1. Nov 12, 2014

Hey guys,

The expression $\partial_{\mu}\partial^{\nu}\phi$ is equal to $\Box \phi$ when $\mu = \nu$. However when they are not equal, is this operator 0?

I'm just curious cos this sort of thing has turned up in a calculation of mine...if its 0 I'd be a very happy boy

2. Nov 12, 2014

ShayanJ

That thing is in fact a second rank tensor which in 1+1 dimensions becomes:

$\partial_\mu \partial^\nu \phi=\eta^{\nu \lambda}\partial_\mu\partial_\lambda \phi=\left( \begin{array}{cc} \eta^{00}\frac{\partial^2 \phi}{\partial x^0 \partial x^0}+\eta^{01}\frac{\partial^2 \phi}{\partial x^0 \partial x^1} \ \ \ \ \ \ \eta^{00}\frac{\partial^2 \phi}{\partial x^1 \partial x^0}+\eta^{01}\frac{\partial^2 \phi}{\partial x^1 \partial x^1} \\ \\ \eta^{10}\frac{\partial^2 \phi}{\partial x^0 \partial x^0}+\eta^{11}\frac{\partial^2 \phi}{\partial x^0 \partial x^1} \ \ \ \ \ \ \eta^{10}\frac{\partial^2 \phi}{\partial x^1 \partial x^0}+\eta^{11}\frac{\partial^2 \phi}{\partial x^1 \partial x^1} \end{array} \right)$

I think its obvious that there is no reason for it to be identically zero.

3. Nov 13, 2014

Matterwave

Usually $\eta$ is used for the Minkowski metric in Cartesian coordinates. Shyan wrote you a general expression for a general metric, in 2 dimensions. Usually we work in 4-dimensions, but by using the Minkowski metric, the expression is much simpler.

It is perhaps easier to deal with $\partial_\mu\partial_\nu\phi$ in which case, this is just the Hessian matrix: http://en.wikipedia.org/wiki/Hessian_matrix (replace f with $\phi$)

4. Nov 13, 2014

Okay thank you. so would you say that $\partial_{\mu}\partial^{\nu}\phi=\partial^{\nu}\partial_{\mu}\phi$?