# How to estimate a star's heliopause?

Is it possible to calculate/estimate the size of a star's heliosphere? If so, how?

I am working on a semi-near future, sci-fi novel. As part of the technology base, humans are able to travel between stars near-instantly, however they must first get to a warp-gate of sorts outside the heliopause (outer edge of the heliosphere) at sub-light speeds. This results in a few months of travel time from the planets to the edge of the solar system, then another few months travel from the edge of the next system to the planets.

My problem is that I don't know if it is possible to calculate the size of a star's heliosphere based on known data. All of the stars I am using are going to be actual known stars. I'm hoping that using the known data of the stars size, type, and luminosity we can determine at least a reasonable estimate. Although the heliosphere should probably be called the helioegg instead, I am really just needing a number that won't be too outrageous. The main reason I need to calculate the distance is actually to then calculate the travel time from the edge of the solar system.

The heliopause of Sol is at roughly 128 AU (give or take) from what I understand, but obviously that is not going to be necessarily even close for every star. It would be interesting if even though Proxima Centauri is the closest star to Sol, if it had a a heliopause distance of like 400 AU it could end up having the longest travel time of all the other local star systems.

The heliopause is a place of equilibrium, where the ram pressure from the solar wind is equal to the pressure of the interstellar medium (ISM). There are a number of sources of pressure in the ISM, but thermal pressure is the main one.1 The ram pressure from a wind with terminal velocity $v_{\infty}$ at a distance $r$ from the a star is $$P_R=n_*(r)mv_{\infty}^2$$ where $n_*(r)$ is the number density of the wind, and $m$ is the mean mass of the particles at the heliopause. It turns out that, by mass conservation, $n_*(r)\propto \dot{M}v_{\infty}^{-1}r^{-2}$, where $\dot{M}$ is the mass-loss rate.2 The thermal pressure of the ISM is $$P_T=n_{I}k_BT$$ where $k_B$ is Boltzmann's constant and $T$ is the temperature. $n_{I}$ is the number density of the interstellar medium. For Sun-like stars, $v\sim400$ km/s is reasonable. $n_{\odot}$ and $n_I$ are about 5 particles per cubic centimeter and 0.01 particles per cubic centimeter, respectively, and $T\sim10^5$ K. Assuming the wind is largely hydrogen at this point, if we set the two equal, we get \begin{align} r_{\text{crit}} & =\left(\frac{n_{\text{crit}}mv_{\infty}^2}{n_Ik_BT}\right)^{1/2}\text{ AU}\\ & \approx311\left(\frac{n_\text{crit}}{5\text{ cm}^{-3}}\right)^{1/2}\left(\frac{v_{\infty}}{400\text{ km/s}}\right)\text{ AU}\\ & =311\left(\frac{\dot{M}}{10^{-14}M_{\odot}\text{ yr}^{-1}}\right)^{1/2}\left(\frac{v_{\infty}}{400\text{ km/s}}\right)^{1/2}\text {AU} \end{align} where $n_{\text{crit}}\equiv n_*(r_{\text{crit}})$. I've used scaling relations such that for the Sun, $r_{\text{crit}}$ is 311 AU.

# Important factors

Some things to note:

• Stellar winds don't all have the same composition, but hydrogen is, by and large, the major component, and the one important factor when it comes to calculating $m$.
• The most important star-dependent variables are $\dot{M}$ and $v_{\infty}$. Note that $r_{\text{crit}}\propto \dot{M}^{1/2}v_{\infty}^{1/2}$. For massive O- and B- stars, winds can have speeds of $\sim2000$ km/s, or more; I think some of the strongest are about $\sim3000$ km/s. This could mean heliopauses of thousands of AU. HD 93129a is a good example, with $v_{\infty}$ of about $\sim3000$ km/s.
• I can try to pull some numbers for Proxima Centauri, but I'll point out that red dwarfs usually don't have strong stellar winds. The interest in the wind of Proxima Centauri is really because Proxima Centauri b orbits close enough to the star that stellar activity - especially flares - could cause severe problems for life.

# Star types

For Sun-like stars, $\dot{M}\sim10^{-14}M_{\odot}$ per year is reasonable, and so you'd get heliopauses of a few hundred AU. For O- and B- type main sequence stars, I'd expect $\dot{M}\sim10^{-6}M_{\odot}$ per year, which is much higher. At the other end of the spectrum - those red dwarfs that are relatively quiet - we'd see maybe $\dot{M}\sim10^{-15}M_{\odot}$ per year. A- and F- main sequence stars might have mass-loss rates a bit higher than the Sun, and perhaps larger terminal velocities.

Off the main sequence, things get more complicated. Red giants - especially asymptotic branch stars, near the end of their lives - have large mass-loss rates that arise via different mechanisms involving dust. These are cool but bright stars; consider $\dot{M}\sim10^{-8}M_{\odot}$ and reasonable fast winds, though not nearly as fast as the hot, massive O-type stars. Additionally, more exotic cases like Be stars that are undergoing mass loss may have larger mass-loss rates. Finally, the young T Tauri stars have mass-loss rates similar to red giants - maybe an order of magnitude or so lower - but their winds are slower than the Sun's.

