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What would happen if the entire inner surface of a Dyson sphere (not neccesarilly a solid shell, maybe a very dense swarm) would be coated into a highly reflective material?

Let's say it can reflect 99.9% at all wavelenghts.

Would it cook the star in its own light?

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    $\begingroup$ This sounds like something there should be an xkcd for, but I can't find it... oh, nevermind, found it, in the blag $\endgroup$ – Baldrickk Jun 6 '17 at 16:19
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    $\begingroup$ and the what-if that was staring me in the face $\endgroup$ – Baldrickk Jun 6 '17 at 16:24
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    $\begingroup$ The fun bit about that XKCD blog and what-if, @Baldrickk is that there's momentum applied to the sun. I recall somewhere (oh, wikipedia) that by restricting the sun's energy output in such a manner you can impart a fairly significant amount of thrust on the entire solar system and turn it into a space ship. Mind, it won't get anywhere interesting any time soon (total delta-v is around 1 meter/second per 50,000 years). $\endgroup$ – Draco18s Jun 6 '17 at 18:36
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    $\begingroup$ We are assuming that the mirrors themselves stay static with regard to their orbital distance. I assume in the real world, the increasing radiation will push the mirrors outwards like giant solar sails (which they effectively are), maybe even to the point of reaching solar escape velocity.... $\endgroup$ – Thucydides Jun 6 '17 at 23:02
  • $\begingroup$ There is a second relevant what-if. Basically you can not raise the temperature beyond some limit - so the stars core temperature would stay the same. $\endgroup$ – bdecaf Jun 7 '17 at 6:11
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Building on L.Dutch's answer:

  • Even if mirrors are very efficient and reflect 99.9% there still remains that residual .1% of energy they absorb.
  • (almost) All energy is prevented from escaping and thus accumulates inside the sphere.
  • Temperature inside the sphere will rise (quasi) linearly.
  • Ditto for both radiation and solar wind.
  • It is not said if the solar wind is allowed to pass, is absorbed by mirrors or is reflected (how?) back.
  • In any case temperature of mirrors will rise and they will start to radiate (Black Body) emitting energy proportional to the fourth power of Temperature (K) (Stefan-Boltzmann law).
  • If mirrors are sturdy enough to withstand temperature and pressure (solar wind) they will come to an equilibrium with power radiated by mirrors equaling the power produced by the star.
  • Temperature of equilibrium will be lower for larger spheres and thus the "color" of the sphere will change (Wien's law).
  • In general the amount of radiation within the sphere, at equilibrium, will be about 1000 times the "normal" radiation (under your assumption a photon has to bounce, on average, 1000 times before getting its fair chance to be absorbed).
  • In extreme cases (small, very sturdy sphere) increment in temperature in the star may be enough to ignite higher level nuclear reactions without need of the normal Exhaust Fuel -> Gravity Contract -> Heat Up -> Ignite "next" Fuel cycle. In that case a Very Anomalous Supernova may result (I doubt your mirrors will withstand that)
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    $\begingroup$ "I doubt your mirrors will withstand that", indeed. $\endgroup$ – r41n Jun 6 '17 at 13:39
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    $\begingroup$ But what if they withstand? $\endgroup$ – Drag and Drop Jun 6 '17 at 14:10
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    $\begingroup$ @DragandDrop: If they withstand you "simply" have a rise in internal temperature, the mirror array becomes brighter and a new (higher) equilibrium point between radiated energy and produced energy is reached. $\endgroup$ – ZioByte Jun 6 '17 at 15:55
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    $\begingroup$ This is actually a plot element in Alastair Reynolds' House of Suns. In the far distant future some human descendants use collections of mirrored ringworlds (termed "stardams") to shield less advanced worlds from nearby suns which are going nova. The humans didn't create the ringworlds; their technology isn't capable of that. They simply repurpose them by collecting them and marshalling them around stars soon to go nova. $\endgroup$ – Necoras Jun 6 '17 at 16:22
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    $\begingroup$ Would the star expand due to higher temperatures and pressures? $\endgroup$ – Tracy Cramer Jun 6 '17 at 23:59
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Light emitted from the star would travel until the mirrors and then would be reflected back, bouncing back and forth. Due to enormous scale of the Dyson sphere, you can neglect cavity effects and related wavelenght selection.

This would build up energy into the sphere, which can only dissipate through the mirrors.

Basically such a configuration would act as a cosmic scale black body (mind I say black body, not black hole).

If your mirrors can withstand the energy contained inside the sphere, the ensemble will behave like a new star with the size of the Dyson sphere, emitting like a black body at the temperature of the mirror.

