I just heard someone mention that photons take 40 thousand years to travel from the centre of the Sun to its surface which is roughly 700,000 kilometres. How is that possible if the speed of light/photons is 300,000 km/second?

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    $\begingroup$ related/possible duplicate: physics.stackexchange.com/q/103987/50583, physics.stackexchange.com/q/352979/50583 $\endgroup$
    – ACuriousMind
    Commented Nov 19, 2020 at 16:53
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    $\begingroup$ It has nothing to do with gravity - they just bounce around. $\endgroup$
    – Fattie
    Commented Nov 19, 2020 at 17:02
  • $\begingroup$ I once estimated the time for a given bit of energy to escape the Sun using the Sun's total thermal energy and the Sun's luminosity. I got an answer of 26 million years. Others commented that my methodology was no good, but I never understood their reasoning. The thread is at the following link if anyone would care to see my method. Thanks! physics.stackexchange.com/questions/364765/… $\endgroup$
    – James
    Commented Nov 20, 2020 at 13:39
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    $\begingroup$ As a quantitative aspect to every answer saying "gravity plays no large role", I calculated (plugging $d\theta=d\phi=d\tau=0$ into the interior Schwarzschild metric and integrating to find the change in coordinate time $t$ for a path from $r=0$ to $r=r_g$) that light from the very center of the sun would take approx. 3.28 seconds to reach the surface of the sun (if it didn't get absorbed), which is about 41.4% longer than if the sun had no mass. Measurable? Maybe. But it's not the main delay. $\endgroup$
    – HTNW
    Commented Nov 20, 2020 at 16:15
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    $\begingroup$ Does this answer your question? Light formed by the sun? $\endgroup$
    – user21820
    Commented Nov 21, 2020 at 3:26

4 Answers 4


Photons (as electromagnetic waves) always travel at the speed of light (in a vacuum and in this case between particle collisions - see below) which is about $\approx 3 \times 10^8 \ \text{ms}^{-1}$. They are being "slowed down", but not the way you think and not because of the suns' gravitational field, of which this has nothing to do with.

More precisely, although the photons travel from the core to the photosphere, which is a certain distance, the path they take is not direct.

Because the sun is dense, a photon emitted at the core will be absorbed by another nearby proton almost immediately, and then that proton will vibrate and re-emit another in a random direction. This happens over and over again many times so that when it reaches the sun's surface, thousands of years have passed. This process is described by what is called a random walk.

The distance that a photon can travel before it is absorbed, is given by what's called the mean free path, given by the equation

$$l = \frac{1}{\sigma n}$$

"where $n$ is the number of target particles per unit volume, and $\sigma$ is the effective cross-sectional area for collision."

As you can appreciate, the number of target particles (protons) will be significantly high making the mean free path extremely small, so the number of collisions is significantly high. in other words, the photon travels a substantial distance from the Suns' core to its surface, which takes thousands of years. After which it takes a measly 9 minutes to reach us,

$^1$ Also be careful with terms like "same" photon, since elementary particles (of a particular type) are indistinguishable (even in principle), so to give any particle a specific identity is technically incorrect. See this article about identical particles.

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    $\begingroup$ "are not the "same" photon", but in the statistical models that give us the time it takes to reach the surface, it is "assumed" to be the same photon. In truth it is energy carried over by the virtual photons in the zillions of interactions that reaches the surface of the sun after a long time. $\endgroup$
    – anna v
    Commented Nov 19, 2020 at 5:50
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    $\begingroup$ Yeah but it isn’t though is it? I think this point should be made since the OP was of the opinion that an emitted photon travels from the centre to the outside unimpeded. $\endgroup$
    – joseph h
    Commented Nov 19, 2020 at 5:56
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    $\begingroup$ @Drjh The language doesn't really work well for quanta. Photons (just like electrons etc.) do not have identity; "the same photon" is essentially meaningless as soon as you lose the ability to factorize "individual" photons. But we can definitely say some interesting things about the "photon emitted from a fusion reaction in the Sun's core" and "photon emitted from the photosphere" - most obviously, they have very different energies, random direction and there's many more "photosphere photons" than there are "fusion photons" (so there's definitely no 1:1 relationship). $\endgroup$
    – Luaan
    Commented Nov 19, 2020 at 13:30
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    $\begingroup$ No, the sun is not extremely dense. The conditions at the center of the sun are quite extreme in terms of pressure and temperature, and the density is quite impressive if one takes into account that the sun is made up mainly of hydrogen (plasma) which under "normal" conditions has a very low density, but in absolute terms the sun has a low density. Indeed the moon has about three times the density of the sun, a fact that, given the curious coincidence that sun and moon disks have almost exactly the same size as seen from the earth, explains why the moon has more tidal effect than the sun has. $\endgroup$ Commented Nov 19, 2020 at 18:03
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    $\begingroup$ @TCooper No, since if that were true the sun would easily beat the moon, as it is $p$ times as far away, $p$ times the linear size, and so $p^3$ times as voluminous, for some large factor $p$ of proportionality (I guess around $300$, thought I'd need to check the numbers). However tidal effect is proportional to the derivative of the vector-valued gravitational strength (or more properly acceleration), which follows an inverse cube law. That precisely cancels the volume effect, so only density remains. $\endgroup$ Commented Nov 19, 2020 at 21:47

Gravity has little to do with it. The sun is a dense plasma and particles of plasma prolifically scatter photons (both elastically and inelastically). Hence the photon does not travel from the center to the surface in a straight line.


I should to clarify that gravitational field changes the wavelength of light (consequently the momentum of the photon in the vacuum) but not the speed of the photon since $$ p = \frac{\hbar}{\lambda} $$ and the speed of light in the vacuum remains absolute according to the relativity principle.

The reason of the apparent slow down of the speed of light in the sun is explained extensively in the previous answers.

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    $\begingroup$ Sure the speed of light in vacuum is always the same locally, but difference between frames of reference in different gravitational fields makes up for a total difference in speed. Not that this is the case with the sun, but light from a very dense neutron star can be significantly Shapiro delayed from the time it leaves the surface until it reaches a far away observer. $\endgroup$
    – lvella
    Commented Nov 20, 2020 at 14:50
  • $\begingroup$ Yes Shapiro delay can be measured for rays passing closely by the sun but it should be negligible in front of the scattering effect. Thank you @Ivella $\endgroup$
    – Appo
    Commented Nov 21, 2020 at 18:53

The energy produced in the core takes that long to reach the surface. It's not the same photon taking that long. Photons are constantly emitted and absorbed, and the energy also spends time as kinetic energy of the electrons and protons in the plasma.

The core gets hot. It takes 40k years (or whatever the correct number may be) for it to cool, working its way though the entire bulk of the sun's material.

  • $\begingroup$ What is "it" in "...40k years (or whatever the correct number may be) for it to cool"? An individual molecule that happens to be in the core? A photon starting out in the core (as a result of fusion)? Something else? It can not be the core itself. $\endgroup$ Commented Nov 21, 2020 at 3:38
  • $\begingroup$ Yes, the core itself. That's how long it takes energy to reach the surface. $\endgroup$
    – JDługosz
    Commented Jan 19, 2021 at 16:23

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