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The radiation of a body is given by the Stefan-Boltzmann equation: $$E= \epsilon \sigma T^4$$ Which means that if you can measure the radiated power $E$, the temperature is $$T=\sqrt[4]{\frac{E}{\epsilon \sigma}}$$ If the emissivity is lower than you think, it follows that you see less emission that you would expect for a given temperature and you would ...

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First of all, you can look at the translation of his paper here. As was already noted Planck firstly discovered the correct blackbody radiation formula by simple interpolation of $R=-\Bigl(\frac{\partial^2 S}{\partial U^2}\Bigr)^{-1}$ where $S$ is entropy and $U$ - mean energy of the oscillator in the bath. He knew that $R=\alpha U$ gives Wien law for ...

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This is from David Bohm's book, Quantum Theory might be a help to you. Not sure it's the complete derivation, but I think it is. also not sure if it's the orginal method P. used to find it, or Bohm's version (probably the latter) (not sure about copyright either, but the book is 70 odd years old :) any problem with that, i will need to delete it

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Whether you see the incoming background radiation as blueshifted depends on your relative motion compared to the large scale Hubble flow. Even in the absence of a black hole you can accelerate in a direction and see blue shift in the forward direction. And that's fundamentally what is happening in the black hole, to stay at a fixed distance from the black ...

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This is simply a statement of conservation of energy. Consider an imaginary surface between plates $k$ and $k+1$. The flux through this surface is: $$J=\sigma T_k^4-\sigma T_{k+1}^4,$$ where the second term is negative because the energy flows in the opposite direction. If the system is in a state of equilibrium, then the temperatures are not changing ...

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The CIE standard colorimetric tables give $\bar x(\lambda)$, $\bar y(\lambda)$ and $\bar z(\lambda)$ as pure numbers, normalized so that $\int\bar x(\lambda)d\lambda$ is the same for the three components. In the end, it boils down to what units you expect your tristimulus signals $X$, $Y$ and $Z$ to have. For dimensionless $\bar x(\lambda)$, $$X=\int ... 17 You are looking at this incorrectly. Pale skin allows the UV to penetrate more deeply than dark skin (that has the melanin in the dead skin cells). Since dark skin individuals absorb the UV in the dead skin layer, it make no difference if it causes DNA damage. 12 Dark skin absorbs UV better than lighter skin. More specifically, melanin absorbs most of the UV radiation so that your skin cells don't have to. 1 Do you mean the flame specifically? The flames are easier to explain as they are a more concrete thing. Flames are gases that are hot enough to emit light in the visible spectrum via thermal radiation. Some proportion of those gases are in the plasma state, but the parts that aren't ionised are also hot enough to emit light in the visible spectrum. 6 I'll let the far greater minds than me at Minute Physics give the bulk of the answer. Turns out that the way you perceive fire is the result of numerous chemical and physical processes. I'll discuss and expand on just two major ones that were mentioned in the video I linked: Atomic Transitions- The distinct colors that you see in a flame are the result ... 3 Fire is a chemical process, usually a self-sustaining oxydation. Things "burn" at elevated temperatures. That burning (oxydation) is exothermic. A sustained fire happens when burning something produces enough heat to cause oxidation of nearby fuel to occur, thereby causing it to burn and produce more heat, which causes more fuel to burn, etc. Most fire ... 3 Fire is a chemical reaction which results in the release of energy in the form of light and heat. 4 Fire is a chemical reaction ,some times hot enough to turn the gas around it into plasma which you can basically think of as a hot ionized gas. You can show that fire can form plasma by putting negatively charged plate on the left of a flame and a positively charged plate on the right of the flame. It will split the flame because the plasma is ionized. 2 Planck's law of black-body radiation can be stated in many different ways, depending on whether one is interested in the spectral energy density per volume or per area. It can also be expressed in terms of radiation wavelength or frequency. The energy of a photon is$$ \epsilon = h\nu = \frac{hc}{\lambda} I will not derive Planck's law here. It can be ...

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In practice, no. In theory, also no. The Universe is filled with photons with an energy distribution corresponding to 2.73 K. Every cm$^3$ of space holds around 400-500 of them. That means that if you place your "stable body" in an ever-so-isolated box, the box itself will never come below 2.73 K, and neither will the body inside. It will asymptotically go ...

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It´s impossible to drop a body temperature to absolute 0K. You must notice that, at the same time the body is radiating energy from its own temperature, it is also receiving temperature from other sources (regardless the distance of the source) like distant stars.

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