What is the minimum size for the Sun?











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How small could our Sun be and still "burn" with nuclear fusion and emit the same wavelengths as the real Sun does?



Edit:



I want the same wavelengths, but at a lower intensity. The goal is to have a small sun inside a huge vessel, such as an O'Neill cylinder.










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  • 2




    Does it need to be a star or can it be a fusion generator powering a light?
    – Nick T
    6 hours ago















up vote
6
down vote

favorite
1












How small could our Sun be and still "burn" with nuclear fusion and emit the same wavelengths as the real Sun does?



Edit:



I want the same wavelengths, but at a lower intensity. The goal is to have a small sun inside a huge vessel, such as an O'Neill cylinder.










share|improve this question









New contributor




user57423 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.










This question asks for hard science. All answers to this question should be backed up by equations, empirical evidence, scientific papers, other citations, etc. Answers that do not satisfy this requirement might be removed. See the tag description for more information.










  • 2




    Does it need to be a star or can it be a fusion generator powering a light?
    – Nick T
    6 hours ago













up vote
6
down vote

favorite
1









up vote
6
down vote

favorite
1






1





How small could our Sun be and still "burn" with nuclear fusion and emit the same wavelengths as the real Sun does?



Edit:



I want the same wavelengths, but at a lower intensity. The goal is to have a small sun inside a huge vessel, such as an O'Neill cylinder.










share|improve this question









New contributor




user57423 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











How small could our Sun be and still "burn" with nuclear fusion and emit the same wavelengths as the real Sun does?



Edit:



I want the same wavelengths, but at a lower intensity. The goal is to have a small sun inside a huge vessel, such as an O'Neill cylinder.







hard-science stars astrophysics






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user57423 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|improve this question









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user57423 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|improve this question




share|improve this question








edited 8 hours ago





















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asked 8 hours ago









user57423

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user57423 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






user57423 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.



This question asks for hard science. All answers to this question should be backed up by equations, empirical evidence, scientific papers, other citations, etc. Answers that do not satisfy this requirement might be removed. See the tag description for more information.




This question asks for hard science. All answers to this question should be backed up by equations, empirical evidence, scientific papers, other citations, etc. Answers that do not satisfy this requirement might be removed. See the tag description for more information.









  • 2




    Does it need to be a star or can it be a fusion generator powering a light?
    – Nick T
    6 hours ago














  • 2




    Does it need to be a star or can it be a fusion generator powering a light?
    – Nick T
    6 hours ago








2




2




Does it need to be a star or can it be a fusion generator powering a light?
– Nick T
6 hours ago




Does it need to be a star or can it be a fusion generator powering a light?
– Nick T
6 hours ago










2 Answers
2






active

oldest

votes

















up vote
11
down vote













The mass of a star is directly related to how hot its surface is, which in turn, is responsible for the wavelengths of light it emits (This is called Black-Body Radiation).



As a main sequence G2V star, the sun has a surface temperature of 5778 K. A smaller main sequence star will be cooler and therefore redder. A larger star will be hotter, and therefore whiter.



The only stars that will emit the same wavelengths as the sun are those that have the same mass.



https://en.wikipedia.org/wiki/Main_sequence



https://en.wikipedia.org/wiki/Black-body_radiation






share|improve this answer





















  • This all looks correct to me, but I'm not sure it really meets the OP's requirement for a hard science answer.
    – HDE 226868
    8 hours ago










  • @HDE226868 To be fair, I've just deleted the science-based tag that I had along with the hard-science tag. So this answer met my requirements as the question was at the time the answer was given.
    – user57423
    7 hours ago










  • @user57423 Looks like I misread what you were saying; my bad. Thanks!
    – HDE 226868
    7 hours ago






  • 1




    You need to note, that atmosphere affects, what light we see at sea-level.upload.wikimedia.org/wikipedia/commons/thumb/e/e7/…
    – Artemijs Danilovs
    7 hours ago










  • It is not just the mass but also the chemical composition. As Hidrogen is converted into Helium and heavier elements, the luminosity of the star changes. As evolutionary models of stars were developed during the 1930s, it was shown that, for stars of a uniform chemical composition, a relationship exists between a star's mass and its luminosity and radius
    – SJuan76
    4 hours ago


















up vote
10
down vote













Black body radiation



The Sun is, approximately, a black body. That means that the light it emits follows a particular spectrum according to Planck's law, with the shape of the spectrum determined solely by the Sun's surface temperature. In particular, the wavelength of peak emission can be found through Wien's law, which is also a function of temperature. Therefore, if we want our new star to emit light just like the Sun does, we need to keep it at the same temperature as the real Sun - about 5800 K.



On the main sequence, there are some simple scaling relationships between mass, radius, temperature and luminosity:
$$Lpropto M^3,quad Rpropto M^{4/7},quad Tpropto M^{4/7}$$
assuming the proton-proton chain reaction is the main source of energy, which is the case. If we want to keep the temperature constant, we need to keep the mass constant, and thus keep the radius constant. Therefore, no main sequence star can be significantly smaller than the Sun and still have the same temperature.



