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Well, it depends how you define "distinct fundamental particle". If you insist that Wigner's classification is what defines a particle, i.e. "particle = irreducible unitary representation of the Lorentz/Poincare group", then the photon is two particles, as you say. But, more commonly, we do not look at the particles like this - particles arise as the ...

6

The notion of temperature is all about how the equilibrium an otherwise isolated system shifts when the system's internal energy changes. So you do not need to worry about whether this internal energy is kinetic, potential, whatever. Actually the temperature is not quite the ensemble average kinetic energy. Your statement is true for an ideal gas and also ...

6

Photons, and in fact all other elementary particles, are not assembled. They exist. We don't know how, or why, they simply do. Photons are the quanta of the electromagnetic field - in quantum field theory, we associate to the classical electromagnetic field particles called photons. For a technical account if how this (roughly) works, read my answer to "The ...

5

In our modern understanding, every electron is thought to be a localized excitation of the electron (or Dirac) (spinor) field $\Psi(x^\mu)$, while every photon is considered to be an excitation of the photon (vector) field $A^\nu(x^\mu)$, which is the quantum field-theoretic counterpart of the classical four-potential. Thus, the answer to your questions ...

5

Basically, absorption lines exist because absorbed photon are not re-emitted in the same direction, so dark lines can be observed. There are various reason causing this. For example, the extra energy can be dissipated as phonon in solid or strongly interacting system. Excited states can also emit multiple low frequency photon if there are meta-stable ...

3

Between the statistical detection limits defined as "quantum efficiency" and the huge dynamic range you're proposing, the answer is very much "no." There are a number of commercial single-photon detectors, all of which come with both their quantum efficiency, or probability of detecting a photon, and their "recovery time," which indicates how long after a ...

3

No, so far no detector featuring such a tremendous dynamic range of the photon number is known. In the optical regime, state-of-the-art photon number resolving detectors can resolve numbers in the single digit range. (This paper has a slightly different take on the topic). In the microwave regime, numbers in the double digit range, or up to $10\,$dB, may be ...

3

If the photon is massless, how can it make an electron change momentum? Because, relativistically, momentum isn't proportional to (invariant) mass? Thus, particles with zero invariant mass can have non-zero momentum.

3

I don't think you can understand how a photon is made without knowing what a photon is, and to understand what a photon is requires understanding quantum field theory - specifically quantum electrodynamics. Quantum field theory is a very odd way of looking at the world, but it works and gives predictions that agree with experiment. In quantum field theory ...

3

Your wording suggests a few misconceptions: It seems you are thinking of light as having a corpuscolar nature (nothing wrong with that, you are in good company). Well it turns out that things just do not work that way. Phenomena like diffraction (to name one) tell us that we cannot describe the behaviour of light thinking of it as composed by (classical) ...

3

Whenever one makes light, one makes zillions of photons, so the way to make photons is the way to generate light: burning wood etc, striking stones, make wires incandescent, as in electric bulbs, etc, fusing nuclei as in the sun and stars and maybe more ways. It is much easier to make zillions of photons. How one photon is made takes us to the realm of ...

2

This is the plot of sunlight, red at ground level. Solar irradiance spectrum above atmosphere and at surface. Extreme UV and X-rays are produced (at left of wavelength range shown) but comprise very small amounts of the Sun's total output power. As all light comes from the sun during daylight this should suffice. One can get the number of photons by ...

2

As an analogy, consider the photon that strike your face and reach my eyes, we say that that photon carries information about your face which then helps me to identify you, You are confusing the individual photons with the electromagnetic wave that is light, which is composed out of a zillion photons. but don't these photons collide midway with air ...

2

charge is an intrinsic property of any particle. we in principle cannot change the intrinsic property of any particle. photons are the carriers of electromagnetic interaction(action at a distance).

2

The photons do not get "stuck" at the event horizon, from their own reference frame they are still traveling at C. The event horizon is simply the point where gravity from the singularity is strong enough that the escape velocity exceeds C. Just above the event horizon, photons can still escape the black hole, and just below it they sink into the ...

2

To my knowledge, there is no discretization of the light wavelength (they form a continuous spectrum). On the other hand, there exists no infinitely narrow absorption "potential". I mean that all transitions of electrons that may correspond to a photon absorption have a finite width. Consequently, the photons have a non-zero probability to get absorbed. ...

