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Can we see emission lines in stars?
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I read that stars show both absorption and emission lines. While how we see absorption lines is clear to me, I don't understand how (and when) we see the emission lines. Based on what I read in several of your threads such as Why don't absorption and emission lines cancel out in our Sun? we should not be seeing emission lines at all, since all the emissions would come from the hotter interior but those can't reach the outside (unless I misunderstood this point), and the only emission we see is blackbody radiation. But then how do we see the few emission lines such as in the spectrum shown here?
astrophysics atomic-physics thermal-radiation spectroscopy stellar-physics
edited 1 hour ago
Qmechanic♦
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Marta
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up vote
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I read that stars show both absorption and emission lines. While how we see absorption lines is clear to me, I don't understand how (and when) we see the emission lines. Based on what I read in several of your threads such as Why don't absorption and emission lines cancel out in our Sun? we should not be seeing emission lines at all, since all the emissions would come from the hotter interior but those can't reach the outside (unless I misunderstood this point), and the only emission we see is blackbody radiation. But then how do we see the few emission lines such as in the spectrum shown here?
astrophysics atomic-physics thermal-radiation spectroscopy stellar-physics
edited 1 hour ago
Qmechanic♦
98.8k121781090
asked 7 hours ago
Marta
161
New contributor
Marta is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
I read that stars show both absorption and emission lines. While how we see absorption lines is clear to me, I don't understand how (and when) we see the emission lines. Based on what I read in several of your threads such as Why don't absorption and emission lines cancel out in our Sun? we should not be seeing emission lines at all, since all the emissions would come from the hotter interior but those can't reach the outside (unless I misunderstood this point), and the only emission we see is blackbody radiation. But then how do we see the few emission lines such as in the spectrum shown here?
astrophysics atomic-physics thermal-radiation spectroscopy stellar-physics
edited 1 hour ago
Qmechanic♦
98.8k121781090
asked 7 hours ago
Marta
161
New contributor
Marta is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I read that stars show both absorption and emission lines. While how we see absorption lines is clear to me, I don't understand how (and when) we see the emission lines. Based on what I read in several of your threads such as Why don't absorption and emission lines cancel out in our Sun? we should not be seeing emission lines at all, since all the emissions would come from the hotter interior but those can't reach the outside (unless I misunderstood this point), and the only emission we see is blackbody radiation. But then how do we see the few emission lines such as in the spectrum shown here?
astrophysics atomic-physics thermal-radiation spectroscopy stellar-physics
astrophysics atomic-physics thermal-radiation spectroscopy stellar-physics
edited 1 hour ago
Qmechanic♦
98.8k121781090
asked 7 hours ago
Marta
161
New contributor
Marta is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 1 hour ago
Qmechanic♦
98.8k121781090
asked 7 hours ago
Marta
161
New contributor
Marta is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 1 hour ago
Qmechanic♦
98.8k121781090
edited 1 hour ago
Qmechanic♦
98.8k121781090
edited 1 hour ago
Qmechanic♦
98.8k121781090
98.8k121781090
asked 7 hours ago
Marta
161
New contributor
Marta is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 7 hours ago
Marta
161
asked 7 hours ago
Marta
161
161
New contributor
Marta is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Marta is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Marta is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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2 Answers
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4
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Light emitted in the interior of a star is largely thermalized by the time it works its way out, and will have a black body spectrum.
The interesting parts of the spectrum is due to molecules and atoms in the corona: the "atmosphere" of the star.
Any gas that is ionized or at least excited to something above the ground state will produce an emission spectrum. Practically all the gas in the first several tens of thousand miles above the star's "surface" will be ionized and emitting light at a high rate, producing emission lines. Consistent with the link you provided, a molecule (or atom) will absorb light only if the light frequency matches an available energy transition in the molecule. If the molecule is already excited above the bottom level of that transition, the transition is not available. However, as long as there is cooler gas at high levels in the stellar atmosphere, there will be absorption lines.
answered 4 hours ago
S. McGrew
4,6962921
add a comment |Â
up vote
1
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In general, emission lines occur in gases that are optically thin, or in atmospheres that have a temperature inversion (i.e. are hotter closer to the observer).
In an optically thin gas, if excited atoms or ions are radiatively deexcited, then the photons produced can escape to an observer. If continuum absorption/emission is unlikely then this leads to an observed emission line. Examples include most nebulae.
