Tag Info

New answers tagged

1

The wiki article you quote is succinct, the photon is an elementary particle in the table of elementary particles of the standard model of particle physics. It is a quantum mechanical entity which means it is described by a wavefunction whose square gives the probability of finding the photon at (x,y,z) at time t. The double slit experiment with a single ...


0

You can increase the intensity (Power/Area) in any one of three ways: increasing the rate of incident photons, decreasing the area, or increasing the energy of the photons. (In each case we take the other two as remaining constant.) Increasing the energy per photon will increase the energy of the emitted electrons, but not the rate of emission. ...


0

The intensity can only be increased by greater energy of the photon. When you increase the energy of the photon, the energy is absorbed into the electron, which consequently flies off with the photon's energy. The more energy the photon has, the more energy the electron has.


1

The general idea is that if you had empty space, with only visible white light moving through it, and you put a line of electrons a few thousand nanometers long, they would start accelerating up and down, maybe in a pattern like this: where each tick of the horizontal axis is something like hundreds of nanometers and the vertical axis is small. This ...


0

White light is composed of photons of varying energies. The photons themselves do not have to have any perpendicular spatial separation as your picture suggests. Rather, it is possible that the photons of different energies are coming along a single path, one after another.


5

If you consider your photon as a point object, it cannot bend its own path. It will always travel on the ridge it creates, speaking in terms of curvature of space. The other idea is possible. Two photons having a momentum, attract each other, trapping each other, like a positronium (typical example for this behavior). In the model of relativity this is ...


0

They have both frequency and wavelength (or wave vector $\mathbf{k}$), but the frequency and the wavelength are not related as in a real photon. Instead, they are independent integration variables of a Fourier integral representing a near field. For example, you can represent a "moving" Coulomb field $V(\mathbf{r}(t))$ as a Fourier integral over $\omega$ and ...


0

Virtual photons are not on the mass shell, that means that their mass can be different than zero in the calculations using the Feynman diagrams. They are not observable in any sense other than as the result of the calculations that need them. They appear because in an interaction quantum numbers are conserved, and these photons are used to carry the quantum ...


2

As I indicated in my comment on Rod's answer, some very powerful lasers do exist - and while their photons don't have much momentum, they do "pack a mean punch". In fact, laser ablation (where a laser beam produces significant local heating and material is ejected at high speed) may just produce the phenomenon needed. Let's take the example of the LLNL ...


2

If a photon [is] massless then it must have no energy This is not the case. One way to think of mass is as nothing more than a convenient name for rest energy. Photons are indeed massless and thus have zero rest energy. This is not an issue because according to special relativity, they do not come with a rest frame. Please note that assuming we denote ...


0

No, a photon does not need to have mass to be able to interact with matter. In fact it is its energy which is important in interactions. For the photoelectric effect the incoming photon must contain enough energy to displace the electrons on the metal of the photodiode. The question of what are photons made of is a pretty deep and difficult question, one ...


1

When people claim that a photon is massless, they mean that a photon has zero rest mass. In special relativity, the formula for the energy of a particle with mass $m$ possessing a momentum $p$ is $$ E = \sqrt{p^2c^2 + m^2c^4}$$ If we set $m = 0$ for a photon, we'll end up with $$E = pc$$ Here the momentum of a photon is described by quantum mechanics ...


0

Generally, there are only two things that can happen in the hypothetical experiment you describe. Either photons are emitted and absorbed in some place (detector, mirror, anything), or they are not emitted (and of course not absorbed). Both cases conserve energy. The latter possibility is the one that is somewhat counter-intuitive but it goes to the heart ...


1

I would recommend the first chapter of Scully and Zubairy, "Quantum Optics". Maxwell's equations are to the photon what the Dirac equation is to the electron. Indeed, one can write down Maxwell's equations in a form that is identical to a zero mass Dirac equation. In quantum optics the electric and magnetic field vectors become vector valued quantum ...


1

The probability of atomic excitation by photon absorption is given in terms of the "cross- section" for single photon absorption. This quantity shows a sharp peak if the photon energy and the atomic energy level difference match (the coupling between the atom and electromagnetic field causes the atomic states to broaden, the natural line width). Thus ...


6

Supposing we could shield ourselves with a perfectly nonabsorbing, reflective shield so that light would perfectly elastically bounce off us, thus preventing high power beams from incinerating us as Anna V's answer validly argues they would. Then the "fundamental" answer to your question is "because light has zero rest mass"; to explain further: the ...


