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Is Sun brighter than what we actually see?
The 2019 Stack Overflow Developer Survey Results Are In
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Why is the sky never green? It can be blue or orange, and green is in between!The Sun Can Make Stuff Hotter Than ItselfRadiation pressure on a Dyson sphereJupiter radiation belt: where >MeV particles comes from?Would it be possible to detect nuclear explosion on exoplanet?What would it be like “inside” a star?How does the sun's surface conduct thermal energy from the convective zone to the corona?Energy transport mechanism in starsWhat will happen if the nuclear reactions in the Sun stop suddenly?Why does the gas in Sun's outer layer not expand?Can we see emission lines in stars?
$begingroup$
I learned from that plasma can reflect radiations of frequency less than that of its own oscillations. If so, considering the plasma in Sun's atmosphere, it should also reflect solar radiations.
That would mean that the radiation emitted from the inner layers of the Sun would be reflected back by the outer layers. So, the only radiation coming out should be the ones generated at the outer layers, for which there is no denser layers of plasma surrounding it. And of course, the ones that have higher frequencies than the plasma in each layer would come out unscathed.
If this is true, most of the radiation generated by fusion will be trapped inside, and what we observe is only a fraction.
Note that the intensity of observable radiation coming out from stars would now mostly depend on the outermost layer. So, wouldn't it be inappropriate to consider stars as Black bodies while determining their temperature and other properties? Is Sun brighter and hotter than what we see from outside?
astrophysics radiation thermal-radiation plasma-physics
$endgroup$
add a comment |
$begingroup$
I learned from that plasma can reflect radiations of frequency less than that of its own oscillations. If so, considering the plasma in Sun's atmosphere, it should also reflect solar radiations.
That would mean that the radiation emitted from the inner layers of the Sun would be reflected back by the outer layers. So, the only radiation coming out should be the ones generated at the outer layers, for which there is no denser layers of plasma surrounding it. And of course, the ones that have higher frequencies than the plasma in each layer would come out unscathed.
If this is true, most of the radiation generated by fusion will be trapped inside, and what we observe is only a fraction.
Note that the intensity of observable radiation coming out from stars would now mostly depend on the outermost layer. So, wouldn't it be inappropriate to consider stars as Black bodies while determining their temperature and other properties? Is Sun brighter and hotter than what we see from outside?
astrophysics radiation thermal-radiation plasma-physics
$endgroup$
1
$begingroup$
You're almost right. As I recall (but would have to double-check), a photon emitted from the Sun's core takes ~1000 years before it escapes from the surface. Not the "same photon", of course, having experienced numerous various-and-sundry processes along the way. But, in any event, the Sun's ultimately in equilibrium (and let's hope it stays that way for a good long time), so everything generated is ultimately emitted.
$endgroup$
– John Forkosh
yesterday
$begingroup$
That is why the last dense opaque layer is called photosphere (indeed at about 5000 K versus a dozen millions K at the core). Outer of the photosphere stars have warmer but less dense not opaque layers. About the figure 1000 in @John Forkosh comment I ear higher values, such as 10000 and more, but in the very last case it was a TV program so I don't know. Already forgot what serious books said :(
$endgroup$
– Alchimista
yesterday
2
$begingroup$
Since there cannot be a energy buildup inside the Sun, the energy flowing across any interior spherical surface should be the same as the energy that eventually comes out. The inner surface has a higher temperature but less surface area and hence radiates less, the outer surface has a lower temperature but a higher surface area. The temperature distribution of the Sun is in such an equilibrium that the energy flux radially outwards is constant.
$endgroup$
– Kevin Selva Prasanna
yesterday
add a comment |
$begingroup$
I learned from that plasma can reflect radiations of frequency less than that of its own oscillations. If so, considering the plasma in Sun's atmosphere, it should also reflect solar radiations.
That would mean that the radiation emitted from the inner layers of the Sun would be reflected back by the outer layers. So, the only radiation coming out should be the ones generated at the outer layers, for which there is no denser layers of plasma surrounding it. And of course, the ones that have higher frequencies than the plasma in each layer would come out unscathed.
If this is true, most of the radiation generated by fusion will be trapped inside, and what we observe is only a fraction.
Note that the intensity of observable radiation coming out from stars would now mostly depend on the outermost layer. So, wouldn't it be inappropriate to consider stars as Black bodies while determining their temperature and other properties? Is Sun brighter and hotter than what we see from outside?
astrophysics radiation thermal-radiation plasma-physics
$endgroup$
I learned from that plasma can reflect radiations of frequency less than that of its own oscillations. If so, considering the plasma in Sun's atmosphere, it should also reflect solar radiations.
That would mean that the radiation emitted from the inner layers of the Sun would be reflected back by the outer layers. So, the only radiation coming out should be the ones generated at the outer layers, for which there is no denser layers of plasma surrounding it. And of course, the ones that have higher frequencies than the plasma in each layer would come out unscathed.
