4
$\begingroup$

All accounts of solar wind I have seen (I am no expert in the topic), seem to refer to it being everywhere a plasma (mainly composed of protons/electrons). For example, I have seen statements about the direction of motion of the plasma at the heliopause (measured by the voyager probes). Maybe this is a misunderstanding from my side.

But given the fact that intensity of solar electromagnetic radiation decreases massively towards the heliopause, my intuition urges me to believe that, as we move away from the sun, an increasing fraction of the protons should recombine with the electrons to form bound hydrogen atoms (or even some molecules?). Especially a very rough calculation (assuming v=300 km/s and s=100 AU) shows that an individual proton takes at least about two years to travel from the sun to the heliopause, a lot of time for reaching the bound ground state.

Are there any graphs or figures that illustrate whether this happens and if yes, to what extent?

I have found a popular article about the topic, but it only talks about the solar wind close to earth, and emphasizes on measurements. What about theory, and what about 100 AU instead of 1 AU?

$\endgroup$
2
  • 1
    $\begingroup$ This seems relevant. $\endgroup$
    – J.G.
    Commented Feb 7, 2023 at 9:25
  • $\begingroup$ Thanks, I have seen this, but it seems to refer only to the outermost boundary of the solar system. My assumption is that the formation of hydrogen starts long before that. $\endgroup$
    – oliver
    Commented Feb 7, 2023 at 9:51

1 Answer 1

7
$\begingroup$

First, look at https://physics.stackexchange.com/a/288810/59023 for recombination rate info and https://physics.stackexchange.com/a/695956/59023 for radial dependence of temperatures in the solar wind plasma. If we assume the expansion of the solar wind is the usual adabatic approximation, then the density should go as ~$r^{-2}$. The recombination time scale goes as $t_{rec} \approx \left( n_{e} \ \alpha \right)^{-1} \sim \sqrt{ \tfrac{T}{n_{e}^{2}} }$. For simplicity, let's assume temperature changes as ~$r^{-1.25}$. Then we have $t_{rec} \approx r^{+1.375}$.

We can change units etc. so that $r$ here is a normalized one. It's not tremendously important but what is important is that this rate changes so slowly that we would need to increase $r$ by a factor of ~150 before the recombination rate would increase by a factor of ~1000. The recombination rate near the Sun is tiny (to the point of being negligible) and the ionization rate is huge. Thus, at ~150 astronomical units or AU (i.e., out past Voyager 1), the recombination is only ~1000 times larger than negligible. It's not zero but it's also not huge, so far as I know.

The point is that recombination does occur and ionization also occurs. The recombined protons are often detected near Earth as energetic neutral atoms or ENAs while the re-ionized protons and alpha-particles are observed as something called pickup ions. The IBEX and IMAP missions focus on ENAs, for example.

$\endgroup$
6
  • $\begingroup$ I am not sure if I misunderstand some part of the argument. But shouldn't it be $t_{rec}\approx \sqrt{r^{-1.25}/r^{-4}}=\sqrt{r^{2.75}}=r^{+1.375}$, which would make the dependency much stronger than $r^{+0.375}$? $\endgroup$
    – oliver
    Commented Feb 7, 2023 at 15:44
  • $\begingroup$ Why do you think density changes as $n \sim r^{-4}$ in the solar wind? $\endgroup$ Commented Feb 7, 2023 at 15:52
  • $\begingroup$ I don't. But the density occurs to the power of two (under the square root) in the recombination time relation you have written. If a power of -2 (the density) is elevated to the power of two, a power of -4 results. Or did I misinterpret something about the formula? $\endgroup$
    – oliver
    Commented Feb 7, 2023 at 15:54
  • $\begingroup$ I think you are missing a step. If you have $\left( \tfrac{ r^{-1.25} }{ r^{-2} } \right)^{1/2}$ = $\left( r^{+0.75} \right)^{1/2}$ = $r^{+0.375}$ $\endgroup$ Commented Feb 7, 2023 at 20:00
  • 2
    $\begingroup$ The recombination rate goes as $\sqrt{ \tfrac{ T }{ n^{2} } }$... Oh yes, I see what I missed. Just a second, I will update things accordingly $\endgroup$ Commented Feb 8, 2023 at 18:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.