Question about d'Alembert operator

[Dixanadu]

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

 

[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.

 

[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 $\phi$)

 

[Dixanadu]

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

 

[pmacias]

The d'Alembert operator, also known as the wave operator or the Laplace-Beltrami operator, is commonly used in physics and mathematics to study wave phenomena and solve differential equations. It is defined as \Box = \partial_{\mu}\partial^{\mu}, where \partial_{\mu} represents the partial derivative with respect to the \mu-th coordinate.

To answer your question, when \mu and \nu are not equal, the d'Alembert operator is not equal to 0. In fact, it represents the second-order differential operator that describes the behavior of waves in a given space. This operator is essential in understanding various physical phenomena, such as sound and light waves, as well as in solving problems in fields such as electromagnetism and general relativity.

Therefore, it is not surprising that the expression \partial_{\mu}\partial^{\nu}\phi is only equal to \Box \phi when \mu = \nu. This is because when \mu and \nu are equal, the operator reduces to the familiar Laplacian operator, which is commonly used in solving differential equations.

I hope this helps clarify your curiosity. Keep exploring and using the d'Alembert operator in your calculations – it is a powerful tool in the world of science and mathematics.