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When light (photon particle) is generated inside the Sun, it takes a long time to bounce around inside to later escape and travel outwards.

Neutrinos escape immediately.

The numbers for the years trapped inside varies so much. I have included some numbers in the title of this question.

Here is a selection of online articles which I found to highlight the numbers:

Energy produced in the form of light keeps bouncing around inside the Sun, as though the Sun were made entirely of mirrors. A particle of light can take more than 30,000 years to reach the surface and escape!

It can take anywhere from a few thousand to a few million years for one photon to escape.

Borexino scientists found measurements using solar neutrinos matched previous measurements using photons, revealing that the sun releases as much energy today as it did 100,000 years ago.

... Process converts 4 million tons of matter into energy. This energy, which can take between 10,000 and 170,000 years to escape the core, is the Source ...

Estimates of the photon travel time range between 10,000 and 170,000 years.

... solar activity has been consistently stable on a 100,000-year scale....

  1. Why so much variation in the estimates?

The Borexino measurements directly use this number to compare/estimate what the sunlight would have been X years ago, but the specific number used is 100,000.

It is one thing to report widely ranging numbers, it is another to use a specific number in calculations to make conclusions.
Like-wise, there are calculations on when the Sun will expand or when the light will dim out, again using some specific number given here.

  1. In this case, why do scientists not calculate with lower bounds and upper bounds, choosing to instead take a specific number?
JRE
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Prem
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    You don't quote what scientists calculate, but what some journalist extracted from their publication. The scientific care and care for limits and error estimates is left out most often - much to most scientists' annoyance when they are then quoted with a single number which explicitly was said initially to be hardly reliable. That said: diffusion time of photons from core to surface can only be estimated, using equation of state and density profiles which have their own error estimates. Depending on what model and you use and what error (68% significance, 97%, 99%...) you get different ranges. – planetmaker Nov 22 '22 at 15:56
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    For some scientific sources, look at "On the Photon Diffusion Timescale for the Sun", whose value of 170,000 years was a correction to $\sim 10^4$ year estimates from earlier astrophysics textbooks. It was later argued in "On the time scale of energy transport in the sun" that the relevant photon diffusion time is actually the solar thermal adjustment timescale of $\sim 10^7$ years, much longer than any of the numbers you cite. – David Bailey Nov 22 '22 at 21:11
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    In a post a few years ago, I thought it would make sense to divide the Sun's total thermal energy by the power output to find the average time for energy to escape the Sun. I got an answer of about 26 million years. I still don't understand what (if anything) is wrong with my method, so please see the following link if you have any comments. https://physics.stackexchange.com/q/364765/90326 – James Nov 23 '22 at 13:24
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    @James I think your result is consistent with Michael Stix's 2003 article "On the Time Scale of Energy Transport in the Sun" and a 2019 ArXiV article "Residence time of energy in Earth's atmosphere and in the Sun" that gives exactly your answer of 26 million years. I'll put a few more details in an answer to your question. – David Bailey Nov 23 '22 at 15:25
  • Yes , it is the energy released , but the "calculation" is using the Photons of today (generated X years ago) versus the Neutrinos of today (generated now) where X is the Contention of the Question. – Prem Nov 24 '22 at 04:30
  • @DavidBailey that is not the photon diffusion time it is the thermal energy diffusion time. The two should not be conflated. The article is behind a paywall, but I do not believe the authors claim the photon diffusion timescale is $\sim 10^7$ years. – ProfRob Nov 28 '22 at 00:28
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    @ProfRob I tried to be clear that there are two separate timescales. I was just pointing out that is some discussion about what is meant by the "diffusion" timescale. What Stix said in the article was "The thermal adjustment takes 100 times longer than the photon- diffusion time derived by Mitalas and Sills (1992). Indeed this longer time scale should be considered as the real time scale of photon diffusion. Photons released at the center of the Sun do not simply random walk as an independent substance." – David Bailey Nov 28 '22 at 01:15
  • @DavidBailey: ProfRob mentioned in the first sentence of his answer below that a given photon does not progress from the core to the surface. That's why I have been thinking of the "photon diffusion time" as the "energy diffusion time" during this discussion. Because photons are flashing into and out of the total Sun energy at all times. Thank you for the Mitalas and Sills reference that seems to agree with my view. – James Nov 28 '22 at 12:17

2 Answers2

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The idea that an identifiable photon "bounces around" inside the Sun and emerges some time later is incorrect.

The photons that are produced in nuclear fusion reactions or indeed by other emission processes (bremsstrahlung etc.) at temperatures of $1.5\times 10^{7}$ K in the solar core are gamma rays and X-rays. Obviously, the radiation we receive from the solar photosphere is mainly in the form of visible and near-infrared light.

What in fact happens is that individual photons are emitted and then absorbed on length scales that can be as short as 1 mm inside the Sun. The exact mean free path of a photon depends on the temperature and density conditions and varies as a function of radius inside the Sun.

