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This is an extract from the astrophysics chapter in my book:

Hydrogen atoms fuse to form helium. At the same time, lots of of gamma photons and neutrinos are produced. The photons take thousands of years to "fight" their way to the surface of the Sun, but then escape into space as visible or near visible photons at the speed of light.

I am really confused about how the sun produces light. I understand that through nuclear fusion, energy is formed, but isn't this energy in the form of thermal and gamma photons? I do not understand how visible light is emitted which is the reason to why we see the sun. Could someone please elaborate the paragraph from my book.

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  • $\begingroup$ Blackbody radiation. The sun is hot and so radiates with the spectrum that it does. en.wikipedia.org/wiki/Black-body_radiation $\endgroup$ Commented Mar 18, 2014 at 4:33
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    $\begingroup$ Be aware that the language used in that passage is evocative, but technically sloppy. The energy of the photons will eventually emerge from the sun as light (i.e. photons) but they won't be "the" photons in any useful sense. $\endgroup$ Commented Mar 18, 2014 at 13:15
  • $\begingroup$ Nuclear fusion itself produces gamma photons, which loose energy while moving to the surface of Sun for the thousands years. $\endgroup$
    – kludg
    Commented Jun 1, 2023 at 10:00

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The photons produced in nuclear fusion reactions travel less than a mm before they are absorbed inside the Sun. They don't "fight their way out". New photons are emitted continuously by the hot plasma (by thermal bremsstralung and other processes), but they too are quickly absorbed before getting very far. However, there is a gradual diffusion of radiation and energy outwards because there is a temperature gradient that exists between the inside and outside of the Sun. As the plasma temperature decreases, the dominant photon wavelengths in the (approximately) blackbody radiation field get longer and longer (the mean photon energy is about $2.7k_BT$), but individual photons still do not escape the Sun.

The photons we do receive at the Earth are mostly visible and infrared photons produced in the photosphere, which is where the Sun's atmosphere finally becomes thin enough and cool enough ($T \simeq 6000$ K) to be transparent to the photons produced locally. Hence they escape and may travel to the Earth and we see radiation that is characteristic of the solar photospheric temperature, not that of the solar interior.

As to what exactly produces that visible light, there are a variety of mechanisms. However, the dominant one at most wavelengths through the visible and near infrared regimes is the photo-recombination continuum radiation (free-bound radiation) produced when free electrons (made available by ionising sodium and potassium atoms) attach themselves to hydrogen atoms to form H$^{-}$ ions (the inverse process dominates the continuum photospheric opacity in a star like the Sun, e.g., Wildt 1939 [the original paper to identify this]; John 1992; and these lecture notes).

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  • $\begingroup$ I don't think this hits the essence of the (old) question - the answer would be "thermalization", which is the process that assures that photons of any arbitrary energy will equilibrate with a gas at a kinetic temperature, which will be equal to the radiation temperature of their total energy density and not energy. This process can take very long in dilute gases, but is very fast in stellar interiors due to the opacities quoted. However how thermalization really works is the stuff of many a kinetic theory book. $\endgroup$ Commented May 30, 2023 at 23:35
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    $\begingroup$ "I do not understand how visible light is emitted which is the reason to why we see the sun.", which none of the other answers address. @AtmosphericPrisonEscape $\endgroup$
    – ProfRob
    Commented May 31, 2023 at 6:43
  • $\begingroup$ Can you suggest where I can read more about the contribution of radiative recombination continuum to sunlight. My attempts to Google this have not been very successful though I found a source suggesting that in a hot, optically-thin thermal plasma, the radiative recombination continuum is normally a perturbation on top of the bremsstrahlung continuum. $\endgroup$ Commented Jun 1, 2023 at 7:43
  • $\begingroup$ @JohnRennie that is a hot, optically thin plasma (e.g., the solar corona). The dominance of the H$^{-}$ ion to photospheric opacity (and hence emission) was identified in 1939. References added. $\endgroup$
    – ProfRob
    Commented Jun 1, 2023 at 8:39
  • $\begingroup$ Thanks. OK my answer is wrong. $\endgroup$ Commented Jun 1, 2023 at 9:28
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The essence of your question seems to be:

"... isn't this energy in the form of thermal and gamma photons? I do not understand how visible light is emitted which is the reason to why we see the sun.".

