# 4 vector notation help

1. Oct 11, 2011

### Onamor

Not a particularly direct question, just something I don't mathematically understand and would very much appreciate help with.

For some scalar field $\phi$, what would $\partial_{\mu} \phi^{*}\partial^{\mu} \phi$ mean in mathematical terms. ie how would I calculate it?

From what I understand its basically $\Sigma_{\mu}\left(\frac{\partial}{\partial x^{\mu}}\phi \right)^{2}$ because of the complex conjugate in the scalar field, and you sum over repeated indexes.

Also, just to ask, why wouldn't I write this $\partial^{\mu} \phi^{*} \partial^{\mu} \phi$? Is it because I wouldn't then be allowed to sum over the $\mu$ index?
Or is it something to do with a contraction being Lorentz invariant?

Thanks for any help, let me know if I haven't been clear.

2. Oct 11, 2011

### vela

Staff Emeritus
$\partial_\mu \phi^* \partial^\mu \phi$ is the same as $\eta^{\mu\nu}\partial_\mu \phi^* \partial_\nu \phi$ where
$$\partial_\mu = \frac{\partial}{\partial x^\mu}$$and repeated indices imply summation, so you have
$$\partial_\mu \phi^* \partial^\mu \phi = - \left\lvert\frac{\partial \phi}{\partial x^0}\right\rvert^2+\sum_i \left\lvert\frac{\partial \phi}{\partial x^i}\right\rvert^2$$
In general, you shouldn't have a repeated index with both raised or both lowered. They should always come one up and one down, otherwise you have a malformed expression on your hands.

3. Oct 12, 2011

### CompuChip

To complement vela's response: the answers are yes and yes.
Summation convention only applies to one upper and one lower index, and the whole idea is that doing this that given some objects behaving properly under Lorentz-transformations, the notation almost forces you into creating new objects behaving properly under Lorentz-transformations, rather than some arbitrary mathematical expression.
As vela shows, it means that if you use the simple trick of "one upper + one lower" index what you are actually doing is making sure you use the spacetime metric in precisely the places you need to get Lorentz-invariance right.

4. Oct 12, 2011

### Onamor

Thank you both, very helpful as always.
Much appreciated.