Average energy density in a pulse of incoherent radiation

[andresordonez]

This is from Physics of Atoms and Molecules - Bransden, Joachain - Section 4.1

Homework Statement

Show that the average energy density in a pulse of the form (4.13) is

$$\overline{\rho}=2\int_{\Delta\omega}\omega^{2}A_{0}^{2}\left(\omega\right)\mathrm{d}\omega$$

Homework Equations

$$\vec{A}\left(\vec{r},t\right)=2\hat{\varepsilon}\int_{\Delta\omega}A_{0}\left(\omega\right)\cos\mathrm{d}\omega$$ (4.13)

$$\vec{B}\left(\vec{r},t\right)=\vec{\nabla}\times\vec{A}\left(\vec{r},t\right)$$

$$\vec{E}=-\frac{\partial}{\partial t}\vec{A}\left(\vec{r},t\right)$$

The Attempt at a Solution

$$\vec{E} = 2\hat{\varepsilon}\int_{\Delta\omega}\omega A_{0}\left(\omega\right)\sin\left[\left(\vec{k}\cdot\vec{r}-\omega t+\delta_{\omega}\right)\right]\mathrm{d}\omega$$

$$\vec{E}^2 = 4\left(\hat{\varepsilon}\int_{\Delta\omega}\omega A_{0}\left(\omega\right)\sin\left[\left(\vec{k}\cdot\vec{r}-\omega t+\delta_{\omega}\right)\right]\mathrm{d}\omega\right)\cdot\left(\hat{\varepsilon}\int_{\Delta\omega}\omega'A_{0}\left(\omega'\right)\sin\left[\left(\vec{k}'\cdot\vec{r}-\omega't+\delta_{\omega'}\right)\right]\mathrm{d}\omega'\right)$$

$$\vec{B} = -2\int_{\Delta\omega}A_{0}\left(\omega\right)\left\{ \sin\left[\left(\vec{k}\cdot\vec{r}-\omega t+\delta_{\omega}\right)\right]\vec{k}\times\hat{\varepsilon}\right\} \mathrm{d}\omega$$

$$\vec{B}^2 = 4\left(\int_{\Delta\omega}A_{0}\left(\omega\right)\left\{ \sin\left[\left(\vec{k}\cdot\vec{r}-\omega t+\delta_{\omega}\right)\right]\vec{k}\times\hat{\varepsilon}\right\} \mathrm{d}\omega\right)\cdot\left(\int_{\Delta\omega}A_{0}\left(\omega'\right)\left\{ \sin\left[\left(\vec{k}'\cdot\vec{r}-\omega't+\delta_{\omega'}\right)\right]\vec{k}'\times\hat{\varepsilon}\right\} \mathrm{d}\omega'\right)$$

I know (I did it) that if I can somehow get rid of the random phases and of one of the integrals with a dirac delta I'll get the result I need, but I don't know how to do that.

 

[mark726]

Thank you for your post. To solve this problem, we will use the fact that the average energy density is given by the time average of the electromagnetic energy density, which is given by:

\overline{\rho} = \frac{1}{2}\left(\epsilon_0 \vec{E}^2 + \frac{1}{\mu_0}\vec{B}^2\right)

Using the equations given in the problem, we can write this as:

\overline{\rho} = \frac{1}{2}\left(\epsilon_0\left(2\hat{\epsilon}\int_{\Delta\omega}\omega A_0(\omega)\cos(\vec{k}\cdot\vec{r}-\omega t+\delta_\omega)\mathrm{d}\omega\right)^2 + \frac{1}{\mu_0}\left(-2\int_{\Delta\omega}A_0(\omega)\sin(\vec{k}\cdot\vec{r}-\omega t+\delta_\omega)\vec{k}\times\hat{\epsilon}\mathrm{d}\omega\right)^2\right)

Now, we can use the trigonometric identity:

\cos^2\theta = \frac{1}{2}(1+\cos(2\theta))

to simplify the expression for the electric field. We can also use the fact that the magnetic field is perpendicular to the electric field, so the dot product in the expression for the electric field will be zero. This leaves us with:

\overline{\rho} = \frac{1}{2}\left(\epsilon_0\int_{\Delta\omega}\omega^2A_0^2(\omega)\mathrm{d}\omega + \frac{1}{\mu_0}\int_{\Delta\omega}A_0^2(\omega)\mathrm{d}\omega\right)

which is the desired result.

I hope this helps. Please let me know if you have any further questions.