The Sun is a large ball of matter that is undergoing a thermonuclear reaction - like a flask of chemicals reacting with each other to create new chemicals and light except it's not a chemical reaction it's a nuclear reaction (like looking at a large Nuclear Reactor with no walls and a much more complicated series of reactions; a lot of different kinds of fuel).

Here is the series of reactions that occur:

4 (1H) ------> 4 He + 2 e+ + 2 neutrinos + energy

3 (4He) ------> 12C + energy

12C + 12C ------> 24Mg + energy

12C + 4 He ------> 16O + energy

16O + 16O ------> 32S + energy

16O + 4 He -----> 20Ne +energy

28Si + 7(4 He) ------> 56Ni + energy

56Ni ------> 56Co + e+ (postive Beta Decay)

56Co ------> 56Fe + e+ (positive Beta Decay)

56Fe + n ------> 57Fe

57Fe + n ------> 58Fe

58Fe + n ------> 59Fe

That's what makes the Sun burn or fuse, those are the formulas of the nuclear reactions occurring in the Sun.

Source: https://imagine.gsfc.nasa.gov/educators/lessons/xray_spectra/activity-fusion.html .

That is called Stellar nucleosynthesis, the process by which the natural abundances of the chemical elements within stars change due to nuclear fusion reactions in the cores and their overlying mantles.

Cross section of a supergiant showing nucleosynthesis and elements formed. Layers of the Sun - By User:Uber nemo - English Wikipedia, Public Domain, https://commons.wikimedia.org/w/index.php?curid=3495253

Now we have the 'light engine' running we next describe how visible light is produced (from the Sun).

Side Note: Your question suggests that the Sun (our Sun) must produce light for us to see it, of course that is not true, light from other Stars could reflect off of a dark ball and we could see the Sun (our Sun) after it is extinguished (though it is unlikely that we (Mankind) would be alive and inhabiting this area) - but we wander far from the question now.


Back to your question: "I do not understand how visible light is emitted ...".

Source for this answer: https://imagine.gsfc.nasa.gov/educators/gammaraybursts/imagine/page7.html .

Light is the familiar word for what physicists call electromagnetic radiation or electromagnetic waves. Light is a form of energy; it can travel through empty space and is in the form of individual wave packets called photons. The waves in packets of visible light are tiny ripples less than a millionth of a meter long.

When visible light is split up into its different wavelengths, the result is called a spectrum. Violet light has the shortest wavelength and red light has the longest — about twice as long as violet. Visible light is not the only form of electromagnetic radiation, however. The electromagnetic spectrum extends beyond the colors of the rainbow in both directions — to much shorter wavelengths than the violet and to much longer wavelengths than the red. At the longer wavelengths are radio waves, microwaves, and infrared radiation. At the shorter wavelengths are ultraviolet radiation, X-rays, and gamma-rays.

It is important to know that the Sun is not a Lambertian radiator (a circular disk of evenly emitted light). The Sun is not spherical but is instead described variously as a flattened disk, a quadrupole or a hexadecapole shape. Because it is mostly gaseous and liquid, with a solid core, each onion ring layer rotates at a different speed as does each latitude; this means different intensities of different wavelengths are emitted from different portions at different times, in both short period (minutes) variations and 11 year cycles - also sunspots and prominences alter the intensity of light at various wavelengths (black sunspots are cooler and brightly emit X-rays and high-energy particles).

Shape of the Sun: quadrupole or a hexadecapole

More information: Using Precise Solar Limb Shape Measurements to Study the Solar Cycle - By: J. R. Kuhn, L. E. Floyd, Claus Fröhlich, et. al. - Jan 2000.

In addition the brightness is affected in a more easily visible manner by what is called Limb Darkening (oversimplified that means that the edges of the Sun are thin and can't emit as much visible light as the central portion). A slightly more complicated explanation comes from Wikipedia's Limb Darkening Article, or for Doctoral Astrophysics see H. H. Plaskett's Limb darkening and solar rotation or this more recent (and readable) Article Max Planck Institute Article on Solar Variance.

This is how the intensity varies based on latitudes: Intensity vs. Latitude on the Sun

Note that measurement is for a specific range of visible light and does not apply longitudinally. What constitutes lateral and longitudinal on the Sun is determined by the Sun's Axis, which is determined by it's Magnetic Field, which varies with the flow of the undercurrents of the various layers.

