The light that takes a hundred thousand years to leave

The Sun’s radius is 696,000 kilometres, and light crosses that in 2.3 seconds. A neutrino made in the core takes exactly that long, because it interacts with almost nothing on the way out.

Energy made as electromagnetic radiation takes very much longer. The photon does not travel; it is absorbed within a fraction of a millimetre, re-emitted in a direction that has forgotten the last one, absorbed again, and so on for something like $10^{20}$ steps. The result is a journey of a hundred thousand years across a distance light crosses in two seconds, and the entire difference is one length appearing squared.

Why the square matters #

A walk of $N$ steps of length $\lambda$ in random directions does not go $N\lambda$. The steps add as vectors, and vectors in random directions largely cancel; what survives is the square root.

The reason is a single line of algebra. The squared displacement after $N$ steps is the sum of $N$ terms of $\lambda^2$ plus $N(N-1)$ cross terms, and the cross terms average to zero because the directions are independent. So the mean square displacement is $N\lambda^2$ and the typical distance is $\sqrt{N}\,\lambda$.

The figure measures that rather than assuming it: over 240 independent walks at each of five lengths, the exponent comes out at 0.503 against an exact half. The single walk drawn beside it is 900 steps long and ends 30 lengths from where it began, which is close to $\sqrt{900}$ and is the whole argument in one picture.

Inverting it gives the number of steps needed to get out of a body of radius $R$: not $R/\lambda$ but $(R/\lambda)^2$. Each step takes $\lambda/c$, so the escape time is

$$t \sim \frac{R^2}{\lambda c}$$

with the free path in the denominator to the first power, because one factor of $\lambda$ has cancelled between the step count and the step duration.

The free path inside a star #

What sets $\lambda$ is the opacity: the cross-section per unit mass for absorbing or scattering radiation. The free path is one over the opacity times the density.

In the Sun’s interior the dominant processes are free–free absorption by electrons passing ions, bound–free absorption from the few remaining bound electrons of heavier elements, and Thomson scattering off free electrons. Together they give a Rosseland mean opacity of a few hundredths of a square metre per kilogram over most of the interior.

At the Sun’s mean density of 1,408 kilograms per cubic metre, an opacity of 0.03 square metres per kilogram gives a free path of about two and a half centimetres. A photon in the solar interior therefore travels a distance a ruler would measure, and does it $10^{20}$ times.

Compare that with the same journey by a carrier that does not interact. A neutrino’s free path in the solar interior is measured in light years, so it never scatters at all and takes the straight line. Same star, same distance, two carriers, and a ratio of about $10^{12}$ in the time — with nothing in the comparison but the free path.

The number the estimate gets wrong #

The uniform-density estimate gives about two thousand years, and the figure usually quoted is around 170,000. That is two orders of magnitude, and where it comes from is worth being explicit about.

A star’s interior falls off in density far more steeply than an atmosphere: the Sun is a hundred and fifty times denser at its centre than on average. That is where the simple estimate goes wrong — a uniform-density calculation puts almost all the path in material of the mean density, and the real star puts almost all of the mass in a small central region where the free path is shortest. The correction is a factor of many thousands, and it goes the way that makes the answer longer.

The Sun is not uniform. Its central density is about 150,000 kilograms per cubic metre, more than a hundred times the mean, and the free path there is correspondingly a hundred times shorter. Since the escape time goes as one over the free path, and since most of the walking has to be done through the dense inner region, the true integral is dominated by the innermost fraction of the radius and comes out far larger than the uniform estimate.

This is the ordinary fate of a one-line estimate applied to a body with a steep profile, and the honest response is to say which part of the answer is trustworthy. The exponent is trustworthy — the escape time genuinely goes as $R^2/\lambda$, and that is what the figure measures. The order of magnitude is trustworthy to within the profile’s dynamic range, which is two decades here. The number is not, and no amount of care with the mean density will fix it, because the mean is the wrong average for a quantity that enters as a reciprocal.

There is a general lesson in that, and it applies well beyond stars. A quantity that enters an answer as a reciprocal cannot be averaged before it is used. The mean of one over the density is not one over the mean density, and when the density varies by two orders the two differ by roughly that much. Any estimate that substitutes a mean into a nonlinear expression is making the same mistake, and the direction of the error is predictable: it always understates a time that is dominated by the slowest region.

Published estimates themselves vary from about 10,000 to a few hundred thousand years depending on exactly what is being computed — the diffusion time for energy, the age of a typical photon, or the time for a perturbation to propagate — and they are not all answering the same question.

What actually comes out #

Calling it a photon’s journey is a convenient shorthand and it is wrong in a way that matters.

A gamma ray made at fifteen million kelvin does not emerge from the Sun. What emerges is a five-thousand-kelvin thermal spectrum, because the photon is absorbed and re-emitted so many times that nothing of its original identity survives — the energy is passed along and the carrier is replaced continually. So “the same photon takes a hundred thousand years” is the wrong picture, and the right one is that the energy takes that long.