If this is true, most of the radiation generated by fusion will be trapped inside, and what we observe is only a fraction.
Note that the intensity of observable radiation coming out from stars would now mostly depend on the outermost layer. So, wouldn't it be inappropriate to consider stars as Black bodies while determining their temperature and other properties? Is Sun brighter and hotter than what we see from outside?
astrophysics radiation thermal-radiation plasma-physics
astrophysics radiation thermal-radiation plasma-physics
edited yesterday
Krishnanand J
asked yesterday
Krishnanand JKrishnanand J
8491325
8491325
1
$begingroup$
You're almost right. As I recall (but would have to double-check), a photon emitted from the Sun's core takes ~1000 years before it escapes from the surface. Not the "same photon", of course, having experienced numerous various-and-sundry processes along the way. But, in any event, the Sun's ultimately in equilibrium (and let's hope it stays that way for a good long time), so everything generated is ultimately emitted.
$endgroup$
– John Forkosh
yesterday
$begingroup$
That is why the last dense opaque layer is called photosphere (indeed at about 5000 K versus a dozen millions K at the core). Outer of the photosphere stars have warmer but less dense not opaque layers. About the figure 1000 in @John Forkosh comment I ear higher values, such as 10000 and more, but in the very last case it was a TV program so I don't know. Already forgot what serious books said :(
$endgroup$
– Alchimista
yesterday
2
$begingroup$
Since there cannot be a energy buildup inside the Sun, the energy flowing across any interior spherical surface should be the same as the energy that eventually comes out. The inner surface has a higher temperature but less surface area and hence radiates less, the outer surface has a lower temperature but a higher surface area. The temperature distribution of the Sun is in such an equilibrium that the energy flux radially outwards is constant.
$endgroup$
– Kevin Selva Prasanna
yesterday
add a comment |
1
$begingroup$
You're almost right. As I recall (but would have to double-check), a photon emitted from the Sun's core takes ~1000 years before it escapes from the surface. Not the "same photon", of course, having experienced numerous various-and-sundry processes along the way. But, in any event, the Sun's ultimately in equilibrium (and let's hope it stays that way for a good long time), so everything generated is ultimately emitted.
$endgroup$
– John Forkosh
yesterday
$begingroup$
That is why the last dense opaque layer is called photosphere (indeed at about 5000 K versus a dozen millions K at the core). Outer of the photosphere stars have warmer but less dense not opaque layers. About the figure 1000 in @John Forkosh comment I ear higher values, such as 10000 and more, but in the very last case it was a TV program so I don't know. Already forgot what serious books said :(
$endgroup$
– Alchimista
yesterday
2
$begingroup$
Since there cannot be a energy buildup inside the Sun, the energy flowing across any interior spherical surface should be the same as the energy that eventually comes out. The inner surface has a higher temperature but less surface area and hence radiates less, the outer surface has a lower temperature but a higher surface area. The temperature distribution of the Sun is in such an equilibrium that the energy flux radially outwards is constant.
$endgroup$
– Kevin Selva Prasanna
yesterday
1
1
$begingroup$
You're almost right. As I recall (but would have to double-check), a photon emitted from the Sun's core takes ~1000 years before it escapes from the surface. Not the "same photon", of course, having experienced numerous various-and-sundry processes along the way. But, in any event, the Sun's ultimately in equilibrium (and let's hope it stays that way for a good long time), so everything generated is ultimately emitted.
$endgroup$
– John Forkosh
yesterday
$begingroup$
You're almost right. As I recall (but would have to double-check), a photon emitted from the Sun's core takes ~1000 years before it escapes from the surface. Not the "same photon", of course, having experienced numerous various-and-sundry processes along the way. But, in any event, the Sun's ultimately in equilibrium (and let's hope it stays that way for a good long time), so everything generated is ultimately emitted.
$endgroup$
– John Forkosh
yesterday
$begingroup$
That is why the last dense opaque layer is called photosphere (indeed at about 5000 K versus a dozen millions K at the core). Outer of the photosphere stars have warmer but less dense not opaque layers. About the figure 1000 in @John Forkosh comment I ear higher values, such as 10000 and more, but in the very last case it was a TV program so I don't know. Already forgot what serious books said :(
$endgroup$
– Alchimista
yesterday
$begingroup$
That is why the last dense opaque layer is called photosphere (indeed at about 5000 K versus a dozen millions K at the core). Outer of the photosphere stars have warmer but less dense not opaque layers. About the figure 1000 in @John Forkosh comment I ear higher values, such as 10000 and more, but in the very last case it was a TV program so I don't know. Already forgot what serious books said :(
$endgroup$
– Alchimista
yesterday
2
2
$begingroup$
Since there cannot be a energy buildup inside the Sun, the energy flowing across any interior spherical surface should be the same as the energy that eventually comes out. The inner surface has a higher temperature but less surface area and hence radiates less, the outer surface has a lower temperature but a higher surface area. The temperature distribution of the Sun is in such an equilibrium that the energy flux radially outwards is constant.
