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26

For visible stars, the answer is no. In Newtonian physics, a star that would pull something travelling at light speed back to itself, i.e. a star for which the escape velocity were $c$, was called a dark star and seems to have been first postulated by the Rev. John Mitchell in a paper to the Royal Society in London in 1783. The great Simon Pierre de Laplace ...


12

Yes the effect is real, potentially at least, but no it's not measurable. As an aside, the redshift of light by its gravitational interaction with (homogeneous and isotropic) dust is exactly what the FLRW metric predicts, but this clearly isn't what the question means. The gravitational field of a beam of light is calculated in the paper On the ...


11

In quantum mechanics a particle can be treated as a wave and a wave can be treated as a particle. This is the notorious wave particle duality. I won't go into this any further here because it's been discussed to death in lots of previous questions. Search this site for wave particle duality if you're interested in finding out more. Anyhow, assuming I ...


11

(I henceforth assume $c= \hbar=1$.) It is forbidden by the four-momentum conservation law. Put yourself in the centre of mass reference frame of the couple of massive particles (electron and positron). There $P_{e\overline{e}} = (2E,\vec{0})$ with $E\geq m_e>0$. Just because four momentum is conserved, this four-momentum must be the same as the one of the ...


10

Gluons and photons are similar in the sense that they are both massless gauge bosons. They do, however, correspond to different gauge symmetries: photons arise due to $\mathrm{U}(1)$ symmetry, while gluons follow from $\mathrm{SU}(3)$. This leads to a different number of particles: there is only one photon, while there are eight different gluons, ...


8

I) At the perturbative/diagrammatic level of photon self-energy/vacuum-polarization $\Pi^{\mu\nu}$ , the photon masslessness is protected by the Ward identity, which in turn is a consequence of - you guessed it - gauge invariance. For the explanation in the setting of QED, see e.g. Ref. 1. Fig. 1: A one-loop contribution to the photon ...


7

Relativistic mass is obsolete. See Why is there a controversy on whether mass increases with speed? . Therefore this is not a question that modern physicists would consider of interest. Furthermore, the usual motivation given for using the relativistic mass convention is that it lets you use Newton's second law without modification, but Newton's second law ...


6

From the perspective of fundamental quantum field theory, gluons and photons are quite similar. Both of them are gauge bosons, meaning that their existence is required by a mathematical mechanism called local gauge invariance. However, as particles, there isn't any particular connection between them. For instance, there's no reason they both have to be ...


4

Comment to the question (v4): It seems relevant to mention that there in principle could be a difference between the universal speed limit constant $c$ (which is usually casually referred to as the speed of light in vacuum), and the actual speed of light in vacuum if the photon has a rest mass, at least from an experimental point of view. Of course, no ...


3

So what about antimatter - since charges are opposite, perhaps it also clumps together to form anti-gravity superpositions. As Red Act says in a comment, gravity is too weak to be important on the scale of individual particles. However charge does group antiparticles together. For example an anti-proton and a positron will form an antihydrogen atom. In ...


3

Yes, the device is called a Faraday mirror and it consists of a normal mirror following a Faraday rotator. The latter is a magneto-optical device that rotates the state of polarization of light passing through it in a non-reciprocal manner. The most well-known application of Faraday rotators is to provide optical isolation. The Faraday effect is wavelength ...


2

Could it have been: $\large l_p = 1.61619926^{-35}m \small \quad \text{ (Planck Length)}$ At any rate, to create a black hole, you simply need enough energy density in a single area that its escape velocity (the speed at which the sums of $E_k$ and $E_p$ are $0$) is larger than the speed of light. As you should know, the photon has no mass. However, its ...


2

If photons transmit the electromagnetic force, which is observable: the photon or the electron? Do we ever directly measure a photon, or do we only measure it's effect on electrons. For example suppose I shine a laser at a wall Let us clear up that photons ( and also electrons) are quantum mechanical elementary particles, and classical electromagnetic ...


1

If you insist on thinking of photons as waves (which is fine as long as you ignore absorption, though you should really think of it as a disturbance in an electromagnetic field), you can more or less think of all of their amplitudes as being equal, and this is why your premise doesn't make sense. More precisely, the amplitude of a single photon isn't ...


1

Some of the other answers have suggested that the amplitude of the photon's wavefunction is well defined, and that it has the same value for any two photons of the same energy. Whatever else we may say, this can't possibly be right. The amplitude of an electromagnetic wave is defined either by its electric field or by its magnetic field. (In a sensible ...


1

Clumps of just anti-matter will have the same gravity field around them as clumps of matter. There was an experiment at either Fermilab or SLAC in the 1970s or early 1980s where the falling of a beam of anti-protons was measured. I was trying to look up details on this a couple years ago, and didn't find it. But I know I read about it long ago. Bottom ...



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