I Trying to get an explicit function(ish)

[Pencilvester]

How would I go about solving for $t_p$ in the following equation:$$t_p - t + vy \cos {(2 \pi \omega t_p )} - vx \sin {(2 \pi \omega t_p )} = 0$$where our input is a point in $ℝ^3$ with coordinates $t$, $x$, and $y$, and where $v$ and $\omega$ are constants. I’m pretty sure it can’t be a function exactly, as I’m pretty sure most, if not all input points will each yield 2 distinct outputs. If it matters to you, $|v| < 1$, but I don’t think that it’s relevant to this problem. And this isn’t any kind of homework problem. I’m not in school, I’m just trying to analyze what a coordinate transformation might look like for going from coordinates of an inertial observer in flat spacetime to coordinates of an observer tracing out a helix in spacetime (or a circle in space), and I’m running into this limitation in my mathematical abilities, so any help would be much appreciated.

 

[mfb]

In general there is no closed solution - you can only find approximations. It is possible to combine the cosine and sine term to a single sine (or cosine) with an additional phase, that makes the problem a bit easier to look at. t_p cannot be too different from t as the other terms cannot get larger than vx or vy, respectively, this gives you a region to look for solutions. Once you found a t_p that is not too far away from a solution tools like Newton's method will work.

 

[WWGD]

Have you tried the implicit function theorem https://duckduckgo.com/?q=implicit+function+theorem&t=hf&atb=v88-7&ia=web ?

Gives conditions under which (at least a ) local expression exists.

 

[Pencilvester]

In general there is no closed solution - you can only find approximations. It is possible to combine the cosine and sine term to a single sine (or cosine) with an additional phase, that makes the problem a bit easier to look at. t_p cannot be too different from t as the other terms cannot get larger than vx or vy, respectively, this gives you a region to look for solutions. Once you found a t_p that is not too far away from a solution tools like Newton's method will work.

Dang. That’s dissatisfying for me. Well at least I didn’t spend much time trying to find a general solution before posting this. Thanks!

 

[Pencilvester]

Have you tried the implicit function theorem https://duckduckgo.com/?q=implicit+function+theorem&t=hf&atb=v88-7&ia=web ?

Gives conditions under which (at least a ) local expression exists.

I had forgotten about that, thanks!