# Tag Info

1

You are correct in one thing: if an atom in an isotropic medium spontaneously emits a photon, it can do so in any direction at all, and the overall emission will be evenly spread over the unit sphere. However, lasers work using stimulated emission, which is slightly different: if an atom is excited, you can induce it to emit its energy by shining an initial ...

4

Have a look at my answer to Slit screen and wave-particle duality because this covers a lot of topics relevant to your question. You're correct that if we imagine the photon as a little ball then if the arms of the interferometer are different lengths the two "halves" of the little ball cannot arrive at the detector at the same time. But this is not how the ...

2

When one says "photon" one is in the quantum mechanical frame. Quantum mechanics does not follow the rules of classical mechanics if one tries to consider the photon one classical entity, like a bag of energy flowing. The photon is a point like elementary particle in the standard model, it has no extent and when it hits the detector it registers at a ...

5

If we assume you are a sphere in space, at the same distance from the sun as Earth, then we can calculate the heat absorbed - and we can calculate how hot you need to be so heat in = heat out (assuming uniform surface temperature, and radiative heat transfer only). For this, we need the Stefan-Boltzmann expression for total emission at a given temperature: ...

0

If a single red photon hits your telescope from the direction of a planet in the Andromeda Galaxy, then all you know that the planet emitted a red photon. Was it caused by a fire? A scattering of starlight through its atmosphere? An Andromedan with a laser pointer? A single photon of light from an unknown source has about as much information as a random ...

2

Suppose you are using a CCD or a photographic plate to record your image. The interaction with the light occurs when the detector absorbs a photon, and this happens at a point. So the image is built up from a collection of points - one for each photon that is detected. In everyday life, e.g. taking pictures with the CCD in your phone, the intensity of the ...

1

I think a simple view is this: The solar cell must have a PN junction, which is a junction between p-type (many holes, no electrons) and n-type (many electrons, no holes) materials. Right where they meet there is actually a "depletion width" within which there is hardly any of either. Within this region, as photons come in they generate electron-hole pairs, ...

8

It's tempting to think of the light as a little ball (the photon), and since little balls have a definite position the little ball has to be in a superposition of a state where it goes through one slit and a state where it goes through the other. However this is not a good description of what actually happens. The light is not a photon, and it's not a wave ...

0

Is it correct to think that the speed of light does not depend on the speed of light source because photons have no mass In a certain sense, yes. The Lorentz transformations guarantee that the speed $c$ is invariant; an object with speed $c$ in one inertial reference frame (IRF) has speed $c$ in all IRFs. But an object with invariant mass $m$ cannot ...

0

the photons will travel at the speed of light relative to both the moving light source and an object in another frame of reference. time dilation will bridges the gap so that both may co exist.

0

photons have inertia. They move in a straight line when no net force act on them,their momentum and energy also remains constant.(Frequency too). Otherwise they donot move in a straight line which can be seen in gravitational lensing.( http://en.m.wikipedia.org/wiki/Gravitational_lens ). So your reasoning is incorrect. But you are indirectly correct because ...

2

Theoretically yes, the laser principle does not consume any material. There is a light source that excites the electrons in the material to higher levels, they deexcite to some intermediate one, here the avalanche of photons appears producing the laser light and leaving the electrons in the ground state. And you can repeat the process without a loss.

2

Q1: For photons of energies much less gamma rays, the quantum mechanical photon-photon interaction is negligible. This is consistent with the classical electrodynamic description where the principle of superposition holds (electromagnetic waves pass through each other unchanged, as well as through electric/magnetic fields). Q2: in reality, charge is defined ...

1

theoretically if its components never wore out then yes. however in practice things do wear out eventually and so no it could not be done in the same way that a perpetual motion machine can work in theory but not in practice.

1

It is the energy of the incident photon. At low Energy (Long wavelengths as in Radio waves and Microwaves, visible light) Rayleigh scattering would dominant, in which the scattered photon would have the same wavelength as the incident photon. At higher energy (shorter wavelengths as in Ultraviolet spectrum), the electron would absorb the energy of the ...