$endgroup$
– Kevin Selva Prasanna
yesterday
$begingroup$
Since there cannot be a energy buildup inside the Sun, the energy flowing across any interior spherical surface should be the same as the energy that eventually comes out. The inner surface has a higher temperature but less surface area and hence radiates less, the outer surface has a lower temperature but a higher surface area. The temperature distribution of the Sun is in such an equilibrium that the energy flux radially outwards is constant.
$endgroup$
– Kevin Selva Prasanna
yesterday
add a comment |
3 Answers
3
active
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$begingroup$
Note that the intensity of observable radiation coming out from stars would now mostly depend on the outermost layer.
Because the material in the star is opaque, it completely depends on the outermost layer. Of course the properties of that layer (such as its temperature) are driven by the energy coming from the interior.
So, wouldn't it be inappropriate to consider stars as Black bodies while determining their temperature and other properties?
It is appropriate for an object that has a spectrum that closely matches a blackbody spectrum. However the only property that describes is the temperature of the visible layer. It doesn't imply anything about the interior and processes that produce and distribute energy. You shouldn't read blackbody and think that means that the (invisible) interior is simple or in some way similar to the exterior.
The sun's interior is much hotter than the exterior (around 15 million Kelvin in the core, compared to the 6000 Kelvin or so at the photosphere). Because it is not visible, I would hesitate to call it "brighter". But you could consider it that way.
$endgroup$
add a comment |
$begingroup$
The following is a nuanced side point but relevant.
Is Sun brighter than what we actually see?
The short answer is yes.
The longer answer is also yes, but for a subtle reason. The solar spectrum actually peaks in green visible light (e.g., see https://physics.stackexchange.com/a/137202/59023). We do not see a green star, however, because our eyes do not filter other frequencies and they have their own nonlinear response function that depends upon frequency. Because the response is not flat in frequency and it does not cover all frequencies, the total brightness "measured" by our eyes is not the same as that emitted by the sun.
I will admit that this is a nuanced point and not directly related to the intention of the original question, at least from what I inferred. However, it is an important point to remember, namely, that our eyes (and ears) do not have flat response functions. We actually see yellow better than green, surprisingly.
Is Sun brighter and hotter than what we see from outside?
Brightness and hotness are somewhat subjective things here if we are asking about what any given human perceives (i.e., because it requires the verbal/written/signed description from the individual in question). Further, as an example, object A can be much hotter than object B but be perceived as dimmer depending the primary frequency at which the object emits radiation and the response of the receiver.
Suppose the receiver is a telescope with spectral line filters and something like a Gaussian response centered on the spectral line peak frequency. If not properly calibrated, a signal could appear to increase then decrease in intensity simply because it scanned the frequency range of the filter. To account for this, one tries to properly calibrate so that things like brightness correspond to some measure of the number of photons (or flux or similar measure) received, i.e., quantify it so it can be compared and tested against other observations.
To address the specific part of your question relating to the "outside" comment, the sun is a perfect illustration of why this criteria is misleading. The core of the sun, as BowlOfRed already stated, is ~4,500 times hotter than the surface (i.e., the photosphere). The outer atmosphere (i.e., the corona) is ~170 times hotter than the surface but you cannot see it as it is so much dimmer than the surface. One needs a coronagraph to view the visible light spectrum of the corona. The corona is so hot it actually radiates in the UV and x-ray frequency range. Again, you could never observe this as it is outside the range of frequencies to which your eye can respond.
So to answer your question, yes, the sun is both brighter and hotter than what you can see. Further, it is hotter further inside AND outside but you cannot see either region with the naked eye (you can see the corona during a total solar eclipse for ~few minutes).
$endgroup$
add a comment |
$begingroup$
Yes, you are looking at the outer layer
All the heat and light from the Sun is emitted by that outermost layer, which is essentially opaque to the processes going on inside it. The solar spectrum is a very good fit for a black body, and thus a black body temperature calculation applies to the surface of the Sun. It's not that hot, at around 5700 K.
Yes, inside it gets hotter. A lot hotter
5700K is not hot enough for fusion, unfortunately. As you correctly guessed, most of the light generated in the core stays there. At the core, the temperature is millions of Kelvin, and more than a hundred times the density of water as a plasma.
Remember the Sun is a big gas/plasma ball
The size of the Sun is governed by the balance of forces; outward pressure from the plasma which wants to expand vs inward pull of gravity. If there was more energy generated, that would tend to make the ball expand - there's no 'surface' to constrain it. So the surface and its properties are entirely a product of the energy being generated within the ball, and in fact largely at the core.
