# Time-average energy density

1. Nov 21, 2014

### physicsjn

1. The problem statement, all variables and given/known data
Given the energy density expression from Jackson
$\frac{1}{2}\big(\mathbf{E}\cdot\mathbf{D}+\mathbf{H}\cdot\mathbf{B}\big)$ (Eq. 6.106)

Show the missing steps to arrive at the time-averaged energy density
$\frac{1}{4}\big(\epsilon\mathbf{E}\cdot\mathbf{E}^*+\frac{1}{\mu}\mathbf{B}\cdot\mathbf{B}^*\big)$ (Eq. 7.13)

2. Relevant equations
See problem above.
$\langle u \rangle_t=\frac{1}{T}\int_0^Tu(t) dt$ (time-averaged quantity)

3. The attempt at a solution
$\langle u \rangle_t=\frac{1}{T}\int_0^Tu(t) dt$
$\langle u \rangle_t=\frac{1}{T}\int_0^T \frac{1}{2}\big(\mathbf{E}\cdot\mathbf{D}+\mathbf{H}\cdot\mathbf{B}\big) dt$
$\langle u \rangle_t=\frac{1}{T}\int_0^T \frac{1}{2}\big(\mathbf{E}\cdot\epsilon\mathbf{E}+\frac{1}{\mu}\mathbf{B}\cdot\mathbf{B}\big) dt$

$\mathbf{E}=\mathcal{E}e^{ik\mathbf{\hat{n}}\cdot\mathbf{x}-i\omega t}$
$\mathbf{B}=\mathcal{B}e^{ik\mathbf{\hat{n}}\cdot\mathbf{x}-i\omega t}$

$\langle u \rangle_t=\frac{1}{T}\int_0^T \frac{1}{2}\big(\mathcal{E}e^{ik\mathbf{\hat{n}}\cdot\mathbf{x}-i\omega t}\cdot\epsilon\mathcal{E}e^{ik\mathbf{\hat{n}}\cdot\mathbf{x}-i\omega t}+\frac{1}{\mu}\mathcal{B}e^{ik\mathbf{\hat{n}}\cdot\mathbf{x}-i\omega t}\cdot\mathcal{B}e^{ik\mathbf{\hat{n}}\cdot\mathbf{x}-i\omega t}\big) dt$

$\langle u \rangle_t=\frac{\omega}{2\pi}\int_0^{\frac{2\pi}{\omega}} \frac{1}{2}\big(\mathcal{E}e^{ik\mathbf{\hat{n}}\cdot\mathbf{x}-i\omega t}\cdot\epsilon\mathcal{E}e^{ik\mathbf{\hat{n}}\cdot\mathbf{x}-i\omega t}+\frac{1}{\mu}\mathcal{B}e^{ik\mathbf{\hat{n}}\cdot\mathbf{x}-i\omega t}\cdot\mathcal{B}e^{ik\mathbf{\hat{n}}\cdot\mathbf{x}-i\omega t}\big) dt$

$T=\frac{2\pi}{\omega}$

$\langle u \rangle_t=\frac{\omega}{2\pi}\int_0^{\frac{2\pi}{\omega}} \frac{1}{2}\big(\mathcal{E}e^{ik\mathbf{\hat{n}}\cdot\mathbf{x}-i\omega t}\cdot\epsilon\mathcal{E}e^{ik\mathbf{\hat{n}}\cdot\mathbf{x}-i\omega t}+\frac{1}{\mu}\mathcal{B}e^{ik\mathbf{\hat{n}}\cdot\mathbf{x}-i\omega t}\cdot\mathcal{B}e^{ik\mathbf{\hat{n}}\cdot\mathbf{x}-i\omega t}\big) dt$

$\langle u \rangle_t=\frac{\omega}{2\pi}\int_0^{\frac{2\pi}{\omega}} \frac{1}{2}\big(e^{-2i\omega t}\big)\big(\mathcal{E}e^{ik\mathbf{\hat{n}}\cdot\mathbf{x}}\cdot\epsilon\mathcal{E}e^{ik\mathbf{\hat{n}}\cdot\mathbf{x}}+\frac{1}{\mu}\mathcal{B}e^{ik\mathbf{\hat{n}}\cdot\mathbf{x}}\cdot\mathcal{B}e^{ik\mathbf{\hat{n}}\cdot\mathbf{x}}\big) dt$

Let $A=\big(\mathcal{E}e^{ik\mathbf{\hat{n}}\cdot\mathbf{x}}\cdot\epsilon\mathcal{E}e^{ik\mathbf{\hat{n}}\cdot\mathbf{x}}+\frac{1}{\mu}\mathcal{B}e^{ik\mathbf{\hat{n}}\cdot\mathbf{x}}\cdot\mathcal{B}e^{ik\mathbf{\hat{n}}\cdot\mathbf{x}}\big)$

$\langle u \rangle_t=\frac{\omega}{2\pi}\int_0^{\frac{2\pi}{\omega}} \frac{1}{2}\big(e^{-2i\omega t}\big)A~dt$

Question: Is the integral $\int_0^{\frac{2\pi}{\omega}}e^{2i\omega t} dt$ equal to zero?
I get 1 -1 = 0, but then I cannot prove what I'm proving with this number. :(

Thanks.

2. Nov 21, 2014

### Staff: Mentor

Yes it is (apart from a prefactor, it is equivalent to $\int_0^{2\pi} e^{i t} dt$)

Your expression for E should be real as there are no complex electric fields. If you take the calculate E everywhere, this problem will disappear.

3. Nov 22, 2014

### physicsjn

Actually, since it was due today, I kinda followed a solution in Wikipedia for Poynting vector (attached is the screenshot of Wikipedia solution. But I replaced the expression with the one used for energy density. In my new solution, the integral above was indeed zero, but there were new trems created by all the crossmultiplying done.I am not sure though if I did it right. But I did managed to arrive at the final expression. But thanks anyways. :)