Consider the simple case of electromagnetic irradiation of a homogeneous isotropic dielectric, neglecting the dispersion of the refractive index. Assuming a transparent medium, the spatial density of forces acting on the dielectric in a static external electromagnetic field can be given as

$$\mathbf{f} = - \nabla p - \nabla \epsilon \dfrac{\langle \mathbf{E}^2 \rangle}{8 \pi} - \nabla \mu \dfrac{\langle \mathbf{H}^2 \rangle}{8 \pi} + \nabla \left[ \left( \rho \dfrac{\partial{\epsilon}}{\partial{p}} \right)_T \dfrac{\langle \mathbf{E}^2 \rangle}{8 \pi} + \left( \rho \dfrac{\partial{\mu}}{\partial{\rho}} \right)_T \dfrac{\langle \mathbf{H}^2 \rangle}{8 \pi} \right] + \dfrac{\epsilon \mu - 1}{4 \pi c} \dfrac{\partial}{\partial{t}}\langle [ \mathbf{E} \times \mathbf{H}] \rangle.$$

$p$ is the pressure in the medium (for a given density $\rho$ and temperature $T$ in zero field.
$\epsilon$ and $\mu$ are the permittivity and magnetic permeability.
$c$ is the speed of light.
The angular brackets denote averaging over a time period far greater than the characteristic alternation period of light.

And my understanding is that $\mathbf{E} \times \mathbf{H}$ is the Poynting vector.

What I don't understand is the squared field terms $\mathbf{E}^2$ and $\mathbf{H}^2$. These field terms are vector fields, and so my understanding is that it is not mathematically valid to take a vector field (or any other vector) to an exponent. So what is meant by $\mathbf{E}^2$ and $\mathbf{H}^2$ in this context?

I would greatly appreciate it if people would please take the time to explain this.


1 Answer 1


The notation of a vector field to an exponent has meaning if we give it meaning; what is very likely meant in this case is $\mathbf{E}^2 = \mathbf{E} \cdot \mathbf{E} = \left|\mathbf{E}\right|^2$, the square of the norm of the vector (so, a scalar --- perhaps the fact that it is still written in bold is not the best choice for clarity).

These terms, then, represent the average square magnitude of the electric and magnetic fields. Notice also that the two sides of the equation then make sense, since we are equating a vector with a vector.

  • $\begingroup$ Thanks for the answer. With regards to the time averaging, is that just $$\dfrac{1}{T} \int_0^T \mathbf{E}^2 \ dt $$? $\endgroup$ Jul 28, 2020 at 15:57
  • 1
    $\begingroup$ You're welcome! Yes, something like that; the precise definition might be slightly different depending on the context, for example, if you want the equation to hold at different times you might want to use a running mean, integrating from $t$ to $t+T$ instead of from 0 to $T$. $\endgroup$ Jul 28, 2020 at 16:49
  • $\begingroup$ Ok. Thanks again! $\endgroup$ Jul 28, 2020 at 16:51

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.