Learn Wien's Displacement Law Derivation with Plainmath's Expert Help

Deeper underlying explanation for color-shift in Wien's Law?
Suppose we have a blackbody object, maybe a star or a metal (although I understand neither of these are actually blackbody objects, to some extent my understanding is they can approximate one). According to Wien's Law, as temperature increases, peak wavelength will decrease, so the color we observe will "blueshift." Despite having reached Wien's Law in my research, my understanding is that this is in fact not an answer to why the blueshift occurs, but just a statement that the blueshift does occur.
So, is there some deeper reason why blackbodies peak wavelength emitted decreases with temperature?

Wien's Displacement Law and reradiation of LWIR by $\mathrm{C}{\mathrm{O}}_2$
Consider Wien's Displacement Law (if I understand correctly): λ = b / T Where, λ = Peak Wavelength b = 0.028977 mK (Wien's constant) T = Temperature
According to this law $\mathrm{C}{\mathrm{O}}_2$ molecules can only absorb and reradiate the Long Wave I.R. frequency (radiated from Earth) in their 15 micrometre (μm) spectrum at a temperature of -80 degrees Celcius. The link with temperature seems crucial to me since the only part of our athmosphere cold enough for this to happen is the Mesosphere at about 50-80 km from Earth. Troposphere and Stratosphere are not cold enough (coldest temperature is about -55 degrees Celsius). In these parts of our atmosphere it is just not cold enough for $\mathrm{C}{\mathrm{O}}_2$ to reradiate I.R. wavelengths back to Earth. Then why is $\mathrm{C}{\mathrm{O}}_2$ considered a greenhouse gas in our Tropo- or Stratosphere? Am I missing out on something here?
In the Mesosphere, however, temperature can drop to over -100 degrees Celsius. In this part of the atmosphere $\mathrm{C}{\mathrm{O}}_2$ can reradiate LWIR back to Earth. But the problem is that the air is so thin, there are hardly any molecules of $\mathrm{C}{\mathrm{O}}_2$
What am I missing here in my reasoning?

What is the most intense wavelength and frequency in the spectrum of a black body?
Planck's Law is commonly stated in two different ways:
$u_{\mathrm{\lambda <!--\; \lambda \; -->}}(\mathrm{\lambda <!--\; \lambda \; -->},T)=\frac{2h{c}^2}{{\mathrm{\lambda <!--\; \lambda \; -->}}^5}\frac{1}{e^{\frac{hc}{\mathrm{\lambda <!--\; \lambda \; -->}kT}}-<!--\; -\; -->1}$
$u_{\mathrm{\nu <!--\; \nu \; -->}}(\mathrm{\nu <!--\; \nu \; -->},T)=\frac{2h{\mathrm{\nu <!--\; \nu \; -->}}^3}{c^2}\frac{1}{e^{\frac{h\mathrm{\nu <!--\; \nu \; -->}}{kT}}-<!--\; -\; -->1}$
We can find the maximum of those functions by differentiating those equations with respect to $\mathrm{\lambda <!--\; \lambda \; -->}$ and to $\mathrm{\nu <!--\; \nu \; -->}$, respectively. We get two ways to write Wien's Displacement Law:
${\mathrm{\lambda <!--\; \lambda \; -->}}_{\text{peak}}T=2.898\cdot <!--\; \cdot \; -->{10}^{-<!--\; -\; -->3}m\cdot <!--\; \cdot \; -->K$
$\frac{{\mathrm{\nu <!--\; \nu \; -->}}_{\text{peak}}}{T}=5.879\cdot <!--\; \cdot \; -->{10}^{10}Hz\cdot <!--\; \cdot \; -->K^{-<!--\; -\; -->1}$
We see that ${\mathrm{\lambda <!--\; \lambda \; -->}}_{\text{peak}}\ne <!--\; \ne \; -->\frac{c}{{\mathrm{\nu <!--\; \nu \; -->}}_{\text{peak}}}$. So what frequency or wavelength is actually detected by an optical instrument most intensely when analyzing a black body? If they are ${\mathrm{\lambda <!--\; \lambda \; -->}}_{\text{peak}}$ and ${\mathrm{\nu <!--\; \nu \; -->}}_{\text{peak}}$, how is ${\mathrm{\lambda <!--\; \lambda \; -->}}_{\text{peak}}\ne <!--\; \ne \; -->\frac{c}{{\mathrm{\nu <!--\; \nu \; -->}}_{\text{peak}}}$?

