- #1

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and want to write ##\partial_\mu\phi^\mu##, it would be ##\partial_0\phi^0 -\partial_i\phi^i##, correct? ##\phi## is a 4-vector.

- Thread starter Maybe_Memorie
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- #1

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and want to write ##\partial_\mu\phi^\mu##, it would be ##\partial_0\phi^0 -\partial_i\phi^i##, correct? ##\phi## is a 4-vector.

- #2

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- #4

George Jones

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$$

\begin{align}

\partial_\mu \phi^\mu &= \partial_0 \phi^0 + \partial_i \phi^i \\

&= g_{\mu \nu} \partial^\nu \phi^\mu \\

&= g_{0 0} \partial^0 \phi^0 + g_{1 1} \partial^1 \phi^1 + g_{2 2} \partial^2 \phi^2 + g_{3 3} \partial^3 \phi^3 \\

&= \partial^0 \phi^0 - \partial^1 \phi^1 - \partial^2 \phi^2 - \partial^3 \phi^3

\end{align}

$$

- #5

George Jones

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A couple of comments:

1) Einstein summation convention often is used only for one up, one index down;

2) the components ##\left\{ g^{\mu\nu} \right\}## only equal the components ##\left\{ g_{\mu\nu} \right\}## for the metric given in the original post, not for general metrics.

- #6

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Ah right, so it should be explicitly written as ##g^{\mu\nu}\partial_\mu\phi_\nu=\partial_0\phi_0 - \partial_1\phi_1 -\partial_2\phi_2 -\partial_3\phi_3##A couple of comments:

1) Einstein summation convention often is used only for one up, one index down;

2) the components ##\left\{ g^{\mu\nu} \right\}## only equal the components ##\left\{ g_{\mu\nu} \right\}## for the metric given in the original post, not for general metrics.

I knew about your second comment. Somehow I've survived a course on Jackson and a differential geometry course yet I'm only really thinking about this stuff now. Thanks for the help!

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