New answers tagged

1

Yes, and it is unavoidable. Let's consider an intrinsic semiconductor for simplicity. If the semiconductor is at the absolute zero of temperature, then all electrons will be in the valence band. At any non-zero temperature there is a chance that some electrons will have been promoted by thermal agitation into the conduction band. These electrons will ...


12

It is a matter of definition of "same". Classically one can define "same" condition of particles by labels stuck on them. Light classically is a wave, and same needs a new definition. We apply the everyday definition by identifying the light beam with the source. The light leaving the sun is the same light arriving on earth. The light reflected from the ...


7

Photons are boson, so it follows the Bose-Einstein statistics which is only true if the particles are truly indistinguishable. If you can distinguish between two photons, then it will follow the classical Boltzmann statistics which is not what happen in experiments. That means photons with same properties are the same. Even in your situation with photon ...


7

Your question is based on the assumption that a photon is a fundamental object i.e. that photons are something we can point to and say here is photon 1, here is photon 2, and so on. The trouble is that quantum field theory particles are somewhat elusive objects. This is particularly so for particles like photons that are their own antiparticles because such ...


1

What are photons? Photons get emitted every time when a body has a temperature higher 0 Kelvin (the absolute zero temperature). All bodies, surrounding us (except black holes) at any time radiate. They emit radiation into the surrounding as well as the receive radiation from the surrounding. Max Planck was the physicist who found out that this radiation has ...


1

To be clear, Maxwell's equations are known as "Lorentz-invariant" equations, which means that they take the same form in every Lorentz-transformed frame of reference. Special relativity actually came about from studying Maxwell's (classical) equations without charges or currents. Then we get: $$\nabla \cdot \mathbf{E}=0$$ $$\nabla \cdot \mathbf{B}=0$$ ...


4

Classical electromagnetism is perfectly compatible with special relativity. In classical E&M, light is an electromagnetic wave and there is generally no useful formulation in terms of particles. The most widely used technique to combine quantum mechanics with special relativity is relativistic quantum field theory. The relativistic QFT that ...


0

The very word, photon, belongs to the quantum mechanical regime. It is one of the elementary particles in the standard model of particle physics. Elementary particles are described with quantum mechanical wave functions, which are complex function. The complex conjugate square gives the probability of finding the particle at (x,y,z,t). In the case of the ...


0

It has to do with the total energy or power of the EM wave you're interested in, as well as the frequency of the wave. As a simple example, a 3mW laser at 500nm wavelength will produce roughly 7.55*10^15 photons per second. From how large this number is, it's not difficult to see how light will usually be made up of an extremely large number of photons. For ...


3

In quantum mechanical domain these type of question does not have meaning. Every single photon is associated with a wave and vice versa. But to talk whether an electromagnetic wave contains a single photon or not is an ambiguous statement. When people say an electromagnetic wave necessarily contains many photons it only means that a incident beam of ...


0

I finally figured it out, at least for the simple case where the two atomic states have the same angular momentum: In this case the photons are always opposite in angular momentum (meaning they are either both left or both right handed). Regarding linear polarization, it depends on the parity of the system. In one case the linear polarizations are 100% ...


2

The color of the photon is related to its frequency $f$, which can be related to the energy of the photon by the expression $E = hf$, where $h$ is Planck's constant. Thus the different colors of the emitted photons describes their different energies. The next step is to determine why specific elements emit certain colors. This has to do with the different ...


0

No, but a mercury lamp does. Each mercury and calcium atom emits two indistinguishable photons out of a single atom, instead of one, like the rest of the atoms. The photons emitted by a single atom at the same time are entangled. Do not expect entangled photons out of atoms that emit a single photon at the time. Also the mercury lamp must not have a ...


1

It does't really make much sense to talk about a tree-level truncation (it helps for calculations, but that's it) or to take the first Feynman diagram as a true representation of reality. By the way, in your $e^- e^- \to e^- e^-$ example, the whole notion of spatial separation is ill-defined since this is a t-channel process. If going from virtual to real ...


0

It doesn't, and IMHO it shouldn't be a laser. Lasers produce light that is (1) coherent, and (2) of gaussian intensity distribution, both of which cause eye strain. The coherence leads to speckle; the gaussian spread means the focussed spot is also gaussian, and your eye keeps trying to improve the PSF. Another reason a laser is bad: it's a very small ...


0

The light from a typical laser emerges in an extremely thin beam with very little divergence. Another way of saying this is that the beam is highly "collimated". An ordinary laboratory helium-neon laser can be swept around the room and the red spot on the back wall seems about the same size at that on a nearby wall. The high degree of collimation ...


1

Semiconductor light emitters are made of such materials, which have quite large index of refraction. This makes it hard for light to exit the emitter — due to Fresnel equations and low index of refraction of air. In a laser the light mostly goes back and forth between two mirrors, and reflections only help the lasing. So the light either exits from a tiny ...