1

You start with $$E' = \frac{E-up}{\sqrt{1-u^2/c^2}} = \\ \frac{E-up}{\sqrt{(1-u/c)(1+u/c)}}.$$ You know that $p = \frac{E}{c}$, so $$\frac{E-uE/c}{\sqrt{(1-u/c)(1+u/c)}} = \\ \frac{E(1-u/c)}{\sqrt{(1-u/c)(1+u/c)}} = \\ \frac{E\sqrt{1-u/c}}{\sqrt{1+u/c}} = \frac{E\sqrt{c-u}}{\sqrt{c+u}},$$ now using $E = hf$, $$f' = f\sqrt{\frac{c-u}{c+u}}. ... 1 To understand how light is affected by gravity, it helps to think of light as energy. So let's ask a basic question: when it come to light, what is energy? By the Planck-Einstein relation, we know$$E = h \nu$$where \nu is the frequency of the light and h is Planck's constant. So when we talk about the energy of light, keep that in mind. Also note ... 0 It sometimes occurs. It is called sonochromism. Here is my source. 0 That is interesting. At the very least, you should be able to tell the difference between the mirror and the world if you put your hand out to it - the mirror should be cold. Also you could blow on the mirror - it will either come back at your face (the atoms in each version of you's breath will bounce off each other) or show up as condensate, depending on ... 1 What they actually measured was not particle behavior. It was just a quantized energy transfer to the probing electrons. That corresponds to the absorption of individual photons, but it doesn't mean the Surface Plasmon Polariton (SPP) field was acting as a particle. It just interacted locally with the electron, as it must. Typically particle-like behavior ... 1 So, I wonder why is it usually said that photons do not interact, or hardly interact? As far as we know photons do not directly interact with each other. Mathematically, this is manifest in the fact that the equations of motion for electromagnetism are linear: given two sources A and B of electromagnetic radiation, the resulting EM field is precisely ... 2 When you have only one electron then \Delta S=0 makes intuitive sense: you can change the angular momentum l of the atom by changing it's internal structure (by pushing the electron in "another orbit" if you will), while you certainly can't change the internal structure of the electron to change s. Would it be possible to change s then you could ... 11 Absolutely. If Compton scattering occurred in first order in e, the only contributing diagram would be the obvious one. Say we're in a frame with the electron initially at rest and an incoming photon in the z direction. Then the electron 4-momentum is$$p^\mu_{\text{in}} = (m,0,0,0)$$while the photon 4-momentum is$$k^\mu_{\text{in}} = ...

5

The E=mc^2 formula only applies to an object at rest, and light is never at rest. You want to use the more general formula: $E^2={m_0}^2c^4+p^2c^2$ Then you can set the mass to zero. $E=pc$ What this says is that light has momentum, which is related to its energy.

1

This is because instead of $$\dfrac{1}{2}mv^2$$ or $$E = mc^2$$ the energy of light is given by $$E = hf$$ Where h is a number called Planck's constant and f is frequency (sometimes v is used) Here is an example, as requested: Imagine red light with $620. nm$ wavelength. The frequency of this light is $0.483$ x $10^{15}Hz$ This makes the energy of a ...

2

We know that solar cells generate electricity by utilizing the energy of the photon, This is an every day language, electricity. It means things electrical in general every day language. but how does it generate electricity forever? What is generated when the photons hit any material, is heat, and the sun's energy is at maximum 1300Watts per ...

-2

But why is there no such interaction between the electrons spin $\vec{s}$ and photons? Actually, maybe there is. See Compton scattering. The incident photon is partially absorbed and decelerated in the vector sense, whilst the free electron moves. IMHO you can visualize this by drawing repeated circles on a piece of paper without lifting your pencil. Now ...

0

An LED is a collection of discrete atoms. When a current flows, an electron excites one every so often. The excited atom decays after a while and emits a photon. Assuming the current is uniform, variations in intensity come from randomness in arrival times and excited state lifetimes. Some time slices have many photons, some fewer. This distribution is ...

1

There is an equation that helps a lot understanding this issue: Fermi's golden rule $$W_{i\rightarrow f}=\frac{2\pi}{\hbar} \left|\left<f\right|H'\left| i\right> \right|^2 \rho$$ It describes the transition rates from one state to another. $\rho$ is the so called Density of States (DOS) of final states. This system has only two states: The initial ...

0

There is indeed a way wherein "one photon" can be thought of as a Maxwellian wave. We are dealing here with the quantized electromagnetic field. If the EM field is in a one photon state then one can compute two vector fields from the EM field's state that: Fulfill Maxwell's equations for freespace propagation and Uniquely define the one photon state. ...

2

The maxwellian wave is an emergent phenomenon from a great multitude of photons with the frequency of the maxwellian wave. This is explained in this blog entry by Lubos Motl. I will give you my experimentalist's interpretation of this: A photon as a quantum mechanical entity has a wavefunction. This wavefunction is a solution of a form of Maxwell's ...

0

It turns out that light can be thought of as a wave and a particle under different conditions. For example, as Cort Ammon described, the Double Slit experiment showed that light had properties of diffraction. On the other hand, the photoelectric effect considers light as packets of light called photons with certain energy $hf$. Another such experiment is the ...

0

Another way of thinking of this is to imagine that space-and-time is an emergent phenomenon, NOT a pre-existent framework that holds photons, be they waves or particles. If space/time emerges for example at the quantum level of individual photons, effectively there may be no immediate space and time there for a wave to propagate through. But at larger ...

