How can a gamma ray can conserve its momentum while travelling from the center towards the surface of the sun? Shouldn't it lose most of its momentum while scattering from very lite hydrogen and helium nuclei? Maybe it scatters from very energetic nuclei?
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$\begingroup$ @PcMan That should be an answer, not a comment. $\endgroup$– J. MurrayCommented Jul 31, 2021 at 11:34
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$\begingroup$ @PcMan You misunderstand me. Posting answers in the comments is explicitly against the policies of the community for several reasons, not least among which that there is no mechanism for quality control or downvoting low quality comments. As such, your comment is likely to be deleted by a moderator. If you are being harassed or treated unfairly on your answers then you can raise a flag for moderation, but restricting your answers to comments is not a viable alternative. $\endgroup$– J. MurrayCommented Jul 31, 2021 at 15:11
4 Answers
A gamma ray photon does not travel from the core of the Sun to the photosphere. The mean free path of a gamma ray photon in the core of the Sun is of order 1 mm. The energy of the photon is absorbed or scattered in such a way that it deposits energy and momentum. This energy is then re-emitted in an almost random direction in the form of other photons with (on average) slightly lower energy at larger radii but in greater numbers.
Energy is conserved in the sense that the luminosity at the photosphere equals the rate of energy generation in the core. Conservation of momentum simply means there is a small radiation pressure term that helps to counteract the weight of the Sun.
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$\begingroup$ The plasma below the photosphere should be opaque to any kind of light. It should be plasma waves that transport the energy, not photons. $\endgroup$– my2ctsCommented Jul 31, 2021 at 12:47
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$\begingroup$ @my2cts radiative energy transport always occurs in all parts of the solar interior. It forms one of the equations of stellar structure. $\endgroup$– ProfRobCommented Jul 31, 2021 at 16:08
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$\begingroup$ Would you describe this radiation as light or as plasma waves? What is the $\kappa$ in the radiative transport equation based on? $\endgroup$– my2ctsCommented Jul 31, 2021 at 20:08
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$\begingroup$ I agree that your answer is the main stream story. However, would you really describe this radiation as (X-ray) light? Note that the energy travels at a surprisingly low speed of about 3 km/year or 100 µ/s, which indicates thermal diffusion of plasma energy. Also how would photons of the order of 3 nm wavelength travel through a very dense hydrogen plasma? $\endgroup$– my2ctsCommented Jul 31, 2021 at 22:05
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$\begingroup$ @my2cts as I have pointed out, the mean free path of a gamma ray photon in the core of the Sun is about 1 mm. Gamma rays are electromagnetic radiation. Alfven and magneto-acoustic waves (I am not sure what else you mean by "plasma waves") play almost no role in energy transport in stellar interiors. $\endgroup$– ProfRobCommented Jul 31, 2021 at 23:33
When you say "a gamma ray" you mean presumably a photon$^1$. Think of photons as continuously being absorbed by one nucleus, and then immediately re-emitted in a random direction, and then off to the next nucleus, absorbed then re-emitted again, and so on etc., until they reach the photosphere and escape.
Each one of these individual processes must conserve energy and so the entire process from core to surface, conserves energy. The directions of emission and absorption result in overall momentum conservation.
$^1$ Photons do not have identity, and so thinking the same photon "travelling from the center to the surface" is not correct terminology.
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$\begingroup$ But can’t the atom absorb some of the momentum of the photon it absorbed by emitting two or more lower frequency photons? $\endgroup$– AndreaCommented Jul 31, 2021 at 10:36
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$\begingroup$ The photons are not absorbed by a nucleus but by a plasma. $\endgroup$– my2ctsCommented Jul 31, 2021 at 12:44
This to be read in addition to the answer by ProfRob.
The photon has no identity further than the energy and momentum it is created with,and the spin direction. It can interact with electric and magnetic fields either elastically keeping its energy-momentum, or inelastically. If inelastically, its whole and energy and momentum may be absorbed in raising the level of electrons about a nucleus, or some of the energy moving off as a "new" photon. The excited atoms will drop to a lower energy, releasing a photon , in a two body reaction, energy and momentum are always conserved. The energy and momentum of the photons created by fusions in the center of the sun is statistically distributed .
This popularisation is wrong and is responsible for your question:
When merging two protons in a nucleus of deuterium to create a helium nucleus, photons are released. This particle, created in the solar core, transmits the light beam to Earth. To send us this photon must traverse the various layers of the Sun. The transit time of a photon of the heart at the surface is between 10 000 and 170 000 years based on collisions.
It talks about a single identifiable particle which is contrary to what we know of photons, as described above.It is the statistical energy and momentum travel of bulk matter that takes the time mentioned. Not an identifiable individual photon.
Here is a description of the energy transfer:
According to current models, random scattering from free electrons in the solar radiative zone (the zone within 75% of the solar radius, where heat transfer is by radiation) sets the photon diffusion time scale (or "photon travel time") from the core to the outer edge of the radiative zone at about 170,000 years.
In the process of heat transfer from core to photosphere, each gamma photon in the Sun's core is converted during scattering into several million visible light photons before escaping into space.
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$\begingroup$ "raising the level of electrons about a nucleus." The suns core is a plasma of ions. Any nuclei have no bound electrons. $\endgroup$– joseph hCommented Jul 31, 2021 at 21:42
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$\begingroup$ @josephh I am answering within the dominant model quoted from wikipedia in the last link in the answer $\endgroup$– anna vCommented Aug 1, 2021 at 4:01
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$\begingroup$ So you’d agree that the core is a plasma of charged nuclei and free electrons? $\endgroup$– joseph hCommented Aug 1, 2021 at 4:49
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$\begingroup$ @josephh charged ions , not bare nuclei , atom missing a number of electrons, and free electrons. It is not a matter of transparency for photons statistically to come out, but of quantum statistics. $\endgroup$– anna vCommented Aug 1, 2021 at 6:46
In addition to @ProRob's answer: Below the photosphere the sun consists of plasma. Plasma is opaque to any kind of photons. The gamma photons are converted into plasma waves which transport the energy to the photon sphere, where they are converted into thermal radiation. In these processes momentum is conserved.