The composition of the star does matter; in fact, according to something called the Vogt-Russell theorem, the mass and composition of a star uniquely determine its properties and evolution. This means that the exact form of the relations above does vary between stellar populations - in part, relying on the mean molecular mass $mu$ - but this still will not make a significant difference here.



Options for a Sun-like star



We could look at subdwarfs, low-metallicity stars that are dimmer than main sequence stars. They're not common, but they exist, lying immediately below the main sequence on the Hertzsprung-Russell diagram. All other G-type stars are either on the main sequence, and thus fairly like the Sun in size, or off the main sequence, as giants or supergiants. Therefore, our options are either G-type subdwarfs or main sequence stars that are slightly cooler than the Sun:





  • Tau Ceti, a main sequence star, might be a good choice; it has a temperature a few hundred Kelvin less than the Sun, and is approximately $0.8R_{odot}$ in size (Texeira et al. 2009).


  • Mu Cassiopeiae A, a subdwarf, has a similarly slightly-cooler temperature and again a radius of about $0.8R_{odot}$ (Boyaijan et al. 2008).


Another star I considered is Groombridge 1830. Like Mu Cassiopeiae, it's a borderline G-type subdwarf, but is about 1000 K cooler - too far from the Sun in terms of temperature.



Low-mass stars and true lower limits



You won't find a star small enough to fit into an O'Neill cylinder - at least, not one that formed by natural means. EBLM J0555-57Ab, a small, late-type red dwarf has a radius 0.85 times that of Jupiter (von Boetticher et al. 2017) - likely too large for your purposes. Bodies smaller than that are likely too low-mass to fuse hydrogen, and would instead be brown dwarfs. Of course, EBLM J0555-57Ab is also likely cool, with a surface temperature far below that of the Sun.



Brown dwarfs - many of which fuse deuterium - are not true stars, and are usually cool, with typical temperatures around 1000 K, producing spectra much different from the Sun's. Some exoplanets may be much hotter than this, with surface temperatures comparable to those of many stars (see e.g. Kepler-70b, with a surface temperature of about 7000 K as per Charpinet et al. 2011). However, those bodies are only hot because they're irradiated by the stars they orbit; on their own, they would not generate that much heat.



White dwarfs



There is a possibility I had completely forgotten about before: a white dwarf. Many white dwarfs are hot, with temperatures up to about 100,000 K or so. However, they do cool - albeit slowly, as they have small surface areas. This cooling takes a long time, but some white dwarfs have become cooler than the Sun. WD 0346+246 is a famous case, with a surface temperature of about 3900 K (Hambley et al. 1997).



This implies that white dwarfs with temperatures like that of the Sun do exist; moreover, they're small. The same group measured WD 0346+246 to have a radius roughly that of the Earth, which is extraordinary - certainly less than that of EBLM J0555-57Ab. The problem, of course, is that white dwarfs don't undergo fusion. Indeed, the degeneracy of the matter inside a white dwarf means that fusion reactions are unstable, and can lead to novae and Type Ia supernovae.



A note on spectral lines



Finally, I should mention that actual stellar spectra are extremely complicated. A black body spectrum only models the global behavior; on smaller scales, there are numerous absorption and emission lines. I wrote in a related answer that Dyson spheres and brown dwarfs could be distinguishable by the presence and absence of various spectral lines.



The same should be true here. Just because another body of some sort is the same temperature as the Sun doesn't mean it would appear identical to the Sun, under spectroscopic scrutiny. I assume you don't care much about this, but it definitely is something to be aware of. Atmospheric composition really does matter.






share|improve this answer























  • Deuterium fusion doesn't require anywhere near as much mass as hydrogen fusion. Could you get the correct spectrum out of a deuterium-burning brown dwarf?
    – Mark
    7 hours ago










  • so, the subdwarfs of the same apprx. R of sol, since lower density, would have lower L & T, right -implying R must be larger to to compensate? (great answer, by the way)
    – theRiley
    6 hours ago












  • @Mark No, you couldn't. Brown dwarfs have surface temperatures of maybe 1000-2000 K at the most, with many around 500-1000-ish K. Maybe there are individual peculiar examples I'm not aware of, but it certainly seems unlikely.
    – HDE 226868
    5 hours ago










  • @theRiley Thanks! Subdwarfs in general will have smaller radii, period. They do span a large temperature range, and can have spectral types up to O - and therefore could actually be decently luminous - but G-type subdwarfs, and those of similar spectral types, will be small and cool, and therefore not as bright as a main sequence star of the same spectral type/temperature.
    – HDE 226868
    5 hours ago











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
11
down vote













The mass of a star is directly related to how hot its surface is, which in turn, is responsible for the wavelengths of light it emits (This is called Black-Body Radiation).



As a main sequence G2V star, the sun has a surface temperature of 5778 K. A smaller main sequence star will be cooler and therefore redder. A larger star will be hotter, and therefore whiter.