What escapes is not a small fraction of the energy being generated
The Sun is in a (broadly) stable state - it's not getting bigger or hotter. And the energy released by fusing nuclei doesn't have any other place to go; it's not going to go back into the heavier nuclei and split them up. So if the Sun is stable, that energy must all be flowing out in one way or another.
We know, then, that the total energy being emitted by the Sun's surface is equal to the amount of energy being released by nuclear interactions at the Sun's core.
We aren't saying the Sun is just a hot blob
By modelling the Sun as a black body, all we are saying is that we can see its spectrum matches that of a generically hot thing at a particular temperature, and from there (and some assumptions about it being basically the same in all directions) we can work out the total power output of the Sun. We haven't had to make any assumptions about what's going on inside the Sun, only observe what reaches us and infer what must be coming out overall.
With this knowledge we can start to infer what must be happening inside
We know some things; how much power the Sun is producing, how big it is, the proportions of atomic species at the surface (from the details of the spectrum), its overall mass, the laws of gravity and the behaviours of plasmas. From that we have to try to model what must be happening inside.
We have inferred, for example, what the dominant nuclear processes inside the Sun must be; we know the temperatures and pressures are not high enough for some cycles, but high enough for others.
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$begingroup$
Note that the intensity of observable radiation coming out from stars would now mostly depend on the outermost layer.
Because the material in the star is opaque, it completely depends on the outermost layer. Of course the properties of that layer (such as its temperature) are driven by the energy coming from the interior.
So, wouldn't it be inappropriate to consider stars as Black bodies while determining their temperature and other properties?
It is appropriate for an object that has a spectrum that closely matches a blackbody spectrum. However the only property that describes is the temperature of the visible layer. It doesn't imply anything about the interior and processes that produce and distribute energy. You shouldn't read blackbody and think that means that the (invisible) interior is simple or in some way similar to the exterior.
The sun's interior is much hotter than the exterior (around 15 million Kelvin in the core, compared to the 6000 Kelvin or so at the photosphere). Because it is not visible, I would hesitate to call it "brighter". But you could consider it that way.
$endgroup$
add a comment |
$begingroup$
Note that the intensity of observable radiation coming out from stars would now mostly depend on the outermost layer.
Because the material in the star is opaque, it completely depends on the outermost layer. Of course the properties of that layer (such as its temperature) are driven by the energy coming from the interior.
So, wouldn't it be inappropriate to consider stars as Black bodies while determining their temperature and other properties?
It is appropriate for an object that has a spectrum that closely matches a blackbody spectrum. However the only property that describes is the temperature of the visible layer. It doesn't imply anything about the interior and processes that produce and distribute energy. You shouldn't read blackbody and think that means that the (invisible) interior is simple or in some way similar to the exterior.
The sun's interior is much hotter than the exterior (around 15 million Kelvin in the core, compared to the 6000 Kelvin or so at the photosphere). Because it is not visible, I would hesitate to call it "brighter". But you could consider it that way.
$endgroup$
add a comment |
$begingroup$
Note that the intensity of observable radiation coming out from stars would now mostly depend on the outermost layer.
Because the material in the star is opaque, it completely depends on the outermost layer. Of course the properties of that layer (such as its temperature) are driven by the energy coming from the interior.
So, wouldn't it be inappropriate to consider stars as Black bodies while determining their temperature and other properties?
It is appropriate for an object that has a spectrum that closely matches a blackbody spectrum. However the only property that describes is the temperature of the visible layer. It doesn't imply anything about the interior and processes that produce and distribute energy. You shouldn't read blackbody and think that means that the (invisible) interior is simple or in some way similar to the exterior.
The sun's interior is much hotter than the exterior (around 15 million Kelvin in the core, compared to the 6000 Kelvin or so at the photosphere). Because it is not visible, I would hesitate to call it "brighter". But you could consider it that way.
$endgroup$
Note that the intensity of observable radiation coming out from stars would now mostly depend on the outermost layer.
Because the material in the star is opaque, it completely depends on the outermost layer. Of course the properties of that layer (such as its temperature) are driven by the energy coming from the interior.
So, wouldn't it be inappropriate to consider stars as Black bodies while determining their temperature and other properties?
It is appropriate for an object that has a spectrum that closely matches a blackbody spectrum. However the only property that describes is the temperature of the visible layer. It doesn't imply anything about the interior and processes that produce and distribute energy. You shouldn't read blackbody and think that means that the (invisible) interior is simple or in some way similar to the exterior.
