Tag Info

New answers tagged

2

For now I will only give you an overview of the ideas involved and show you how you should interpret the idea of a "local realistic theory" that cannot exist at the microscopic scale. Once you've read it, and if you feel you need more mathematical rigor to be convinced, then I will draw you step by step the proof of Bell's inequality (it is not the only one ...


0

I don't know of polarizing x-ray beam splitters, and I suspect they don't exist, but I'm always very interested to be shown wrong. It appears from a cursory search that working x-ray interferometers have been built, and are conceptually more similar to neutron interferometers than to optical interferometers. For neutrons there are polarizing supermirrors, ...


3

Yes it is possible but as BarsMonster points out it isn't like an optical mirror. X-ray reflectors are used in the construction of nuclear weapons and are critical to increasing the yield. How they work is the initial fusion reaction releases high energy radiation, this is then reflected back into the reaction mass increasing the energy levels of the ...


7

Unfortunately X-ray and gamma mirrors are impossible to build the way you think - mainly because there is much less interaction with the matter comparing to UV - it will go through all materials commonly used for making mirrors. Even for EUV light (wavelength of 13.5nm) building effective mirrors is a royal pain. As wavelength of X-Rays is very small (down ...


0

That in mind, does this mean the light can be bent around or does the light loose energy when its waves rotate? This answer expands on the previous one by user3814483 . You must be familiar with the fact that light is composed of alternating electric and magnetic fields propagating in space whether there is a medium or not. The electromagnetic ...


1

The Faraday effect occurs because the phase speed of left-circularly polarized waves can be different than right-circularly polarized waves in certain media. You can think of a linearly polarized electromagnetic wave as a superposition of left and right hand circularly polarized waves. You can visualize how a left and right hand circularly polarized ...


3

That's right, running away from a gamma source fast enough would shift them into the visible portion of the spectrum. It goes without saying that he'd have to run quite fast: $$\frac{\lambda_{\rm obs}}{\lambda_{\rm emit}} = \sqrt{\frac{1+\frac{v}{c}}{1-\frac{v}{c}}}$$ Picking rough round numbers for gamma radiation at $\lambda=10\;\rm pm$ and visible at ...


1

Does it need to "slow down" to the new speed of light? Or does a new photon get generated? One has to keep clear in this case the difference between a photon and an electromagnetic wave. An individual photon is an elementary particle, its wavefunction is given by the quantum mechanical form of Maxwell's equations and the square of this wave function ...


8

The cosmic microwave background has a redshift of about $1100$, see here for instance. Keep in mind that the "surface of last scattering" that gives rise to the CMB in fact existed everywhere in space, it's just that the photons currently reaching us have $z\sim1100$.


2

From your question, I guess the double mirror configuration is just an example you thought of. I suppose your question actually is about if a photon can be trapped. Then basically yes. A device able to confine electromagnetic wave or light or photon is called cavity. You should understand a photon does not necessarily means a propagating plane wave. It can ...


0

The best complex dielectric mirrors, see http://en.wikipedia.org/wiki/Perfect_mirror may reflect up to 99.999 percent of the incident energy. The loss is about 1/100,000, so after 100,000 reflections, the total intensity decreases $e=2.718$ times or so. If the distance between the mirrors is 3 meters, the light travels 3 meters times 100,000 = 300,000 ...


0

Yes & No, You can however create an perfect mirror, which does not absorb any of the photons energy however its simply not possible or even feasible at this time to create such a device without energy being conserved in the photon, but it will however it will loose its energy due to Gravitational red shift after a long-time or Red shift due to moving in ...


4

There are many ways to detect X-rays. I will list just a few that are (or have been) used in medical imaging. Essentially there are several strategies, but it always involves stopping the radiation and using the energy released to effect a change - chemical or electrical - that can be detected. Photographic film. Exposure to Xrays has a similar effect to ...


0

I disagree with the other answer. I am pretty sure optical fibers (and other dielectric waveguides) do not have a cutoff wavelength. This is, in theory and within the limits of transparency of the medium, source beam quality, alignment, divergence, etc. All wavelengths can propagate! They do however have a single-mode cutoff wavelength, which is the ...


2

Light (all the electromagnetic radiations) is something like raindrops-each little lump of light is called photon-and if the light is all one color, all the "rain-drops" are the same size.$_1$ The size makes photons of visible light and other waves different. Credits: $_1$ Richard Feynman-QED, The strange Theory of light and Matter.


11

Yes, X-ray, UV, and even radio-waves ares made of photons. The differences is the Energy (or equivalently the wavelength). See the picture of the Electromagnetic spectra . The different nomination comes from the time of the discovery. Youre eyes can see the visible part. the radiowaves can be observes with antenna etc... The only differences is the way we ...


2

Any optical fibre - and any optical waveguide in general - has a cutoff wavelength (and therefore a cutoff frequency or a cutoff energy) because wether light is confined in the film region (guided modes) or escapes to the substrate (substrate-radiation modes) depends on the propagation constant, $\beta$, which is related to frequency trough the dispersion ...


0

John's answer is clear for the ensemble of photons that make up the electromagnetic wave If you are really asking how individual photons end up making the classical electromagnetic wave, whether reflected or not, you have to dig in further in to quantum electrodynamics. Lubos Motl has in his blog an entry of how classical waves emerge from a large ensemble ...


