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For simplicity, lets use our Sun as the star.

Next, a dyson sphere is built around it (completely enclosed solid sphere, not a dyson swarm or network, etc.). Handwave how it's built, it's not relevant to the question.

Not sure if the diameter of the dyson sphere makes a difference, but in case it does, lets say its inner surface is about 1AU from the Sun. (sorry Earth, you've been replaced).

Again, for simplicity, lets assume no other celestial bodies remain inside the sphere (Mercury, Venus, other small bodies, etc., completely used up to build the sphere, or otherwise intentionally removed)

Now, the supposed purpose of a dyson sphere is to collect the energy from the star, and put it to good use doing ... something. But shortly after completion of this one, its builders (for whatever reason) flipped the switch from "collect" to "reflect". So instead of collecting the energy from the Sun, it's reflecting most of it back.

What about efficiency? Well, silver and can be up to 95% reflective, at least across the visible spectrum. Aluminum comes close at about 90%. I don't have numbers for materials reflecting non-visible spectrum energy, but since we're talking about dyson sphere level technology, lets handwave the composition of the reflective surface, and just say it reflects 90% of all forms of energy that strike the surface. Regardless of what form of energy it is, the surface reflects 90% of it back at the sun.

The remaining 10% of the energy, the part not reflected, is absorbed (collected) by the sphere, converted to heat, and then radiated into space on the outside of the Dyson sphere, and has no ill effect on the sphere itself.

Does this cause any changes in the star?

Best answers will include details of changes, if any, and explanations of why they take place, and the time-scales during which the changes manifest themselves. Or, explanations of why no changes take place.

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    $\begingroup$ Should we assume that Dyson sphere is made of a material that maintains its integrity and 90% reflectivity no matter how high it is heated? $\endgroup$
    – Alexander
    Jan 13 at 19:34
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    $\begingroup$ @Alexander Yes, that was the intention of the comment about radiating the 10% to space, and no other effects on the sphere. I've edited for clarity. $\endgroup$
    – Harthag
    Jan 13 at 19:35
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    $\begingroup$ Naively, if a reaction produces heat and 90% of that heat is contained within the system would the system not heat up until a point where the 10% of the heat of the system lost is the same as the heat produced by the reaction? $\endgroup$
    – JonSG
    Jan 13 at 20:58
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    $\begingroup$ Photon produced at the star's fusion core takes upwards of millions of years to reach the surface and you are just catching them and throwing them back in so it's like a greenhouse but only at the surface ;D $\endgroup$
    – user6760
    Jan 14 at 0:18
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    $\begingroup$ If my maths is right, the Sun would take up about 0.002% of the total area of a 1 AU sphere. Pure intuition say such a sphere is either not going to have that much of an effect, or it will take really long to have an effect. But I'm no astronomer. $\endgroup$
    – NotThatGuy
    Jan 14 at 21:28

8 Answers 8

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The star will very slowly reach a new equilibrium.

The energy being reflected back will lead to an increase in the star's temperature, starting with the photosphere (which is opaque). The photosphere will become hotter, but there's next to no fusion going on there. The outer layers of the star will swell, and the total energy output will increase (this energy is just the reflected-back energy that gets reflected forward again because of Stefan-Boltzmann's law).

Actually I think we can calculate this: the new equilibrium for the photosphere requires that the energy coming in from the Sun's insides is dissipated, but the dissipation just went down 90%. So the sun must radiate ten times as much, which means its temperature must increase 1.77 times, to about 11200 K (a blue-white star). Its volume should then approximately double (I'm playing fast and dirty with Clapeyron's equation for ideal gases, ignoring the gravity component altogether), which means its radius should increase roughly by 25%.

These two effects are really opposite (when the radiative surface increases, the radiated energy per unit surface must go down, so the temperature also needs to go down. So the final result will be a temperature of less than 11200 K - and a radius increase of less than 25%, but with back-of-the-envelope calculations I am unable to say how to split the difference).

The Dyson sphere, meanwhile, is intercepting more and more power, and will ultimately intercept what was, before, the full output of the star (it will still be 10% of the total output, but the total output has gone up by a factor of 10).

Slowly, the star will grow and become a strange blue-white demi-giant, while it will take hundreds of thousands of years for the extra heat to make its way to the star's core below the radiative layer.

When that happens, the solar core also will begin to swell (not so much actually), which will reduce its density. As a result, fusion events become less likely, until the power output will decrease, in a self-limiting process.

