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

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

$p$ 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.

It is said that, by expressing $\langle E^2 \rangle$ through $I$ (the light intensity) and introducing the refractive index $n = \sqrt{\epsilon}$, we can transform the striction force equation to

$$\mathbf{f}_{\text{str}} = \nabla \left[ \left( \rho \dfrac{\partial{\epsilon}}{\partial{\rho}} \right)_T \dfrac{\langle \mathbf{E}^2 \rangle}{8 \pi} \right] = \nabla \left[ \left( \rho \dfrac{\partial{n}}{\partial{\rho}} \right)_T \dfrac{I}{c} \right].$$

I'm trying to understand how exactly we get $\nabla \left[ \left( \rho \dfrac{\partial{\epsilon}}{\partial{\rho}} \right)_T \dfrac{\langle \mathbf{E}^2 \rangle}{8 \pi} \right] = \nabla \left[ \left( \rho \dfrac{\partial{n}}{\partial{\rho}} \right)_T \dfrac{I}{c} \right]$. I've been doing a lot of research to try and understand this, but I'm stuck.

My best attempt is as follows. As said here, in optics, the time-averaged value of the radiated flux is technically known as the irradiance, more often simply referred to as the intensity. The Wikipedia article for intensity says that, if $I$ is the local intensity (I'm not completely sure if this is the correct assumption for our case), then we have that $I = \dfrac{cn \epsilon_0}{2}|E|^2$, where $\epsilon_0$ is the vacuum permittivity. And so, if we assume that $\langle \mathbf{E}^2 \rangle = |E|^2$ (which seems to be true, given the answer here), then we get that $|E|^2 = \dfrac{2I}{cn \epsilon_0}$, and so $\nabla \left[ \left( \rho \dfrac{\partial{\epsilon}}{\partial{\rho}} \right)_T \dfrac{\langle \mathbf{E}^2 \rangle}{8 \pi} \right] = \nabla \left[ \left( \rho \dfrac{\partial{n^2}}{\partial{\rho}} \right)_T \dfrac{I}{4 \pi c n \epsilon_0} \right]$. But it is not clear how one proceeds from here.

Some other potentially relevant facts that I found during my research are as follows:

  • According to the article on irradiance (different from the article on intensity), $E_{{\mathrm {e}}}={\frac {n}{2\mu _{0}{\mathrm {c}}}}E_{{\mathrm {m}}}^{2}\cos \alpha ={\frac {n\varepsilon _{0}{\mathrm {c}}}{2}}E_{{\mathrm {m}}}^{2}\cos \alpha$. If we let that $\cos(\alpha) = 1$ for our case, then this might be relevant.
  • The article on vacuum permittivity states that $\varepsilon _{0}={\frac {1}{\mu _{0}c^{2}}}$, where $\mu_0$ is the vacuum permeability.
  • This page on "energy density, flux and power" has numerous relevant-looking facts that include $E$ and time-averaged values, and look like they could potentially cancel the necessary factors, such as $4\pi$ or $8\pi$, somehow.

I would greatly appreciate it if people would please take the time to explain exactly how we get from $\nabla \left[ \left( \rho \dfrac{\partial{\epsilon}}{\partial{\rho}} \right)_T \dfrac{\langle \mathbf{E}^2 \rangle}{8 \pi} \right]$ to $\nabla \left[ \left( \rho \dfrac{\partial{n}}{\partial{\rho}} \right)_T \dfrac{I}{c} \right]$.

  • 1
    $\begingroup$ The formula you are trying to prove is written in gaussian units (en.wikipedia.org/wiki/Gaussian_units), and equations that contain $\epsilon_0$ and $\mu_0$ are written in SI units. $\endgroup$
    – atarasenko
    Commented Aug 4, 2020 at 0:37
  • $\begingroup$ @atarasenko Oh wow, I think you're right. Thank you for that. Looking through that Wikipedia article, I think I was on the right track with $\nabla \left[ \left( \rho \dfrac{\partial{\epsilon}}{\partial{\rho}} \right)_T \dfrac{\langle \mathbf{E}^2 \rangle}{8 \pi} \right] = \nabla \left[ \left( \rho \dfrac{\partial{n^2}}{\partial{\rho}} \right)_T \dfrac{I}{4 \pi c n \epsilon_0} \right]$! But which formula in particular should I be using? The closest that I can see might be $\mu _{0}\varepsilon _{0} = 1/c^{2}$, but this still doesn't get us what we need. Do you know enough to post an answer? $\endgroup$ Commented Aug 4, 2020 at 0:57

1 Answer 1


Using the conversion formula from SI to gaussian units $E^{G}=\sqrt{4\pi\epsilon_0}E^{SI}$ (see Table 1 in https://en.wikipedia.org/wiki/Gaussian_units), the formula for intensity is transformed to: $$ I=\frac{cn\epsilon_0|E^{SI}|^2}{2}\rightarrow I=\frac{cn|E^{G}|^2}{8\pi} $$ For a monochromatic linearly polarized wave with amplitude $E_0$, $\left<\mathbf{E}^2\right>=E_0^2/2$, and $$ I=\frac{cn\left<\mathbf{E}^2\right>}{4\pi} $$ $$ \rho\left(\frac{\partial\epsilon}{\partial\rho}\right)_T\frac{\left<\mathbf{E}^2\right>}{8\pi}=\rho\cdot 2n\left(\frac{\partial n}{\partial\rho}\right)_T\cdot\frac{4\pi I}{8\pi cn}=\rho\left(\frac{\partial n}{\partial\rho}\right)_T\frac{I}{c} $$

  • $\begingroup$ Thanks a lot for taking the time to post this! $\endgroup$ Commented Aug 4, 2020 at 3:41

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