# Tag Info

## New answers tagged photons

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 ...

0

Energy is most definitely conserved in the case of gravitation-ally red shifted (GRS) photons. The sun is 4.6 billion years old and has energy output equal to 3.8x10+26 watts. If 1% of the energy output is lost to GRS, an enoumous quantity of energy is missing. If the lost energy were simply hanging out in the surrounding gravitational field, it should be ...

1

Your starting place should be E.O. Kane "Theory of Photoelectric Emission from Semiconductors', Phys. Rev. 127(1) 131-141 (1962). This is done from a density of states perspective, considering both volume and surface states. The following paper in Phys Rev, G.W. Gobeli and F. G. Allen, 'Direct and Indirect Excitation Processes in Photoelectric Emission ...

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: $$Ox + z ... 0 The photoelectric effect (one core electron in, one photon in; one energetic free electron out, no photon out) is weaker at higher input photon energies because the output electron roughly needs to take all of the photon's momentum, but the photon does not give enough energy to do so. At lower photon energies, however, the momentum distribution of the input ... 4 With photons the physical attributes of the pattern are connected to the frequency/energy of the emitted light, which is a wave with wavelength in 400-700nm, right? Yes, with photons the double-slit pattern is not surprising, since photons are excitations of a specific mode of radiation and the modes obey the usual classical wave equation, which of course ... 1 It is understood how the process works. In a nutshell, the energy of a photon is converted to the mass of an electron and positron as is given by Einstein's E = mc^2. This is spelled out as follows: γ → e^− + e^+ There are two musts for this: The photon must have a higher energy than the sum of the rest mass energies of a positron and electron for ... 0 The energy momentum equation for a particle of mass m, energy E and momentum p is:$$ E^2=p^2c^2+m^2c^4 $$so if m=0 then E = pc. For a particle of mass m moving with velocity v we have the standard relations E=\gamma mc^2 and p=\gamma mv. The above equations imply v=𝑐. However, as that implies \gamma=\infty which implies 𝐸=\infty ... 0 There is no "point of view of the photons", you can't attach a frame of reference to them. Best if you imagine the photons as waves, and as these waves are propagating with c. The classical time dilatation, length contraction formulas (you know, everywhere the \frac{1}{\sqrt(1-\frac{v^2}{c^2})} in them) are defined only for macro-sized objects going ... 0 There is quite a bit of ambiguity in the question(s), so let me start by substituting electro-magnetic (EM) wave for "light." Then, the "universal speed limit," is the speed at which EM waves propagate in "space." The reason I use space (not vacuum), is because it is the characteristics of space (u_o, \epsilon_o) that determine the speed of propagation of ... -1 Interesting question. Electrons have electric charge and are therefore a source of electric field. An atom can be thought as a series of energy levels in which electrons can be in. When the electron transit between these levels, the energy will be emitted as electromagnetic radiation, since the fundamental interaction involved here is the electromagnetic ... 1 The atomic model developed starting from the light spectrum emitted by the hydrogen atom. It was known that hydrogen was one proton and one electron. The Balmer series or Balmer lines in atomic physics, is the designation of one of a set of six named series describing the spectral line emissions of the hydrogen atom. The Balmer series is calculated using ... 0 The photons generated in a laser optical cavity have frequencies imposed by the cavity resonances. Thus, when changing the refractive index in the cavity the cavity resonances change and the generated photons have different frequencies. Once generated the photons do not change frequency but the cavity refractive index change induces the change of the ... 1 I'm afraid that linear polarization is not as interesting an example as you may have hoped. First, the answer: in quantum optics, whether or not a quantum state exhibits linear polarization is independent of the photon count for that state. A single-photon state can be linearly polarized. Now, the explanation: in quantum electrodynamics (QED) it is ... 1 @Holger Fiedler Hi, Thanks for your comment on my answer here. Here I would like to say that I am an experimental physicist and working on the ultrashort coherent XUV radiation. I have first hand experience with the interference of XUV radiation using double slits. Your question is very genuine. Long ago I have thought over this question for quite a some ... 0 Everything you did is right.$$\frac{p}{4 \pi r^2 c hf}$$has dimensions \frac{1}{[Length]^3} You can see this by the following dimensional argument: p:\frac{[Energy]}{[Time]} \frac{1}{hf}:\frac{1}{[Energy]} \frac{1}{r^2 c}: \frac{[Time]}{[Length]^3} Combining all these together, you see that only the \frac{1}{[Length]^3} factor remains. ... 6 Correction, a single photon does not have a circular polarization. It has spin +1 or -1 to the direction of its motion. Qualitatively Left and right handed circular polarization, and their associate angular momenta. The way the classical wave emerges from the quantum mechanical level of photons is given in this blog entry, and it needs quantum ... 3 Let us look at coherent states$$ |\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^{-(|\...

