Apr 27th 2011, 09:05 PM #1 Member   Join Date: Nov 2010 Posts: 40 Please help in vector potential involve line integral with log. This is a new question, I am just using the old thread because the tittle applys. I want to find the vector potential A at origin due to a current segment I(t)=kt flowing along x axis from -ba. This mean $I_{(t)}=\hat x kt$ from -b to -a on the left of the origin. $\vec A_{(\vec r,t)} = \hat x \frac{\mu_0 k}{4\pi}\int _{-b}^{-a} \frac {(t-\frac {\eta}{c})}{\eta} dx \;\hbox { where }\;\eta = |x|$ $\vec A_{(\vec r,t)} = \hat x \frac{\mu_0 k}{4\pi}\int _{-b}^{-a} \frac {t}{|x|} dx -...........= \hat x \frac{\mu_0 k}{4\pi}ln|x|_{-b}^{-a} -........= \hat x \frac{\mu_0 k}{4\pi} ln(\frac a b)-.........$ I did not write the second part because that is not part of the question. But the book said it is: $\vec A_{(\vec r,t)} =\hat x \frac{\mu_0 k}{4\pi} ln(\frac b a)-.........$ Please help me on this, thanks Alan
Apr 28th 2011, 03:42 AM   #2

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 Originally Posted by yungman This is a new question, I am just using the old thread because the tittle applys. I want to find the vector potential A at origin due to a current segment I(t)=kt flowing along x axis from -ba. This mean $I_{(t)}=\hat x kt$ from -b to -a on the left of the origin. $\vec A_{(\vec r,t)} = \hat x \frac{\mu_0 k}{4\pi}\int _{-b}^{-a} \frac {t}{|x|} dx -...........= \hat x \frac{\mu_0 k}{4\pi}ln|x|_{-b}^{-a} -........= \hat x \frac{\mu_0 k}{4\pi} ln(\frac a b)-.........$ I did not write the second part because that is not part of the question. But the book said it is: $\vec A_{(\vec r,t)} =\hat x \frac{\mu_0 k}{4\pi} ln(\frac b a)-.........$ Please help me on this, thanks Alan
(I had to edit your quote because the system is a bit feral when it comes to how many images a post can have.)

$\int \frac{dx}{|x|} = sgn(x) \cdot ln|x| = -ln|x|$<--For negative x.
(Plus the arbitrary constant, of course.) This is the source of the extra negative sign outside the ln() function.

-Dan
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Apr 28th 2011, 07:41 AM   #3
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 Originally Posted by topsquark (I had to edit your quote because the system is a bit feral when it comes to how many images a post can have.) Your integral is slightly off. $\int \frac{dx}{|x|} = sgn(x) \cdot ln|x| = -ln|x|$<--For negative x. (Plus the arbitrary constant, of course.) This is the source of the extra negative sign outside the ln() function. -Dan
Is sgn(x) the sign of x?

Is this because if x>0 then |x|=x and d|x|/dx=1. But if x<0, then |x|=-x and d|x|/dx=-1. So if we do substitution of U=|x|, dU=sgn(x)dx?

So $\int \frac {dx}{|x|} = sgn(x)\int {dU}{U}= sgn(x)ln|x|$.

Thanks

Alan

Last edited by yungman; Apr 28th 2011 at 07:45 AM.

Apr 28th 2011, 08:42 AM   #4

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 Originally Posted by yungman Is sgn(x) the sign of x? Is this because if x>0 then |x|=x and d|x|/dx=1. But if x<0, then |x|=-x and d|x|/dx=-1. So if we do substitution of U=|x|, dU=sgn(x)dx? So $\int \frac {dx}{|x|} = sgn(x)\int {dU}{U}= sgn(x)ln|x|$. Thanks Alan
Exactly. Note that (integral)dx/|x| will involve ln|x|. This might have been another problem with your result, but since both -a and -b are negative and it is (-b)/(-a) that enters the logarithm this does not matter.

-Dan
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Apr 28th 2011, 05:00 PM   #5
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 Originally Posted by topsquark Exactly. Note that (integral)dx/|x| will involve ln|x|. This might have been another problem with your result, but since both -a and -b are negative and it is (-b)/(-a) that enters the logarithm this does not matter. -Dan

Can you give an example of this? I don't quite get what you saying here.

Thanks

Apr 28th 2011, 05:49 PM   #6

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 Originally Posted by yungman Can you give an example of this? I don't quite get what you saying here. Thanks
Sure. Let a and b be positive. Then
$\int_{-a}^b \frac{dx}{x} = ln \left | \frac{b}{-a} \right |$

In this case we need to beware of that negative sign on the -a. The absolute value gets rid of it so we have
$\int_{-a}^b \frac{dx}{x} = ln \left | \frac{b}{a} \right |$

Similar to your problem, if we have
$\int_{-a}^{-b} \frac{dx}{x} = ln \left | \frac{-b}{-a} \right |$

the negative signs cancel out anyway, so there is nothing to worry about.

-Dan
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 Apr 28th 2011, 06:04 PM #7 Member   Join Date: Nov 2010 Posts: 40 I am confused again, my original question is like your last example: $\int_{-b}^{-a} \frac{dx}{x} = ln \left | \frac{-a}{-b} \right |$ How can it be your original answer that I thought I understand that $\int_{-b}^{-a} \frac{dx}{x} = ln \left | \frac{b}{a} \right |$ Thanks Alan Last edited by yungman; Apr 28th 2011 at 06:07 PM.
Apr 28th 2011, 06:07 PM   #8

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 Originally Posted by yungman How do you post the equation again? The method you showed don't work anymore!!! I am confused again, my original question is like the last example $\int_{-a}^{-b} \frac{dx}{x} = ln \left | \frac{-b}{-a} \right |$
Right. I don't mean to make too much out of this. All I'm saying is that the two negative signs cancel so you don't need to worry about the absolute value. Nothing more than that.

To post an equation using LaTeX copy the following line:
[img]http://latex.codecogs.com/png.latex?*LaTeX commands go here*[/img]

-Dan
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Apr 28th 2011, 06:09 PM   #9
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 Originally Posted by topsquark Right. I don't mean to make too much out of this. All I'm saying is that the two negative signs cancel so you don't need to worry about the absolute value. Nothing more than that. To post an equation using LaTeX copy the following line: [img]http://latex.codecogs.com/png.latex?*LaTeX commands go here*[/img] -Dan
I modified the last post, please read it again. Somehow the Latex work now!!!

Apr 29th 2011, 07:33 AM   #10

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 Originally Posted by yungman Note: I have removed your quote. Apparently there is a limit to the number of tex statements we can put in the posts and the ones in the quote count.
Oh, I'm sorry. First we have to change this back to the original integral:
$\int_{-b}^{-a} \frac{dx}{|x|} = sgn(x) ln \left | \frac{-a}{-b} \right | = - ln \left ( \frac{a}{b} \right )$

This last step follows from both limits on the integral are negative, x is negative so sgn(x) = -1. And the negative signs in the ln cancel. Also, we can remove the absolute value bars since both a and b are positive.

Now recall that
$- ln(x) = ln(x^{-1}) = ln \left ( \frac{1}{x} \right )$

$\int_{-b}^{-a} \frac{dx}{|x|} = - ln \left ( \frac{a}{b} \right ) = ln \left ( \left [ \frac{a}{b} \right ] ^{-1} \right )$

$= ln \left ( \frac{b}{a} \right )$

-Dan
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