I have seen various estimates of this which vary widely (like 10,000 to 10,000,000 years) and want to know if there is an accepted answer?

I am not interested in explanations about the identity of photons and absorption and reemission other obscure matters. I want to know in essence what is the total radiant energy inside the Sun divided by the annual rate at which it comes out. That will serve my purpose. References if possible please.

Those are two different, but related, questions?

1) How long does it take light to travel from the core to the surface?
2) Total radiance divided by this time? [Why do you want this one?]

The first is known as the random walk. I think the BA once noted a low estimate of 18,000 years. However, I think this time is based on a simplified model where the Sun's density is held as a constant, which it is not.

The Sun is in a very stable equilibrium with its own gravity. The variation of the total amount of radiation, decade after decade, is only about 0.1%. [UV variance can be as much as 30%, however.]

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I really only want one. I just figured that the 2nd way is stated so that there is no wiggle room. People in this forum often give answers that bring in too many extraneous factors.

You are wanting total radiance per random walk time?

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This paper (http://adsabs.harvard.edu/abs/1992ApJ...401..759M) says that the diffusion time is 170,000 years. I don't know how they get that, but it gibes with estimates I've heard.

Thanks Ken, that is nearer to the middle of the range, but it concerns me that the results I have seen are spread over 3 orders of magnitude. I can see that random walks is tricky with short path lengths. That is why I also phrased the question in terms of the radiation content of the Sun and the rate at which it comes out. It is in fact the radiation content that I am actually most interested in. I would have thought that it should be known quite accurately from the temperature and density profile of the Sun at various depths.

It is, it's just not easy to find it summarized. Interior models of the Sun are very reliable, and can give you the total radiant energy and gas kinetic energy inside the Sun, I would think to 1% accuracy or so.

Here is a short report that calculates the random walk to be 10,000 years. It uses an average mean free path and an average density for the Sun. They mention that another simplification factor, which is removed from the computation, is the brief time a photon expires before re-emittance. This should explain why 10,000 years is on the very low end of the time scale for the Random Walk, I assume.

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That's just a toy calculation, there's no reason to take the result seriously. The diffusion time is actually more like 170,000 years, and that still doesn't count the time "spent waiting between emissions". If you include that time (which is a lot), you get not the diffusion time, but rather what is known as the "Kelvin-Helmholtz time". The diffusion time is essentially the radiant energy in the Sun at any moment divided by the luminosity of the Sun, and the K-H time is the total energy in the Sun (mostly in hot gas, and much more than in the radiation) divided by the luminosity. The latter is about two orders of magnitude higher than the former.

Rtomes, why do you want to know any more accurately than 'a long time'?

If I understand this, you are saying the Sun has enough residual power to maintain its current radiation flux for roughly another 100 years. Is this right?

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The other way round-- we multiply the 170,000 by two orders of magnitude to get the time the Sun can maintain its radiation flux. So it's about 10 million years, IIRC. In other words, the fusion in the core of the Sun could end forever right now, and we'd hardly notice a thing for many thousands of years. It wouldn't be an obviously different object for millions of years.

[Ah, I quickly realized I was way off but couldn't corect.]

Is the difference the heat content between the two terms?

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The difference between the 170,000 year "diffusion time" and the ~10 million year "Kelvin-Helmoltz time" (which is normally defined in terms of gravitational energy, but this is related by factors like 2 that I'm not concerned with here) has to do with what energy you are tracking. If you are only tracking radiant energy, you get the former, but if you are also tracking the heat in the gas, you get the latter. So yes, I think that's what you mean by the heat content of the two.

That is interesting, thanks Ken.

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My students learn in the intro to astrophysics class that the characteristic photon diffusion time scales as:

t_diff = <N> * <l> / c
or about 3 (R/<l>)^2 * <l> /c, where <N> is the mean number of interactions per escape, <l> is some usefully typical photon mean free path, c is the speed of light in the vacuum, and R is the star's radius.

This works out to be:

154,000 yrs * (R/R_sun)^2/(<l>/1 mm), where the characteristic photon mean free path is given in units of millimeters (more appropriate than cm in our Sun), essentially what KenG said it was, in the case of our Sun.

