# Squared vector field terms $\mathbf{E}^2$ and $\mathbf{H}^2$?

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.

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).
• Thanks for the answer. With regards to the time averaging, is that just $$\dfrac{1}{T} \int_0^T \mathbf{E}^2 \ dt$$? – The Pointer Jul 28 '20 at 15:57
• 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 $T$. – Jacopo Tissino Jul 28 '20 at 16:49