How to derive full solar sprectrum from gamma rays

In summary, the gamma rays that are generated in the core of the Sun are transformed into the full spectrum of visible light photons that we see from Earth.
  • #1
jordankonisky
41
3
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?
 
Astronomy news on Phys.org
  • #2
jordankonisky said:
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?

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.
 
  • #3
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
 
  • #4
drvrm said:
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.
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.
 
  • #5
Ken G said:
So the chromosphere is actually hotter than the photosphere, and tends to produce emission rather than absorption.

but the temperatures are of the order of 4300 degreeK of the chromosphere - lower than photosphere...
 
  • #6
No, that's what I'm saying, that is not correct.
 
  • #7
  • #8
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.
 
  • Like
Likes drvrm
  • #9
  • #10
DrSteve said:
The temperature profile of the chromosphere is non-trivial.
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.
 
  • Like
Likes Drakkith and DrSteve
  • #11
my2cts said:
Within 3/4 of the solar radius energy is transported out by radiation.
Not being an expert, I would expect this radiation to be in the form plasma waves.
In astronomy, "radiation" almost always means just plain light (not necessarily visible light), as it does here.
 
  • #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?
 
  • #13
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.
 
  • #14
Ken G said:
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.

I understand what you reply is about, but I don't see how it addresses my question.
 
  • #15
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).
 
  • Like
Likes Drakkith
  • #16
jordankonisky said:
I just don't understand how physically the full spectrum of photon energy at each temperature would be generated?

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 ...
 
  • #17
drvrm said:
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 ...
Thanks do much for the clear explanation.
 
  • Like
Likes drvrm

1. How do you derive the full solar spectrum from gamma rays?

The full solar spectrum can be derived by analyzing the energy levels and wavelengths of the different types of radiation emitted by the sun. This includes not only gamma rays, but also x-rays, ultraviolet light, visible light, infrared radiation, and radio waves. By measuring and plotting the intensity of each type of radiation, we can create a full spectrum that shows the range of energies and wavelengths present in the sun's radiation.

2. What instruments are needed to derive the full solar spectrum?

To accurately measure and analyze the full solar spectrum, scientists use a variety of instruments including spectrometers, telescopes, and detectors. These instruments are designed to detect and measure different types of radiation, and are often used in combination to provide a complete picture of the sun's spectrum.

3. Why is it important to derive the full solar spectrum?

Deriving the full solar spectrum is crucial for understanding the physical processes that occur in the sun and for studying the effects of solar radiation on Earth. It also allows us to better understand the composition and structure of the sun, as well as to make predictions about solar activity and potential impacts on our planet.

4. Can the full solar spectrum change over time?

Yes, the full solar spectrum can change over time due to fluctuations in the sun's activity and changes in its composition. For example, during periods of increased solar activity, the intensity of certain types of radiation may be higher, resulting in a slightly different spectrum. Additionally, changes in the sun's magnetic field and the presence of sunspots can also affect the full solar spectrum.

5. How is the data from the full solar spectrum used in scientific research?

The data from the full solar spectrum is used in a wide range of scientific research, including solar physics, atmospheric science, and climate studies. By analyzing the different components of the solar spectrum, scientists can learn more about the sun's energy output, the effects of solar radiation on Earth's atmosphere, and how the sun's activity impacts our climate. This information is also important for developing technologies and strategies to mitigate potential risks associated with solar flares and other solar events.

Similar threads

Replies
12
Views
2K
  • Astronomy and Astrophysics
Replies
1
Views
2K
  • Sci-Fi Writing and World Building
Replies
6
Views
653
  • Sci-Fi Writing and World Building
Replies
2
Views
1K
  • Astronomy and Astrophysics
Replies
5
Views
1K
  • Astronomy and Astrophysics
Replies
1
Views
1K
Replies
5
Views
1K
  • High Energy, Nuclear, Particle Physics
Replies
12
Views
2K
  • Advanced Physics Homework Help
Replies
8
Views
853
  • Astronomy and Astrophysics
Replies
4
Views
8K
Back
Top