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I'm starting to study electromagnetic waves and as i understand, an electromagnetic wave projects a varying electric field. This electric field can in turn give forces of repulsion/attractions to the electrons and protons it passes very close to. Why doesn't it violate the law of conservation of energy?

Aayush Agrawal
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The electromagnetic field itself contains energy distinct from the energy of charged bodies, the energy in a given volume of empty space can be found by integrating the energy densities $\frac{1}{2}\epsilon E^2$ and $\frac{1}{2} \frac{B^2}{\mu}$ over the region. When the EM fields increase the kinetic energy of charged particles, there is a corresponding decrease in the energy of the EM field in that region, so total energy is unchanged. The general proof that any combination of fields and charges obeying Maxwell's equations will conserve energy is known as Poynting's Theorem, proved for example on pages 346-348 of Introduction to Electrodynamics, Third Edition by David J. Griffiths, or on this page from physicspages.com

Hypnosifl
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  • Can you "prove" it? I mean, in several words about physical phenomena instead of just a statement "so the total energy is unchanged". – Vladimir Kalitvianski Nov 23 '14 at 17:57
  • I added some references to the general proof, known as Poynting's Theorem. – Hypnosifl Nov 23 '14 at 18:19
  • Now it is better, but I wanted another thing (which is implicitly present in the Poynting's Theorem): to calculate correctly the energy balance, we must take into account the the total field, not only the external one. – Vladimir Kalitvianski Nov 23 '14 at 18:32
  • If we're talking about a collection of point charges this wouldn't be an issue, would it? And for extended continuous charge distributions, then as you said, it's already implicit in Poynting's theorem that you're taking into account the field inside the volume occupied by the charge distribution as well as the field outside it. – Hypnosifl Nov 23 '14 at 19:10
  • On the contrary, for a collection of point particles, it is an issue. Even for one charged particle, there are no CED equations conserving the total energy exactly. – Vladimir Kalitvianski Nov 23 '14 at 19:20
  • Ah, so you're specifically concerned with the fact that the field is singular at the exact location of a point particle? It may be that Poynting's theorem breaks down in this case--are you speculating that this might be the case, or do you have a specific analysis or reference that says it does? – Hypnosifl Nov 23 '14 at 19:37
  • I found an example of an energy calculation for two point charges using the energy in the field on p. 3 of this pdf, it says you do get two "self-energy" terms whose volume integral is infinite, but these terms don't change as the charges move, if you just want to figure out the change in energy due to motion of the charges you can ignore these terms and just look at the "interaction energy" term. But this is an electrostatic case where charges move slowly enough that waves can be neglected, not sure if this generalizes. – Hypnosifl Nov 23 '14 at 19:50
  • The Poynting's theorem holds, even in this case, but the problem now is in implying a physical solution to the equation system. In electrostatics the constant energy term is inessential indeed, but for a relativistic case we cannot write the mechanical equation with the right radiation reaction force. We only write an approximate equation, that's the problem. – Vladimir Kalitvianski Nov 23 '14 at 20:15
  • Maybe you should ask a new question which is more specifically about this issue of the radiation reaction force in a scenario involving relativistic velocities, others might know more about what the existing literature has to say about it than I do. Also, do I understand correctly that you're saying the problem is with finding an exact solution in certain cases, without disputing the general proof that the solution for all possible initial conditions must obey energy conservation, whether or not we know how to find the solution in each specific case? – Hypnosifl Nov 23 '14 at 20:34
  • There is no need to ask an additional question here. It is an old "eternal" problem in CED. No, the problem is not in finding solutions to our equations. Those solutions exists and are non physical. The real problem is in figuring out the right equations. They are still unknown. – Vladimir Kalitvianski Nov 23 '14 at 20:51
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An electromagnetic wave doesn't violate the conservation of energy. Maxwell's electromagnetism does. When you apply the Maxwell's equations to two beams of light of the same frequency, collinear, out of phase 180 degree, occurs the phenomenon of destructive interference. In this special case the electric field of one wave is canceled with the electric field of the other wave; the same happens with the magnetic fields. Both fields become zero. Following Maxwell the energy is contained only in those fields.

Let make the math: Let us have these two waves: Be $E$ = electric field; $B$=magnetic field

$$E_1 = E_m \sin (kx - \omega t), \quad E_2 = E_m \sin (kx - \omega t + p)\\ B_1 = B_m \sin (kx - \omega t), \quad B_2 = B_m \sin (kx - \omega t + p) $$

$E = E_1 + E_2$ and $B = B_1 + B_2$

\begin{align} E &= E_m \sin (kx - \omega t) + E_m \sin (kx - \omega t + p) \\ B &=B_m \sin (kx - \omega t) + B_m \sin (kx - \omega t + p) \end{align}

But, $\sin (kx - \omega t + p) = - \sin (kx – \omega t)$ ,

Then, $E = 0$ and $B = 0$ and,

\begin{align} U_T& = U_E + U_B \\ &= \frac12 \epsilon_0E^2 + \frac12\mu_0B^2 \\ &= 0 \end{align}

According to Maxwell equations, in destructive interference of two collinear beams out of phase 180 degree, the energy density, which is a function of the electric and magnetic fields, is destroyed too. In this way, the principle of conservation of energy is violated.

Kyle Kanos
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luis fondeur
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