# Tag Info

16

The problem is that you are confusing light intensity with energy of a single photon. The photoelectric effect requires a certain energy per photon to work. But low light intensity just means fewer photons come - you can actually see the grain if the conditions are too dark: every pixel can get ~10 photons or less... and yet still, each photon that comes has ...

7

Yes, but you'll have to go really, really fast. And even then, don't worry about the photons. The relation between velocity $v$ and the observed and "true" wavelength $\lambda_\mathrm{obs}$ and $\lambda$ of the light is $$\lambda_\mathrm{obs} = \sqrt{\frac{1-v/c}{1+v/c}} \lambda.$$ If you consider optical (i.e. visible to humans) light with a wavelength ...

3

Absolutely it can - and it happens all the time. If you excite an atom, it can go through various "stages" of decay back to the ground state - with each drop in energy resulting in an emission of radiation. This happens during photosynthesis: see this page from which I copy this image: As you can see, there are multiple paths for the energy to be lost by ...

3

The electron-positron pair has a center-of-mass reference frame where the momentum is 0. Obviously, there exists no one-photon system with positive energy which has 0 momentum, as the energy-momentum relation for a photon is $E = p c$.

3

Your question is the one that Einstein pondered for long time and from which Special Theory of Relativity was born. He wondered what could happen if you travel at the speed of light how would you see a ray light. The problem was that according to Maxwell's Electrodynamics, explained light as oscillating $E$ and $B$ vectors along space and time, so as a ...

3

When people say that photons always travel the speed of light, they meant in terms of reference frames. And in fact, even through water, photons do travel the speed of light. The only reason why they appear to travel more slowly is because they "run into" molecules, and the path is not straight. In a vacuum, there are no molecules to hinder the light, and it ...

3

What happens to photons when they hit our eye? Some of them pass through the iris and are focussed by the lens onto the retina where they are absorbed by rods or cones. where do they end up? Some of them end up absorbed by rods/cones, some by other tissues, some are reflected (c.f. red-eye in photography). why our eye don't get overheated? ...

3

The electromagnetic potential $A^\mu$ is a four-vector, and hence transforms in the fundamental representation of $\mathrm{SO}(1,3)$, i.e. $A^\mu\mapsto \Lambda^\mu_\nu A^\nu$ where $\Lambda$ is the usual 4x4 matrix associated to a Lorentz transformation. Your question seems fundamentally confused about the difference between the field and the particle. The ...

2

Think about it this way. If you had a radioactive atom it might decay and emit a photon. And you don't know when. So there is a whole range of times for when a nearby photon detector might go off. But there is also an issue of coherence. Which is a specific technical term, not a word people use for emotional purposes. If you have a short coherence length ...

2

Building on a comment by CuriousOne (who, honestly, should leave off commenting since he only ever writes answers in then anyway): A photon is not an object in and of itself. A photon is an excitation in a quantum field, which is not localized but fills space. In the double slit experiment you have an emitting source, a mask with two slits, and a ...

2

In a static spacetime, there is (by definition) a timelike Killing vector field $\xi^\mu$, which implies that geodesics with four-velocity $u^\nu$ have a conserved quantity $\epsilon = -g_{\mu\nu}\xi^\mu u^\nu$. For example, in Schwarzschild spacetime, this is $$\epsilon = \left(1-\frac{2M}{r}\right)\frac{\mathrm{d}t}{\mathrm{d}\lambda}\text{,}$$ where ...

2

Light does not travel at the same speed in all mediums. In water the speed of light is less than that in vacuum.

2

The formula E=h*f is a quantum mechanical formula for the elementary particle called a photon. The photon travels with velocity c in vacuum. If it is not a vacuum it will interact with appropriate quantum mechanical interactions, and in the interval between interactions it will be traveling with velocity c. If its interaction is elastic, as happens in ...

2

From the theory of Thompson scattering (see http://farside.ph.utexas.edu/teaching/em/lectures/node96.html ) we know that a charged particle of mass m interacting with a plane wave electromagnetic field of Strength $E_0$ and frequency $\omega$ has an effective dipole moment of magnitude $$d=\frac{e^2E_0}{m\omega^2}$$ Note that the dipole moment scales ...

2

There is light all around the lamp, and all around you (except of course for the points that light cannot reach because of walls or whatever). You just only perceive the light that comes into your eyes. The light that comes directly from the lamp is generally more intense than that reflected by the objects around you, so that you tend to ignore the latter. ...

