New answers tagged

4

A comment about coherence in general I find the definition of coherence as some sort of "unrelated phase" problematic for a couple of reasons: This formulation somewhat implies that coherence is discrete, i.e. there is incoherent and coherent. Of course that is not true, you can have a continuum partially coherent states. But what quantity are you going ...


2

Apart from the heating due to sound absorption, as per the comment by HolgerFiedler, I don't think you will find a mechanism that can radiate due to polarization effects in the medium. Any EM radiation would be at the frequency of your acoustic waves. With the difference between the speed of sound and the speed of light, that would be very-long-wave ...


0

There are lots of emission and absorption lines – at different energy scales. Atomic spectra, vibrational spectra, rotational spectra, and so on. For each of them, one may discuss millions of molecules in principle (most of the large number are organic molecules). The amount of data to cover "everything" is clearly unrealistically huge. But any subset of ...


0

The electric and magnetic fields are always in-phase if the wave can be treated as a plane wave (which simply means it cannot be too close to the source), and in vacuum or any medium with linear response, such as air. Boundary conditions of wave guides change this relationship, and must be solved for each specific case. If the wave guide is large enough, ...


1

The answer to this question is a lot like the answer to Why does the comb attract the pieces of papers if they're neutral? I'm guessing that your comb, which will be negatively charged, is not evenly charged across its width. So even though you are holding the center of the comb above the vane, the electric field between the comb and the ground (plane) ...


0

The four (metal) vanes of a Crooke's are attached to an axle. There is very little friction between the axle and its supports so a very small torque applied to the vanes would produce a noticeable change in the rotation of the vanes. My suggestion is that charges are induced on the vanes by the charged comb. Thus there is a net force of attraction between ...


0

Detecting the emitted photons from a charge accelerating in Earth's gravity looks like a hopeless task, given that the radiated power is tiny (see formula 9 in Anna's answer). But we can try to exploit the fact that the number of emitted photons always diverges no matter how small the acceleration. While very low energy photons (so-called "soft photons") ...


1

Clarifications There is a difference between a solar flare and a the phenomena that cause things like the 1989 Quebec blackout and/or the 1970s New York blackout (I think it was 1972 but do not recall off hand). The latter phenomena are called coronal mass ejections (CMEs) because they actually involve large amounts (i.e., upwards of billions of tons) of ...


1

Just to confirm that CuriousOne is correct as regards direct radiation damage to your body. From Solar Flares Solar flares are gigantic explosions associated with sunspots, caused by the sudden release of energy from “twists” in the sun’s magnetic field. They are intense bursts of radiation that can last for anywhere from minutes to hours. Solar flares ...


1

There is some misunderstanding here. whether a freely falling charge radiate photons, how strongly and relative to which frame of reference it does or does not radiate if you mean a charge in free fall. In this calculation:, from the conclusion It is found that the "naive" conclusion from the principle of equivalence - that a freely falling ...


3

The radiation emitted by an accelerated charge depends on the boundary conditions on the fields at infinity. When one takes this into account properly, then accelerated observers will agree with inertial observers about the emitted radiation (after trivial transforms are applied). Any treatment which purports to show that in the accelerated observer's frame ...


2

Assuming that Classical Electrodynamics (Maxwell's Equations) holds, the answer is that the inertial observer would see the radiation while the non-inertial observer would NOT. The question you are asking is basically the following paradox: https://en.wikipedia.org/wiki/Paradox_of_a_charge_in_a_gravitational_field This paradox has been analyzed and ...


9

The force of radiation pressure is $F=P/c$ for absorbed radiation or $F=2P/c$ for reflected radiation, with $P$ the power and $c$ the speed of light. If you want to generate 750 N of force, you need (a) a radiation source of 100 GW and (b) a mirror that reflects so well that nor the mirror, nor you will be vaporized in an instant. Update: CuriousOne ...


0

Stainless steel is steel with > 10.5% chromium. Since chromium has a lower atomic number than iron, it isn't obvious that adding it would improve the absorption coefficient of the alloy - which depends in part on density.


1

The oscillating fields of the electromagnetic wave just add linearly with the static electric or magnetic fields. Nothing much happens really, and the wave goes on its merry way.


-1

Iron-57 is a stable isotope , so it does not emit any gamma ray . You measure a prompt gamma from a nuclear reaction ( n , gamma ) .


2

I don't know the origin of either convention, so I am reluctant to make any absolute statements, but as an experimenter the question I ask myself is "What material has I learned is present in the sample?" or perhaps "What material do I bring into the lab to observe this line?". The answer to these questions would be Berylium-7 in the former case and ...


