# Tag Info

## New answers tagged photons

0

Schroedinger solved this problem in 1927. The trick is to realize that in quantum mechanics, momentum is proportional to wave number (or inverse wavelength). So the COM system for an electron and a photon is the system where they have the same wavelength. The beauty of doing this in a COM system is that you have the incoming electron and the outgoing ...

-1

Because a photon is a packet of electromagnetic energy, so it must be interacted by the same field or any of electric or magnetic field. As nucleus has a certain electric field by which photon's field actually interacts and the result of this interaction is pair production-to conserve energy momentum and charge.

1

The laws of conservations of momentum and energy combined forbid the reaction $$e^- + \gamma \rightarrow e^-$$ (Go ahead and do the math, is simple and enlightnening). But a completely different story is: $$e^- + \gamma \rightarrow e^- + \gamma$$ Where the incoming photon has a different energy that the outcoming one. And also, you can have an ...

0

"This illustration shows a Feynman diagram of a simple example of scattering. Two fast-moving electrons (e-) deflect each other via the electromagnetic force (represented in the diagram by the squiggly line of a photon)." @dmckee So that would be electrons interacting with each other via interaction with a virtual photon? ...

0

When you say the particles cannot interact, yes it will take at least 100 photons to excite all the particles. You could have one particle absorb a photon, then radiate a lower energy photon that is absorbed by another particle but you have ruled that out by saying no interaction.

0

"Other sources say that photons create electron-hole pairs.": Electrons have mass. Creating them even if possible would need a great deal of energy. So this is not what happens when a photon creates electric current in a substance.

1

Does it gain infinite momentum before it crosses the horizon? Momentum is frame dependent so, when asking for the momentum, one must specify according to whom? Since the Schwarzschild metric is independent of time, the time component of the four-momentum of freely falling particle is constant. $$p_0 = -E$$ Now, imagine that the particle is at some ...

3

When you're asking a question about general relativity you need to state what coordinates you want to use. This isn't just a mathematical nicety - as you'll see shortly, the different coordinate systems attached to different observers will describe very different behviours. The obvious interpretation of your question is to ask what happens when an observer ...

1

Here is another hypothetical (i.e. extremely impracticable) answer to your question that is rather interesting (althgough Aksakal's Answer is likely to be a bit more practical!). You have to imagine yourself to be a very deft light-catcher with mirrors (I can't help thinking here of Mozart the Light Catcher). You trap light in the box by suddenly (within ...

0

short answer is yes from a certain energy level to another. it's not that certain, every photon will have slightly different energy. it's a long story, but excited state of the atom is influenced by the vacuum around, so the spectral peak will always have some width. look at this spectrum this is gas inside the lamp. each peak has a width. when the ...

1

The light will die out quickly. Think of playing B tone on a string tuned to A. It's pretty much the same thing. Also, 3m wave is not light, it's VHF used in TV UPDATE: In the sound analogy, if you attach B tone generator to A-tuned string, as @WetSavannaAnimal suggested, there will be a B tone wave on a string, but it will be only at and around the point ...

1

Usually the Newtonian limit is described as taking $v << c$ but a much better way to express it is saying that the kinetic energy is much less than the rest energy $$\frac{1}{2}m v^2 << m c^2$$ of course this runs into trouble when we talk about photons since we don't have a well defined concept of velocity, in the Newtonian sense. This is ...

1

Like KsdLingen said a photon does not really have a length or size. You could ague that this is due to its wave-particle duality (a concept from quantum mechanics). The wavelength of a photon, $\lambda$, indicates what distance it will travel in vacuum while its electromagnetic field completes one period. The direction of these fields are always ...

2

The wavelength of light, and for any wave in general, is measured along the direction of propagation. It has every bit of the intuitive meaning that the wavelength of a water surface wave does. One of the most meaningful ways to visualize light is as an oscillation of the electric and magnetic fields over space and time: (Image source) The electric ...

1

The 'length' is indeed measured in the direction the wave is traveling. A lightwave is a transverse wave, consisting of an electric and a magnetic field. A photon can be imagined as a localized wave. Length and shape of a photon are meaningless concepts

1

For a particle of fixed mass $m$ moving in a fixed gravitational potential $\phi(\vec{r})$ the motion is independent of the mass of the particle. The equations are $$\vec{F}=-m\nabla\phi$$ and $$\vec{F} = \frac{d\vec{p}}{dt} = m \frac{d\vec{v}}{dt}$$ It's clear that the $m$'s cancel when combining these equations. So from this point of view it doesn't ...

2

When you say "without altering the actual momentum of it" is that really true? $$E^2 = p^2c^2 + m^2c^4$$ so for a photon $E = pc$, since rest mass is zero. Now according to your first "traditional" calculation of m, we would have $E = pc = m_1c^2$, and therefore $p=m_1c$, where $m_1$ is mass according to the first "traditional" calculation. For your ...

