# Temporal average

1. May 2, 2010

### fluidistic

1. The problem statement, all variables and given/known data
The temporal average of a function f(t) over an interval T is given by $$\langle f(t) \rangle=\frac{1}{T} \int _{t}^{t+T} f(t') dt'$$.
Let $$\tau =\frac{2 \pi}{\omega}$$. Calulate the temporal average of $$\langle \sin ^2 (\vec k \cdot \vec r - \omega t) \rangle =\frac{1}{2}$$, when $$T=\tau$$ and $$T >> \tau$$.

2. Relevant equations
Already given in the problem description.

3. The attempt at a solution
I don't understand the problem. Aren't they asking me to prove that $$\langle \sin ^2 (\vec k \cdot \vec r - \omega t) \rangle =\frac{1}{2}$$ for $$T=\tau$$ and to calculate $$\langle \sin ^2 (\vec k \cdot \vec r - \omega t) \rangle$$ for $$T>> \tau$$? If so, it's really not clear. If not, could you please explain with a bit more details? I can't start the problem since I don't understand it.

2. May 2, 2010

### jdwood983

Looks to me like the question is asking you to show that

$$\left\langle \sin^2\left[\vec{k}\cdot\vec{r}-\omega t\right]\right\rangle=\frac{1}{2}$$

for both $T=\tau$ and $T\gg\tau$.

3. May 2, 2010

### fluidistic

Ok thanks a lot. I'll give some tries.

4. May 3, 2010

### gabbagabbahey

The first case, $T=\tau$ can be easily done with a useful trick....if you graph sin^2(u) and cos^2(u) over a full period, you should see that the area under them is equal and so

$$\int_{u_0}^{u_0+2\pi}\sin^2(u)du=\int_{u_0}^{u_0+2\pi}\cos^2(u)du=\frac{1}{2}\left(\int_{u_0}^{u_0+2\pi}\sin^2(u)du+\int_{u_0}^{u_0+2\pi}\cos^2(u)du\right)=\frac{1}{2}\int_{u_0}^{u_0+2\pi}du=\pi$$

For the second case, I think you will want to use the double angle formula, and then expand your result in a Taylor series.