Wave Mechanics help-for optics course

[nissanztt90]

Im using the integral table laws of sins/cosines and for the integral of cos^2 I am using the integral of cos^2 = 1/2(x + sinxcosx)...after simplifying everything to what i believe is correct i end up with...

$$\frac{1}{2} (2\pi + sin(2\pi - \frac{\omega r}{c})cos(2\pi - \frac{\omega r}{c}) - sin(\frac{\omega r}{c})cos(\frac{\omega r}{c}))$$

 

[gabbagabbahey]

Huh?! I've never seen that trig ID before (because it is incorrect). I've seen $cos^2(x)=\frac{1}{2}(1-cos(2x)$ before though. There is an easier way though; the integral of cos^2 over a full period represents the area under cos^2 from 0 to T. Due to symmetry, the area under sin^2 will be the same so

$$\int_0^T cos^2(\omega(t-r/c))dt=\frac{1}{2} \left(\int_0^T cos^2(\omega(t-r/c))dt +\int_0^T sin^2(\omega(t-r/c))dt \right)$$

$$=\frac{1}{2} \int_0^T (cos^2(\omega(t-r/c)) + sin^2(\omega(t-r/c)))dt=\frac{1}{2} \int_0^T (1)dt=\frac{T}{2}$$

$$\Rightarrow \left< cos^2(\omega(t-r/c)) \right> _T=\frac{1}{2}$$

I think you are probably supposed to take a spatial average of P_rad as well, in that case, just integrate the r,theta and phi dependent terms over the volume of a sphere and divide by the volume.

 

[nissanztt90]

Thats because its not a trig identity, it was the integral, although i did it incorrectly anyway.

So the power simplified is..

$$\frac{q^2 \omega^4 z^2_0}{12\pi\epsilon_0 c^3} * sin^2(\frac{2\pi n}{L} rsin\theta cos\phi + a)$$

There is no mention of a spatial average in the hw, nor in my notes.

 

[gabbagabbahey]

Well, the question asks for the average energy per unit time. To me, that means spatial and temporal average. It makes no sense that the average power would depend on your position in the radiation field.

 

[nissanztt90]

I will ask the professor tomorrow what exactly he is looking for. The only thing i have in my notes for average power (that is the same as average energy per unit time right?) is:

power = $$\frac{q^2}{6\pi\epsilon c^3} *a(t-\frac{r}{c})$$

Looking over my notes I think we did do the spatial average during the derivation of the above formula...part the derivation consists of what you described on the previous page.

Then he gave us the average power formula

power = $$\frac{q^2}{6\pi\epsilon c^3} *<a(t-\frac{r}{c})>_T$$