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I saw a proof that shows that there is no such a thing as magnetism. I think the fault in the proof is with simply connected regions. Proof is as follows:

One of Maxwell’s equations tell us that $$\nabla\cdot \mathbf{B}=0$$ where $\mathbf B$ is a magnetic field. Then using the divergence theorem, we find

$$\iint_S\mathbf{B}\cdot\mathbf{\hat{n}} dS=\iiint_V\nabla\cdot \mathbf{B}dV=0.$$

Because $\mathbf B$ has a zero divergence, we know that there exists a vector function, call it $\mathbf A$, such that

$$\mathbf{B}=\nabla \times\mathbf{A}.$$

Combining these two equations, we get

$$\iint_S\mathbf{\hat{n}}\cdot \nabla \times\mathbf{A}dS=0.$$

Next we apply Stokes’ theorem and the above result to find

$$\oint_c\mathbf{A}\cdot \mathbf{\hat{t}}ds=\iint_S\mathbf{\hat{n}}\cdot \nabla \times\mathbf{A}dS=0$$

Thus we have shown that the circulation of $\mathbf A$ is path independent. It follow that we can write $\mathbf A$ as $\mathbf{A}=\nabla\psi$ where $\psi $ is some scalar function.

Since curl of gradient of a function is zero, we arrive at the remarkable fact that

$$\mathbf{B}=\nabla\times \nabla \psi=0;$$

that is, all magnetic fields are zero.

Where is the mistake?

Dale
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Young Kindaichi
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1 Answers1

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The mistake is here:

Next we apply Stokes’ theorem and the above result to find $$\oint_c\mathbf{A}\cdot \mathbf{\hat{t}}ds=\iint_S\mathbf{\hat{n}}\cdot \nabla \times\mathbf{A}dS$$

The integral on the right is over a closed surface. So there is no boundary over which to do the path integral on the left.

Dale
  • 99,825