Thanks Ken, George and SpacemanSpiff.

Am I right in understanding that the Kelvin-Helmhotz time is the time that gravitational energy would make the Sun shine? The fact that this is about 10 million years does seem to explain the 10 million year period that I saw. I guess either that explanation was not good or I got mixed up. Anyway that does seem to narrow the answer down a lot and indicate that the 10,000 year time frame was simply based on too long a mean free path in the model and that 170,000 years is the correct vicinity. Thanks.

One further question related to this. Would I be correct in saying that this means that at any moment around 10^-7 of the Sun's mass is actually made up of radiant energy?

I guess that the choice of word "mass" there might be contentious (equally so if "energy" is used). So I think I need to add that a large bunch of photons bouncing about for a long time more or less in place do have mass in the sense that the ensemble is not in motion. If someone can suggest better words to describe this then that would also be appreciated.

It would be more correct to say that it is the time it could make the Sun shine without dramatically changing its size and attributes. The Sun could shine for much longer on its way to becoming a cooling white dwarf.
I think sometimes the distinction is not made clearly enough.

Yes, I think that is the right conclusion here.

Let's look at that. I'd say that at any time about 1/100 of its internal energy is radiant, but the rest is related to gas pressure at a temperature of about 10 million K, which corresponds to protons moving at 1/1000 of the speed of light. So you have 100 times more energy in the proton motions, and 1 million times more energy in the proton rest mass than in their motions, so yes I get maybe 10^-8 but that's not accurate enough to rule out 10^-7.I think it would be fair to call this the "gravitational mass" of the Sun, for example. We are dropping factors of 2 and so forth right and left, but I think it's fair to say that up to 1 part in 10 or 100 million of the gravity that the Sun produces comes from the radiant energy in there at any moment.

Hi Ken

Yes, the statement that the gravity comes from that energy is a nice way of putting it. I am also interested in the question of what proportion of the gravitational energy also comes from kinetic energy due to motion of matter (i.e relativistic mass increase). If I remember when I worked this out once before it was small compared with the radiant energy, but it would be nice to have this confirmed.

Just to confirm how I got 10^-7, I assumed that the life of the Sun is about 10^10 years and that it converts 0.7% of its mass to radiant energy (although I suppose some goes to neutrinos) as a result of H --> He fusion. I used the 170,000 years to work out what proportion of the energy is in the Sun at any one time, surprisingly 0.0017% of what it ever produces. Of course there are some other processes after that, but I am guessing that the present rate of conversion is about an average for 10^10 years. That gave me an answer of 1.19*10^-7 which I figure is accurate to about 1 digit anyway.

Regards
Ray

No, the heat in the gas is about 100 times more than the radiant energy, so it will also contribute similarly more to the gravitational mass.
That might be a factor of 3 or something too high-- not all the H goes to He, only the H in the core.
That's valid for a rough estimate. It's probably just a little high overall, but my calculation was also rough so it could easily be between the two.

Oh wow!

Now I don't know how to derive this exactly, but had a shot anyway. I get that the velocity of nuclei near the centre of the sun are about 400 km/s (which I am roughly calculating based on what I think the velocity relationship is to temperature and atomic mass) which makes their relativistic mass increase only about 2 parts in a million. That is only about 20 times my estimate of radiation and I suppose 200 times yours. OK that sounds about right. But that doesn't allow for electrons which I suppose are doing more like 18,000 km/s which increases their mass by something of the order of 0.3% but they are only 1/1836 of the mass so that makes roughly the same amount again? Total nearer 3 to 4 millionths. Of course the nuclei and electrons away from the centre are going slower, so correct answer maybe still 2 millionths.

I can't help noticing that the answer of about .000002 of the mass being gravitational as you put it, or relativistic mass if you like is very similar to the factor GM/c^2/r for the Sun that determines light bending and such which comes to .00000214 and wondering if this is purely coincidence.

Yes, that sounds reasonable. it's just a rough estimate, but it shouldn't be too far off.

For what it's worth, Allen's Astrophysical Quantities says ...

• Work required to dissipate solar matter to infinity: 6.6x10⁴⁸ erg.
• Sun's total internal radiative energy: 2.8x10⁴⁷ erg.

