# Metric tensor

1. Dec 22, 2014

### Maybe_Memorie

Some subtleties of the metric tensor are just becoming clear to me now. If I take $g_{\mu\nu}=diag(+1,-1,-1,-1)$
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. Dec 22, 2014

### dextercioby

Not really. The minus only comes in once you you use the metric somewhere, but with $\partial_{\mu}\phi^{\mu}$ if you use the contravariant components of phi, then there's no use of the metric tensor.

3. Dec 22, 2014

### Maybe_Memorie

Ah I see. So in that case, $\partial_0\phi^0 + \partial_i\phi^i=\partial_\mu\phi^{\mu}=g^{\mu\nu}\partial_\mu\phi_\nu=\partial_0\phi_0 - \partial_i\phi_i$ ?

4. Dec 22, 2014

### George Jones

Staff Emeritus
Expanding on what dextercioby wrote, and using the metric given in the original post:
\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. Dec 22, 2014

### George Jones

Staff Emeritus

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. Dec 22, 2014

### Maybe_Memorie

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

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!