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In astronomy, the background from the camera itself is called "dark current" and is removed by first taking an exposure with the shutter closed for, say, half an hour, and then subtract those counts from the real observations, normalized to the exposure time of a given image. Sometimes, if you're bored at the telescope due to bad weather, you can even take ...

10

By comparing the signal to the background. Suppose you get 10 IR photons from the camera and lens background but an extra 5 from the source then you can still detect the source. There is a whole science of signal processing to detect signals much fainter than the background. Especially in IR astronomy.

1

Higher energies are less probable, so there are less electrons emitting high frequency photons. The intensity is proportional to amplitude squared, in the wave description, and proportional to number of photons in the particle description

3

I will address the question from the current body of the question, asking for $\operatorname{Tr}(S^2\rho)$ on a system of $N$ spin-1/2 particles in the thermal state $\rho = \frac1Ze^{-\beta S_z}$. This has little to do with the photons in the question title. The total spin operator splits into single-particle terms and two-particle cross terms.  S^2=\...

2

Apples and oranges. The first is the photon density in a volume whose radiation field is in thermal equilibrium. The second is the rate at which photons pass a unit area regardless of the source of the radiation. They are both correct, but they describe different things. BTW, it's much better to tell us what is in a document instead of asking us to look ...

1

Every object with nonzero temperature (so, every object) radiates EM radiation The spectrum of a black body (that is, a body that absorbs all light at all wavelengths) is given by Planck's law. Generally, nothing is completely black (except for empty space, which is why cosmic microwave background has a very nice black body radiation spectrum). In general, ...

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Edit: Please note important Caveat #2 at the bottom. The Russian wikipedia page for Kirchhoff's law of thermal radiation is simpler and shorter than the English version, however it contains the answer to the question, which is absent in the English version. Translation follows: Bodies, whose absorptivity is frequency-independent are called "gray bodies"....

2

If you can see through it then it it is not black body radiation. A blackbody emitter must absorb light of all wavelengths that is incident upon it and must be in thermal equilibrium. i.e. It must be "optically thick" at all wavelengths. It may have a spectrum similar to the Planck function if the partial absorptivity is independent of wavelength. This is ...

1

In a way, these are intrinsic properties of a material that depend on temperature. The temperature of the object dictates what a black body curve looks like and which wavelengths are emitted in what amounts. The emissivity of a material is not always uniform across all wavelengths, which means at different temperatures, you might get different emission ...

2

One possibility is so called optical parametric amplification (OPA), which basically splits a photon with frequency $\omega_0$ in two photons with frequency $\omega_i$ and $\omega_s$ with $\omega_0 = \omega_i+\omega_s$. This however only works nicely for high intensities i.e. pulsed light sources.

2

I could imagine a molecule which has two states $E_1$ and $E_2$ above his groundstate with $|E_1| < |E_2|$. A photon could excite the molecule to $E_2$ and the molecule would first fall into $E_1$ and then into the ground state. If $E_2$ is something like two times $E_1$ you would have two emitted photons with a smaller wavelength than the one photon ...

0

The study of quantum mechanics has taught us that asking questions about things that "really" happen even though they can't be observed is a recipe for making nonsensical or physically meaningless statements. Physics, like all healthy sciences, is an attempt to describe things that can (at least in principle) be observed. Numerous valid general-...

1

As you correctly mentioned, using a classical approach leads to the wrong result, in which the distribution goes asymptotically to infinity as the frequency grows, this is called the ultraviolet catastrophe. It is true that classically each mode will have the same energy,but most of the energy in a natural vibrator will be in the smaller wavelengths and ...

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