# Tag Info

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Photons are blue-shifted when attracted by gravity (I mean - moving towards a mass, not moving at right angles to the gravitational field like in an orbit). They can't go faster, but their energy goes up.

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You don't feel acceleration. When onboard the ISS, you are accelerating towards the earth (down) due to gravity: if you didn't, you would just fly away from the planet. Because you and the ISS are accelerating exactly the same way, you don't feel a thing. You don't feel a force if it's accelerating you: you feel pressure caused by opposing forces. Here ...

6

Ignore the diagram. It's complex because it includes all the various bits of electronics required for the experiment and that confuses the issue. The experiment is just a variant of Mössbauer spectroscopy. The experiment uses a $^{57}$Fe source that emits a gamma ray with an energy of 14.4 keV. Since this energy matches the spacing in the energy levels of ...

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Thus, say there was a radio wave with a wavelength of 1 meter. Could this radio wave then dodge around object B and still hit object A, assuming object B is smaller than 1 meter, say a basketball? Waves of large wave lengths can indeed 'wash around' an object that is sufficiently smaller than the wave lengths, a phenomenon called diffraction. Have a ...

3

Most likely, your garage door opener operates at a frequency of 315 MHz. Multiplying by Planck's constant, that means each photon has energy of about $2\times 10^{-25}$ joules. Most likely, your garage door opener operates at about $1/10$ of a watt (or less, per comments below). So each second, it emits 1/10 of a joules of energy. That's $(2/10) ... 3 That the numbers of photons far, far outweighs the number of neutrinos can also be determined as follows. One of the main contributors of energy production in the Sun's core is the proton-proton chain reaction, which produces one neutrino and about$26.7\:\mathrm{MeV}$of energy per helium atom produced. This colossal amount of energy is gradually ... 3 To talk about acceleration in space is a little bit dangerous without exact definition. One has to separate free fall and acceleration from an impulse. Imagine, you are inside the ISS during an orbit correction. The impulse from the rocket engine you could feel, you get some weight, and and this is an acceleration. In all other time you are weightless and ... 3 A system can absorb a photon if the energy of the photon matches an excitation in the system. So the hydrogen atom can absorb a photon if its energy matches one of the frequencies in the hydrogen spectral series. A proton is a composite object and it does have a spectral series. However the excited states of the proton involve rearrangements of the energy ... 3 Does a proton have a "bandgap"? If yes, what happens when a photon is absorbed by a proton? For single protons, as in a plasma , there exists Compton scattering . The photon transfers part of its energy to the proton and scatters off at a lower energy/frequency, the proton taking up the energy-momentum balance. This is a continuous spectrum, from very ... 2 For two perfect laser sources without noise (infinite narrow linewidth) and of same amplitude, you could achieve this when both sources have the same frequency and same polarization. You need to control the length of one pathway to control the relative phase to$\phi_0=\pi$:$E=E_1+E_2=E_0 (sin(\omega t)+sin(\omega t+\phi_0))$The problem in general will ... 2 Actually A.Zee's book on "QFT in a nutshell" has a very nice explanation on this on chapter I.5, I will breifly sketch it (this is a very rough skeptch), $$Z=\int DA e^{iS(A)} =e^{iW(J)}$$ where W(J) is given by, $$W(J)=-1/2 \int \int d^4xd^yJ(x)D(x-y)J(y)$$ where D(x-y) is the photon propogator and J(x) and J(y) refer to two lumps of matter Plugging in ... 2 As dmckee says in a comment, the proof is ridiculously simple. Suppose we work in the centre of momentum frame so the total momentum is zero. The particle comes in with some momentum$p$and the antiparticle comes in with the opposite momentum$-p$, and the two annihilate. Suppose the annihilation produced a single photon. The momentum of a photon is:$$p ... 2 The concept of a photon comes from quantum field theory, but the radiation from your remote control is well in the classical regime, so we need to figure out how to go from the quantum field theory to the classical field theory. Now, the way we do that is by forming so called minimum uncertainty states. There's a Heisenberg uncertainty principle for fields ... 2 (1) it's the energy in the sense that the photon oscillates at a certain frequency. (2) i'm not sure you can physically explain a light wave. More light is just more photons, more energy is photons with higher frequencies. (3) when it comes to the particle nature of light the photon has a frequency. It frequently oscillates through positive and negative ... 2 Firstly, entanglement isn't magic. And it is next to meaningless to just say something is entangled. Entangled just means not factorizable. But is the lack of factorizability from the spatial degrees of freedom? From the polarization degrees of freedom? Is it super close to factorizable? Is it maximally entangled? Just saying it is entangled isn't really a ... 2 If you want to understand the photoelectric effect and energy of light you have to be very very careful to keep the terminology correct. The energy of a electromagnetic wave (1 photon) is completely determined by its frequency. The intensity of the electromagnetic wave(s) is how many photons per second are hitting on some area of surface. You can't ... 2 Consider this nomenclature for simplicity: Photon 1 is the photon emitted by the spaceship when it was at rest with respect to the outside observer and Photon 2 is the photon emitted by the spaceship when it is moving at a uniform speed with respect to the outside observer. Since the mechanism through which the photon is emitted from the spaceship is not ... 2 Yes. It is much easier to think of this in terms of conservation of momentum: Because light (and electromagnetic radiation in general) has momentum, you will have to gain momentum in the opposite direction to conserve total momentum --- just like if you were to throw the flashlight. It is difficult to think of this in terms of forces because we tend to ... 2 Yes, the requirement for at least two photons is because a single photon would violate conservation of momentum. See my answer to Particle anti-particle annihilation and photon production for a (very simple!) proof of this. Annihilation can produce more than three photons. In fact the decay of ortho-positronium to two photons is forbidden, and it (mostly) ... 1 In optics, you rather speak of intensity of light which is the energy per time and per surface. The brighter the light the more intensity it has. The energy transported by a light beam per minute is proportional to the squared amplitude of a wave or the number of photons times their energy.$E\propto |E_0|^2$for waves and$E=n \cdot E_{photon}=n \cdot h ...

