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It's tempting to think of the light as a little ball (the photon), and since little balls have a definite position the little ball has to be in a superposition of a state where it goes through one slit and a state where it goes through the other. However this is not a good description of what actually happens. The light is not a photon, and it's not a wave ...

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Absolutely. If Compton scattering occurred in first order in $e$, the only contributing diagram would be the obvious one. Say we're in a frame with the electron initially at rest and an incoming photon in the $z$ direction. Then the electron 4-momentum is $$p^\mu_{\text{in}} = (m,0,0,0)$$ while the photon 4-momentum is $$k^\mu_{\text{in}} = ... 9 The claim that during the experiment they let the detector on but did not stored the data so it showed waves, only when they stored the data it showed as particle. is inaccurate and inconsistent with quantum mechanics. In a double-slit experiment, any device that can even in principle provide which-way information will destroy the interference ... 8 Even if the laser had perfectly reflecting, i.e. lossless, mirrors at either end of the cavity, and both ends were sealed so no light could escape it would still require a continual power input. That's because excited atoms/molecules can decay by mechanisms that don't involve a photon e.g. collisional de-excitation. The lost energy goes into heating up the ... 5 The E=mc^2 formula only applies to an object at rest, and light is never at rest. You want to use the more general formula: E^2={m_0}^2c^4+p^2c^2 Then you can set the mass to zero. E=pc What this says is that light has momentum, which is related to its energy. 5 If we assume you are a sphere in space, at the same distance from the sun as Earth, then we can calculate the heat absorbed - and we can calculate how hot you need to be so heat in = heat out (assuming uniform surface temperature, and radiative heat transfer only). For this, we need the Stefan-Boltzmann expression for total emission at a given temperature: ... 4 Have a look at my answer to Slit screen and wave-particle duality because this covers a lot of topics relevant to your question. You're correct that if we imagine the photon as a little ball then if the arms of the interferometer are different lengths the two "halves" of the little ball cannot arrive at the detector at the same time. But this is not how the ... 4 You are correct in one thing: if an atom in an isotropic medium spontaneously emits a photon, it can do so in any direction at all, and the overall emission will be evenly spread over the unit sphere. However, lasers work using stimulated emission, which is slightly different: if an atom is excited, you can induce it to emit its energy by shining an initial ... 2 A laser is just a thin slice from the spectrum of light. Is it more efficient compared to the visible spectrum of light? It depends on the frequency of the laser and how efficiently the solar panel can turn light of that frequency into electrical energy. If a solar panel would operate better/best with light of a certain frequency, using a laser with that ... 2 It would certainly require a material that allows electron release from energies lower than those of the visible spectrum. The energy of a wave is given by E=hf where h is the planck constant (6.63 x 10^-34) and f is the frequency. The wavelengths of IR light range from 0.001 m to 750 x 10^-9 m. (Hyperphysics.com, infrared) Using this knowledge you can get ... 2 When one says "photon" one is in the quantum mechanical frame. Quantum mechanics does not follow the rules of classical mechanics if one tries to consider the photon one classical entity, like a bag of energy flowing. The photon is a point like elementary particle in the standard model, it has no extent and when it hits the detector it registers at a ... 2 Suppose you are using a CCD or a photographic plate to record your image. The interaction with the light occurs when the detector absorbs a photon, and this happens at a point. So the image is built up from a collection of points - one for each photon that is detected. In everyday life, e.g. taking pictures with the CCD in your phone, the intensity of the ... 2 Using a process called interference, we can find wavelength, because the way that waves interfere is reliant of wavelength. Interference is based off of two key principles of waves: they are made up of peaks and troughs. When troughs overlap, they go lower. When peaks overlap, thy go higher. When a peak meets a trough, they cancel. Of course, the positions ... 2 When you have only one electron then \Delta S=0 makes intuitive sense: you can change the angular momentum l of the atom by changing it's internal structure (by pushing the electron in "another orbit" if you will), while you certainly can't change the internal structure of the electron to change s. Would it be possible to change s then you could ... 2 Q1: For photons of energies much less gamma rays, the quantum mechanical photon-photon interaction is negligible. This is consistent with the classical electrodynamic description where the principle of superposition holds (electromagnetic waves pass through each other unchanged, as well as through electric/magnetic fields). Q2: in reality, charge is defined ... 2 Theoretically yes, the laser principle does not consume any material. There is a light source that excites the electrons in the material to higher levels, they deexcite to some intermediate one, here the avalanche of photons appears producing the laser light and leaving the electrons in the ground state. And you can repeat the process without a loss. 2 The maxwellian wave is an emergent phenomenon from a great multitude of photons with the frequency of the maxwellian wave. This is explained in this blog entry by Lubos Motl. I will give you my experimentalist's interpretation of this: A photon as a quantum mechanical entity has a wavefunction. This wavefunction is a solution of a form of Maxwell's ... 2 We know that solar cells generate electricity by utilizing the energy of the photon, This is an every day language, electricity. It means things electrical in general every day language. but how does it generate electricity forever? What is generated when the photons hit any material, is heat, and the sun's energy is at maximum 1300Watts per ... 2 What they actually measured was not particle behavior. It was just a quantized energy transfer to the probing electrons. That corresponds to the absorption of individual photons, but it doesn't mean the Surface Plasmon Polariton (SPP) field was acting as a particle. It just interacted locally with the electron, as it must. Typically particle-like behavior ... 1 To understand how light is affected by gravity, it helps to think of light as energy. So let's ask a basic question: when it come to light, what is energy? By the Planck-Einstein relation, we know$$E = h \nu$$where \nu is the frequency of the light and h is Planck's constant. So when we talk about the energy of light, keep that in mind. Also note ... 1 You start with$$ E' = \frac{E-up}{\sqrt{1-u^2/c^2}} = \\ \frac{E-up}{\sqrt{(1-u/c)(1+u/c)}}. $$You know that p = \frac{E}{c}, so$$ \frac{E-uE/c}{\sqrt{(1-u/c)(1+u/c)}} = \\ \frac{E(1-u/c)}{\sqrt{(1-u/c)(1+u/c)}} = \\ \frac{E\sqrt{1-u/c}}{\sqrt{1+u/c}} = \frac{E\sqrt{c-u}}{\sqrt{c+u}}, $$now using E = hf,$$ f' = f\sqrt{\frac{c-u}{c+u}}. ...