2

The trick is that because the "slit" is infinitely wide, you shouldn't work in the far-field approximation (Fraunhofer difraction integral which leads to fourier optics), but with distances computed to the square order (Fresnel diffraction integral). The results are appropriately named Fresnel integrals (this time this is a name of a special analytical ...

2

Electromagnetic (EM) radiation between 400nm and 700nm in wavelength is the same thing as light. There is no difference. Neither is there a distinct difference between light, ultraviolet, x-rays, gamma rays, infrared, microwaves, or radio waves. Those names are all just human convention to specify EM radiation in certain frequency ranges. And regardless ...

2

What we call reflection is in reality a more complicated process than bouncing a ball to a wall. For the part of the electromagnetic radiation that we call visual light and for low densities of this light the surface electrons are responsible for the absorption and re-emission of this photons. So yes, mirror will gain momentum and the photons will lose ...

2

The easiest way to measure the energy of a photon, is to make a reaction using the photoelectric effect. The photon hits a surface, knocks out an electron, and the electron can be prevented from carrying charge away from the surface by putting a small attracting voltage onto that surface (this is called the 'stopping potential'). It is an experiment usually ...

1

Yes it does. For example a (perfectly) black surface would absorb all photons, and hence the radiation pressure would be given by: $P_R=\frac{E_f}{c}$ Where $E_f$ is the photon energy flux and $c$ is the speed of light. A (perfect) mirror reflects all the light, and so you would get double the momentum transfer, and hence double the radiation pressure. ...

1

The classical light beam, an electromagnetic wave, emerges from zillions of photons which travel with velocity c and build it up. The energy of a photon is E=h*nu, where h the Planck constant, and nu is the frequency which will appear in a classical wave built up by this energy photons. The way this happens is explained mathematically here, but is not ...

1

It is convenient to use wavelength as a unit of measurement when it comes to photons but really it starts with the frequency that the photon is oscillating at. We know the speed of light and if you diffract it from one surface to another we can measure the distances using Pythagorean's Theorem to determine the frequency or wavelength. A photon oscillating at ...

1

Yes it will be affected by the car's movement (if you are outside the car), since momentum is conserved. If you are moving with the car, the source is not moving relative to you, so the photon is moving down straight. PS from the outside point of view, it will seem like the photon traveling diagonally covers more distance. But since the speed of light ...

1

In the particular setup that you show, where both the car with the photon source and the surface with the detector move in the same direction at the same speed, the result is the same regardless of the emitted object is a ball or a photon: it will hit the detector. This is best understood by a transformation of reference frames: Instead of looking at the ...

1

Your overall efficiency will be well less than 50%. There are several factors at play here: To generate a positron with photons, you also have to generate an electron in order for the quantum symmetry to be preserved. When you generate a pair this way, they end up with half the energy each. If you are only using one of those for your transmission vessel, ...

1

Neither image displays diffraction. They both illustrate shadowing. The second photo was apparently shot on a clear sunny day. All of the rays come from the sun. Some are blocked, those that aren't go straight down to the floor. The second was probably taken on a hazy day. Rays originate from the haze; they come from all directions, and make their ...

1

You are describing a spherical wavefront i.e. light radiating outwards from a point source with spherical symmetry. But suppose you have two such point sources near enough to each other that their wavefronts overlap. Now your expansion model has to have space expanding simultaneously in opposite directions. Consider also that a wavefront can be any shape. ...

1

Elementary particles do not have consciousness, individuality or volition. They follow the rules of the boundary value solutions of the quantum mechanical equations they obey. The relativistic quantum mechanical mathematics have zero mass particles moving at velocity c, and in all valid frames massive particles move at velocities less than c. It is the ...

1

let's assume we have 2 rays moving at 180°. If the container ( and the background ) move instead of the photons, the container would have to move in the direction of both rays. But, it's impossible since we assumed 2 antiparallel rays. How can container move forward and backward simultaneously.

1

The atom absorbs the photon that kicks up an electron to an excited state, and it is the atom that will emit a photon when it de-excites. Not the electron. Is the invariant mass of an atom higher when the electron is in an excited state? Take the hydrogen atom. The ground state energy is at -13.6eV. This means that the mass of hydrogen is less than the ...

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