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?
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asked Nov 23, 2014 at 11:22
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2 Answers
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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
answered Nov 23, 2014 at 17:17
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$\begingroup$ Can you "prove" it? I mean, in several words about physical phenomena instead of just a statement "so the total energy is unchanged". $\endgroup$ Commented Nov 23, 2014 at 17:57
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$\begingroup$ I added some references to the general proof, known as Poynting's Theorem. $\endgroup$ Commented Nov 23, 2014 at 18:19
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$\begingroup$ 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. $\endgroup$ Commented Nov 23, 2014 at 18:32
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$\begingroup$ 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. $\endgroup$ Commented Nov 23, 2014 at 19:10
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$\begingroup$ 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. $\endgroup$ Commented Nov 23, 2014 at 19:20
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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.
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1$\begingroup$ Your assessment is a little flawed, perhaps this Q&A will help. $\endgroup$ Commented Feb 12, 2015 at 17:55