# 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?

Astronomy Asked on December 7, 2021

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 $$v_g=cmu$$, 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 $$t(nu)=int_0^dfrac{dl}{v_g}-frac{d}{c}$$ 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 $$nuggnu_p$$, and after some algebra, we find that $$t(nu)=frac{1}{c}int_0^dleft(1+frac{nu_p^2}{2nu^2}right)dl-frac{d}{c}=frac{e^2}{2pi m_ec}frac{1}{nu^2}int_0^dn_edl$$ 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}{2pi m_ec}frac{1}{nu_1^2}int_0^dn_edl-frac{e^2}{2pi m_ec}frac{1}{nu_1^2}int_0^dn_edl\ &=frac{e^2}{2pi m_ec}left(int_0^dn_edlright)left[frac{1}{nu_1^2}-frac{1}{nu_2^2}right] end{aligned} If we define the dispersion measure by $$text{DM}equivint_0^dn_edl$$ and define it in units of $$text{cm}^{-3};text{pc}$$, write the chunk of constants in front in units of $$text{GHz}^2;text{pc}^{-1};text{cm}^3;text{ms}$$: $$frac{e^2}{2pi m_ec}=4.15;text{GHz}^2;text{pc}^{-1};text{cm}^3;text{ms}$$ 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 $$t_1-t_2=4.15left(frac{text{DM}}{text{cm}^{-3};text{pc}}right)left[left(frac{nu_1}{text{GHz}}right)^{-2}-left(frac{nu_2}{text{GHz}}right)^{-2}right];text{ms}$$ which has the proper units.

In short: If you write your frequencies in gigahertz and your dispersion measure in parsecs per cubic centimeter, the formula will give you a time delay in milliseconds.

Answered by HDE 226868 on December 7, 2021

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