# EM Wave - Where Does the Energy Go?

## Main Question or Discussion Point

For an EM wave in vacuum, we know the energy density is given by

1/2 e E^2 for the electric field, with a similar expression for the magnetic (e is permittivity of vacuum). E^2 implies that the energy oscillates as a cosine squares function if we represent the E field as E_0.cos [ kx - wt].

But since cosine squared goes between 0 and 1, where does the energy go when it is at 0?? It cant be stored in the magnetic field since they are in phase in vacuum, so where does it "go"? I hope you can understand my question!

Secondly, in Poyntings theorem for the conservation of energy with EM waves, the kinetic energy density is given by E.J. Does this current density refer to a current that the E field has generated AND/OR one that was already there?

Cheers guys!!

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Hello, I'm only following an introductory course on electromagnetism myself (Griffiths-level), so I'm no authority, but I'll do my best to try and help!

Not a very physically insightful answer, but I believe the standard answer on your question is "The amplitude is not zero everywhere at the same time", i.e. when you're at a certain x and you wait for the t so that cos(kx-wt) = 0, this just means the energy has "moved" to another x' where cos(kx' - wt) is NOT zero.

I'm not sure how you'd draw the distinction between a current generated by your electric field or not. Anyhow, the J in the formula is simply any current (density) that's there (in that point), no matter how it got there.
Too maybe help demistify the "E.J" expression:
we know the electromagnetic power on a particle is $$\vec F_{\textrm{electromagnetic}} \cdot \vec v = q(\vec E + \vec v \times \vec B) \cdot \vec v = q \vec E \cdot \vec v$$
and this gives us for the power density $$\rho \vec E \cdot \vec v$$, and $$\rho \vec v$$ is exactly what we call $$\vec j$$.
So the E.J is just an indication of the (derivative of the) work delivered by the electromagnetic force on some charge, no matter how that charge got there

Bill_K
It doesn't go anywhere. The energy in a sinusoidal electromagnetic wave just happens to be zero some places and nonzero others. All of it travels forward with the speed of light.

You must be thinking that the first cycle of the wave somehow has the job of passing its energy back to the next cycle to maintain it, but that's not the case. Each cycle is independent of the others. You could have just one cycle! The amplitude is an arbitrary function, just as the amplitude of a sound wave is an arbitrary function or in any other solution of the wave equation. All that Maxwell's Equations tell you is that E and B are transverse, that B is perpendicular to E with amplitude related to dE/dt, and that the whole thing moves forward with velocity c.

sophiecentaur