# Formulation for the fluence from energy-density

I'm struggling to find a mathematical expression for the total fluence of a traveling electromagnetic wave (through vacuum).

Though its not really a physics problem rather than a problem with basic math, I guess...

I'm pretty sure that I'm just being extraordinarily stupid...

As an example, consider the following sketch of a unit volume with a unit surface area $$A$$. The wave travels to the right.

Now suppose, that we know the energy density $$u(t)$$ at every instant in time. Then the energy flux $$\Phi(t)$$ being the energy $$E(t)$$ crossing the area $$A$$ per unit time $$\Delta t$$ is simply $$\Phi(t)=\frac{E(t)}{A\Delta t}=u(t) c_0$$ where $$c_0$$ is the vacuum speed of light.

The total energy, which has crossed the same area within some time-interval $$t_1-t_0$$ is then the time-integral of the above expression

$$F=\int_{t_0}^{t_1} u(t)c_0 \,\mathrm dt ,$$

having units of $$\mathrm{J/m²}$$

Now consider the energy density to be a function of time and $$y$$-coordinate, so that every volume-element $$\,\mathrm dA\cdot c_0\,\mathrm dt$$ (2nd sketch) contains a different amount of energy

The total energy within each "slice" is $$E(t,y)=u(t,y)\cdot \,\mathrm dA\cdot c_0 \,\mathrm dt$$

This is as far as I can get... How can I find an equivalent expression for the total fluence $$F$$ as above? (Not just $$F(y)$$ but the total $$F$$, i.e. integrated along $$y$$ as well)

As an example, consider the following sketch of a unit volume with a unit surface area $$A$$.

If $$A$$ is a unit area, the volume isn't equal to $$1$$, but to $$c\,\mathrm d t$$.

Then the energy flux $$\Phi(t)$$ being the energy $$E(t)$$ crossing the area $$A$$

Then the energy flux $$\Phi(t)$$ being the energy $$E(t)$$ crossing the unit area

How can I find an equivalent expression for the total fluence $$F$$ as above?

The total energy per unit area is still

$$F(y)=\int\limits_{t_0}^{t_1} u(t,y)c_0 \,\mathrm dt$$

Let $$z_0$$ be the thickness of the cuboid and write $$\mathrm dA$$ as

$$\mathrm dA = z_0 \, \mathrm dy$$

The total energy per area $$\mathrm dA$$ is therefore

$$F_{\mathrm dA} = F(y)\,\mathrm dA = \left(\int\limits_{t_0}^{t_1} u(t,y)c_0 \,\mathrm dt\right) (z_0 \, \mathrm dy)$$

Now you can get the net energy passing through the entire unit right side by integrating from $$0$$ to $$y_0$$ where $$y_0$$ the height of the cuboid.

$$F = \int\limits_{0}^{y_0} F(y)\,\mathrm dA = \int\limits_{0}^{y_0} \left(\int\limits_{t_0}^{t_1} u(t,y)c_0 \,\mathrm dt\right)\,z_0\,\mathrm dy$$

Since the right side is a unit area,

$$z_0y_0 = 1\Rightarrow y_0 = \frac 1 {z_0}$$

Therefore

$$F = \int\limits_{0}^{\frac 1 {z_0}} \left(\int\limits_{t_0}^{t_1} u(t,y)c_0 \,\mathrm dt\right)\,z_0\,\mathrm dy$$

$$F$$ is the net energy passing through the right side (of the unit area) of the cuboid.