# Tag Info

2

No, it doesn't mean that the photons don't go any further. It means that when the temperature gradient inside the star reaches a threshold, the gas becomes convectively unstable. Heat is transferred more efficiently by moving parcels of gas than by transferring photons from hotter regions to cooler regions. So, the photons continue to diffuse outwards, but ...

-1

At any particular point in space, there is only one value for the electric field. Of course, if multiple electromagnetic fields are overlapping at that point, then their electric field components are added together to yield the total electric field (this is because electromagnetic fields combine with linearity). When multiple electromagnetic waves overlap ...

1

Don't think about individual single photons bouncing around. The energy (heat) diffuses slowly, exchanging energy with neighboring particles by various means (smashing into each other and radiation that only makes it as far as the next particle before being absorbed). Different photons are emitted and absorbed over and over, as well as physical collisions ...

1

The polarization states can be effectively defined in terms of the Stokes operators $\hat{S}_i$ for $i=[0,3]$ $\hat{S}_0 = \hat{a}^{\dagger}_{x}\hat{a}_{x}+\hat{a}^{\dagger}_{y}\hat{a}_{y}$ $\hat{S}_1 = \hat{a}^{\dagger}_{x}\hat{a}_{x}-\hat{a}^{\dagger}_{y}\hat{a}_{y}$ $\hat{S}_2 = \hat{a}^{\dagger}_{x}\hat{a}_{y}+\hat{a}^{\dagger}_{y}\hat{a}_{x}$ ...

1

Of course linear polarization is an observable, we measure it. Linear polarization along an axis x, and perpendicular to x, are the eigenstates of the observable "polarization along x". In fact, by measuring polarization along x, we measure the projection of the electric field along the axis x.

0

Well, the initial problem was not so glorious... The problem existed with the so-called "blackbody radiation". There was for one, energy distribution per frequency as derived by classical mechanics, which yielded the Rayleigh-Jeans formula for blackbody radiation: $\rho_T(\nu) = \frac{8 \pi \nu^2 kT}{c^2} d\nu$ ($\nu$ here is the frequency. For some ...

1

1) There exists the classical electromagnetic wave as described by Maxwell's equations. 2) The photoelectric effect showed that these electromagnetic waves are composed of photons, with energy E=h*nu , where nu is the frequency of the classical wave. Single photon experiments have been performed by limiting the intensity of the beam to one photon at the ...

2

The interaction between two charged partices occurs through a change of momentum. All very well, but the next question is how we calculate the momentum exchange, and this is where quantum field theory comes in. The interaction between two electrons is described as a disturbance in the quantum fields involved. Quantum field theory gives us an expression for ...

0

For Wheeler's delayed choice experiment, the 'choice' of a photon is said to be delayed from the reference frame of the experimenter. There are a couple of things to point out here. First the delayed 'choice' is not an actual thing (e.g. the photon doesn't go backwards in time through the slit and make a different choice), it arises from a common ...

0

Briefly, no. Charged particle interaction is fundamentally a non-instantaneous photon exchange. The interaction can be written or expanded as the ''naive'' instantaneous interaction ($e^2/r$) plus a photon exchange portion which contains exactly a $-e^2/r$ term and a non-instantaneous (retarded in time) term. The author calls the first term the Coulomb ...

1

Strictly speaking a photon cannot be localized and the single particle "wavefunction" (as well as it's corresponding position operator $\hat{r}$) only exists in an approximate sense. The reason for this is quantum electrodynamics (QED), which is the theory that contains photons, is a quantum field theory (QFT) rather than the (non-relativistic) quantum ...

0

Each photon has an oscillating electric and a magnetic component. In vacuum the components are always perpendicular to each other and also perpendicular to the motion of the photon (see this sketch). When Ppotons are unpolarized, these oscillations are equally distributed in 360 °. A polarization grid can rotate about 50% of the photons so that they can pass ...