The sun's interior is much hotter than the exterior (around 15 million Kelvin in the core, compared to the 6000 Kelvin or so at the photosphere). Because it is not visible, I would hesitate to call it "brighter". But you could consider it that way.
answered yesterday
BowlOfRedBowlOfRed
18.1k22746
18.1k22746
add a comment |
add a comment |
$begingroup$
The following is a nuanced side point but relevant.
Is Sun brighter than what we actually see?
The short answer is yes.
The longer answer is also yes, but for a subtle reason. The solar spectrum actually peaks in green visible light (e.g., see https://physics.stackexchange.com/a/137202/59023). We do not see a green star, however, because our eyes do not filter other frequencies and they have their own nonlinear response function that depends upon frequency. Because the response is not flat in frequency and it does not cover all frequencies, the total brightness "measured" by our eyes is not the same as that emitted by the sun.
I will admit that this is a nuanced point and not directly related to the intention of the original question, at least from what I inferred. However, it is an important point to remember, namely, that our eyes (and ears) do not have flat response functions. We actually see yellow better than green, surprisingly.
Is Sun brighter and hotter than what we see from outside?
Brightness and hotness are somewhat subjective things here if we are asking about what any given human perceives (i.e., because it requires the verbal/written/signed description from the individual in question). Further, as an example, object A can be much hotter than object B but be perceived as dimmer depending the primary frequency at which the object emits radiation and the response of the receiver.
Suppose the receiver is a telescope with spectral line filters and something like a Gaussian response centered on the spectral line peak frequency. If not properly calibrated, a signal could appear to increase then decrease in intensity simply because it scanned the frequency range of the filter. To account for this, one tries to properly calibrate so that things like brightness correspond to some measure of the number of photons (or flux or similar measure) received, i.e., quantify it so it can be compared and tested against other observations.
To address the specific part of your question relating to the "outside" comment, the sun is a perfect illustration of why this criteria is misleading. The core of the sun, as BowlOfRed already stated, is ~4,500 times hotter than the surface (i.e., the photosphere). The outer atmosphere (i.e., the corona) is ~170 times hotter than the surface but you cannot see it as it is so much dimmer than the surface. One needs a coronagraph to view the visible light spectrum of the corona. The corona is so hot it actually radiates in the UV and x-ray frequency range. Again, you could never observe this as it is outside the range of frequencies to which your eye can respond.
So to answer your question, yes, the sun is both brighter and hotter than what you can see. Further, it is hotter further inside AND outside but you cannot see either region with the naked eye (you can see the corona during a total solar eclipse for ~few minutes).
$endgroup$
add a comment |
$begingroup$
The following is a nuanced side point but relevant.
Is Sun brighter than what we actually see?
The short answer is yes.
The longer answer is also yes, but for a subtle reason. The solar spectrum actually peaks in green visible light (e.g., see https://physics.stackexchange.com/a/137202/59023). We do not see a green star, however, because our eyes do not filter other frequencies and they have their own nonlinear response function that depends upon frequency. Because the response is not flat in frequency and it does not cover all frequencies, the total brightness "measured" by our eyes is not the same as that emitted by the sun.
I will admit that this is a nuanced point and not directly related to the intention of the original question, at least from what I inferred. However, it is an important point to remember, namely, that our eyes (and ears) do not have flat response functions. We actually see yellow better than green, surprisingly.
Is Sun brighter and hotter than what we see from outside?
Brightness and hotness are somewhat subjective things here if we are asking about what any given human perceives (i.e., because it requires the verbal/written/signed description from the individual in question). Further, as an example, object A can be much hotter than object B but be perceived as dimmer depending the primary frequency at which the object emits radiation and the response of the receiver.
Suppose the receiver is a telescope with spectral line filters and something like a Gaussian response centered on the spectral line peak frequency. If not properly calibrated, a signal could appear to increase then decrease in intensity simply because it scanned the frequency range of the filter. To account for this, one tries to properly calibrate so that things like brightness correspond to some measure of the number of photons (or flux or similar measure) received, i.e., quantify it so it can be compared and tested against other observations.
To address the specific part of your question relating to the "outside" comment, the sun is a perfect illustration of why this criteria is misleading. The core of the sun, as BowlOfRed already stated, is ~4,500 times hotter than the surface (i.e., the photosphere). The outer atmosphere (i.e., the corona) is ~170 times hotter than the surface but you cannot see it as it is so much dimmer than the surface. One needs a coronagraph to view the visible light spectrum of the corona. The corona is so hot it actually radiates in the UV and x-ray frequency range. Again, you could never observe this as it is outside the range of frequencies to which your eye can respond.
So to answer your question, yes, the sun is both brighter and hotter than what you can see. Further, it is hotter further inside AND outside but you cannot see either region with the naked eye (you can see the corona during a total solar eclipse for ~few minutes).
$endgroup$
add a comment |
$begingroup$
The following is a nuanced side point but relevant.
Is Sun brighter than what we actually see?
