# I don't understand the notation (tensor?, not even sure)

1. Aug 15, 2013

### fluidistic

1. The problem statement, all variables and given/known data
In order to show that if $(\vec E (\vec x, t), \vec H (\vec x, t))$ is a solution to Maxwell's equation then $(\vec E (\vec x -\vec L, t), \vec H (\vec x-\vec L, t))$ is also a solution, my professor used a proof and a step I do not understand.
Let $\vec x' =\vec x-\vec L$.
At one point he wrote that since $\frac{\partial \vec H}{\partial x^i}=\frac{\partial x^{'j}}{\partial x^i} \frac{\partial \vec H}{\partial x^{'j}} =\delta _i^j \frac{\partial \vec H}{\partial x^{'j}}=\frac{\partial \vec H}{\partial x^{'j}}$ then $\vec \nabla _{\vec x '} \times \vec H (t,\vec x ')=\vec \nabla _{\vec x} \times \vec H (t, \vec x ' (\vec x))=\vec \nabla _{\vec x} \times \vec H(t, \vec x -\vec L)=\vec \nabla \times \vec H_L (t, \vec x)$.

I don't understand anything between the 2 red words. It looks like there's a weird kronecker's delta as well as partial derivatives of the magnetic fields. But what exactly are i and j (components of the H field?), I am not sure. And what do these partial derivatives have to do with the rotor?

2. Aug 15, 2013

### CompuChip

The first equality is the chain rule in multiple dimensions: given a function $\vec u(\vec x)$,
$$\frac{\partial f(\vec u(\vec x))}{\partial x^i} = \sum_{j} \frac{\partial f}{\partial u^j} \frac{\partial u^j}{x^i}$$
You can leave of the summation sign if you agree that repeated indices will be summed over.
Then, since in this case $u^i = x^i + \text{const.}$, the partial derivative $\frac{\partial u^i}{\partial x^j}$ will be 1 if i = j (e.g. $u^1 = x^1 + \text{const.}$ so $\frac{\partial u^1}{\partial x^1} = 1$ and $\frac{\partial u^1}{\partial x^2} = \frac{\partial u^1}{\partial x^3} = 0$). Hence the Kronecker delta appearing in the second equality.
For the final equality, I think you made a typo, and it should read $\frac{\partial \vec H}{\partial x'^i}$ with i instead of j; hopefully you see why that is (hint: in the expression with the Kronecker delta there is still summation over j but all the terms but one vanish).

The rotation is defined in terms of the components, $\nabla_{\vec x}$ is a differential operator with components like $\frac{\partial \vec H^3}{\partial x^2} - \frac{\partial \vec H^2}{\partial x^3}$.
If the vector notation confuses you, try writing it out in components. That usually requires a bit less understanding, at the expense of slightly more ink.
What you need to do is consider $\nabla_{\vec x'}$ which has differentiation with respect to the components of x' and rewrite it to an expression with $\nabla_{\vec x}$ which has partial differentation with respect to the components of x (unprimed). The identity between the red words tells you how these are related.

3. Aug 15, 2013

### voko

That should have been $\frac{\partial \vec H}{\partial x^{'i}}$ in the end. This is nothing but the application of the chain rule and the very simple dependence between $x$ and $x'$. I think that should answer the rest of your question, too.

4. Aug 15, 2013

### fluidistic

Thank you guys I get it now.
I didn't realize he used Einstein's notation so I was even confused on the dimension of the vector field H he was working on (though the curl should have indicated me 3d).
Problem solved. Thanks.