5

A ray of light is a geometrical line describing the propagation of an electromagnetic wave. The electromagnetic wave is composed of zillions of photons each with a tiny momentum. The momentum is not large enough to sense an impact, it is pico newtons even for a laser beam. Lasers can have very high energy and momentum, but like knives, they cut soft tissue ...


1

Your sources maybe talking about the efficiency with which electromagnetic radiation is produced compared with the dissipation of energy in the antenna. All accelerating charges will produce electromagnetic waves. In your case an alternating current means that charges are being accelerated. In general, the electromagnetic fields produced will have a number ...


1

I believe it's the integral of intensity over time. $$\text{Lifetime intensity}=\int_0^\infty{I(t)dt}$$ "Intensity" in the instantaneous light output - the total light you get is that intensity from excitation until final decay. That said - it seems that the figure 2 in the paper you referenced in your link is using inconsistent labeling - because in the ...


0

No, unfortunately I believe that the identification with time is wrong. It is clearer to have a mode index for the creation operator $a_k^\dagger$ each of which refers to some spatial mode. For free space, you could choose plane $k$, but don't have to. Now each mode, if you choose it to be an eigenmode of the Hamiltonian, has a well-defined energy $\omega$ ...


2

Since this was not stated yet, I would just like to give my stance on it. All fundamental particles can be seen as excitations of fields. This is true for photons, electrons, neutrinos, etc. Do these fields need a medium in which they propagate? Not as far as we can tell. Everything we see and experience are excitations of these fields, a single one of ...


4

It may be useful to start this explanation from the origin of a light wave: an oscillating charge. Start with the idea that a stationary charge is surrounded by an electric field, then imagine wiggling that charge up and down. Now the field lines will turn to wiggles instead of straight lines. Those wiggling field lines are the electromagnetic waves we call ...


2

The electron on an atom gets excited to a higher level when some how the energy is transferred to the electron. But I can't understand it. The way we currently understand in physics this interaction is exactly like that: a photon transfers its energy to the atom and as a consequence one of the electrons goes to a corresponding exited state. And this can ...


1

Einstein once compared the photon with a famous person (sorry I forgot the name) who changed confession at young age and returned to its initial confession before he died: Light behaves as a photon at the starting point and at the end point, and it behaves like a wave during its travel. By the way, the light wave is not going up and down, it is not a ...


3

The light we see with our eyes is electromagnetic radiation, very well modeled by Maxwell's equations. Electromagnetic waves can be imagined as a self-propagating transverse oscillating wave of electric and magnetic fields. This 3D animation shows a plane linearly polarized wave propagating from left to right. Note that the electric and magnetic fields ...


0

Light waves are exactly a theoretical explanation of light radiation. Propagation of waves of electromagnetic fields is a good theory that works for low frequencies, but as Einstein showed (and was Nobel prized for) the photoelectric effect can only be explained if electromagnetic radiation is emitted as directed quanta of energy. I guess that experienced ...


0

As a practical matter we generally use these devices in cases where the photon arrivals are random. That is the mean rate may be known, but the actual arrivals are distributed according to a exponential time law rather than being periodic. In those case, counting statistics usually dominate the uncertainty (in low background detectors shot-noise can ...


3

It is certainly possible. It depends, as others have said, on the detector. But it also depends on the detection electronics, and the techniques used to do the measurement. Common sources of noise are shot noise, dark current noise, statistical fluctuations in the detection mechanism, and thermal noise in the detection electronics. Which of these are ...


2

Since your question is purely theoretical I assume you are referring to the intrinsic noise limits. This limit is called shot noise limit. Generally photon detection is complicated by the fact that photons arrive (are generated) in a random fashion. This is because the photon emission/absorption events occur independently and there is no communication ...


1

I read what is written in the physicsforums.com/threads/ that you indicate. So, you speak of the electron as of a QUANTUM particle. That means that it has a linear momentum of the same order of magnitude as that of the particles with which you want to test its movement. If the linear momentum of those particles were much smaller, s.t. the collision with ...


0

You can consider the photon as being a quantum mixture of the "more fundamental" V-photon and the neutral W to produce the photon and the Z instead. Electroweak mixing Keeping with standard model, particles are not fundimental at all — fields are. And elecromagnetism isn't a primary field, but arises out of the more abstract B and W fields which ...


2

One serious proposed theory that is now not widely held to be correct ("debunked") is the neutrino theory of light, wherein the photon was postulated to be a neutrino/anti-neutrino pair. This was taken seriously in the 1930s by people like Max Born and Ralph Kronig. See the Neutrino Theory of Light Wikipedia Page. One reason for beliefs along these lines is ...


