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I was surprised when I first heard that the energy produced at the Sun's core takes a long time to escape the Sun. The process is often explained as a photon traveling a "drunkard's walk" based on the mean free path within the layers of the Sun.

One such NASA estimate is linked here. It mentions estimates ranging from 4,000 years to several million years.

It occurred to me that maybe the diffusion time could better be estimated by dividing the total thermal energy of the Sun by the power output of the Sun. One estimate of the thermal energy of the Sun is 3.09 x 10^41 J given in the accepted answer of this question. On Wikipedia, I found that the Sun's luminosity is 3.828 x 10^26 W.

Dividing, we get a time of 8.07 x 10^14 seconds, or about 26 million years. I guess this is how long it would take the Sun to cool completely if it stopped producing energy, but kept radiating at its current rate.

Does my calculation make sense? 26 million years is above the usual estimates of energy diffusion times.

EDIT BELOW:

I want to explain why I thought that the simple division of thermal energy by luminosity would give a reasonable estimate of the energy diffusion time.

Assume that each bit of energy produced in the Sun's core travels radially outward at a steady pace until reaching the surface. I think it's clear that, given the inputs above, each bit of energy would require 26 million years to escape.

Now, due to the "random walk" nature of the diffusion process, each bit of energy will take a random amount of time to escape. My thinking is that the average escape time would still have to be 26 million years to maintain the Sun's (fairly constant) thermal energy and luminosity.

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  • $\begingroup$ Your reasoning is correct, and this is a very nice trick because it gives a good ballpark estimate with an argument which is accessible to undergraduates (and potentially highschoolers). $\endgroup$ Nov 24 at 13:10
  • $\begingroup$ The reasoning is not correct because the radiation field only holds about 1% of the Sun's thermal energy and is continually replenished. $\endgroup$
    – ProfRob
    Nov 28 at 0:31

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How long it takes the sun to radiate X amount of power is only a quantification of the radiated power in different units of X/s instead of joules/sec.

In this case, in units of the the approximate suns thermal energy per second.

Imagine the core is very small so the thermal energy you consider is then much smaller. Consequently the time you will estimate will be much smaller. But a smaller core should mean a longer journey to the surface. This case reveals that your equation takes no account of the journey through the extent of the body beyond its small core.

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  • $\begingroup$ But if I perform the calculation with a smaller core as you suggest, then the calculated time will reflect only the amount of time it takes to escape the core, not to escape the Sun. I think I must be missing your point. Also, see my addition to my original post. Thank you. $\endgroup$
    – James
    Oct 25, 2017 at 11:57
  • $\begingroup$ No, it will reflect the amount of time it takes to move the energy present in the core via the luminosty of the current sun, with the current solar radius. $\endgroup$
    – JMLCarter
    Oct 25, 2017 at 21:36
  • $\begingroup$ For a moment, instead of discussing energy moving from the Sun's core to the surface, let's discuss water traveling from the left end of a 10 meter long garden hose to the right end. If the hose holds 10 kg (1 kg/m) of water and the flow rate is 1 kg/s, then the flow time of a single molecule of water is on average 10 seconds. Are you suggesting that this is invalid because we get a different answer if we only consider the first 1 meter of the hose with a 1 second flow time? $\endgroup$
    – James
    Oct 26, 2017 at 11:11
  • $\begingroup$ You have to inject water into the hose at, some points to simulate energy production, rather than energy (water) content. It's the rates of flow that are important, not the amount of energy (water) in the hose. $\endgroup$
    – JMLCarter
    Oct 26, 2017 at 17:09
  • $\begingroup$ I intended in the analogy that the left end of the hose would have water injected at 1 kg/s, just as the Sun's core has an energy production rate equal to the Sun's luminosity. For the hose, the amount of water in the hose is absolutely important when calculating the flow time of a water molecule (time = mass in hose divided by mass flow rate). $\endgroup$
    – James
    Oct 26, 2017 at 17:21
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I believe your calculation makes sense.

Michael Stix's 2003 article "On the Time Scale of Energy Transport in the Sun" states that "the time scale of energy transport in the Sun is the Kelvin–Helmholtz time scale, of order $3 \times 10^7$ years, roughly 100 times longer than the photon-diffusion time estimated by Mitalas and Sills (1992). The difference corresponds to a factor $U_{gas}/U_{rad}$, the ratio of thermal energy density to radiation energy density. Thus the heat transport, even when mediated by photons, is slowed down by the large heat capacity of the star."

Your value of $2.6\times 10^7$ years also happens to be the precise value calculated for "energy residence time in the Sun" in a 2019 ArXiV article "Residence time of energy in Earth's atmosphere and in the Sun" by Carlos Osácar, Manuel Membrado, and Amalio Fernández-Pacheco. A revised version was later published as "Residence time of energy in the atmosphere", but without this precise value.

