Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

What is the D'Alembert operator

  1. Apr 30, 2006 #1
    I've seen two different text books write two different expressions for this, what is the proper D'Alembert Operator?
     
  2. jcsd
  3. Apr 30, 2006 #2

    robphy

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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 http://en.wikipedia.org/wiki/D'Alembert_operator .
     
    Last edited: Apr 30, 2006
  4. Apr 30, 2006 #3
    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?
     
  5. Apr 30, 2006 #4

    robphy

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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.
     
  6. Apr 30, 2006 #5
    Just as a side point, it's invariant since

    [tex]\square = \partial_\mu \partial^\mu[/tex]
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: What is the D'Alembert operator
  1. What is this? (Replies: 1)

  2. Projection operator (Replies: 1)

  3. Inverse of an operator (Replies: 1)

Loading...