This means that the initial flare-up will be reabsorbed, and the star will grow larger, dimmer and redder, reducing the output being reflected back. This oscillation will be very slow (deca-millennial or slower) and will converge to a dimmer, longer-lived star with a reduced burn rate, but an artificially higher temperature and a spectrum similar to its previous one (it will of course fall outside the Hertzsprung-Russel diagram). I want to say that the star's life will be extended by the same factor of 10 that drives the backreflection, but of course I don't know.

enter image description here

If the back-reflection is raised slowly from 0% to the full 90%, slowly enough that the process never overshoots, then I expect that the star will slowly and linearly progress to the end state, as described by this paper [3.4] ("a sufficiently insulating Dyson sphere could put enough energy back into its central star that the star might expand, cool, and dim.").

I suspect that the final equilibrium will see a marginally swollen core and a whiteyellow-red sub-giant star, its temperature artificially higher, but with the life expectancy of a yellow main-sequence star or even longer.

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    $\begingroup$ I think we can safely assume that the star will swell to radius more than 1AU $\endgroup$
    – fraxinus
    Jan 14 at 7:42
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    $\begingroup$ Stars have negative heat capacity (this can be proven using the virial theorem). Therefore when energy is added to a star using a Dyson sphere, it will cool down rather than heating up. More information at this Physics.SE question: Explanation for negative specific heat capacities in stars? $\endgroup$
    – Thorondor
    Jan 14 at 8:31
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    $\begingroup$ @N.Virgo The question only specifies 90% efficient reflector. The surface will lose temperature (and lose a lot of it, because it so huge area), and thus layers just under it will be cooler than layers near the core. $\endgroup$
    – hyde
    Jan 14 at 10:16
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    $\begingroup$ @Thorondor that holds if the core increases its density and contracts, which directly increases the gravitational potential energy. But if thermal energy is added from the outside, the gravitational potential energy is not directly altered. $\endgroup$
    – LSerni
    Jan 14 at 14:07
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    $\begingroup$ @LSerni The virial theorem works both ways: if thermal energy is added from the outside, the star decreases its density and expands. Here's a more specific source (which also has a lot of other interesting information about Dyson spheres): arxiv.org/pdf/2006.16734.pdf $\endgroup$
    – Thorondor
    Jan 15 at 2:27
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This was recently studied and modeled. Link: here.

Excerpt from the Conclusion below, but TLDR is: reflecting a lot back onto a small star makes it bigger, cooler, and hence live longer. Big stars are unaffected.

Irradiated stars expand and cool. A Dyson sphere may send a fraction of a star's light back toward it, either by direct reflection or thermal reemission. This returning energy can be effectively transported through convective zones but not radiative zones. So, it can have strong impacts on low-mass main-sequence stars with deep convective zones which extend to the surface. It causes them to expand and cool, slowing fusion and increasing main-sequence lifetimes. For higher-mass stars with little to no convective exterior, the returned energy cannot penetrate far into the star and therefore has little effect on the star's structure and evolution, besides some surface heating.

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    $\begingroup$ That's incredible -- the final publishing was the same day as the Stack Overflow post! $\endgroup$
    – LSerni
    Jan 18 at 12:03
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A star is a gas ball, It will expand

What will happen to a star being heated? Sane thing as to any star / ideal gas that has increased temperature. It will expand until new equilibrium is met.

The time scale will be just a matter of how fast energy transfers between the outer star components. Of concern would be how fast will the non reflected energy heat up the Dyson sphere, and cause the sphere to fail.

As to the rest of the system: It will warm up until a failure mode is reached. Few structures can exist at photosphere temperatures.

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You might extinguish the star.

https://www.astronomy.ohio-state.edu/ryden.1/ast162_4/notes14.html

How does a star's natural thermostat work? Consider what would happen if you increased the fusion rate in a star's core:

-(1) Core temperature increases

-(2) Core pressure increase

-(3) Core expands

-(4) Core density & temperature decrease

-(5) Fusion rate decreases

In the OP scenario, the star is heated by the sphere. The star will expand and become less dense. Less density means decreased fusion. At a low enough density fusion cannot be sustained. The gas cloud will remain hot and the sphere could keep it hot for a long time.

