1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Propagation of gravitational waves

  1. Oct 11, 2014 #1
    Stumbled upon this problem lately. Maybe someone could help me clarify some subtleties I do not see?

    1.
    Consider the propagation speed ##c## of periodic surface of gravity waves with wavelength ##\lambda## and amplitude ##a## in water of depth ##H##. Let ##\rho_{a}## and ##\rho_{w}## be the density of air and water. Ignoring viscosity and surface tension, use diminutional analysis to find an expression for the propagation speed ##c##. Since ##\frac{\rho_{a}}{\rho_{w}} \approx 0.001## the density of air is usually ignored. How does your expression simplify in this case? What about in the small amplitude limit ##a \rightarrow 0##?

    2. Relevant equations: none


    3. The attempt at a solution:

    1. I expressed ##c## as follows:

    ##c=f(H, \rho_{a}, \rho_{w}, t, a, \lambda)##, where ##H## is the depth of water, ##t## is time.

    I used the ##Buckingham-\pi## Theorem.

    2. I non-dimensionalized ##\pi_1## as follows: ##\pi_1 = \frac{\lambda}{a}##.

    3. Now, ##c^* = f_1(\rho_{a}^*, \rho_{w}^*, t, \frac{\lambda}{a})## - I divided by ##a## which has units of length. So now we have the following dimensions in the variables of ##f_1##:

    ##[\rho_{a}^*] = ML^{-4}##
    ##[\rho_{w}^*] = ML^{-4}##
    ##[t] = T##
    And ##\frac{\lambda}{a}## is dimensionless.
    Where M is in kilograms, L is in meters, T is in seconds.

    So, ##\pi_2 = c^{*\alpha}\rho_a^{*\beta}\rho_w^{*\gamma} t^{\delta}## (is this correct?)

    4. Now, as per the ##Buckingham-\pi## Theorem, we have

    ##[\pi_2]=T^{-\alpha+\delta} M^{\beta + \gamma} L^{-4\beta - 4\gamma}##

    So,

    ##-\alpha + \delta = 0##
    ##\beta + \gamma = 0##
    ##-4\beta - 4\gamma = 0##

    Thus ##\alpha = \delta##, ##\beta = -\gamma##. And now we have:

    ##\pi_2 = (ct)^{\alpha}(\frac{\rho_{a}}{\rho_{w}})^{\beta}##.

    So we could express the dimensionless equation as

    ##F(\frac{\lambda}{a}) = ct(\frac{\rho_{a}}{\rho_{w}})^{\beta}##. (I omitted the *'s).

    I think there is or are some errors in my deduction because if we use the same method that I described but eliminate the density of air or the amplitude of waves, we get ##{\pi_2=1}##.

    Thank you for your attention!
     
    Last edited: Oct 11, 2014
  2. jcsd
  3. Oct 16, 2014 #2
    Thanks for the post! Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?
     
  4. Oct 17, 2014 #3
    Have you thought about any other physical parameters for your analysis? For instance, does the earths gravity play a part in your analysis? (You can answer this by considering how gravity waves in fluid propagate if you were on a planet with little gravity etc.)

    In Buckingham Pi analysis, you can always over fit the dimensionful quantities, then by interrelations between them exclude them in your refinement.
     
  5. Oct 21, 2014 #4
    I have now solved this problem. I had actually misinterpreted the problem, as I thought it was about relativistic waves rather than water waves :)
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Propagation of gravitational waves
  1. Gravitational waves (Replies: 0)

Loading...