How is phase defined in a binary orbit with eccentricity?

Are orbital phases of the black spot in the figure measured from some angle or time/period?

Usually, how is phase=0 defined? It is relative to the line of apses perhaps with phase=0 at periastron?

Any of the anomalies below is defined as orbital phase? Please specify reference papers or books.

E: eccentric anomaly

M: mean anomaly

u: true anomaly Source

• I'm really not sure what you're asking. Have you read here en.wikipedia.org/wiki/Kepler_orbit#Development_of_the_laws ? May 6 '21 at 21:52
• You mean true anomaly is what I want? May 7 '21 at 1:36
• I don't know what you want, you've not made it clear at all. May 7 '21 at 13:27
• It is better now? May 7 '21 at 16:04
• Yes, I think that's much better. Thank you. I gave +1 May 7 '21 at 16:08

Moreno et al. in their paper Eccentric binaries: Tidal flows and periastron events  define the orbital phase as from -0.5 to 0.5, where periastron is at phase 0, and apastron is at -0.5 and 0.5. Orbital phase here is $$\phi = \dfrac{t}{P}$$, where time $$t=0$$ at periastron and $$P$$ is the orbital period. This definition allows them to plot the behavior of several binary star systems with different periods on the same x-axis. Here is a figure from their paper: We don't necessarily have to choose periastron as "phase=0." For example, as we observe an eclipsing binary star system, the light curve will appear to pulse. This is an effect due to the reduced overall system luminosity as one star passes in front of the other.

Here is an example from Dan Bruton's website: One can see that the length of 1 unit of phase is equal to the orbital period. Here, the deep dips in magnitude correspond to when the lower luminosity star passes in front of the more luminous star, and the more shallow dips correspond to when the higher luminosity star passes in front of the less luminous star. Since the light measurements are somewhat noisy, the technique of folding is often used. To fold a light curve, one segments the data set along the integer phase boundaries and then adds all the light curves together, to increase the signal to noise ratio.

Here is a corresponding wikipedia gif of the Beta Lyrae system using the CHARA array: The origin (phase=0) in the above example is chosen to coincide with the occultation of one star by the other. In general, the choice of origin will be made to highlight whatever feature the astronomer is interested in. The eccentricity of the orbit won't drive the choice of origin for this technique. Instead, the choice of origin is driven by the light curve itself, which is a function of both the orbital system and the location of the observer.

• This isn't (as I understand it) wholly what the question is asking about, the binary doesn't need to be eclipsing, and your answer doesn't address how phase is determined during the orbit of an eccentric binary. e.g. Where is phase 0.25? May 7 '21 at 18:35
• @ProfRob Very insightful comment, as usual. Edited. May 7 '21 at 19:42
• @ConnorGarcia t/p, you mean the phase is proportional to time even if eccentricity!=0? May 8 '21 at 8:55
• I reverse my previous comment. This is the only way that a phase can be defined observationally, where you may not know the eccentricity. May 9 '21 at 8:31

Usually, how is phase=0 defined? It is relative to the line of apses perhaps with phase=0 at periastron?

The true anomaly, $$\theta$$, is the angle between the current location of the orbiting particle and its location in the orbit at which it is closest to the central body (called the periapsis/periastron). The word "phase" is used variously in physics and astronomy, but based on your question, you may mean $$\theta$$ since it is defined as being zero at periastron.

Are orbital phases of the black spot in the figure measured from some angle or time/period?

It ultimately depends on how you want to compute the trajectory/orbit of the particle (i.e., the black "spot" in the OP image) that is moving about a primary body (i.e., the star in the OP image). Alternate values can be used instead of $$\theta$$, for example $$\displaystyle M$$, the mean anomaly.

• The mean anomaly is proportional to time. The phase is also intended to be defined to be proportional to time and has nothing to do with eccentricity? May 8 '21 at 8:44
• The true anomaly is not directly proportional to time. The true anomaly is related to the eccentric anomaly by $\cos\theta = (\cos E - e)/(1 - e\cos E)$. May 8 '21 at 17:40