The second case arises because if you have a hot layer on top of a cool layer, then the light we receive from the "photosphere" will come from a hotter layer at the wavelength of a strong radiative transition (closer to the observer) compared with the continuum. As the hotter layer is brighter, we see an emission line. The exact opposite case to the formation of an absorption line, which arises when light at the central wavelength of a transition arises from a higher, cooler layer.
A temperature inversion requires a way of non-radiatively depositing heat in the hotter layer, otherwise radiative diffusion would eliminate it. An example is a stellar chromosphere, which is above and hotter than the photosphere, and heated by magnetic fields. This can produce high temperature emission lines.
answered 2 hours ago
Rob Jeffries
66.6k7131224
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
Light emitted in the interior of a star is largely thermalized by the time it works its way out, and will have a black body spectrum.
The interesting parts of the spectrum is due to molecules and atoms in the corona: the "atmosphere" of the star.
Any gas that is ionized or at least excited to something above the ground state will produce an emission spectrum. Practically all the gas in the first several tens of thousand miles above the star's "surface" will be ionized and emitting light at a high rate, producing emission lines. Consistent with the link you provided, a molecule (or atom) will absorb light only if the light frequency matches an available energy transition in the molecule. If the molecule is already excited above the bottom level of that transition, the transition is not available. However, as long as there is cooler gas at high levels in the stellar atmosphere, there will be absorption lines.
answered 4 hours ago
S. McGrew
4,6962921
add a comment |Â
up vote
4
down vote
Light emitted in the interior of a star is largely thermalized by the time it works its way out, and will have a black body spectrum.
The interesting parts of the spectrum is due to molecules and atoms in the corona: the "atmosphere" of the star.
Any gas that is ionized or at least excited to something above the ground state will produce an emission spectrum. Practically all the gas in the first several tens of thousand miles above the star's "surface" will be ionized and emitting light at a high rate, producing emission lines. Consistent with the link you provided, a molecule (or atom) will absorb light only if the light frequency matches an available energy transition in the molecule. If the molecule is already excited above the bottom level of that transition, the transition is not available. However, as long as there is cooler gas at high levels in the stellar atmosphere, there will be absorption lines.
answered 4 hours ago
S. McGrew
4,6962921
add a comment |Â
up vote
4
down vote
up vote
4
down vote
Light emitted in the interior of a star is largely thermalized by the time it works its way out, and will have a black body spectrum.
The interesting parts of the spectrum is due to molecules and atoms in the corona: the "atmosphere" of the star.
Any gas that is ionized or at least excited to something above the ground state will produce an emission spectrum. Practically all the gas in the first several tens of thousand miles above the star's "surface" will be ionized and emitting light at a high rate, producing emission lines. Consistent with the link you provided, a molecule (or atom) will absorb light only if the light frequency matches an available energy transition in the molecule. If the molecule is already excited above the bottom level of that transition, the transition is not available. However, as long as there is cooler gas at high levels in the stellar atmosphere, there will be absorption lines.
answered 4 hours ago
S. McGrew
4,6962921
Light emitted in the interior of a star is largely thermalized by the time it works its way out, and will have a black body spectrum.
The interesting parts of the spectrum is due to molecules and atoms in the corona: the "atmosphere" of the star.
Any gas that is ionized or at least excited to something above the ground state will produce an emission spectrum. Practically all the gas in the first several tens of thousand miles above the star's "surface" will be ionized and emitting light at a high rate, producing emission lines. Consistent with the link you provided, a molecule (or atom) will absorb light only if the light frequency matches an available energy transition in the molecule. If the molecule is already excited above the bottom level of that transition, the transition is not available. However, as long as there is cooler gas at high levels in the stellar atmosphere, there will be absorption lines.
answered 4 hours ago
S. McGrew
4,6962921
answered 4 hours ago
S. McGrew
4,6962921
answered 4 hours ago
S. McGrew
4,6962921
answered 4 hours ago
S. McGrew
4,6962921
4,6962921
add a comment |Â
add a comment |Â
up vote
1
down vote
In general, emission lines occur in gases that are optically thin, or in atmospheres that have a temperature inversion (i.e. are hotter closer to the observer).
