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

Modes with boundary conditions

  1. Feb 22, 2008 #1
    If I have a finite boundary, say of length L. Is it possible to demonstrate that if I were to allow all possible CONTINUOUS values of a wave to exist (with unit amplitude) then deconstrutive interference destroys all waves except those with wavelength:


    Where n =0,1....

  2. jcsd
  3. Feb 22, 2008 #2
    Do you mean a superposition of ALL continous functions on L ...? Supposing this makes any sense I would expect the sum to be the zero function, since for each function in your sum, there is also its negative ....How would you expect to get multiple possible result (n=0,1...) for a certain sum of all continous functions..?
    Last edited: Feb 22, 2008
  4. Feb 22, 2008 #3


    User Avatar
    Staff Emeritus
    Science Advisor

    The difficulty is that you haven't said what boundary conditions you are applying. Are you assuming, without saying so, that you are considering only waves that are 0 at 0 and L?
  5. Feb 22, 2008 #4
    Even if so, the sum would still be zero.
  6. Feb 22, 2008 #5
    Destructive interference is not what destroys the waves as energy cannot be created or destroyed. Without losses, either waves transmitted outside the boundary, or attenuation within the boundary, the waves will persist forever, and the signal will remain the same amplitude. The typical boundary condition used for black body radiation is a box with high conductivity. In which case waves which do not have nodes at the boundary cause a current which dissipates energy.
  7. Feb 23, 2008 #6
    No, just a superposition of all wavelengths (greater than zero oviously) with no boundary conditions.

    We usually impose that the wave function goes to zero at 0 and L (Dirichlet BC's), but I want to see if its possible to demonstrate this is the case physically...eg those waves that dont go to zero at the boundaries will interfere deconstructively with all the others that dont go to zero at the boundary when we take all frequencies from zero to infinity.
    Last edited: Feb 23, 2008
  8. Feb 23, 2008 #7

    Precisely not, in fact.

    I'm hoping to show that the boundary conditions (wave goes to zero at 0 and L) have a physical origin.
    Last edited: Feb 23, 2008
  9. Feb 23, 2008 #8
    I understand that, and still think that this superposition will be zero.

    What you are talking about are Dirichlet boundary conditions. Von neumann B.C. means prescribing the normal derivative at the boundary (in this case, just the ordinary derivative)
  10. Feb 23, 2008 #9
    No, good point but I certainly appreciate that. I'm trying to understand from a physical perspective why only standing waves form between Casimir plates.

    You see, typically the quantum vacuum has the freedom to take any frequency from zero to [tex]\infty[/tex], but in the presence of boundaries, the degrees of freedom are modified and only standing waves form. I'm hoping to understand what it is about the boundaries that 'inhibits' the vacuum, and if perhaps is some consequence of interference.
  11. Feb 23, 2008 #10
    Me too! I want to show it mathematically.

    Yes, I 'knew' that. Slip of the brain...thanks.

    Last edited: Feb 23, 2008
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Modes with boundary conditions
  1. Boundary vs. Boundary (Replies: 8)