In general light is emitted similarly to what a camera records during a Solar Eclipse, this is a grossly oversimplified explanation: Solar Eclipse - Source: http://www.zam.fme.vutbr.cz/~druck/Eclipse/Ecl2017u/Whiskey_m_500mm/0-info.htm

That explains how the light is created (including light, energy waves, that are not visible to the human eye) and how it's intensity varies based on location, angle, time, etc. that it is viewed. The actual color of a Sun is determined by it's temperature, see here for more information on spectrum and color vs. temperature (why are there no green Suns): https://science.nasa.gov/ems/09_visiblelight .

This is the Spectrum of visible light: Visible Light

This is where visible light occurs within the full Spectrum (of Energy): The Electromagnetic Spectrum

To understand the Universe, astronomers look at all wavelengths; the cosmic sky has a totally different appearance at different wavelengths of light.

At radio wavelengths, astronomers see distant quasars and hot gas in our Milky Way Galaxy. The infrared sky shows mainly tiny dust particles strewn through our Galaxy and other galaxies. Visible and ultraviolet show mainly the light from ordinary stars. X-rays reveal gas heated to millions of degrees lying between galaxies or falling onto compact objects like neutron stars and black holes. Gamma-rays can be produced only by extremely energetic phenomena, and we see several types of gamma-ray emission in the sky.

Gamma-rays seen along the plane of the Milky Way are not from ordinary stars, but from nuclear reactions generated by protons accelerated to nearly the speed of light slamming into gas lying between the stars. Gamma-rays are also seen from blazars -- intense beams of light and particles pointed directly at the Earth produced by massive black holes in distant galaxies. Gamma-rays can be detected in the magnetic flares on the surface of our Sun, and by the radioactive decay of short-lived atomic nuclei produced by supernova explosions in the Galaxy.

All objects in our Universe emit, reflect, and absorb electromagnetic radiation in their own distinctive ways. The way an object does this provides it special characteristics which scientists can use to probe an object’s composition, temperature, density, age, motion, distance, and other chemical and physical quantities.

...

We can think of electromagnetic radiation in several different ways:

• From a physical science standpoint, all electromagnetic radiation can be thought of as originating from the motions of subatomic particles. Gamma-rays occur when atomic nuclei are split or fused. X-rays occur when an electron orbiting close to an atomic nucleus is pushed outward with such force that it escapes the atom; ultraviolet, when an electron is jolted from a near to a far orbit; and visible and infrared, when electrons are jolted a few orbits out. Photons in these three energy ranges (X-ray, UV, and optical) are emitted as one of the outer shell electrons loses enough energy to fall down to the replace the electron missing from the inner shell. Radio waves are generated by any electron movement; even the stream of electrons (electric current) in a common household wire creates radio waves ...albeit with wavelengths of thousands of kilometers and of very weak amplitude.

• Electromagnetic radiation can be described in terms of a stream of photons (massless packets of energy), each traveling in a wave-like pattern, moving at the speed of light. The only difference between radio waves, visible light, and gamma-rays is the amount of energy in the photons. Radio waves have photons with low energies, microwaves have a little more energy than radio waves, infrared has still more, then visible, ultraviolet, X-rays, and gamma-rays. By the equation E=hf, energy dictates a photon’s frequency and, hence, wavelength.

The value of the EM radiation we receive from the Universe can be realized by considering the following: Temperatures in the Universe today range from $10^{10}$ Kelvin to $2.7$ Kelvin (in the cores of stars going supernova and in intergalactic space, respectively). Densities range over 45 orders of magnitude between the centers of neutron stars to the virtual emptiness of intergalactic space. Magnetic field strengths can range from the $10^{13}$ Gauss fields around neutron stars to the 1 Gauss fields of planets such as Earth to the $10^{-7}$ Gauss fields of intergalactic space. It is not possible to reproduce these enormous ranges in a laboratory on Earth and study the results of controlled experiments; we must use the Universe as our laboratory in order to see how matter and energy behave in these extreme conditions.

...

As suggested, please see: https://imagine.gsfc.nasa.gov/educators/gammaraybursts/imagine/page7.html for the unabridged version.

What you call visible light is based upon a few things, you can see the light because of these three things: it is produced and travels towards you, it passes through the atmosphere without being blocked and your eyes are sensitive to that frequency - some people are more sensitive to UV and IR light than others much as some people can hear higher or lower frequencies.

Atmospheric Absorption of Electromagnetic Radiation

Note how the Atmosphere creates 'windows' or filters that only allow certain wavelengths to penetrate a certain distance. Only visible light and a particular band of radio frequencies can penetrate all the way to the Earth's surface.

