# How to determine arrival time delay given dispersion measure?

In an online textbook, the following formula is given to calculate the arrival time delay between two frequencies, $$v_1, v_2$$ in a dispersed radio pulse:

$$t_1 - t_2 = 4.15 \cdot DM [(v_1/\text{GHz})^{-2} - (v_2/\text{GHz})^{-2}]$$

Where does the $$4.15$$ come from, and what exactly is “GHz”? Do I divide by 1 GHz or?

• Can you provide a link to where you found the equation online? That would help provide more context for your question. – Eric Jensen Jul 24 '20 at 1:12

As is the case with light traveling through any medium, radio waves traveling through space experience refraction, which reduces their speed. A wave of infinite frequency will experience no refraction, meaning that we can compute how much a given wave will be delayed compared to such an infinite-frequency wave as it moves through outer space. The group velocity of a radio wave is determined by $$v_g=c\mu$$, with $$\mu$$ the refractive index defined by $$\mu=\sqrt{1-\left(\frac{\nu_p}{\nu}\right)^2}$$ with $$\nu$$ the frequency of the wave and $$\nu_p$$ the plasma frequency, a fundamental property of the medium, which depends on its density: $$\nu_p=\sqrt{\frac{e^2n_e}{\pi m_e}}$$ with $$e$$, $$m_e$$ and $$n_e$$ the charge, mass and number density of electrons. (You're probably more familiar with refractive indices for light that humans can see, but they're important in a much wider portion of the electromagnetic spectrum!) In the ISM, we'd expect to see plasma frequencies in the kilohertz range - much lower than the frequencies of radio waves from whatever source we're observing, which are likely in the hundreds of megahertz to tens of gigahertz range. The delay a wave of frequency $$\nu$$ experiences over a distance $$d$$ is $$t(\nu)=\int_0^d\frac{dl}{v_g}-\frac{d}{c}$$ where the second term requires no integral as the refractive index of a wave of infinite frequency is always $$\mu=1$$ regardless of the medium. We can use a binomial approximation for the refractive index of our physical finite-frequency wave, since $$\nu\gg\nu_p$$, and after some algebra, we find that $$t(\nu)=\frac{1}{c}\int_0^d\left(1+\frac{\nu_p^2}{2\nu^2}\right)dl-\frac{d}{c}=\frac{e^2}{2\pi m_ec}\frac{1}{\nu^2}\int_0^dn_edl$$ Now consider two waves of frequencies $$\nu_1$$ and $$\nu_2$$. The difference in arrival times is \begin{aligned}t(\nu_1)-t(\nu_2)&=\frac{e^2}{2\pi m_ec}\frac{1}{\nu_1^2}\int_0^dn_edl-\frac{e^2}{2\pi m_ec}\frac{1}{\nu_1^2}\int_0^dn_edl\\ &=\frac{e^2}{2\pi m_ec}\left(\int_0^dn_edl\right)\left[\frac{1}{\nu_1^2}-\frac{1}{\nu_2^2}\right] \end{aligned} If we define the dispersion measure by $$\text{DM}\equiv\int_0^dn_edl$$ and define it in units of $$\text{cm}^{-3}\;\text{pc}$$, write the chunk of constants in front in units of $$\text{GHz}^2\;\text{pc}^{-1}\;\text{cm}^3\;\text{ms}$$: $$\frac{e^2}{2\pi m_ec}=4.15\;\text{GHz}^2\;\text{pc}^{-1}\;\text{cm}^3\;\text{ms}$$ and write out frequencies in units of $$\text{GHz}$$, then we indeed recover the expression you give - with the time delay in units of milliseconds. It would be a lot clearer if we rewrote it as $$t_1-t_2=4.15\left(\frac{\text{DM}}{\text{cm}^{-3}\;\text{pc}}\right)\left[\left(\frac{\nu_1}{\text{GHz}}\right)^{-2}-\left(\frac{\nu_2}{\text{GHz}}\right)^{-2}\right]\;\text{ms}$$ which has the proper units.