1

(sorry, I couldn't write this in the comment section) Have you met the postulates of quantum mechanics? Here is a summary of them http://vergil.chemistry.gatech.edu/notes/quantrev/node20.html Postulate 3 says if an observable has associate a (hermitian) operator, the only values we would observe for one photon the spin-angular momentum are the eigenvalues ...

1

Use the equation E=hc/λ. In case you don't know, h is Planck's constant and λ is the wavelength in nano meters. convert electron volts to joules (E) using 1 electron volt = 1.60217662 × 10^-19 joules. Planck's constant is 6.62607004 × 10^-34 m2 kg / s

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If your photon has not enough energy to excite the electron then it will just not be absorbed and will pass by, and if you have an electron with an excess energy, it will be absorbed and a photon with the excess energy will be automaticaly emitted and the electron will jump to an excited state. So yeah in your case, you might have a photon with $0.1 \space ... 1 Unfortunately PMTs don't do that, at least not cleanly. One of the crucial performance characteristics for photon counting is the peak-to-valley ratio of the PMT, which you get by measuring a pulse-height distribution. PMTs produce dark noise, which means they indicate "photons" that aren't there and they can't cleanly tell the difference between one and two ... 1 The work function of the photocathode is such that it can eject no more than one electron for each photon. Well, it's possible that very energetic photons can eject two, but that process is rare, and can be suppressed with filters. So that's the sense that it's said that PMTs detect single photons. But only 30% to 70% of photons in the proper energy ... 1 This isn't really how it works. A photon doesn't interact with a single electron, it interacts with the entire molecule. Suppose you take the example of ozone photolysis to$O_2$and an oxygen atom. We can do a calculation for ozone and come up with a series of molecular orbitals, then put two electrons in each orbital. So far so good. But if you remove an ... 1 Your assumption of a perfect mirror is something that will make this difficult to answer (also assuming you can close the door fast enough to keep some amount of the light in before it escapes). If the mirror is perfect and no light escapes, then it should contain the light, bouncing around forever. In reality, a small portion of the light is absorbed upon ... 1 so the Hamiltonian has a degenerate eigenspace of dimension two? Yes. If that's correct, then all states$|x\rangle,|y\rangle, |R\rangle,|L\rangle$are stationary (do not change in time), Yes. so even in the right-hand circular polarization state$|R\rangle$there is no rotation of the polarization. Yes. So the polarization must come ... 1 Suppose$a\overline{a}\rightarrow\gamma$is possible for a particle$a$with a definite nonzero mass,$p_a^2=m^2>0$("mostly-minus" metric,$c=1$). Conservation of momentum implies$p_\gamma=p_a+p_{\overline{a}}\implies p_\gamma^2=m_\gamma^2=0=p_a^2+p_{\overline{a}}^2+2p_a p_{\overline{a}}=2m^2+2 p_a \cdot p_\overline{a}$However, the scalar product on ... 1 Your attempt is right since$1=\frac{h}{\lambda{P}}$. Then you say that you substitute for$P=\frac{h}{\lambda}$then how you get next relation with$P$included? You must replace$P$by$\frac{h}{\lambda}$and your equation become$1=\frac{h\lambda}{h\lambda}$and this gives$1=1\$ . So, this relation have no physical significance. But mathematicaly this ...

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