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It is the energy of the incident photon. At low Energy (Long wavelengths as in Radio waves and Microwaves, visible light) Rayleigh scattering would dominant, in which the scattered photon would have the same wavelength as the incident photon. At higher energy (shorter wavelengths as in Ultraviolet spectrum), the electron would absorb the energy of the ...

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theoretically if its components never wore out then yes. however in practice things do wear out eventually and so no it could not be done in the same way that a perpetual motion machine can work in theory but not in practice.

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This is because instead of $$\dfrac{1}{2}mv^2$$ or $$E = mc^2$$ the energy of light is given by $$E = hf$$ Where h is a number called Planck's constant and f is frequency (sometimes v is used) Here is an example, as requested: Imagine red light with $620. nm$ wavelength. The frequency of this light is $0.483$ x $10^{15}Hz$ This makes the energy of a ...

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There is an equation that helps a lot understanding this issue: Fermi's golden rule $$W_{i\rightarrow f}=\frac{2\pi}{\hbar} \left|\left<f\right|H'\left| i\right> \right|^2 \rho$$ It describes the transition rates from one state to another. $\rho$ is the so called Density of States (DOS) of final states. This system has only two states: The initial ...

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the photons will travel at the speed of light relative to both the moving light source and an object in another frame of reference. time dilation will bridges the gap so that both may co exist.

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Is it correct to think that the speed of light does not depend on the speed of light source because photons have no mass In a certain sense, yes. The Lorentz transformations guarantee that the speed $c$ is invariant; an object with speed $c$ in one inertial reference frame (IRF) has speed $c$ in all IRFs. But an object with invariant mass $m$ cannot ...

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So, I wonder why is it usually said that photons do not interact, or hardly interact? As far as we know photons do not directly interact with each other. Mathematically, this is manifest in the fact that the equations of motion for electromagnetism are linear: given two sources A and B of electromagnetic radiation, the resulting EM field is precisely ...

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I think a simple view is this: The solar cell must have a PN junction, which is a junction between p-type (many holes, no electrons) and n-type (many electrons, no holes) materials. Right where they meet there is actually a "depletion width" within which there is hardly any of either. Within this region, as photons come in they generate electron-hole pairs, ...

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What I'm asking is, has someone measured, that at one moment of time the peak of magnetic component is in the same distance from source as the peak at the electric field. That should be not easy because this peaks are moving with c. You're asking the wrong question. Scientists don't measure the differences in the locations of the peaks of two signals. ...

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