# Electrodynamical oblivion

1. Nov 26, 2005

### armandowww

I was searching for some proof on macroscopical theory about lattice behaviour under lightning. I found a passage in which the average in time of optical power is given by average of electrical field square times constant dielectric ratio four pi! How could I explain this concept?
Regards

2. Nov 26, 2005

### Gokul43201

Staff Emeritus
I'm not sure I understand what you are talking about - it would help if you provided a reference or link to the passage.

In general, the energy density of an electric field is half the number you stated.

$$\frac{d{\cal E}}{dV} = \frac{1}{2} \epsilon E^2~~(SI~Units)$$

$$\frac{d{\cal E}}{dV} = \frac{1}{8\pi} \epsilon_r E^2~~(Gaussian~Units)$$

3. Nov 26, 2005

### armandowww

I've just forgot to say you I'm in gaussian units... but your 8 is my 4

4. Nov 26, 2005

### armandowww

Mayhap... the reason is this: you told about an energy density at a given instant, while the time average of my waves (PLANE waves) has brought new root-square-two terms (2 terms because E is the base of a power 2) forward in my formula...
Is the Poynting vector involved in anyway?
Thank you

5. Nov 26, 2005

### Gokul43201

Staff Emeritus
$$E_{RMS} = \frac{E_0}{\sqrt{2}}$$

That would only make things worse.

$$\frac{d{\cal E}}{dV} = \frac{1}{8\pi} \epsilon_r E_{RMS}^2 = \frac{1}{16\pi} \epsilon_r E_0^2$$