1. Limited time only! Sign up for a free 30min personal 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!

One-dimensional Uniform plane wave

  1. Oct 1, 2014 #1
    to solve one dimensional uniform plane wave, why assumed E = F1(z-ct) + F2(z+ct) as a general solution of second order differential equation and in trigonometric form particular function assumed to be Ey =Sin B(z+mt).. Is there any other method like Laplace solution or something where i don't have to assume this... ..
    upload_2014-10-1_11-16-56.png
    Is there solution exist without assuming anything for this function.
     
  2. jcsd
  3. Oct 3, 2014 #2

    vanhees71

    User Avatar
    Science Advisor
    2016 Award

    Let's write the equation as
    [tex]\partial_1^2 f(x_1,x_2)-\partial_2^2 f(x_1,x_2)=0.[/tex]
    You can always set [itex]x_1=c t[/itex] to get it in physical units of time.

    This is the wave equation in 1 spatial dimension. Now the trick is to rewrite this in terms of "light-cone coordinates",
    [tex]\xi=x_1+x_2, \quad \eta=x_1-x_2.[/tex]
    Now you have, according to the chain rule,
    [tex]
    \frac{\partial}{\partial x_1} f=\frac{\partial \xi}{\partial x_1} \frac{\partial f}{\partial \xi} +\frac{\partial \eta}{\partial x_1} \frac{\partial f}{\partial \eta}=\frac{\partial f}{\partial \xi} + \frac{\partial f}{\partial \eta}.[/tex]
    In the same way you derive step by step the second derivatives [itex]\partial_1^2[/itex] and [itex]\partial_2^2[/itex] in terms of the derivatives with respect to the light-cone variables [itex]\xi[/itex] and [itex]\eta[/itex].

    At the end the wave equation reads
    [tex]\partial_{\xi} \partial_{\eta} f=0.[/tex]
    This is now very easy to integrate. The vanishing of the partial derivative with respect to [itex]\xi[/itex] means that the function is only dependent on [itex]\eta[/itex]. Thus you have
    [tex]\partial_{\eta} f=g(\eta).[/tex]
    But this implies that
    [tex]f(\xi,\eta)=f_1(\eta)+f_2(\xi), \quad f(\eta)=\int \mathrm{d} \eta g(\eta).[/tex]
    There's an additional function [itex]f_2(\xi)[/itex], because the partial derivative of a function wrt. to [itex]\eta[/itex] determines this function only up to a function, indepenent of [itex]\eta[/itex], i.e., it can only depend on [itex]\xi[/itex].

    Now you can rewrite this equation in terms of the old coordinates, i.e.,
    [tex]f(x_1,x_2)=f_1(x_1-x_2) + f_2(x_1+x_2).[/tex]
    Here [itex]f_1[/itex] and [itex]f_2[/itex] are indeed arbitrary functions that are determined by appropriate initial and boundary conditions.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: One-dimensional Uniform plane wave
  1. Plane waves (Replies: 3)

Loading...