The short answer is yes.
The longer answer is also yes, but for a subtle reason. The solar spectrum actually peaks in green visible light (e.g., see https://physics.stackexchange.com/a/137202/59023). We do not see a green star, however, because our eyes do not filter other frequencies and they have their own nonlinear response function that depends upon frequency. Because the response is not flat in frequency and it does not cover all frequencies, the total brightness "measured" by our eyes is not the same as that emitted by the sun.
I will admit that this is a nuanced point and not directly related to the intention of the original question, at least from what I inferred. However, it is an important point to remember, namely, that our eyes (and ears) do not have flat response functions. We actually see yellow better than green, surprisingly.
Is Sun brighter and hotter than what we see from outside?
Brightness and hotness are somewhat subjective things here if we are asking about what any given human perceives (i.e., because it requires the verbal/written/signed description from the individual in question). Further, as an example, object A can be much hotter than object B but be perceived as dimmer depending the primary frequency at which the object emits radiation and the response of the receiver.
Suppose the receiver is a telescope with spectral line filters and something like a Gaussian response centered on the spectral line peak frequency. If not properly calibrated, a signal could appear to increase then decrease in intensity simply because it scanned the frequency range of the filter. To account for this, one tries to properly calibrate so that things like brightness correspond to some measure of the number of photons (or flux or similar measure) received, i.e., quantify it so it can be compared and tested against other observations.
To address the specific part of your question relating to the "outside" comment, the sun is a perfect illustration of why this criteria is misleading. The core of the sun, as BowlOfRed already stated, is ~4,500 times hotter than the surface (i.e., the photosphere). The outer atmosphere (i.e., the corona) is ~170 times hotter than the surface but you cannot see it as it is so much dimmer than the surface. One needs a coronagraph to view the visible light spectrum of the corona. The corona is so hot it actually radiates in the UV and x-ray frequency range. Again, you could never observe this as it is outside the range of frequencies to which your eye can respond.
So to answer your question, yes, the sun is both brighter and hotter than what you can see. Further, it is hotter further inside AND outside but you cannot see either region with the naked eye (you can see the corona during a total solar eclipse for ~few minutes).
$endgroup$
The following is a nuanced side point but relevant.
Is Sun brighter than what we actually see?
The short answer is yes.
The longer answer is also yes, but for a subtle reason. The solar spectrum actually peaks in green visible light (e.g., see https://physics.stackexchange.com/a/137202/59023). We do not see a green star, however, because our eyes do not filter other frequencies and they have their own nonlinear response function that depends upon frequency. Because the response is not flat in frequency and it does not cover all frequencies, the total brightness "measured" by our eyes is not the same as that emitted by the sun.
I will admit that this is a nuanced point and not directly related to the intention of the original question, at least from what I inferred. However, it is an important point to remember, namely, that our eyes (and ears) do not have flat response functions. We actually see yellow better than green, surprisingly.
Is Sun brighter and hotter than what we see from outside?
Brightness and hotness are somewhat subjective things here if we are asking about what any given human perceives (i.e., because it requires the verbal/written/signed description from the individual in question). Further, as an example, object A can be much hotter than object B but be perceived as dimmer depending the primary frequency at which the object emits radiation and the response of the receiver.
Suppose the receiver is a telescope with spectral line filters and something like a Gaussian response centered on the spectral line peak frequency. If not properly calibrated, a signal could appear to increase then decrease in intensity simply because it scanned the frequency range of the filter. To account for this, one tries to properly calibrate so that things like brightness correspond to some measure of the number of photons (or flux or similar measure) received, i.e., quantify it so it can be compared and tested against other observations.
To address the specific part of your question relating to the "outside" comment, the sun is a perfect illustration of why this criteria is misleading. The core of the sun, as BowlOfRed already stated, is ~4,500 times hotter than the surface (i.e., the photosphere). The outer atmosphere (i.e., the corona) is ~170 times hotter than the surface but you cannot see it as it is so much dimmer than the surface. One needs a coronagraph to view the visible light spectrum of the corona. The corona is so hot it actually radiates in the UV and x-ray frequency range. Again, you could never observe this as it is outside the range of frequencies to which your eye can respond.
So to answer your question, yes, the sun is both brighter and hotter than what you can see. Further, it is hotter further inside AND outside but you cannot see either region with the naked eye (you can see the corona during a total solar eclipse for ~few minutes).
answered yesterday
honeste_viverehoneste_vivere
8,19731777
8,19731777
add a comment |
add a comment |
$begingroup$
Yes, you are looking at the outer layer
All the heat and light from the Sun is emitted by that outermost layer, which is essentially opaque to the processes going on inside it. The solar spectrum is a very good fit for a black body, and thus a black body temperature calculation applies to the surface of the Sun. It's not that hot, at around 5700 K.