2

This answer is in addition to the answer by Sofia, on the classical electromagnetic wave, i.e. light, and replies to the title. Do Photons Move in a Wave Like Pattern Photons build up the electromagnetic wave. They are elementary particles, and as such are governed by quantum mechanical equations and rules. The rules applied, there is consistency ...

2

It is not correct saying that no force is applied. A photon carries momentum see PE here so on reflection there is momentum transfer. This is the idea behind laser propulsion discussed here. Concerning the speed it is even more complicated. The fact that light gets reflected usually requires an abrupt change in the index of refraction. To get reflected, ...

2

We do consider that energy. It reduces the amount of energy needed for the electron to be freed from the surface. An analogy: If a satellite is already in orbit, you need less energy to make it escape earth's gravity than if you started with the satellite on a launch pad on earth. The energy of the satellite in orbit is like the energy of the electron ...

1

The Cosmic Microwave Background includes photons that will not be absorbed before the universe inflates to the point where there is nothing to hit ever again. If parts of the last scattering were somehow barred from releasing that energy, I think we would notice. The photon carries energy. Particles do that. How is it any different from the electron, which ...

1

Of course light has a gravitational pull since photons carry the frequency-dependent energy $e=h\cdot f$, which is equivalent to mass via $m=e/c^2$. If you had a system where no radiation could get out, but in, and you'd pump a constant 100 Watt of light into the system, after 28 million years the system would have gained 1 kg (see example)

1

There are two things. More photons means a brighter beam. power (Energy/sec) is proportional to the number of photons/sec. Photons with shorter wavelengths and higher frequencies have more energy. That is a bluer beam has more power. So $P = nh\nu$ where n is the number of photons/sec.

1

As you can see the spectrum at the top of the atmosphere is continuous, with some saw tooth excesses, but still continuous. The absorption does create a saw tooth pattern, even so there is continuity. To dispel doubts here is the sun spectrum showing continuity and absorption spectra Solar spectrum with Fraunhofer lines as it appears visually. ...

1

At the present time in physics, the photon is an elementary particle, a member of the table of elementary particles that are the basis for the standard model of physics. Elementary particles included in the Standard Model It has zero mass, zero charge and spin one, and it is the gauge boson of the electromagnetic interactions. In all electromagnetic ...

1

The radiative transfer equation is a simplified model for describing light transfer. Of course it is possible to derive the radiative transfer equation by the Boltzmann equation for a photon density function $f(x,t)$: $$\partial_t f(x,t) + v_x \partial_x f(x,t) = (\partial_t f(x,t))_{coll}.$$ Here, the term $(\partial_t f(x,t))_{coll}$ is the gain and ...

1

By the uncertainty relation between energy and time, when the energy is known with precision, the time at which the photon was emitted is totally unknown. So, in the Fock state $|1\rangle$ we cannot know the phase because we don't know when it was emitted. Also, since we cannot measure a photon without disturbing it, if we measure it once at a time $t_0$ we ...

1

Probably too late now, but I have to answer this. There is indeed a way to predict the direction of a scattered Lyman $\alpha$ photon. The answer depends on whether the scattering takes place in the core or the wings of the line. In the core (i.e. closer to the line center than about 3 Doppler widths), we can use the dipole approximation, so the phase ...

1

The are two approaches about how light moves, and they are both correct. One is geometrical optics that describes the way light passes through lenses or is reflected by mirrors, etc. It refers to the light as moving according to straight lines. But there is another way to treat the light, I'd say a deeper approach. Light is indeed waves. Did you hear about ...

1

Yes, this is actually often used in a spectroscopic technique called REMPI -- see the image on this wikipedia page https://en.wikipedia.org/wiki/Resonance-enhanced_multiphoton_ionization There are some important physics techniques that rely on interaction with two photons -- two photon spectroscopy (http://cua.mit.edu/8.421_S06/Chapter9.pdf). Some other ...

1

Conservation of energy cannot be violated - so you cannot have a single interaction in which a photon of energy $E$ results in the emission of a photon of energy $E + \Delta E$ unless there is another source of energy. It is conceivable that an atom in an already excited state could result in such a phenomenon; and of course there is two-photon interaction ...

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