1

It's an issue of contrast; in the classical wave experiment there is plenty of data, and the contrast between the peaks and valleys is very clear; but when you are counting one-by-one the pixelation remains obvious. Pixelation can be reduced by (a) more gray levels in each pixel, and (b) more pixels per unit of area. You can simulate this by taking off ...


5

This is one of the first examples of energy levels for electrons within the atom! If we take the Bohr model, which imagines that electrons circle the nucleus on set orbits Each of these orbits has a corresponding energy. The electrons are more stable at lower energy levels, and thus, prefer to be there. When you provide energy to the electrons (in the ...


2

The spatial coherence is due to the fact that even for a single emitted photon it's the same wave that reach the 2 slits. I'm noot too sure what you mean by that. Spatial coherence has nothing to do with photons, it comes from the apparent size of the source as seen by the observer. Every source you might want to use in an interference experiment (a ...


2

As for your last question, a similar experiment has been done though it doesn't involve a double-slit. It's called the Michelson experiment, and using mirrors it tests the interference pattern created by light when the light-waves are combined with time-delayed versions of themselves. By changing the distance of one of the mirrors, the time-delay can be ...


0

You can get it as "clean" as you want - use sensors that only respond to a narrow wavelength (eg something so unlikely that it almost must be from your source...not IR for example!), set up your experiment so that the source will, statistically, emit a single photon every day or two to avoid the possibility that you're getting interference from the source ...


2

The key is in your words "to ... appear". I believe that it's a perceptual issue with how your brain processes the two kinds of images: a smooth rendering or a pixelated rendering. There is another possibility. In order to be sensitive to single photons, the detector is also going to be sensitive to very low-level noise. An image taken with a bright ...


17

Very reasonable question. I will try to answer it in an intuitive way. If you have a scattering medium, photons are reflected in random directions; but when you have a refractive medium, something else happens. The photon is not absorbed and re-emitted: instead, the photon interacts with the electrons in the medium, and since these electrons are somewhat ...


0

The part of your question about the non-visible light gives quite surprising (to me at least) results! : Compare the power from the sun with the cosmic microwave background ("CMB"), taking both as blackbody radiators: The temperatures are 2.725 and 5778 K respectively (wikipedia :)) The solid angle of the CMB is obviously $4\pi$. The diameter of the sun ...


1

Any two points in spacetime are linked by a four-vector that physicists conventionally write as $(x^0, x^1, x^2, x^3)$, where $x^0$ is normally the timelike dimension and the other three components are spatial. If we use the usual Cartesian coordinates in flat spacetime we'd generally write the four-vector as $(t, x, y, z)$. In this case suppose the light ...


-5

Why have our eyes not evolved to see “gluons”? Because gluons are virtual particles rather than real particles. See the Wikipedia article: "Although in the normal phase of QCD single gluons may not travel freely, it is predicted that there exist hadrons that are formed entirely of gluons — called glueballs. There are also conjectures about other ...


-2

'zero sum game' Thermalization spreads the absorbed photon energy so rapidly throughout the thousands of surrounding molecules so rapidly that there is virtually undetectable heating rate (increase in T2) and the energy returned to the GH molecule is tiny and will take a long time to statistically accumulate enough energy in any given GH molecule to re-emit ...


-1

A photon that escapes from black hole's neighborhood does work on the black hole. The photon causes the black hole not to be in the photon's gravity well after the photon has escaped. In other words the photon increases the potential energy of the black hole. The following paragraph may not be science, I just want to say something sane, as opposed to ...


0

This is just ordinary potential energy from first semester physics -- when the photon is close to the black hole, it's deep in the potential well. As it goes away from the black hole, it picks up gravitational potential energy, so therefore, it must lose kinetic energy. For a photon, the kinetic energy is given by the Planck formula $E = hf$, so the photon ...


0

Since due gravitational time dilation something takes forever to fall inside a black hole from the perspective of an outside observer, there is nothing in it yet that could come out. The light which is emitted from outside the horizon does reach an outside observer, but since it hasn't yet fallen in, it technically doesn't escape from inside the black ...


0

To put it simply, there is an evolutionary advantage to be able to see the objects around you using photons, but there would be no particular survival advantage to be able to see gluons, even if they weren't essentially confined to the nucleus.


0

There are several broadening mechanisms, and you have to know how they "add" together. Since a Voigt profile is the convolution of a Gaussian and Lorentzian profile, you rightly calculated both widths rather than just the overall width of the Voigt peak. For the Lorentzian portion, the width is the arithmetic sum of the individual widths: $ \Delta ...