Yes, inside it gets hotter. A lot hotter
5700K is not hot enough for fusion, unfortunately. As you correctly guessed, most of the light generated in the core stays there. At the core, the temperature is millions of Kelvin, and more than a hundred times the density of water as a plasma.
Remember the Sun is a big gas/plasma ball
The size of the Sun is governed by the balance of forces; outward pressure from the plasma which wants to expand vs inward pull of gravity. If there was more energy generated, that would tend to make the ball expand - there's no 'surface' to constrain it. So the surface and its properties are entirely a product of the energy being generated within the ball, and in fact largely at the core.
What escapes is not a small fraction of the energy being generated
The Sun is in a (broadly) stable state - it's not getting bigger or hotter. And the energy released by fusing nuclei doesn't have any other place to go; it's not going to go back into the heavier nuclei and split them up. So if the Sun is stable, that energy must all be flowing out in one way or another.
We know, then, that the total energy being emitted by the Sun's surface is equal to the amount of energy being released by nuclear interactions at the Sun's core.
We aren't saying the Sun is just a hot blob
By modelling the Sun as a black body, all we are saying is that we can see its spectrum matches that of a generically hot thing at a particular temperature, and from there (and some assumptions about it being basically the same in all directions) we can work out the total power output of the Sun. We haven't had to make any assumptions about what's going on inside the Sun, only observe what reaches us and infer what must be coming out overall.
With this knowledge we can start to infer what must be happening inside
We know some things; how much power the Sun is producing, how big it is, the proportions of atomic species at the surface (from the details of the spectrum), its overall mass, the laws of gravity and the behaviours of plasmas. From that we have to try to model what must be happening inside.
We have inferred, for example, what the dominant nuclear processes inside the Sun must be; we know the temperatures and pressures are not high enough for some cycles, but high enough for others.
$endgroup$
add a comment |
$begingroup$
Yes, you are looking at the outer layer
All the heat and light from the Sun is emitted by that outermost layer, which is essentially opaque to the processes going on inside it. The solar spectrum is a very good fit for a black body, and thus a black body temperature calculation applies to the surface of the Sun. It's not that hot, at around 5700 K.
Yes, inside it gets hotter. A lot hotter
5700K is not hot enough for fusion, unfortunately. As you correctly guessed, most of the light generated in the core stays there. At the core, the temperature is millions of Kelvin, and more than a hundred times the density of water as a plasma.
Remember the Sun is a big gas/plasma ball
The size of the Sun is governed by the balance of forces; outward pressure from the plasma which wants to expand vs inward pull of gravity. If there was more energy generated, that would tend to make the ball expand - there's no 'surface' to constrain it. So the surface and its properties are entirely a product of the energy being generated within the ball, and in fact largely at the core.
What escapes is not a small fraction of the energy being generated
The Sun is in a (broadly) stable state - it's not getting bigger or hotter. And the energy released by fusing nuclei doesn't have any other place to go; it's not going to go back into the heavier nuclei and split them up. So if the Sun is stable, that energy must all be flowing out in one way or another.
We know, then, that the total energy being emitted by the Sun's surface is equal to the amount of energy being released by nuclear interactions at the Sun's core.
We aren't saying the Sun is just a hot blob
By modelling the Sun as a black body, all we are saying is that we can see its spectrum matches that of a generically hot thing at a particular temperature, and from there (and some assumptions about it being basically the same in all directions) we can work out the total power output of the Sun. We haven't had to make any assumptions about what's going on inside the Sun, only observe what reaches us and infer what must be coming out overall.
With this knowledge we can start to infer what must be happening inside
We know some things; how much power the Sun is producing, how big it is, the proportions of atomic species at the surface (from the details of the spectrum), its overall mass, the laws of gravity and the behaviours of plasmas. From that we have to try to model what must be happening inside.
We have inferred, for example, what the dominant nuclear processes inside the Sun must be; we know the temperatures and pressures are not high enough for some cycles, but high enough for others.
$endgroup$
add a comment |
$begingroup$
Yes, you are looking at the outer layer
All the heat and light from the Sun is emitted by that outermost layer, which is essentially opaque to the processes going on inside it. The solar spectrum is a very good fit for a black body, and thus a black body temperature calculation applies to the surface of the Sun. It's not that hot, at around 5700 K.
Yes, inside it gets hotter. A lot hotter
5700K is not hot enough for fusion, unfortunately. As you correctly guessed, most of the light generated in the core stays there. At the core, the temperature is millions of Kelvin, and more than a hundred times the density of water as a plasma.
Remember the Sun is a big gas/plasma ball
The size of the Sun is governed by the balance of forces; outward pressure from the plasma which wants to expand vs inward pull of gravity. If there was more energy generated, that would tend to make the ball expand - there's no 'surface' to constrain it. So the surface and its properties are entirely a product of the energy being generated within the ball, and in fact largely at the core.
