# Converting Watts, Distributed Over An Area, To Energy Density

1. Mar 29, 2012

### jaketodd

Is there a way to get energy density from watts, with the resultant energy density not involving time, as watts do?

Thanks,

Jake

2. Mar 29, 2012

### mrspeedybob

You have not supplied enough information.

First of all energy density is not well defined in your question. Depending on the context it could mean energy per unit volume or energy per unit mass. You stated that you are starting with power per unit area so (J/t)/m2 or J/tm2. Assuming that by energy density you mean energy per unit volume the units are J/m3 so in order to equate the 2 you need to be able to equate time to meters. In other words, you need the velocity of the energy.

3. Mar 30, 2012

### QuantumPion

Watt is a unit of power, and is equal to a joule per second. If you have a power and want to know an energy density, you need to multiply the power by some interval of time and divide by the area or volume to which the power is distributed.

4. Mar 30, 2012

### jaketodd

Thanks guys...

The watts travel at the speed of light. Also, the volume is m$^{3}$. Does this help get an answer?

Thanks,

Jake

5. Mar 30, 2012

### jaketodd

I'm looking for a conversion of watts, given that they travel at the speed of light, over a predefined area, which I could, perhaps, multiply by. In other words, I'm trying to get the energy density.

Thanks,

Jake

6. Mar 31, 2012

### K^2

So you have either RF radiation or some other massless boson field flux through a given area. Easy.

$$\rho_E = \frac{P}{A c}$$

Edit: Note that it gives you J/m³, which is what you want, if I'm not mistaken.

7. Mar 31, 2012

### jaketodd

Excellent. However, I think I shot myself in the foot by saying "area." I meant volume. I assume the 'A' in your equation is for two-dimensional area, and not volume? And, yes, $\stackrel{J}{m^{3}}$ is what I want.

EDIT: Actually I think what you gave me might work for my purposes, but it would be nice to have answered this latest query.

Thanks!

Jake

Last edited: Mar 31, 2012
8. Mar 31, 2012

### K^2

Power flux through volume doesn't make sense. Power flux through a give surface section does. That's what you need to know to find the average energy density of whatever's flowing past.

Think about it in terms of analogous situation with electrical current. The quantities correspond as follows.

$$P \rightarrow I$$
$$A \rightarrow A$$
$$\rho_E \rightarrow \rho_q$$
$$c \rightarrow v$$

Where v is drift velocity, and I is total current flowing through cross section. Then you know that current is given by this expression.

$$I = \rho_q A v$$

Rearrange the terms, and you get the same thing.

9. Mar 31, 2012

### jaketodd

Like I said in the edit, I think what you have provided will suffice for me, and I thank you greatly!

Jake