- #1
schrodingerscat11
- 89
- 1
Homework Statement
Given the energy density expression from Jackson
[itex]\frac{1}{2}\big(\mathbf{E}\cdot\mathbf{D}+\mathbf{H}\cdot\mathbf{B}\big)[/itex] (Eq. 6.106)
Show the missing steps to arrive at the time-averaged energy density
[itex]\frac{1}{4}\big(\epsilon\mathbf{E}\cdot\mathbf{E}^*+\frac{1}{\mu}\mathbf{B}\cdot\mathbf{B}^*\big)[/itex] (Eq. 7.13)
Homework Equations
See problem above.
[itex]\langle u \rangle_t=\frac{1}{T}\int_0^Tu(t) dt[/itex] (time-averaged quantity)
The Attempt at a Solution
[itex]\langle u \rangle_t=\frac{1}{T}\int_0^Tu(t) dt[/itex]
[itex]\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[/itex]
[itex]\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[/itex]
[itex]\mathbf{E}=\mathcal{E}e^{ik\mathbf{\hat{n}}\cdot\mathbf{x}-i\omega t}[/itex]
[itex]\mathbf{B}=\mathcal{B}e^{ik\mathbf{\hat{n}}\cdot\mathbf{x}-i\omega t}[/itex]
[itex]\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[/itex]
[itex]\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[/itex]
[itex]T=\frac{2\pi}{\omega}[/itex]
[itex]\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[/itex]
[itex]\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[/itex]
Let [itex]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)[/itex]
[itex]\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[/itex]
Question: Is the integral [itex] \int_0^{\frac{2\pi}{\omega}}e^{2i\omega t} dt [/itex] equal to zero?
I get 1 -1 = 0, but then I cannot prove what I'm proving with this number. :(
Thanks.