# Tag Info

7

I would tell them to re-read and understand that paper, and know that few spectroscopists would disagree with it. The point is that far too many people use the word "photon" without knowing what a photon really is or under what context the word can be used. For the vast majority of applications a semi-classial conception of the radiation field is adequate. ...

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 ...

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 ...

4

The issue here, I believe, is not existence of photons, but the fact that people may choose terminology and concepts they find appealing. The word photon has been coined long ago for an idea that is quite far from the current views on light and the meaning of the word has been evolving many decades. Its current use in textbooks and papers is quite broad ...

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 ...

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 : ...

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, ...

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 ...

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 ...

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 ...

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 ...

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

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 ...

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

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 ...

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

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 $<E>=<ψ|E|ψ>$.

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 ...

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 ...

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 ...

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

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.

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

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

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

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 ...

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 ...

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 ...

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 ...

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