3

Spontaneous parametric down-conversion converts a single incoming photon to two outgoing photons. I think the article is saying that that if you measure one photon coming out there must be a second photon as well. The author is referring to the second as a heralded photon in the sense that measurement of the first photon is a sign that the second (heralded) ...


1

What makes you think an electron reflects photons as you have drawn? Electrons scatter photons in any direction, although not uniformly. (Examples: Thomson scattering, Rayleigh scattering, X-ray crystallography) The electron may absorb the photon for an arbitrary period of time, changing momentum and thus position, then release a photon of a different ...


0

It has been said that your second experiment is impossible. For me it is possible but can't be used to predict the position of the electron and its momentum. First consider that your "classic experiment" is true. In this experiment, the important fact is that one can't obtain the momentum of the electron when the photon angle is measured. Now, consider ...


4

Your observation is linked to the "Optical window in biological tissue". Like you already suspected, the absorption of blue light in tissue is higher than the absorption for red light. Best read the related wikipedia article, where all relevant effects are nicely illustrated. http://en.wikipedia.org/wiki/Optical_window_in_biological_tissue


8

First of all, the uncertainty principle is more than just disturbance of observation. From the Wikipedia article "Uncertainty principle": Historically, the uncertainty principle has been confused with a somewhat similar effect in physics, called the observer effect, which notes that measurements of certain systems cannot be made without affecting ...


2

The problem with your set up is that you have ignored the quantum mechanical nature of the photons. The photons are subject to the uncertainty principle as well as the electron, and so there is no way to send in 2 photons with precisely the same momentum at precisely the same time, and we can't guarantee they will scatter of in precisely the same way. The ...


0

Your classic experiment 1) is quantum mechanical. Electrons and photons are elementary particles and interact individually as quantum mechanical entities. The plot shows the elastic scattering of photon on electron, a computable process quantum mechanically thus the angular distribution is known. As we are talking quantum mechanics there is a probability ...


1

It is a tensor product. (At least it always was when I encountered such notations, I can't speak with authority about SPDC specifically) Let $\mathcal{H}_1$ be the Hilbert space of polarization states for a single photon. Then the space of states for a two photon system is $\mathcal{H}_2 = \mathcal{H}_1 \otimes \mathcal{H}_1$, and the state you consider in ...


2

Photons generate what we call Radiation Pressure. From wikipedia, http://en.wikipedia.org/wiki/Radiation_pressure, we get the equation: $$ P_{absorb} = \frac{E_f} {c} cos\space\alpha\\ \text{and} \space P_{reflect}=\frac{2E_f} {c} cos^2\space\alpha $$ Where:$P_{absorb}$ is the Radiation Pressure on an absorptive surface (in Pascals). $P_{reflect}$ is the ...


2

With appropriate lab equipment, you can derive extremely narrow pulse shapes. A typical setup involves splitting the incoming beam of light and interfering it with itself. By shifting one of the path lengths, you can observe the change in the diffraction pattern and calculate the pulse width. This won't work, of course, for a single event. For that case, ...


-1

When anything moves at the speed of light, all of our physical models break down. If you were to watch a spaceship speed up to the speed of light, you would see a clock on the ship slow down and come to a complete stop when it hit the speed of light (assuming you could even see it at this point). The ship would also contract so much in the direction of ...


1

This falls into the "terahertz band" which has been getting a lot of attention recently. I'm not an expert in that area, but I can suggest you search "terahertz optics". I note that Newport Corporation has a line of terahertz components, along with some notes on their use. And there are many more vendors.


1

It is not hard to imagine a toy universe in which different fundamental forces propagate at different speeds. However, a necessary consequence of that would be violations of lorentzian symmetry, and the ability to triangulate a preferred rest frame. Although I don't see a theoretical reason why these speeds need be the same (I might be missing something ...


0

Photons are absorbed and then re-emitted by the electrons in the material. I'm not sure what you mean by 'negligible'. To physicists working in optics, the absorption-to-re-emission stage is negligible, but it is studied by those working in attosecond physics.


3

There are several things that can happen, depending on the nature of the obstacle. The simplest case is if, classically, we think that the obstacle is not too much smaller than the wavelength of the light. Then, the light wave can be reflected, absorbed, or diffracted. Most people are familiar with reflection, the light more or less bounces off the ...


1

This is where virtual particles come into play. http://youtu.be/K6i-qE8AigE?t=3m23s Essentially you can think of these virtual particles as temporary photons as carriers that dont exactly behave ver well with conservation of energy. The field is full of these non-conservative carriers for a very brief instant as a function of the mass of the carrier ...


0

Well photons do not really interact, especially when seen as a thermal gas. Intuitively you can see across the room despite all the reflection from the walls because the photons coming from across the room do not interact with other ones reflected by other walls that cross its path. If you look at field theory you find that photons interact very weakly. ...


2

It's better to think of the deflection of the photon as an effect of its travel through curved spacetime. You can generally choose to analyze the problem in the rest frame of the massive object. In that case spacetime is curved in all directions around the object, and so the photon's path is deflected both as it approaches and as it recedes. If you want to ...


1

Any object that has mass just has a gravitational field around it and light (or particles, photons) just get refracted when it enters this field.



Top 50 recent answers are included