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    $\begingroup$ that is called a supernova, the star cools down after it has expended all the fuel in its core. blue stars bun through their fuel faster but they don't cool off until they have used it all up. supernova happen because a star has used up all its fuel and starts to collapse drastically increasing temperature and pressure. the core is expanding in this scenario but the percentage of the star that counts as core also drastically expands. $\endgroup$
    – John
    Jan 13 at 23:01
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    $\begingroup$ @John - how is a supernova cooling down and collapsing in any way like the star in the OP being heated up and expanding? They seem to me diametric opposites. $\endgroup$
    – Willk
    Jan 13 at 23:24
  • $\begingroup$ then you may ant to familiarize yourself with how supernova work. seem my answer. $\endgroup$
    – John
    Jan 14 at 0:16
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    $\begingroup$ @John This star hasn't spent its fusion fuel, so it can't go supernova. $\endgroup$
    – hyde
    Jan 14 at 8:55
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    $\begingroup$ @john Nope. You are heating the core a bit, expanding it, decreasing but not stopping the rate of fusion. $\endgroup$
    – PcMan
    Jan 16 at 20:16
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The star will burn hotter, faster until the Dyson sphere essentially IS a star.

For this I will assume the Dyson sphere is sufficiently advanced magic so it can reflect all or most the stars light and particles back and not explode in the a cloud of gas. All or most has essentially the same effect it just changes the degree.

So here is how a star works, in the core the heat and pressure is high enough to sustain a fusion reaction but only in the core. that is what is supplying the heat so a stars temperature is controlled by its size, it is a balance between the light pushing out and gravity pushing in.

But in your star the star is the same size but way hotter, the interior temperature quickly (millions of years, which is quickly for a star) rises until far more of the stars mass can undergo fusion. this is because there is nowhere else for energy to go, photons just bounce back at the star until all their energy is absorbed. So your star is essentially ALL core, this will burn through its fuel astoundingly fast, your star may only last a few million years before it burns out. but all this heat means photon pressure also makes your star expand, the math is tricky but likely until the Dyson sphere essentially IS the outer surface of the star.

Normally something like this only happens to a lesser degree in a supernova, when a star runs out of fuel in its core and collapses, this drastic rise in pressure causes much of the rest of the stars mass (which is mostly stull fusible material) to suddenly undergoes fusion, which produces so much heat and light it blows the star apart, but your star can't explode, the outward pressure is not just countered by gravity but also the reflected light and even reflected particles. This photon pressure will be less than the pressure pushing out but not that much less when you add gravity. You have created a stellar pressure cooker.

This is where the amount of reflectivity matters, at a 100% physics breaks down, it is a closed system and sooner or latter all the mass inside gets converted into energy. If it is anything less than 100% the Dyson sphere temperatures rises until it reaches equilibrium. Until it is emitting as much heat as it is absorbing, which means it ranges from hotter than the hottest star to merely as hot as the surface of a blue star depending on reflectivity.

Eventually your star runs out of fuel, collapsing, but unlike a normal star it may not have enough fuel left to go supernova, so it may just quietly burndown into a a white dwarf or neutron star* quietly emitting low energy light for the rest of the lifetime of the universe.

  • it may turn into a neutron star because although normally a star that size would not form a neutron star, it also blows away most of its mass in the supernova stage, this star does not, ALL of the remaining mass collapses inwards which may be enough.

Now if your civilization is clever they build another bigger Dyson sphere around it which will harvest way more energy, albeit for a much shorter amount of time.

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Mathematical stellar constructs only exist in mathematics

(NOTE: as indicated in the comments, this is a comment on your design rather than an answer. I put this because your question carries a science-based tag.. the construct you propose cannot exist)

In a static Dyson sphere, very little light will be reflected onto the star itself

A Dyson sphere has a radius much larger than the star's radius. Rays reflecting back into the star perfectly only exist in mathematics, with a perfectly centered, perfect sphere. A Dyson won't have these properties.

Any imperfection in the Dyson's shape will cause rays to go the wrong way and bounce around, loosing 10% at every bounce, eventually releasing 100% of the energy onto your mirrors. That will be the case for a significant portion, when the orbit covered by the Dyson becomes only slightly decentric, the focus point will shift away from the star.

Agree with Alexander's comment it will get very hot in these mirrors, you'll need a very special material.

You'll need precise surface and orbit control

Imagine a planet-like superstructure in orbit around the sun, which is able to spawn a Webb-quality parabolic mirror with a diameter of about 0.5 AE, in orbit around the sun. You could focus that mirror somewhere, e.g. to do your experiment ("can I blow up my star") or on a moon, to harvest the energy, or onto a far away planet, to raise its temperature. But you'd need propulsion for that, it should be aligned precisely. You can't do these things without a dynamic orbit correction system.

