What is the D'Alembert operator

  • Context: Graduate 
  • Thread starter Thread starter SeReNiTy
  • Start date Start date
  • Tags Tags
    D'alembert Operator
Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
4 replies · 14K views
SeReNiTy
Messages
170
Reaction score
0
I've seen two different textbooks write two different expressions for this, what is the proper D'Alembert Operator?
 
Physics news on Phys.org
It's the wave operator
[tex]\square = \left[-\frac{\partial^2}{\partial t^2} + \nabla^2 \right][/tex], written in rectangular coordinates, that appears in the wave-equation. Some write [tex]\square^2[/tex] and some write it with an overall opposite sign.

See "[URL .
 
Last edited by a moderator:
robphy said:
It's the wave operator
[tex]\square = \left[-\frac{\partial^2}{\partial t^2} + \nabla^2 \right][/tex], written in rectangular coordinates, that appears in the wave-equation. Some write [tex]\square^2[/tex] and some write it with an overall opposite sign.

See "[URL .

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

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

[tex]\square = \partial_\mu \partial^\mu[/tex]