The only stars that will emit the same wavelengths as the sun are those that have the same mass.



https://en.wikipedia.org/wiki/Main_sequence



https://en.wikipedia.org/wiki/Black-body_radiation






share|improve this answer





















  • This all looks correct to me, but I'm not sure it really meets the OP's requirement for a hard science answer.
    – HDE 226868
    8 hours ago










  • @HDE226868 To be fair, I've just deleted the science-based tag that I had along with the hard-science tag. So this answer met my requirements as the question was at the time the answer was given.
    – user57423
    7 hours ago










  • @user57423 Looks like I misread what you were saying; my bad. Thanks!
    – HDE 226868
    7 hours ago






  • 1




    You need to note, that atmosphere affects, what light we see at sea-level.upload.wikimedia.org/wikipedia/commons/thumb/e/e7/…
    – Artemijs Danilovs
    7 hours ago










  • It is not just the mass but also the chemical composition. As Hidrogen is converted into Helium and heavier elements, the luminosity of the star changes. As evolutionary models of stars were developed during the 1930s, it was shown that, for stars of a uniform chemical composition, a relationship exists between a star's mass and its luminosity and radius
    – SJuan76
    4 hours ago















up vote
11
down vote













The mass of a star is directly related to how hot its surface is, which in turn, is responsible for the wavelengths of light it emits (This is called Black-Body Radiation).



As a main sequence G2V star, the sun has a surface temperature of 5778 K. A smaller main sequence star will be cooler and therefore redder. A larger star will be hotter, and therefore whiter.



The only stars that will emit the same wavelengths as the sun are those that have the same mass.



https://en.wikipedia.org/wiki/Main_sequence



https://en.wikipedia.org/wiki/Black-body_radiation






share|improve this answer





















  • This all looks correct to me, but I'm not sure it really meets the OP's requirement for a hard science answer.
    – HDE 226868
    8 hours ago










  • @HDE226868 To be fair, I've just deleted the science-based tag that I had along with the hard-science tag. So this answer met my requirements as the question was at the time the answer was given.
    – user57423
    7 hours ago










  • @user57423 Looks like I misread what you were saying; my bad. Thanks!
    – HDE 226868
    7 hours ago






  • 1




    You need to note, that atmosphere affects, what light we see at sea-level.upload.wikimedia.org/wikipedia/commons/thumb/e/e7/…
    – Artemijs Danilovs
    7 hours ago










  • It is not just the mass but also the chemical composition. As Hidrogen is converted into Helium and heavier elements, the luminosity of the star changes. As evolutionary models of stars were developed during the 1930s, it was shown that, for stars of a uniform chemical composition, a relationship exists between a star's mass and its luminosity and radius
    – SJuan76
    4 hours ago













up vote
11
down vote










up vote
11
down vote









The mass of a star is directly related to how hot its surface is, which in turn, is responsible for the wavelengths of light it emits (This is called Black-Body Radiation).



As a main sequence G2V star, the sun has a surface temperature of 5778 K. A smaller main sequence star will be cooler and therefore redder. A larger star will be hotter, and therefore whiter.



The only stars that will emit the same wavelengths as the sun are those that have the same mass.



https://en.wikipedia.org/wiki/Main_sequence



https://en.wikipedia.org/wiki/Black-body_radiation






share|improve this answer












The mass of a star is directly related to how hot its surface is, which in turn, is responsible for the wavelengths of light it emits (This is called Black-Body Radiation).



As a main sequence G2V star, the sun has a surface temperature of 5778 K. A smaller main sequence star will be cooler and therefore redder. A larger star will be hotter, and therefore whiter.



The only stars that will emit the same wavelengths as the sun are those that have the same mass.



https://en.wikipedia.org/wiki/Main_sequence



https://en.wikipedia.org/wiki/Black-body_radiation







share|improve this answer












share|improve this answer



share|improve this answer










answered 8 hours ago









Arkenstein XII

1,581219




1,581219












  • This all looks correct to me, but I'm not sure it really meets the OP's requirement for a hard science answer.
    – HDE 226868
    8 hours ago










  • @HDE226868 To be fair, I've just deleted the science-based tag that I had along with the hard-science tag. So this answer met my requirements as the question was at the time the answer was given.
    – user57423
    7 hours ago










  • @user57423 Looks like I misread what you were saying; my bad. Thanks!
    – HDE 226868
    7 hours ago






  • 1




    You need to note, that atmosphere affects, what light we see at sea-level.upload.wikimedia.org/wikipedia/commons/thumb/e/e7/…
    – Artemijs Danilovs
    7 hours ago










  • It is not just the mass but also the chemical composition. As Hidrogen is converted into Helium and heavier elements, the luminosity of the star changes. As evolutionary models of stars were developed during the 1930s, it was shown that, for stars of a uniform chemical composition, a relationship exists between a star's mass and its luminosity and radius
    – SJuan76
    4 hours ago


















  • This all looks correct to me, but I'm not sure it really meets the OP's requirement for a hard science answer.
    – HDE 226868
    8 hours ago










  • @HDE226868 To be fair, I've just deleted the science-based tag that I had along with the hard-science tag. So this answer met my requirements as the question was at the time the answer was given.
    – user57423
    7 hours ago










  • @user57423 Looks like I misread what you were saying; my bad. Thanks!
    – HDE 226868
    7 hours ago






  • 1




    You need to note, that atmosphere affects, what light we see at sea-level.upload.wikimedia.org/wikipedia/commons/thumb/e/e7/…
    – Artemijs Danilovs
    7 hours ago










