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

  1. Aug 15, 2013 #1

    fluidistic

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    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. jcsd
  3. Aug 15, 2013 #2

    CompuChip

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    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.
     
  4. Aug 15, 2013 #3
    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.
     
  5. Aug 15, 2013 #4

    fluidistic

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    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.
     
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