So the total gravitational energy of the sun appears to be about a factor of 30 in excess of the total radiant energy. Assuming the sun maintains a constant energy output of 3.845x10³³ erg/sec, the total internal radiative energy by itself would last about 2,300,000 years. But in any "realistic" scenario you have to dissipate a chunk of the gravitational energy as well, so the sun would really last somewhat longer under the constant luminosity assumption, which is likely not a very good assumption anyway.

As for how long it takes a photon to make the trip from the center to the surface, the numbers in this thread, on the order of 150,000 years are all much smaller than the numbers usually quoted in allegedly reliable sources that I have seen, which range from 500,000 to 1,000,000 years. Unfortunately, the one reliable source I have on hand (Foukal's Solar Astrophysics) talks about radiative diffusion, but does not quote a number, and I am too lazy at the moment to try to figure it out myself. My guess is that the numbers on the order of 150,000 are a bit short due to simplifying assumptions, but I can't offer an allegedly superior number at the moment. But I do think that we have enough in the thread to point out that there is not any one "generally accepted" number, only a "generally accepted" order of magnitude.

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Yeah, it probably depends on how one defines the diffusion time, it's not such a precise concept. I actually think rtomes approach is a good one, that the average diffusion time may be usefully defined as the total radiant energy divided by the luminosity. With your numbers, that comes out about 7 x 10¹³ seconds, or about 2 million years. The Kelvin-Helmholtz time would then be about 60 million years.

Thanks Tim and Ken.

I just realized that I was a bit confused about the gravitational energy component or the relativistic energy of the matter. Are these the same thing? Sure, if you let all the matter fall in from infinity it will make the Sun hot, but not necessarily exactly as hot as it is. The thing that I am really trying to get (I think) is the relativistic mass content of the Sun. I had been assuming that this was mainly in radiation because the matter was not going fast enough. But of course the matter has a lot more mass, so even a small relativistic factor comes out to a lot.

I guess that I am back somewhere near where I started in having a variable range for the correct answer on the radiation, but understanding a bit more about the background of that variation. It was also very valuable to get a fix on the matter relativistic or gravitational component which is now really more important for my purposes than the radiation question. Well at least I think I have the correct question now. ;-)

When my present ATM thread comes to an end in the next week I will probably start another one in which the answers to these questions are very relevant. It concerns the sunspot cycle.

Interesting paper, and a brief one, finally.

It appears they argue for a mfp of 0.9 mm average, which is considerably less than the references that used averages ranging from 5 to 10 mm. As a result, these older values yield erroneous random walk times from 3,000 to 30,000 years.

Their result, as you said, is 170,000 years, which is the time from the center of the core to the top of the radiative zone. The short time through the convective zone is considered negligible.

I'm curious if it might be possible to measure the diffusion rate someday when separate regions of varibable star disks are observable. If the He^- zone location is known, then there should be a time dealy in the light pulse due to the CLV. Since we could see deeper into the central region of the disk, the light pulse would appear slightly sooner. [Just some loose thinking.] [Added: and some very loose sentence structure. ]

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I think the relativistic energy that goes into the gravity includes the rest masses and the kinetic energy, plus the radiant energy, minus the gravitational energy. The kinetic energy tends to be about half the gravitational energy, so it sounds like the gravitational mass of the Sun is actually a tiny bit less than the total rest mass of its constituents.
Infalling would tend to make the Sun twice as hot as it is, the rest of the heat going into light that was lost from the system (the nuclear fusion doesn't heat anything up, it is in equilibrium with what is lost after the Sun reaches its current size). It's not clear if the number Tim Thompson quoted for the work needed includes the internal kinetic energy or if it is just the external work you'd need to add, but my guess is it's the former, so that would mean that the internal kinetic energy is about half that amount, or about 3 x 10⁴⁸ erg. That would also make it about 10 times the radiant energy, not 100 times as I tended to recall. But I have no reason to doubt those numbers.

Yes, the speed is not the best way to think about it, consider the kinetic energy. That is at least 10 times the radiant energy. But there's also the negative gravitational correction, which is even higher.