<p>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 <span class="math-container">$v_g=c\mu$</span>, with <span class="math-container">$\mu$</span> the refractive index defined by
<span class="math-container">$$\mu=\sqrt{1-\left(\frac{\nu_p}{\nu}\right)^2}$$</span>
with <span class="math-container">$\nu$</span> the frequency of the wave and <span class="math-container">$\nu_p$</span> the <em>plasma frequency</em>, a fundamental property of the medium, which depends on its density:
<span class="math-container">$$\nu_p=\sqrt{\frac{e^2n_e}{\pi m_e}}$$</span>
with <span class="math-container">$e$</span>, <span class="math-container">$m_e$</span> and <span class="math-container">$n_e$</span> 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 <span class="math-container">$\nu$</span> experiences over a distance <span class="math-container">$d$</span> is
<span class="math-container">$$t(\nu)=\int_0^d\frac{dl}{v_g}-\frac{d}{c}$$</span>
where the second term requires no integral as the refractive index of a wave of infinite frequency is always <span class="math-container">$\mu=1$</span> regardless of the medium. We can use a binomial approximation for the refractive index of our physical finite-frequency wave, since <span class="math-container">$\nu\gg\nu_p$</span>, and after some algebra, we find that
<span class="math-container">$$t(\nu)=\frac{1}{c}\int_0^d\left(1+\frac{\nu_p^2}{2\nu^2}\right)dl-\frac{d}{c}=\frac{e^2}{2\pi m_ec}\frac{1}{\nu^2}\int_0^dn_edl$$</span>
Now consider two waves of frequencies <span class="math-container">$\nu_1$</span> and <span class="math-container">$\nu_2$</span>. The difference in arrival times is
<span class="math-container">$$
\begin{aligned}t(\nu_1)-t(\nu_2)&amp;=\frac{e^2}{2\pi m_ec}\frac{1}{\nu_1^2}\int_0^dn_edl-\frac{e^2}{2\pi m_ec}\frac{1}{\nu_1^2}\int_0^dn_edl\\
&amp;=\frac{e^2}{2\pi m_ec}\left(\int_0^dn_edl\right)\left[\frac{1}{\nu_1^2}-\frac{1}{\nu_2^2}\right]
\end{aligned}$$</span>
If we define the <em>dispersion</em> measure by
<span class="math-container">$$\text{DM}\equiv\int_0^dn_edl$$</span>
and define it in units of <span class="math-container">$\text{cm}^{-3}\;\text{pc}$</span>, write the chunk of constants in front in units of <span class="math-container">$\text{GHz}^2\;\text{pc}^{-1}\;\text{cm}^3\;\text{ms}$</span>:
<span class="math-container">$$\frac{e^2}{2\pi m_ec}=4.15\;\text{GHz}^2\;\text{pc}^{-1}\;\text{cm}^3\;\text{ms}$$</span>
and write out frequencies in units of <span class="math-container">$\text{GHz}$</span>, 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
<span class="math-container">$$t_1-t_2=4.15\left(\frac{\text{DM}}{\text{cm}^{-3}\;\text{pc}}\right)\left[\left(\frac{\nu_1}{\text{GHz}}\right)^{-2}-\left(\frac{\nu_2}{\text{GHz}}\right)^{-2}\right]\;\text{ms}$$</span>
which has the proper units.</p>
<p>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.</p>

CC BY-SA 4.0
2020-12-21T17:41:55.063Z
2020-12-21T19:10:51.327Z
2020-12-21T19:10:51.327Z
HDE 226868
<p>In an online textbook, the following formula is given to calculate the arrival time delay between two frequencies, <span class="math-container">$v_1, v_2$</span> in a dispersed radio pulse:</p>
<p><span class="math-container">$$t_1 - t_2 = 4.15 \cdot DM [(v_1/\text{GHz})^{-2} - (v_2/\text{GHz})^{-2}]$$</span></p>
<p>Where does the <span class="math-container">$4.15$</span> come from, and what exactly is “GHz”? Do I divide by 1 GHz or?</p>

CC BY-SA 4.0
2020-07-24T01:12:21.570Z
Can you provide a link to where you found the equation online?  That would help provide more context for your question.
Eric Jensen
<astrophysics><radio-astronomy><stellar-astrophysics><pulsar><plasma-physics>
How to determine arrival time delay given dispersion measure?
PerplexedDimension
# How to determine arrival time delay given dispersion measure?

(Asked by: username_0 on 2020-07-23)

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?

**Tags:** astrophysics, radio-astronomy, stellar-astrophysics, pulsar, plasma-physics
> **Comments:**
> **username_1**: Can you provide a link to where you found the equation online?  That would help provide more context for your question.  

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## 1 answers

### Answer by username_2 on 2020-12-21. Score: 2

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.

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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