  • It is not just the mass but also the chemical composition. As Hidrogen is converted into Helium and heavier elements, the luminosity of the star changes. As evolutionary models of stars were developed during the 1930s, it was shown that, for stars of a uniform chemical composition, a relationship exists between a star's mass and its luminosity and radius
    – SJuan76
    4 hours ago
















This all looks correct to me, but I'm not sure it really meets the OP's requirement for a hard science answer.
– HDE 226868
8 hours ago




This all looks correct to me, but I'm not sure it really meets the OP's requirement for a hard science answer.
– HDE 226868
8 hours ago












@HDE226868 To be fair, I've just deleted the science-based tag that I had along with the hard-science tag. So this answer met my requirements as the question was at the time the answer was given.
– user57423
7 hours ago




@HDE226868 To be fair, I've just deleted the science-based tag that I had along with the hard-science tag. So this answer met my requirements as the question was at the time the answer was given.
– user57423
7 hours ago












@user57423 Looks like I misread what you were saying; my bad. Thanks!
– HDE 226868
7 hours ago




@user57423 Looks like I misread what you were saying; my bad. Thanks!
– HDE 226868
7 hours ago




1




1




You need to note, that atmosphere affects, what light we see at sea-level.upload.wikimedia.org/wikipedia/commons/thumb/e/e7/…
– Artemijs Danilovs
7 hours ago




You need to note, that atmosphere affects, what light we see at sea-level.upload.wikimedia.org/wikipedia/commons/thumb/e/e7/…
– Artemijs Danilovs
7 hours ago












It is not just the mass but also the chemical composition. As Hidrogen is converted into Helium and heavier elements, the luminosity of the star changes. As evolutionary models of stars were developed during the 1930s, it was shown that, for stars of a uniform chemical composition, a relationship exists between a star's mass and its luminosity and radius
– SJuan76
4 hours ago




It is not just the mass but also the chemical composition. As Hidrogen is converted into Helium and heavier elements, the luminosity of the star changes. As evolutionary models of stars were developed during the 1930s, it was shown that, for stars of a uniform chemical composition, a relationship exists between a star's mass and its luminosity and radius
– SJuan76
4 hours ago










up vote
10
down vote













Black body radiation



The Sun is, approximately, a black body. That means that the light it emits follows a particular spectrum according to Planck's law, with the shape of the spectrum determined solely by the Sun's surface temperature. In particular, the wavelength of peak emission can be found through Wien's law, which is also a function of temperature. Therefore, if we want our new star to emit light just like the Sun does, we need to keep it at the same temperature as the real Sun - about 5800 K.



On the main sequence, there are some simple scaling relationships between mass, radius, temperature and luminosity:
$$Lpropto M^3,quad Rpropto M^{4/7},quad Tpropto M^{4/7}$$
assuming the proton-proton chain reaction is the main source of energy, which is the case. If we want to keep the temperature constant, we need to keep the mass constant, and thus keep the radius constant. Therefore, no main sequence star can be significantly smaller than the Sun and still have the same temperature.



The composition of the star does matter; in fact, according to something called the Vogt-Russell theorem, the mass and composition of a star uniquely determine its properties and evolution. This means that the exact form of the relations above does vary between stellar populations - in part, relying on the mean molecular mass $mu$ - but this still will not make a significant difference here.



Options for a Sun-like star



We could look at subdwarfs, low-metallicity stars that are dimmer than main sequence stars. They're not common, but they exist, lying immediately below the main sequence on the Hertzsprung-Russell diagram. All other G-type stars are either on the main sequence, and thus fairly like the Sun in size, or off the main sequence, as giants or supergiants. Therefore, our options are either G-type subdwarfs or main sequence stars that are slightly cooler than the Sun:





  • Tau Ceti, a main sequence star, might be a good choice; it has a temperature a few hundred Kelvin less than the Sun, and is approximately $0.8R_{odot}$ in size (Texeira et al. 2009).


  • Mu Cassiopeiae A, a subdwarf, has a similarly slightly-cooler temperature and again a radius of about $0.8R_{odot}$ (Boyaijan et al. 2008).


Another star I considered is Groombridge 1830. Like Mu Cassiopeiae, it's a borderline G-type subdwarf, but is about 1000 K cooler - too far from the Sun in terms of temperature.



Low-mass stars and true lower limits



You won't find a star small enough to fit into an O'Neill cylinder - at least, not one that formed by natural means. EBLM J0555-57Ab, a small, late-type red dwarf has a radius 0.85 times that of Jupiter (von Boetticher et al. 2017) - likely too large for your purposes. Bodies smaller than that are likely too low-mass to fuse hydrogen, and would instead be brown dwarfs. Of course, EBLM J0555-57Ab is also likely cool, with a surface temperature far below that of the Sun.



Brown dwarfs - many of which fuse deuterium - are not true stars, and are usually cool, with typical temperatures around 1000 K, producing spectra much different from the Sun's. Some exoplanets may be much hotter than this, with surface temperatures comparable to those of many stars (see e.g. Kepler-70b, with a surface temperature of about 7000 K as per Charpinet et al. 2011). However, those bodies are only hot because they're irradiated by the stars they orbit; on their own, they would not generate that much heat.



