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I How to derive full solar sprectrum from gamma rays

  1. Sep 9, 2016 #1
    I understand that the fusion of hydrogen to helium in our sun's core generates gamma rays. My question is how are these gamma rays transformed to the full spectrum of photons that we that we observe from earth?
     
  2. jcsd
  3. Sep 9, 2016 #2
    it may be wrong to think that gamma rays are responsible for visible spectrum of the Sun -gamma rays may be responsible for heating the star to a high temperature at the core...
    In the Sun, a continuum/black body spectrum is produced by the part of the Sun that we see as its surface called Photosphere.

    The temperature of photosphere is about 5800 degree K and so, by Wien’s Law, we can see that the bulk of its energy is emitted as visible light photons.

    Outside of the photosphere is a more diffuse, cooler ( at 4500K) level known as the Chromosphere. Atoms within the chromosphere absorb some of the photons, causing absorption lines (Dark lines) in the solar spectrum.
     
  4. Sep 9, 2016 #3

    mathman

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    Te gamma rays at the core collide with protons, electrons, etc. that are the constituents of the sun. It takes a long time for the photons produced in the core to reach the surface, having undergone very many collisions and losing energy along the way.

    http://www.astronoo.com/en/articles/journey-of-the-photon.html
     
  5. Sep 9, 2016 #4

    Ken G

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    You're thinking of the upper layers of the photosphere, which get cooler since temperature drops as the light diffuses out. But once you get to a low enough density that the light just streams out, there is no longer any need for the temperature to continue to drop, and in fact what actually happens (kind of like in the Earth's atmosphere) is the temperature starts to rise again. So the chromosphere is actually hotter than the photosphere, and tends to produce emission rather than absorption.
     
  6. Sep 9, 2016 #5
    but the temperatures are of the order of 4300 degreeK of the chromosphere - lower than photosphere...
     
  7. Sep 9, 2016 #6

    Ken G

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    No, that's what I'm saying, that is not correct.
     
  8. Sep 10, 2016 #7

    DrSteve

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  9. Sep 10, 2016 #8

    Drakkith

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    I don't recommend trying to imagine a bunch of gamma rays propagating out of the core, gradually losing energy as they go, until they become black-body photons radiated from the Sun's surface. The gamma rays are absorbed soon after they are created and this energy can then be transformed into something else. For example, right after absorbing a gamma ray, an electron or ion may then collide with another particle, producing many photons of a lower frequency.
     
  10. Sep 10, 2016 #9
  11. Sep 10, 2016 #10

    Ken G

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    Indeed, in fact it depends a lot on how one defines the chromosphere. It seems the Wiki is using the definition of everything above the optical depth of 2/3 point, which is what some call the "photosphere" because it is the average depth of last emission of sunlight. That's not a very physically relevant definition, however-- it treats the "photosphere" as if it was the surface of a sphere, but the "chromosphere" is a spherical shell of finite thickness-- not very consistent terminology. Instead, a more meaningful definition is to look at the physics, and call the "photosphere" the region where the energy equation is dominated by radiative diffusion of sunlight (so, above the convection zone), and the "chromosphere" the region where the energy equation is dominated by mechanical heating (be it wave or magnetic heating) and radiative cooling from lines. The crossover between those types of physics occurs as the density drops and the gas becomes very easy to heat mechanically rather than by absorption of photons, but the mechanical heating also produces what is called a "temperature inversion" similar to what we see in Earth's atmosphere. Indeed, an alternative less physical but pretty simple definition of the chromosphere is everything above the temperature minimum. That definition would mean that the chromosphere starts out rather cool, but very rapidly gets much hotter as you go up. In any event, the chromosphere is characterized by being hot, rather than by being cool.

    This last point is even more true if we go away from static, fixed-temperature models, and start to treat the chromosphere more realistically as a highly dynamical layer of the Sun. Since it is characterized by mechanical heating, and this heating tends to be rather sudden and varied in space and time, a more sophisticated view of the chromosphere is that it is comprised of many stochastically distributed mini versions of solar flares. Then instead of a spherical shell, we think of the chromosphere as comprised of dynamical "spicules", which are like hot fingers protruding from the Sun in regions where the heating is going on, combined with cooler gas that is raining back down having had its heating shut off. So it's a complex and dynamical layer, that would indeed by most definitions include some fairly cool gas, but overall it is characterized by low-density gas that has been mechanically heated to temperatures significantly above those of the solar surface. This is why it glows red from Balmer alpha emission (a type of emission you don't get much of at the temperature of the surface of the Sun), and that red glow is the "chromo" in chromosphere.
     
  12. Sep 10, 2016 #11

    Ken G

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    In astronomy, "radiation" almost always means just plain light (not necessarily visible light), as it does here.
     
  13. Sep 10, 2016 #12
    Thanks to all of you for your comments which really helped. However I do have a question about Wien's law. As I understand it the law, the distribution of wavelengths is solely a function of temperature, but independent of the nature of the underlying physical entities. But suppose we heat a plasma of consisting of only protons and electrons. Would Wien's law still apply? I just don't understand how physically the full spectrum of photon energy at each temperature would be generated?
     
  14. Sep 10, 2016 #13

    Ken G

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    It's one of the amazing thing about thermal physics, there are a lot of situations where the behavior you see is just the most likely given the energy constraints-- the details don't matter. You can think of it as a kind of general case of the central limit theorem, wherein you can expect a certain type of distribution (in that case, Gaussian instead of Planckian) simply because it is the one that happens in the most ways, so it is the most likely-- regardless of the details of the system that is creating it. So just like if I looked at the height of a random collection of people at a given age, I expect a Gaussian distribution even if I know nothing of the physics of growing or the biology of DNA, similarly you can expect a Planckian radiation field coming from a "blackbody" (a system that interacts strongly with light) at a given temperature, regardless of the details of what the system is made of or how it actually does interact with light.
     
  15. Sep 10, 2016 #14
    I understand what you reply is about, but I don't see how it addresses my question.
     
  16. Sep 10, 2016 #15

    Ken G

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    The answer was, it doesn't matter how the plasma makes that light, it only matters that it is a thermal process coming from a plasma in a narrow temperature range that is strongly interacting with light. That's all you need to get that spectrum. The details of the interaction are quite varied and depend on which layer of the Sun you are talking about-- hotter ionized layers have free electrons scattering and absorbing photons (the scattering is called Thomson scattering, the absorption and emission is called bremsstrahlung), lower neutral layers have a lot of emission when a neutral hydrogen captures an extra electron and becomes an H- ion. The potential complexity is such that it is a fortunate thing indeed that we can understand a thermal spectrum without knowing the range of possible processes involved, but it sounds like your question is about those processes, so that's the list of the most important ones. Of course there's also emission in normal neutral hydrogen lines, but they tend to be in rather narrow wavelength regimes, except for UV wavelengths shorter than the Lyman continuum (where you get continuum emission when protons and electrons come together to make neutral hydrogen).
     
  17. Sep 11, 2016 #16
    i think when one looks up the 'wien's distribution' its an emperical formulation on the basis of observations of intensity distribution of a 'black-body' and for theoretical understanding we go to the Planck's radiation law based on oscillators....suppose one artificially arranges some particles at some thermal state ...the whole spectrum may not be generated ...just logically the emission mechanism must be there for the radiation to come out and contribute....one should employ simple logic ....
     
  18. Sep 13, 2016 #17
    Thanks do much for the clear explanation.
     
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