What is the D'Alembert operator

  • Context: Graduate 
  • Thread starter Thread starter SeReNiTy
  • Start date Start date
  • Tags Tags
    D'alembert Operator
Click For Summary

Discussion Overview

The discussion revolves around the D'Alembert operator, specifically its definition and notation as encountered in different textbooks. Participants explore its role in the wave equation and the variations in its expression.

Discussion Character

  • Technical explanation
  • Debate/contested

Main Points Raised

  • Some participants assert that the D'Alembert operator is the wave operator expressed as \(\square = \left[-\frac{\partial^2}{\partial t^2} + \nabla^2 \right]\), with variations in notation.
  • One participant mentions using a squared version of the operator with a factor of \(1/(c^2)\), suggesting that the squared operator is equivalent to the unsquared one.
  • Another participant acknowledges the omission of the wave speed in their initial expression and clarifies that the operator can also be written as \(\square = \left[-\frac{1}{c^2}\frac{\partial^2}{\partial t^2} + \nabla^2 \right]\).
  • There is a mention of the wave equation being represented in different notational forms, either as \(\square \phi=0\) or \(\square^2 \phi=0\), indicating a notational preference rather than a conceptual difference.
  • A side point is raised about the invariance of the operator, expressed as \(\square = \partial_\mu \partial^\mu\).

Areas of Agreement / Disagreement

Participants do not reach a consensus on a single expression for the D'Alembert operator, as multiple notational forms and interpretations are presented. The discussion reflects differing preferences and interpretations rather than a unified agreement.

Contextual Notes

There are unresolved aspects regarding the definitions and notational conventions of the D'Alembert operator, including the role of the wave speed and the implications of using squared versus unsquared forms.

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]
 

Similar threads

  • · Replies 9 ·
Replies
9
Views
6K
  • · Replies 16 ·
Replies
16
Views
4K
  • · Replies 25 ·
Replies
25
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 30 ·
2
Replies
30
Views
3K
  • · Replies 8 ·
Replies
8
Views
1K
  • · Replies 16 ·
Replies
16
Views
2K