White dwarfs



There is a possibility I had completely forgotten about before: a white dwarf. Many white dwarfs are hot, with temperatures up to about 100,000 K or so. However, they do cool - albeit slowly, as they have small surface areas. This cooling takes a long time, but some white dwarfs have become cooler than the Sun. WD 0346+246 is a famous case, with a surface temperature of about 3900 K (Hambley et al. 1997).



This implies that white dwarfs with temperatures like that of the Sun do exist; moreover, they're small. The same group measured WD 0346+246 to have a radius roughly that of the Earth, which is extraordinary - certainly less than that of EBLM J0555-57Ab. The problem, of course, is that white dwarfs don't undergo fusion. Indeed, the degeneracy of the matter inside a white dwarf means that fusion reactions are unstable, and can lead to novae and Type Ia supernovae.



A note on spectral lines



Finally, I should mention that actual stellar spectra are extremely complicated. A black body spectrum only models the global behavior; on smaller scales, there are numerous absorption and emission lines. I wrote in a related answer that Dyson spheres and brown dwarfs could be distinguishable by the presence and absence of various spectral lines.



The same should be true here. Just because another body of some sort is the same temperature as the Sun doesn't mean it would appear identical to the Sun, under spectroscopic scrutiny. I assume you don't care much about this, but it definitely is something to be aware of. Atmospheric composition really does matter.






share|improve this answer























  • Deuterium fusion doesn't require anywhere near as much mass as hydrogen fusion. Could you get the correct spectrum out of a deuterium-burning brown dwarf?
    – Mark
    7 hours ago










  • so, the subdwarfs of the same apprx. R of sol, since lower density, would have lower L & T, right -implying R must be larger to to compensate? (great answer, by the way)
    – theRiley
    6 hours ago












  • @Mark No, you couldn't. Brown dwarfs have surface temperatures of maybe 1000-2000 K at the most, with many around 500-1000-ish K. Maybe there are individual peculiar examples I'm not aware of, but it certainly seems unlikely.
    – HDE 226868
    5 hours ago










  • @theRiley Thanks! Subdwarfs in general will have smaller radii, period. They do span a large temperature range, and can have spectral types up to O - and therefore could actually be decently luminous - but G-type subdwarfs, and those of similar spectral types, will be small and cool, and therefore not as bright as a main sequence star of the same spectral type/temperature.
    – HDE 226868
    5 hours ago















up vote
10
down vote













Black body radiation



The Sun is, approximately, a black body. That means that the light it emits follows a particular spectrum according to Planck's law, with the shape of the spectrum determined solely by the Sun's surface temperature. In particular, the wavelength of peak emission can be found through Wien's law, which is also a function of temperature. Therefore, if we want our new star to emit light just like the Sun does, we need to keep it at the same temperature as the real Sun - about 5800 K.



On the main sequence, there are some simple scaling relationships between mass, radius, temperature and luminosity:
$$Lpropto M^3,quad Rpropto M^{4/7},quad Tpropto M^{4/7}$$
assuming the proton-proton chain reaction is the main source of energy, which is the case. If we want to keep the temperature constant, we need to keep the mass constant, and thus keep the radius constant. Therefore, no main sequence star can be significantly smaller than the Sun and still have the same temperature.



The composition of the star does matter; in fact, according to something called the Vogt-Russell theorem, the mass and composition of a star uniquely determine its properties and evolution. This means that the exact form of the relations above does vary between stellar populations - in part, relying on the mean molecular mass $mu$ - but this still will not make a significant difference here.



Options for a Sun-like star



We could look at subdwarfs, low-metallicity stars that are dimmer than main sequence stars. They're not common, but they exist, lying immediately below the main sequence on the Hertzsprung-Russell diagram. All other G-type stars are either on the main sequence, and thus fairly like the Sun in size, or off the main sequence, as giants or supergiants. Therefore, our options are either G-type subdwarfs or main sequence stars that are slightly cooler than the Sun:





  • Tau Ceti, a main sequence star, might be a good choice; it has a temperature a few hundred Kelvin less than the Sun, and is approximately $0.8R_{odot}$ in size (Texeira et al. 2009).


  • Mu Cassiopeiae A, a subdwarf, has a similarly slightly-cooler temperature and again a radius of about $0.8R_{odot}$ (Boyaijan et al. 2008).


Another star I considered is Groombridge 1830. Like Mu Cassiopeiae, it's a borderline G-type subdwarf, but is about 1000 K cooler - too far from the Sun in terms of temperature.



Low-mass stars and true lower limits



You won't find a star small enough to fit into an O'Neill cylinder - at least, not one that formed by natural means. EBLM J0555-57Ab, a small, late-type red dwarf has a radius 0.85 times that of Jupiter (von Boetticher et al. 2017) - likely too large for your purposes. Bodies smaller than that are likely too low-mass to fuse hydrogen, and would instead be brown dwarfs. Of course, EBLM J0555-57Ab is also likely cool, with a surface temperature far below that of the Sun.



