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48

Here is a quick & simple answer until professionals arrive. On the Standard Model, it is zero. This $< 1.10^{-18} \frac{\mathrm{eV}}{c^2}$ is an experimental upper limit (i.e. if it has a rest mass, because of physics beyond the Standard Model, it must be smaller as this value). This value is very small, compare to the estimated rest mass of the ...

47

It's the second one: the reason the speed $299792458\ \mathrm{m/s} = c$ is special is because it's the universal speed limit. Light always travels at the speed $c$, whatever that limit may be. The reason there is a "universal speed limit" at all has to do with the structure of spacetime. Even in a universe without light, that speed limit would still be ...

18

Even though there is a single photon in a volume of your choice the light is still a wave. An experiment was performed which proved this. In this experiment a Michelson interferometer was set up and the incident light is so weak that only one photon was in the whole setup at a time. A photographic plate was used to detect the interference pattern. Now just ...

16

By definition of spin $S$ it is a positive integer number or zero. Not to confuse with the spin projection possible values $S_z$, which may run from $-S$ to $S$.

15

We can't measure to infinite precision; so even if a particle had in fact zero mass we couldn't experimentally measure it to the infinite precision needed to justify this; which is why certain amount of judgement is called for, and that judgement is made in the context of a theoretical framework. The second point to make is that all particles with zero rest ...

12

Photons, as each massless particles, are characterized not by spin (which is defined as total angular momentum at rest, and mathematically corresponds to irreducible representation of the little group of representation), but by helicity $\lambda$ - the projection of total angular momentum on the direction of motion. Actually, the Casimir operator, which ...

11

Massless particles with spin do not have a "$S_z = 0$" state because they actually do not have spin like massive particles do. They have helicity, which is the value of the projection of the spin operator onto the momentum operator. The reason for this is the representation theory of the group of spacetime symmetry, the PoincarĂ© group. To understand this, ...

10

There are indeed massless particles. As of 2015 there were two known massless particles (both gauge bosons): the photon (carrier of electromagnetism) and the gluon (carrier of the strong force). It should be noted, however, that gluons are never observed as free particles, since they are confined within hadrons. Gravitons (if discovered) would be another ...

9

Is it possible that two photons move in parallel, on the same trajectory - having the same wavelength, but differ in phase? Photons are quantum mechanical entities/particles listed in the elementary particle table which is the basis of the standard model of particle physics. When measured individually, it just gives an (x,y,z,t) , by interacting with some ...

8

Even if nothing propagated at the speed $c$, it would still be a universal speed limit, and we could still measure it. In fact, it's not impossible that light has a (very tiny) mass in reality. If it does, that wouldn't change anything about special relativity. It would make teaching it even more of a nightmare than it already is, because we'd have to deal ...

6

Above all, speed of light is the speed of propagation of fields through space. While light may be slowed down when crossing matter, fields (electromagnetic fields, gravity) are always propagated at c. One of the consequences is the "speed limit for causality" mentioned by DavidZ and the speed limit for transmission of information.

6

Correction, a single photon does not have a circular polarization. It has spin +1 or -1 to the direction of its motion. Qualitatively Left and right handed circular polarization, and their associate angular momenta. The way the classical wave emerges from the quantum mechanical level of photons is given in this blog entry, and it needs quantum ...

5

Spin 1 just means that the spin in any direction can assume values out of {-1,0,1}. The 0 is only possible for massive particles, so the photon can have spin -1 or +1. That's like clockwise and anticlockwise circular polarization

4

In Quantum Field Theory the one particle states are defined as the states of an irreducible unitary representation of the Poincare Group. If this was not true, there would be states of a reducible representation that would not be connected by a Poincare transformation. These states are rather different particles. The Casimirs If we have an irreducible ...

4

The photon is an elementary particle in the standard model of particle physics. It does not have a wavelength. It is characterized in the table as a point particle with mass zero and spin one. Its energy is given by E=h*nu, where nu is the frequency of the classical electromagnetic wave which can be built up by photons of the same energy. This is where ...

4

Okay, so first, here's another good explanation for a photon's "mass": ...the word "mass" has been used in two different ways in physics. One was the way Einstein used it in $E=mc^2,$ where mass is really just the same thing as energy (E) but measured in different units. This is the same "m" that you multiply velocity by to find momentum (p), and thus is ...

