astronomyspectroscopystarstemperature
The Hertzsprung-Russell diagram that categorizes star types uses the letter codes O, B, A, F, G, K, and M to indicate a star's temperature/color. Hottest (blue) is O and coolest (red) is M.
What do the letters themselves relate to?
For example, why is "O" used for hottest/blue?
1) If one object is appearing blue and other object red. That doesn't always mean blue object is hotter than red? Is that correct? If yes then how is it so? I am sorry but I'm bit confused over it.
Yes it's correct: when an object is glowing blue when burning that means it has more temperature than an object glowing red. It's because an object which is glowing blue has more energy. Energy of an object of temperature $T$ (in kelvins) is calculated using this equation: $E=kT$ where $k$ is Boltzmann's constant. Energy of a light with frequency $f$ is calculated using this equation: $E=hf$ where $h$ is plank's constant. So spectral radiance of emitted light with frequency $f$ from object with temperature $T$ is calculated using $\beta(T) = \frac{2hf^3}{c^2}\frac{1}{e^{\frac{hf}{kT}}-1}$. For derivation See Plank's Law. The higher the frequency of light the more energy it has and it looks 'more blue' and the lower the frequency the less energy and it looks 'more red'.
Here's very good image of "light" (light is part of spectrum we see (visible spectrum), full spectrum is called electro-magnetic spectrum, but for simplicity let's call it just light) spectrum:
Here's a very good diagram of emitted light with wavelength $\lambda$ from an object with temperature $T$ (in kelvins). (NOTE: for light with frequency $f$, wavelength is $\lambda = \frac{c}{f}$ where $c$ is the speed of light.)
2) With the "Color Temperature" concept on Wikipedia. They say 1,700 K for match flame. But 15,000–27,000 K for a clear blue poleward sky.
This confuses me. The sky appears blue, does that mean its hot? But it isn't right? It's colder. I am not sure if I'm able to frame it properly. It's a bit confusing to me.
Is this something like a glowing object vs reflecting object? The sun is white but Earth is blue. But sun is hotter than the earth. It's the sun radiating light but earth is just scattering it.
You can calculate spectral radiance. Spectral radiance is like an (spectral radiance isn't fully intensity, it's just like an intensity) intensity of emitted light with frequency $f$ from object with temperature $T$ (in kelvins) of emitted light from an object with temperature $T$ (in kelvins) using this equation:$$\beta(T) = \frac{2hf^3}{c^2}\frac{1}{e^{\frac{hf}{kT}}-1}$$ Earth is blue because it reflects light, not emits it. But the Sun doesn't reflect light (it's a blackbody, but not perfect), it just emits it, that's why it's hotter.
An object at 1000K glows red because that is basically dominating the radiation that it is emitting that we can see. Most of the radiation it emits emerges in the infrared part of the spectrum to which our eyes are not sensitive.
Your edited question asks about the appearance of hot metals. The chart below is an example of a "colour-temperature" chart for steel (from here) and roughly matches the second sentence of your question.
Hot metals do approximate to blackbodies, but to get some idea of what colour a relatively cold (i.e. $<2000$ K) black body will look like, you need to convolve the blackbody flux distribution with the response of the eye. In the case of objects at around 1000 K, the peak of the blackbody function, in terms of wavelength, is at around 3 microns - way below where our eyes are sensitive. This means at these temperatures your eyes are only sensitive to the short wavelength Wien tail of the blackbody distribution. As the temperature increases, the Wien tail in the visible part of the spectrum increases and light at bluer wavelengths become bright enough to excite a response in the eye. The radiation must always be weighted so that there is more red light than blue, but because the eye's response peaks at shorter wavelengths, I assume that when a metal reaches 1500 K or so, there is enough yellow light in the Wien tail to stimulate the eye into perceiving a yellow colour.
Stars are hotter. At 3000 K the peak of the blackbody distribution is just redward of what we can perceive. The stars emit light across the whole of the visible range, but weighted towards the red. Given the above discussion about metals we might expect such stars to appear almost white. A number of people have gone to the trouble of working out what colours stars should be, as viewed by the eye. In fact, stars are not strongly coloured at all, because at temperatures of 3000K and above, the black-body(ish) radiation from stars covers wavelengths across the entire range of the eye's sensitivity. The descriptions of "red star" and "yellow star" really refer to where the peak of the blackbody flux distribution lies, rather than their true visible appearance.
Best Answer
When astronomers started to get spectra of stars and began classifying them, the initial classification was based on the strength of the Balmer absorption lines in the spectra. The Balmer lines are created by electons in hydrogen atoms that are currently in the second energy level (N=2) absorbing energy and jumping up to higher levels. The stars with the strongest lines were given a designation of A, the second strongest B, etc. on down the line.
Later, the association with temperature was made and many of the classes were combined and reordered based on the temperature of the star. However, the main letter designations were kept even though they were now "out of order". So the letters actually correspond to the relative strength of the Hydrogen absorption lines.
The reason the O stars, which are hottest, are so far down on the scale is that they are so hot that Hydrogen is fully (or very nearly fully) ionized. If the atom has been ionized, it by definition doesn't have any electrons in the N=2 level since it doesn't have any electrons at all. Since there are no electrons in the proper level to form the line, the line is weak as it is only created when a chance recombination occurs.
The cooler M stars, at the other end of the main sequence, also show very week Hydrogen lines. The reason for this is that they are too cool to even get the electrons excited enough to get up to the N=2 energy state. If there are no electrons in N=2, there is no way to get the electron to absorb energy and jump higher and form the Balmer lines. What hydrogen lines are seen here are due to the few higher velocity atoms on the long tail of the velocity distribution. When they collide, they have enough energy to get their electrons up to N=2.
As the temperature increases, the energy available to excite the atoms increases as well. As you move up from M class stars you get to spectral types K, G, F, and then A, each one increasing in strength of the lines. The A stars, at about 10,000 K, are just the right temperature to push almost all the hydrogen atoms into the higher N=2 state where they can then be knocked higher and produce the Balmer series. Since there are a lot of electrons in N=2, you get strong lines. The next spectral class up is B, which has weaker lines because at this point you've started to ionize the hydrogen atoms and therefore have fewer atoms with electrons in the N=2 state. And then comes O stars which are nearly fully ionized and so are very week in the Balmer series.