At every absorption the quantum ceases to exist, and what is re-emitted is a new one drawn from the local thermal distribution. Since that distribution cools steadily outward — from 15 million kelvin at the centre to 5,800 at the surface — the energy arrives as an enormous number of low-energy photons rather than as the handful of gamma rays that started. Roughly a thousand visible photons leave for each gamma ray absorbed.

So what crosses the Sun is energy, and it does so by diffusion. That is exactly the same statement as heat diffusing down a temperature gradient, with radiation as the carrier instead of molecules, and the mathematics is identical: a flux proportional to a gradient, and a coefficient built out of a mean free path and a speed.

The practical consequence is the one that makes stellar structure possible to compute. Since the transport is diffusive, the interior is very close to local thermal equilibrium everywhere, and the radiation field at each depth is a Planck spectrum at the local temperature to a part in $10^{10}$. The whole of stellar interior physics rests on that, and it rests on the free path being short.

The thin layer where it stops #

Diffusion is a good description as long as a photon’s free path is short compared with the distance over which conditions change. Near the surface that stops being true, abruptly.

The surface of a star is where the optical depth measured inward reaches about one — that is the definition, and it is the whole of what “surface” means for an object with no solid boundary anywhere. Everything about what is seen is decided in that thin layer: below it photons are trapped and above it they leave freely, and the transition happens over a few hundred kilometres out of seven hundred thousand.

The photosphere is defined as the layer from which a photon has a fair chance of escaping without another interaction — optical depth of order one measured outward. It is a few hundred kilometres thick out of 696,000, so a star has an edge that is sharp to a part in a thousand, which is why the Sun looks like a disc with a boundary rather than a fuzzy ball.

Everything a telescope sees comes from that layer. The interior is completely invisible: no photon carries information from below it, because none survives the trip. That is why the neutrinos matter so much observationally — they are the only messengers that leave the core intact, and the solar neutrino problem of the 1970s was the first direct test of anything happening down there.

The thinness of that layer relative to the star is a direct consequence of the free path being short. A photosphere is about one free path thick by definition, and the free path near the surface is a few hundred kilometres because the density there has fallen by many orders from its interior value. So the sharpness of a star’s edge is the same number, read at the other end of the density profile — and a star with a much more extended atmosphere, such as a red supergiant, has an edge that is correspondingly indistinct and a radius that depends on which wavelength it is measured at.

The abruptness also means the diffusion approximation has to be abandoned in a region where it is doing most of the observable work. Model atmospheres solve the full radiative transfer equation there rather than a diffusion equation, and matching the two descriptions across the transition is a substantial part of stellar modelling.

Three different questions with three different answers #

Part of the spread in published figures is not disagreement but different questions, and separating them is worth doing because each has a different use.

Asking when a diffusing profile first reaches a point, when half of it has arrived, and when a marked particle typically arrives are three different questions with three different answers. That is the honest reason the quoted numbers for this problem range over four orders of magnitude in the literature: they are answers to different questions, and most sources do not say which one they are answering.

How long does a perturbation take to reach the surface? This is the diffusion time in the strict sense — the timescale on which a change in the core’s luminosity would show up outside — and it is what the $R^2/\lambda c$ estimate is about. It matters because it says the Sun’s surface brightness cannot respond to anything happening in the core on any timescale a civilisation would notice, which is a comfort when arguing about solar variability.

How long does a given quantum of energy take? This is a question about a marked particle rather than about a profile, and it has a longer answer, because a random walk revisits: energy that has diffused halfway out has a substantial chance of wandering back in. The distinction is the same one that makes a random walk in low dimensions recurrent — the walk keeps returning, and the expected number of returns is what stretches the answer.

And how old is the energy leaving now? This asks for a distribution rather than a number, and the distribution is broad: some of what leaves this second entered the radiative zone recently and some has been in there for a great deal longer than the mean. Quoting a single figure for it is quoting the mean of a distribution with a very long upper tail, which is the least informative summary available.

A survival curve is the shape any “how long does this take” question has when the underlying process is random, and its mean is not its typical value. For a broad distribution the average is dominated by rare long excursions, so a photon that “typically” escapes in one time may have a mean escape time much longer — and quoting either without saying which is what produces the disagreement above.

All three are computed from the same free path, and all three are dominated by the dense interior. What differs is which moment of the same walk is being reported, and a figure quoted without saying which is not wrong so much as unattributed.

Where else the square turns up #

The escape-time argument has nothing to do with stars specifically. It is the general behaviour of anything transported by many small random steps.

The same square-root law governs a dye molecule in still water, a neutron in a reactor and energy leaving a star. What differs between them is only the free path and the speed — the mathematics is identical, and that is why an estimate made for one can be checked against a measurement made on another. It is also why the answer is so sensitive to the free path: the time goes as its reciprocal, and the free path is the quantity hardest to know.

A neutron in a moderator does the same thing, and the reactor designer’s version of $\lambda$ is what decides the size of a critical assembly. A dye molecule in still water takes hours to cross a centimetre, which is why stirring exists. Light in fog takes the same walk on a smaller scale, which is why a fog is bright and opaque rather than dark: the photons are not absorbed, only turned round many times.

