# Tag Info

1

The phosphors lining the glass tube of a fluorescent light do a pretty good job of smearing the atomic mercury line emission spectrum into something closer to black body radiation. There could be a phosphor mix that accomplishes what you want...

1

When you scatter light off of a material there is a photon-phonon interaction which will shift the photon frequency depending on the phonon energy (Raman scattering, Brillouin scattering). The effect is quite small, however. How much broadening do you need? Rayleigh scattering through a warm, high density gas will probably go quite far in messing up the ...

1

This might not be feasible for your setup, but you could try rapidly rotating your light source, which would Doppler broaden your spectral lines. You're correct in that the speed would need to be a substantial with respect to the speed of light. As an example, if your frequency is 500nm let's say. If you'd like it to spead out on the order of a single ...

2

In 1995 Willis Lamb published a provocative article with the title "Anti-photon", Appl. Phys. B 60, 77-84 (1995). As Lamb was one of the great pioneers of 20th century physics it is not easy to dismiss him as an old crank. He writes in the introductory paragraph: "The photon concepts as used by a high percentage of the laser community have no scientific ...

7

Both the wave theory of light and the particle theory of light are approximations to a deeper theory called Quantum Electrodynamics (QED for short). Light is not a wave nor a particle but instead it is an excitation in a quantum field. QED is a complicated theory, so while it is possible to do calculations directly in QED we often find it simpler to use an ...

3

In this link there exists a mathematical explanation of how an ensemble of photons of frequency nu and energy E=h*nu end up building coherently the classical electromagnetic wave of frequency nu. It is not simple to follow if one does not have the mathematical background. Conceptually watching the build up of interference fringes from single photons in a ...

1

In Heisenberg picture states (kets) do not depend on time. But the so-called effective wave function here must clearly depend on time, seeing how it is defined. Think of nonrelatvistic QM, where the value $\psi\left(\mathbf{x},t\right)$ of a wave function at some point in space and time does not depend on whether you're working in Schrödinger or Heisenberg ...

3

If basic symmetry and homogeneity assumptions about the Universe hold, then yes, all massless real particles (see Anna V's answer for virtual particles must travel at a universal constant $c$, the speed of a massless particle, in all frames of reference. Given these basic symmetry and homogeneity assumptions, one can derive the possible co-ordinate ...

5

Another way to say this: Speed of photon, graviton, gluon all equal to c? or Whether all massless particles necessarily have the same speed? You must not have been introduced to the concept of a virtual particle: In physics, a virtual particle is a transient fluctuation that exhibits many of the characteristics of an ordinary particle, but that ...

1

I thought the same thing for a long time. I wondered why gluons don't fly out of the nucleus at the speed of $c$. The difference is that photons don't interact with other photons and gravitons don't interact with other gravitons. They can move around and pass through each other. On the other hand, gluons do interact with each other. In fact, gluons form ...

1

In many cases the particle interpretation is perfectly right. It's known as the path integral formulation. Basically, what you do is consider that when a particle travels from point $A$ to point $B$, it goes through every possible trajectory (including back and forth in time) all at the same time! In fact, every paths have the same probability, they just ...

2

Photons have some conditions to have an evanescent wave, e.g. total internal reflection. Suppose we have some material with index of refraction $n_1$ and a layer of another material, with smaller $n_2<n_1$. At some angle we'll see total internal reflection, i.e. when the light totally reflects, but leaves some exponentially decaying trails in layer with ...

0

The potential barrier problem and solution in quantum mechanics is discussed within the solutions of Schrodineger's equation in which there exist potentials, and the solutions of the equations with the boundary conditions give the wave function of a particle, i.e an entity with a mass. In addition it is a non relativistic equation.Thus in this framework: ...

0

You are probably referring to Bose-Einstein condensates, commonly abbreviated BEC. If not, please comment and clarify your question. The article you likely have heard of is Hau et al., Nature 397, 594-598 (1999): "Light speed reduction to 17 metres per second in an ultracold atomic gas," which translates to about 38 mph. In fact, you will find that the ...

6

It is incorrect to say that the energy of a string directly gives us the mass of the particle. While it is true that more the oscillations on the string, higher the mass, the relation between the oscillations and the mass it not that of a simple proportionality. What's really happening is that the string has some energy $E$ (due to oscillations on it) and a ...

2

Yes, you are basicaly right. The Poynting vector gives you the momentum of the the EM wave. At the quantum level, it is an operator of the form (see page 7 of http://www.physics.usu.edu/torre/3700_Spring_2013/What_is_a_photon.pdf) : $$\hat{\mathbf{P}} =\sum_\mathbf{k} \mathbf{k}\, \hat a^\dagger_\mathbf{k} \hat a_\mathbf{k}$$ for a given polarisation (here ...

0

I think you also have to consider that in the real world, the photon would travel a less linear path than the neutrino. This is due to things like gravitational lensing and any particles the photon interacts with any particles. Thinking of the super nova, how long does it take a photon to get from the center of a star to the outer most layers vs. a neutrino? ...

0

One common way that this happens is through spontaneous parametric down-conversion. From Wiki: an important process in quantum optics, used especially as a source of entangled photon pairs, and of single photons. A nonlinear crystal is used to split photons into pairs of photons that, in accordance with the law of conservation of energy, have combined ...