Spruit, in "Theory of Solar Irradiance Variations", emphasizes the difference between the thermal and diffusive timescales, and gives as an example the behaviour of a chunk of aluminium suspended in space and heated from the inside. The thermal time scale is how long the chunk takes to reach thermal equilibrium with with the heat radiated from the surface equal to the internal heating power. This depends on the total heat capacity of the chunk. The diffusive time scale is how long it takes the temperature of different parts of the chunk to equilibrate, which depends on the thermal conductivity of the metal. This is much shorter than the thermal time scale because aluminium is such a good thermal conductor. Similarly, the Sun's diffusive timescale is much shorter than its thermal timescale because of the very large turbulent diffusivity (effectively the "conductivity") in the Sun's convection zone.

As you know, the comments and discussion for the question "Sun Light takes 1,000/30,000/100,000/170,000/1,000,000 years bouncing around inside to then reach the Earth" are somewhat relevant.

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  • $\begingroup$ Very interesting, thanks! I'm going to have to give some more thought to the aluminum chunk analogy. At first blush it doesn't make sense to me. Is it saying that the diffusive time scale is the time at which the temperatures have stopped changing? If so, that would be thermal equilibrium, which is also the definition of the thermal time scale. $\endgroup$
    – James
    Nov 23 at 15:41
  • $\begingroup$ Maybe there is some confusion because your definition of "thermal time scale" above doesn't seem to be the same as the "thermal time scale" defined in the link you provided in the other question... en.wikipedia.org/wiki/Thermal_time_scale $\endgroup$
    – James
    Nov 23 at 15:46
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    $\begingroup$ @James My paraphrasing of Spruit may not be perfect. I think the analogy is that if one location is heated, the whole chunk starts to warm quickly due to conduction, but it takes a longer time for the whole chunk to reach the temperature where input and output (radiative) power are in balance. So a power change affects temperature quickly, but it takes a while for the temperature to stop changing. So if the interior of the sun turned off, we'd see the effects in $\sim 10^5$ years, but the sun wouldn't "turn off" for $\gtrsim 10^7$ years. $\endgroup$ Nov 23 at 15:59
  • $\begingroup$ Your last sentence above cleared up my confusion. Thanks! $\endgroup$
    – James
    Nov 23 at 16:07
  • $\begingroup$ Maybe I'm just confused by terminology. Wikipedia says that the thermal time scale is the "time it takes for a star to radiate away its total kinetic energy content at its current luminosity rate", which is what I calculated in my post above. Whereas the definition from Spruit above seems to be the time required for the star to reach steady state if the internal heating power changes. $\endgroup$
    – James
    Nov 23 at 16:16
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An alternative to diffusion models is to make a estimate of the Virial relaxation time for some model of the sun.

I use a particularly spherical and bovine model in class sometimes (simple enough to be written up in a few pages and done at the board in under an hour), and it give numbers like 100,000 years for the time it would take the luminosity to drop by 1% in the event that fusion suddenly shut off in the core.

This should be taken with a significant degree of skepticism as the model involved is brain dead and was selected for it's tractability rather than for physical fidelity, but it is striking that the result falls in the middle of the range of results from diffusion model and because it predicts a mean temperature that is similarly reasonable (about $10^7\,\mathrm{K}$).

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  • $\begingroup$ So, to be clear, the OP's model, is the same as or similar to your own "bull" model? i.e. simple enough that people can think they understand it, simulating the experience of learning (and teaching)? $\endgroup$
    – JMLCarter
    Oct 25, 2017 at 0:19
  • $\begingroup$ The Virial theorem connects the gravitational binding energy and the temperature (because that is a measure of mean kinetic energy), so there is more physics content in a Virial based model than in the OP's simple power model because it enforces an equilibrium state. The model I use in class is too simple by half because it assumes uniform temperature and density, but equilibrium arguments are often surprisingly robust and that seems to be the case this time. $\endgroup$ Oct 25, 2017 at 22:10
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The random walk argument gives the timescale for a photon/radiation, $\tau_{\rm rad}$, to diffuse out of the star assuming it is absorbed and remitted immediately in random directions.

This is quite different to the thermal energy diffusion timescale, $\tau_{\rm therm}$ - the timescale for most of the thermal energy to diffuse out of a star.

That is because in most stars, a tiny minority of the thermal energy is held in the radiation field. Roughly speaking $$\frac{\tau_{\rm therm}}{\tau_{\rm rad}} \sim \frac{\rm Thermal\ energy\ in\ the\ star}{\rm Thermal\ energy\ in\ the\ radiation}$$

A Sun-like star in which fusion is turned off will replenish the energy in its radiation field about a hundred times before it cools.

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  • $\begingroup$ Thanks for your answer! I'm not a physicist, but I had some heat transfer in college... just enough to be dangerous probably. Can the "radiation" and the "thermal energy" really be talked about separately? I know that a photon travels a short distance, gets absorbed, and then another photon gets emitted. So isn't any given photon just part of the overall "thermal energy" of the Sun? $\endgroup$
    – James
    Nov 28 at 0:58
  • $\begingroup$ @James yes - the energy density of the gas and the radiation can be separated. One depends on $T$, the other on $T^4$. When you calculated the thermal energy of the sun you neglected the thermal energy in the radiation. That's ok, because it is only a 1% addition. In high mass stars it is comparable or even greater. $\endgroup$
    – ProfRob
    Nov 28 at 6:38

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