Ilmari Karonen
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 1d comment Light's inverse square law: Does it require a minimum distance from the source? Honestly, I suspect the "problems" are mostly in your understanding of physics (and in particular, the scales at which quantum-mechanical effects are relevant). As for your specific question just above, I'm not sure what you're trying to mean by "area of probability" (it's not a standard term that I know of), but if you just mean "area", then yes, that's true by basic geometry: a photon in free space propagates as a spherical wave front (or part of one), and the area of a sphere increases proportionally to the square of its radius. 1d comment Light's inverse square law: Does it require a minimum distance from the source? ... In any case, the people testing emergency vehicle warning lights are certainly not concerned about photons or wave functions or quantum mechanics in any way -- for such practical applications involving macroscopic incoherent light sources (i.e. not lasers), the classical optics I'm using are more than accurate enough. Ditto for the flashlight and cardboard cutout image your linked to in earlier comments; no quantum mechanics (or even wave optics) needed for that, either. 1d comment Light's inverse square law: Does it require a minimum distance from the source? You could certainly calculate the time evolution of the quantum mechanical wave function for a single photon emanating from a point source, and find that its square (i.e. the expected probability of observing the photon at a particular location) does obey the inverse square law (after which you can forget about quantum mechanics, and just use the inverse square law you derived). But that's a lot of work to derive a basic classical result, not to mention that trying to visualize it quickly gets into tricky territory like wave-particle duality. 1d comment Light's inverse square law: Does it require a minimum distance from the source? This is all classical ray optics, no photons (or even explicit waves) involved at all. I'm effectively taking the inverse square law for negligibly small sources as given (either as an empirical postulate, or as derived from a lower-level theory), and calculating how much a light source with a non-negligible spatial extend will deviate from it at close distances. (I'm also implicitly assuming that light intensity is an additive scalar quantity, which is not strictly true for coherent light sources due to wave interference.) Apr 4 comment Why do travelling waves continue after amplitude sum = 0? @curiousdannii: jpa does have a point; the dynamics on either side of the midpoint would look exactly the same even if you fastened the midpoint to an immobile wall. (In fact, this is how one can show that a wave hitting a fixed boundary will produce an inverted reflection.) So we effectively have two equivalent descriptions of the same motion; one in which the waves pass through each other and combine linearly, and one in which they never interact except through the midpoint, which never moves. Mar 28 comment How to interpret the units of the dot or cross product of two vectors? @Ilja: I'm not quite sure what you're asking. Rotating something by 90° surely is a physically meaningful operation; I don't see how it needs any kind of "interpretation". If you like, you can take a stick and physically rotate it 90°, in any plane of your choosing, to see that it's indeed possible and physically meaningful in real physical space. Mar 27 comment How to interpret the units of the dot or cross product of two vectors? It is, in fact, possible to interpret the dot product geometrically. See my answer for some examples. Feb 25 comment How does a spinning object “know” that it is spinning? I suspect you may have hit the source of the confusion here. Given that, apparently, the laws of physics are intrinsically invariant with respect to position, orientation and velocity (= the time derivative of position), it's somewhat surprising that they're not invariant with respect to angular velocity (= the time derivative of orientation). But of course, an extended object with non-zero angular velocity necessarily (either breaks apart or) experiences non-zero centripetal acceleration, which is also absolute. Feb 22 comment Why doesn't an electron ever hit (and stick on) a proton? -1, this doesn't seem like a useful way to answer the question. An electron does not orbit the nucleus in a classical orbit like the moon orbits the earth, and that misconception is precisely the source of the OP's confusion. That's not something you should be reinforcing in the first paragraph. Jan 17 comment How am I able to stand up and walk down the aisle of a flying passenger jet? This does not seem to answer the question in any meaningful sense. Dec 28 comment Why is there an electric field in a wire even though it is a conductor? @1110101001: The fact that some current/voltage source (e.g. a battery, a generator, etc.) is constantly moving electrons the other way, and thus replenishing the field as fast as it decays. Disconnect the source, and the field does disappear very quickly (limited only by the resistance of the wire). Dec 25 comment Do raindrops spin as they fall? True, that part of your answer is correct -- the shape of a raindrop is not the one with minimum drag among all possible shaped. (It also doesn't seem to have anything to do with the question, but it is correct.) Dec 25 comment Do raindrops spin as they fall? Raindrops are not actually "teardrop-shaped", and do not have a tapered tail. This is just a common myth. The actual shape of a falling raindrop is a slightly flattened sphere, not entirely unlike, say, a round bread roll. Dec 5 comment In terms of forces and kinematics, why does a projectile, thrown forwards, bounce forwards? Note that, if the ball is spinning fast enough when it hits the ground, it can bounce backwards. That's probably not the case you're asking about, but I just wanted to point out that the direction of the bounce depends on both the velocity and the rotation of the ball. You can't treat those as independent, unless you want to neglect friction completely. Nov 19 comment Could we launch a missile from a planet with the mass of Jupiter? Note that a planet with the radius of Earth and the mass of Jupiter would have a mean density of $1.7 \times 10^6\,{\rm kg/m^3}$. That's over 75 times the density of solid osmium, making such a planet pretty much physically impossible. (White dwarfs and neutron stars achieve far higher densities, around $10^9\,{\rm kg/m^3}$ and above, but there doesn't seem to be any way to have a planetary-size body with a density like your calculations imply.) A (slightly) more reasonable scenario might be a Jupiter-mass planet with the same density as Earth. Nov 17 comment Why can't I see the blue color scattered by the lower atmosphere of the earth? You sure you can't? You may just need to be in the right place (clear dry air + distant backdrop = mountains, typically) to notice it clearly. Oct 28 comment If UV radiation 1 cm away from the halogen bulb is equal to Sun's radiation, what is the level of radiation 1 meter away? Good point, a 1 cm filament at a distance of 1 cm is in the transition region where neither 1/r nor 1/r² is a very good approximation of the falloff. But still, the range from 1 cm to 1 m lies mostly in the 1/r² region, so the intensity at 1 m will be about 1/10,000 = 0.01% of the intensity at 1 cm. It certainly won't be anywhere near 1/100 = 1%. Oct 27 comment If UV radiation 1 cm away from the halogen bulb is equal to Sun's radiation, what is the level of radiation 1 meter away? The $1/r$ falloff applies to a long cylinder (i.e. one significantly longer than your distance from it). A light bulb filament, at distances of over $1\,{\rm cm}$, is well approximated by a point source (i.e. the $1/r^2$ falloff rule applies). Oct 19 comment Will the box move? If the surface is not frictionless, it's possible to obtain a non-zero net velocity using method 3. In fact, an even better method would likely be for the man to repeatedly walk from one end of the box to the other, and run back. Or just stand in place and shift his weight back and forth, slowly in one direction and quickly in the other. Sep 8 comment Do solar systems typically spin in the same direction as their galaxy? "note, the solar system is not drawn to scale compared with the Galaxy" ...or compared with itself, for that matter.