What is the difference between Wien's Displacement Law for peak frequency vs peak wavelength?
While doing research for a high school report I came across the fact that WDL actually has two forms, one for peak frequency and one for peak wavelength, and that these two forms are not the same and can not be used interchangeably.
My question is why peak frequency isn't the same as peak wavelength? That is, since wavelength is directly determined by frequency (since frequency = speed of light divided by wavelength), there is a one-to-one correspondence between a given wavelength and its frequency. Therefore why doesn't a peak in frequency correspond to a peak in wavelength, and visa versa (meaning that the two forms of WDL could be used interchangeably)?
I know that this question was already posted elsewhere, but I did not understand the answers. Since I am a high school student, complicated terminology can fly right over my head, so I would greatly appreciate it if someone could take the time to explain it clearly and simply (i.e. no monster equations).

Wien's fifth power Law and Stephan Boltzmann's fourth power laws of emissive power
Wien's fifth power law says that emissive power is proportional to the temperature raised to the fifth power. On the other hand, the Stefan–Boltzmann law says emissive power is proportional to the temperature raised to the fourth power. How can both of these be true?

Wien's Displacement Law for real bodies
It is known that for perfect blackbodies,
$\mathrm{\lambda <!--\; \lambda \; -->}T=c$
where $\mathrm{\lambda <!--\; \lambda \; -->}=\text{peak wavelength}$
$T=\text{Absolute temperature}$
$c=\text{Wien's constant}$
But this is for perfect blackbodies only, which have no theoretical existence. Does a similar formula exist for real bodies, which expresses $\mathrm{\lambda <!--\; \lambda \; -->}T$ in terms of its emissitivity $\mathrm{ϵ<!--\; ϵ\; -->}$? I googled it, but found no relevant results.

Area under Wien's displacement graph
Why does the area under Wien's displacement graph give Stefan-Boltzmann law for a black body?
I couldn't find any proof of this. (I could just find this expression). I am not aware of the function of Wien's displacement graph as well (I just know that it is between Intensity and wavelength emitted by a black body).
Is there a mathematical way to prove this?

Frequency and Wavelength peak for Wien's displaement law of a blackbody
This is a question relating to Wien's displacement law for the Planck function. As we all know frequency and wavelength are related to the speed of light by:
$\mathrm{\nu <!--\; \nu \; -->}\mathrm{\lambda <!--\; \lambda \; -->}=c$
However, why is it that:
${\mathrm{\nu <!--\; \nu \; -->}}_{\mathrm{p}\mathrm{e}\mathrm{a}\mathrm{k}}{\mathrm{\lambda <!--\; \lambda \; -->}}_{\mathrm{p}\mathrm{e}\mathrm{a}\mathrm{k}}\ne <!--\; \ne \; -->c$
Any explanations would be very much appreciated.
To all of the people wanting to know where this statement came from. It hasn't come from anywhere specific, is it a well known fact of the Planck function. ${\mathrm{\lambda <!--\; \lambda \; -->}}_{\mathrm{p}\mathrm{e}\mathrm{a}\mathrm{k}}=0.290T^{-<!--\; -\; -->1}$ cm K and ${\mathrm{\nu <!--\; \nu \; -->}}_{\mathrm{p}\mathrm{e}\mathrm{a}\mathrm{k}}=5.88\times <!--\; \times \; -->{10}^{10}T$ Hz K${}^{-<!--\; -\; -->1}$

