# Tag Info

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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 ...

5

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:

4

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 ...

3

That could be a positron annihilation peak. If the halo of your beam is scraping somewhere there will be a variety of interactions between the particles and matter. If it is an electron beam then many of the interactions will be electromagnetic showers which will produce (energy allowing) some $e^+$--$e^-$ pairs. When the positrons annihilate they will ...

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 ...

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 ...

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

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 ...

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

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, ...

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 ...

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 ...

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 ...

2

Photoproduction is a process where something is produced by the interaction of a high energy photon. Something like $$\gamma + p \to p + \pi^0 \,.$$ Experimentally it is useful because the electromagnetic vertex is well understood, and photon taggers allow the creation of incident photons with well know energy and momentum.

1

First, what is excitation? "Excitation is an elevation in energy level above an arbitrary baseline energy state". So, for atom to get excited, it needs to absorb some amount of energy, which is described by energy levels for each element. When atom "come in contact with photons", it is usually photon having certain energy that, if enough to move shell's ...

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 ...

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 ...

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 ...

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

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 ...

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

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 ...

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