Infinitisemal magnetic flux density

[Vegeta]

These Q's are probably simple for those of you who have/completed a course in Vector Calculus. But I'm only a higschool/secondaryschool student, so I haven't.

1. I'm not sure how the infinitisemal magnetic flux density $d\textbf{B}$ from a wire element $d\textbf{l}$ with a current $I$, which is
$$d\textbf{B}=\frac{\mu_0I}{4\pi}\frac{d\textbf{l}\times\textbf{x}}{|\textbf{x}|^3}$$
Can be written as
$$\textbf{B}(\textbf{x})=\frac{\mu_0}{4\pi}\int\textbf{J}(\textbf{x}')\times\frac{\textbf{x}-\textbf{x}'}{|\textbf{x}-\textbf{x}'|^3}\,d^3x'$$
Is it by using the definition of the current density,
$$I=\iint_A \textbf{J}\bullet\textbf{n}\,dA$$
and insert this in the equation? but I can't figure it out, hope someone can help me on this one.

2. I know that
$$\frac{\textbf{x}-\textbf{x}'}{|\textbf{x}-\textbf{x}'|^3} = -\nabla\left(\frac{1}{|\textbf{x}-\textbf{x}'|}\right)$$
It is then stated that by the use of this, you get
$$\textbf{B}(\textbf{x})=\frac{\mu_0}{4\pi}\,\nabla\times\int\frac{\textbf{J}(\textbf{x}')}{|\textbf{x}-\textbf{x}'|}\,d^3x'$$
Is this because
$$-\nabla\left(\frac{1}{|\textbf{x}-\textbf{x}'|}\right)\times\textbf{J}(\textbf{x}')=\frac{-1}{|\textbf{x}-\textbf{x}'|}\left(\nabla\times\textbf{J}(\textbf{x}')\right)$$
But what happens to the minus sign? Is it because that be take the gradient with respect to $\textbf{x}$ which is negative, and therefore when we take the curl of the current density we have to take it with respect to $\textbf{x}'$ which is negative, and therefore the gradient would be positive?

The book I'm using is
John David Jackson, Classical Electrodynamics, 3. ed
This subject can be found under 5.3, page 178.

 

[StatusX]

For the first one, consider the magnetic field generated by some differential volume element d³x containing current density $\vec J$. First we need to find the total current flowing in d³x. Assume we have axes lined up so that z is parellel to J. This is only for convenience, as we will switch back to coordinate free notation at the end. Then d³x=dxdydz, and the total current is $I = \vec J dx dy$. Then $I d\vec l= \vec J dx dy dz = \vec J d^3 x$. The rest should follow pretty easily from this.

 

[Vegeta]

Thanks, I think I understand it now.
But what about my 2. question. Am I correct on that statement?

 

[StatusX]

That isn't the correct product rule. You want:

$$\vec \nabla \times (f \vec A ) = f \vec \nabla \times \vec A + \vec \nabla f \times \vec A$$

Then you substitute two of these terms in for the third (the one you have) in your integral. One of them can be pulled outside the integral over x', since it is a derivative over x, and the other can be shown to vanish. Don't forget that the order of the cross product can be reversed at the cost of a minus sign, ie, $\vec A \times \vec B = -\vec B \times \vec A$.

 

[Vegeta]

O yeah that correct I can't just do such operations as in my first post.
But when I use that like

$$\textbf{A}\times (-\nabla f) = \nabla f \times\textbf{A} = \nabla\times(f\textbf{A}) - \nabla f\times\textbf{A} = \nabla\times(f\textbf{A}) + \textbf{A} \times\nabla f$$

Is that correct?
And if so, how do you get

$$\int_V\nabla\times(f\textbf{A}) + \textbf{A} \times\nabla f\,dV=\nabla\times\int_Vf\textbf{A}\,dV$$

 

[quasar987]

You're in high school and reading Jackson for your personal amusement? I'm intimidated. :shy:

 

[StatusX]

Actually, that's not correct either. You have the same term on both sides. Once you do have the correct formula, for one of the terms you can pull the curl outside the integral, since the curl is a derivative with respect to x and the integral is over x'. This is an simple extension of the single variable rule:

$$\int \left( \frac{\partial }{\partial y} f(x,y) \right) dx = \frac{d}{dy} \int f(x,y) dx$$

The other term will vanish because J(x') does not depend on x.

