# Thus it is proved that there is no such a thing as magnetism [duplicate]

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?

• That's the same. I didn't find it when I searched for similar problems. Sep 7 '20 at 4:02
• By the way, where did you see this proof?
– Dale
Sep 7 '20 at 12:19
• It's on the book name : Div, Curl and all that Chapter IV section Problem. Sep 7 '20 at 12:40

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