And the same square is why diffusion is slow at large scales and fast at small ones. Doubling the distance quadruples the time, so a process that is instantaneous across a cell is glacial across a room. Biology is organised around that fact, and so is the Sun.

Where the walk is abandoned for something faster #

The outer third of the Sun does not do any of this, and the reason is worth following because it is decided by a comparison rather than by a new mechanism.

Diffusion carries a given flux only by maintaining a temperature gradient, and the shorter the free path the steeper that gradient has to be. Where the opacity climbs, the required gradient climbs with it. Meanwhile the gas will tolerate only so steep a gradient before it becomes unstable: a parcel displaced upward expands and cools along its own adiabat, and if the surroundings cool faster with height than the parcel does, the parcel finds itself lighter than its new neighbours and keeps rising. That comparison — the gradient radiation demands against the gradient the gas can sustain — is Schwarzschild’s criterion, and where radiation loses it, the gas overturns.

The Sun loses it at 0.713 of its radius, a figure now measured by helioseismology rather than inferred. Two things conspire there. Hydrogen begins to recombine, which raises the opacity by orders of magnitude and makes the radiative gradient steeper still; and the energy that goes into ionisation rather than into temperature makes the adiabatic gradient shallower. Both push the same way.

Above that depth energy is carried by moving matter, and it is carried enormously faster. The granules visible on the solar surface are the tops of convection cells about a thousand kilometres across, rising at a kilometre or two a second and lasting a few minutes; the whole convection zone turns over in a matter of weeks. Two hundred thousand kilometres that a diffusing photon would have found the slowest part of the journey are crossed in less time than the Sun takes to rotate.

That is worth holding against the essay’s headline number. The hundred thousand years is not the time from the core to the sky; it is the time from the core to the base of the convection zone, after which the last third of the way is essentially free.

The reservoir, and what the luminosity is actually set by #

There is a second timescale for the Sun that is easy to confuse with this one and is asking a different question again.

Take the total thermal energy stored in the Sun and divide it by the rate at which the Sun radiates. The answer is about thirty million years — the Kelvin–Helmholtz time, and the length of time the Sun could go on shining if nothing were replenishing it. It is three hundred times longer than the diffusion time, because the star holds vastly more energy than is in transit through it at any moment.

That number has a history. Kelvin and Helmholtz proposed gravitational contraction as the Sun’s power source, which gives exactly this figure for its age, and it was in flat contradiction with the geologists and with Darwin, who needed hundreds of millions of years for the record they were reading. The conflict stood for half a century and was settled by a source nobody had — nuclear fusion.

The two timescales together say something about a star that is not obvious. The reservoir outlasts the leak by a factor of hundreds, so the rate at which energy escapes is not controlled by the rate at which it is made. It is the other way round: the luminosity is a transport property. How fast energy gets out is fixed by the opacity and the size, and the core’s temperature then settles at whatever value makes fusion supply exactly that much — because if fusion ran faster the core would expand and cool, and if it ran slower the core would contract and heat.

So the answer to why the Sun shines as brightly as it does is not in the nuclear physics. It is in the number this essay has been computing, and the nuclear reactions are the part of the arrangement that adjusts.

What the pictures cannot show #

The walk drawn is in two dimensions, and the star’s is in three. The square-root law is the same in any number of dimensions above one, and the coefficient differs; the figure’s slope is the claim being made and the coefficient is not.

The steps are all the same length. Real free paths are exponentially distributed, and the mean square displacement then depends on the mean square free path rather than on the square of the mean. For an exponential distribution that is a factor of two, which is small next to the two orders of magnitude the density profile contributes.

Opacity is treated as one number. It is a strong function of temperature, density and composition, rising sharply where a species is partially ionised, and the bumps in that function are what drive the pulsations of Cepheid variables.

Scattering and absorption are treated as one process. They are not the same: Thomson scattering changes a photon’s direction and not its energy, while a bound-free absorption destroys it and lets a new one be emitted at the local temperature. Both shorten the free path and only the second thermalises, and which dominates decides how quickly the radiation field forgets where it came from. The distinction shows up as polarisation near the solar limb, which is one of the few observational handles on the layer.

And nothing here transports energy by convection. Over the outer thirty per cent of the Sun’s radius radiation is not the dominant carrier at all: the gas becomes unstable to overturning and convection takes over, moving energy far faster than diffusion could. The hundred-thousand-year figure is a statement about the radiative interior.

The ladder from here #

Later rungs on this anchor: the Rosseland mean and why an average weighted for a flux is the right one rather than a simple mean; the radiative transfer equation and its moments, which is where the diffusion approximation comes from and where it fails; the Eddington approximation and the temperature structure of an atmosphere; and the opacity bump that drives Cepheid pulsation, which is this same coefficient behaving non-monotonically.

The neighbouring ladders are how far a neutrino gets, which is the same star with the free path at the other extreme, how far a molecule gets, where the free path is measured rather than inferred, and the equation that only runs forwards, which is what a random walk looks like when it is written as a partial differential equation.