Poynting vector in dielectric

1. May 13, 2015

Wminus

Hi.

According to classical electromagnetism (and common sense) the intensity of a beam of light entering a dielectric medium should remain constant. Hence the length of the poynting vector must remain constant.

But how do you derive mathematically the last point? Because if you just replace $c$ with $v=c/n$ and $\epsilon_0$ with $\epsilon = \epsilon_0 n^2$ and $E$ with $E/n^2$ you get into trouble when trying to transform the poynting vector.

Let's say you have light entering glass from vacuum with $n = \sqrt{\epsilon/\epsilon_0}$. => Before: $<S_{vac}> = \frac{c^2 \epsilon_0}{2} E_{vac} B_{vac}$. After: $<S_{glass}> = \frac{(c^2/n^2) (\epsilon_0 n^2)}{2} (E/n^2) B = \frac{(c^2) (\epsilon_0}{2} (E_{vac}/n^2) B_{vac} \neq< S_{vac}>$

All thoughts on this are highly appreciated.

EDIT: fixed typo

Last edited: May 13, 2015
2. May 13, 2015

Wminus

Am I being unclear perhaps?

3. May 13, 2015

crador

Does the magnetic field not change as well?

4. May 14, 2015

Wminus

ahh yes, $B = B_{vac} /n$? But things still don't work out!

5. May 14, 2015

DrDu

I can't follow you. S=ExH. In the optical region, $\mu=\mu0$. The change of E differs for the component parallel and perpendicular to the surface. For parallel E (normal incidence) E doesnt change, so S doesnt either.
In general you have to consider also reflection.

6. May 14, 2015

crador

What Dr.Du said (I'll be honest, I didn't follow you at first either). A bound charge on the surface of the dielectric will create a discontinuity on the perpendicular component of the field. The parallel component cannot and does not have a discontinuity, or else you would violate conservation of energy (make a little charged wheel spanning the surface and it will be continually accelerated -- perpetuum mobile). So the "normal intensity" is preserved.

7. May 14, 2015

Wminus

But doesn't the E field change by $E/\epsilon_r= E/n^2$ in a dielectric, where $n$ is the refractive index because $\epsilon_0 \rightarrow \epsilon_0 \epsilon_r$?

In case of no reflection, I thought you could just replace $\epsilon_0$ with $\epsilon_0\epsilon_r = \epsilon_0 n^2$ everywhere in the vacuum poynting vector to get the poynting vector in a material. Was I wrong?

8. May 14, 2015

DrDu

That's true only for the E field normal to the surface.

9. May 14, 2015

Wminus

But you just said in #5 the E field normal to the surface doesn't change?

10. May 14, 2015

crador

No, he said for normal incidence of light, the parallel component doesn't change. The normal component changes. If it were not so we would create an opportunity for perpetuum mobile at the surface of the dielectric -- surely you will agree that that is an unphysical result.

11. May 14, 2015

Wminus

OK I see. Thanks for clearing up the confusion.

12. May 14, 2015

crador

No problem, your intuition steered you in the right direction anyway. Oftentimes that is more valuable than having the correct answer from prior study -- for example when you investigate something that hasn't been studied before.

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook