Hot answers tagged

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. ...


19

It is a matter of definition of "same". Classically one can define "same" condition of particles by labels stuck on them. Light classically is a wave, and same needs a new definition. We apply the everyday definition by identifying the light beam with the source. The light leaving the sun is the same light arriving on earth. The light reflected from the ...


14

As many have said, the inverse square law applies to point-sources. These are idealized light sources which are sufficiently small compared to the rest of the geometry that their size is of no importance. If a light source is larger, it is typically modeled as a collection of idealized light sources, potentially using integration. The exact definition of ...


12

Your question is based on the assumption that a photon is a fundamental object i.e. that photons are something we can point to and say here is photon 1, here is photon 2, and so on. The trouble is that quantum field theory particles are somewhat elusive objects. This is particularly so for particles like photons that are their own antiparticles because such ...


12

The inverse square law applies to point sources. For extended sources becomes accurate at distances that are large compared to the size of the source. At large distances the source looks like a point. What "large" means depend on the application. In the case of light fixtures, the Illuminating Engineering Society and other organizations have made ...


12

Not really. The "speed of light" has very little to do with light; it is built into the actual geometry of spacetime independent of what matter fills it. In particular, $\epsilon_0$ and $\mu_0$ don't tell us anything physical about the vacuum; looking at the (simplified) expressions $$E = \frac{1}{4\pi \epsilon_0} \frac{q}{r^2}, \quad B = ...


11

Photons are boson, so it follows the Bose-Einstein statistics which is only true if the particles are truly indistinguishable. If you can distinguish between two photons, then it will follow the classical Boltzmann statistics which is not what happen in experiments. That means photons with same properties are the same. Even in your situation with photon ...


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 ...


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 ...


7

EM waves are a special case of electromagnetic radiation, where typically the source is periodic, or near enough that there is a carrier wave, as with radio and television. Maxwell's equations support a "sourceless" electromagnetic wave, as if it has existed forever. Also see Why do we think of light as a wave? Let's consider two cases of electromagnetic ...


6

The inverse square law applies to point sources. A real emergency light is not a point source, and therefore the law appears to not apply at close distances, because any real point is at a varying distance from different parts of the emergency light.


6

When a high energetic photon (like the gamma or X ray photon) hit a charged particle like an electron, due to inelastic collision, the photon loses some energy and the electron get scattered. The energy lost by the photon will be equal to the energy gained by the scattered electron. This process of inelastic scattering of electron by a photon is called ...


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 ...


4

Classical electromagnetism is perfectly compatible with special relativity. In classical E&M, light is an electromagnetic wave and there is generally no useful formulation in terms of particles. The most widely used technique to combine quantum mechanics with special relativity is relativistic quantum field theory. The relativistic QFT that ...


4

Higher energy gamma and longer wavelength radio? Keep in mind that the different 'kinds' are merely human labeling conventions for a spectrum that is continuous in the mathematical sense. There is no feature of "radio" that distinguishes it objectively from microwaves. We just pick a boundary on the basis of some technological limitations that apply when ...


4

A better way to think of it is "speed of causality". That's the fastest any cause-and-effect will spread over space. With nothing to cause it to go slower, changes to electric and magnetic fields will occur at that speed. No coincidence that changes to spacetime (causing gravity) propigate at the same speed. You really need to show how Minkowski spacetime ...


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 ...


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 ...


3

Whether waves == radiation is somewhat a question of semantics, and thus a bit subjective, however... In general wave equations support radiation, but not all solutions to a wave equation are radiative. Evanescent solutions are also called waves ("evanescent waves"), yet typically not considered to be radiation since they do not propagate in three ...


3

The inverse square law says that the intensity of incident light falls off in proportion to the inverse of the square of the distance from the light source. The important word here is "the distance" — the inverse square law implicitly assumes that all parts of the light source are at the same distance from the measurement point, or at least ...


3

No, each color in the spectrum has a characteristic frequency. Every light source has a so called spectrum of frequencies. The relative intensity of these frequencies determines what color you see (or not). For example, the sun looks yellow because it's peak intensity is in the yellow wavelength. White light comes from a source consisting of a very broad ...


3

In quantum mechanical domain these type of question does not have meaning. Every single photon is associated with a wave and vice versa. But to talk whether an electromagnetic wave contains a single photon or not is an ambiguous statement. When people say an electromagnetic wave necessarily contains many photons it only means that a incident beam of ...


3

Definition of resistivity: Electrical resistivity (also known as resistivity, specific electrical resistance, or volume resistivity) is an intrinsic property that quantifies how strongly a given material opposes the flow of electric current. A low resistivity indicates a material that readily allows the flow of electric current. So you are using the ...


3

This is a common misconception about what boundary conditions do and how they do it (for example here). You discussed two types of boundary conditions, Neumann and Dirichlet. In Neumann boundary conditions, we impose that the derivative of the variable normal to the boundary is specified, generally to be zero. With Dirichlet, we impose the value that the ...


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 ...


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 ...


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 ...


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 ...


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 ...



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