If your mirrors are not so sturdy they will simply evaporate and join the stellar wind.

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    $\begingroup$ "...join the stellar wind." +1 poetic $\endgroup$ – Pureferret Jun 6 '17 at 12:49
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    $\begingroup$ If one of the mirror faces allowed an opening, that would make for one hell of a laser!! $\endgroup$ – steverino Jun 6 '17 at 15:01
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    $\begingroup$ @steverino It wouldn't really be a laser as the light wouldn't be coherent. $\endgroup$ – JAB Jun 6 '17 at 15:52
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And now, to do away with pointless talk and actually do some science.

Theoretical body which absorbs all the light is called black body, theoretical body which doesn't absorb all the light but absorption efficiency doesn't depend on wavelength is called grey body. Object for which absorption depends on wavelength is called coloured body.

Curious property of grey bodies is that they not only absorb light less effectively than black bodies, they also emit light less effectively, assuming same temperature.

Energy emission per unit of surface for grey body is: $$ j = \epsilon * \sigma * T^4[\frac{W}{m^2}] $$ where $\epsilon$ is absorptivity/emissivity, $\sigma$ is Stefan-Boltzmann constant and $T$ is temperature in Kelvins. Square brackets contain dimensions.

Your theoretical sphere will emit: $$ P = 2*4\pi R^2\epsilon\sigma T^4 [W] $$ Where $R$ is radius of sphere. Notice factor of 2 at the start. That's because it will emit to the outside and to the inside (I deliberately wrote it as $2*4$ instead of just $8$).

Meanwhile, star emits: $$ P_s = 4\pi r^2\sigma t^4 [W] $$

Considering that entire shell is reflective, we can assume that reflected star light does not fall on other parts of the shell and instead returns to star to be fully absorbed (stars are with good approximation black bodies). However, internal emission of the shell with be into half-full spatial angle, thus we can't make such assumption.

Star will absorb back all the emitted light which falls on it, but shell will again absorb only $\epsilon$. Since from each infinitesimal part of the shell star obscures only part of the full angle we can see that star will absorb $\frac{\pi r^2}{2\pi R^2}=\frac{r^2}{2R^2}$ of total internal emission. This means that $\epsilon(1-\frac{r^2}{2R^2})$ will be absorbed by shell while $(1-\epsilon)(1-\frac{r^2}{2R^2})$ bounce again, thus star will again absorb $\frac{r^2}{2R^2}(1-\epsilon)(1-\frac{r^2}{2R^2})$. This looks like a geometric sequence with first term of $a=\epsilon(1-\frac{r^2}{2R^2})$ and multiplicative factor of $q=(1-\epsilon)(1-\frac{r^2}{2R^2})$. Since obviously $q<1$ sum of the sequence converges. Summing from 0 to infinity we get: $$ A_{shell}=\frac{\epsilon(1-\frac{r^2}{2R^2})}{1-(1-\epsilon)(1-\frac{r^2}{2R^2})} $$ Now we need to calculate same series for absorption by star and we will be able to calculate total fraction of internal emission absorbed by star to internal emission absorbed back by shell. This time we get $a=\frac{r^2}{2R^2}$ and $q=(1-\epsilon)(1-\frac{r^2}{2R^2})$, thus sum is: $$ A_{Star}=\frac{\frac{r^2}{2R^2}}{1-(1-\epsilon)(1-\frac{r^2}{2R^2})} $$

Since obviously all the internal emission has to be absorbed over course of infinite bounces, $A_{Shell}+A_{Star}=1$ has to be true. And indeed it is, verifying that no mistakes were made.

Thus, over infinite reflections of internal emission, shell will absorb back: $$ A_{Shell}*P=\frac{\epsilon(1-\frac{r^2}{2R^2})}{1-(1-\epsilon)(1-\frac{r^2}{2R^2})} 4\pi R^2\epsilon\sigma T^4 $$ While star will absorb: $$ A_{Star}*P=\frac{\frac{r^2}{2R^2}}{1-(1-\epsilon)(1-\frac{r^2}{2R^2})} 4\pi R^2\epsilon\sigma T^4 $$

Thus total power absorbed by shell will be: $$ A_{Shell}P+\epsilon P_s=\frac{\epsilon(1-\frac{r^2}{2R^2})}{1-(1-\epsilon)(1-\frac{r^2}{2R^2})} 4\pi R^2\epsilon\sigma T^4 + \epsilon 4\pi r^2\sigma t^4 $$ Which for equilibrium has to be equal to total emitted power: $$ P = 2*4\pi R^2\epsilon\sigma T^4 $$ Combining those equations we get T as a function of t,r,R and $\epsilon$: $$ T=t\sqrt{\frac{r}{R}}\sqrt[4]{\frac{1}{2-\frac{\epsilon(1-\frac{r^2}{2R^2})}{1-(1-\epsilon)(1-\frac{r^2}{2R^2})}}}=t\sqrt{\frac{r}{R}}\sqrt[4]{\frac{\epsilon (1-\frac{r^2}{2R^2})+\frac{r^2}{2R^2}}{\epsilon (1-\frac{r^2}{2R^2})+2\frac{r^2}{2R^2}}} $$

