- #1
georg gill
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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?
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?
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