Is the irradiance (or intensity) of an electromagnetic wave given by \begin{equation} I = \left<\|\mathbf{S}\|\right>_{T} \end{equation} or by \begin{equation} I = \|\left<\mathbf{S}\right>_{T}\| \; , \end{equation} with $\mathbf{S} = \mathbf{E} \times \mathbf{H}$ being the Poynting vector and $\left<\right>_{T}$ denoting the time average?

In other words, is it the time average of the Poynting vector's norm or the norm of the time-averaged Poynting vector?

First update:
Also textbooks seem to have a diverging view on this question. "Optics" from E. Hecht teaches the first variant, while "Fundamentals of Photonics" from B.E.A. Saleh and M.C. Teich presents the second one referring to "Principles of Optics" from M. Born and E. Wolf.

In addition, Born and Wolf emphasize that $\mathbf{S} \cdot \mathbf{n}$ with $\mathbf{n}$ being the normal of the considered surface (irradiance is power per area) is the relevant quantity. So maybe the best description is \begin{equation} I = \left<\mathbf{S} \cdot \mathbf{n}\right>_{T} \; ? \end{equation}


2 Answers 2


As you point out, it is important to consider not the norm but the component normal to the surface under consideration. This allows you to consider actual flows of energy and eliminates the muddles you'd get, for example, from superposed waves going in different directions.

That said, the time averages $\langle \mathbf{S}\cdot\mathbf{n}\rangle_T$ and $\langle \mathbf{S}\rangle_T\cdot\mathbf{n}$ are equal. This is because the average is a linear function, and so is the dot product.

The situations where you don't get such equalities is when you could have cancellations driving some averages down but not others: thus $\langle \cos^2(\omega t)\rangle_t\neq\langle \cos(\omega t)\rangle_t^2$, as the latter is zero because of cancellations that are not present in the former. For a linear function, though, which must preserve sign changes, this can't happen.


I think the correct expression is more accurately $\textrm{I} = \langle \| \vec{S}\| \rangle$, if you average before taking the norm, you could get zero intensities which are still radiating energy.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.