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How can I obtain a phase-folded phase - velocity diagramme from a time series of radial velocity data when it is known that the periodicity changes as $\frac{{d}\omega}{{d}t}$?

Without $\frac{{d}\omega}{{d}t}$ it is:

p = ((T-Tref)/P)%1.0

$T$ is a time

$T_{ref}$ is a reference time

$P$ is a period

$p$ is fold phase

Many thanks

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  • $\begingroup$ Welcome to SE. I think your question would profit if you could at least briefly explain your notation. IMHO a bit more context would not hurt either (what do you mean with 'phase'? Do you mean 'phase fold'?) $\endgroup$
    – planetmaker
    Mar 17, 2022 at 9:04
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    $\begingroup$ Yes, I mean phase fold. $\endgroup$
    – Carly
    Mar 17, 2022 at 9:16
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    $\begingroup$ Using your formula, what is t'? It would be incredibly large and the units do not correspond. $\endgroup$
    – Carly
    Mar 17, 2022 at 9:48
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    $\begingroup$ Thanks and do you have a reference for this relation. $\endgroup$
    – Carly
    Mar 17, 2022 at 10:35
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    $\begingroup$ Uhm... I'm suggesting only a simple coordinate transform from $t$ (you call it $T$) to $\omega$. This assumes that you know your period $P$ - but that is required anyway when you want to phase-fold your data. And tbh, it doesn't matter at which time of the phase you start as long as you cover exactly $2\pi$. The definition of what time equals phase = 0 is completely arbitrary. Mind the relation that period and frequency are related: $P = 2\pi/\omega$. $\endgroup$
    – planetmaker
    Mar 17, 2022 at 14:49

1 Answer 1

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If $\omega(t)$ is $$ \omega(t) = \omega(T_0) + \int_{T_0}^{t} \frac{d\omega}{dt}\ dt$$ then the phase $\phi(t)$ is $$ \phi(t) = \phi(T_0) + \int_{T_0}^{t} \omega(t)\ dt$$

And your phase-folded ($0 \rightarrow 1$)) coordinate would be $$p = \left(\frac{\phi(t) - \phi(T_0)}{2\pi}\right) - {\rm floor}\left(\frac{\phi(t) - \phi(T_0)}{2\pi}\right)\ , $$ where floor truncates a decimal to an integer.

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