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What would happen to the Sun if you would reflect, in whatever way, all the outgoing electromagnetic radiation (Solar winds can be neglected)?

Deschele Schilder
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This is the same as asking what would happen if the sun couldn't get rid of the heat it generates. In that case, the heat builds up, the sun's temperature goes up, and in response the sun expands a bit. This expansion causes the fusion reactions in the sun's core to slow down, which slows down the rate of heat generation. But if none of that heat can escape, the sun will get hotter and it will continue to expand, and the fusion rate in its core will continue to decrease. At some point in this process the sun goes blown up so much in size and the temperature in its core falls so far that fusion can no longer occur and the sun's power source is shut down.

niels nielsen
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Some facts (see also the picture below):

The circular core where fusion (the source of the emergent energy) takes place has a radius of about 175 000 kilometers. The radius of the whole Sun is about 700 000 kilometers. Because of the enormous pressure, the plasma density in the core is about 150 times the density of water. The temperature here is 15 000 000 Kelvin.

It takes tens of thousands of years for energy produced in the core to reach the photosphere, the outside of the Sun which we can see, and which has a depth of only 1000 kilometers. The temperature inside the photosphere is about 5700 Kelvin. The density here is about $3\times10^{-4}(\frac{kg}{m^3})$.

In the radiative zone, the density drops from $2\times 10^4 \frac{kg}{m^3}$ (about the density of gold) down to only $20(\frac{kg}{m^3})$ [less than the density of water, about $10^3 (\frac{kg}{m^3})$] from the bottom to the top of the radiative zone. The temperature falls from 7,000,000° Kelvin to about 2,000,000° Kelvin over the same distance, about 300 000 kilometers.

The remaining 225 000 km consists mainly out of the convective zone. Convective motions carry heat quite rapidly to the surface. The fluid expands and cools as it rises. At the visible surface, the temperature has dropped to 5,700 Kelvin and the density is only $2\frac{kg}{m^3}$ (about 1/10,000th the density of air at sea level).

Now, the total energy contained in the Sun, $E_{sun}=m_{sun}c^2=1,74\times 10^{47}(J)$. Its binding energy can be neglected. The total fusion potential is about $1,2\times 10^{44}(J)$, so it's safe to say that the Sun's mass stays the same during its lifetime.

Now, considering these facts, what would happen if the energy radiated away would stay inside the Sun? Picture it. The photosphere and convective zone would heat up, which makes these layers expand, obviously. Too tittle though to have any influence on the core's density. The non-radiated heat will reach the core after tens of thousands of years, which is practically in zero time compared to the Sun's lifetime. After the core has absorbed the energy it will heat up, but its temperature will rise insignificantly.

To sum it up: In practice, we can see no difference between the time fusion takes when the radiating heat would be reflected back, so nothing really changes though. When fusion processes have died out, the Sun will follow the typical scenario of a star with the Sun's mass.

enter image description here

Deschele Schilder
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  • From https://en.wikipedia.org/wiki/Gravitational_binding_energy "According to the virial theorem, the gravitational binding energy of a star is about two times its internal thermal energy in order for hydrostatic equilibrium to be maintained." Hydrogen fusion converts about 0.7% of the hydrogen mass to light, which is certainly negligible, and the Sun will only burn about half of its hydrogen. It currently loses a little more mass through light than it does through solar wind. However, it will lose roughly half its mass over its lifetime, mostly during the red giant phases. – PM 2Ring Aug 12 '20 at 07:43
  • I meant the binding energy can be neglected when compared to the Sun's total energy ($E=M_{sun}c^2). Also, until fusion ceases to occur the energy loss of the Sun is negligible in comparison to that energy (i.e. the Sun's mass doesn't decrease significantly during this periob (before the process of becoming a red giant). – Deschele Schilder Aug 12 '20 at 09:08
  • This isn't correct. If the star is allowed to come into a new equilibrium with a larger radius, then the virial theorem tells you that the interior will be cooler. i.e. The core temperature falls on timescales longer than the thermal timescale of the envelope. The photosphere also becomes cooler as the star expands, since $R^2 T^4$ is lower because of the lower energy generation at the core. – ProfRob Aug 12 '20 at 16:14
  • @RobJeffries This is more than compensated for by the distribution of densities. – Deschele Schilder Aug 12 '20 at 17:24
  • And even if the luminosity were to stay the same, it is obvious that the photosphere must cool if it expands. – ProfRob Aug 12 '20 at 17:35
  • @RobJeffries The potential fusion energy will be completely used before the Sun starts to behave like a star comparable to the Sun in the Russel-Hertzsprung diagram. Of course, the photosphere expands, lust like part of the convective zone, but not enough to have any influence whatsoever on the Sun's core. So all fusion that can take place does take place, in the same timespan as the Sun radiating the produced energy away. – Deschele Schilder Aug 12 '20 at 18:41
  • @RobJeffries Do you really think that the if the relatively low value of the emitted energy during the Sun's lifetime will change anything? – Deschele Schilder Aug 12 '20 at 18:46