Problem about the derivation of divergence for a magnetic field

[georg gill]

Summary:: I am trying to derive that the divergence of a magnetic field is 0. One of the moves is to take the curl out of an integral. Can someone prove that this is addressable

Biot Savart's law is

$$B(r)=\frac{\mu _0}{4\pi} \int \frac{I(r') \times (r-r')}{|r-r|^3}dl'=\frac{\mu _0}{4\pi} \int \frac{J(r') ds \times (r-r')}{|r-r|^3}dl'=\frac{\mu _0}{4\pi} \int \frac{J(r') \times (r-r')}{|r-r|^3}dV'$$

We use:

$$\frac{(r-r')}{|r-r'|^3}=-\nabla \frac{1}{|r-r'|}$$

$$B(r)=-\frac{\mu _0}{4\pi} \int J(r') \times \nabla \frac{1}{|r-r'|}dV'$$

We use the identity

$$-J \times (\nabla \Psi)=\nabla \times (\Psi J) - \Psi (\nabla \times J)$$

$$B(r)=\frac{\mu _0}{4\pi} \int \nabla \times \{\frac{ J(r')}{|r-r'|} - \frac {1}{|r-r|}\nabla \times J(r') \}dV'$$

$$\nabla \times J(r')$$ is 0 since we take the derivative of r on a function of r'.$$B(r)=\frac{\mu _0}{4\pi} \int \nabla \times \frac{ J(r')}{|r-r'|} dV'$$

My question. The next move is to take the curl outside the integral so that one obtains

$$B(r)=\frac{\mu _0}{4\pi}\nabla \times \int \frac{ J(r')}{|r-r'|} dV'$$

Indeed this is a nice move as one obtains the formula for the magnetic vector potential. And also gives that the divergence of the magnetic field is 0 since divergence of the curl is 0. But my problem is: How can one prove that one can take the curl outside the integral when the curl inside the integral has r as a variable?

 

[marcusl]

If this is a homework problem, then you should have posted it in the homework forum and have filled out the template. Since you have shown your work, however, I'll point out that you have been sloppy with your variables; the denominators should have $|r-r'|$, and the integral is over $dV'$ thus operating on the primed variable $r'$. The curl, on the other hand, is unprimed and operates on $r$. Integral and curl can be interchanged because they operate on different variables.

 

[georg gill]

If this is a homework problem, then you should have posted it in the homework forum and have filled out the template. Since you have shown your work, however, I'll point out that you have been sloppy with your variables; the denominators should have $|r-r'|$, and the integral is over $dV'$ thus operating on the primed variable $r'$. The curl, on the other hand, is unprimed and operates on $r$. Integral and curl can be interchanged because they operate on different variables.

I have tried to update the prime notation. Should I move it to homework?

The problem is that I want a proof for why integral and curl can be intechanged because they operate on different variables. I have been given the argumentation before. But I need a proof. For example if you start with the integral it seems that you would integrate the current density and also the denominator which would give an ln function or something. After that you would have to derivate this ln function back to $\frac{1}{r-r'}$. If you go the other something else migh happen. And also why does this laways hold. There has to be a mathematical proof for this. For example I have a mathematical proof for the fact that $\frac{\partial^2 f}{\partial x \partial y}=\frac{\partial^2 f}{\partial y \partial x}$?

 

[marcusl]

Proof? I'm not a mathematician and don't do proofs. As for a demonstration, you could try it in one dimension: write the integral as a Riemann sum in the limit that $\Delta z=0$ and the derivative as the usual limit of differences (the basic definition from Calc I) and notice that they operate on different variables so you can interchange terms.

 

[PeroK]

Indeed this is a nice move as one obtains the formula for the magnetic vector potential. And also gives that the divergence of the magnetic field is 0 since divergence of the curl is 0. But my problem is: How can one prove that one can take the curl outside the integral when the curl inside the integral has r as a variable?

In terms of a proof, all the multi-variable cases reduce to the single variable case where you have: $$g(x) = \int f(x, y) \ dy \ \ \text{and} \ \ \frac{d}{dx}g(x) = \int \frac{\partial}{\partial x} f(x, y) \ dy$$ And what you are really doing is reversing the order of two limits: the integral and the derivative. In general, if $f$ is a well-behaved function, then you can do this. There's a lot more here (including proofs) if you are interested:

https://en.wikipedia.org/wiki/Leibniz_integral_rule

 

[georg gill]

If this is a homework problem, then you should have posted it in the homework forum and have filled out the template. Since you have shown your work, however, I'll point out that you have been sloppy with your variables; the denominators should have |r−r′|, and the integral is over dV′ thus operating on the primed variable r′. The curl, on the other hand, is unprimed and operates on r. Integral and curl can be interchanged because they operate on different variables.

The problem is that I want a proof for why integral and curl can be intechanged because they operate on differnet variables. I have been given the argumentation before. But I need a proof. For example if you start with the integral

In terms of a proof, all the multi-variable cases reduce to the single variable case where you have: $$g(x) = \int f(x, y) \ dy \ \ \text{and} \ \ \frac{d}{dx}g(x) = \int \frac{\partial}{\partial x} f(x, y) \ dy$$ And what you are really doing is reversing the order of two limits: the integral and the derivative. In general, if $f$ is a well-behaved function, then you can do this. There's a lot more here (including proofs) if you are interested:

https://en.wikipedia.org/wiki/Leibniz_integral_rule

Thanks for the proof!