# Metric tensor

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.

dextercioby
Homework Helper
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.

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## ?

George Jones
Staff Emeritus
Gold Member
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}

George Jones
Staff Emeritus
Gold Member
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## ?

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.