While reading "Physics of Solar Cells" by Würfel, I came across an amusing statement:

An interesting aspect arises if we had to pay for solar energy, but could also get a refund for energy returned to the sun (which would not be unjustified, since it would actually prolong the sun's lifetime).

How exactly does returning energy to the Sun prolong its lifetime? Why does the returned energy somehow "conserve" the Sun's fuel? Shouldn't it actually slightly increase the mass of the Sun and thus cause fusion at its core to happen faster, thereby decreasing the lifetime of the Sun?

Furthermore, I don't see any reason to expect this effect is specific to solar cells; any surface will reflect some light back to the Sun. Given that efficient solar cells actually have very high absorptivity, then surely any surface underneath it would have radiated more energy to the Sun were it not covered by the solar cell, and thus solar cells on the Earth's surface cause an implicit decrease in the Sun's lifespan? In this case, only highly reflective surfaces such as mirrors would work.

Finally, for fun, what order of magnitude life extension can we expect from this mechanism, assuming the Earth perfectly reflects the approximately $\mathrm{170\ PW}$ it receives back to the Sun, as if it were an ideal plane mirror?

  • $\begingroup$ For some funny contrast, here's a satirical advertisement against the usage of solar energy. $\endgroup$ Jun 23, 2015 at 16:40
  • $\begingroup$ Re How exactly does returning energy to the Sun prolong its lifetime? It doesn't. It shortens the Sun's life. By an immeasurably small amount. This was yet another dumb parenthetical remark in a book. $\endgroup$ Jun 23, 2015 at 18:10
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    $\begingroup$ @DavidHammen Not so fast. Returning energy into the top of the photosphere changes the energy balance in a similar way to sunspots. This probably/possibly does extend the life of the Sun. I have looked at this problem before and may provide a full answer later. $\endgroup$
    – ProfRob
    Jun 23, 2015 at 19:21
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    $\begingroup$ I am with Rob on this one. Raising the temperature of the sun will lower the density of the core and reduce total power output. If it wasn't like this, then there would be a positive feedback and sun-like main sequence stars would go supernova. Having said that, solar panels do not necessarily return energy to the sun. One would have to look very carefully at the effective albedo of Earth due to solar panels. $\endgroup$
    – CuriousOne
    Jun 23, 2015 at 21:09
  • $\begingroup$ @CuriousOne That's an interesting argument which I never considered. However, I recall reading that the fusion rate in the core of stars is very sensitive to temperature, and Wikipedia states that fusion power generated by the triple-alpha process goes as $\sim T^{40}$. Does that mean the solar core density is dependent on temperature to an even stronger power than that? That's quite impressive in and of itself. $\endgroup$ Jun 23, 2015 at 21:31

1 Answer 1


Reflecting energy back into the photosphere using mirrors is equivalent to asking what happens if we artificially increase the opacity of the photosphere - akin to covering the star with large starspots - because by reflecting energy back, you are limiting how much (net) flux can actually escape from the photosphere

The phenomenon could be treated in a similar way to the effects of starspots. The canonical paper on this is by Spruit & Weiss (1986). They show that the effects, for a star like the Sun that has a relatively thin convective envelope on top of a large radiative interior, 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, the stellar structure remains the same as does the surface temperature. As a small fraction of the flux from the the Sun is returned, the net luminosity at infinity will be decreased.

On longer timescales, in a star like the Sun, the luminosity will tend to stay the same because the nuclear burning core is almost unaffected by what is going on in the thin convective envelope. To lose the same luminosity it turns out that the radius increases and the photosphere gets a little hotter. In this case, the radius squared times the photospheric temperature to the fourth power will increase to make sure that the luminosity at infinity 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 reflected back into the photosphere.

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 core temperature and luminosity do drop slightly in response to the increased radius.

So to first order, the energy generation rate in the Sun stays almost the same and as the amount of fuel is invariant, then the lifetime stays the same. To second order, reflecting 10% of the energy back will end up in a $\sim 1\%$ reduction in energy generation and thus a 1% increase in lifetime.

You ask what happens if $1.7\times 10^{17}$ W is reflected back into the Sun? Well this is only $4\times10^{-10} L_{\odot}$. So by interpolation this will increase the lifetime by a fraction of $\sim 4\times 10^{-11}$, corresponding to an increase in the main sequence lifetime of less than 1 year!

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    $\begingroup$ So the claim in the book turns out to be true, with quite a bit more nuance behind it than I expected. Thanks for the answer, it was interesting, precise, and simple to follow! $\endgroup$ Jun 24, 2015 at 20:51

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