Energy proportional to Amplitude squared?

[SamRoss]

Does anyone know a simple derivation that explains why the energy of a wave is proportional to the amplitude squared?

[jtbell]

Does "simple" allow for calculus?

A wave is a collection of simple harmonic oscillators. The energy of a SHO equals the potential energy at maximum displacement. The PE at maximum displacement is the work done by an external force in pushing the oscillator out from the equilibrium position to maximum displacement. The external force acts against the oscillator's internal force which obeys Hooke's Law F = -kx.

\int_0^A {F(x)dx} = \int_0^A {kx dx} = \frac{1}{2} kA^2

[vin300]

Why doesn't this work for EM waves ?

[Vagn]

Why doesn't this work for EM waves ?

Why should it? What constant would you introduce in place of k, bearing in mind that it must have the same unit, and there isn't a spring constant for an EM wave.

[SamRoss]

\int_0^A {F(x)dx} = \int_0^A {kx dx} = \frac{1}{2} kA^2

Thanks for the quick reply. I was actually thinking of an EM wave, though. Do you know a derivation for that as well?

[jtbell]

In general, not just in EM waves, the energy density of an electric field E is proportional to E^2, and the energy density of a magnetic field B is proportional to B^2. Introductory textbooks usually derive these by considering the external work it takes to charge up a parallel-plate capacitor so that it has a field E between the plates:

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/capeng.html

or to increase the current flowing through a long solenoid (inductor) so it has a field B inside:

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/indeng.html

[SamRoss]

cool, thanks

[DaleSpam]

My favorite site on this topic is:
http://farside.ph.utexas.edu/teaching/em/lectures/node89.html