11
$\begingroup$

We know that we can focus radiation of sun and can burn a paper. If we think of this thought experiment, will that happen? If someone constructs a concave mirror on the sun and concentrate radiation of other part of sun to a point, will that point be hotter that outside (everything is on the sun)? I guess this can violate some laws of energy transfer or thermodynamics, but I am not sure. Any ideas?

Isn't this something like a machine is running, I am getting work out of it and supplying back to the machine to 'accelerate' it? I am not expert in physics, but intuitively thinks that this may not be possible.

Improve this question
$\endgroup$
  • 3
    $\begingroup$ Note that this can be recast as "If I wrap insulation around a heat source, will it get hotter than if I don't?". $\endgroup$ – keshlam Nov 16 '14 at 17:00
  • $\begingroup$ @keshlam But a star can change in size and can also change it's internal energy generation rate. You are over-simplifying the problem. $\endgroup$ – ProfRob Nov 16 '14 at 22:08
  • $\begingroup$ You cant concentrate infrared radiation in a room of constant temperature to heat up one portion (thermo 2nd law). HOWEVER, the sun's surface is NOT in a constant temperature environment, it gets heat from hotter regions below. The mirror will make it hotter, just like aliexpress.com/item/… $\endgroup$ – Kevin Kostlan Nov 16 '14 at 23:08
  • 1
    $\begingroup$ @RobJeffries: Yes, I'm simplifying it. However, I think Kevin's point is the same one I'm making. Independent of all the other things a star can do, if heat is being generated and can't escape it must become hotter than if heat is generated and allowed to escape. Longterm that additional heat might be enough to cause expansion and reduce the surface temperature again, or to change the sequence the star follows, or whatever... but on human timescales, the region would get hotter thru retaining heat in exactly the same way that it would get hotter if we pumped additional heat into it. $\endgroup$ – keshlam Nov 16 '14 at 23:17
  • $\begingroup$ @Keshlam. On "human timescales" the heat would be distributed throughout the Sun's convection zone and there would be negligible change in the surface temperature. The mirror would need to be in place for $>10^5$ years to see the surface heat up. $\endgroup$ – ProfRob Nov 16 '14 at 23:36
11
$\begingroup$

If it was possible to reflect the energy back at the sun, yes, the location where the energy strikes will become hotter. If fact, if you could insulate the sun from radiating energy, then the sun would get even hotter.

Improve this answer
$\endgroup$
  • $\begingroup$ This turns out to be true (if you keep the mirror in place for a long time), but is not at all obvious. In fact if the Sun had a deeper convection zone it would not be true. You need to argue why the excess luminosity is not reradiated by an increase in radius. $\endgroup$ – ProfRob Nov 16 '14 at 9:00
  • 1
    $\begingroup$ I disagree about it being not obvious. It's trivially true with conservation of energy. The sun is a big ball of matter that gradually releases energy trapped within matter via a self regulating process governed by its energy density, which is ultimately governed by the rate at which it radiates energy. If you could somehow meaningfully reduce the rate at which energy radiated, it would heat up and find a new equilibrium at a higher temperature, since the energy it was producing would have nowhere to go (or rather, would have to wait longer to escape). $\endgroup$ – Wug Nov 16 '14 at 20:25
  • $\begingroup$ I'm having trouble fitting a response within the character limit, but I argue that for sunlike stars with a stable equilibrium, the stability of the equilibrium does guarantee that the decrease in energy loss will result in an equilibrium with a higher latent energy (which, for a star, would map directly to a higher average temperature). Parts of the star, such as the outer atmosphere, might be cooler, but the total energy trapped inside the star would be greater. This reasoning only applies if you don't raise the temperature enough to lose equilibrium, in which case the sun would come apart. $\endgroup$ – Wug Nov 16 '14 at 22:20
  • $\begingroup$ @Wug No, it really isn't obvious. Another solution for equilibrium is to be found by increasing the radius and decreasing the core temperature (by the virial theorem). The intrinsic luminosity falls whilst the unheated surface stays at the same temperature. This is what would happen in a star with a thicker convection zone. As I say, not obvious and certainly not trivial. $\endgroup$ – ProfRob Nov 16 '14 at 22:21
  • $\begingroup$ But if you do that, you won't be in equilibrium because the lower energy output of the core will fail to heat the star, which will cool and contract. $\endgroup$ – Wug Nov 16 '14 at 22:23
11
$\begingroup$

LDC3 and Kitchi addressed your main question, but I'd like to comment on your second paragraph.

Isn't this something like a machine is running, I am getting work out of it and supplying back to the machine to 'accelerate' it? I am not expert in physics, but intuitively thinks that this may not be possible.

Actually, we do this all the time! Electricity generators in power plants, for example, need to consume some electricity to work (this is called excitation). To start them, external power is needed, but once they're going we just feed some of the output power back on itself.

It doesn't violate any principle. Don't confuse output power with generated power. When there is feedback, for instance, we'll just have the former smaller than the latter, like: $$P_\mathrm{output}=P_\mathrm{generated}-P_\mathrm{feedback}$$

Improve this answer
$\endgroup$
  • $\begingroup$ Only wound rotor synchronous machines need excitation current to start. Granted, all generators at power stations are wound rotor synchronous machines, but most other electric machines are asynchronous or permanent magnet synchronous types which don't need external excitation. $\endgroup$ – ntoskrnl Nov 15 '14 at 19:06
  • $\begingroup$ @ntoskrnl that's what I had in mind, power plants. Thanks for pointing out, I'll amend it. $\endgroup$ – André Chalella Nov 15 '14 at 21:15
7
$\begingroup$

The route to the answer is somewhat anti-intuitive. By reflecting some of the Sun's energy back towards the sun at a point you are effectively reducing the flux of energy that can emerge from the photosphere and escape.

