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Nov
14
comment Why aren't EM waves directionless?
@Jean-JacqduPlessis It's a bit surprising that waves with tiny wavelengths ($500\mathrm{nm}$ or so) behave almost exactly like rays over distances of meters. But they do. A water surface from throwing a rock into a pond will cause, what, maybe $30$ wavelengths to be visible at a moment? But with visible light, you're dealing with more like $10^7$ wavelengths in a comparable distance. At that scale, things are more like rays.
Nov
12
comment How to reconcile these two approximations?
@Energizer777 blog.stackoverflow.com/2011/07/…
Nov
11
revised How to reconcile these two approximations?
added 241 characters in body
Nov
11
accepted Spin $\frac{3}{2}$ representation in Georgi's book?
Nov
11
comment Tension in string question
If the tensions weren't different, what would the net torque on the pulley be?
Nov
7
revised Looking for clarification on superposition
added 714 characters in body
Nov
7
answered Looking for clarification on superposition
Nov
6
awarded  Good Answer
Nov
6
comment Physical interpretation of Fourier $[x(t)]$ where $x(t)$ is the position of mass $m$ as a function of time?
Just a check re your comments. Do you mean the Fourier transform or the Fourier series? They require different methods to tackle and I interpret them in different ways. A Fourier Transform is like $\mathcal{F}[1]=\int_{-\infty}^\infty e^{-i \omega x} dx=2\pi\delta(\omega)$. A Fourier series is like $f(x)=\sum_{-\infty}^\infty A_n e^{i \omega n x}$.
Nov
5
comment Derivation of $E=pc$ for a massless particle?
Why not start with Maxwell's equations and a plane wave?
Nov
5
comment The Lagrangian of a Rocket
Yeah, as stated in the OP
Nov
5
comment The Lagrangian of a Rocket
Wouldn't your equation imply (ignoring potential) $m(t)\ddot{q}=-\dot{q} \dot{m}(t)$, as opposed to the correct equation $m(t)\ddot{q}=-u_0\dot{m}(t)$?
Nov
5
awarded  Nice Answer
Nov
4
revised Why doesn't a spinning object in the air fall?
edited body
Nov
4
comment Why doesn't a spinning object in the air fall?
@rb612 you could relate $T$ to $v$ by saying that the acceleration of the object, $v^2/R$, has to be caused by the acceleration due to tension, $T\cos(\theta)/m$. Then you get $mv^2/R=T\cos(\theta)$ and you could solve for the equilibrium value of $\theta$!
Nov
4
comment Why doesn't a spinning object in the air fall?
@rb612 well, there IS no force countering the tension. There is a force $T\cos(\theta)$ in the horizontal plane towards your hand (if you hold your hand still and the thing circles around), and there are forces $T\sin(\theta)$ and $-m g$ along the vertical axis. Those are the only forces acting on the object you're spinning around.
Nov
4
answered Why doesn't a spinning object in the air fall?
Oct
29
comment Exact closed form solution to the quantum harmonic oscillator
Wow, there's no typo in that equation? Can you give the page number and edition number?
Oct
28
comment Explicit calculation using metric tensor
What definition of $\Lambda$ have you been given? Usually the lorentz group is defined as the set of linear operators that leave $\eta$ invariant, in which case it's trivial. You could also rewrite your equation as a matrix equation, which is MUCH easier to work with. (You end up with $S=A\eta A$ where $A=\Lambda^{-1}$ and these are just matrices)
Oct
22
comment Understanding elastic collisions of objects with the same velocities
Oh! I misread it and misread your question. @Ovi the author of that quote surely meant that you're in a frame where one particle is at rest! That is the usual result.