# What is the D'Alembert operator

1. Apr 30, 2006

### SeReNiTy

I've seen two different text books write two different expressions for this, what is the proper D'Alembert Operator?

2. Apr 30, 2006

### robphy

It's the wave operator
$$\square = \left[-\frac{\partial^2}{\partial t^2} + \nabla^2 \right]$$, written in rectangular coordinates, that appears in the wave-equation. Some write $$\square^2$$ and some write it with an overall opposite sign.

See http://en.wikipedia.org/wiki/D'Alembert_operator .

Last edited: Apr 30, 2006
3. Apr 30, 2006

### SeReNiTy

I write the sqaured version but with a 1/(c^2) factor in it, so the sqaured operator is the same as the unsqaured one?

4. Apr 30, 2006

### robphy

Oops, I left off the wave speed (which is sometimes absorbed into the variables for convenience)
$$\square = \left[-\frac{1}{c^2}\frac{\partial^2}{\partial t^2} + \nabla^2 \right]$$. Thanks for pointing that out.

Some write the wave equation
$$\left[-\frac{1}{c^2}\frac{\partial^2}{\partial t^2} + \nabla^2 \right] \phi=0$$
as
$$\square \phi=0$$, and some as $$\square^2 \phi=0$$. It's a notational thing.

5. Apr 30, 2006

### masudr

Just as a side point, it's invariant since

$$\square = \partial_\mu \partial^\mu$$