Here's an interesting "proof" that there is no such thing as magnetism. I know the answer but I love this so much I had to ask it here. It's a great way to confuse people!

As we all know, $$\nabla \cdot\vec{B} =0$$ Using the divergence theorem, we find $$ \iint_S \vec{B} \cdot \hat{n} \, dS = \iiint_V \nabla \cdot \vec{B} \, dV = 0$$ Since $\vec{B}$ has zero divergence, there exist a vector function $\vec{A}$ such that $$\vec{B} = \nabla \times \vec{A}$$ Combining the last two equations, we get $$\iint_S \hat{n} \cdot \nabla \times \vec{A} \, dS = 0$$ Applying Stokes' theorem, we find $$\oint_C \vec{A} \cdot \hat{t} \, ds = \iint_S \hat{n} \cdot \nabla \times \vec{A} \, dS = 0$$ Therefore, $\vec{A}$ is path independent and we can write $\vec{A} = \nabla \psi$ for some scalar function $\psi$. Since the curl of the gradient of a function is zero, we arrive at: $$\vec{B} = \nabla \times \nabla \psi = 0,$$ which means that all magnetic fields are zero, but that can't be!

Can you see where we went wrong?

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    $\begingroup$ Not advanced trick. What do you love about it? $\endgroup$
    – Asphir Dom
    Aug 25, 2013 at 23:19
  • $\begingroup$ This might only confuse people that don't realize you shouldn't throw away important information in the intermediate stages of a computation... $\endgroup$
    – Michael
    Aug 26, 2013 at 6:24
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    $\begingroup$ @AsphirDom With all due respect, I didn't ask for your opinion, as long as some people like it, it's interesting. Besides, no expert was born an expert, there are people who are still learning. $\endgroup$
    – user20250
    Aug 26, 2013 at 7:31
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    $\begingroup$ Don't take my comment personally. I am just curious to find out what we can LEARN from it? $\endgroup$
    – Asphir Dom
    Aug 26, 2013 at 9:43
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    $\begingroup$ $$\int\!\!\!\!\!\int \!\!\!\!\!\!\! \bigcirc\text{?}$$ $\endgroup$ Sep 7, 2013 at 1:41

3 Answers 3


Note that $\partial V=S$, so that

$$\tag{1} C~=~\partial S~=~\partial^2V~=~\emptyset$$

is the empty set. (Topologically, the boundary of a boundary is empty, or equivalently, the boundary operator $\partial^2=0$ squares to zero.) On the other hand, the circulation

$$\tag{2} \Gamma~=~\oint_{C=\emptyset}\vec{A}\cdot d\vec{r}~=~0$$

along the empty curve $C=\emptyset$ vanishes identically for any vector field $\vec{A}$. In particular, one can not conclude from (2) that the magnetic potential $\vec{A}$ should be a gradient field.

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    $\begingroup$ So, does all this comes from topology? $\endgroup$ Aug 30, 2013 at 0:30
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    $\begingroup$ “the boundary of a boundary is zero!” - John Archibald Wheeler $\endgroup$ Aug 30, 2013 at 2:09
  • $\begingroup$ @LarryHarson Yup! I suspect the OP has too little difficulty telling the difference between a doughnut and a coffee mug :) ! Or, more relevant to this question, the lack of homoemorphism between a closed sphere and a punctured one and the fallacious substitution of one for the other is what the fault is. $\endgroup$ Aug 31, 2013 at 1:52

Here's something from wikipedia:

"This classical Kelvin–Stokes theorem relates the surface integral of the curl of a vector field F over a surface Σ in Euclidean three-space to the line integral of the vector field over its boundary ∂Σ."

The catch in the statement of Stokes theorem is "boundary".So if V is a 3D volume,then it's boundary will be it's 2D surface which will be closed as V is a 3D volume.Now here's the catch,the boundary of a closed surface will not exist or will be null.(This has been pointed out mathematically by Qmechanic as $C=\partial S=\partial^2 V=\phi$).

So in extremely simplified terms it's like concluding $a=0$ from $a \times 0=0$.


Adding a closed boundary $C$ around the volume generally splits $S$ into two surfaces $S_1$ and $S_2$ for which Stoke's theorem holds:

\begin{align*} \oint_C \vec{A} \cdot \hat{t} \, ds &= \iint_{S_1} \hat{n} \cdot \nabla \times \vec{A} \, dS\tag{1}\\ \oint_C \vec{A} \cdot \hat{t} \, ds &= \iint_{S_2} \hat{n} \cdot \nabla \times \vec{A} \, dS\tag{2} \end{align*}

I can use these two ways to show where your argument is wrong:

enter image description here

  1. The left hand sides of (1) and (2) must be the negative of one another because the right hand sides sum to zero, and therefore $$\oint_C (\vec{A} - \vec{A}) \cdot \hat{t} \, ds = \iint_{S_1} \hat{n} \cdot \nabla \times \vec{A} \, dS + \iint_{S_2} \hat{n} \cdot \nabla \times \vec{A} \, dS = 0$$ So you should have used $\vec{A}-\vec{A}$ in place of $\vec{A}$ in your application of Stokes theorem and concluding arguments, to give the trivial result $\vec B - \vec B = 0$
  2. Let $S_1\rightarrow S,\,S_2\rightarrow 0$ as $C\rightarrow 0$. Then $(1)$ reduces to your application of Stokes theorem which reduces to $\vec{A}\cdot\vec 0 = 0$, making your following arguments irrelevant.
  • $\begingroup$ Nice surfaces. :) $\endgroup$ Nov 11, 2016 at 23:10
  • $\begingroup$ @ArturodonJuan haha, thanks :) This was before I learned how to use Inkscape, so I used Microsoft word and their predefined 3d- shapes. Maybe that explains the lack of upvotes for my obviously correct answer ;) $\endgroup$ Nov 14, 2016 at 2:50

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