current moving in the positive x-direction goes from x'=-x too x'=0; and the current travelling in the negative x-direction goes from x'=x to x=0. That gives you two integrals (two sections of current) which you can combine to get the desired form.

    However, looking only at the figure I would have assumed that the current go from -infinity to zero and infinity to zero; so I recommend you clarify this point with your professor first!
     
  6. Feb 23, 2009 #5
    hi ive managed that. and ive found A_y and A_z=0 as no current in that direction

    i'll post my working just now
     
  7. Feb 23, 2009 #6
    you have a region with i in positive x direction and x' from -infinity to 0. here xi goes from +infinity to x. and a region in which i is in negative x direction but with x' from 0 to +infinity. here xi goes from x to -infinity.

    xi=x-x'

    so in first region [itex]d \xi = dl'[/itex] as [itex]x' <0[/itex] and in second region [itex]d \xi=-dl'[/itex]. here dl' is a scalar as im about to take x component of A!

    [itex]A_x=\frac{\mu_0 I}{4 \pi} \int_{+\infty}^x \frac{d \xi}{(\xi^2+y^2+z^2)^{\frac{1}{2}}} + \frac{\mu_0 (-I)}{4 \pi} \int_x^{-\infty} \frac{-d \xi}{(\xi^2+y^2+z^2)^{\frac{1}{2}}}=-\frac{\mu_0 I}{4 \pi} \int_{-x}^{-\infty} \frac{d \xi}{(\xi^2+y^2+z^2)^{\frac{1}{2}}} - \frac{\mu_0 (-I)}{4 \pi} \int_{-\infty}^x \frac{-d \xi}{(\xi^2+y^2+z^2)^{\frac{1}{2}}}[/itex]
    so
    [itex]A_x=-\frac{\mu_0 I}{4 \pi} \int_{-x}^x \frac{d \xi}{(\xi^2+y^2+z^2)^{\frac{1}{2}}} [/itex]

    does this look acceptable?
     
  8. Feb 23, 2009 #7

    gabbagabbahey

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    Looks good to me :smile:
     
  9. Feb 23, 2009 #8
    couple of questions about this:

    1, is it ok to say [itex]d \xi=dl'[/itex] in region 1 and [itex]d \xi=-dl'[/itex] in region 2?

    2,in the first integral i pull out a negative sign by changing the limits from i.e. is it true that [itex]\int_{\infty}^x=-\int_{-x}^{-\infty}[/itex]

    3, in the second integral i go from [itex]\int_{x}^{-\infty}=-\int_{-\infty}^{x}[/itex]
    is this ok? it seems to disagree with my question 2?
     
  10. Feb 23, 2009 #9
    i then evaluate the integral and get [itex]A_x=-\frac{\mu_0 I}{4 \pi} [\ln{(x+(x^2+y^2+z^2)^{-\frac{1}{2}})}-\ln{(-x+(x^2+y^2+z^2)^{-\frac{1}{2}})}][/itex]
    i need to show that [itex]\frac{\partial{A_x}}{\partial{y}}=-\frac{\mu_0 I}{4 \pi}[-\frac{2xy}{\sqrt{x^2+y^2+z^2}(y^2+z^2)}][/itex]

    ive pretty much been trying to show this for the best part of eight hours and am nearing the stage of wanting to kill myself!
     
  11. Feb 23, 2009 #10

    gabbagabbahey

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    Let's start by writing your original integrals as:

    [tex]A_{x,1}=\frac{\mu_0 I}{4 \pi} \int_{-\infty}^0 \frac{dx'}{((x-x')^2+y^2+z^2)^{\frac{1}{2}}}[/tex]

    and

    [tex]A_{x,2}=\frac{-\mu_0 I}{4 \pi} \int_{0}^{\infty} \frac{dx'}{((x-x')^2+y^2+z^2)^{\frac{1}{2}}}[/tex]

    Where dl'=dx'i

    Okay with this part?

    if [itex]\xi=x-x'[/itex], then [itex]d\xi=-dx'[/itex] and [itex]\xi[/itex] goes from [itex]x-(-\infty)=+\infty[/itex] to [itex]x-0=x[/itex]

    [tex]\implies A_{x,1}=\frac{-\mu_0 I}{4 \pi} \int_{\infty}^x \frac{d\xi}{(\xi^2+y^2+z^2)^{\frac{1}{2}}}[/tex]

    On the other hand, you could define [itex]\xi=x'-x[/itex] instead,and then [itex]d\xi=+dx'[/itex] and [itex]\xi[/itex] goes from [itex](-\infty)-x=-\infty[/itex] to [itex]0-x=-x[/itex]

    [tex]\implies A_{x,1}=\frac{\mu_0 I}{4 \pi} \int_{-\infty}^{-x} \frac{d\xi}{(\xi^2+y^2+z^2)^{\frac{1}{2}}}[/tex]

    Follow?

    No, this is not alright. What is true is that [itex]\int_{\infty}^x=-\int_{x}^{\infty}[/itex]

    I had assumed that you had gone from [itex]x\to -x[/itex] and [itex]-\infty \to \infty[/itex] by making the transformation (substitution) [itex]\xi \to -\xi[/itex] which would have been acceptable.

    Yes, this part is acceptable.

    You need to realize that [itex]\xi[/itex] is essentially a dummy variable snce it is being integrated over and so you don't have to have it represent the exact same quantity in both integrals. It is easiest if you define [itex]\xi=x-x'[/itex] for one of the integrals and [itex]\xi=x'-x[/itex] for the other ----which is what I had thought you did when you said:

     
  12. Feb 23, 2009 #11
    yeah that's what i was meaning but just couuldnt figure out how to put it in maths

    do you have any ideas for the very last bit of the question?
     
  13. Feb 23, 2009 #12

    gabbagabbahey

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    This is just straight integration and differentiation. For the integration, try the substituion [itex]u=\xi+\sqrt{\xi^2+y^2+z^2}[/itex]
     
  14. Feb 23, 2009 #13
    yeah got that now cheers. what about the last part as in what happens if [itex]z^2 >>x^2+y^2[/itex] bit?
     
  15. Feb 23, 2009 #14
    oh and the substitution you recommend in post 12, how do u show [itex]du=d \xi[/itex]?
     
  16. Feb 23, 2009 #15

    gabbagabbahey

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    Try writing your general expression for B in terms of the dimensionless variable [tex]w=\frac{x^2+y^2}{z^2}[/itex] then Taylor expand it about w=0.
     
  17. Feb 23, 2009 #16

    gabbagabbahey

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    You don't because it's not true.

    If [itex]u=\xi+\sqrt{\xi^2+y^2+z^2}[/itex] , then

    [tex]du=\left(1+\frac{\xi}{\sqrt{\xi^2+y^2+z^2}}\right)d\xi=\frac{\sqrt{\xi^2+y^2+z^2}+\xi}{\sqrt{\xi^2+y^2+z^2}}d\xi=\frac{u}{\sqrt{\xi^2+y^2+z^2}}d\xi[/tex]
     
  18. Feb 24, 2009 #17
    ok is there some physical explanation of why you chose xi=x-x' in one region and xi=x'-x in the other - i.e. something other than "to make the maths work"?

    how did you know to use this trick?
     
  19. Feb 24, 2009 #18

    gabbagabbahey

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    No physical reason; but it's not really a 'trick' either. Its a simple mathematical substitution. You treat each integral separately, and so you can use different substitutions for different integrals. It turns out that if you use xi=x-x' for one and xi=x'-x for the other, both your integrals end up in the same form and can therefor be combined in the desired way.
     
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