0

Look into Wave-particle duality. It is a major part of Quantum Mechanics which answers your question. A quick summary: light is not just a wave, not just a particle. In some situations it behaves like a wave; in others it behaves like a particle. In some situations, neither wave nor particle sufficiently describe reality. A solid example of this is the ...

0

Good discussion of this question here. In short: because momentum p imparted by light with energy E is given by $E = pc$ then if we compare a photon "shot" to a bullet with the same energy the momentum generated by a photon gun is on the order of $10^{-6}$ the momentum of the traditional gun. But felt recoil is a tricky subject, because it incorporates ...

1

The negative potential works against the $K$ of the ejected electrons. So at very high negative voltages there is no photocurrent. As the voltage moves closer to zero, we hit the point where $f_2$ produces a current. If it's able to produce a current, this means that the electrons are ejected with a high kinetic energy, and thus the work function that binds ...

0

The simplified answer for this is that gravity basically means that you are in elevator which is accelerating: even if you are standing on the ground. So if you take that constant gravity field is pretty much the same as a constant acceleration, then if a beam of light bends downward in an accelerating elevator, it must bend downward if it is in an constant ...

2

In the two-dimensional rubber sheet visualization, it is wrong to think that things fall towards the massive object because they are "rolling down the hill" of the curved spacetime. There is no perpendicular gravity pulling things down into the well. What happens is that you are moving along your world line at a constant velocity, "into the future at one ...

0

Yes, photons are affected by those curves. They also curve space-time due to something called a stress-energy tensor. Remember that the energy of a photon is given by $$E = hf$$, so photons do have energy. This energy lets them be affected by gravity.

0

"Do two photons traveling in opposite directions emit gravitational waves?" Does virtual graviton count as gravitational waves? for sure there are gravitational effects so call it gravitational field. And two photons can interact indirectly through other fields. "If so, does it mean that any volume filled with photon gas will eventually degrade into ...

1

When you say the stopping voltage determines the kinetic energy of the electrons you have that the wrong way round. For a given kinetic energy of photoelectron you must apply a potential to bring it to rest. So it is true that increasing the intensity of the incident light will increase the photocurrent as you have more electrons ejected from the metal, ...

2

A photon that strikes the metal either has enough energy to release a valence electron from its bond with the atoms in the metal, from a certain frequency and up, or its energy is just not enough to eject the electron from the metal. The surplus of energy over the energy needed to eject the electron is used as kinetic energy of the electron and determines ...

2

I'm not sure about the “pick it up with a laser” part, but let's simplify by assuming that we can hit the golf ball from below the ground with photons. Let's further assume that the mass of the ball is $m_{ball} = 46\,\mathrm g$ and we want to hurl it five yards away ($\approx 4.6\,\mathrm m$). If the ground is level and we neglect air drag and wind, the ...

3

Hurl it several yards away Rather than a simple yes/no, let's figure out some numbers - as usual I work with "round" numbers. Mass of golf ball = 50 gram "several yards" = 5 meters Projectile launched at 45° with initial velocity $v$ travels for a time $$t = 2\frac{v}{\sqrt{2}g}$$ during which time it will travel a distance x = v_x t = ...

-1

If the ball was completely reflective, yes. The concept was discussed e.g. here http://en.wikipedia.org/wiki/Solar_sail edit: e.g.- radiation pressure is what keeps Sun from collapse

-1

Well, photons behave both as a wave and as particles, and the photons in coming from the beam of light "enlighten" the particles in front of you.

1

If the source is far away, light acquires a certain degree of coherence. Have a look at the Van Cittert–Zernike theorem, also in wikipedia: "the wavefront from an incoherent source will appear mostly coherent at large distances". The resulting fringes are different for different colors, but any color has max for straightforward direction. So, you see the ...

1

We know that light is massless so why does a black hole's gravity attract light? Because gravity doesn't just attract objects with mass. It alters the path of light too. Because gravity is caused by a concentration of energy which "conditions" the surrounding space, altering its metrical properties, whereupon we talk about spacetime curvature. But note that ...

1

The Coulomb "interaction" appears as the answer to a very specific question, even in QED, that is "what is the correction to the ground state energy of the electromagnetic field if two charges $q_1$ and $q_2$ are pinned at specific locations separated by a distance $r_{12}$?" . The answer to that question is exactly $\Delta E_0 = \frac{q_1q_2}{4 \pi ... 1 A photon has a rest mass of nought (where the rest mass$m$is the Lorentz-invariant quantity in the four-momentum's Minkowski norm squared$E^2/c^2 - p^2 = m^2 c^2$). However, a lightfield of energy$E$gravitates and itself has a gravitational source equivalent to a mass$E/c^2\$. Also, a system of photons has a nonzero rest mass (see reference), as does ...

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