# Tag Info

64

In short, the answer is: because gluons behave in a way that makes them useless for this purpose. To understand why, let's back up a little and look at how photons are useful, and then see how gluons behave differently. We (animals pretty broadly) evolved to see photons because they allow us to move around in and respond to our environment more ...

17

Very reasonable question. I will try to answer it in an intuitive way. If you have a scattering medium, photons are reflected in random directions; but when you have a refractive medium, something else happens. The photon is not absorbed and re-emitted: instead, the photon interacts with the electrons in the medium, and since these electrons are somewhat ...

12

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 ... 11 It is not a good idea to see a Feynman diagram as some sort of collision process really happening. The diagram is just a term in the perturbative expansion of a quantum mechanical transition amplitude (in other words, a nice "graphical" way to represent a bunch of integrals). The only actual observed objects are two incoming photons with a certain energy, ... 10 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 ... 9 Virtual particles are not real. They come, as I've said in many answers on this site, from a naive interpretation of Feynman diagrams which should not be taken as an actual, exact description of how the physics works. The actual description of an interaction in the quantum field theory is more complicated than "photons are exchanged". In particular, ... 9 Would you dig a ditch with a surgeons scalpel? Yes, quantum mechanics ultimately underlies all physical observations but the mathematical expressions for large dimensions with respect to \hbar become cumbersome and are replaced by the simplest ones for the appropriate study. Thermodynamics, for the study of bulk matter, blends smoothly with quantum ... 9 The fact that the theory is not gauge invariant implies that all degrees of freedom of A_\mu must have physical meaning: This is not the theory of photons where only transverse degrees of freedom make sense. This way you must tackle some non-trivial issue like the negative norm associated with temporal modes. This could be avoided by adding a mass to ... 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 ... 6 This comes from classical electrodynamics, there is no need to go to Standard model theory or quantum electrodynamics for this. The simple answer is that electric potentials, like electric fields, are just a way of characterizing the way charged particles interact with each other. So, charged objects create voltage analogous to the way that they create ... 6 This might not be quite the answer you are looking for, but one useful way to think about it is: the accelerating electron emits photons because nothing forbids it from doing so. By definition, because an electron has electric charge it is coupled to the electromagnetic field, and is able to produce excitations in this field which we can call photons. This ... 6 First of all, note that different authors disagree what should be the Coulomb potential V in d spatial^1 dimensions. Here we will assume that it satisfies Gauss's law, i.e.$$\tag{1} V(r)~\propto~\left\{\begin{array}{rcl} r^{2-d} &\text{for}& d~\neq~ 2, \\ \ln(r)&\text{for}& d~=~2. \end{array}\right.$$Here we will only discuss the ... 6 Huygens's Wave Theory is what you call a first order scalar diffraction theory of light. So what does it describe and what does it fail to describe? First order means that electromagnetic effects like induced currents in surfaces etc. are ignored. These can be described by solving Maxwell's equations for the same system instead of working with the wave ... 5 I believe it would be incorrect to term this as "photon-splitting". What is really happening is that a photon is being annihilated and subsequently two new photons are created by means of a non-linear optical process. Such a process that is routinely harnessed in quantum optics laboratories is spontaneous parametric downconversion (SPDC). The reason why ... 5 There are very strict limits on the mass of the photon already, so it would only affect our understanding of physics on the largest scales. The cosmologists would have some hard thinking to do, for instance. However, contrary to a comment, it would not affect relativity beyond requiring us to reconsider the usual name for c: not "the speed of light" but ... 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 ... 5 Feynman diagrams are most definitely not a representation of what's going on between the particles. Feynman diagrams are simply a tool to help you remember formulas: if you want to calculate the probability that two electrons will scatter off each other in so-and-so angle, you draw all possible diagrams with two incoming electrons and two outgoing electrons ... 4 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 ... 4 It fails when the photon number is small. Since the electromagnetic field can never be zero because of the third law of thermodynamics this automatically couples the temperature, the effective volume and the photon number to each other. As a result it is experimentally impossible to do experiments with single photons at low frequencies because we can't ... 4 .Your title asks about the electric field. The content is about the electromagnetic waves, two different entities. Electromagnetic waves emerge from an innumerable number of single photons. As one cannot have water waves with just a few molecules but need of the order of 10^{23} (avogadros number) one cannot measure electromagnetic waves if the photons ... 4 So how is it possible to have a quantized outcome from a symmetric continuous event? Easily. So easily that I'll describe the easiest example to me. Which is to describe what happens when a Stern-Gerlach device interacts with a spin 1/2 particle. You could have a particle with any spin whatsoever, but no matter what single particle state you pick it ... 3 The propagator as defined by the time-ordered 2-point VEV ⟨0|T[ψ^(x)ψ^(x+y)]|0⟩ is clearly not invariant under the local gauge transformation$$A_{\mu}(x)\to A_{\mu}(x)+\partial_\mu\Lambda(x)\ \ \ ; \ \ \ \psi(x) \to \exp(ie\Lambda(x)\psi(x)$$since the VEV acquires an extra phase factor \exp(ie\Lambda(y)) under the transformation. Schwinger, among many ... 3 From a classical point of view, if you look at the field lines created by charge at rest or at constant velocity you'll see straignt lines. Now if the charge changes its speed (i.e accelerating), a "ripple" will appear and propagate along those field lines. Check the animation on this page :http://www.tapir.caltech.edu/~teviet/Waves/empulse.html. You'll ... 3 Photons come with chirality, so you should consider angular momentum conservation as well. For 1\gamma \to 2\gamma scattering, this will not be possible. (I'm assuming production of collinear photons only; it's obvious when two are not collinear, energy and momentum conservation will be violated) 3 Given you've read only QED, this is a highly astute question. Conservation laws in the quantum world work a little differently from classical conservation grounded on Noether's theorem (there is a kind of quantum analogue in the Ward-Takahashi identity). If a quantum entity has state |\psi\rangle, then conserved quantities are measurement means defined ... 3 I know that photons are quantized, they are not continuous. Photons are not quantised, nor are they continuous. They are the charge carriers of the electromagnetic field as arising in quantum field theory. An accelerated charge generates an electromagnetic field whose carriers are, in turn, the photons whose energy might be quantised. So how is it ... 3 Consider for the shake of simplicity a free neutral scalar field \phi. Passing to the second quantization picture, it is a operator valued distribution$$C_0^\infty(M;\mathbb R) \ni f \mapsto \phi(f)$$where M is Minkowski spacetime and \phi(f) is a densely defined symmetric operator on the Hilbert space$$F_+(\cal H) = \mathbb C \oplus \cal H ...

3

If you had an electron in a Coloumb potential, completely isolated from everything else, then indeed the atom would be in a stationary state and there would be no transitions. In fact, since, the lifetime of a state is inversely proportional to the width of the transition line, so the fact that the electron is in a stationary state and cannot decay also ...

3

You do not need to invoke QM at all. Measurements are often normally (Gaussian like) distributed due to random errors unavoidable on each individual measurement. Actually, your result $d_1-d_2= 0.381 \pm 0.524$ is consistent with the difference being zero (because the error is larger than the obtained average value)

3

Think I got it: One uses the anti-commutation relationship: $$\gamma^\nu \epsilon_\nu \gamma^\mu k'_\mu = 2\epsilon \cdot p - \gamma^\mu k'_\mu \gamma^\nu \epsilon_\nu$$ And then uses the fact that the spinors satisfy the Dirac equation, i.e: $$\bar{u} \gamma^\mu k'_\mu \approx 0$$ So that we are only left with $2\epsilon \cdot k'$ in the numerator. ...

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