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

[fluidistic]

Homework Statement

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?

 

[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.

 

[voko]

$\frac{\partial \vec H}{\partial x^i}=\frac{\partial x^{'j}}{\partial x^i} \frac{\partial \vec H}{\partial x^{'j}} =\delta _i^j \frac{\partial \vec H}{\partial x^{'j}}=\frac{\partial \vec H}{\partial x^{'j}}$

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.

 

[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.

 

[paranormal]

I can explain the notation and the reasoning behind it. First, the notation used is called tensor notation, which is commonly used in physics and mathematics to represent vector and tensor quantities. In this case, $\vec E$ and $\vec H$ are both vector fields, which means they have a magnitude and direction at every point in space.

The proof your professor presented involves a change of coordinates, specifically from $\vec x$ to $\vec x' = \vec x - \vec L$. This change of coordinates is represented by the equation $\frac{\partial \vec H}{\partial x^i}=\frac{\partial x^{'j}}{\partial x^i} \frac{\partial \vec H}{\partial x^{'j}}$. This equation is known as the chain rule in multivariable calculus and it allows us to express the partial derivative of a function with respect to one set of coordinates in terms of the partial derivatives with respect to another set of coordinates. In this case, $i$ and $j$ represent the components of the vector fields in the original and new coordinates, respectively.

The Kronecker delta, represented by $\delta_i^j$, is a mathematical symbol that takes a value of 1 when $i=j$ and 0 otherwise. In this case, it is used to simplify the equation and show that the partial derivative with respect to the new coordinates is simply equal to the partial derivative with respect to the old coordinates.

The reason why the partial derivatives of the magnetic field are important in this proof is because Maxwell's equations involve the curl of the electric and magnetic fields. The curl, also known as the rotor, is a mathematical operation that takes a vector field as an input and produces another vector field as an output. In this case, the equation $\vec \nabla \times \vec H_L (t, \vec x)$ represents the rotor of the magnetic field in the new coordinates. By showing that this is equal to the rotor of the original magnetic field, the proof demonstrates that the solution is still valid in the new coordinates.

In summary, the notation and equations used in the proof may seem confusing at first, but they are necessary to show that the new coordinates do not affect the validity of the solution to Maxwell's equations.