2

In all the theories for which we have experimental verification the photon is a fundamental particle. However it's a different type of particle to the electrons that quarks that make up the matter around us, so you can't make a tuxedo from it. Particles are divided into fermions and bosons. Matter (e.g. tuxedos) is made up from fermions, while bosons create ...


3

If your electron is very fast you can track its trajectory in a ionization chamber. But if the electron is slow, its wave-like (quantum) character enters into the scene. About absorption and re-emission of a photon, please check if such things are possible, i.e. write the laws of conservation of energy and momentum as if these two bodies are billiard balls, ...


2

In a given volume, we can have light throughout, such that there is no space with no light in it (with the electron which is to be seen). Note that in this view you can hardly talk of photons as particles localized somewhere and somehow bouncing around. If you consider a given volume with a given amount of electromagnetic radiation in it you are ...


-2

If photon got no mass, you got first particle which got no dual nature. I don't know who teaching this photon got no mass. If you say it got no rest mass when travelling at speed of light than fine. Remember, if you reduced speed of light even in vacuum if you can, it will got rest mass tiny winy how much you reduced speed. How you can, nobody got idea. But ...


3

In the elementary particle framework, when an elementary particle meets another elementary particle, it is called scattering. Passing through would mean that the two particles continue on their way, momentum and energy unchanged. Not passing through is called interaction. This is the elementary particle table that is part of the standard model : ...


0

Einstein's introduce the concept of photon to describe the photoelectric effect and proposed that it has energy = hν and has mass approximately zero but has a momentum. So it being an amount of energy it obviously can pass through each other. In case of interference phenomena we may say that there is a redistribution of photon energy. Interference phenomena ...


0

The idea of a photon as a "particle" in the sense of our everyday use of the word, that is, as a small round hard thing, is useful for many purposes. But it is only a model, and like all models it resembles the true object in many ways, but does a poor job representing other aspects. A model airplane has no engine, and possibly no moving parts at all. ...


-1

The answer is clearly "yes" because if you accept that all electromagnetic radiation is made up of photons then the "size" of photons in the radio spectrum are quite large. If they did not pass through each other radio communications would be almost impossible. However, at high energies there can occur photon-photon interactions


2

Your single photon pulse wave function is an element of the first Fock layer (the zeroth is the vacuum layer) of the quantised Maxwell field Fock space. The electric field is still an operator but you can obtain its expectation value as $<E>=<ψ|E|ψ>$.


1

Please believe me that if the experiment were done not with two pairs of entangled photons, but with two pairs of entangled electrons, the results were similar, except that the wave-function of two entangled fermions is antisymmetrical instead of symmetrical. So, the light velocity plays no role here. I recommend you look at the equation (2) in the article ...


0

You didn't say how big is the mirror. Assume the photon is in the visible domain, and the mirrors are classical object, i.e. big. Such objects will remain insensitive to the linear momentum lost by the photon, because the recoil velocity of the mirror is practically zero. The reflection of the photon will be elastic, no energy imparted to the mirror, ...


0

This part is incorrect: When photons are reflected, they impart a small amount of their momentum/energy into the reflector and are red shifted. Momentum is always conserved. If in the frame of reference you choose, the photon accelerates and increases the momentum of the mirror, then the reflected photon will have less momentum (or less eneergy). In ...


1

I think the chosen answer is misleading and therefore will add my two cents of the euro. Photons are elementary particles, that is they are part of the building blocks of matter and energy in the standard model of particle physics. Elementary particles are quantum mechanical entities, not classical particles or waves. Quantum mechanical entities have ...


1

I believe the other current answer is wrong on several levels. Consider the cross section limit in Mie (spherical particle) scattering divided by the geometric cross section of the sphere. The limiting value is 2 for large spheres. What this means is that photons that don't 'hit' the sphere are scattered by the sphere. Since Mie scattering is easily ...


1

The question is correct in that the relativistic (or lorentz-invariant) time-energy uncertainty relation is a bit different (but still there is) For example here is a pre-print Lorentz-Invariant Time-Energy Uncertainty Relation for Relativistic Photon, arxiv Abstract: The time-energy uncertainty relation is discussed for a relativistic massless ...


1

The relation says that it would take about $\Delta t$ time to measure the energy with an error of order $\Delta E$. $\Delta t$ is not the photon time, it is the time in the observer's (laboratory) frame of reference.


0

Under most circumstances a bean splitter will split light. Calling light a photon makes it seem like a particle and it is but is also is a wave. A particle cannot be divided by a beam splitter but a wave can and all QM particles are also waves. Light will act like a wave if you test it to see if it is a wave and it will act like a particle if you test it ...


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 ...



Top 50 recent answers are included