2,778 reputation
113
bio website
location
age
visits member for 11 months
seen 1 hour ago

17h
reviewed Leave Open How can there be anything “beyond” the CMB?
18h
answered How can there be anything “beyond” the CMB?
1d
comment Why does a particle feel no force at radii greater than itself?
+1 Huh. Honestly, even with the title of the question, this interpretation didn't even occur to me. But you are very likely correct that this is where the source of OP's confusion.
1d
reviewed No Action Needed Astronomy Olympiad Books
1d
awarded  Custodian
1d
reviewed Close Humans surviving in space
1d
answered Why does a particle feel no force at radii greater than itself?
1d
reviewed No Action Needed Yellow object between δ Ori and η Ori
2d
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