I had Originally thought of using Nicoll-Dyson beams to propel probes and small ships from the Milky Way to as many galaxies as they can reach to turn stars in those galaxies into Shkadov thrusters to return to our galaxy.

The problem with Shkadov thrusters is they have extremely slow initial acceleration due to the mass of the star but once its around a billion years a star like our sun can have moved 35,000 light-years and will be moving at 20 km/s by then.

Nicoll-Dyson beams also have an issue of the beam spreading and being less effective propelling a ship once we get into inter-galactic distances. So I thought since they are arriving home on Shkadov thrusters they could also have left the Milky Way in the same way, at least your massive fuel source isn't millions of light-years away.

For the return journey the majority of stars will be red dwarfs due to their trillions of years life span but they are extremely slow moving stars due to their low energy output so a massive star may be the best bet as even though they are much more massive and harder to move their energy output is magnitudes higher but the major problem with these stars is their short life span.

What makes this calculation even harder is that the two galaxies are moving towards each other and it could be asked why bother traveling there when we will collide in 4 billion years but I would like to arrive in Andromeda long before the merger.

Is there a type of star that could get to Andromeda as a Shkadov thruster long before the merger or could even a Nicoll-Dyson beam using the correct star type propel a probe or small ship all the way to Andromeda in the shorter travel times I am after?

Edit: someone had done some calculations and they said it is not possible to reach Andromeda with a massive star, their calculations put our suns output as being able to reach in 20 billion years and a 10 solar mass star reaching in 1 billion years, both travel times being far longer than their life span but they did have an interesting suggestion of riding the supernova blast for the remaining journey.

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    $\begingroup$ 'As soon as possible'? A metric that is very hard to qualify. $\endgroup$ Commented Jun 30, 2020 at 15:41
  • $\begingroup$ @JustinThymetheSecond Well I did have a question is was going to add which is: When would be the best time to leave, should you leave as soon as possible to arrive in the fastest time, or would waiting till the speed the galaxies are moving together increases actually be he best time to leave? $\endgroup$
    – user69935
    Commented Jun 30, 2020 at 15:45
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    $\begingroup$ @JustinThymetheSecond The aim is to arrive as soon from now as possible using a single star, and as mentioned in the answers the stars output vs life span is the issue. $\endgroup$
    – user69935
    Commented Jun 30, 2020 at 22:02
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    $\begingroup$ Solar system: "That's it I'm moving out when I grow up!". Milky: "Oh no you don't, Sun." $\endgroup$
    – user6760
    Commented Jul 1, 2020 at 1:57
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    $\begingroup$ The thing is, the star doesn't have to last the entire journey. Conservation of momentum. Once the star accelerates you to a certain velocity, you keep going at that velocity. So 'best' could imply using up the star as quickly as possible, for the greatest speed. But do you want to decelerate at the end, or do you want any part of the star to be left over? Artillary shells do not need to decelerate at the end of their trip, so 'best time' for them is different than 'best time' for an airplane. $\endgroup$ Commented Jul 1, 2020 at 4:37

4 Answers 4


You want to use light to push the star.

The more light the star emits, the more push it can produce. But to produce more light the star needs more mass, which will affect your acceleration. Where is the sweet spot?

According to Wikipedia, the mass-luminosity relationship can be written as

${L \over L_{S}}=p({M\over M_s})^q$


  • if $M<0.43M_s$ then $q=2.3, p=0.23$
  • if $ 0.43M_s < M<2M_s$ then $q=4, p=1$
  • if $ 2M_s < M<55M_s$ then $q=3.5, p=1.4$
  • if $M>55M_s$ then $q=1, p=32000$

If we assume that thrust is proportional to luminosity, the above can give us the dependence between thrust and mass and thus allow us to calculate the maximum acceleration we can get, assuming that in the non relativistic regime we have $a=F/m$

We get that

$a = {p L_s \over {M_s}^q}M^{q-1} $

Finding the maximum vs M of the above function will give you the optimum thruster.

As a crude engineer I have plotted a chart of the acceleration vs the mass of the star, resulting in the following chart

acceleration vs star mass

Which tells that the best thruster is a star with 55 solar masses. Bigger than that will not give you more acceleration.

If you are interested in maximum deltaV instead that in the maximum acceleration, you have now to combine the thrust with the amount of time it can act, given by the star lifetime.

This table gives an indication of a star lifetime based on its mass

lifetime vs mass

conveniently computed into a table where $deltaV = a\cdot time$, you get the following

delta v

It is evident that the maximum deltaV will be provided by a star with 60 solar masses: a lot of push for a very short amount of time.

