Integral limit trouble

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?
 

gabbagabbahey

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we have [itex]\mathbf{A(r)}=\frac{\mu_0 I}{4 \pi} \int dV' \mathbf{\frac{dl'}{|r-r'|}}[/itex]
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?

then if we pick [itex]\mathbf{r'}[/itex] on the x axis, y'=z'=0
How do you justify picking [itex]\mathbf{r'}[/itex] to be on the x axis? What physical problem is this in regards to?

x' goes between [itex]-\infty[/itex] and [itex]+\infty[/itex] btw
Why is that? What is the entire problem?
 
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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!
 
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
 
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?
 

gabbagabbahey

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Looks good to me :smile:
 
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?
 
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!
 

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?

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?
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?

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]
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.

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?
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:

so in first region [tex]d\xi =dl'[/tex] and in second region [tex]d\xi =-dl'[/tex]
 
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?
 

gabbagabbahey

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

gabbagabbahey

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yeah got that now cheers. what about the last part as in what happens if [itex]z^2 >>x^2+y^2[/itex] bit?
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.
 

gabbagabbahey

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oh and the substitution you recommend in post 12, how do u show [itex]du=d \xi[/itex]?
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]
 
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?
 

gabbagabbahey

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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?
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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