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I'm trying to learn about relativistic electrodynamics on my own, and I am struggling with derivatives of the 4-potential and index (Einstein) notation.

I think I understand expressions such as $\partial_\mu A^\mu$. The index is repeated and is once up and once down, so I would expand the sum as: $\partial_0 A^0 + \partial_1 A^1 + \partial_2 A^2 + \partial_3 A^3$, which gives me a scalar.

  1. How am I to interpret something like this: $(\partial_\mu A_\nu)(\partial^\mu A^\nu)$ ? We are summing over the two indices this time, which is fine. What confuses me is that we are taking a covariant derivative of a covariant vector. Does one need to "convert" $A_\nu$ in the first term to contravariant, like so: $(\partial_\mu A^\rho \eta_{\nu\rho})(\partial^\mu A_\sigma\eta^{\nu\sigma})$?

    I guess my doubts arise from the fact that I see a covariant vector as being an entirely different object from a contravariant one. The covariant derivative $\partial_\mu = \frac{\partial}{\partial x^\mu}$ differentiates with respect to the components of the contravariant vector $x$. So I don't understand how such an operation can be applied to a vector that isn't also contravariant.

  2. How should I interpret terms such as $(\partial_\mu A^\mu)^2$ ? Is it just $\left(\partial_0 A^0 + \partial_1 A^1 + \partial_2 A^2 + \partial_3 A^3\right)^2$ or is there something else going on?
  3. According to some textbook, $(\partial_\mu \phi)^2 = \eta^{\mu\nu}\partial_\mu \phi\partial_\nu\phi$, but I don't understand why. For me $\partial_\mu \phi$ is just the derivative of a scalar $\phi$ with respect to some (unspecified) component $\mu$ of a contravariant 4-vector $x$. Instead, judging from the right-hand side, it is to be interpreted as a vector $(\frac{\partial}{\partial x^0},\boldsymbol{\nabla})\phi$ which is then squared. Is it just sloppy notation, or am I being stupid?

Thanks.

EDIT:

  1. Are the following then true? $$\frac{\partial}{\partial(\partial_\mu A_\nu)} \left(\partial_\mu A_\nu\right) = 1$$ $$\frac{\partial}{\partial(\partial_\mu A_\nu)} \left(\partial^\mu A^\nu\right) = 0$$
  2. Can I also raise and lower the indices of a partial derivative?
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yes, you can definitely raise and lower the indices of a partial derivative, but $\partial_{\mu}$ is the one that you know and love. –  Jerry Schirmer Oct 12 '12 at 22:08
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1 Answer 1

up vote 3 down vote accepted

1) You have the right idea about $\partial_{\mu}A^{\mu}$ and 2) $(\partial_{\mu}A^{\mu})^2$.

For the rest of it, you're goign to have to be careful about covariant and contravariant indices of $\partial_\mu$ and $A^{\mu}$. Since you are already going with $A$ having vector indices, I think you should stick to that as your "base" version of $A$ for now. Now, you correctly seem to note that lowering and raising is done with $\eta_{\mu\nu}$ and its inverse $\eta^{\mu\nu}$ (which are the same matrix for Mikowskian coordinates). Thus, you will find that an expression like $A_{\mu}A^{\mu} = \eta_{\mu\nu}A^{\mu}A^{\nu} = -(A^{0})^{2} + (A^{1})^{2} + (A^{2})^{2} + (A^{3})^{2}$, and you can extend this to most of the other examples you cite.

3) applying $\partial_\mu$ to a scalar $\phi$ gives you a result $\partial_{\mu}\phi$ that is best interpreted as a one form, not as a scalar--that's because you have the free index $\mu$ floating around, and if you change coordinates, you will have to apply the chain rule to get the right answer for your new $\partial_{\mu}\phi$. this is why you have to apply the inverse metric to "square" $\partial_{\mu}\phi$.

4) If you are taking variations of vector valued quantities, you should leave a dummy index for your variational term:

$\begin{equation} \frac{\delta}{\delta A^{\mu}}A^{\nu} = \delta^{\nu}{}_{\mu} \end{equation}$

This will give you terms like

$\begin{equation} \frac{\delta}{\delta A^{\alpha}} A^{\mu}A_{\mu} = \frac{\delta}{\delta A^{\alpha}}(\eta_{\mu\nu}A^\mu A^{\nu}) = \eta_{\mu\nu}\delta^{\mu}_{\alpha}A^{\nu} +\eta_{\mu\nu}A^{\mu}\delta^{\nu}_{\alpha} = 2 A_{\alpha} \end{equation}$

which should seem pretty reasonable from your calculus I based intuition. Similarly, you should probably think of $\frac{\delta}{\delta \partial_{\alpha}A_{\beta}}\partial_{\mu}A_{\nu} = \delta^{\alpha}_{\mu}\delta^{\beta}_{\nu}$. If you're taking a variation of the up version of a quantity with respect to the down version of the quantity, just use the metric to raise/lower the target before taking the variation, and fix everything after the fact using raising and lowering conventions and the Kroneker deltas.

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Thank you so much, that was most helpful and I am finally starting to makes sense of it. I have added a #4 to my questions, I hope I am not abusing your kindness. –  kdfc Oct 12 '12 at 21:48
    
@kdfc: there's a lazy attempt at an answer. –  Jerry Schirmer Oct 12 '12 at 22:08
    
Again, thank you very much! –  kdfc Oct 12 '12 at 22:30
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