# Tag Info

9

But what we never seem to see is why the electron and positron move the way that they do. Saying "they move like they do because of the force on them" doesn't explain anything at all. It's a non-answer. The equation of motion for charge particle (electron,positron) in magnetic field is  m\frac{d}{dt}\left(\frac{\mathbf ...

9

After the hypothetical split, 2 photons with the same energy would be propagating at an angle ok with momentum conservation. Then there would be a rest frame where the angle is 180 degrees. Now if you stay in this restframe and go back in time before the split, your single photon would be at rest. However, that is not possible: According to relativity, speed ...

8

A photon is an elementary particle. As much elementary and as much particle as the electron . A single elementary particle has a fixed mass and cannot emit another particle without violating energy conservation, because its mass is fixed. In the center of mass of a massive elementary particle, electron, there is no energy for an emission , for a ...

5

Virtual particles are not real. A virtual particle is essentially defined by being associated to a propagator. It is, formally, nothing more than such a propagator. The idea of "virtual particle" doesn't even exist before you notice that you can draw pretty Feynman diagrams as a succinct representation of the way QFT amplitudes are calculated. This is ...

3

There are two factors at play here. The Lorentz force which causes the paths to bend with a radius proportional to the particles velocity and with a sense that dependent on both the particles charge and the direction of the particles velocity. In high energy (compared to $m_e$ events) such as the one pictured, the particles are nearly co-linear at the ...

2

The violation of gauge invariance by this term is the "only" reason why it's never written down – as long as we define the word "only" to include all other reasons that may be shown to be "physically equivalent" to gauge symmetry. Gauge symmetry is extremely important and its violation would make a similar theory inconsistent, especially at the quantum ...

2

In classical electrodynamics, the process of how much light refracts, passing through the glass, and how much light reflects, is determined by the Huygens-Fresnel principle. This principle, named after Christiaan Huygens and Augustin-Jean Fresnel, is a method of analyzing the wave propagation patterns of light, especially in diffraction and refraction. It ...

2

Photons come with chirality, so you should consider angular momentum conservation as well. For $1\gamma \to 2\gamma$ scattering, this will not be possible. (I'm assuming production of collinear photons only; it's obvious when two are not collinear, energy and momentum conservation will be violated)

1

But I want to know, if an incoming photon is unable to excite an electron, then why not all the photons pass through glass? i.e. photons should not reflect off glass, all the photons should pass through glass. While I don't know the exact details of reflection off glass (it is related to solid state physics from what I recall, with plasmon and such ...

1

If you want to think of a free on-shell particle as a mode that propagates like a soliton far from interactions then you can think of it as a thing that interacts with the vacuum so as to make more of itself. But really you should stick to asking questions the formalism is designed to answer. The vacuum is not a thing like a house that provides space for ...

1

If you had a laser you wouldn't see it unless it was aimed at your eye (ouch). Or if there is dust or such around for it to scatter off of. And scattering is the key. If you want to see something then it either has to get to your eye or it needs to deflect something towards your eye. If you have a beam of electrons you could try to get something to ...

1

Photons and electrons are elementary particles and their behavior is predicted by quantum mechanical equations. Quantum mechanics predicts probabilities for a reaction to happen. Photon photon interactions have very very small probabilities of happening. Thus photons pass through each other for all measurable purposes at low energies ( light and below ...

1

However, I'm having trouble understanding this formulation when applying it to a free field. A free field is just a cavity with infinitely large boundaries, right? That would seem to imply that for each frequency, there is only one harmonic oscillator. A free field is a field that obeys homogeneous Maxwell equations (with charge and current density ...

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