4

There are some issues with the experimental setup you proposed (apart from the fact that when its temperature is lowered the gas would become a liquid and then a solid - if it's not $^4$He: in that case it will stay a liquid). Let's see why. In the picture above, I've sketched your experimental setup. The black box must be impermeable to matter in order ...


4

It won't work because your perfect vacuum is permeated by the cosmic background radiation, which itself is only asymptotically reducing to zero with the expansion of the universe. Trying to exclude the cosmic background radiation backs you into the infinite steps that forms the basis of the third law again. Also, using a container results in quantum ...


3

(the following answer is included essentially in "The Feynman LECTURES ON PHYSICS-Mechanics, Radiation & Heat ,Vol. 1, 26-3 Fermat's principle of least time.) Suppose you are at point A in the land and a screaming girl is at point B in the sea. You can run with a speed $\:v_{1}\:$ on the land greater than the speed $\:v_{2}\:$ you can swim in the ...


10

You seem to make the implicit assumption that your vessel is placed in an environment that does not emit any thermal radiation, i.e. is already at 0 K temperature. The temperature of your container will asymptotically decrease to 0 K but will never actually reach it. Assuming black-body radiation, fixed heat capacity $c$, and sufficient thermal ...


0

If you are talking about a spherically symmetric em wave created by a point source then the intensity of the wave (namely the power transferred through unit area) drops according to the inverse square law (this follows because the total power radiated over a sphere of fixed radius remains the same, so as you increase the size of the sphere the power per unit ...


0

Charged particle is accompanied with EM radiation (has field that falls with distance as $1/r$) when it moves with acceleration. This can be shown to be a consequence of Maxwell's equations, well-verified and reliable part of physics. It is immaterial whether the charged particle is a point or an extended body. None of that depends on the value that ...


1

Wavelength is used as a convenience. It's much easier to imagine a photon with a 500 nm wavelength than to comprehend a photon oscillating 600 trillion times per second. But in reality that's all it is is a photon moving at the speed of light and oscillating 600 trillion times per second as it goes along. The photon completes one cycle every 500 nm. Many on ...


2

Hardware hacker CNLohr did a nice time lapse collection of signal strength mapping a 4 foot square in his house, and then a 3d cube with the help of a CNC router table. I saw it on hackaday, his project is here: https://hackaday.io/project/4329-wifi-power-mapping And he links to a cool video here: https://www.youtube.com/watch?v=aqqEYz38ens It pretty much ...


1

The far-field strength, $B$, of a dipole magnetic field has an angular dependence and hence so does the energy density (proportional to $B^2$). If the field simply rotated with the magnet then regions of high/low energy density would also rotate. Suppose the magnet is rotating once per second. Then at distances >>1 light-second we would have energy moving ...


1

Hint : A key to the solution is what is meant by the complex wave 3-vector $\:\mathbf{k}\:$. This vector is not any complex 3-vector in $\: \mathbb{C}^{3}\:$ $$ \mathbf{k} \ne \left(k_{1}, k_{2}, k_{3} \right) \in \mathbb{C}^{3}, \:\:\text{that is with} \:\: k_{\rho} \in \mathbb{C} \tag{a-01} $$ but $$ \mathbf{k}=\left(k_{1}, k_{2}, k_{3} \right) ...


6

I think your mental picture is pretty close to accurate, as long as you bear a few things in mind: First, the wavelength of the wireless signals are much longer than visible light. At 2.4GHz, the wavelength is 12.5cm. Just imagine that the waves are about half a foot long (if you have 5GHz wireless, the waves are half as long). So you can get some ...


3

If the absorptivity of a medium really was discrete, then there would be no way it could emit blackbody radiation. The defining characteristic of a blackbody is that it absorbs light of all frequencies that are incident upon it (and that it is in thermal equilibrium). There is a close relationship (a direct proportionality) between the Einstein absorption ...


1

As you said, vibrational and rotational transitions are also possible, and I believe that the energy differences involved there are enough to have a quasi-continuous absorption spectrum in most real-life scenarios (of course, in real life you will never have perfect absorption at all wavelengths). In the following picture, you can see two energy wells ...


0

A black-body can absorb all the radiation falling on it(light at all wavelengths) and appears black when cold. When it gets heated it can emit radiation at all wavelengths like a heated piece of metal. The hotter it gets the higher the photon frequency (energy) and so a shorter wavelength. Hotter objects emit more total radiation per unit surface area. This ...


0

The edge of the doors and area around the door on the microwave oven side when closed during operation act like microwave notch filter. The notch filter is tuned for 2450 Mhz. Any deformation of the door by a dent of a insignificant amount could cause the microwave to leak rf energy as it changes the notch frequency. If you are in front of it at the ...


