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I have trouble understanding the intensity of electromagnetic waves. I already looked at this article and this question but didn't understand completely, where it is given that,

$$\text{Intensity} = \frac{\epsilon_0}{2} |\vec E|_{RMS}^2 V + \frac{1}{2\mu_0} |\vec B|_{RMS}^2 V = \epsilon_0 E_\textrm{peak}^2 V$$

Is this the correct formula?

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  • $\begingroup$ A side note: Be careful and use these formulas only in the case of the linear polarised wave. In the case of the circular polarisation, sines represent only projections of vectors, and the notion of peak value is not applicable since values are always the same. The average of the magnitude of the Poynting vector is the most general definition. $\endgroup$
    – firtree
    Commented May 13, 2013 at 7:19

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Intensity generally refers to a power per area (energy per area per time). For an electromagnetic wave, you can find its intensity by computing the magnitude of the Poynting vector, and in most circumstances taking its time average.

For a plane wave and using SI units, the time-averaged intensity comes out to $\frac{1}{2} c \epsilon_0 E_0^2$ where $E_0$ is the peak electric field. This is almost the same as the formula in the article you linked to in your question, but the time-averaging contributes the factor of 1/2.

The $V$ in the formula from the phys.SE question you've linked refers to a volume, which you might use if wish to compute the entire E&M energy contained within said volume, but is not needed for expressing the intensity.

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I think that this is half the correct result, for a sinusoidal wave the RMS value squared is half the peak value, and the magnetic field is smaller than the electric by a factor $1/c=\sqrt(\mu_0 \epsilon_0)$

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  • $\begingroup$ Magnetic field has the same energy. It is harder to see in SI but it is obvious in CGS where we have $|\vec{E}|=|\vec{B}|$ in the free wave, and they contribute to energy with the same factors. Just check all factors carefully. $\endgroup$
    – firtree
    Commented May 13, 2013 at 5:45
  • $\begingroup$ yes, that's my answer $\endgroup$
    – Ikiperu
    Commented May 13, 2013 at 5:58

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