  • $\begingroup$ Thanks, do you know the life-span for a 55 solar mass star? the calculations are bit over my head. $\endgroup$
    – user69935
    Commented Jun 30, 2020 at 14:28
  • $\begingroup$ I just read that for a 40 solar mass star its life span is a millions years? so that wouldn't reach Andromeda. $\endgroup$
    – user69935
    Commented Jun 30, 2020 at 14:36
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    $\begingroup$ @RandySavage, included also the lifetime of the star into the calculation $\endgroup$
    – L.Dutch
    Commented Jun 30, 2020 at 15:50
  • $\begingroup$ I think our answers differ because of your assumption that $L\propto F$ and that the proportionality factor is constant. While the former is true, I think that just using that ignores that $L=Fv$, and so there's a quadratic relationship between $L\Delta t/M$ and $v$, not the linear one you're proposing. I guess we also differ in that you've considered using stars that are quite massive and I've implicitly ignored them as too volatile (and I say this for the benefit of anyone reading the answers and wondering why they're so different!). $\endgroup$
    – HDE 226868
    Commented Jun 30, 2020 at 16:11
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    $\begingroup$ It looks like you dropped a factor of 1000 in the sub-3-solar-mass star lifetimes in the delta-V table. This changes the results. $\endgroup$
    – AI0867
    Commented Jul 1, 2020 at 9:12

Ideally a star of $6\text{-}8$ solar masses.

Very massive stars are not the best choice, for two reasons. The first one is that these stars tend to be quite violent during their lives, with strong stellar winds and sometimes energetic non-thermal radiation, like x-rays. Adding shielding to a megastructure like a Shkadov thruster might be possible, but it's a pain. Plus, after some millions of years, if the star is heavier than 8 solar masses, it'll explode in a supernova, and there's a very good chance that your thruster will simply be destroyed in intergalactic space.

The second reason is that for stars of above $2M_{\odot}$, the final velocity a star can produce throughout the entirety of its lifetime is essentially independent of mass, for a reasonable mass-luminosity relation.$^{\dagger}$ We can actually do these calculations by simply invoking conservation of energy, following the method of Hooper 2018, who applied to to propelling stars using energy gathered by Dyson spheres. The argument there is one of conservation of energy. The final velocity $v$ after a thruster has operated for time $\Delta t$ is, for stars of $M>2M_{\odot}$, $$v=0.034c\;\left(\frac{\Delta t}{1\;\text{Gyr}}\right)^{1/2}\left(\frac{M}{2M_{\odot}}\right)^{1.25}\left(\frac{\eta}{1}\right)^{1/2}$$ where $\eta$ is some efficiency factor. Let's assume our star will die before we reach Andromeda, an assumption I think should hold for all stars of $M>6M_{\odot}$$^{\ddagger}$. The lifetime of the star scales as $\tau\propto M^{-2.5}$, and so if we assume that $\Delta t=\tau$, we see that the mass dependence for $v$ actually drops right out!

Let's assume, then, that the mass of the star is unimportant for the stars of the masses we're interested in. I then argue that we should pick a star in the range $6M_{\odot}<M<8M_{\odot}$. Why? There are a couple of reasons:

  • A star more massive than $8M_{\odot}$ will undergo a supernova prior to arriving at Andromeda.
  • A star less massive than $6M_{\odot}$ will not have had its full energy tapped by the time it arrives at Andromeda.
  • A less massive star lives for a longer period of time, and therefore it can be a source of auxiliary power for other thruster functions for longer.
  • Stars in this mass range are much less likely to have outbursts and eruptions than the massive stars others have argued for.

In short, pick a star of moderate mass, and you'll reach Andromeda efficiently and, most importantly, without having been incinerated by a supernova.

$^{\dagger}$L.Dutch notes a break in the mass-luminosity relation for $M>55M_{\odot}$, though I'm not sure that this is widely-used, and at any rate, these stars are extremely rare.

$^{\ddagger}$I got this value by assuming that all stars of $M>2M_{\odot}$ reach terminal speeds of $v_{\text{max}}\approx0.045c$ (which you can see by a quick calculation using the above formula) and would have mean speeds of approximately half that. The travel time to Andromeda is then roughly 114 million years, and a star of mass $M=6M_{\odot}$ would leave the main sequence after that time - I neglect main sequence evolution.

  • $\begingroup$ Thanks, do you know how long the journey would take for a 20 solar mass and at what point in the journey it will go supernova? I was imagining an epic scene with loads of supernova and red giants filling the sky of the galaxy they were arriving to. $\endgroup$
    – user69935
    Commented Jun 30, 2020 at 15:01
  • $\begingroup$ But would a 9 solar mass take much longer to make the same distance? does the 57 million years include the star going supernova and coasting the rest off the momentum? $\endgroup$
    – user69935
    Commented Jun 30, 2020 at 15:49
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    $\begingroup$ @RandySavage I once again made an incorrect assumption about the time it would take - it should be greater than that by a factor of 2. The total travel time does take into account the time spent coasting, yes, for those stars that would die after reaching Andromeda. $\endgroup$
    – HDE 226868
    Commented Jun 30, 2020 at 15:59
  • $\begingroup$ Note you'll need the star to decelerate before reaching the destination, so "jettisoning" is not really an option, is it ? $\endgroup$ Commented Jun 30, 2020 at 20:27
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    $\begingroup$ @StephenG The OP specifically mentioned "probes and small ships" in the question, so I suspect the payload is actually quite low. Nothing unfeasible about, say, storing energy from the star along the way to power a set of ion thrusters or some such - it's peanuts in comparison to the star's total output. The only issue is storing electricity, and I would be surprised if that's a serious problem. $\endgroup$
    – HDE 226868
    Commented Jun 30, 2020 at 21:14

Instead of using a Shdakov thruster, use... The Caplan thruster! A hypothetical megastructure that essentially acts as an immense rocket, shooting stuff one way to propel yourself the other way. This requires a basic Dyson swarm first. Since your civilisation can construct Shdakov thrusters only through probes, I'm going to assume that they have the capability to create a Dyson swarm.

