<p>If <span class="math-container">$\omega(t)$</span> is
<span class="math-container">$$ \omega(t) = \omega(T_0) + \int_{T_0}^{t} \frac{d\omega}{dt}\ dt$$</span>
then the phase <span class="math-container">$\phi(t)$</span> is
<span class="math-container">$$ \phi(t) = \phi(T_0) + \int_{T_0}^{t} \omega(t)\ dt$$</span></p>
<p>And your phase-folded (<span class="math-container">$0 \rightarrow 1$</span>)) coordinate would be
<span class="math-container">$$p = \left(\frac{\phi(t) - \phi(T_0)}{2\pi}\right) - {\rm floor}\left(\frac{\phi(t) - \phi(T_0)}{2\pi}\right)\ ,   $$</span>
where floor truncates a decimal to an integer.</p>

CC BY-SA 4.0
2022-03-17T23:04:33.793Z
2022-03-17T23:04:33.793Z
ProfRob
<p>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 <span class="math-container">$\frac{{d}\omega}{{d}t}$</span>?</p>
<p>Without <span class="math-container">$\frac{{d}\omega}{{d}t}$</span> it is:</p>
<pre><code>p = ((T-Tref)/P)%1.0
</code></pre>
<p><span class="math-container">$T$</span> is a time</p>
<p><span class="math-container">$T_{ref}$</span> is a reference time</p>
<p><span class="math-container">$P$</span> is a period</p>
<p><span class="math-container">$p$</span> is fold phase</p>
<p>Many thanks</p>

CC BY-SA 4.0
2022-03-17T09:04:36.763Z
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'?)
planetmaker
CC BY-SA 4.0
2022-03-17T09:16:21.280Z
Yes, I mean phase fold.
Carly
CC BY-SA 4.0
2022-03-17T09:26:14.897Z
I made another edit in which I tried to capture the uniqueness here that the change in periodicity is the crucial point. Please revert or change, if my edit changes your question in a bad way.
planetmaker
CC BY-SA 4.0
2022-03-17T09:27:33.370Z
It would also improve the question if you told us what you tried and what sources you checked for a solution yourself. Had you tried to create a new time base to a t' = $ t / d\omega/dt$ or converting time directly into omega as you know the relation $t(\omega)$ - and do the phase-folding only as last step once you have a amplitude - omega relation (which is not yet folded)
planetmaker
CC BY-SA 4.0
2022-03-17T09:48:33.820Z
Using your formula, what is t'? It would be incredibly large and the units do not correspond.
Carly
CC BY-SA 4.0
2022-03-17T10:31:42.707Z
t is time. If you have RV data, you have a date and time attached to your measurements. Yes, t is large, but the offset is arbitrary and does not matter. That's why you often see in these time plots time denoted as JD-2415743 or similar. Similar it does not matter if $\omega$ is offset by a huge multiple of $2\pi$ (or 360°).
planetmaker
CC BY-SA 4.0
2022-03-17T10:35:43.347Z
Thanks and do you have a reference for this relation.
Carly
CC BY-SA 4.0
2022-03-17T14:49:42.707Z
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: $P = 2\pi/\omega$.
planetmaker
<radial-velocity><eclipsing-star-systems>
How to phase fold data when periodicity change is known as d$\omega$/d$t$?
Carly
# How to phase fold data when periodicity change is known as d$\omega$/d$t$?

(Asked by: username_0 on 2022-03-17)

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

**Tags:** radial-velocity, eclipsing-star-systems

> **Comments:**
> **username_1**: 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'?)  
> **username_0**: Yes, I mean phase fold.  
> **username_1**: I made another edit in which I tried to capture the uniqueness here that the change in periodicity is the crucial point. Please revert or change, if my edit changes your question in a bad way.  
> **username_1**: It would also improve the question if you told us what you tried and what sources you checked for a solution yourself. Had you tried to create a new time base to a t' = $ t / d\omega/dt$ or converting time directly into omega as you know the relation $t(\omega)$ - and do the phase-folding only as last step once you have a amplitude - omega relation (which is not yet folded)  
> **username_0**: Using your formula, what is t'? It would be incredibly large and the units do not correspond.  
> **username_1**: t is time. If you have RV data, you have a date and time attached to your measurements. Yes, t is large, but the offset is arbitrary and does not matter. That's why you often see in these time plots time denoted as JD-2415743 or similar. Similar it does not matter if $\omega$ is offset by a huge multiple of $2\pi$ (or 360°).  
> **username_0**: Thanks and do you have a reference for this relation.  
> **username_1**: 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: $P = 2\pi/\omega$.  

---

## 1 answers

### Answer by username_2 on 2022-03-17. Score: 4 | Accepted answer

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.


---

