Tag Info

New answers tagged

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 ...


0

Generically, if I have a particle which has potential energy $\phi(x,y,z)$, then the force on that particle will be given by ${\vec F} = - {\vec \nabla}\phi$. So, generically, the motion of particles will "try" to minimze the potential energy. In particular, the only points where the particle will not move will be those points where $\nabla \phi = 0$, or, ...


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

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 ...


4

Gravitational redshift is extremely small for stars, it has no significant impact on cosmic redshift measurements of galaxies. The gravitational redshift of light emitted by a star is of the form $$ z = \frac{1}{\sqrt{1 - \frac{GM}{Rc^2}}} - 1 \approx \frac{GM}{2Rc^2}, $$ where $R$ is (approximately) the radius of the star. For the Sun, this is of the order ...


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 ...


0

This question is unanswerable using current loop quantum gravity, which does not yet have a completely consistent way to couple gravity to matter.


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 ...


9

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, ...


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 ...


0

Wonderful question! And it relies on one central concept: virtual particles. Particles that do not necessarily exist, but explain (or would explain) phenomena wonderfully. By assuming that the particles (electrons, protons, etc) are exchanging photons, we can predict their behavior with great accuracy. Imagine that two protons are people, and they are ...


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 ...


-1

I haven't done this in a long time, but my understanding from Feynman's QED is that the speed of a photon is unknown (Heisenberg) - photons traveling in a vacuum are around the speed of light +-, but at any instant the speed differs due to uncertainty. The photons going faster and slower than light speed cancel in the same manner that photons bouncing off a ...


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 ...


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 ...


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 ...


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 ...


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 ...


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 ...


0

I'd like to give a different point of view than the one provided in Qmechanic's answer. The reason is not because of gauge invariance. Indeed, gauge invariance is just a statement of redundancy and it can't possibly have any physical consequences. My answer is instead the following: the photon is massless because it has just 2 degrees of freedom while ...


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 ...


1

Binocular vision has already been discussed, but it left out an important aspect. A single eye is sensitive to distance. The shape of the lens changes to focus on near/far objects. The reason this is needed is that our pupil has finite size and cannot be modeled as a pinhole. The same physics is going on here as in a lens of a camera focusing on an ...


3

To have depth perception two eyes are needed. Our two eyes are some distance apart which causes the photons from an object to arrive at slightly different angles. The brain then reconstructs the depth field from these differences. Similarly, we can figure out how far nearby stars are by using images made by a telescope at two different times of the year, ...


2

There is no edge of the universe. The standard model for cosmology is based on the FLRW metric. It is what happens when you assume the universe is homogeneous and apply general relativity. In this model the universe could have a finite volume that's not growing too fast, where paths just loop back on themselves. A photon in such a universe would keep ...


0

Will an unhindered (un-scattered) photon go to the edge of the universe? To add to the answer of @RedAct : If you are thinking of the three dimensional part of the universe we observe now, it is an instantaneous universe, i.e. time t has one value. In this sense we are at the edge of the expanding universe, the right cutoff at 13.8 billion years in the ...


1

The answer probably depends on how that question is interpreted. The universe is expanding. The ultimate fate of the universe isn't known for sure, but the growing consensus among cosmologists is that the universe will probably continue to expand forever. If that's the case, then a photon that leaves the Earth now will never catch up to what is currently ...


1

Lets attempt some answers: Both can happen, a quantum transtion can be associated with a photon exchange or a photon exchange can be associated with a quantum transition (this is just 2 ways to state the conservation of energy in these cases) Photons do not have mass but they do have momentum. There are some approaches in physics which associate a virtual ...


0

I have personally measured many photons, certainly more than one. Not once did a photon tell me that it was the same one that I measured an hour earlier. :-) That time stands still for a photon is not true, by the way. Photons simply make a full rotation from the time-like to the space-like coordinate axis. To a photon "when" becomes "where". You can see ...


3

You are using the concept of time in a mixture between Newtonian ideas and Relativistic. It is true, that for the photon time is slowed to a standstill, however, for an observer who has mass, time still flows, one can measure simultaneously two separate photons, that are not causally connected and know these are two separate entities since having measured ...


1

Yours is a subtle question with a rather subtle answer. From the way you ask the question, you seem to be thinking of the photon as a little billiard ball. It is not. It is an excitation of a quantum field, which is described very differently from a "particle" in the classical sense. Even so, the short answer, in some very subtle ways, is that indeed there ...


4

It's not necessarily true that most of the photons that strike a wall will be absorbed and turned into heat. The whitest white paints can have a light reflectance value of up to about 85%. There isn't a "wavelength corresponding to white color". An ideal white surface reflects as much as possible of all wavelengths in the visible spectrum. That sounds ...


0

1) No, substances almost never completely absorb photons. Otherwise you could not see them. In case a substance would absorb all photons (which is quite hard to achive intentionally) it would be pitch black even if you shine arbitrarily strong light on it (-> black-body). 2) It will be reflected back and forth, but only a finite amount of time. This is ...


1

Almost always, when photons hit matter or interact with it, they are not reflected in the way a billiard ball bounces off a billiard table edge. Rather, they are absorbed, the absorber rises into a metastable state, and then a new photon is emitted on the decay of the metastable state. Sometimes, though, when photons undergo an interaction with a lone ...


3

Almost but not quite. Qualitatively the spectrum is the same with the $1/n^2$ spacing, but the scale of the spectrum is set by the reduced mass $\mu$, $$\mu = \frac{1}{\frac{1}{m_l}+ \frac{1}{m_p}}$$ where $m_p$ is the proton mass and $m_l$ is the lepton (muon or electron) mass. Since $m_p \approx 2000 m_e$, it is not a large error to take $\mu = m_e$ for an ...


1

Photon photon interaction, which is what a collision will mean, is practically non existent as it is higher order and in the context of this question, light sources, non existent in reality. If we reach gamma ray energies then particles will be produced but this has nothing to do with this question. There will be interference patterns as whenever coherent ...


2

I think what you say is correct. By saying it should be unitary, he means it should be time-reversible. It is indeed time-reversible if before reaching the detector 2 it is in the same state as the initial state. It starts off with the superposition, and then only vertical polarization in the top part and horizontal in the lower part of the arms, which ...



Top 50 recent answers are included