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70

A short summary of the paper mentioned in another answer and another good site. Basically planes fly because they push enough air downwards and receive an upwards lift thanks to Newton's third law. They do so in a variety of manners, but the most significant contributions are: The angle of attack of the wings, which uses drag to push the air down. This ...


49

There are zero contradictions between quantum mechanics and special relativity; quantum field theory is the framework that unifies them. General relativity also works perfectly well as a low-energy effective quantum field theory. For questions like the low-energy scattering of photons and gravitons, for instance, the Standard Model coupled to general ...


49

I think it's a great question, and enjoyed it very much when I grappled with it myself. Here's a picture of some of the forces in this scenario.$^\dagger$ The ones that are the same colour as each other are pairs of equal magnitude, opposite direction forces from Newton's third law. (W and R are of equal magnitude in opposite directions, but they're acting ...


46

From Stick and Rudder by Wolfgang Langewiesche, page 9, published 1944: The main fact of all heavier-than-air flight is this: the wing keeps the airplane up by pushing the air down. It shoves the air down with its bottom surface, and it pulls the air down with its top surface; the latter action is the more important. But the really ...


43

The answer is no. The pole would bend/wobble and the effect at the other end would still be delayed. The reason is that the force which binds the atoms of the pole together - the Electro-Magnetic force - needs to be transmitted from one end of the pole to the other. The transmitter of the EM-force is light, and thus the signal cannot travel faster than the ...


29

Since you asked for an explanation appropriate to an non-specialized audience, maybe this will do: "A Physical Description of Flight; Revisited" by David Anderson & Scott Eberhardt. It is a revision of the earlier "A Physical Description of Flight" (HTML version).


29

Since general relativity is a local theory just like any good classical field theory, the Earth will respond to the local curvature which can change only once the information about the disappearance of the Sun has been communicated to the Earth's position (through the propagation of gravitational waves). So yes, the Earth would continue to orbit what ...


28

Simplest, you say? There are two that strike me as being simple to demonstrate. Luckily someone on the internet has already spent some time to help us here to make these easy to illustrate: Shadows differ from place to place: Eratosthenes carried out this experiment to determine the circumference of the Earth, already assuming its spherical shape; ...


26

The keywords here are Rayleigh scattering. See also diffuse sky radiation. But much more simply, it has to do with the way that sunlight interacts with air molecules. Blue light is scattered more than red light, so during the day when we look at parts of the sky that are away from the sun, we see more blue than red. During sunset or sunrise, most of the ...


24

First, let me emphasize something that is being covered by a thick layer of misinformation in the media these days: it is totally premature to conclude whether the LHC will see SUSY or not. The major detectors have only collected 45/pb (and evaluated 35/pb) of the data. The "slash pb" should be pronounced as "inverse picobarns". The LHC is designed to ...


22

The other answers provide a first-order approximation, assuming uniform density (though Adam Zalcman's does allude to deviations from linearity). (Summary: All the mass farther away from the center cancels out, and gravity decreases linearly with depth from 1 g at the surface to zero at the center.) But in fact, the Earth's core is substantially more dense ...


18

Gravitational influences do propagate at the speed of light, not instantaneously. The question of what would happen if the Sun instantly disappeared is actually a funny one in general relativity. The equations of general relativity imply as a mathematical consequence that energy must be locally conserved. Therefore, there is no valid solution to the ...


17

This answer is nothing more than a variation of Sklivv's answer. I simply wish to discuss some quantitative ideas following from Sklivv's answer and discuss what I understand (from an aerospace engineering friend) to be a common conceptual mistake - that the application of "mere surface effects" and "application of Bernoulli's principle" is wrong. These ...


14

Sure you can! This is actually a simple but very interesting result, and it is usually shown in quantum mechanics courses. It's called the Ehrenfest theorem, and I won't prove it here but I'll copy the result from Sakurai Modern Quantum Mechanics (1991). You can check the mathematical details there, or in many other books. If you have a hamiltonian with the ...


13

All observations are consistent with standard GR so far, but I don't think the speed of gravity, in particular, has ever been measured. Experimental measurements of the speed of gravity was quite a controversy a few years ago when a paper came out claiming that the speed of gravity was very close to $c$ as measured by the Shapiro delay. To see papers on the ...


