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3

Photons can come from a variety of sources, some of which are indeed transitions of electron levels (or nucleus levels or molecular levels), which are indeed discrete (it is also not because of wobbling). The most common way of having photons of arbitrary wavelengths is via Bremsstrahlung radiation. This is the radiation obtained when a charge is ...


0

The inverse square law is "simply" a statement of the fact that a diverging cone of "energy"/particles/stuff will have a cross sectional area that increases with the square of the distance. This is a result of basic geometry and the fact that area is proportional to length squared. As the stuff-beam impacts over an increasing area as distance ttravelled ...


2

SUMMARY: This is a very good question. In a lossless medium, fundamentally the answer to your question is "no, an individual ray does not lose energy in propagating" because it represents a plane wave (in photon language, a momentum eigenstate), whose intensity does not vary as it propagates. Intensity information is encoded in the flux density of rays ...


2

Photons come with chirality, so you should consider angular momentum conservation as well. For $1\gamma \to 2\gamma$ scattering, this will not be possible. (I'm assuming production of collinear photons only; it's obvious when two are not collinear, energy and momentum conservation will be violated)


2

One single photon is not enough for you to see objects. If there would be just one photon coming to your eye, you would first see only darkness (before the photon arrives), then you would notice a dim point for very short time (when the photon hit your retina), followed by darkness again. You can see objects because there is a constant flow of huge amount ...


2

A photon is an elementary particle with zero mass, moves always with the velocity of light c, and has energy given by E=h*nu . It has spin +1 or -1 and its wavefunction also has a polarization which will build up the polarization of the emergent from many photons classical electromagnetic wave. Its energy is a property that characterizes the emitting ...


0

The electron doesn't absorb the photons, it scatters them. The energy absorbed by an interaction depends on the scattering angle - this can be determined using the Compton formula for the wavelength of the scattered photon. And the electron tends to scatter more at different angles (proportional to the Thomson differential cross section). From this I found ...


9

After the hypothetical split, 2 photons with the same energy would be propagating at an angle ok with momentum conservation. Then there would be a rest frame where the angle is 180 degrees. Now if you stay in this restframe and go back in time before the split, your single photon would be at rest. However, that is not possible: According to relativity, speed ...


0

This paradox is an artifact of describing the mirror classically. Quantum mechanics is itself free from paradoxes, and it is easy to see that this paradox vanishes the moment you describe the mirror as a quantum mechanical object. A simple way to see what goes wrong is to apply the uncertainty relation. Suppose that we have a freely floating mirror and we ...


-1

An entangled state is a joint state for both the particles. For instance if you had electrons then when you write $\left|\uparrow\downarrow\right\rangle -\left|\downarrow\uparrow\right\rangle$ for instance the minus sign in the middle there is a definite phase between the two parts. And the joint state itself has an overall phase. So for instance ...


1

The answer depends in part on the Z of the material you are looking at. This is something you can easily verify by looking at the XCOM database To generate an example, I entered "single element, Z=25" (manganese) and selected plots for different types of interaction in the range up to 10 MeV. The result looks like this: As you can see, the photoelectric ...


0

I am not totally sure of this answer, which is why I asked the question. However, I think the answer is that only relatively short wavelengths can pass through an annular aperture. Specifically, I think that if the outer radius of the annulus is R and the width is W, where W << R, the maximum wavelength that passes through is approximately 2R ...


8

A photon is an elementary particle. As much elementary and as much particle as the electron . A single elementary particle has a fixed mass and cannot emit another particle without violating energy conservation, because its mass is fixed. In the center of mass of a massive elementary particle, electron, there is no energy for an emission , for a ...


3

There's two main things to consider - energy and absorption charasteristics of different photon wavelengths. The Sun emits a lot of energy, obviously. Even at Earth's distance from the Sun, the energy concentration is still far from negligible - when this energy impacts your body and is absorbed, it mostly causes heating (a bit complicated by wavelength, ...


1

There are multiple "kinds" of photons - different wavelengths have different effects on you. X-ray works somewhere around the 1nm range of the spectrum. It is ionizing radiation which can mostly go through soft materials but can harm cells when passing. So you usually get only the minimal needed amount of photons to create the image and not much more. The ...


20

In addition to the answer from @MichaelS, you need to consider where the energy from each source is deposited: Sunlight energy is deposited on/in the skin where there are numerous nerve endings. An increase in skin temperature is "measured" and your brain is aware of it. X-ray energy which is absorbed by the body is mainly absorbed by bones and some ...


28

X-rays do warm you up. It's just that the X-rays are more dangerous per photon (they can do major damage to cells and DNA, and are known to cause tumors and cancer), so they limit the amount of time you're exposed to the bare minimum needed for a clear image. The total energy from standing in the sunlight for several seconds is much higher than the energy ...


1

When there is a resonance in electron response, both refractive index and absorption change rapidly: the refractive index has a "jiggle" in the vicinity of the resonance, like this sketch (adapted from this earlier answer by John Rennie - but I disagree with the "n=1" label so I cut it off...: As you can see there is higher refractive index at the low ...


2

Although there are already some excellent answers, I believe they are a little complex. Please allow me to offer a simplistic answer. Let me start with the analogy of sound waves and the ear. The sound enters the ear and causes certain cilia to vibrate in response to the frequency and amplitude of the sound wave. Similarly a photon (as a wave), enters the ...


2

Photons are energy. When a photon hits your retina, that energy is absorbed and converted to electrical energy in your optic nerve.


6

Light from all over the place hits your eyeball fairly randomly. The lens forces light from a specific angle to hit a specific part of the retina. This HowStuffWorks article shows how the mechanics of that work. The only major differences between camera lenses and eyeball lenses is that we can dynamically alter the shape of the lens to focus on different ...


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Photons can be created and destroyed freely, since they don't have charge or mass. Turn on a light, and you create many photons. Any body (made of atoms) not at absolute zero temperature will spontaneously emit photons. They are consumed just as easily. Most any bit of bulk matter will absorb a photon in the electrons on the surface, transforming the energy ...


4

Imagine a spring-loaded trap with a hole that's sized such that only a particular size of object can enter the hole and trigger the trap. The molecules involved in vision are like that trap, with a bond having an electron energy gap tuned to the visible frequencies of light, encapsulated in a specialized protein that transforms the absorbed energy into a ...


-15

Shortly, the energy of the photon goes over to the electron. But energy is a vague concept. In material sense, could the photon, or better, the electron's electric field and the electron's magnetic field be quantized? I developed a model with two different quanta. Photons, electrons, positron's, protons, neutrons, ... are made from this quanta. Photons are ...


1

No. The other photon might even be forbidden to produce a pair over by itself all by itself since there might be no nucleus over by it. The other photon doesn't have to copy what the first one does. But many things could happen to the entanglement. And that is partly because there are many ways the photons could have been entangled. For instance, you could ...


28

From the wiki article on color vision as an illustration of how photons are absorbed: Perception of color begins with specialized retinal cells containing pigments with different spectral sensitivities, known as cone cells. In humans, there are three types of cones sensitive to three different spectra, resulting in trichromatic color vision. Each ...


1

No. Being entangled does not mean they mirror everything the other does. It only means certain properties are in an inseparable state. Destroying one of the pair would end this state


1

You cannot increase the amount of energy that the lamp has already emitted. Not unless you add more energy in some other way. For typical household lamps the easiest way to double the apparent illuminance of a specific area (a newspaper for example) is to note that some light is being emitted in directions that do not directly illuminate your desired area. ...


1

You need to increase the power into the light. How easy that is depends on the light. If you have a 60W incandescent, the easiest way is to buy a 100W bulb. Depending on the fixture, the extra heat may be a problem. If you have a CF or LED bulb, you may again be able to find a higher wattage equivalent. For a standard incandescent, without replacing ...


0

Lubos's answer covers very well and in a general way the different possibilities for reactants and products in the reactions of subatomic particles and nucleons. The basic equation is Energy of photons = energy of reactants - energy of products. Due to the enormous energies involved it is actually unlikely that visible light will be released, most ...


6

This is a special example of "what will happen" under given circumstances. Almost all of physics – and natural science – is about answering such questions. But they're really very many very different questions and one must be a little bit more specific about what the question is. Your general question "what forms of energy will result" is so general that it ...


0

When these terms are used, light is been pictured as an electro-magnetic wave. Thinking of the microwave example, inside what you would have is like a sea where light is equivalent to the undulations. So in this picture, the wave-length is the distance between two consecutive wave peaks. Larger wave-length implies more separation between these peaks. Also ...


1

You can make beams of light that have orbital as well spin angular momentum and they can go through an annular aperture. So circular versus linear isn't enough. Linear has the phase advance orthogonal to the advancement with the polarization at a fixed angle. Circular has the polarization rotate but with an annular filter you can give the wavefront a twist ...


2

To answer your questions: Yes, fibreoptics transfer light. Maybe. I'll discuss that now Fibreoptics are strands of glass, they're CRAP at going around corners, I mean seriously crap, communications fibre is VERY THIN. Even then it can't go around bends well, they test it at every stage during laying. However with communications stuff the path matters ...