2

Light never completely behaves as a particle. Light never completely behaves as a wave. As pointed out by hsinghal, the Michelson interferometer showed that, even at the "single photon" level, we still see wave behaviors. These behaviors are well modeled by quantum mechanics, which treats light as neither a pure wave nor a pure particle. As you "add ...

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 ...

2

I commented on the question but on advise, I am promoting this to an answer. A particle moving in a straight line can have non zero orbital angular momentum if the origin of the coordinate system doesn't lie on that straight line. There can also be helical modes through which light can have non zero orbital angular momentum. The first one is origin dependent,...

1

See the lookback time to redshift relation in https://en.m.wikipedia.org/wiki/Redshift You can ignore inflation if you get redshifts, temperatures and universe size (radius, scale) ratios between now and times in the past after inflation. For recombination the relations of 1+z to the scale ratios and temperature ratios are linear and direct . So for T(then)/...

0

Visualizations are not easy when one is dealing in the relativistic regime. One has to do the mathematics. In this case define angular momentum within special relativity. This has been done but needs to read through the maths. Intuition for special relativity rests on the equivalence of mass/energy. The photon has zero mass and angular momentum can have ...

1

Well, I'm not going to tell you my opinion, because that would be irrelevant to the actual science. But, what I can do is assess your premises and conclusions. What you should note first, and I'm unsure if you know this or not already, is that the four dimensions of spacetime are the three spatial dimensions and time. We can define a velocity through the ...

0

When you discuss polarization of matter due to the propagation of electromagnetic waves you usually consider low-energy photons, meaning photons of long wavelenghts. Given that the inter-lattice spacing $a$ is a few angströms, if you consider for instance visible light of $\lambda \sim$ a few $100$ nm, you have $\lambda >> a$. As the wavevector goes ...

-1

The least you may require of something to be able to call it a particle is that it can be established where it is when, that it interacts, however weakly, with the objects in its environment. Since according to relativity theory a massless particle must move at the speed of light and at that speed there passes no time at all, the particle -its state- is ...

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 ...

2

Take a look online at the "Dirac three polarizer experiment" for a more comprehensive answer.The key points are well explained by the following illustrations from "http://alienryderflex.com/polarizer/" keep in mind that these illustrations are based upon an initial horizontal polarizer, not a vertical one as stated above, but the concept is the same either ...

0

Anything moving at speed of light loses its reference system. This is the reason why your calculation yields strange results. Everything is multiplied by zero so that not only photons (which indeed have no length) but also long distances are reduced to zero.

1

I am only a layman, so don't take this answer seriously. This length contraction formula, and the whole Special Relativity in its original form, is for macro-sized, non-quantummechanical objects. Thus, the formulas work if you want to calculate the size of a spaceship nearing the speed of the light. And not if you want to calculate the size of photons with ...

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 ...

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 ...

-10

Articles published in Science and Nature say the speed of light is not constant: http://science.sciencemag.org/content/347/6224/857 "Spatially structured photons that travel in free space slower than the speed of light" Science 20 Feb 2015: Vol. 347, Issue 6224, pp. 857-860 http://www.nature.com/nature/journal/v406/n6793/full/406277a0.html Nature 406, ...

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.

0

The numerical value of $c$ does not have any fundamental significance. Rather it is the number we get based on the experimental fact (according to the number & unit system employed) . If some alien civilization ended with some different value of $c$ compared to us. Even that is not a problem. They will reach the conclusion that this is upper bound 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 ...

0

There are many ways a photon can interact with matter, but since the photon is a quantum mechanical entity, one has to define matter in the quantum regime. In dimensions commensurate to h_bar matter is composed of atoms in various combinations. One way of interacting with matter can be seen here: Atoms are modeled by electrons in orbitals around a ...

0

Consider the thing within second quantization form: the Hamiltonian contains term like $b^\dagger a_e+\text{h.c}$, which describes the annihilation of a excitation would produce an photon, and its conjugate procedure (which you cares more): absorbing a photon and becoming excited. This kind of Hamiltonian is derived, in principally, from ...

0

Particles are represented by fluctuations in fields in quantum field theory. So if you have a photon and an electron you will have a corresponding fluctuation in each field. The two fields interact with each other and so the fluctuation in the photon field can influence the fluctuation in the electron field and disappear. If you want an analogy think of ...

-1

I will try in a very general sense, and then you can use your imagination, because books may not answer this for you. Every interaction takes place via some kind of force. When we push a car, we transfer our energy into the car, but the energy first coverts to force, and then goes into the car. Same way, the photon must transform into a tiny force which ...

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) ...

1

The text "any material particle … by analogy with photons" shows that the author is talking about massless photons in contrast to mass particles. That means that his following formulations are for the best imprecise. $E = mc^2$ applies to mass which is rest energy, and it does not apply to other types of energy. Apparently, the author wanted to say: "...

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 ...

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 ...

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 ...

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 ...

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 ...

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 ...

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