12

I guess so - I mean, as far as I know, there's no law of physics that strictly prohibits those "exotic" states from being realized. As long as the state exists and can be reached by some path from the "center" of the state space where the likely states are, there should be a nonzero (not even infinitesimal, really) probability of accessing it. But for a ...


12

The whirl is due to the net angular momentum the water has before it starts draining, which is pretty much random. If the circulation were due to Coriolis forces, the water would always drain in the same direction, but I did the experiment with my sink just now and observed the water to spin different directions on different trials. The Coriolis force is ...


10

Matt Reece gives a good answer, but one additional area of tension that seems worth mentioning is the problem of time. The role of time in quantum theory is quite different from general relativity. For a review of some of the issues involved, see Canonical Quantum Gravity and the Problem of Time. C. J. Isham. "Recent Problems in Mathematical Physics", ...


10

You can find the shortest and easiest derivation of this result in the paper where it was released by Einstein himself (what better reference can you find?) in 1905. It is not the main paper of Special Relativity, but a short document he added shortly afterwards. A. Einstein,Ist die Trägheit eines Körpers von seinem Energieinhalt Abhängig?, Annalen der ...


9

Short Answer You've hit upon the quirk that the SI and CGS systems not only measure electric charge with different units, but also assign them different dimensionality. In SI, the Ampere is a base unit. Amperes are not made out of anything else - they are primitive, like meters, kilograms, and seconds. One Ampere is one Coulomb per second, so the unit of ...


9

There are solutions to Einstein's field equations, which have closed timelike curves. For example Godel's solution. Would that constitute time travel, if you can reach the same point on your world line in finite time? One objection may be that such solution do not describe the universe, but examples as the Tipler's cylinder suggest that at least in theory we ...


9

It is mathematically possible to create some instances in which an object goes back in time relative to some observer. For example, simply going faster than light causes such an effect, but of course, speed of light is the limit for any massive object. While it is mathematically possible, there are many paradoxes caused by time travel to past, unless you ...


9

Hwlau is correct about the book but the answer actually isn't that long so I think I can try to mention some basic points. Path integral One approach to quantum theory called path integral tells you that you have to sum probability amplitudes (I'll assume that you have at least some idea of what probability amplitude is; QED can't really be explained ...


9

No. In relativity you cannot consider extended objects to be infinitely "stiff" - they must bend and stretch, as real objects do. When you move one end of the steel rod, it makes part of it bend and stretch which exerts a force on the next section which makes that move and which makes a new part bend and stretch and so on and so on until you reach Alpha ...


9

The most important problem that supersymmetry solves is the hierarchy problem: why is the weak scale, which determines the rate of beta decay or the masses of the W and Z bosons, so much smaller than the Planck scale, which is related to the strength of the gravitational force? In other words, why is the weak force so strong, compared to gravity? The real ...


8

Assuming spherically symmetric mass distribution within Earth, one can compute gravitational field inside the planet using Gauss' law for gravity. One consequence of the law is that while computing the gravitational field at a distance r < R (with R being the radius of the Earth), one can ignore all the mass outside the radius r from the center ...


7

Here is an interesting site about this idea and similar ideas: http://math.ucr.edu/home/baez/physics/Relativity/SpeedOfLight/FTL.html#3 Essentially the problem with this idea is there are no such thing as perfectly rigid bodies. So as you push, it sends a little compression wave through the material, which travels at the speed of sound in the material, as ...


7

Sitting for a while by the seashore ought to make it clear the Earth isn't flat, even if you don't happen to see a ship go over the horizon. The edge of the discworld Earth would have to be just a few miles away, and there's no way that the entire, circular world would fit inside the circle that the ocean horizon seems to make. Humans have not just known ...


6

Yes the dimension is different. In SI the current (A) a base unit independent from length (m), mass (kg) and time (s) because we choose to, but in CGS Gaussian unit this is not (1 unit of current = 1 g1/2 cm3/2 s-2), by setting $\epsilon_{0,SI} = \frac1{4\pi}$. This also leads to some perhaps unintuitive results, like the unit capacitance in CGS Gaussian is ...



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