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If I believe right, blue flames are hotter than red ones and if I recall correctly it's because blue tend to ultraviolet. Then why is the sun, which's capable of warming whole planets, red instead of blue?

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    $\begingroup$ The Sun is not red, the Sun is white. We perceive it as red (or orange or yellow, depending on the time of day) because of Earth's atmosphere. You can read a bit more about it here. $\endgroup$ – L. Werneck Dec 18 '17 at 16:33
  • $\begingroup$ The Sun is about as white as white can get. When Venus or the Moon rises, they start as dim yellowish or reddish objects, getting whiter and brighter as they rise above the horizon. When the Sun rises, it too starts as a dim (compared to high noon) yellowish or reddish object and gets whiter and brighter as it rises above the horizon. You can look at Venus or the Moon when they are high in the sky. You can't look at the Sun, even for the briefest of moments, once it gets above a few degrees above the horizon. $\endgroup$ – David Hammen Dec 18 '17 at 17:19
  • $\begingroup$ That the sun is capable of warming whole planets is not only dependent on its temperature (which is the cause of its radiation). Imagine two suns shining on the same planet - the planet heats up double as much now, but the temperature of the objects shining on it is the same as before. $\endgroup$ – Steeven Dec 18 '17 at 17:27
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    $\begingroup$ You are conflating temperature with heat output. The Sun puts out enough heat to warm entire planets--planets that are millions of miles away. But that doesn't mean that the temperature of the Sun has to be terribly high. The surface of the Sun is a mere 5700 Kelvin--just slightly hotter than the filament of a tungsten halogen light bulb. The reason that the Sun can warm whole planets is that the Sun is HUUUUUUUGE! And the entire surface of it is radiating at that 5700 K temperature. $\endgroup$ – Solomon Slow Dec 18 '17 at 21:14
  • $\begingroup$ I know this is a physics SE, but: the Sun is white for the same reason the atmosphere is transparent: we evolved so that that was the case. Any other solution should have been out-competed. $\endgroup$ – JEB Dec 18 '17 at 23:52
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The surface of the sun is around 5,700 degrees. The sun radiates all colors of the visible spectrum. The sun radiates like a black body and its spectrum is shown below.

enter image description here

The sun does emit blue light but it also emits red and the other colors with almost the same intensity. A mixture of all the colors results in white light. As the temperature decreases, the peak moves to the right and the sun would start to appear red. Likewise, as you said, if the sun was hotter, the peak would move left and the sun would start to appear blue as the peak approached the ultraviolet part of the spectrum.

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The comparison with flames is a bit misleading. Flames look yellow when they have have particles of unburnt carbon in them, and they look blue when there is no unburnt carbon present. The amount of unburnt carbon decreases as the flame gets hotter, so blue flames are indeed (usually) hotter than yellow flames. But this is not a fundamental property of the flame. It's due to the unburnt carbon.

The Sun is not a flame as nothing is burning in it. Instead it is a plasma i.e. it is a gas of ionised hydrogen and helium atoms, and the colour is described by Wien's displacement law. The temperature of the surface is around 5,700K and in fact it is more or less white. We tend to see it as yellowish because the atmosphere scatters blue light, but if you look at the Sun from outside the Earth's atmosphere it looks white.

You can get blue stars, but to have a significant blue colour the stars have to be very hot indeed. The surface temperature needs to be in the range 10,000 - 20,000 K or 2 to 4 times hotter than the Sun. The amount of energy emitted by a star of temperature $T$ is proportional to $T^4$ (see the Stefan Boltzmann law) so if you doubled the temperature of the Sun (and kept it the same size) it would emit 16 times as much heat and rapidly roast us all to a crisp!

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  • $\begingroup$ The peak wavelength is described by Wien's displacement law but I would argue the color (which is a result of not just the peak but all the wavelengths) is better described by the Planck Law. $\endgroup$ – Floris Dec 18 '17 at 17:36
  • $\begingroup$ To add, the determining factor that makes some stars blue is their mass. Very massive stars are blue. See the Wikipedia article on O-type stars. These are the most massive of the main sequence star (O-B-A-F-G-K-M, from most massive to least). For comparison the Sun is G-type. $\endgroup$ – Allure Dec 18 '17 at 21:57
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The light of the sun contains a mixture of wavelengths - as given by the Planck law. The spectral distribution looks like this:

enter image description here

As you can see, the intensity is biased slightly towards the blue/violet part of the spectrum. However, as we evolved, we have learnt to interpret that mix of colors as "white". Strangely enough, when you are in a room illuminated by tungsten (incandescent) lights, when you see a piece of white paper you will have no trouble identifying it as "white" also. Yet the objective color is much more red/yellow (this is why cameras these days have AWB - automatic white balance - and why you used to have a "tungsten filter" that you would put on your film-based camera if you wanted to shoot slide film indoors: if you didn't, the images would come out looking horribly red).

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  • $\begingroup$ "the intensity is biased slightly towards the blue/violet part of the spectrum" is true only for wavelength-based spectral density. If you instead consider frequency-based density, you may be surprised to see that the peak is in IR, 883 nm instead of the 502 nm you get with wavelength-based density. So this sentence in your answer doesn't make too much sense. $\endgroup$ – Ruslan Mar 23 at 7:59
  • $\begingroup$ @Ruslan that’s interesting and i may generate that plot to convince myself; but I was just interpreting the graph as drawn (“as you can see”...) which used wavelength. I am not sure why anyone would want to plot frequency along X, then claim “the peak is at 883 nm” rather than “at 400 THz”. How would you rephrase my answer? $\endgroup$ – Floris Mar 23 at 11:11
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    $\begingroup$ I think there's no need to speak about bias at all. You might just want to note that spectral power density is not constant, but that's all what can be told about it for the purpose of this post. As for "generate the plot", there are some plots (and more discussion of human eye sensitivity vs solar spectrum) here: oceanopticsbook.info/view/light_and_radiometry/level_2/… $\endgroup$ – Ruslan Mar 23 at 13:20
  • $\begingroup$ Actually, "not constant" would also not say anything: a constant distribution in one variable won't be constant in the other. $\endgroup$ – Ruslan Mar 23 at 13:45
  • $\begingroup$ Right - you really need to define what bandwidth you have to integrate for "blueness" and "redness". Bottom line: our eyes evolved to consider anything that is "roughly like sunlight" as white; but in fact we are always recalibrating our white point depending on the color (temperature) of the illumination. Thus we will see things as white in tungsten illumination, even when a (uncompensated) photograph of the same thing would look distinctly yellow. $\endgroup$ – Floris Mar 23 at 13:48

protected by Qmechanic Dec 18 '17 at 20:09

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