A Caplan thruster is a space station-like megastructure pointing towards the sun that draws on energy from the Dyson swarm and gathers solar matter, powering nuclear fusion which ejects particles from its 'thruster' at around 1% the speed of light. A secondary thruster fires a second jet of particles at the sun, pushing it forward so the power of the primary thruster doesn't cause the Caplan megastructure to impact the Sun.

To quote from the paper, which I will link.

'A jet with the mass loss rate m and average speed (v) gives the sun an acceleration of m(v)/M⊙'

To maximise the acceleration, or a, you must increase m and (v) without m being large enough to impact the lifespan of the star.

The Caplan thruster uses immense electromagnetic fields to gather hydrogen and helium from the sun, since it requires millions of tons of fuel a second. However this sparse interstellar matter is not enough to power the Caplan thruster alone. This is where we will use the Dyson swarm. The swarm will focus sunlight onto the star itself, heating these areas to incredible temperatures and causing millions upon millions of tons of matter to rise from the star, which will be funneled into the Caplan thruster using its electromagnetic fields.

The helium and hydrogen are separated, where the helium is used in thermonuclear fusion reactors, with primary thruster expelling radioactive oxygen at a billion degrees. The secondary thruster works by using particle accelerators to fire the collected hydrogen back at the sun, balancing out the Caplan thruster to prevent it crashing into the surface. The star can be moved 50 light years in only one million years.

The use of stellar matter will also extend the lifespan of the star, since smaller stars undergo fusion at a slower rate.

If we assume a perfectly efficient Dyson Swarm, in only 5 Megayears the star could reach velocities of up to 200km/s as opposed to the 20km/s that Shdakov thrusters reach after an even longer period of time, however the mass loss rate limits the usage of the star to 100 megayears of use before the star becomes impacted enough as to limit performance and shrink.

It's more viable to redirect the star onto the trajectory you wish for it to travel, firing the Caplan thruster for only 10 megayears in that direction.

I know this isn't directly answering your question but I think that a Caplan thruster is currently the best way to go about stellar engines>

Link to the paper: link

  • $\begingroup$ I hadn't even heard of this type of thruster, thats really cool, thanks. $\endgroup$
    – user69935
    Commented Jun 30, 2020 at 23:52

An unnatural one.

L.Dutch's answer is a good start. If you just want to find a natural star to ride along with, something in 55-60 solar mass range is fine. And indeed, that is a good place to start....

But you can do much better than just finding a natural star and riding along. After all, you've already got the technology to build a Shkadov thruster, and you've got millions of years and a whole stellar system of resources to continue developing.

Stars increase in luminosity throughout their lifespans, as the core becomes more compact and fusion gets faster. The final supernova is kind of just the endpoint of that continuous process... and kind of a huge waste, as well.

If you can lift material off the star as it ages during the journey, you can arrest the luminosity increase and extend its lifetime. That mass then has a variety of uses. You can use it as reaction mass to improve your propulsion efficiency and get to Andromeda faster. You can use it to slowly build a companion star that will provide additional power output and improved thrust. Or you can save it to feed back into the original star later when it starts to actually run out of fuel.

  • $\begingroup$ Thanks, I had thought about lifting to control the star but I wasn't sure how the lower mass star later in the journey would effect the speed, Having a secondary star would look really cool, do you think they would start binary orbiting and cause trouble to the ship and direction of travel? $\endgroup$
    – user69935
    Commented Jun 30, 2020 at 23:51
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    $\begingroup$ @RandySavage They would have to be built to orbit each other. As long as the secondary artificial star is built to thrust in the same direction, that should not cause any problems. $\endgroup$ Commented Jul 1, 2020 at 0:12
  • $\begingroup$ I am not sure if the stellar composition won't have some influence on parts of that process. The lifted material will be higher in heavy elements and not be the best start for a new star. On the other hand, at that level you might as well fully manage the nuclear reaction, i.e. lift out the used up fusion products and occasionally throw in some brown dwarfs full of hydrogen you bring along. $\endgroup$
    – mlk
    Commented Jul 1, 2020 at 16:38
  • $\begingroup$ @mlk That's an excellent start for a new star--higher metallicity means it will burn brighter from the start, giving you a better power-to-mass ratio. $\endgroup$ Commented Jul 1, 2020 at 18:23

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