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Oct
3
comment Time Dilation - How does it know which Frame of Reference to age slower?
@gwho Conceptually, inertial means nonaccelerating. Direction does not matter for time dilation between inertial frames. In spacetime, inertial worldlines (paths) are straight, and accelerated ones are bent, so not only is there an asymmetry, but since the length of the worldline is the time measured by a clock on it, having inertial and accelerated observers register different times is no more mysterious than the length of a line segment being different some other curved line connecting the endpoints. Beyond that, I suggest asking a separate question or looking for ones on twin paradox.
Aug
21
comment Am I right in saying that $gmm/r$ is motion due to potential energy?
I'm not going to say that. My point is that integral does not diverge. It's just $\sqrt{1-r_s/r}\big\vert_{r_s}^R$. Hence the suggestion of $F\,\mathrm{d}s$ instead, because that would be with respect to a 'radial distance'.
Aug
21
comment Am I right in saying that $gmm/r$ is motion due to potential energy?
What's the $F$ you're integrating? If $F$ is the static force, you don't get $-\infty$ at the horizon. Although you would for $F\,\mathrm{d}s$, where $\mathrm{d}s = \mathrm{d}r/\sqrt{1-2M/r}$, or something like that. Maybe some other $F$, too.
Aug
21
answered Am I right in saying that $gmm/r$ is motion due to potential energy?
Aug
16
comment Unknown letter ℑ used in an equation
This is a variant of a Fraktur-typeface J, i.e. $\mathfrak{J}$ (edit: it's a J, not I), which is used for the imaginary part of the complex number, also commonly denoted $\mathrm{Im}$, and contrasted to the real part $\mathfrak{R}$ or $\mathrm{Re}$, etc.
Aug
15
comment How accurate are constants in cgs units?
The Gaussian gravitational constant is $$k\equiv\sqrt{G} = 0.01720209895\,\mathrm{AU}^{3/2}\,\mathrm{day}^{-1}\,\mathrm{M}_\odot^{-1/2}$$ which has been introduced with this value in 1809 and has a much higher precision than either $G$ or the solar mass alone. Also, the quantity $\mu = GM$ of a body is usually called the standard gravitational parameter.
Aug
14
comment How accurate are constants in cgs units?
Coulomb's constant is an exact defined value in both SI and Gaussian-cgs units.
Aug
14
comment Is it possible to derive Lorentz transformation equation without Einstein's postulates?
@rob I defined the "parabolic" trig functions by the equation given. That makes them completely trivial because the the power series cuts off after the linear term, $\mathrm{sinp}\,\lambda = \lambda$, etc., but I phrased it in this way to carry the analogy to Euclidean rotations ($i^2 = -1$ with $e^{i\phi} = \cos\phi + i\sin\phi$) and Lorentz boosts ($j^2 = +1$ with $e^{j\alpha} = \cosh\alpha + j\sinh\alpha$).
Aug
10
comment Confusion about proper time
To be precise, the proper time between any two spacetime points is path-dependent regardless of acceleration or curvature, but in flat spacetime inertial motion provides a specific path between (timelike-separated) points that we conventionally select to be the standard.
Aug
7
comment Is everything moving at c in a c unit circle
@CaptainGiraffe: Lorentz transformations rotate along a hyperbola in spacetime in the exact same way that Euclidean rotations rotate along a circle in space. The Lorentz boost is just an addition of hyperbolic angles $\alpha = \tanh^{-1}(v/c)$. Beyond that, I'm not clear on what your question is.
Aug
7
comment Is everything moving at c in a c unit circle
@Danu: I know; that's why I said it was a hyperbola first.
Aug
7
comment Is everything moving at c in a c unit circle
The relativistic "circle" (locus of points a set interval from a given point) is actually a hyperbola, or hyperboloid in $3+1$ dimensions, because the Lorentzian analogue of the Pythagorean theorem has a minus sign in it.
Aug
5
comment Interpretation of the wave function in quantum mechanics
@ArmenAghajanyan: $S$ is the phase of the complex number that $\Psi$ returns multiplied by $\hbar$ ($S/\hbar$ is the phase itself). See Euler's formula to see that every complex number $z$ can be written in the form $z = re^{i\phi}$, where $r$ is the modulus and $\phi$ is the phase. This is the polar form of the complex number.
Aug
5
answered Interpretation of the wave function in quantum mechanics
Jul
22
comment Massless particles and the speed of light - New? Theories of existence
Your assumptions about "matter based mass" do not make much sense, since in theory one could have a composite massive particle made entirely of massless particles, if those massless particles are not exactly unidirectional. Glueballs are an example of one such hypothetical particle. There is also no such thing as "pure energy"--energy is a property that things have. As for the rest, I'm sorry, but I can't make much sense of it at all.
Jul
18
comment Finding the metric tensor from the Einstein field equation?
Spherical symmetry alone determines the Schwarzschild ansatz. Kerr was working from Petrov classification (on non-Type-I vacuum spacetimes) and proved that all Killing vectors are one of two types. The only unsubstantiated guess was a convenient form for an asym. timelike Killing vector; otherwise, the discovery was rather rigorous. Notably, there was no claim that this was solution was unique until a decade later, proven by Robison. I'm not saying this comment-answer is wrong (as typical presentations go), but relativists aren't nearly as morally loose as it might appear at first glance.
Jul
13
revised Computing the Christoffel symbols with the geodesic equation
added 35 characters in body
Jul
12
comment Computing the Christoffel symbols with the geodesic equation
The main trouble is that your first equation is actually incorrect, because of product rule and differentiation of $\nu$.
Jul
12
answered Computing the Christoffel symbols with the geodesic equation
Jul
12
comment Is the apparent lack of (Ricci) curvature in the Schwarzschild metric due to a choice of coordinates?
"Shouldn't there always be curvature in the presence of mass... ?" -- There is no curvature inside a spherically symmetric shell: spacetime is flat there. So it's not true that the presence of mass necessarily guarantees that spacetime is everywhere non-flat.