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Prove curl grad $\phi = 0$ for all scalar fields. $\phi$ (i.e. $\nabla \times \nabla \phi \ 0$ using Einstein summation notation only, no term by term expansion.

Well, first step is to set up in notation form: $\epsilon_{ijk}$$d\over x_j$$d\phi\over dx_k$

In which the next step, I use the identity of what $\nabla\phi$ is, which would be
$d\phi\over dx$x + $d\phi\over dy$y + $d\phi\over dz$z

then do

$\nabla \times ($$d\phi\over dx$x + $d\phi\over dy$y + $d\phi\over dz$z)

in which I show that the curl is 0, because each term is the partial of a component has no component of the partial.

Is this a valid proof technique?

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1 Answer 1

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Of course both ways are equally valid to prove the desired equation. But your approach is not really making use of Einstein summation.

For example writing $\nabla \phi$ as $\frac{\partial\phi}{\partial x}e_x+\frac{\partial\phi}{\partial y}e_y+\frac{\partial\phi}{\partial z}e_z$ is already something you do not have to to when using Einstein summation.

But you already had the right idea by writing

$$(\nabla \times \nabla \phi)_i = \varepsilon_{ijk} \frac{\partial}{\partial x_j} \frac{\partial\phi}{dx_k}.$$

You could now proceed by changing the order of derivation and applyeaing $\varepsilon_{ijk} = - \varepsilon_{ikj}$. By exchanging names of the indices $j$ and $k$ you will then get an equation of the form $X = -X$, hence $X = 0$.

Note that you would do the same when expanding, e.g. $$ (\nabla \times (\frac{\partial\phi}{\partial x}e_x+\frac{\partial\phi}{\partial y}e_y+\frac{\partial \phi}{\partial z}e_z))_x = \frac{\partial}{\partial y}\frac{\partial\phi}{\partial z} - \frac{\partial}{\partial z}\frac{\partial\phi}{\partial y}$$

which yields $0$ by changing the order of derivation in one of the summands. I must admit that I do not really understand your last statement but note that it is not necessarily the case that applying $\frac{\partial}{\partial x_i}$ on $\frac{\partial \phi}{\partial x_j}$ yields $0$ for $i \neq j$.

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