we know that a waveform of electromagnetic wave can be decomposited into sum of a series of monochromatic waves, so is the total energy of this EMW also equal to the sum of energy of these monochromatic waves? Seems not? Due to the amplitude^2 of total wave is not equal to the sum of amplitude^2 of all monochromatic component, is it right? If they are not equal, how can this decomposition reasonable? If they are not equal, then, if a short pulse EMW propagates in medium and separated into many small segments of monochromatic waves back an forth due to the dispersion effect, so where is the energy difference (between the original EMW and the sum of these monochromatic waves) going, is it disappeared??
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$\begingroup$ this is a good question: clear, reasonable and answerable. It does not deserve a down vote $\endgroup$– JEBCommented Nov 30, 2023 at 17:24
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$\begingroup$ Can you give an example of a case where " the amplitude^2 of total wave is not equal to the sum of amplitude^2 of all monochromatic component"? $\endgroup$– The PhotonCommented Nov 30, 2023 at 17:37
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1$\begingroup$ Also you might want to research Parseval's Theorem. $\endgroup$– The PhotonCommented Nov 30, 2023 at 17:37
1 Answer
Here's the easy case:
$$ E(x, t) = A_1 e^{i(k_1x-\omega_1t)} + A_2 e^{i(k_2x-\omega_2t)} $$
with $\omega_i=k_ic$ and $A_i$ complex, to account for phase differences.
Now square it:
$$ P = EE^* = ||A_1||^2 + ||A_2||^2 + (A_1A_2^*e^{i(k_1x-\omega_1t)}e^{-i(k_2x+\omega_2t)}+ A_2A_1^*e^{i(k_2x-\omega_2t)}e^{-i(k_1x+\omega_1t)}) $$
With $I=A_1A_2^*$, the interference term is
$$ \Delta(x, t) = Ie^{i(k_1-k_2)x}e^{-i(\omega_1-\omega_2)t} + I^*e^{i(k_2-k_1)x}e^{-i(\omega_2-\omega_1)t}$$
With $\Phi(x) = Ie^{i(k_1-k_2)x}$: $$ \Delta(x, t) = \Phi(x)e^{-i(\omega_1-\omega_2)t} + [\Phi(x)e^{-i(\omega_1-\omega_2)t}]^*$$
Edit: It's better to write $\Delta k = k_2-k_1$ and $\Delta \omega = \omega_2-\omega_1$, then:
$$ \Delta(x, t)=2\cos{(\Delta kx-\Delta \omega t)}$$
so averaged over time: $$ \bar\Delta(x)=2\langle\cos{(\Delta kx-\Delta \omega t)}\rangle_t $$ $$ \bar\Delta(x)= 2\big(\cos{\Delta kx}\langle \cos{\Delta \omega t}\rangle_t + \sin{\Delta kx}\langle \sin{\Delta \omega t}\rangle_t\big )= 0 $$
Regarding dispersion, that does move energy around. For instance, a chirped pulse (a linear frequency ramp for a finite length pulse $\tau$) has energy spread over a time $\tau$. In a matched linear dispersive medium, all that energy can be concentrated into a time on the order one over the pulse bandwidth.
This is why sonar and radar send out long chirps and convert them into short pings. In the optical domain, see Mourou and Strickland's Nobel Prize.
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1$\begingroup$ thank you for your response, and it is very helpful. $\endgroup$– WeinCommented Nov 25, 2023 at 19:59
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$\begingroup$ @Ghoster ty. I had all kinds of typos, that why I didn't do the time average...I ran out of gas. I may add it later. $\endgroup$– JEBCommented Nov 30, 2023 at 17:20