The "100,000 year" figure is what you get if you assume indeed that the absorption of a photon is immediately followed by the re-emission of a (different) photon in a random direction - a so-called "random walk process". It is easy enough to show that the time taken for such a random walk to emerge at the solar surface is $$ \tau = \frac{R^2}{lc}\ ,$$ where $R$ is the solar radius and $l$ is the mean free path of a photon. If you reverse-engineer this equation you will see that $l=0.5$ mm corresponds to $\tau= 10^5$ years.

On average, a photon that is emitted from a larger radius within the Sun will have a slightly lower energy, because of the temperature gradient. This effectively transfers energy from the inside to the outside and is called radiative diffusion.

The reason that there is variation in the "100,000 year" number is because of different assumptions about what to use for the average mean free path of a photon; which as I say, varies considerably with depth inside the Sun - from $<0.1$ mm in the core to 2-3 mm further out. It is also because it is not correct to say that each photon travels a set distance $l$ and is then absorbed. In reality there is a distribution of distances, which result in additional numerical factors in this back-of-the-envelope calculation. There is also the issue that convective heat transport rather than radiative diffusion becomes the more dominant heat transfer process in the outer 20% (by radius) of the Sun.

However, it is approximately correct to say that the neutrino flux we see today will be reflected in the photospheric luminosity of the Sun in about $10^5$ years, where I am deliberately using one significant figure. Even this is debatable. I think this would represent the timescale on which the photospheric luminosity would begin to change. For the star to reach some new equilibrium luminosity would take a longer Kelvin-Helmholtz time.

ProfRob
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  • Supposing it were only the process of photon diffusion that carried energy to the surface. But the sun has convection which will bring energy up at some rate. A quick search didn't find that rate. But even if it was only 1 km/hr, it would bring heat to the surface in under a century. – Boba Fit Nov 22 '22 at 19:17
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    The Sun only has convection in it's outer 20% (by radius) @BobaFit Assuming this would reduce the overall energy transport time from the core by about 20% because the heat still has to get to the base of the convection zone. – ProfRob Nov 22 '22 at 20:29
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    "in about $10^5$ years, where I am deliberately using one significant figure" - is that one sig fig, or zero? – user2357112 Nov 23 '22 at 03:30
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    @user2357112 I don't understand your question. One significant figure is one sig fig. $10^5 = 1\times 10^5$. As opposed to $1.0\times10^5$ which would be 2 sig figs. – ProfRob Nov 23 '22 at 07:54
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    @ProfRob: Is the 1 actually significant, or is this a pure order-of-magnitude approximation? (As another way to think about it, if you learned that the actual value was more like $3 \times 10^5$, would you think "woah, I was way off", or "looks like I was about right"?) – user2357112 Nov 23 '22 at 08:26
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    My guess is it's right to within a factor of 2. @user2357112 in the sense of the lag between a change in fusion rate and the photospheric luminosity beginning to change. It would be an interesting MESA project. – ProfRob Nov 23 '22 at 08:41
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    Thanks for another valuable answer. You mention convection. One obvious question is whether in relatively opaque materials like the Sun's interior (I assume that short free paths can be translated as "opaque") heat conduction is relevant. Obviously, given the large distance from the core to the photosphere, it would take a long time as well until changes in the core would propagate to the surface (if we ignore things like sun quakes or even collapses resulting from hypothetical changes in the core, which would probably propagate faster). – Peter - Reinstate Monica Nov 23 '22 at 10:09
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    So, do you know of any estimates how energy transfer is distributed between the trinity of radiation, conduction and convection, averaged over the (very different) layers? Would it be possible or make sense at all to separate them? Because the photon scattering heats the matter which may conduct and convect heat away elsewhere where in turn later photons are emitted again -- in effect the energy transport to the surface may be a constant mix of all three, varying by depth because of different conductivity, convection intensity and free photon path lengths. – Peter - Reinstate Monica Nov 23 '22 at 10:10
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    @Peter-ReinstateMonica of course all 3 are in operation. For the Sun, radiative diffusion dominates in the inner 80%, then convective transport is dominant in the outer 20% with obviously some grey area at the boundary where both are ~equally important. Conduction is not usually important until the density/temperature combination starts to produce electron degeneracy - in lower mass stars, white dwarfs, brown dwarfs, end-state helium cores etc. – ProfRob Nov 23 '22 at 10:56
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    @ProfRob because random walk has an R ² term, reducing the effective radius by 20% will reduce the time by (about) 35%. – Martin Bonner supports Monica Nov 23 '22 at 15:09
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Actually, the entire idea about a photon being emitted and then bouncing around until it reaches the surface, is hilarious: Each nuclear reaction inside the sun's core produces energies in the range of several $MeV$, both in kinetic energy and gamma rays. The photons that emerge from the surface have their wavelength maximum around $500nm$, which corresponds to energies around $2.4eV$. As such, for each gamma ray produced inside the core, millions of photons are emitted from the sun's surface.

Where did all those extra photons emerge from? There is no such thing as a correspondence between those gamma rays and surface photons, other than that they are part of the unceasing transformation of radiative energy into kinetic energy and back.

cmaster - reinstate monica
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