Brown dwarfs - many of which fuse deuterium - are not true stars, and are usually cool, with typical temperatures around 1000 K, producing spectra much different from the Sun's. Some exoplanets may be much hotter than this, with surface temperatures comparable to those of many stars (see e.g. Kepler-70b, with a surface temperature of about 7000 K as per Charpinet et al. 2011). However, those bodies are only hot because they're irradiated by the stars they orbit; on their own, they would not generate that much heat.



White dwarfs



There is a possibility I had completely forgotten about before: a white dwarf. Many white dwarfs are hot, with temperatures up to about 100,000 K or so. However, they do cool - albeit slowly, as they have small surface areas. This cooling takes a long time, but some white dwarfs have become cooler than the Sun. WD 0346+246 is a famous case, with a surface temperature of about 3900 K (Hambley et al. 1997).



This implies that white dwarfs with temperatures like that of the Sun do exist; moreover, they're small. The same group measured WD 0346+246 to have a radius roughly that of the Earth, which is extraordinary - certainly less than that of EBLM J0555-57Ab. The problem, of course, is that white dwarfs don't undergo fusion. Indeed, the degeneracy of the matter inside a white dwarf means that fusion reactions are unstable, and can lead to novae and Type Ia supernovae.



A note on spectral lines



Finally, I should mention that actual stellar spectra are extremely complicated. A black body spectrum only models the global behavior; on smaller scales, there are numerous absorption and emission lines. I wrote in a related answer that Dyson spheres and brown dwarfs could be distinguishable by the presence and absence of various spectral lines.



The same should be true here. Just because another body of some sort is the same temperature as the Sun doesn't mean it would appear identical to the Sun, under spectroscopic scrutiny. I assume you don't care much about this, but it definitely is something to be aware of. Atmospheric composition really does matter.






share|improve this answer























  • Deuterium fusion doesn't require anywhere near as much mass as hydrogen fusion. Could you get the correct spectrum out of a deuterium-burning brown dwarf?
    – Mark
    7 hours ago










  • so, the subdwarfs of the same apprx. R of sol, since lower density, would have lower L & T, right -implying R must be larger to to compensate? (great answer, by the way)
    – theRiley
    6 hours ago












  • @Mark No, you couldn't. Brown dwarfs have surface temperatures of maybe 1000-2000 K at the most, with many around 500-1000-ish K. Maybe there are individual peculiar examples I'm not aware of, but it certainly seems unlikely.
    – HDE 226868
    5 hours ago










  • @theRiley Thanks! Subdwarfs in general will have smaller radii, period. They do span a large temperature range, and can have spectral types up to O - and therefore could actually be decently luminous - but G-type subdwarfs, and those of similar spectral types, will be small and cool, and therefore not as bright as a main sequence star of the same spectral type/temperature.
    – HDE 226868
    5 hours ago













up vote
10
down vote










up vote
10
down vote









Black body radiation



The Sun is, approximately, a black body. That means that the light it emits follows a particular spectrum according to Planck's law, with the shape of the spectrum determined solely by the Sun's surface temperature. In particular, the wavelength of peak emission can be found through Wien's law, which is also a function of temperature. Therefore, if we want our new star to emit light just like the Sun does, we need to keep it at the same temperature as the real Sun - about 5800 K.



On the main sequence, there are some simple scaling relationships between mass, radius, temperature and luminosity:
$$Lpropto M^3,quad Rpropto M^{4/7},quad Tpropto M^{4/7}$$
assuming the proton-proton chain reaction is the main source of energy, which is the case. If we want to keep the temperature constant, we need to keep the mass constant, and thus keep the radius constant. Therefore, no main sequence star can be significantly smaller than the Sun and still have the same temperature.



The composition of the star does matter; in fact, according to something called the Vogt-Russell theorem, the mass and composition of a star uniquely determine its properties and evolution. This means that the exact form of the relations above does vary between stellar populations - in part, relying on the mean molecular mass $mu$ - but this still will not make a significant difference here.



Options for a Sun-like star



We could look at subdwarfs, low-metallicity stars that are dimmer than main sequence stars. They're not common, but they exist, lying immediately below the main sequence on the Hertzsprung-Russell diagram. All other G-type stars are either on the main sequence, and thus fairly like the Sun in size, or off the main sequence, as giants or supergiants. Therefore, our options are either G-type subdwarfs or main sequence stars that are slightly cooler than the Sun:





  • Tau Ceti, a main sequence star, might be a good choice; it has a temperature a few hundred Kelvin less than the Sun, and is approximately $0.8R_{odot}$ in size (Texeira et al. 2009).


  • Mu Cassiopeiae A, a subdwarf, has a similarly slightly-cooler temperature and again a radius of about $0.8R_{odot}$ (Boyaijan et al. 2008).


Another star I considered is Groombridge 1830. Like Mu Cassiopeiae, it's a borderline G-type subdwarf, but is about 1000 K cooler - too far from the Sun in terms of temperature.



Low-mass stars and true lower limits



You won't find a star small enough to fit into an O'Neill cylinder - at least, not one that formed by natural means. EBLM J0555-57Ab, a small, late-type red dwarf has a radius 0.85 times that of Jupiter (von Boetticher et al. 2017) - likely too large for your purposes. Bodies smaller than that are likely too low-mass to fuse hydrogen, and would instead be brown dwarfs. Of course, EBLM J0555-57Ab is also likely cool, with a surface temperature far below that of the Sun.