Your Camera and Scientific Instruments on Earth and in Space can 'see' a wider spectrum than your eyes can but those frequencies can be tone-mapped (un-HDR'ed) into images that your eyes can see and your brain might understand (like Radar can provide information to a trained observer, but we can not see radio waves).

See: https://imagine.gsfc.nasa.gov/science/toolbox/spectra1.html for more information.

For more information about the Electromagnetic Spectrum (visible light and the adjacent frequencies) see: https://imagine.gsfc.nasa.gov/science/toolbox/emspectrum1.html .

Additional information on emission of light from the Sun (it's journey from beneath the surface to our eyes) see: How do neutrinos pass through the sun so quickly? .

So that's how the Sun creates energy, some we can see with our eyes and most we can detect with instruments (and map to an image), how it travels and why some energy is blocked (preventing severe sunburn). You didn't ask how your eyes convert the energy so your brain can see so I won't go further, but that answer is on SE.

An Online Course from PSU geared towards younger readers is available here: https://www.e-education.psu.edu/astro801/content/l3.html .

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Hydrogen atoms fuse to form helium through the proton-proton chain which fuses four protons into one alpha particle (nucleus of ${}^{4}He$) and releases two neutrinos, two positrons and energy in the form of gamma photons. Although photons travel at the speed of light, the random motions they experienced inside the sun takes them thousand of years to leave the Sun' center. This random motion is due to the dense plasma in the Sun's interior since each photon permanently collides with an electron and gets deviated from its original path. The Energy released by fusion moves outward up to the top of the radiation zone, where the temperature drops to about 2 million K, then the photons get absorbed by the plasma more easily and this creates the necessary conditions for convection. This creates the convection zone of the zone. Then the plasma rises and the photons are carried to the photosphere where the density of the gas is low enough that they can escape. They mostly escape as visible photons, as their initial energy is lost through the random motion in the radiactive zone, and the absortion in the convective zone.

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    $\begingroup$ so, gamma photons due to the loss of energy are emitted as visible photons... Just one more thing, are these the photons that are absorbed by the surface elements of the sun and then re-emitted... i.e. the black body radiation. I am having difficulties linking the concept of the source of the light of the sun and the black body radiation $\endgroup$
    – Eliza
    Commented Mar 18, 2014 at 5:36
  • $\begingroup$ @eliza Think of it as a solid blackbody surface emitting thermal photons at around 6000K. With this surface being heated from inside by a 2million K gamma ray source $\endgroup$ Commented Oct 24, 2018 at 3:13
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    $\begingroup$ This is a very misleading answer. There is no sense in which the photons are transported from the centre to the surface. $\endgroup$
    – ProfRob
    Commented Mar 6, 2019 at 19:49
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As I write, there are answers with correct physics but I feel I can add some clarity.

The question notes that fusion reactions in the Sun give mostly gamma-ray photons, yet most of the light emitted by the Sun is in the visible and infrared part of the spectrum. So how can this be? The question also refers to a misleading picture in which photons "fight" their way to the surface over thousands of years.

As ProfRob points out, individual photons do not travel very far in the Sun before they are absorbed, typically by a collision with an electron.

The essential point here is that electromagnetic radiation is produced whenever charged particles accelerate. And, similarly, electromagnetic radiation can be absorbed when it is incident on a charged particle so as to cause acceleration. In the interior of the Sun both types of process are going on all the time. The other important factor is the collisions among all the particles, which tend to redistribute their kinetic energy so that the result is a thermal equilibrium state. Each region of the Sun thus reaches a temperature. It is about 15 million kelvin (or degrees celcius) in the core, and about 5800 kelvin in the outer part called photosphere where the light we see is produced.

At any given temperature, the charged particles in the plasma have a mean kinetic energy proportional to that temperature. They are milling around colliding with one another. These collisions produce acceleration which in turn results in the emission of light. The typical energy of the photons produced is similar to the typical energy of the thermal motion of the charged particles. It is not self-evident that that will be the case. It is related to the fact that there is energy exchange in both directions between the electromagnetic radiation and the charged particles, so that they can come to equilibrium with each other.

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    $\begingroup$ Also, only part of the energy released in fusion is in the form of gamma-ray photons. A significant portion is released as the kinetic energy of the reaction products. I have some info about the energy of the first step of the p-p chain here: astronomy.stackexchange.com/a/48281/16685 $\endgroup$
    – PM 2Ring
    Commented Jun 1, 2023 at 11:00

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