How can Wien's Displacement Law be 'changed' to a version for frequency?
Wien's Displacement Law stated that for a blackbody emitting radiation,
${\mathrm{\lambda <!--\; \lambda \; -->}}_{max}={\displaystyle \frac{1}{T}}$
where $T$ is the temperature of the body and ${\mathrm{\lambda <!--\; \lambda \; -->}}_{max}$ is the maximum wavelength of radiation emitted.
Due to the relationship between wavelength, frequency and the speed of light, a value of maximum wavelength would give a value of minimum frequency, and vice versa.
I then saw on the Wikipedia page for Wien's Displacement Law that
$f_{max}={\displaystyle \frac{\mathrm{\alpha <!--\; \alpha \; -->}{k}_{B}T}{h}},$
where $\mathrm{\alpha <!--\; \alpha \; -->}=\mathrm{2.82...}$, $k_B$ is Boltzmann's Constant, $T$ is the temperature of the body and $h$ is Planck's Constant.
How can this relationship for maximum frequency be shown?

If a CMB photon traveled for 13.7 billion years (- 374,000 years) to reach me. How far away was the source of that CMB photon when it first emitted it?
My attempt to solve this question was to use the following assumptions:
1.Temperature of CMB photon today is 2.725 K (will use value of 3 K here)
2.Temperature of CMB photon when it was first emitted is 3000 K
3.A factor of x1000 in temperature decrease results in a factor of x1000 in wavelength increase. (According to Wien's displacement law)
Does this mean that the source of the CMB photon that just reached me today, was actually 13.7 billion light years / 1000 = 13.7 million light years away from me when it first emitted the photon?

Why 8–15 µm is considered "thermal infrared" if typical room temperature kT is 48 µm?
According to Wikipedia:
"Long-wavelength infrared (8–15 µm, 20–37 THz, 83–155 meV): The "thermal imaging" region, in which sensors can obtain a completely passive image of objects only slightly higher in temperature than room temperature - for example, the human body - based on thermal emissions only and requiring no illumination such as the sun, moon, or infrared illuminator. This region is also called the "thermal infrared"."
However, using $\frac{hc}{\mathrm{\lambda <!--\; \lambda \; -->}}=k_{\mathrm{B}}T$, the temperature range 288–308 K (15–35 °C) is equivalent to 50–46.7 µm, while 8–15 µm is equivalent to 1800–960 K (using the same equation).

Derivation of Wien's displacement law
I know this might be a silly question, but is it necessary to know Planck's Law in order to show that ${\mathrm{\lambda <!--\; \lambda \; -->}}_{max}\propto <!--\; \propto \; -->\frac{1}{T}$? If you set
$u(\mathrm{\lambda <!--\; \lambda \; -->},T)=\frac{f(\mathrm{\lambda <!--\; \lambda \; -->}T)}{{\mathrm{\lambda <!--\; \lambda \; -->}}^5}$
then
$\frac{\mathrm{\partial <!--\; \partial \; -->}u}{\mathrm{\partial <!--\; \partial \; -->}\mathrm{\lambda <!--\; \lambda \; -->}}=\frac{1}{{\mathrm{\lambda <!--\; \lambda \; -->}}^5}\frac{\mathrm{\partial <!--\; \partial \; -->}f}{\mathrm{\partial <!--\; \partial \; -->}\mathrm{\lambda <!--\; \lambda \; -->}}-<!--\; -\; -->\frac{5}{{\mathrm{\lambda <!--\; \lambda \; -->}}^4}f=0$
$\frac{\frac{\mathrm{\partial <!--\; \partial \; -->}f}{\mathrm{\partial <!--\; \partial \; -->}\mathrm{\lambda <!--\; \lambda \; -->}}}{f}=5\mathrm{\lambda <!--\; \lambda \; -->}$
But I am stuck here because if I integrate the L.H.S.
$\int <!--\; \int \; -->\frac{\frac{\mathrm{\partial <!--\; \partial \; -->}f}{\mathrm{\partial <!--\; \partial \; -->}\mathrm{\lambda <!--\; \lambda \; -->}}}{f}d\mathrm{\lambda <!--\; \lambda \; -->}=\log <!--\; \; -->(f(\mathrm{\lambda <!--\; \lambda \; -->}T))=\frac{5}{2}{\mathrm{\lambda <!--\; \lambda \; -->}}^2$