Brown dwarfs - many of which fuse deuterium - are not true stars, and are usually cool, with typical temperatures around 1000 K, producing spectra much different from the Sun's. Some exoplanets may be much hotter than this, with surface temperatures comparable to those of many stars (see e.g. Kepler-70b, with a surface temperature of about 7000 K as per Charpinet et al. 2011). However, those bodies are only hot because they're irradiated by the stars they orbit; on their own, they would not generate that much heat.



White dwarfs



There is a possibility I had completely forgotten about before: a white dwarf. Many white dwarfs are hot, with temperatures up to about 100,000 K or so. However, they do cool - albeit slowly, as they have small surface areas. This cooling takes a long time, but some white dwarfs have become cooler than the Sun. WD 0346+246 is a famous case, with a surface temperature of about 3900 K (Hambley et al. 1997).



This implies that white dwarfs with temperatures like that of the Sun do exist; moreover, they're small. The same group measured WD 0346+246 to have a radius roughly that of the Earth, which is extraordinary - certainly less than that of EBLM J0555-57Ab. The problem, of course, is that white dwarfs don't undergo fusion. Indeed, the degeneracy of the matter inside a white dwarf means that fusion reactions are unstable, and can lead to novae and Type Ia supernovae.



A note on spectral lines



Finally, I should mention that actual stellar spectra are extremely complicated. A black body spectrum only models the global behavior; on smaller scales, there are numerous absorption and emission lines. I wrote in a related answer that Dyson spheres and brown dwarfs could be distinguishable by the presence and absence of various spectral lines.



The same should be true here. Just because another body of some sort is the same temperature as the Sun doesn't mean it would appear identical to the Sun, under spectroscopic scrutiny. I assume you don't care much about this, but it definitely is something to be aware of. Atmospheric composition really does matter.






share|improve this answer














Black body radiation



The Sun is, approximately, a black body. That means that the light it emits follows a particular spectrum according to Planck's law, with the shape of the spectrum determined solely by the Sun's surface temperature. In particular, the wavelength of peak emission can be found through Wien's law, which is also a function of temperature. Therefore, if we want our new star to emit light just like the Sun does, we need to keep it at the same temperature as the real Sun - about 5800 K.



On the main sequence, there are some simple scaling relationships between mass, radius, temperature and luminosity:
$$Lpropto M^3,quad Rpropto M^{4/7},quad Tpropto M^{4/7}$$
assuming the proton-proton chain reaction is the main source of energy, which is the case. If we want to keep the temperature constant, we need to keep the mass constant, and thus keep the radius constant. Therefore, no main sequence star can be significantly smaller than the Sun and still have the same temperature.



The composition of the star does matter; in fact, according to something called the Vogt-Russell theorem, the mass and composition of a star uniquely determine its properties and evolution. This means that the exact form of the relations above does vary between stellar populations - in part, relying on the mean molecular mass $mu$ - but this still will not make a significant difference here.



Options for a Sun-like star



We could look at subdwarfs, low-metallicity stars that are dimmer than main sequence stars. They're not common, but they exist, lying immediately below the main sequence on the Hertzsprung-Russell diagram. All other G-type stars are either on the main sequence, and thus fairly like the Sun in size, or off the main sequence, as giants or supergiants. Therefore, our options are either G-type subdwarfs or main sequence stars that are slightly cooler than the Sun:





  • Tau Ceti, a main sequence star, might be a good choice; it has a temperature a few hundred Kelvin less than the Sun, and is approximately $0.8R_{odot}$ in size (Texeira et al. 2009).


  • Mu Cassiopeiae A, a subdwarf, has a similarly slightly-cooler temperature and again a radius of about $0.8R_{odot}$ (Boyaijan et al. 2008).


Another star I considered is Groombridge 1830. Like Mu Cassiopeiae, it's a borderline G-type subdwarf, but is about 1000 K cooler - too far from the Sun in terms of temperature.



Low-mass stars and true lower limits



You won't find a star small enough to fit into an O'Neill cylinder - at least, not one that formed by natural means. EBLM J0555-57Ab, a small, late-type red dwarf has a radius 0.85 times that of Jupiter (von Boetticher et al. 2017) - likely too large for your purposes. Bodies smaller than that are likely too low-mass to fuse hydrogen, and would instead be brown dwarfs. Of course, EBLM J0555-57Ab is also likely cool, with a surface temperature far below that of the Sun.



Brown dwarfs - many of which fuse deuterium - are not true stars, and are usually cool, with typical temperatures around 1000 K, producing spectra much different from the Sun's. Some exoplanets may be much hotter than this, with surface temperatures comparable to those of many stars (see e.g. Kepler-70b, with a surface temperature of about 7000 K as per Charpinet et al. 2011). However, those bodies are only hot because they're irradiated by the stars they orbit; on their own, they would not generate that much heat.



White dwarfs



There is a possibility I had completely forgotten about before: a white dwarf. Many white dwarfs are hot, with temperatures up to about 100,000 K or so. However, they do cool - albeit slowly, as they have small surface areas. This cooling takes a long time, but some white dwarfs have become cooler than the Sun. WD 0346+246 is a famous case, with a surface temperature of about 3900 K (Hambley et al. 1997).