1

If $\Delta \lambda$ is much smaller compared to either $\lambda_1$ or $\lambda_2$(it doesn't really matter, it should be much smaller than both), then we can make the following approximation: $$|\Delta E|= \left|\Delta \left( \frac {hc}{\lambda}\right)\right|\approx\left|\frac {hc\Delta\lambda}{\lambda^2}\right|$$

5

The field lines in your drawing are not the trajectories of photons. The field lines show the direction of the force on a test magnetic dipole. The force, and therefore its direction, is mediated by virtual photons (or can be described that way) but those photons will travel in straight lines just like ordinary photons.

0

If we assume that mirrors will leak some energy, then is it possible to put objects such as a photo multiplier tube (in combination with a mirror) and adjust it in such a way that only the amount of energy lost by reflection of the mirror is recovered and sent back to the other mirror. This cannot be done even as a thought experiment. Photomultipliers ...

0

A particle's interaction(with anything it can interact with) can be thought of as ,it making a measurement (of the physical quantity associated with the interaction-eg electric field in case of charged particles interaction)and acting accordingly This isn't the correct way to think about interactions, and as you later point out this (kind of) violates ...

1

The best account for photon emission when an electron drops to a lower eigenstate is the Wigner-Weisskopf Model for spontaneous emission, see this paper from the Photonics group at ETH Zürich and the co-efficients for this model can be calculated by standard quantum electrodynamics. I explain this model further and give references in my answer here. A ...

2

Under normal circumstances, what you are seeing is the steady state condition where the rate of absorption is equal to the rate at which energy is conducted away or radiated away again, so the material doesn't heat up. As I understand, unless a material fluoresces, the de-excitation happens in the infrared. Also, you should realize that just because a ...

1

I wonder if you're getting mixed up with propagation of waves in a physical medium like a string. If you have a wave travelling on a string then it has a velocity along the string, but the string is also oscillating normal to its length. So if you stretched the string along the $z$ axis, as the wave travelled along the string (i.e. the $z$ axis) the string ...

0

A photon is the quantized unit of the electromagnetic field. If you have en electromagnetic wave propagating in the x-direction, this must consist of a magnetic field and an electric field oscillating perpendicularly to the direction of travel, and to each other, i.e. in the y and z directions. If you have a wave with a frequency of, as an example, 50Hz, it ...

0

You're confusing the process of quantization with the wave-nature of propagating electromagnetic fields. When you look at a Electromagnetic waves as photons, this means you don't look at their wave-characteristics, and you consider them as particles travelling with the speed of light, and those particle could "hit" electrons and knock them our of the atom ...

11

The only stars you can reliably see are ones that are spewing enough photons at your eyeballs to appear stable. Any star which is so dim that photons entering your eye can literally be counted one by one, simply will not register in your vision, because your eye's retina is not sensitive enough. So your question is basically embroiled in observer bias; it ...

23

Although I agree with all three of the above answers let me present a slightly different perspective on the problem. It's tempting to think of the light from the star as a flood of photons that behave like little bullets. However this is oversimplified because a photon is a localised object i.e. we observe a photon when something interacts with the light ...

45

The answer is simple: Yes, stars really do produce that many photons. This calculation is a solid (though very rough) approximation that a star the size of the sun might emit about $10^{45}$ visible photons per second (1 followed by 45 zeros, a billion billion billion billion billion photons). You can do the calculation: If you're 10 light-years away from ...

8

Allow me to channel something akin to the anthropic principle here. You can only see the stars that have a lot of photons reaching your eye. If a star were so far away that photons were reaching your eyes only occasionally then the star would be too dim for you to see in in the first place. Even if you could see the photons, the star would appear to ...

5

A star radiates in all directions. You would still see the star regardless of the number of steps you take to any side, just not the same photons. A laser radiates in only one direction (or in a very small cone). If you took a large enough step to the side (larger than the angular size of the emitted beam) so as to exit this cone, then you would no longer ...

1

In the specific case of slowing light with a Bose-Einstein condensate there will be a limit because the slowing of the light is due to an interaction of the light with the BEC to form a polariton. If you put too much energy in you'll destroy the BEC and it will stop slowing the light. Offhand I don't know what the limit is, but it will be a very small amount ...

7

I will turn my comment into an answer, because the question in the header: Do photons age? is very anthropomorphic , and physics is a discipline that discourages interpreting data by use of the anthropic principle. The photon is an elementary particle. Aging is not a verb to be used with elementary particles in general because a) they have no ...

18

Photons don't have a rest frame, since in all inertial frames they must go at the speed of light. So the following statement: By that logic, photons don't age in a vacuum state as, to us, the time stops for them. is meaningless because one really can't talk about proper time for a photon. However, in a medium, their speed decreases, Nope. The ...

-2

photons can get absorbed by electrons of course! good questions for 1 and 2 it is infinte like Ron said for 3 it is more the acceleration of the elctron that emits photons. Speed is relative. But if you set up an experiment and give higher speed to some elctrons then they will some time get stonger acceleration and thus emit more photons 4- confusing ...

0

As other answers have mentioned light in classical electricity and magnetism is described very well by the solutions of Maxwell's equations which combine electrostatics and magnetism and describe a traveling wave of energy propagating at a speed c. This speed is not arbitrary but, as it comes out from the equations, depends on the electric and magnetic ...

0

Light travels as a wave and interacts as a particle so to be pedantic there is no way to answer the questuon.

7

Your confusion arises because the light is not a photon and it's not a wave. It's a quantum field, specifically the photon field, and this quantum field can interact with other matter in particle like or wave like ways. My preferred way of thinking about this is that the photon is the unit of interaction of the quantum field with something else, so the ...

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