What escapes is not a small fraction of the energy being generated
The Sun is in a (broadly) stable state - it's not getting bigger or hotter. And the energy released by fusing nuclei doesn't have any other place to go; it's not going to go back into the heavier nuclei and split them up. So if the Sun is stable, that energy must all be flowing out in one way or another.
We know, then, that the total energy being emitted by the Sun's surface is equal to the amount of energy being released by nuclear interactions at the Sun's core.
We aren't saying the Sun is just a hot blob
By modelling the Sun as a black body, all we are saying is that we can see its spectrum matches that of a generically hot thing at a particular temperature, and from there (and some assumptions about it being basically the same in all directions) we can work out the total power output of the Sun. We haven't had to make any assumptions about what's going on inside the Sun, only observe what reaches us and infer what must be coming out overall.
With this knowledge we can start to infer what must be happening inside
We know some things; how much power the Sun is producing, how big it is, the proportions of atomic species at the surface (from the details of the spectrum), its overall mass, the laws of gravity and the behaviours of plasmas. From that we have to try to model what must be happening inside.
We have inferred, for example, what the dominant nuclear processes inside the Sun must be; we know the temperatures and pressures are not high enough for some cycles, but high enough for others.
$endgroup$
Yes, you are looking at the outer layer
All the heat and light from the Sun is emitted by that outermost layer, which is essentially opaque to the processes going on inside it. The solar spectrum is a very good fit for a black body, and thus a black body temperature calculation applies to the surface of the Sun. It's not that hot, at around 5700 K.
Yes, inside it gets hotter. A lot hotter
5700K is not hot enough for fusion, unfortunately. As you correctly guessed, most of the light generated in the core stays there. At the core, the temperature is millions of Kelvin, and more than a hundred times the density of water as a plasma.
Remember the Sun is a big gas/plasma ball
The size of the Sun is governed by the balance of forces; outward pressure from the plasma which wants to expand vs inward pull of gravity. If there was more energy generated, that would tend to make the ball expand - there's no 'surface' to constrain it. So the surface and its properties are entirely a product of the energy being generated within the ball, and in fact largely at the core.
What escapes is not a small fraction of the energy being generated
The Sun is in a (broadly) stable state - it's not getting bigger or hotter. And the energy released by fusing nuclei doesn't have any other place to go; it's not going to go back into the heavier nuclei and split them up. So if the Sun is stable, that energy must all be flowing out in one way or another.
We know, then, that the total energy being emitted by the Sun's surface is equal to the amount of energy being released by nuclear interactions at the Sun's core.
We aren't saying the Sun is just a hot blob
By modelling the Sun as a black body, all we are saying is that we can see its spectrum matches that of a generically hot thing at a particular temperature, and from there (and some assumptions about it being basically the same in all directions) we can work out the total power output of the Sun. We haven't had to make any assumptions about what's going on inside the Sun, only observe what reaches us and infer what must be coming out overall.
With this knowledge we can start to infer what must be happening inside
We know some things; how much power the Sun is producing, how big it is, the proportions of atomic species at the surface (from the details of the spectrum), its overall mass, the laws of gravity and the behaviours of plasmas. From that we have to try to model what must be happening inside.
We have inferred, for example, what the dominant nuclear processes inside the Sun must be; we know the temperatures and pressures are not high enough for some cycles, but high enough for others.
answered 19 hours ago
Phil HPhil H
870413
870413
add a comment |
add a comment |
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-astrophysics, plasma-physics, radiation, thermal-radiation
1
$begingroup$
You're almost right. As I recall (but would have to double-check), a photon emitted from the Sun's core takes ~1000 years before it escapes from the surface. Not the "same photon", of course, having experienced numerous various-and-sundry processes along the way. But, in any event, the Sun's ultimately in equilibrium (and let's hope it stays that way for a good long time), so everything generated is ultimately emitted.
$endgroup$
– John Forkosh
yesterday
$begingroup$
That is why the last dense opaque layer is called photosphere (indeed at about 5000 K versus a dozen millions K at the core). Outer of the photosphere stars have warmer but less dense not opaque layers. About the figure 1000 in @John Forkosh comment I ear higher values, such as 10000 and more, but in the very last case it was a TV program so I don't know. Already forgot what serious books said :(
$endgroup$
– Alchimista
yesterday
2
$begingroup$
Since there cannot be a energy buildup inside the Sun, the energy flowing across any interior spherical surface should be the same as the energy that eventually comes out. The inner surface has a higher temperature but less surface area and hence radiates less, the outer surface has a lower temperature but a higher surface area. The temperature distribution of the Sun is in such an equilibrium that the energy flux radially outwards is constant.
$endgroup$
– Kevin Selva Prasanna
yesterday