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Apr
13
comment What experiment would disprove string theory?
@DarenW Ancient Greek philosophers did mention atoms, but their atoms have nothing to do with what we know as atoms today, because άτομο means something that cannot be divided in constituents and atoms certainly can. One could say that quarks are the atoms of the philosophers, but nobody is certain. Also, their theories were not just constrained by deficiencies in terms of experimental methods, but they mostly were too vague to be of any practical use for anything at all. Not to mention that no matter what one philosopher would say, another would claim the opposite. That said, I'm Greek!
Apr
13
comment Relation between shm and circular motion
They are $-a\cos t\hat{x}$, $-a\sin t\hat{y}$.
Apr
13
comment Moment of inertia of a sphere
@DylanDang Yes, what David Z is saying is that $dm$ in the notes is not infitesimal mass in dimension 3, the notation is confusing. If you write this as $dI = 2\pi \rho dz \int_0^r r^{\prime 3} d r'$, it should be clear.
Apr
13
comment Relation between shm and circular motion
Yes, I tried to make things simple, so that the idea is easier to convey. This is the reason the equations above have $\omega =1$, so that we don't need to set initial conditions that might seem arbitrary at first sight.
Apr
13
answered Relation between shm and circular motion
Apr
12
awarded  Citizen Patrol
Apr
12
awarded  Commentator
Apr
12
comment Reason why $F^{\mu\nu}F_{\mu\nu}$ and $\tilde{F}_{\mu\nu}F^{\mu\nu}$ are Lorentz invariant
@RobinEkman Yes, that's why I'm used to combining the two, depending on which leads to simpler expressions. For example, I've never done contractions using the Hodge dual and I prefer to replace it with the expression in terms of tensors.
Apr
12
answered Reason why $F^{\mu\nu}F_{\mu\nu}$ and $\tilde{F}_{\mu\nu}F^{\mu\nu}$ are Lorentz invariant
Apr
11
comment Force on a point charge q inside a cavity in an uncharged conductor
@Tobias Thanks. If you change the distribution of induced charges, this will affect the charge in the cavity before the induced charges settle down in their new distribution, because the former propagates at the speed of light and the latter at lower speed. You don't need to freeze the system arbitrarily. Also, Earnshaw's theorem implies that the configuration with the charge in the cavity is always unstable, I'm not sure where I implied that this is a stable system.
Apr
11
answered Reflections in Rearview Mirror
Apr
10
answered Force on a point charge q inside a cavity in an uncharged conductor
Apr
7
answered Integral form of Gauss's law for magnetism from Stokes' theorem?
Apr
6
answered Quantum mechanics as classical field theory
Apr
5
revised Formalism to deal with discontinuous potentials in classical mechanics (hard wall, hard spheres)
Added another way to derive the property of the potential without using conservation of energy
Apr
5
comment Are signal fronts in a beam not at rest to each other?
You still don't understand what I'm telling you. Synge writes about measuring the elapsed time between sending and receiving a photon, how do you measure that on a photon?
Apr
5
comment Are signal fronts in a beam not at rest to each other?
You're confused. The point is that there must exist an observer to measure something at rest, it is not that the wave fronts do not exchange photons that makes what you ask not make sense at all. The measurement process includes sending and receiving photons, but what use would that be even if photons could do that, since a photon cannot measure the distance? To make a measurement you'll need at least a clock and a device to send and receive photons, how can a photon on its own do that? Even Synge says so, the curves must be time-like, why do you keep ignoring this?
Apr
4
comment Equivalence between Hamiltonian and Lagrangian Mechanics
$\dot{q} = \partial H / \partial p$ is valid regardless of whether $\dot{q}$ is a derivative or not. It could be any function, but if it isn't a derivative of $q$, then this is not the equation of Hamilton.
Apr
3
answered Do Maxwell's Equations overdetermine the electric and magnetic fields?
Apr
3
revised Equivalence between Hamiltonian and Lagrangian Mechanics
I mixed up the $p$ and $\dot{q}$, corrected this to match Arnold and the OP