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I have a question about the following exercise. 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 $L$. Now in principle I know how to do this problem, by using the following relations, $$\frac{\partial^2W}{\partial\Omega\partial\omega} = 2|\vec{A}(\omega)|^2, \qquad \vec{A}(t) = r\sqrt{\epsilon_0c} \vec{E_{rad}}.$$ 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!

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In case anyone else runs into this problem, with some outside help I managed to find a solution. If one makes the substitution $t_r = t-\frac{r}{c}$ in the last integral, the full expression reads, $$\vec{A}(\omega) \sim e^{i\omega\frac{r}{c}} \int_{-\infty}^{\infty} e^{i\omega t_r} e^{-\frac{k}{m}t_r} dt_r.$$ So now the two integrals only differ by a phase shift factor $e^{i\omega\frac{r}{c}}$, which does not influence the spectral density which only depends on the absolute value. I did not realise this at first since the derivative $\frac{dt}{dt_r}$ usually introduces extra factors. In this case however the observer is far away so $r$ is independent of time. Then the integral transforms much more easily.

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