In an optically thin gas, if excited atoms or ions are radiatively deexcited, then the photons produced can escape to an observer. If continuum absorption/emission is unlikely then this leads to an observed emission line. Examples include most nebulae.
The second case arises because if you have a hot layer on top of a cool layer, then the light we receive from the "photosphere" will come from a hotter layer at the wavelength of a strong radiative transition (closer to the observer) compared with the continuum. As the hotter layer is brighter, we see an emission line. The exact opposite case to the formation of an absorption line, which arises when light at the central wavelength of a transition arises from a higher, cooler layer.
A temperature inversion requires a way of non-radiatively depositing heat in the hotter layer, otherwise radiative diffusion would eliminate it. An example is a stellar chromosphere, which is above and hotter than the photosphere, and heated by magnetic fields. This can produce high temperature emission lines.
answered 2 hours ago
Rob Jeffries
66.6k7131224
add a comment |Â
up vote
1
down vote
In general, emission lines occur in gases that are optically thin, or in atmospheres that have a temperature inversion (i.e. are hotter closer to the observer).
In an optically thin gas, if excited atoms or ions are radiatively deexcited, then the photons produced can escape to an observer. If continuum absorption/emission is unlikely then this leads to an observed emission line. Examples include most nebulae.
The second case arises because if you have a hot layer on top of a cool layer, then the light we receive from the "photosphere" will come from a hotter layer at the wavelength of a strong radiative transition (closer to the observer) compared with the continuum. As the hotter layer is brighter, we see an emission line. The exact opposite case to the formation of an absorption line, which arises when light at the central wavelength of a transition arises from a higher, cooler layer.
A temperature inversion requires a way of non-radiatively depositing heat in the hotter layer, otherwise radiative diffusion would eliminate it. An example is a stellar chromosphere, which is above and hotter than the photosphere, and heated by magnetic fields. This can produce high temperature emission lines.
answered 2 hours ago
Rob Jeffries
66.6k7131224
add a comment |Â
up vote
1
down vote
up vote
1
down vote
In general, emission lines occur in gases that are optically thin, or in atmospheres that have a temperature inversion (i.e. are hotter closer to the observer).
In an optically thin gas, if excited atoms or ions are radiatively deexcited, then the photons produced can escape to an observer. If continuum absorption/emission is unlikely then this leads to an observed emission line. Examples include most nebulae.
The second case arises because if you have a hot layer on top of a cool layer, then the light we receive from the "photosphere" will come from a hotter layer at the wavelength of a strong radiative transition (closer to the observer) compared with the continuum. As the hotter layer is brighter, we see an emission line. The exact opposite case to the formation of an absorption line, which arises when light at the central wavelength of a transition arises from a higher, cooler layer.
A temperature inversion requires a way of non-radiatively depositing heat in the hotter layer, otherwise radiative diffusion would eliminate it. An example is a stellar chromosphere, which is above and hotter than the photosphere, and heated by magnetic fields. This can produce high temperature emission lines.
answered 2 hours ago
Rob Jeffries
66.6k7131224
In general, emission lines occur in gases that are optically thin, or in atmospheres that have a temperature inversion (i.e. are hotter closer to the observer).
In an optically thin gas, if excited atoms or ions are radiatively deexcited, then the photons produced can escape to an observer. If continuum absorption/emission is unlikely then this leads to an observed emission line. Examples include most nebulae.
The second case arises because if you have a hot layer on top of a cool layer, then the light we receive from the "photosphere" will come from a hotter layer at the wavelength of a strong radiative transition (closer to the observer) compared with the continuum. As the hotter layer is brighter, we see an emission line. The exact opposite case to the formation of an absorption line, which arises when light at the central wavelength of a transition arises from a higher, cooler layer.
A temperature inversion requires a way of non-radiatively depositing heat in the hotter layer, otherwise radiative diffusion would eliminate it. An example is a stellar chromosphere, which is above and hotter than the photosphere, and heated by magnetic fields. This can produce high temperature emission lines.
answered 2 hours ago
Rob Jeffries
66.6k7131224
edited 2 hours ago
answered 2 hours ago
Rob Jeffries
66.6k7131224
answered 2 hours ago
Rob Jeffries
66.6k7131224
answered 2 hours ago
Rob Jeffries
66.6k7131224
66.6k7131224
add a comment |Â
add a comment |Â
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