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No he's right ill leave a light turned off for days and it still glows in pitch black and this is noticed at several places iv lived/owned


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This is the spectrum of black body radiation, classical, and quantum calculations. As the temperature decreases, the peak of the black-body radiation curve moves to lower intensities and longer wavelengths. The black-body radiation graph is also compared with the classical model of Rayleigh and Jeans. So as you see the wavelengths are in the x axis ...


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Discreteness of energy reproduces experimental results in a satisfying manner. If you assume discreet spectrum you get to preform summation instead of integration. To be precise, you get to sum geometric series which converges. After that, you get the factor of integration of the form 1/(exp (hν/kT) − 1). This now is OK. If you dont have discreteness ...


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If you try to analyze the power radiated by a black body and assume a continuous spectrum of energy, you find that the black body must emit an infinite amount of energy. This is known as the ultraviolet catastrophe and was solved by Max Planck assuming that energy is emitted in discrete units of $hv$ and not in a continuous spectrum.


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Assuming no energy input, the lifetime of a black hole is related to its mass by: $$ T = \frac{5120\pi G^2}{\hbar c^4}M^3 $$ There is a nice summary of the derivation of the lifetime on this web site. I make the condition assuming no energy input because for large black holes the Hawking temperature is less than the temperature of the cosmic microwave ...


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You can pump heat from cold objects to hot objects if you pay some more energy (that's what your refrigerator is doing) and that doesn't violate second law of thermodynamics. You should note as you heat object, its thermal radiation will increase. Intensity (that is power per unit surface area) of thermal radiation is proportional to $T^4$ so when the ...


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Yes. There are a few ways this happens. Ultimately, it's an energy budget issue. There is less "mass" being radiated, and more "energy" being radiated. Of course, they're fundamentally related. Here are some of the ways this can happen: Loss of energy due to tidal dissipation Radiative Heat Transfer / Radiant heat loss Atmospheric particle escape ...


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All bodies with temperatures higher than 0Kelvin radiate away electromagnetic energy according to black body radiation. The power emitted goes according to the Stefan-Boltzmann law. The law states that the total energy radiated per unit surface area of a black body across all wavelengths per unit time (also known as the black-body radiant exitance or ...


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The spectrum of the black body radiation is given by Planck's law. The total amount of radiation is given by the Stefan-Boltzmann law. In principle there is no shortest wavelength, because the radiated intensity remains non-zero at arbitrarily small wavelengths. However, at the low wavelength end of the spectrum the radiated intensity falls exponentially ...


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The equation that you have is right, but actually a little more complicated than you need for this problem. A point source is infinitely small and you don't need to integrate over it . This problem can be handled without working through integrals formally. Consider a full sphere around the point (rather than the partial surface that you've got). In that ...


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The book's "derivation" is unusual. Usually, Plank's energy distribution is derived by considering the energy in a large cavity, which tells you about the energy that would be emitted through a small hole in the cavity. That energy has the blackbody spectrum. I think the book is trying to whitewash away a bunch of details by talking about "oscillators ...


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There is also, a gas of photons in the right side. It was trapped there when you assembled your box, and since you are assuming a perfectly zero emissivity, these photons must be perfectly reflected from all surfaces. That means they are blue shifted if the wall moves toward the right and red-shifted if the wall moves toward the left. Result: If you ...


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The Rayleigh-Jeans law states that the spectral radiance of a black body at temperature $T$ is given as $$ B(\nu,T) = \frac{2k_B T}{c^2}\nu^2$$ The total radiance is then $$ B_\text{tot}(T) = \int_0^\infty B(\nu,T)\mathrm{d}\nu$$ but the integral $\int_0^\infty x^2\mathrm{d}x$ does not converge. Worse, it is infinte, i.e. $\lim_{k\to\infty}\int_0^k ...


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Planck published several works on the theory of blackbody radiation based on different ideas, but generally the use of integer counting of energy he meant to be used for the energy of material oscillators. He did not believe the quantization applied to light itself - he assumed Maxwell's theory with its differential equations and derived his spectral ...


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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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