# Specific cases

I looked around and found instances where $\dot{M}$ and $v_{\infty}$ have been observed, and made an estimate of $r_{\text{crit}}$: $$\begin{array}{|c|c|c|c|c|c|}\hline \text{Star} & \text{Stellar type} & \text{Mass }(M_{\odot}) & \dot{M}(M_{\odot}\text{ yr}^{-1}) & v_{\infty}(\text{km/s}) & r_{\text{crit}}(\text{AU})\\\hline \text{HD 93129Aa}^1 & \text{O2} & 95 & 2\times10^{-5} & 3200 & 3.93\times10^{7}\\\hline \tau\text{ Sco}^2 & \text{B0} & 20 & 3.1\times10^{-8} & 2400 & 1.32\times10^{6}\\\hline \sigma\text{ Ori E}^3 & \text{B2} & 8.9 & 2.4\times10^{-9} & 1460 & 2.91\times10^{5}\\\hline \alpha\text{ Col}^2 & \text{B7} & 3.7 & 3\times10^{-12} & 1250 & 9.52\times10^{3}\\\hline \text{Deneb}^4 & \text{A2} & 20 & 10^{-6} & 225 & 2.3\times10^{6}\\\hline \text{Sun} & \text{G2} & 1 & 10^{-14} & 400 & 311\\\hline \text{Proxima Centauri}^5 & \text{M6} & 0.12 & 10^{-13} & 550 & 1332\\\hline \end{array}$$ 1Cohen et al. (2011)
2Cohen et al. (1997)
3Krtička et al. (2006)
4Aufdenberg et al. (2002)
5Wargelin & Drake (2002)

Now, HD 93129Aa and Deneb are supergiants, so they're off the main sequence, but their properties here shouldn't be too far off from main sequence stars of the same spectral type. Deneb's mass-loss rate is maybe a bit high in comparison to main sequence A stars. Also, I'm slightly skeptical of the value for HD 93129Aa's heliopause, so it's possible that other factors play a role - for instance, thermal pressure could indeed be important in its hot wind. Additionally, some M dwarfs have higher stellar winds and mass-loss rates because of flares and other activity.

1 We can disregard ram pressure, as the ISM is, by and large, slow-moving. Likewise, magnetic fields are typically not important. Similarly, we can neglect thermal pressure in the stellar wind; even though winds may have temperatures of several million K, ram pressure is more important.
2 Specifically, the density $\rho_{\odot}(r)$ (related to number density by $\rho_{\odot}(r)=mn_{\odot}(r)$) is given by $$\rho_{\odot}(r)=\frac{\dot{M}}{4\pi r^2v(r)}$$ where $\dot{M}$ is the mass loss rate and $$v(r)=v_{\infty}\left(1-\frac{R_*}{r}\right)^\beta$$ with $R_*$ being the radius of the star. We typically assume $\beta\approx1$, but at $r\gg R_*$, we can say that $v(r)\approx v_{\infty}$.

• Since we know that then heliosphere is more of an egg shape and shortest distance to Sol heliopause is around 128 AU instead of the 311 AU that you calculated, might it be safe to assume that the radius you are calculating is more accurately an average radius of the whole heliosphere, rather than shortest distance to closest side? I'm mainly wondering because the numbers don't quite match what else I have found, but it mostly seems to be an estimated range. Sol's seems to be between 100- 450 AU, so both numbers fit just fine. – TitaniumTurtle Jun 28 '18 at 2:03

The heliosphere, in layman terms, is defined by the equilibrium between the star wind and the galactic wind.

Once the star wind is not strong enough to blow away the galactic wind, there we have the heliopause, bordering the heliosphere.

Therefore, if you know:

• Star wind velocity
• Star wind velocity reduction rate
• Galactic wind velocity
• Relative motion of the star with respect to the surrounding star

you should be able to estimate the extension of the heliosphere for the star in question.

L.Dutch lays out the numbers you need to consider, having done a little reading on stellar winds it would appear that most stars in the A-K section of the main sequence, might be expected to have similar sized heliopauses in a similar area of the galaxy. Cooler and giant stars have much smaller heliopauses and hot white and blue stars much much larger ones due to the extremely high escape velocities of their surfaces. Based on that the hardest nearby star to ship to and from would actually be Sirius rather than any of the three nearby Centauri stars. Barnard's Star should be a very quick transit though.

• Not that this goes into a ton of detail, but wouldn't the trinary nature of the Centauri stars give it a more pronounced heliosphere? – TitaniumTurtle Jun 24 '18 at 17:42
• @TitaniumTurtle Maybe and then again because of how far apart they are, particularly Proxima is something like 0.15L.Y. from the Alpha binary, maybe not too, it would almost certainly mess with the shape of the heliopause so possibly you'd have to drop below light-speed earlier to avoid shape fluctuations. – Ash Jun 24 '18 at 17:48