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Nov
11
comment What really allows airplanes to fly?
The bound vortex moves from front to back above the wing, and back to front below the wing, so the net velocity of the air is greater above the wing than below. See en.wikipedia.org/wiki/Kutta_condition But this doesn't require a particular wing shape (other than the sharp trailing edge). Mostly the reason airplane wings are shaped the way that they are is to minimize drag while still achieving lift - not because the shape is necessary to generate lift.
Nov
11
comment What really allows airplanes to fly?
@MarkFoskey I think it's misleading to think of the Bernoulli principle as causing the reduction of pressure above the wing. Bernoulli's principle tells us that decreasing pressure occurs simultaneously with increasing velocity - and indeed if we know the velocity we can calculate the pressure - but this is kind of obscuring why the velocity is higher above the wing. For wings with a sharp trailing edge moving with a positive angle of attack, a vortex will initially form at the back of the wing, and produce a corresponding bound vortex of air circulating around the wing.
Nov
5
comment Is a non-degenerate wavefunction real or complex?
It might be better to say that the wavefunction "can be chosen to be real". For instance in John Rennie's example you could also have $\psi_r = \psi_i$, in which case the state is not degenerate, but instead we have $\psi = (1+i)\psi_r$. You can choose to drop the overall phase factor.
Nov
3
comment I have trouble understanding work
For example, if you're behind a cart pushing it then you are giving energy to the cart, but if you stand in front of a cart as it moves towards you and push against it then you are taking energy away from the cart.
Nov
3
comment I have trouble understanding work
Then you want to consider the force that you're exerting. Of course, when you exert a force on something, it exerts a force back on you (Newton's 3rd law), but the direction of the force matters here, because we're taking the dot product of the force with the displacement. So for instance if the force is in the same direction as the movement then you do positive work, whereas if the force is in the opposite direction of the movement then you do negative work.
Nov
3
comment I have trouble understanding work
That depends whether you're talking about the work done by you or the work done on you.
Nov
3
answered I have trouble understanding work
Nov
2
comment What makes energy content of a body harder to accelerate it?
That step is only necessary if you want to recover the non-relativistic formula for kinetic energy. In relativity, kinetic energy is $\gamma m c^2 - m c^2$ (where $\gamma = \frac{1}{\sqrt{1-v^2/c^2}}$). If Einstein's audience knew that formula, he could have just said "Look, $E$ plays the role of $mc^2$ and skipped the approximation - but he'd only just published his first paper on special relativity 3 months earlier. (Also available online: fourmilab.ch/etexts/einstein/specrel/www )
Nov
2
comment What makes energy content of a body harder to accelerate it?
@Weezy Taylor series expansion at $v/c = 0$, keeping the dominant term.
Nov
2
answered What makes energy content of a body harder to accelerate it?
Nov
2
awarded  Yearling
Nov
2
answered Will my mass increase in my perspective while approaching near the speed of light?
Nov
2
answered Forward-scattering off a potential well
Nov
2
answered What does the Hamiltonian do in the Heisenberg picture?
Oct
31
awarded  Revival
Oct
30
comment What really allows airplanes to fly?
That said, the wing shape is not so important; it makes some difference, but even perfectly flat wings can produce lift with the right angle of attack.
Oct
30
comment What really allows airplanes to fly?
@MarkFoskey Yes the air does move faster above the wings than below - in fact, the difference in speed is in general greater than the equal time assumption would imply. And yes, this does correspond to a change in pressure, in accordance with the Bernoulli effect. Given the actual airflow velocities, the Bernoulli effect is consistent with the aerodynamics of flight; the common error is in determining the airflow velocity from the fallacious equal time assumption.
Oct
30
answered How to determine the sign of the s-wave scattering length?
Jul
20
comment Why friction is zero when wheel slip is zero?
I can't resist pointing out the amusing typo in my last comment. I meant "a real-world elastic tire", not "a wheel world elastic tire".
Jul
20
comment Why friction is zero when wheel slip is zero?
+1 Rick and thank you for the discussion on my answer above. For those who don't want to read our whole back-and-forth, the key point is for tires with elasticity (any real-world tire) there is compression and thus V is always a bit less than R omega. See here