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This is a great question. I feel it necessary to point out the level of study and understanding that go behind asking this question. Well done! Here's the way I understand it. You analysis is flawless; in a radiation dominated universe, $a\propto\sqrt t$. That said, it is not correct to interpret this as the photons exerting some sort of pressure that ...


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The impact of beam hardening makes most of these attempts at "exact" (closed form) calculations very difficult. High energy radiation scatters - this takes it out of the "forward" beam but it continues "broadly in the same direction" and may even scatter back in the forward direction. You need to keep this is mind when you analyze the problem. Typically you ...


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According to classical electromagnetic theory... I think this is the key assumption you are building your question on, and for atoms/nucleons/electrons/everything smaller this assumption just doesn't hold true. All these objects have to be described with quantum mechanics, so there is no trajectory of a localized charge or something alike - all you're ...


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Among other reasons stated in the answers and the commments, here is another very important reason for why we care about blackbody radiation:


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Blackbody radiation is characteristic of every object in thermodynamic equilibrium and black bodies at constant uniform temperature. At any temperature objects emit thermal radiation. EM radiation is emitted because inside the object, due to thermal motion of particles charged particles/dipoles start to oscillate, electromagnetic radiation is emitted ...


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Leaving aside the fact that I don't see why one would make that assumption (what are its mathematical consequences in the derivation?), The assumption of cavity is an assumption of physical arrangement for which it is justified to use boundary conditions that are necessary supplement to the equations in order to arrive at the desired results ...


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Carbon 14 has a mass of 14, not 12.


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Yes, infrared radiation which is invisible to human eyes but still radiates heat. This is how they make thermal cameras, they are using your body's infrared radiation which is detected by the camera and forms an image based on visible light.


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Three different ways (at least). 1) Use a detector, like a photodiode or photomultiplier, which responds to UV by producing electrons when hit by photons even if those photons are not visible, 2) Use a detector, like a bolometer, which measures the temperature rise produced when UV energy is absorbed, or 3) Coat a visible-only detector with a phosphor ...


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It's common to measure spectra in the non-visible region and UV or infra-red spectrometers are available off the shelf. Google for many, many manufacturers of the kit. Also see the Wikipedia articles on UV spectroscopy and infrared spectroscopy. Typically UV and IR spectrometers will use a diffraction grating to disperse the light and a photomultiplier tube ...


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How did Lyman discover his series in hydrogen atom? He was directed into spectroscopy by his advisor. At the time the equipment was pretty poor for spectral measurements and much of his time he spent trying to get good spectral wavelength measurements. Part of the measurement error ended up coining the term "Lyman ghosts" in the spectral lines due to ...


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This is a coupled linear differential equation. Just write it in a matrix form as $\dot{X} = M X$, where $X$ is a vector formed by $a$ and $b$, and $M$ is a matrix. The solution is similar to the one dimensional case, but you will get the exponential of the matrix $M$ in the solution. You can get the components of $\exp (M)$ by using the Taylor expansion of ...


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A good analogue of how there does not exist a single point in our universe that is the unique point where the Big Bang happened is the expanding balloon . I will stretch it further to include the photon background: Suppose that we have a solid ball of elastic very hot material, and somehow we can blow air at its center. The ballon forms and the surface ...


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You are assuming the Big Bang happened at a point, so the CMB is a shell of radiation expanding outwards from that point. However the Big Bang happened everywhere so every point in the universe is a source of the CMB. The CMB radiation we are detecting today comes from regions of the universe that were about 13.8 billion light years away at the moment the ...


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This question is closely related to the question "If photon energies are continuous and atomic energy levels are discrete, how can atoms absorb photons?". While this technically isn't an exact duplicate of the link above, a similar conceptual explanation applies, namely that for complicated strongly-coupled quantum systems (like a hot chunk of metal), the ...


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Anytime a particle slows down or speeds up (non-relativistic). The particle loses some energy as light. http://en.wikipedia.org/wiki/Larmor_formula This could be in X-ray, UV, IR, Visible ect... This could be for any reason: particle-particle deflections, magnetic field and particle interactions, curved non uniform motion, electric field particle ...


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I do not trust Dr Bussards scaling. He did not have enough data to make those scaling claims. The University of Sydney (Gummersall, 2013) scaled some factors in their simulations: A. Current in the rings (Amps or AmpTurns) B. Size of the rings (Meters) C. Energy of electrons (KeV) But, they were looking at how many electrons were trapped - not fusion ...


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I'm by no means an expert on this, but since I found this question interesting I searched in the infinite vastness of the internet for a hint and came up with three links: Number 1. (search for the word "celebrated") Number 2. eq. 1.10 on p.6 (warning: it's a.) a pdf and b.) it's in german) Number 3. eq. 2.5 on p.7 (same warning) Have I interpreted ...


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This is a link with the conclusions from the experiment and which observations lead to which conclusions. The Conclusion When Rutherford mathematically investigated the results he proposed a model that explained the results that Geiger and Marsden obtained. The fact that the vast majority of the alpha particles got straight through led Rutherford to ...


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When they show a decay with a certian amount of energy, this energy is net of the masses of the particles. So you get Co = $\beta$ + Ni + 0.31 MeV, the energy is attached to the ejected beta particle.


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Gamma emission is emission of a photon upon a nucleus transitioning from an excited state to a lower or ground state of the same nucleus. The number of neutrons and protons in the nucleus is exactly the same before and after the gamma photon is emitted. Beta decay results from a nucleus having too few or too many neutrons relative to the number of protons ...



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