What if a Dyson swarm would be built with Webb's mirrors ?

Say the construction would be done and you have the resources and personnel, to blow up your sun safely. You'd use a Dyson swarm with Webb quality servo-driven individual mirror segments and very precise control, hanging around the sun, with a diameter of 2AE (hand waive hand waive)

The control can compensate for decentering and shape errors. It has to be dynamic. There's no single vertical angle either.. Your Dyson was built as a rotating swarm, so it will flatten out, while it rotates. In any case, the mirror angles are to be adjusted, to compensate for that, you don't have to move the surface, only adjust the mirror angles. All kinds of compensations are needed. The swarm gets deformed, when gas giants orbits are near. Suppose everything can be solved..

You'd get a heat feedback loop.. you may not need to maintain it for long

I could proceed an attempt to actually answer this question now.. say you would have a sun that gets a relevant part of its own energy back. Ok what would happen.. you have a certain helium/hydrogen balance in your sun, that balance will shift, more hydrogen will be burnt by your star and it will get hotter. How long would it take, before the star is "burnt out". In fact you invoke a sliding slope, a feedback loop, because the mirrors keep working. The star is getting hotter and hotter. Inside, the helium is used to build heavier atoms. I can't provide the formulas, but it could be happening quite fast. You'll blow up your sun.

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    $\begingroup$ The question specifies that a fictional Dyson sphere reflects the light back on our sun, and your answer is, No it doesn't? $\endgroup$
    – D.J. Klomp
    Jan 13 at 20:29
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    $\begingroup$ Ugh, you can always cover the interior of the sphere (or whatever surface) in small cube-corner mirrors. This way all the rays of light will be reflected back parallel to themselves. No need for perfect spheres etc. -- the envelope can have any shape. $\endgroup$
    – AlexP
    Jan 13 at 20:37
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    $\begingroup$ if you build a dyson sphere, you necessarily need active attitude control, so we can take that as a given. $\endgroup$
    – ths
    Jan 13 at 21:32
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    $\begingroup$ @D.J.Klomp It says it reflects light at the sun, it does not specify that 100% of the light hits the sun. I can throw a rock at target, but that does not mean I will hit. What Goodie is saying is that yes, you can reflect light back at the star, but a significant quantity of that light likely to miss. $\endgroup$
    – Nosajimiki
    Jan 13 at 22:06
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    $\begingroup$ I like how this answer assumes that building a Dyson sphere is perfectly plausible... but asserts that aiming a mirror to reflect light back at the sun is scientifically impossible. $\endgroup$
    – NPSF3000
    Jan 14 at 6:09
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So, after some quite careful thinking, this is what I come up with. It's a bit boring, but it is what it is.

The luminosity of the star inside the Dyson sphere will increase 10 times, by becoming bigger and hotter on the surface. Thus 10 times more energy will hit the Dyson sphere, of which 10% escape, resulting in the same outside luminosity as the sun has now, matching the almost unchanging energy output in the star's core, as required by the star's mass to keep gravitational collapse at bay.

The temperature of the outside of the Dyson sphere will be approximately the natural temperature at 1 AU, which is 0°F or -18°C in our solar system currently (Earth surface is warmer only because of the atmosphere), so its luminosity will be in quite cool infrared wavelengths. That's still 200+ Kelvins above the galactic background of course, so should be very visible as a very curious object from far away with an infrared telescope.

Inside the sphere, the outer layers of the star would do a lot of what happens in a red giant. That is, energy the star radiates must increase, which happens by two methods: increased surface area or increased temperature. Normally a sun mass star will cool when this happens (that's why they become red), but in this case the extra energy is coming from outside, so I'll make an assumption that the surface temperature actually stays about the same. So only surface area increases, by increasing radius.

According to this Wikipedia page, The luminosity of Sun-mass star going through its red giant phase can go from 2.2x to 2802x the original luminosity, with corresponding radius increase from 2x to 179x. So, the 10x luminosity needed for us is in the very low end of that range. Doing a quick calculation, to increase the surface area of a sphere by 10x, we need to increase radius by 3.14x. So here we see how it doesn't much matter if the temperature stays the same, because if it becomes hotter, the star will be a bit smaller, and if it becomes cooler, the star will be a bit larger, but still small enough to fit inside the sphere. For comparison, 1 Au is about 217 solar radiuses.

