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In reference to the question, "Why is energy in a wave proportional to amplitude?" Ben Crowell answered with the answer attached. The original post is linked here. I did not respond to the original post because I am not able to, and I cannot contact Ben Crowell directly.

In his explanation he says that since the energy of the wave doesn't depend on phase, it should be expected that only even terms should occur in the expansion. Assuming we're modeling as a harmonic oscillator (which is why I think there's a phase independence), why would only the even terms occur in this expansion because of phase independence?

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    $\begingroup$ I was not allowed to because the number of points I have aren't enough, per the error message from the site. Also, you can't directly message anyone on here. $\endgroup$
    – Shocked
    Jul 6, 2018 at 9:04
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    $\begingroup$ @sammygerbil in reasonably certain in pointed it out to you previously, but commenting globally requires 50 rep, which OP clearly does not have. Also, asking questions in this manner is perfectly acceptable. $\endgroup$
    – Kyle Kanos
    Jul 6, 2018 at 11:05
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    $\begingroup$ @sammygerbil He is no more available in the site :) $\endgroup$ Feb 13, 2022 at 5:58

1 Answer 1


I think the explanation has to do with the Taylor expansion of a complex function. If $E\propto|A|$ and $E$ admits a Taylor expansion, then $E=a_0+a_1A+a_2A^2+...$ where $A$ is in general a complex number. The (magnitude) square(d) of $A$ then is a real number and hence every other even power of $A$. However, if for example $A$ is complex, one can write it as $A=|A|e^{i\phi}$, where $\phi$ is some phase. Since the energy should be independent of the phase, such terms should be absent from the expansion. The same goes for terms dependent on odd powers of $A$. This is my guess. I hope it helps

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    $\begingroup$ I can just see repetition of the texts(which Ben said). I can't see any clear explanation here. $\endgroup$ Feb 13, 2022 at 14:22

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