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?

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.

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