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For a monochromatic plane wave: $$\mathbf E = \mathbf E _0e^{i(\mathbf k \cdot \mathbf r -\omega t)},\qquad \mathbf H = \dfrac{\mathbf B}{\mu _0}= \mathbf H_0e^{i(\mathbf k \cdot \mathbf r-\omega t)},$$ it's straightforward to show, using Maxwell's equations, that the Poynting vector $\mathbf S = \mathbf E \times \mathbf H$ is given (in MKS units) by$$\mathbf S = \varepsilon _0cE_0^2 \cos ^2 (\mathbf k \cdot \mathbf r - \omega t)\mathbf n,$$ where $\mathbf n = \mathbf k /k$ is the direction of propagation of the wave, so that the average energy flux in that direction is $$I= \overline {\mathbf S \cdot \mathbf n}=\lim _{T\to \infty} \intop ^T \mathbf S \cdot \mathbf n \text dt\propto E_0^2.$$

This relation is usually (compare, for example 1) assumed as valid. That is, the irradiance is defined (or assumed to be) proportional to the average of the square of the electric field (1 puts directly $I = \overline {\mathbf E \cdot \mathbf E ^*}$) even when the field is the superposition of two plane harmonic waves.

Now, if the two waves (say $\mathbf E _0i, \mathbf k _i$ for $i=1,2$) share the same direction of propagation, I see no problem in the definition. But when the two directions are different, it's easy to see that the magnitude of the Poynting vector isn't strictly proportional to $(\mathbf E_1+\mathbf E_2)^2$. Moreover the direction of $\mathbf S$ is a quite complicated function of $\mathbf n _1 ,\mathbf n _2, \mathbf E _1, \mathbf E _2$ (in the case of two linearly polarized waves one has: $$\frac{1}{\varepsilon _0 c}\mathbf S = E_1 ^2 \mathbf n _1 + E_2 ^2 \mathbf n _2 + (\mathbf n _1 +\mathbf n _2)\mathbf E _1 \cdot \mathbf E _2 - (\mathbf n _1 \cdot \mathbf E _2 ) \mathbf E _1 - (\mathbf n _2 \cdot \mathbf E _1) \mathbf E _2.$$ As an extreme example, if one has two waves travelling in opposite direction, with $\mathbf k _2 = -\mathbf k _1$ and $\mathbf E _1 = \mathbf E _2,$ the flux of energy is clearly zero, while $I=4E^2$ according to the above definitions.

On the other hand, from the above expression, it is seen that if $\mathbf n _2 = \mathbf n _1 +\delta \mathbf n$, the difference $\mathbf S - \varepsilon _0 c E^2\mathbf n$ is $O(\delta \mathbf n).$

To sum up, what is the meaning of the relation $I \propto E^2$? In particular is it

  1. An approximation of the exact law (see above). Or
  2. Is it a definition? In this case, why is it a useful one?

Thank you in advance.


1 Grant R. Fowles, “Introduction to modern optics”.

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  • $\begingroup$ I am not sure that it is OK to equate intensity with flux. If you absorb all the energy of the EM wave at a given point the energy will be proportional to the time averaged amplitude of the E field. When you start to do vector addition of Poynting vectors you see the net flow of energy but I don't think that is the same thing as intensity. $\endgroup$
    – Floris
    Commented Mar 28, 2015 at 14:31
  • $\begingroup$ @Floris, the question asks about irradiance, which is defined in terms of Poynting flux. What do you mean by intensity? $\endgroup$ Commented Mar 28, 2015 at 14:33
  • $\begingroup$ @JánLalinský Intensity = energy per unit area. It does not have a direction unlike irradiance. I see that OP used the symbol $I$ (which I associate with intensity) for irradiance. As you said in your answer these become the same thing with simple wave fronts; the source of much confusion (including, it seems, mine). $\endgroup$
    – Floris
    Commented Mar 28, 2015 at 14:38
  • $\begingroup$ How do you calculate energy per unit area? That should give zero for any EM field. $\endgroup$ Commented Mar 28, 2015 at 14:46

1 Answer 1

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The author confused things in section 3.1.

In macroscopic EM theory of radiation (radiometry), technically irradiance at a given point of a plane is defined as time average of the normal component of the Poynting vector (normal to the plane):

$$ I = \overline{\mathbf S \cdot \mathbf n} $$

As you have shown, this is function of $|\mathbf E|^2$ only in some cases, like single plane wave; it also seems reasonable for isotropic thermal radiation. For general EM field, however, irradiance cannot be expressed as function of $|\mathbf E|^2$ only.

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  • $\begingroup$ So I'm assuming that the proportionality refers to special cases where the wave directions are nearly the same - indeed this is the case with several interferometric devices. $\endgroup$
    – pppqqq
    Commented Mar 29, 2015 at 9:22

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