3

I understand that one can measure a single photon being absorbed using a photomultiplier tube or CCD. Can one measure a single photon being emitted by monitoring the current through an LED or the recoil of an emitting ion? The photon is a particle. It will have particle interactions, i.e. scattering off electrons and/or the spill over electric and ...

0

The master equation used in metamaterial calculations is usually phrased in its most general form as the eigenvalue equation $$\Theta \mathbf{H}=\left(\frac{\omega}{c}\right)^2\mathbf{H}$$ where $$\Theta=\nabla\times\frac{1}{\epsilon(\mathbf{r})}\nabla\times$$ is a linear operator which typically has some translational symmetry property due to some sort of ...

3

Yes, there is a physical significance. The longitudinal mode $A^0$ is pure gauge, it does not propagate (in other words, the equation of motion for $A^0$ is a constraint [Gauss Law], not an equation of motion and it's canonical momenta is identically 0 , meaning we cannot impose canonical commutation relations on it). Some of the spatial modes do propagate, ...

3

There are a couple of issues you might want to consider. Firstly there is the slightly boring one that we physicists measuring the mass of the black hole are outside it, and from this position the photon never reaches the event horizon let alone crosses it. I don't want to go into this here since the subject has been flogged to death in numerous questions ...

3

Please tell me what I did wrong It takes General Relativity (GR) to describe black holes and, in GR, energy conservation is, well, subtle. From John Baez's Relativity FAQ "Is Energy Conserved in General Relativity?": In special cases, yes. In general — it depends on what you mean by "energy", and what you mean by "conserved". So, in general, ...

3

I'm not sure I've understood your question but I think you're asking if a big wave can have wave-features on its large features. If so, sure, why not? You can add waves of different frequencies to achieve results like:

1

String theories respect symmetries of the 4d Poincare group, including those that result in special relativity. As such, faster than light particles are expected to be absent in nature, if string theory is correct.

2

In particle interactions the total number of particles is not conserved. For example in a collision in the LHC two photons collide and many hundreds of particles are created in the collision. There are still some conserved quantities, for example lepton number is still conserved so you cannot just create an electron. You need to create an electron and ...

0

First: momentum is conserved, energy is conserved, electric charge is conserved, and a few other things. Particles and particle number can, in some instances, flit into and out of existence. The photon is absorbed by the electron. I.e. the photon ceases to exist, its energy and momentum have to go somewhere (the electron and the bulk material). In the ...

0

Have you read the wiki link electrons are only dislodged by the photoelectric effect if light reaches or exceeds a threshold frequency, below which no electrons can be emitted from the metal regardless of the amplitude and temporal length of exposure of light. To make sense of the fact that light can eject electrons even if its intensity is low, Albert ...

3

The other answers to the effect that one needs big optics to see fine detail are indeed true for are true for conventional imaging optics that sense the electromagnetic farfield or radiative field i.e. that whose Fourier component at frequency $\omega$ can be represented as a linear superposition of plane waves with real-valued wave-vectors ...

7

Photons or cosmic rays don't (normally) emit gravity waves. Consider the comparison with radio waves. A moving electron doesn't emit radio waves. It has to be accelerating to emit EM radiation. Specifically radio waves are only emitted when there is a changing dipole moment. So you wouldn't expect a particle moving at constant velocity (photon or ...

3

In light of this, why do photons traveling from the most distant reaches of the observable universe not lose energy due to the gravitational radiation they must emit? There is a misconception here in "gravitational radiation they must emit" . There does not yet exist a unified theory of elementary particles and the three interactions well described by ...

3

If you read the wikipedia article on orbital angular momentum of light you will see that in the first place it is a classical electromagnetic concept, where the light has a vorticity, i.e. a helical motion around the axis of the vortex. When one goes to the quantum detail of photons one can define an OAM against this classical axis for each photon in this ...

-2

Since this question is about how a photon can travel at light speed and yet have no mass, I will answer by saying that photons having no mass is precisely why they can travel so fast, and without mass, it becomes intangible for anything to make it go slower.

1

It's better to know about wave particle duality before going to your question. Lets know what Broglie says in his noble lecture (December 12, 1929): This is the extracted passage which makes an attempt to say the importance of both wave and particle nature. The existence of a granular structure of light and of other radiations was confirmed by ...

0

You are confusing terms. Photons have energies, and waves have frequencies. Generally (and not so accurately), a wave function can be expressed as a superposition of mutually orthogonal wave function, called eigen-functions, each associated with an eigen-value (energy in your case), so that when a measurement (yours, or some interaction with an appropriate ...

Top 50 recent answers are included