Find Solutions to Helical Worldline's Surfaces of Simultaneity

[Pencilvester]

For an inertial frame in flat spacetime with cartesian coordinates, and a particle in that frame whose worldline is a helix (moving in a circle at constant speed in x-y plane), given an arbitrary event with coordinates $t$, $x$, and $y$, (we won’t worry about $z$) how would I go about determining the coordinate time(s), $t$, at which the particle’s surface of simultaneity contains the event? I was working the problem this way: we have a separation vector, $\mathbf v \equiv \mathbf x - \mathbf p$, where $\mathbf x$ is the event and $\mathbf p$ is the position(s) of the particle, which can be parameterized by $t_p$:
$\mathbf p(t_p) = \begin{pmatrix} t_p \\ r \cos {\theta} \\ r \sin {\theta} \end{pmatrix}$ and where $\theta = 2 \pi \omega t_p$ and $\omega$ is angular frequency. We also have the time-like basis one-form for the particle, ω⁰ (of course unrelated to the angular frequency), which can also be parameterized by $t_p$:
ω⁰$(t_p) = \begin{pmatrix} -\gamma & -v \gamma \sin {\theta} & v \gamma \cos {\theta} \end{pmatrix}$. So we know that iff an event $\mathbf x$ is in the particle’s surface of simultaneity, then $\langle$ω⁰ | $\mathbf v \rangle = 0$. I find that this gives the equation $t_p - t + vy \cos {\theta} - vx \sin {\theta} = 0$ which only leaves me the option of using Newton’s method on an event by event basis. Is it simply not possible to derive a general solution for this problem where you plug in an event, and out pops a $t_p$ (or two)?
Also, how does one get a boldfaced omega using LaTeX? I tried \mathbf {\omega} and \mathbf ω , but neither actually made the omega look boldfaced.

 

[PeterDonis]

how would I go about determining the coordinate time(s), $t$, at which the particle’s surface of simultaneity contains the event?

There is no "surface of simultaneity" in the sense I think you mean for a particle whose worldline is a helix. By which I mean, there is no way to slice up spacetime into "surfaces of simultaneity" for the particle such that (a) each surface is orthogonal to the particle's worldline at the event at which it intersects that worldline, and (b) every event in the spacetime lies on one and only one surface of simultaneity.

In other words, what looked to you like a simple question is actually opening a very large can of worms.

 

[Ibix]

Dolby and Gull have a method of assigning coordinates based on radar that will work here. It won't necessarily give the coordinates you were looking for, bit it will systematically and uniquely assign coordinates. I suspect the result will be transcendental, I'm afraid, and there has to be some kind of screw discontinuity at the origin.
https://arxiv.org/abs/gr-qc/0104077

As Peter says, there is no unique way to do this. I believe this realisation started Einstein on the path to GR.

 

[Pencilvester]

there is no way to slice up spacetime into "surfaces of simultaneity" for the particle such that (a) each surface is orthogonal to the particle's worldline at the event at which it intersects that worldline, and (b) every event in the spacetime lies on one and only one surface of simultaneity.

I did realize this fact, but I was just hoping there might be a way to get a nice general solution that maybe included a $\pm$ or something that would yield two distinct positions of the particle for any given event. But judging by your response, mine was a false hope. So does it really just come down to using Newton’s method to get an approximation for each event?

 

[Pencilvester]

So does it really just come down to using Newton’s method to get an approximation for each event?

Never mind, just saw Ibix’s response.

 

[PeterDonis]

just saw Ibix’s response.

Yes, Dolby & Gull radar coordinates are one way to do it, but for the case you're looking at, if you're willing to accept that the surfaces of simultaneity won't be orthogonal to the worldline at every intersection point (which you have to accept anyway), then a simple way of choosing coordinates is just to choose the coordinates of the inertial frame in which the center of the circle the particle is moving in is at rest. Those are the coordinates in which you wrote down the parameterization of the particle's worldline in the OP. In these coordinates, events with the same time coordinate $t$ are simultaneous, and that's all there is to it.

 

[Pencilvester]

Thanks for stopping me wasting my own time, @PeterDonis and @Ibix!