1

By Wigner's general procedure of representing the little group, the $\theta(\Lambda,p)$ is the angle of rotation associated to the massless little group element $L(\Lambda,p)\in\mathrm{SE}(2) = \mathrm{SO}(2)\ltimes\mathbb{R}^2$ fulfilling $$ L(\Lambda,p) = l^{-1}(\Lambda k)\Lambda l(k)$$ where $l(k)$ is the Lorentz transformation carrying the null vector ...


2

A wave plate is a passive component, and can be modelled as a unitary operator on the quantum state. The state is a superposition of left and right circular polarized photons, and the operator gradually alters the relative phases. The total distance then determines the final polarization state.


2

Short answer: because the Sun emits photons, not gluons. Having a long range sense is vital for finding food and recognizing predators. Seeing light and forming an image of our surroundings is one of the three long range senses we have (the others are hearing and smelling). Gluons are extremely short range; they don't even exist as naked particles. How ...


64

In short, the answer is: because gluons behave in a way that makes them useless for this purpose. To understand why, let's back up a little and look at how photons are useful, and then see how gluons behave differently. We (animals pretty broadly) evolved to see photons because they allow us to move around in and respond to our environment more ...


0

Newton's 1st law states, as you rightly say, that a force-free body maintains its state of motion. (This also holds in relativity.) Newton's 2nd law states, that if there is a net force on the body, the body will accelerate in the direction of that force, $\vec{F}=m\vec{a}$. (This will be altered a bit in relativity, but for simplicity, let's stick with ...


1

Science never answers "why" questions, so in a strict sense there is no such explanation, but one can try to triangulate where we stand, at the moment. In classical physics space, time and the existence of massive bodies are inexplicable pre-physical facts. Inertia then becomes an observed property of massive bodies that allows to differentiate them by ...


-1

Answer:1 Mass is just a form of potential energy of a gravity wave of relativistic bent time space and light is a gravitational wave of bent space, when it travels at the velocity of C, where time relativistically slows down. At that velocity (with time dilation) it makes a gravity wave into the photon. because it is traveling relativistic-ally it is slowing ...


0

Other answers paraphrase it well in technical terms. It might be easier to see if you remember that when two particles interact they must do so in a way so that the momentum, energy, spin, etc. are conserved. After the interaction the two particles still remain in a superposition state but if you measure one of them after an interaction you can find out ...


1

Okay, so I am taking this question to mean what is the lowest-energy photon that can be individually detected. This is certainly an interesting technological question. I can't give an authoritative answer, but the lowest energy detectors I am familiar with is at the CMB microwave background energy of ~ $3Kk_B$, which corresponds to a wavelength of about 5 ...


0

Cosmic rays have the highest energy particles that have ever been detected. An upper limit has been deduced for the energy of photons: From direct observations of the longitudinal development of ultra-high energy air showers performed with the Pierre Auger Observatory, upper limits of 3.8%, 2.4%, 3.5% and 11.7% (at 95% c.l.) are obtained on the fraction ...


2

There is a great misunderstanding here. The photon is an elementary particle (the gamma in the table is the photon) of spin one and mass zero. This has been validated innumerable times in nuclear physics, atomic physics and particle experiments. There is no question about it. Special relativity treats the four-vector of a particle. In this framework the ...


-2

Light is a gravitational wave of bent space, it travels at the velocity of C, where time slows down. At that velocity (with time dilation) it makes a gravity wave into the photon. Mass is nothing more than the potential energy of a gravity wave of relativistic bent space. because it is traveling relativistic-ally it is slowing time and bending space. So ...


-3

I think the answer could be just as simple as that mass of any body i just a projection of that system vibrating. Consider a particle vibrating about a fixed position which starts gaining speed but first when it had zero speed or small vibration frequency its body vibrations were restricted to small particle area but when it start vibrating its projection ...


0

A negative charged ion like Boron is towards P end of Junction. When it gains a photon , electron from $B^-$ is excited from valence band to conduction band. This creates a hole at lattice point where electron was formerly present. Free electron moves towards N doped side which is positively charged. ...


1

The point dipole is an approximation from classical physics - note that it also involves an infinite field strength in its center, where the field amplitude is not differentiable. I think such a source is not compatible with the common approach to quantum mechanics. If you take such a very small, subwavelength source, it is true that the evanescent near ...


0

The photoelectric work function is primarily a surface effect, and for a given metal will vary significantly by crystal face. Note the variations given for silver, with the lowest, 4.26 eV, being from the polycrystalline form. Modelling of efficiency is complicated; from a macroscopic viewpoint one has the skin depth of the metal by wavelength, which ...


0

Mainstream physics has described the microcosm of molecules, atoms, elementary particles with the theory of quantum mechanics, and in particular the quantum mechanical standard model of elementary particles, and it has a mathematical form, a Lagrangian. . In this Lagrangian the elementary particles, including the photon, are entered as point particles with ...



Top 50 recent answers are included