Additionally, if the Dyson sphere is not a near-perfect sphere, most of the radiation reflected back will actually miss the star in the middle (it's still so very small, radius 3x the original vs Dyson sphere radius is over 200x). Accounting for that the outer layers of the star would actually heat much less, when each extra reflection loses 10% of the energy. But maybe we are to assume, that the 10% loss actually accounts for this effect too.

So... Boring result is, the star inside the Dyson Sphere will just expand its outer layers when those are heated by the reflected radiation, until they are big enough to radiate away both the normal energy output of the star, and the extra reflected energy. In a sun-mass star this does not change the evolution of the star significantly, as far as its core is concerned. As discussed in the comments, the pressure at the center of the star may get less as the star expands, but the expanded portion is basically the atmosphere of the star and very low density, so only a tiny portion of the total mass of the star, so this does not matter. There is always room for the outer layers to expand as much as is needed, and the core doesn't really care, it'll just keep doing fusion to resist gravity.

Of course any planets left between the Dyson sphere and the star would get quite baked, and any star ships traveling inside the sphere would need to be able to travel that close to a luminous star, which many star ships might not be equipped to do, unless the ship mass does not matter and they can have as much heat shielding and radiators as needed.

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    $\begingroup$ Hotter star makes for larger star. Larger star makes for less gravitational pressure on the core. Less pressure on the core = less dense core = slower rate of fusion. $\endgroup$
    – PcMan
    Jan 16 at 20:15
  • $\begingroup$ @PcMan I don't see that, intuitively. Forces are symmetric, you can't "lift" the pressure of the atmosphere without an equal downward force, keeping the internal pressure the same. And shape shouldn't really matter for pressure at the core, only mass above (=all of it). It's a bit like saying if you heat up Earth's atmosphere (for example via increased solar flux) so it expands a bit (while keeping it's mass the same or accounting for it), the pressure at the bottom of Mariana Trench will get less, because atmosphere expanded. Doesn't seem right to me. $\endgroup$
    – hyde
    Jan 17 at 5:44
  • $\begingroup$ @PcMan Or, if you created a lead ball with same mass as Jupiter, that giant lead ball would have higher pressure at the center, doesn't sound right. Note that here the heat comes from outside. If anything, the radiation pressure of the reflected energy from the Dyson mirror here will increase the pressure a bit, requiring a higher rate of fusion to resist it (but I assume this extra radiation pressure to be negligible). $\endgroup$
    – hyde
    Jan 17 at 5:48
  • $\begingroup$ @hydra sure you can lift the pressure on the core by lifting the atmosphere high enough. Simply because part of that atmosphere is now further away from the gravitation of the core. It may mass the same as before, but it weighs a lot less, thus imposes much less pressure... You are used to thinking on Earth, where the top and bottom of the atmosphere are within 0.5% of the same distance from the core, and thus gravitational difference between top and bottom are (almost) negligible. For a gasball like a star, this is very different $\endgroup$
    – PcMan
    Jan 17 at 5:48
  • $\begingroup$ @PcMan But the area the pressure is applied to also larger, which should result in equal compressive force. But maybe this a question for physics.stackexchange.com or something, "Does temperature, and thus size, of a big ball of hydrogen in gravitational equilibrium, affect the pressure at its center?" $\endgroup$
    – hyde
    Jan 17 at 5:52
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We have a pretty much natural example of what happens if a star is forced to radiate 10 times more heat. It is called

Red giant phase.

The star surface has to radiate more. But it cannot get any hotter, because the surface gravity and the surface temperature of a star are somewhat connected. So the star swells. The surface gravity goes down, so does the surface temperature. The star swells some more and at some point settles.

The core loses some of the pressure and some of the nuclear reaction rate, because much of the envelope is now far away and attracted less. But this effect is minor because most of the stellar mass is in the core anyway and it contributes to the pressure much more.

p.s. and @Goodies is right, your mirror will have to be impressively precise and even then a great deal of the light will be reflected not to the star, but to another part of the mirror instead. At least, until your star swells enough.

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    $\begingroup$ Exactly. Except, unlike with the Red giant, the core has by no means run out of hydrogen to fuse, thus preventing the massive core pressures and densities that allow helium (and later on carbon) fusion. It becomes a red giant, and stays there for a very very long time. Longer than the original star's lifetime would have been. $\endgroup$
    – PcMan
    Jan 16 at 20:18
  • $\begingroup$ @PcMan agree. This could be called "responsible stellar energy harvesting". $\endgroup$
    – fraxinus
    Jan 17 at 8:47

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