2

The photon as an elementary particle is special: the quantum mechanical wave equation whose solutions squared will describe the probability of finding the photon at a given phase space point is the Maxwell equation that describes the classical electromagnetic wave, except it is the potential form of it and it acts as an operator on the wave function of ...

1

The formal analogy between a mode of the radiation field and a particle in a harmonic potential stems from the fact that both systems have the Hamiltonian (in appropriate units) $$H = \frac{1}{2}P^2 + \frac{1}{2}\omega^2 X^2,$$ where the variables $X$ and $P$ obey canonical commutation relations $[X,P] = \mathrm{i}\hbar$. For the radiation field, these ...

0

Whether or not your idea produces thrust depends on at least one of the following statements being true: The laser doesn't travel with the vehicle. The photon eventually leaves the mirror structure. Forget the structures holding the mirrors and consider each component as the photon interacts with it. In all cases, I assume each emission and reflection ...

0

There exists the KISS rule ( keep it simple stupid), and I will. Let us take an excited potassium atom and solve the Schrodinger equation for the transitions available. This will give for this atom a wavefunction, in principle from -infinity to +infinity in space and time. The square of this wavefunction will give the probability for a photon to fall on a ...

0

This is a perfect example of the collapse of the wave function. The "potassium" atom makes a transition from higher energy to lower energy, sending out a speherical wavefront into the universe. One hundred years later, that spherical wavefront crosses the photomultiplier tube. Since the energy of that wave is spread over hundreds of square light years, there ...

0

The reason is because there are no particles at the 1eV level to absorb the energy. In the ground state, all the particles are at the lowest level.

1

The photon is not deflected by the charge. It's deflected by the spacetime curvature. The spacetime curvature is related to the stress-energy tensor (by Einstein's equation), so if you change the stress-energy tensor you change the curvature. A static black hole, described by a Schwarzschild metric, has a very simple stress-energy tensor since it's ...

0

It seems contradictory that red shifted light has lost energy yet total energy is conserved (where did the energy go?). The trick to understanding this is knowing that the energy measured depends on the frame of reference you are measuring from. Consider a ball flying towards you quite fast and hitting you in the head. From your perspective it has a lot ...

1

Red shifts happen because of various causes. there also exist blue shifts: Conversely, a decrease in wavelength is called blueshift and is generally seen when a light-emitting object moves toward an observer or when electromagnetic radiation moves into a gravitational field. Now on redshifts: Some redshifts are an example of the Doppler effect, ...

2

This link says why photon electric field interactions are "trivial" : In a vacuum, the classical Maxwell's equations are perfectly linear differential equations. This implies – by the superposition principle – that the sum of any two solutions to Maxwell's equations is yet another solution to Maxwell's equations. For example, two beams of light pointed ...

1

No. Light is a disturbance of electromagnetic field and from particle aspect, photons aren't charged particles. Assuming you're asking about light in vacuum (that's when you can see sole effects caused by electromagnetic field), the external electromagnetic field can't affect light because it's a linear field (you can Google to know what linearity means). ...

1

Given two like charges -- two electrons for example -- does moving them farther apart release a photon? Photons are released when charged particles are accelerated. . Moving them apart may generate photons if there is acceleration. If the velocity is constant, no radiation. Electrons in conduction bands of metals are in a quantum mechanical state and ...

0

The idea that gravity is specifically a force that masses exert on each other is actually false, although it works as a very good model for many applications. But actually, (as far as we can tell) the way that gravity works is it actually causes the space-time to bend. So gravity isn't causing light to bend, rather the light is travelling on a straight path ...

1

Here is a suggestion of how to produce the state in your second formula. Choose to make holes in a screen, NOT at the two points A and B of which I spoke in my first answer, but at two other points, e.g. on top of the upper cone, and on the bottom of the lower cone. Then, pass the upper beam through a beam splitter $BS_1$. Block the transmitted wave, let ...

0

Since the field of each charged particle extends to infinity, the fields of two particles are "in contact" with each other (no "communication" is necessary). When the charges are not equal (+ & -), the fields "cancel" each other along the line connecting their centers. This causes the attraction of the particles. When the charges are the same (+ & + ...

