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

2-D Waves: Relation of power with radius

  1. Jul 29, 2007 #1
    As I understand, power is directly propotional to amplitude squared for all waves. In the 2-D case, how will power depend on 'r'?

    For plane waves, the wavefunction is exp(ik.r) and multiplying it with its complex conjugate gives a constant amplitude and thus a constant power (I think). But who says that all 2-D waves are plane waves!

    In special cases, power can be constant, but I guess, in general, it must have a dependence on 'r'. I read in AP French's book 'Vibrations and Waves', somewhere in the last chapter where he discusses the solutions of wave equation in two and three dimensions, that far from the source the amplitude falls off as 1/sqrt(r). So Amplitude squared would be 1/r and thus power will fall as 1/r.

    Is that correct? :confused:
  2. jcsd
  3. Jul 29, 2007 #2


    User Avatar
    Homework Helper

    um... nobody says that. Indeed, plane-waves with respect to the distance from some origin (e.g., [tex]e^{ikr}[/tex]) are not solutions to the wave-equation in more than one dimension; the scattered (or radiation) part of the wave may have the form of a plane wave *multiplied by another function*, but never a plane-wave alone in more than one-dimension.

    As you stated, in two dimensions the scattered wave does infact have the form
    I.e., a plane wave in radial distance multiplied by one over the squareroot of the distance.

    sure. Thus in two dimensions the power radiated is always finite, not infinite and not zero, as it should be. Similarly in three dimensions having radiation fall-off like 1/r leads to finite power radiated, and in one dimension having radiation fall off like 1/r^0=1 leads to finite power radiated.
  4. Jul 30, 2007 #3
    Thank you.

    Correct me if I am wrong: Power falls off as
    1/r^2 in 3-D
    1/r in 2-D
    Constant w.r.t. 'r' in 1-D?

    You've been very helpful. I've been confused about this question for a while now!
  5. Jul 30, 2007 #4


    User Avatar
    Homework Helper

    Yeah, I believe so. What you want to look at is the Poynting vector
    \vec S=\vec E \times \vec H\;.
    If you integrate this over a surface it tells you the energy which flows through the surface per unit time (power). One can thus figure out how fast the radiated power falls off by reasoning thusly:

    In three dimensions consider a very large spherical surface which encloses a radiating object. The energy flowing through the surface is all due to radiation because we have chosen it to be very far away. The area of the surface is [tex]4\pi r^2[/tex] thus the energy radiated must fall off as [tex]1/r^2[/tex].

    Similarly in two dimensions we consider a very large circle whose "surface" length is
    [tex]2\pi r[/tex] thus in two dimensions the energy radiated falls off as 1/r.

    Similarly in one dimension we consider a "surface" (which is just two points) very far away whose "area" doesn't change with distance thus the energy radiated doesn't change with distance.

    You can work out further cases in higher (non-physical) numbers of space dimensions...

  6. Aug 3, 2007 #5
    That is very helpful. I never knew one could think of the relationship between power and raidus that way.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: 2-D Waves: Relation of power with radius
  1. Effiency of Wave Power (Replies: 2)

  2. Basic 2-D Collision (Replies: 2)