2

The violation of gauge invariance by this term is the "only" reason why it's never written down – as long as we define the word "only" to include all other reasons that may be shown to be "physically equivalent" to gauge symmetry. Gauge symmetry is extremely important and its violation would make a similar theory inconsistent, especially at the quantum ...


1

The origin of this equation is reasonably well explained in Abramowicz (1991). If you take a relativistically expanding enevelope and only consider Thomson scattering, then as the electron scattering cross-section in the co-moving frame $\sigma_T$ is independent of frequency, then the mean free path of a photon in the co-moving frame is independent of the ...


1

Laser Mirrors are used for beam-steering in demanding laser applications. Laser Mirrors are Optical Mirrors that have been designed for specific laser types or wavelengths.Optics’ Laser Mirrors feature dielectric coatings that have been optimized for high reflectance at specific laser wavelengths. Edmund Optics’ dielectric coatings typically feature greater ...


8

While photons can in principle form a black hole, the black hole will not be massless. The mass of the black hole will be related to the energy of the photons that went into it by Einstein's famous equation $E = mc^2$. The black hole will be a regular black hole, and classically it has an infinite lifetime. Once you include Hawking radiation the black hole ...


0

As with most of the questions in this forum, the answer is right in the same section of the book where you read about the problem. Why don't you go back and find it? First you talk about a photon hitting an electron, and then you suddenly change to a photon hitting a photon, a completely different problem (photons don't interact with other photons, only ...


1

Photons and electrons are elementary particles and their behavior is predicted by quantum mechanical equations. Quantum mechanics predicts probabilities for a reaction to happen. Photon photon interactions have very very small probabilities of happening. Thus photons pass through each other for all measurable purposes at low energies ( light and below ...


1

If you had a laser you wouldn't see it unless it was aimed at your eye (ouch). Or if there is dust or such around for it to scatter off of. And scattering is the key. If you want to see something then it either has to get to your eye or it needs to deflect something towards your eye. If you have a beam of electrons you could try to get something to ...


4

To someone outside it looks the same as a regular, massive black hole. Classically the lifetime is the lifetime of the universe. It might merge with another black hole. It might last forever. It might meet a singularity in a big crunch if the whole universe contracts to a singularity. If you are worried that it can't decay by Hawking radiation because an ...


3

To get an understanding on quantum field theory issues, you have to understand the difference between virtual particles and real particles. Virtual particles, in contrast to real particles, are a mathematical construct inspired by the Feynman diagrams used to describe interactions. These diagrams start with real particles, i.e. particles that have the mass ...


2

This situation would lead to an accumulation of the photons inside this sphere. Quantum mechanics has no prohibitions for bosons, like photons, to occupy the same space under the same quantum numbers, so the photons would just build up continuously. But at no moment any of them should create other particles, since this would require interactions if charges ...


0

This is a typical matrix for an optically active element that rotates the light polarisation. A cuvette of water with sugar will do the job. Proportional to the sugar concentration, you can obtain arbitrary wave rotation. Note that the U matrix has imaginary eigenvectors (1+i)/sqrt(2) and (1-i)/sqrt(2). Accordingly, unlike λ/4 plates, the eigenwaves of such ...


1

Let's just say that given a sufficient density of photons, you would expect things like pair production to occur with some probability. However, these massive particles will occur as particle-antiparticle pairs, thus won't "accumulate" since they'll annihilate each other if there are too many of them. They might also annihilate with your mirrored sphere. As ...


2

Photons have two degrees of freedom, the helicity. But they are not an ideal gas with equation of state $$PV = NkT,$$ so the usual derivation of the adiabatic exponent does not apply. You need to use the equation of state $$U = PV$$ which is valid for any ultra-relativistic gas. You can derive it in the same way as the ideal gas law -- by considering ...


0

In a dilute gas, the photon density should be the same as inside an evacuated black box of the same temperature (independent of the gas density). Edit: In other words, the massive particles in the Sun have a temperature based on their kinetic energies. The photons must be distributed as black-body radiation at the same temperature. The density of photons in ...


2

They do radiate light randomly in all directions -- in the object's own reference frame. But not from the Sun's reference frame. The effect in general is called "relativistic beaming." Here's the clearest derivation that I know. Take the Pauli matrices $\sigma_i$ and adjoin the identity matrix to them as $\sigma_0 = I$. Now take a four-vector $v^\mu$, and ...


0

Do photons experience every moment in time and position in space simultaneously? No. LIke WillO said, photons don't experience anything. Would it be more correct to say that a photon, traveling at the speed of light, would experience all points in time simultaneously, and therefore be everywhere at once? No it wouldn't. A photon is emitted from ...



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