4

With photons the physical attributes of the pattern are connected to the frequency/energy of the emitted light, which is a wave with wavelength in 400-700nm, right? Yes, with photons the double-slit pattern is not surprising, since photons are excitations of a specific mode of radiation and the modes obey the usual classical wave equation, which of course ...

3

The upper limit you mention reflects the hypothesis that photons in vacuum could have some tiny rest mass. But it seems to be more important that c is the velocity of massless particles such as photons in vacuum. However, there is no real vacuum in the universe: Not only that even in outer space you will always find some interstellar atoms. But also, the ...

3

The argument in your book is heuristic, that is it presents a reasonable justification for the de Broglie wavelength but it is not a proof. The correct expression for the de Broglie wavelength is: $$\lambda = \frac{h}{p} \tag{1}$$ For a massive particle in the non-relativistic limit the momentum is given by: $$p = mv$$ so we can write equation (1) ...

3

The photon is an elementary particle and as you correctly state it just has a spin and an energy, which has been identifiedas E=h*nu, where nu is the frequency of the classical electromagnetic wave which is built up from a huge number of same energy photons. As a quantum mechanical particle the photon has a wavefunction. This wave function is a solution of ...

3

Let us look at coherent states $$|\alpha\rangle~=~e^{-|\alpha|^2/2}e^{\alpha a^\dagger}|0\rangle$$ $$e^{-|\alpha|^2/2}\sum_{n=0}^\infty \frac{(\alpha)^n (a^\dagger)^n}{n!}|0\rangle$$ If you have a classical system it means overlap between states is small. We then look at over lap $\langle\alpha'|\alpha\rangle$ $$\langle\alpha'|\alpha\rangle~=~e^{-(|\... 3 In a rechargeable battery, two types of reversible chemical reactions take place: Oxidation reaction: in which a chemical, referred to as the reducing agent (Re) is oxidised by donating electrons:$$Re \to Re^{z+} + z e^-$$Reduction reaction: in which a chemical, referred to as the oxidising agent (Ox) is reduced by receiving electrons:$$Ox + z ...

3

Yeah, pretty much. When you increase the amplitude of a light wave, you are essentially just sending more photons of the same kind. The energy of each photon is $hf$ (though with some allowance for wavepackets, where each photon will come with as a probability distribution over a range of frequencies), so if you increase the energy flux you need to increase ...

2

According to quantum electrodynamics, the most accurately verified theory in physics, a photon is a single-particle excitation of the free quantum electromagnetic field. More formally, it is a state of the free electromagnetic field which is an eigenstate of the photon number operator with eigenvalue 1. The single-particle Hilbert space of the photon ...

2

Concerning massless particles, don't forget that the spacetime of their lightlike worldline is empty (= zero). That means that the point of emission and of absorption are adjacent in spacetime, even if the space interval between them measures billions of lightyears. By consequence, there is no problem for the transmission of particle characteristics for ...

2

Apples and oranges. The first is the photon density in a volume whose radiation field is in thermal equilibrium. The second is the rate at which photons pass a unit area regardless of the source of the radiation. They are both correct, but they describe different things. BTW, it's much better to tell us what is in a document instead of asking us to look ...

2

Yes, pair production is possible for quarks in the same way that for electrons or muons. But there are two caveats: QCD has a nasty property called confinement: quarks themselves can't exist isolated in nature, they must be inside a meson or baryon. In fact, they usually produce jets of such particles. Electrons are much lighter than quarks, and of course, ...

2

If one understand "carry particle" in the meaning that the particles move with the same velocity as the photons, than clearly no: Photons by itself could not carry particles. Photons move with the velocity of light $c$ and massive particles could not be accelerated to this velocity. This is both an observation and the basis of the Special and General ...

2

Good question :-) In the "ancient" times there was this old problem about the particle-wave duality. At the time, there was a Pilot wave theory, on which the particles and the waves are also different entities in some interaction. But it didn't live too long. Probably it failed some sophisticated experiments. On the current theory, there are fields, and ...

2

But how we can say that photon pass through the second filter if we know that after passing the first one it's polarized vertically? For me there is probability 0 to passing second filter if that filter isn't polarized vertically. Polarizing filters don't just discard photons, they change the polarization of photons. This is simply true, by experiment. If ...

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