Deriving Wien's displacement law (Zettili)
According to Zettili, we can derive Wien's displacement law from Planck's energy density
$\stackrel{~<!--\; ~\; -->}{u}(\mathrm{\lambda <!--\; \lambda \; -->},T)=\frac{8\mathrm{\pi <!--\; \pi \; -->}hc}{{\mathrm{\lambda <!--\; \lambda \; -->}}^5}\frac{1}{e^{hc/kT\mathrm{\lambda <!--\; \lambda \; -->}}-<!--\; -\; -->1}$
where $\mathrm{\lambda <!--\; \lambda \; -->}$ is the wavelength and $T$ is the temperature. Taking the derivative of this, setting it equal to zero, and rearranging terms, one gets
$\frac{\mathrm{\alpha <!--\; \alpha \; -->}}{\mathrm{\lambda <!--\; \lambda \; -->}}=5(1-<!--\; -\; -->e^{-<!--\; -\; -->\mathrm{\alpha <!--\; \alpha \; -->}/\mathrm{\lambda <!--\; \lambda \; -->}})$
where $\mathrm{\alpha <!--\; \alpha \; -->}=hc/(kT)$. To solve this, Zettili goes on to say that we can write
$\frac{\mathrm{\alpha <!--\; \alpha \; -->}}{\mathrm{\lambda <!--\; \lambda \; -->}}=5-<!--\; -\; -->\mathrm{ϵ<!--\; ϵ\; -->}$
so that the previously mentioned equation becomes
$5-<!--\; -\; -->\mathrm{ϵ<!--\; ϵ\; -->}=5-<!--\; -\; -->5e^{-<!--\; -\; -->5+\mathrm{ϵ<!--\; ϵ\; -->}}$
Apparently, $\mathrm{ϵ<!--\; ϵ\; -->}\approx <!--\; \approx \; -->5e^{-<!--\; -\; -->5}$ is a good answer. My first question is why do we use $5-<!--\; -\; -->\mathrm{ϵ<!--\; ϵ\; -->}$ to rewrite the equation? What method is this? I've never seen that before and other resources just say that they got the values by solving numerically. My second question is why $\mathrm{ϵ<!--\; ϵ\; -->}\approx <!--\; \approx \; -->5e^{-<!--\; -\; -->5}$ when there is an extra factor of $e^{\mathrm{ϵ<!--\; ϵ\; -->}}$ in the equation? Why can we ignore that factor?

If I stood next to a piece of metal heated to a million degrees, but in a perfect vacuum, would I feel hot?
A friend of mine told me that if you were to stand beside plate of metal that is millions of degrees hot, inside a 100% vacuum, you would not feel its heat. Is this true? I understand the reasoning that there is no air, thus no convection, and unless you're touching it, there's no conduction either. I'm more so asking about thermal radiation emitted by it.

Energy of photon given temperautre
I have a question asking to find the energy of a photon emitted from mater of temperature $T$. If the question had asked frequency, it would have clearly been that $E_{photon}=h\cdot <!--\; \cdot \; -->f$. I do know that classically, the energy of an oscillator is $E=K_b\cdot <!--\; \cdot \; -->T$. Is it okay to use this formula for this question?

