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comment Hulse Taylor Binary Pulsar - How is “Cumulative Periastron Time shift” calculated?
@steveOw It's not assumption; it's just vocabulary. If $P$ were constant, then $1/P$ is orbital frequency; in time interval $\Delta t$, the number of (or fraction of) orbits completed is $\Delta t/P$. $P$ actually varies continuously, but that's what calculus is for: over infinitesimal $\mathrm{d}t$, the fraction of orbit traversed is $\mathrm{d}t/P$, and integration is adding them. Thus, $\int\mathrm{d}t/P$ is the exactly correct orbital count, regardless of whether one wishes to call its $2\pi$ scaling 'mean anomaly' or not. If prefer not, just replace 'mean anomaly' with 'orbital phase'.
Nov
17
comment Hulse Taylor Binary Pulsar - How is “Cumulative Periastron Time shift” calculated?
I don't know what you mean byt $\mathrm{d}t$ fixed. If $\dot{P}\neq 0$, then $M(\tau)\neq 2\pi\tau/P(\tau)$ (not necessarily constant, just nonzero). Otherwise, ok.
Nov
16
comment Hulse Taylor Binary Pulsar - How is “Cumulative Periastron Time shift” calculated?
Over a small time interval $\mathrm{d}t$, the fraction of orbit traveled is $\mathrm{d}t/P$. That's why mean motion is $M(t) = \int_0^t 2\pi\,\mathrm{d}t'/P(t')\neq 2\pi t/P$ (unless $\dot{P} = 0$). Then $\dot{M} = 2\pi/P$ by fundamental theorem of calculus, at all times, not just $t = 0$. Mean motion is not a physical angle, but it doesn't have to be: the orbital frequency is $1/P$, angular frequency (mean motion) $n\equiv 2\pi/P$, so the orbital phase (mean anomaly) is its integral over time.
Nov
16
comment Hulse Taylor Binary Pulsar - How is “Cumulative Periastron Time shift” calculated?
I've followed up on the PPN calculations for the $\dot{P} = KP^{-5/3}$ formula, and while your $K$ is a function of $e$, $e$ is itself not constant (should have checked sooner). But this makes very little difference because the rate of change of $e$ turns out to be quite negligible.
Nov
16
revised Hulse Taylor Binary Pulsar - How is “Cumulative Periastron Time shift” calculated?
failed to realize K is not actually a constant; corrected
Nov
16
revised Hulse Taylor Binary Pulsar - How is “Cumulative Periastron Time shift” calculated?
Cleaned up a bit and explained small time shift
Nov
16
revised Hulse Taylor Binary Pulsar - How is “Cumulative Periastron Time shift” calculated?
Added exact constant-K periastron shift formula
Nov
16
revised Hulse Taylor Binary Pulsar - How is “Cumulative Periastron Time shift” calculated?
added 1 character in body
Nov
16
comment Hulse Taylor Binary Pulsar - How is “Cumulative Periastron Time shift” calculated?
@steveOw $\dot{M} = 2\pi/P\,\implies\,\ddot{M} = -2\pi\dot{P}/P^2$. You have an extra factor of $2$.
Nov
16
revised Hulse Taylor Binary Pulsar - How is “Cumulative Periastron Time shift” calculated?
added 1528 characters in body
Nov
15
comment Hulse Taylor Binary Pulsar - How is “Cumulative Periastron Time shift” calculated?
@steveOW $M(t) = \int_0^t\frac{2\pi}{P(t')}\,\mathrm{d}{t'}$ ensures that if the first periastron is when $M(t) = 0$, then the next is when $M(t) = 2\pi$, etc. If $P$ is constant, then $M(t) = 2\pi t/P$; otherwise, no.
Nov
15
revised Hulse Taylor Binary Pulsar - How is “Cumulative Periastron Time shift” calculated?
added 1491 characters in body
Nov
15
answered Hulse Taylor Binary Pulsar - How is “Cumulative Periastron Time shift” calculated?
Nov
15
answered What are some applicable problems with the correct usage of G?
Nov
15
comment What are some applicable problems with the correct usage of G?
If you measure the stars to be $3.54\times 10^{20}\,\mathrm{ly}$ apart using parallax, or any other method for that matter, then at that point you should know you've made a mistake and re-do the measurements.
Nov
14
comment Is the dark energy between the moon and Earth measurable in any capacity?
+1, though it would be better to say that it would have tiny effects that for cosmological constant are time-independent. Amusingly, orbits in the Kottler spacetime (isolated spherically symmetric body with Λ) have an effective potential that's just the Schwarzschild potential plus $-\Lambda r^2/6$, which is exactly what one would get by putting in $-M_\Lambda/r$, where $M_\Lambda$ is the mass-equivalent of dark energy enclosed at that radius by the Euclidean formula. So by a fortuitous coincidence, the calculation is even more correct than one might expect, at least for spherical bodies.
Nov
14
comment What is the long term fate of the gas giants?
Jupiter can generate its own heat through the Kelvin–Helmholtz process: as it radiates heat, it contracts, thus converting gravitational potential to thermal. Negative heat capacity is characteristic of gravitationally bound systems. The present Jovian intrinsic luminosity is $L_\text{J} = 8.7\times 10^{-10}$, so the thermal timescale is $$\tau_\text{KH}\sim\frac{GM^2}{RL_\text{J}}\sim 10^{11}\,\mathrm{yr}\text{.}$$ We used the present value, so indications of the long-term future are uncertain. But the point is that it's not obvious that one can dismiss internal heat generation so quickly!
Nov
14
comment How would you calculate expansion rate of universe in Shanon Entropy?
Probability distribution of what? The supposition that cosmic expansion is translatable to some entropy runs into problems with the fact that a cosmologically contracting universe is allowed by the laws of physics, including the second law of thermodynamics.
Nov
14
comment How would you calculate expansion rate of universe in Shanon Entropy?
How are any of those things simpler? $74.3\,\mathrm{km/s/Mpc}$ has a clear and simple meaning that has no direct connection to atoms or Shannon entropy. Are you perhaps thinking that cosmic expansion means the universe gets more atoms? If so, you're mistaken, but either way, clarifying your question would be best, because it's completely unclear what you're talking about.
Nov
13
answered Why all the photos from 67p are black and white?