Mass Equivalence of Photon Content Inside the Sun?

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If it takes anywhere between 5,000 to a couple of hundred thousand years (various internet sources have various values) for photons generated in the sun’s core to reach its surface and radiate out, what is the estimated mass equivalence of all these photons making their way out from the core to the surface?

Is it a significant percentage of the sun’s total mass?

And does this ratio of photon mass equivalence to total star mass vary with stellar evolution and/or star type?


IH
 

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  • #2
Drakkith
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If it takes anywhere between 5,000 to a couple of hundred thousand years (various internet sources have various values) for photons generated in the sun’s core to reach its surface and radiate out
I don't like this description. Photons are created and destroyed all the time inside the Sun, and as far as I know you would not be able to identify a single photon and trace it outwards from the core to the photosphere. A newly created photon should be absorbed almost immediately in the core, which destroys it. The hot charged particles then generate new photons of various frequencies when they interact, which are also quickly absorbed. The net effect is a slow propagation of energy outwards from the core to the photosphere and into space, with more photons of lower frequencies being generated on average as the distance from the core increases and the temperature decreases. So one photon eventually turns into hundreds, thousands, or more.

Note that this ignores other forms of energy transfer like convection and conduction. Both of these occur as well, but at different rates in different regions.

what is the estimated mass equivalence of all these photons making their way out from the core to the surface?
Good question. I think it would be easier to ask how much internal energy the Sun contains and then to convert that energy into a mass equivalent. Unfortunately I don't how much internal energy the Sun contains. :frown:
 
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  • #3
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I don't like this description. Photons are created and destroyed all the time inside the Sun, and as far as I know you would not be able to identify a single photon and trace it outwards from the core to the photosphere. A newly created photon should be absorbed almost immediately in the core, which destroys it. The hot charged particles then generate new photons of various frequencies when they interact, which are also quickly absorbed. The net effect is a slow propagation of energy outwards from the core to the photosphere and into space, with more photons of lower frequencies being generated on average as the distance from the core increases and the temperature decreases. So one photon eventually turns into hundreds, thousands, or more.

Note that this ignores other forms of energy transfer like convection and conduction. Both of these occur as well, but at different rates in different regions.



Good question. I think it would be easier to ask how much internal energy the Sun contains and then to convert that energy into a mass equivalent. Unfortunately I don't how much internal energy the Sun contains. :frown:

Thanks for your clarification Drakkith, intuitively what you wrote makes more sense than what I read here and there and wrote in this present thread.

Will try to find some source for the sun’s internal energy but this seems no to be a practical figure so finding it won’t be easy...


IH
 
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Orodruin
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I don't like this description.
You are being to kind to it. I would even go so far as calling it bogus. Also, if photons created in the core did escape the Sun we see from Earth would not be a black body spectrum at the temperature of the Sun's surface. Instead, the energy released in the thermonuclear processes in the Sun's centre is quickly absorbed as heat and the heat transfer to the Sun's surface occurs through various processes.
 
  • #5
Bandersnatch
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Is it a significant percentage of the sun’s total mass?
Rather than focusing on the dubious picture of a single photon travelling to the surface, one can reframe the question to: is the mass equivalent of all the photons present in the Sun's interior at any one time a significant fraction of its total mass?

If you take the current solar luminosity, multiply it by its lifetime on the main sequence, and then convert it to mass equivalent, you'll end up with less than 0.1 of a percent (1/1000th) of its total mass.
That is to say, towards the end of its life, the Sun will have converted only a tiny fraction of its mass to energy.

This already tells you that whatever the proportion of photons streaming through the solar interior at any one time is, its mass equivalent must be essentially negligible when compared to the mass of matter the Sun is made of.
 
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Rather than focusing on the dubious picture of a single photon travelling to the surface, one can reframe the question to: is the mass equivalent of all the photons present in the Sun's interior at any one time a significant fraction of its total mass?

If you take the current solar luminosity, multiply it by its lifetime on the main sequence, and then convert it to mass equivalent, you'll end up with less than 0.1 of a percent (1/1000th) of its total mass.
That is to say, towards the end of its life, the Sun will have converted only a tiny fraction of its mass to energy.

This already tells you that whatever the proportion of photons streaming through the solar interior at any one time is, its mass equivalent must be essentially negligible when compared to the mass of matter the Sun is made of.

I find that amazing. But then we are only talking of photons, right? With solar emission of neutrinos and the particulate bit of the solar wind, the sun would have lost a much bigger percentage of its mass over its lifetime, would it not?


IH
 
  • #7
Vanadium 50
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Everyone is overthinking things.

The binding energy of 4He is 7 MeV/nucleon. That means that if you burned all the hydrogen to helium and captured all the energy, the mass equivalent of the energy is 7 MeV/ 938 MeV (the proton's mass) or 0.075%. Call it 10-3. It would take 1011 years to do this, so if the sun contains 104 years of light, the light is 10-10 times the mass of the sun.
 
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  • #8
I would attempt to familiarize yourself with how the sun generates it's energy. Photons are a byproduct of processes such as the proton-proton-chain and CNO cycle. You could then solve the question you are asking by doing some multiplication and subtraction of the total valuesvalues. Also understand that photons do take an extremely long time to radiate outwards from the core due to the intense magnetic fields inside (and out) the sun. Photons are only created inside the core where fusion happens, not anywhere else because nowhere else is fusion happening.
 
  • #9
http://www.live-counter.com/sun/

Does this help?

the sun "burns" about 564 million tons hydrogen per second, resulting in 559.7 million tons of helium. The loss of mass, about 4.3 million tons per second, is transformed into energy. But don't worry, it's only 0.0000000000000000002 percent of the sun's entire mass.
 
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davenn
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Photons are only created inside the core where fusion happens, not anywhere else because nowhere else is fusion happening.
I cannot agree with that,
firstly, you DONT NEED fusion to create photons … our electric lights wouldn't work if that was the case and a zillion other examples.
we get photon production from all over the surface of the sun from its features … filaments, sunspots, prominences … not forgetting, the Corona !!!!
else it would be pointless me pointing my photon collector ( telescope) at the sun and getting pictures like this …..

180617 Ha Cap_015.jpg



180707 Cap024-2.jpg






Dave
 

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Drakkith
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Like davenn said, photons are not solely generated from fusion reactions. The Sun is composed of an enormous number of hot, charged particles and every time these particles undergo acceleration they release photons. Since each and every collision or close pass between these particles is an acceleration event, an enormous number of photons are created everywhere in the Sun every second.
 
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davenn
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