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This is quite an interesting problem in astrophysics so I thought it would be a good idea to ask here so we can archive the solution for future reference.

Consider a pulsar that emits pulses of frequencies $\omega_1$ and $\omega_2$. Due to the interstellar space, these pulses will each take a different time to arrive to Earth, thus we will observe a delay between signals.

If we know the dispersion relation $\epsilon(\omega)$, how can one find the delay between these two signals? For example, let's consider the most simple dispersion relation:

$$\epsilon(\omega)=1-\frac{\omega_p^2}{\omega^2}$$

where $\omega_p$ is the plasma frequency (for simplicity, let's take it as constant).

My idea is to start from the wave equation for a wave packet in a dispersive media and take an arbitrary distance $L$ for the wave to travel, and calculate the times $t_1$ and $t_2$ for the wave to take using each frequency, and somehow introduce the dispersion relation somewhere on the wave equation. However, while I know the evolution for a wave packet $\psi(x,t)$, I don't know how to find the time it takes to travel a distance $L$.

Edit: I found the solution using a less complicated method. We know that the wave number $k$ follows the relation,

$$k^2=\epsilon \mu\frac{\omega^2}{c^2}$$

Substituting the dispersion relation and considering non-magnetic media ($\mu=1$),

$$k^2=\frac{\omega^2}{c^2}(1-\frac{\omega_p^2}{\omega^2})$$

Thus we get,

$$\omega^2=k^2c^2+\omega_p^2$$

Taking the derivative and remembering the definition of group velocity,

$$\frac{d\omega}{dk}=\frac{kc^2}{\omega}=v_g$$

Substituting $k$ in the previous expression, we have:

$$v_g=c\sqrt{1-\frac{\omega_p^2}{\omega}}$$

If we consider that the wave travel a distance $L$, each pulse takes a time:

$$t_1=\frac{L}{v_1}=\frac{L\omega_1}{c\sqrt{\omega_1^2-\omega_p^2}}$$

$$t_2=\frac{L}{v_2}=\frac{L\omega_2}{c\sqrt{\omega_2^2-\omega_p^2}}$$

And thus the time difference is a function of the distance and frequencies,

$$\Delta t=\frac{L}{c}\left ( \frac{\omega_1}{\sqrt{\omega_1^2-\omega_p^2}} - \frac{\omega_2}{\sqrt{\omega_2^2-\omega_p^2}} \right )$$

If anyone can tell me it's correct so I can post it as an answer to my question, and thus close this thread.

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Your second logic - calculating group velocity via a dispersion relation - is reasonably sound and does produce the correct answer, although I have two minor caveats:

  1. We can assume that the plasma frequency $f_p\ll f$ (apologies for using different notation) given the typical properties of the interstellar medium. The electron number density is on the order of $n_e\simeq0.05\;\text{cm}^{-3}$, and the $f_p\propto n_e^{1/2}$ relationship puts the plasma frequency around 1 kHz, whereas pulsar astronomers usually observe above the ~100 MHz range (typically at least 400 MHz, but instruments like LOFAR are probing below that). Since $(f_p/f)\ll1$, we can invoke the binomial approximation and simpify your expression to $$\frac{1}{v_g}=\frac{1}{c}\left(1-\frac{f_p^2}{f^2}\right)^{-1/2}\simeq\frac{1}{c}\left(1+\frac{1}{2}\frac{f_p^2}{f^2}\right)$$ A way to present this that's a bit more intuitive is to explicitly calculate the frequency-dependent refractive index $\mu$ and use the relation $v_g=\mu c$.
  2. The ISM varies in density, so in reality we have to integrate the number density (encoded in the plasma frequency) over the path length to find the delay rather than just assuming a uniform distribution. Compared to a wave of infinite frequency (which would experience no dispersion), a wave of frequency $f$ experiences a delay $$t=\left(\int_0^d\frac{\mathrm{d}l}{v_g}\right)-\frac{d}{c}=\frac{e^2}{2\pi m_ec}\frac{\int_0^d n_e\mathrm{d}l}{f^2}\propto f^{-2}$$ and we typically define the dispersion measure $\text{DM}$ and dispersion constant $\mathcal{D}$ to be $$\text{DM}\equiv\int_0^dn_e\mathrm{d}l,\quad\mathcal{D}\equiv\frac{e^2}{2\pi m_ec}$$ Your expression for the time delay then becomes $$\Delta t=\mathcal{D}\cdot\text{DM}\left[\frac{1}{f_1^2}-\frac{1}{f_2^2}\right]$$ which is a little bit nicer.
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