1

Spontaneous Parametric Down Conversion is a process that produces pairs of photons. The process goes like this: a strong beam of ultra-violet (UV) photons, is sent upon a down-conversion crystal (placed inside your black-box). Inside the crystal, the UV photon is SPLIT into two photons, named (I don't know why) SIGNAL and IDLER. In general the two photons ...

0

Does the photon add to the mass of the black hole an amount of mass m = e/c^2, where e is photon energy? Yes. If so, is it the energy the photon had at the beginning, or the energy the blue shifted photon has as it crosses the horizon? The energy the photon had in the beginning. As such, it will reach the horizon with no energy at all. He will ...

0

The speed of the wave = wavelength x frequency, say the frequency is 6.0 x 10^14 Hz calculate the wavelength. NOw use the de Broglie wavelength, Wavelength = h/MV use E=hv to get the energy than, Use E=MC^2 to get the mass Plug that back in to the de Broglie eq you get the same answer

0

hv is the energy per photon, the power radiated is essentially this times the number of photons emitted per second, so a constant frequency antenna with variable power is emitting a variable number of photons per unit time.

0

No, the equal and opposite reaction in this case is handled by the electrons. As you may know photons do not exist by themselves in lasers and the like but are created when an electron is excited and later relaxes by releasing a photon, the opposite impulse of that photon is handled by the electron, so you have billions of billions of tiny electrons handling ...

1

As amplification: The emission of photo-electrons is a quantum effect: one electron receives the energy from a single photon. If the energy of that photon, proportional to its frequency, is sufficient to overcome the work function, then photo-electrons will be produced. If the frequency is too low, then no photo-electrons will be produced. Increasing ...

4

The work function is the difference between the energy of the most excited electrons in the conduction band and the vacuum. It's the energy required to take an electron from the conduction band and remove it to outside the metal. To a first approximation the work function is independant of the intensity of the light, because it's an intrinsic property of ...

1

Now per QED, electrical charges interactions are effected by photons. Suppose you are one of the two charges. How do you know to attract or repel the other charge? You want something that does not exist - intuitive picture of physical process within a theory which is a demonstration of how far can one go with mathematisation of experience and ignoring ...

0

The attraction of unlike charges and the repulsion of like charges is an experimental observation that has to be included in any model of electromagnetic reactions When talking of photons one is in the quantum mechanical regime. As in classical electrodynamics the sign of the charge defines the potential, attractive or repulsive, between the two charges, ...

0

Photons are observed as radiation with a given spin and other properties. As such, it does exist. But according to the principles of time dilation and length contraction, from their (hypothetical) proper point of view, they would be reduced to a single momentum. Their proper time would be zero, the distance of their geodesic (and also the spacetime ...

1

Photon counting statistics cannot always be explained by classical fields. In these experiments, the state of the field is monitored continuously by a photodetector. I believe these represent one of the clearest experimental demonstrations of the quantum nature of the radiation field. For example, in observing the emission of photons from a single atom, ...

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 ...

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. ...

0

Once the gammas are produced, they do not carry particular information. There is just a certain probability, measured by a differential cross section, that, if they scatter again, they produce $e^+,\,e^-$ as in the initial pair.

0

Disclaimer: in this answer “photon” refers to an excitation of electromagnetic field, not to a fundamental particle as understood in QED and Standard Model. Possibly unexpectedly for some readers, a photon can have its rest frame in a homogeneous medium that moves w.r.t. the frame with speed equal to the light speed in this medium (that is less than c ...

0

I am a bit lost in your problems with $mc^2$, but in the special relativity for a massless particle your identity can be proved as follows. First observation is $$v_x = |v| \frac{p_x}{|p|} = c \frac{p_x}{|p|},$$ as far as for a photon the speed is always speed of light, $|v|=c$. Then $$\sum p_xv_x = \frac{c}{|p|} \sum p_x^2.$$ Then one can observe that ...

Top 50 recent answers are included