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Ok, here's my take on calculating the color of a blackbody, or any spectrum in fact: Disclaimer: I'm not a color theorist, and there may be more accurate methods. But the result, shown in the bottom, looks about right. Spectrum First note that since color is a function of the relative intensity in various wavelength bands, it doesn't matter whether we ...


7

Without any other information, you cannot distinguish between the two effects. $$ T = T_0 (1 + z) $$ A blackbody spectrum of temperature $T$ is identical to a blackbody spectrum of temperature $T_0$ with redshift $z$. For stellar/galactic radiation, we can use the fact that the radiation is not a perfect blackbody. For the CMB, we can use the fact that ...


5

The material at the surface of a white dwarf is not degenerate. The "visible" surface is defined as where the optical depth exceeds some threshold and this will occur at a low enough density that even at a few hundred kelvin, the ratio of the Fermi energy to the thermal energy is too low for significant degeneracy. In addition, at these temperatures, the ...


4

Black holes can "eat" Hawking radiation, but the radiation has to be directed towards the black hole before it can be "eaten". Black holes are pretty small, and space is very very very big. So as a black hole evaporates, and gives off" Hawking radiation, some may eventually fall into another black hole, but most will not. It will just travel out into space, ...


3

The wavelength of the CMB increases linearly with the scale factor $a$, which is defined as 1 today, so the "sound" of the CMB has dropped an octave when the Universe has doubled in size (in all three directions), i.e. when $a = 2$, which will happen when it is roughly 25 billion years old (see e.g. Fig 1 of Davis & Lineweaver 2004). The magnitude of ...


3

That is what is done. This is shown in an old xkcd comic https://xkcd.com/54/ The curve shows the distribution of frequencies in the CMB, and by using the marked value of the maximum you can determine the value of T, the apparent (red-shifted) temperature of the CMB


3

It would be good to specify what "majority of radiation" means. If it means "radiative energy", then this premise is surely wrong. To find the energy emitted per wavelength, take the Planck function $B(T, \lambda)$ at a certain temperature $T$ and multiply with the wavelength $\lambda$. You will find that the short-wavelength radiation always dominates for a ...


3

Here's what I got for $\nu B_{\nu}(T)$ using Python, which is easy enough to read that maybe you can check against your Excel calculation. I also plotted just $B_{\nu}(T)$ for comparison. Neither looks like a straight line and they shouldn't. If you print out $\log_{10}(k_B T / c)$ you get about 16.2 which is right in the middle of your range. They've ...


2

What you need to understand is that a black hole is a region of space defined by an event horizon. The key point about an event horizon is that if you're inside the region it surrounds you cannot get out. But it's only the stuff inside that is restricted in this absolute way. If something is outside (even just outside) it can, in principle, happily orbit ...


2

Let's answer the question from the point of view of a stable neutron star. The luminosity is $$L_0 = 4\pi R_0^2 \sigma T_{0}^4$$ in the inertial frame of the neutron star surface (assume a non-spinning object). For a distant observer, the radius of the neutron star is (e.g. see Haensel 2001) $$R_{\infty}= R_0 \left(1 - R_s/R_0\right)^{-1/2}\, ,$$ the ...


2

A blackbody spectrum is continuous. As Dean comments, there are no true blackbodies. But this is not really the issue here: It's true that most emitted photons from, say, a star, which is quite close to a blackbody, is the result of some discrete process. E.g. an electron is brought to an excited state by a collision, the atom de-excites, and a photon with ...


1

Þe olde goode dimensional analysis can help. The Hawking radiation is essentially a quantum field effect of the same nature as the Unruh effect. Quantum fields “live” in a distorted space-time of the hole and do not need to know anything about gravitation. Like for the Unruh effect (having $c/a$ as its characteristic time), there is a natural time scale for ...


1

In a sense, you're right. Vacuum energy over a surface area is proportional to the surface area so larger area means more of the particle-antiparticle occurrences and that's the source of Hawking radiation, that is, if you don't mind the virtual particle explanation, which is the only explanation I sort of understand. I look at the equations and my brain ...


1

Part 1 On the one hand, ${\rm 1\, ergs\cdot cm^{-2}\cdot s^{-1} \cdot sr^{-1}\cdot Hz^{-1}} = 10^{-3} {\rm J}/({\rm m^2} \cdot {\rm sr})$ has basic dimensions ${\rm mass} \cdot {\rm time}^{-2} \cdot {\rm angle}^{-2}$ and is a radio brightness. On the other hand, there is ${\rm 1 MJy}\cdot{\rm sr}^{-1} = 10^{-20} {\rm J}/({\rm m^2} \cdot {\rm sr})$ which has ...


1

Black hole don't allow anything to escape, if the matter/light is inside the event horizon (boundary which describes the point of no return). The emission radiation that is referred in the question is not within the event horizon, so it can escape. This is caused by friction within matter which is swirling in to the black hole.


1

The best way to do this is to pick a particular astronomical instrument or science mission and go to their website. Almost all astronomy projects are required to release their data to the public and they almost always link to their data from the website for that mission/instrument. For example, the Sloan Digital Sky Survey which produces spectra for a huge ...


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