# How to phase fold data when periodicity change is known as d$\omega$/d$t$?

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

• 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'?) Mar 17, 2022 at 9:04
• Yes, I mean phase fold. Mar 17, 2022 at 9:16
• Using your formula, what is t'? It would be incredibly large and the units do not correspond. Mar 17, 2022 at 9:48
• Thanks and do you have a reference for this relation. Mar 17, 2022 at 10:35
• 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: $P = 2\pi/\omega$. Mar 17, 2022 at 14:49

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.