Deriving Planck's Constant from Wien's Displacement Law
So I'm reading an introductory book on Quantum Theory (David Park, 3rd ed.) and I am having trouble with the following question:
"According to Wien's displacement law, the wavelength ${\mathrm{\lambda <!--\; \lambda \; -->}}_m$ at which blackbody radiation at temperature T has its maximum intensity is given roughly by ${\mathrm{\lambda <!--\; \lambda \; -->}}_{m}T\simeq <!--\; \simeq \; -->$ 3 mm K. Assuming that the quantum energy at this temperature is of the order of kT where k is Boltzmann's constant, estimate the value of Planck's constant."
My attempt at a solution is as follows:
${\mathrm{\lambda <!--\; \lambda \; -->}}_{m}T\simeq <!--\; \simeq \; -->$ 3 mm K $,$$\mathrm{\lambda <!--\; \lambda \; -->}=\frac{c}{\mathrm{\nu <!--\; \nu \; -->}}$$\to <!--\; \to \; -->$$\frac{cT}{\mathrm{\nu <!--\; \nu \; -->}}=3\cdot <!--\; \cdot \; -->{10}^{-<!--\; -\; -->3}mK$$\cdot <!--\; \cdot \; -->(\frac{k}{c})\to <!--\; \to \; -->$$\frac{kT}{\mathrm{\nu <!--\; \nu \; -->}}=1.38\cdot <!--\; \cdot \; -->{10}^{-<!--\; -\; -->34}Js$
$(\frac{1}{2}m{v}^2{)}_{max}=k{T}_{max}=h\mathrm{\nu <!--\; \nu \; -->}-<!--\; -\; -->e\mathrm{\varphi <!--\; \varphi \; -->}$
$\frac{k{T}_{max}}{\mathrm{\nu <!--\; \nu \; -->}}+\frac{e\mathrm{\varphi <!--\; \varphi \; -->}}{\mathrm{\nu <!--\; \nu \; -->}}=h$
$1.38\cdot <!--\; \cdot \; -->{10}^{-<!--\; -\; -->34}+\frac{e\mathrm{\varphi <!--\; \varphi \; -->}}{\mathrm{\nu <!--\; \nu \; -->}}=h$
At this point I get stuck, the orders of magnitude and units seem to be right but I'm not sure what to invoke/what is given in the question that can solve/eliminate this last term to get h, or if I'm even on the right track.

How do quasars produce light?
Wikipedia states that the radiation of quasars is partially nonthermal (i.e. not blackbody).
Well, what percentage of the total radiation isn't black body?
For some reason Wikipedia doesn't give a number for the temperature of the gas of the accretion disc of a quasar. I had to search for a while and tried many keywords on Google to get an answer of 80 million Kelvin. If someone knows better and/or can give me a range instead of a single number that would be nice.
Based on that number and an online calculator of wien's displacement law I got that quasars emit blackbody radiation mainly on the x-ray and gamma ray section of the spectrum. Is that true?

Inferring Properties of Stars from Masses and Radii
I have two questions related to inferring properties of stars from their masses and radii.
1.What properties of a star's spectrum could we deduce? In particular, do all stars emit like black bodies? If so, I'm guessing that one can calculate the star's effective temperature and use Wien's displacement law to find the peak of the black body spectrum.
2.How would the mass and radius influence Type 1 X-ray bursts?

Physics of an infrared thermometer
The thing about infrared thermometers that bugs me is how can you get the same temperature reading regardless of the distance to the object. Shouldn't there be a difference when measuring from two different standing points since energy flux density decreases with $\frac{1}{distanc{e}^2}$ and infrared thermometers work by focusing IR light on a thermopile, which then results in decreased (when measuring from further away) absorbed energy and therefore lower temperature and finally lower voltage across thermopile. Is there something I am getting wrong about this, or do IR thermometers make use of some other physics law like Wien's displacement law, by somehow measuring ${\mathrm{\lambda <!--\; \lambda \; -->}}_{peak}$ to determine the temperature?

Showing Wien's Displacement Law from Wien's Law
Does anyone know how I would show that $\mathrm{\lambda <!--\; \lambda \; -->}\ast <!--\; \ast \; -->T$ is constant, using only Wien's Law? That $\mathrm{\rho <!--\; \rho \; -->}(\mathrm{\lambda <!--\; \lambda \; -->},T)=1/{\mathrm{\lambda <!--\; \lambda \; -->}}^5\ast <!--\; \ast \; -->f(\mathrm{\lambda <!--\; \lambda \; -->}T)$ I differentiated, but all I could get was $\mathrm{\lambda <!--\; \lambda \; -->}T=5f(\mathrm{\lambda <!--\; \lambda \; -->}T)/f^{\prime }(\mathrm{\lambda <!--\; \lambda \; -->}T)$, which I don't think means it's necessarily a constant.