# Another way to find the energy diffusion time in the Sun?

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.

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.

• 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. – James Oct 25 '17 at 11:57
• 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. – JMLCarter Oct 25 '17 at 21:36
• 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? – James Oct 26 '17 at 11:11
• 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. – JMLCarter Oct 26 '17 at 17:09
• 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). – James Oct 26 '17 at 17:21

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 $10^7\,\mathrm{K}$).

• 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)? – JMLCarter Oct 25 '17 at 0:19
• 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. – dmckee Oct 25 '17 at 22:10