Spectral density of radiative electric field

[Decimal]

Homework Statement

Given the following radiative field, $$\vec{E}_{rad}(\vec{r},t) = K \frac{\sin{\theta}}{r} e^{-\frac{k}{m} t_r},$$ with $t_r = t-\frac{r}{c}$ the retarded time far away from the radiating particle, and $K,k,m$ constants. The exercise is to show that the spectral angular density can be written as follows, $$\frac{\partial^2W}{\partial\Omega\partial\omega} = L \sin^2\theta f(\omega), \qquad f(\omega) = \frac{1}{\omega^2 + (\frac{k}{m})^2}.$$ I have to then give the correct expression for the constant $L$.

Relevant Equations

$$\frac{\partial^2W}{\partial\Omega\partial\omega} = 2|\vec{A}(\omega)|^2, \qquad \vec{A}(t) = r\sqrt{\epsilon_0c} \vec{E_{rad}}.$$ Here $\vec{A}(\omega)$ is the fourier transform of $\vec{A}(t)$. I use the notation also adopted in Jackson section 14.5.

So I have to find an expression for $\vec{A}(\omega)$, $$\vec{A}(\omega) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} \vec{A}(t)e^{i\omega t} dt.$$ This point is where my confusion comes up. In the answer sheet they integrate over the retarded time $t_r$, so the integral is, $$\vec{A}(\omega) \sim \int_{-\infty}^{\infty} e^{i\omega t_r} e^{-\frac{k}{m}t_r} dt_r.$$ However I feel like I could have just as well integrated over the observers time $t$ as follows, $$\vec{A}(\omega) \sim \int_{-\infty}^{\infty} e^{i\omega t} e^{-\frac{k}{m}(t-\frac{r}{c})} dt.$$ To my eye these two expressions are not equivalent. I think the distinction is that the second expression gives the spectral density as observed by the observer at distance $r$, and the first expression gives the spectral density as actually emitted by the particle. In the exercise however they do not differentiate between the two and just ask for the spectral density.

So here is my confusion. If the first expression is generally correct and the second one is not, why? Or is my idea correct and should the exercise have been more specific? Any help would be greatly appreciated!

 

[TSny]

In the answer sheet they integrate over the retarded time $t_r$, so the integral is, $$\vec{A}(\omega) \sim \int_{-\infty}^{\infty} e^{i\omega t_r} e^{-\frac{k}{m}t_r} dt_r.$$ However I feel like I could have just as well integrated over the observers time $t$ as follows, $$\vec{A}(\omega) \sim \int_{-\infty}^{\infty} e^{i\omega t} e^{-\frac{k}{m}(t-\frac{r}{c})} dt.$$

Suppose you make the substitution $t_r = t-\frac{r}{c}$ in the second integral above. How does this compare with the first integral? Note that the spectral density depends only on $|A(\omega)|$.

 

[Decimal]

Suppose you make the substitution $t_r = t-\frac{r}{c}$ in the second integral above. How does this compare with the first integral? Note that the spectral density depends only on $|A(\omega)|$.

Ah yes I see what you mean. Making this substitution just introduces a constant phase factor $e^{iw\frac{r}{c}}$, which disappears when calculating the spectral density. I guess where I messed up was not realizing that the derivative of $t_r$ in this case is trivial since $r$ is time independent. Thank you!