2

There are several kinds of "UV Light." Only kind I know of that "looks purple" is a so-called black light. Black lights emit UV that is very close to the top-end of the visible spectrum. The designers try to minimise the visible radiation so that it won't wash out the light emitted by fluorescent substances in the field, but it's hard to filter all of the ...


2

The uv light often originates from excited Mercury atoms which are in the bulb. Excited Mercury atoms also produces violet and blue visible light.


0

You probably managed to get only one hotspot of maximum wave amplitude hitting your coffee while two is more usual. Putting your cup off-centre on the rotating dish would probably get the more average heating you expected.


0

Here is an argument that Casimir force is really van der Waals force, and not a force that originates from vacuum energy: http://lanl.arxiv.org/abs/1605.04143 In short, vacuum energy originates from the pure electromagnetic term in the Hamiltonian, which does not have any explicit dependence on matter degrees of freedom and hence cannot generate any forces ...


42

It's more like the walls were semi-transparent glass, if you want to imagine it as light (and even then, you neglect diffraction effects). It would actually be better to imagine it as sound! But this seems to be exactly what you're looking for: http://arstechnica.com/gadgets/2014/08/mapping-wi-fi-dead-zones-with-physics-and-gifs/


21

It's hard to access how 'accurate' an analogy is (i.e. how is this being quantified?). But, I think, there is a simple - better analogy: WiFi is more like sound in a house. The transmitter is a speaker. If its a good, loud speaker, you will still easily be able to hear it in the next room - through a wall. A few walls inbetween and it gets very faint. ...


0

Your understanding of a real caustic (I presume you call it real as opposed to the imaginary caustic that you mention later) is correct. First the easy part: an imaginary caustic is a caustic located on the extension of the light rays beyond the optical system from which they arrive. For instance, in the presence of a convex lens, imaginary caustics may ...


3

In Wikipedia the caustic is defined as follows. In optics, a caustic is the envelope of light rays reflected or refracted by a curved surface or object, or the projection of that envelope of rays on another surface. You can think of the envelope of a family of curves as a curve that is a tangent to each of them. Here is a diagram on page 60 of "A Treatise ...


0

I think I've figured out why. It's simply because the critical density $n_{crit}$ is a function of $\omega$. The critical density of the plasma is that which is required for the EM wave frequency to equal the plasma frequency $\omega_p$, so is dependent on the frequency of the radiation. So $$\omega=ck\left(1-\frac{n_e}{n_{crit}(\omega)}\right)^{-1/2}$$ ...


0

When the phase velocity is a constant (with respect to wavelength), the group velocity will indeed be equal to it, as you yourself have shown. What you've got wrong here is the assumption for this case that the product of the phase velocity and group velocity equal the square of the speed of light, which can be true in other cases but not for plasma ...


-2

I have been wandering that same question for 40 years...I have asked all kinds of doctors and researchers, only to see a glaze form over their eyes. The question, I think, is not whether it exists, but how do you measure it. The problem is figuring out the machine that can measure it. And then, where to measure, because I think there will be not only an ...


2

High pressure gases do not produce a true continuum, and they do not become closer to the blackbody case with increasing pressure. The individual emission lines only become broader with pressure and make the emission spectrum appear continuous because the lines start to overlap. The number of molecular transitions does not change with pressure. There is a ...


1

I'm not sure about that particular equation. There's always issues with how quantities are defined, and signs of particular values. To understand what that equation is saying exactly, you have to look at the derivation. So to solve this problem, I'm going back to some basic electromagnetic relations for a plane wave. $$E = c B$$ $$c = ...


-1

There are no known transistors or oscillators yet produced that can handle 10^19 frequancies, we are still experimenting at terahertz range 10^12 or thereabouts so the simple answer is no.


0

It is clearly acausal that for a photon to be emitted it must be absorbed. That's action at a distance. That is not what time symmetry means. That view is known to not be correct, and there are no problems with advanced and retarded potentials, it is known what they mean and how to deal with them, and it is not this. Wheeler and Feynmans paper is clearly ...


3

This sounds like the "retrocausation" in the Wheeler–Feynman absorber theory. Since the only invariant quantity in relativity is the relativistic interval, which is zero along light like curves, all "place-instants" of photon's existence are technically not separated from each other in the (pseudo) metric, and hence causal, sense. This means that photon ...


1

You haven't seen these effects because your eyes are rather insensitive to wavelength changes. A moderately resolving spectrometer can detect these changes quite easily. It's called "Stark effect" and it can be observed in atomic spectroscopy: https://en.wikipedia.org/wiki/Stark_effect. For magnetic fields the analog is called "Zeeman effect": ...



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