Unfortunately, for star it's more complicated. Simplified equilibrium requires that temperature raises enough so that total emission is equal to original star emission plus reflected starlight plus absorbed internal shell emission. In practice, it will increase temperature, increasing rate of fusion, which increases internal power generation, increasing temperature even further. I can not at this point make predictions on this. So I will continue with grossly oversimplified equilibrium conditions. Thus, in grossly oversimplified conditions, star temperature has to raise so that following are true:

$$ P'_s=P_s+P_s(1-\epsilon)+P_s(1-\epsilon)^2+...+A_{star}P=\frac{P_s}{\epsilon}+A_{star}P $$ Term $\frac{P_s}{\epsilon}$ represents infinite series of starlight bouncing from shell, being absorbed by star, emitted again, bounced, absorbed and so on.

Which after using expressions, using expression for T(t) and simplifying a bit: $$ t'^4=\frac{t^4}{\epsilon}+\frac{\frac{r^2}{2R^2}}{1-(1-\epsilon)(1-\frac{r^2}{2R^2})}\epsilon \frac{\epsilon (1-\frac{r^2}{2R^2})+\frac{r^2}{2R^2}}{\epsilon (1-\frac{r^2}{2R^2})+2\frac{r^2}{2R^2}} t^4 =\frac{t^4}{\epsilon}+ \frac{\frac{r^2}{2R^2}}{\epsilon (1-\frac{r^2}{2R^2})+2\frac{r^2}{2R^2}}\epsilon t^4=t^4(\frac{1}{\epsilon}+\frac{\frac{r^2}{2R^2}}{\epsilon (1-\frac{r^2}{2R^2})+2\frac{r^2}{2R^2}}) $$ Which means that simplified equilibrium temperature of star will be: $$ t'=t\sqrt[4]{\frac{1}{\epsilon}+\frac{\frac{r^2}{2R^2}}{\epsilon (1-\frac{r^2}{2R^2})+2\frac{r^2}{2R^2}}} $$ And final temperature of shell will be T'=T(t'): $$ T'=t'\sqrt{\frac{r}{R}}\sqrt[4]{\frac{\epsilon (1-\frac{r^2}{2R^2})+\frac{r^2}{2R^2}}{\epsilon (1-\frac{r^2}{2R^2})+2\frac{r^2}{2R^2}}}=t\sqrt[4]{\frac{1}{\epsilon}+\frac{\frac{r^2}{2R^2}}{\epsilon (1-\frac{r^2}{2R^2})+2\frac{r^2}{2R^2}}} \sqrt{\frac{r}{R}}\sqrt[4]{\frac{\epsilon (1-\frac{r^2}{2R^2})+\frac{r^2}{2R^2}}{\epsilon (1-\frac{r^2}{2R^2})+2\frac{r^2}{2R^2}}} $$

Now it's just a trivial matter of calculating unimportant details. Feel free to put in whatever values you want.

Obviously, you can calculate external emission of the shell to know how much power will that pseudo-star output. Simply use $P=4\pi\sigma\epsilon T'^4$.

EDIT:

Disclaimer: expression $\frac{r^2}{2R^2}$ comes from assumption that shell is significantly larger than star. If you want shell to be merely slightly larger, replace it with $\frac{r^2}{R^2}$

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  • $\begingroup$ You've got my up doot. $\endgroup$ – mtheorylord Jun 7 '17 at 22:59
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It is also conceivable that the mirrors are not absorbing the remaining 0.1% of the radiation, but are letting it through. In that case, the intensity of the radiation outside the sphere won't change, while the intensity inside the sphere will increase 1000-fold (assuming the mirrors are able to withstand it).

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    $\begingroup$ At equilibrium it will not matter. Internal radiation will rise till the amount of energy escaping equals the amount of energy produced. In your hypothetical case it would be direct radiation and the mirrors would remain cool, otherwise the energy would heat the mirrors til they radiate the same amount of energy. Spectrum would be completely different and transient would be different (radiation would increase linearly and not with T^^4), so, observing from a safe distance, You can tell what's actually happening. $\endgroup$ – ZioByte Jun 6 '17 at 16:03

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