The global effect of this on the Sun must be similar to that of blocking the flux at the photosphere - in other words, similar to the effects of sunspots. The local effect on the temperature structure will of course be completely different, because sunspots are places where the photospheric temperature is much (1000 K) cooler than the unspotted photosphere. Here, you would be creating a hotspot, nevertheless, the flux emerging from the surface and escaping to infinity would be lower than it would for a star of the same radius and effective temperature where there was no mirror.

The local effects really would be quite local. Convective energy transfer is very effective just below the photosphere, so the excess energy is redistributed on a local convective turnover timescale (five minutes).

The global effects can be treated in a similar way to the effects of sunspots. The canonical paper on this is by Spruit & Weiss (1986). They show that the effects have a short term character and then a long term nature. The division point is the thermal timescale of the convective envelope, which is of order $10^{5}$ years for the Sun.

On short timescales the nuclear luminosity of the Sun is unchanged, there will be an additive effect due to the hot spot on the surface, but the stellar structure remains the same as does the surface temperature. As about half the flux from the hotspot goes into the Sun and only half goes into space, the net luminosity at infinity (after subtracting that blocked by the mirror) will be lower, whilst the flux at the mirror will increase.

On longer timescales, the luminosity will tend to stay the same because the nuclear burning core is unaffected by what is going on in the thin convective envelope. However, roughly half the flux reflected by the mirror can't escape from the star. To lose the same luminosity it turns out that the radius increases and the photspheric area not affected by the reflected beam (the "unspotted region") gets a little hotter. In this case, the radius squared times the photospheric temperature will increase to make sure that the luminosity observed beyond the mirror stays the same - i.e. by $R^2T^4(1 - \beta) = R_{\odot}^2 T_{\odot}^4$, where $\beta$ is the fraction of the solar luminosity intercepted by the mirror.

The calculations of Spruit et al. (1986) indicate that for $\beta=0.1$ the surface temperature increases by just 1.4% whilst the radius increases by 2%. Thus $R^2 T^4$ is increased by a factor 1.09. This is not quite $(1-\beta)^{-1}$ because the luminosity does drop slightly.

So yes, if you keep the mirror there for longer than $10^5$ years you will increase the temperature of the Sun, but perhaps not by as much as you would have thought.

Further edit:

The above discussion is true for the Sun because it has a very thin convection zone and the conditions in the core are not very affected by conditions at the surface. As the convection zone thickens (for example in a main sequence star of lower mass), the response is different. The increase in radius becomes more pronounced; to maintain hydrostatic equilibrium the core temperature decreases and hence so does the nuclear energy generation. The luminosity of the star falls and the surface temperature stays roughly the same.

This is why I have made comments on other answers here, because although they correctly state that the Sun will get hotter, it is not obvious that this should be so and indeed would not be so for a lower mass star.

Improve this answer
$\endgroup$
3
$\begingroup$

There won't be a violation of thermodynamics because you are not creating energy from nothing. The total energy of the system is still conserved, it is just fed back into the system.

Here's what will probably happen - The concave mirror will not be perfectly reflecting, so will reflect something like $99\%$ of the incident energy. This energy (although very large) is not enough to ignite fusion in the sun.

The point at which the beam is focused will be between the corona (where the gas is very sparse) and the core (where the gas is dense), and will heat up particles in that vicinity to slightly more than their local surrounding temperature.

The small pocket of gas that gets heated by the mirror will be slightly hotter, and hence slightly less dense than the surrounding, and this will set off a convenction loop. The sun already has a convection zone so placing a mirror won't really do much to it.

Improve this answer
$\endgroup$
  • $\begingroup$ Certainly no fusion. If the mirror reflects 99% then you'll establish a new equilibrium where the sun radiates 100x the energy, 99% is reflected back and 1% is absorbed in the mirror. Radiant energy goes up at the 4th power of temperature so the temperature must rise the 4th root of 100x to balance it--3.16x the temperature, 18,271K. You'll get a UV-color star. $\endgroup$ – Loren Pechtel Nov 16 '14 at 0:19
  • 1
    $\begingroup$ Gregory Benford and Larry Niven have a sci-fi book called 'Bowl of Heaven' in which a bowl shaped structure is pulled along behind a star. The bowl itself has a significant portion mirrored which is reflected back at the star and causes that spot to heat up and flare. The consistent asymmetric flares on the star then propel the entire star and bowl system. (article on the idea and physics behind it). $\endgroup$ – user20936 Nov 16 '14 at 4:10
  • $\begingroup$ Right! BTW, practical mirrors (so far designed) have reflectivity only around 85% or something. Thermodynamics always holds, that the efficiency is never 100% :) $\endgroup$ – Waffle's Crazy Peanut Nov 16 '14 at 7:30
  • $\begingroup$ @Loren Petchel That is not what would happen. In the case of such a large perturbation the Sun would expand, the nuclear burning would decrease and it is even possible that the surface would become cooler (more luminosity does not equal hotter surface). $\endgroup$ – ProfRob Nov 16 '14 at 9:05
  • $\begingroup$ @LorenPechtel - Sorry, I should have made it clear. In my answer, I'm considering this hypothetical mirror to be a fraction of the size of the sun. i.e., only reflecting a few % (at most) of the radiation emitted. $\endgroup$ – Kitchi Nov 16 '14 at 10:38
1
$\begingroup$

Yes. Instead of allowing the energy to radiate into space, you're containing it and sending it back to the source. Think fire in a room vs. fire outside: outside the heat gets lost to the environment but in a room it stays and warms the room to a much higher temperature than the fire outside warmed the surrounding air.

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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