This implies that white dwarfs with temperatures like that of the Sun do exist; moreover, they're small. The same group measured WD 0346+246 to have a radius roughly that of the Earth, which is extraordinary - certainly less than that of EBLM J0555-57Ab. The problem, of course, is that white dwarfs don't undergo fusion. Indeed, the degeneracy of the matter inside a white dwarf means that fusion reactions are unstable, and can lead to novae and Type Ia supernovae.



A note on spectral lines



Finally, I should mention that actual stellar spectra are extremely complicated. A black body spectrum only models the global behavior; on smaller scales, there are numerous absorption and emission lines. I wrote in a related answer that Dyson spheres and brown dwarfs could be distinguishable by the presence and absence of various spectral lines.



The same should be true here. Just because another body of some sort is the same temperature as the Sun doesn't mean it would appear identical to the Sun, under spectroscopic scrutiny. I assume you don't care much about this, but it definitely is something to be aware of. Atmospheric composition really does matter.







share|improve this answer














share|improve this answer



share|improve this answer








edited 3 hours ago

























answered 8 hours ago









HDE 226868

61.9k12215399




61.9k12215399












  • Deuterium fusion doesn't require anywhere near as much mass as hydrogen fusion. Could you get the correct spectrum out of a deuterium-burning brown dwarf?
    – Mark
    7 hours ago










  • so, the subdwarfs of the same apprx. R of sol, since lower density, would have lower L & T, right -implying R must be larger to to compensate? (great answer, by the way)
    – theRiley
    6 hours ago












  • @Mark No, you couldn't. Brown dwarfs have surface temperatures of maybe 1000-2000 K at the most, with many around 500-1000-ish K. Maybe there are individual peculiar examples I'm not aware of, but it certainly seems unlikely.
    – HDE 226868
    5 hours ago










  • @theRiley Thanks! Subdwarfs in general will have smaller radii, period. They do span a large temperature range, and can have spectral types up to O - and therefore could actually be decently luminous - but G-type subdwarfs, and those of similar spectral types, will be small and cool, and therefore not as bright as a main sequence star of the same spectral type/temperature.
    – HDE 226868
    5 hours ago


















  • Deuterium fusion doesn't require anywhere near as much mass as hydrogen fusion. Could you get the correct spectrum out of a deuterium-burning brown dwarf?
    – Mark
    7 hours ago










  • so, the subdwarfs of the same apprx. R of sol, since lower density, would have lower L & T, right -implying R must be larger to to compensate? (great answer, by the way)
    – theRiley
    6 hours ago












  • @Mark No, you couldn't. Brown dwarfs have surface temperatures of maybe 1000-2000 K at the most, with many around 500-1000-ish K. Maybe there are individual peculiar examples I'm not aware of, but it certainly seems unlikely.
    – HDE 226868
    5 hours ago










  • @theRiley Thanks! Subdwarfs in general will have smaller radii, period. They do span a large temperature range, and can have spectral types up to O - and therefore could actually be decently luminous - but G-type subdwarfs, and those of similar spectral types, will be small and cool, and therefore not as bright as a main sequence star of the same spectral type/temperature.
    – HDE 226868
    5 hours ago
















Deuterium fusion doesn't require anywhere near as much mass as hydrogen fusion. Could you get the correct spectrum out of a deuterium-burning brown dwarf?
– Mark
7 hours ago




Deuterium fusion doesn't require anywhere near as much mass as hydrogen fusion. Could you get the correct spectrum out of a deuterium-burning brown dwarf?
– Mark
7 hours ago












so, the subdwarfs of the same apprx. R of sol, since lower density, would have lower L & T, right -implying R must be larger to to compensate? (great answer, by the way)
– theRiley
6 hours ago






so, the subdwarfs of the same apprx. R of sol, since lower density, would have lower L & T, right -implying R must be larger to to compensate? (great answer, by the way)
– theRiley
6 hours ago














@Mark No, you couldn't. Brown dwarfs have surface temperatures of maybe 1000-2000 K at the most, with many around 500-1000-ish K. Maybe there are individual peculiar examples I'm not aware of, but it certainly seems unlikely.
– HDE 226868
5 hours ago




@Mark No, you couldn't. Brown dwarfs have surface temperatures of maybe 1000-2000 K at the most, with many around 500-1000-ish K. Maybe there are individual peculiar examples I'm not aware of, but it certainly seems unlikely.
– HDE 226868
5 hours ago












@theRiley Thanks! Subdwarfs in general will have smaller radii, period. They do span a large temperature range, and can have spectral types up to O - and therefore could actually be decently luminous - but G-type subdwarfs, and those of similar spectral types, will be small and cool, and therefore not as bright as a main sequence star of the same spectral type/temperature.
– HDE 226868
5 hours ago




@theRiley Thanks! Subdwarfs in general will have smaller radii, period. They do span a large temperature range, and can have spectral types up to O - and therefore could actually be decently luminous - but G-type subdwarfs, and those of similar spectral types, will be small and cool, and therefore not as bright as a main sequence star of the same spectral type/temperature.
– HDE 226868
5 hours ago










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