 

[Physics Monkey]

Hi Vegeta,

The thing your looking at is $$\int \textbf{J}(\textbf{x}') \times \frac{\textbf{x} - \textbf{x}'}{|\textbf{x} - \textbf{x}'|^3} \, d^3 x'$$, right? The trick is to write the second part as a gradient like this: $$\int \textbf{J}(\textbf{x}') \times \frac{\textbf{x} - \textbf{x}'}{|\textbf{x} - \textbf{x}'|^3} \, d^3 x'= \int \textbf{J}(\textbf{x}') \times \left( - \nabla \frac{1}{|\textbf{x} - \textbf{x}'|}\right) \, d^3 x'$$ where the derivative is with respect to $$\textbf{x}$$ not $$\textbf{x}'$$. Now by swtiching the order of the cross porduct you get $$\int \left( \nabla \frac{1}{|\textbf{x} - \textbf{x}'|}\right) \times \textbf{J}(\textbf{x}') \, d^3 x' = \int \nabla \times \left( \frac{\textbf{J}(\textbf{x}') }{|\textbf{x} - \textbf{x}'|}\right) \, d^3 x'$$ since $$\textbf{J}$$ doesn't depend on $$\textbf{x}$$. Now you can take the gradient outside the integral. A good thing to remember is that $$|\textbf{x} - \textbf{x}' |$$ is symmetric in its arguments, so its an often used trick to switch which argument you're differentiating with respect to.

 

[Vegeta]

Ohh now I see why StatusX said "Actually, that's not correct
either.", that's because in post 5 I wrote

$$\textbf{A}\times (-\nabla f) = \nabla f \times\textbf{A} = \nabla\times(f\textbf{A}) - \nabla f\times\textbf{A}$$

Hehe that makes no sense, it was a typing error. What I meant was

$$\textbf{A}\times (-\nabla f) = \nabla f \times\textbf{A} = \nabla\times(f\textbf{A}) - f\left(\nabla\times\textbf{A}\right)$$

Which should be correct, because this is the formula Statusx said I
should use;
$$\vec \nabla \times (f \vec A ) = f \vec \nabla \times \vec A + \vec \nabla f \times \vec A$$

Now with $\textbf{A}=\textbf{J}(\textbf{x}')$ and $f = 1/|\textbf{x}-\textbf{x}'|$, under the integral we then have

$$\nabla\times\left(\frac{\textbf{J}(\textbf{x}')}{|\textbf{x}-\textbf{x}'|}\right) - \frac{1}{|\textbf{x}-\textbf{x}'|}\left(\nabla\times\textbf{J}(\textbf{x}')\right)$$

Now you say that the curl(J(x')) = 0, why? I know you say that this
is because J doesn't depend on x, but is that general in
vectorcalculus?

I can't see any of those LaTeX generated equations, only in TeX
code, but that's ok. But I can't see the TeX code either on your
last equation Physics Monkey.

Then the curl can be taken out side the integral because the curl is
taken with respect to x and the integral is with respect to x', and
as StatusX says, "This is an simple extension of the single variable
rule"

$$\int \left( \frac{\partial }{\partial y} f(x,y) \right) dx = \frac{d}{dy} \int f(x,y) dx$$

Therefor I get

$$\nabla\times\int\frac{\textbf{J}(\textbf{x}')}{|\textbf{x}-\textbf{x}'|}\,d^3x'$$

So is my statements correct now?

 

[StatusX]

Now you say that the curl(J(x')) = 0, why? I know you say that this is because J doesn't depend on x, but is that general in
vectorcalculus?

J(x') is a constant with respect to x. Keep in mind the distinction between x and x': you pick a point x where you want to know the field, then for every other point x', you find the contribution to the field at x that comes from the current at x', and then you integrate over all x' to get the total field at x. The contribution from x' to the field at x depends on the current at x' as well as the distance from x to x'. But the current at a point x' doesn't change when you change the point x at which you are evaluating the field. Thus, d(J(x'))/dx=0 (that's abusing notation a little, but you get the idea). I think everything else looks ok, although it's hard to tell with the latex not working.

 

[Vegeta]

Thanks, I think I understand now. Because curl(J(x')) is like a derivative to J(x'), and since J doesn't depend on x, as you both said, then the "derivative" of J with respect to x must be 0.
Btw how long will it be, before the LaTeX will be working?