Equations for observed distance/velocity in SR

[PeterDonis]

I was merely exploring this topic on my own and wanted to try to solve for some equations relevant to questions I had about these things, and was excited to have come up with what I was looking for.

It would have been helpful if you had posted how you derived the expressions in the OP, rather than just posting them with no derivation. Not only are they not the standard length contraction/time dilation formulas (which you know), but they are not the standard relativistic Doppler formulas either. So it's not clear exactly what they're supposed to mean.

There is another, deeper issue with what you are trying to do. Basically, it looks like you are trying to construct spacetime coordinates using the relativistic Doppler-shifted observations directly, instead of adjusting them for light travel time as is done with standard SR inertial coordinates. But if you don't adjust for light travel time, you can't construct a valid coordinate chart at all: you do not get a one-to-one mapping between coordinates and events. You have shown some examples of this in your posts. Without a valid coordinate chart, any interpretation of your quantities as times and distances is extremely problematic.

 

[Arkalius]

Ok, I've worked out a derivation that I can show people who are interested. Let's start with an image depicting the scenario:

This is 2d space (no time axis). The observer observes an object at the observed position which has since moved to the current position with velocity ##v##. The value ##d## indicates the object's current distance from the observer, while ##d_o## is the observed distance (the path the light from the observed position took to reach the observer). The angle ##\alpha## represents the angle between the relative velocity vector and the observer (or object, depending on your perspective). The angle ##\alpha_o## is the observed version of that angle by the observer. We note that the distance between the observed position and current position is velocity times the amount of time the light took to reach the observer, so it is equal to ##v d_o / c##. We can use ##\beta = \frac v c## here to end up with ##\beta d_o##.

We can solve for ##d_o## using the law of cosines. This gives us:
$$\begin{align} d^2_o & = d^2 + \beta^2d^2_o - 2d \beta d_o \cos \left(\pi-\alpha\right) \\
0 & = d^2_o - \beta^2d^2_o - 2d \beta d_o \cos \alpha - d^2 \\
0 & = d^2_o \left(1 - \beta^2 \right) - 2d \beta d_o \cos \alpha - d^2 \end{align}$$
(We used ##\cos \left(\pi - \alpha \right) = -\cos \alpha## in here). We now have a quadratic equation in ##d_o## and we can use the quadratic formula to solve it. This gives us:
$$d_o = \frac {2d \beta \cos \alpha \pm \sqrt {4 d^2 \beta^2 \cos^2 \alpha + 4d^2 \left(1-\beta^2 \right)}} {2 \left( 1-\beta^2 \right)}$$
We can cancel out the 2's (and 4's in the radical) easily, and also pull out ##d## from the whole equation. We also note that ##\frac 1 {1 - \beta^2} = \gamma^2## so we can use that as well. With that, we are at:
$$\begin{align} d_o & = d \gamma^2 \left(\beta \cos \alpha \pm \sqrt{\beta^2 \cos^2 \alpha - \beta^2 + 1} \right) \\
d_o & = d \gamma^2 \left(\beta \cos \alpha \pm \sqrt{\beta^2 \left( \cos^2 \alpha - 1 \right) + 1} \right) \\
d_o & = d \gamma^2 \left(\beta \cos \alpha \pm \sqrt{\beta^2 \sin^2 \alpha + 1} \right) \end{align}$$
We note that the value in the radical will always be greater than or equal to 1, and the value of ##\beta \cos \alpha## will always be less than 1, so only the positive square root gives us a sensible answer (otherwise we have a negative distance), so we use the positive only. And this gets us to my original equation: $$\frac {d_o} d = \gamma^2 \left(\beta \cos \alpha + \sqrt{1 + \beta^2 \sin^2 \alpha} \right)$$

To get ##\alpha_o## we can use the law of sines:
$$\begin{align} \frac {\sin \left( \alpha - \alpha_o \right)} {\beta d_o} & = \frac {\sin \left( \pi - \alpha \right)} {d_o} \\
\sin \left( \alpha - \alpha_o \right) & = \beta \sin \alpha \\
\alpha - \alpha_o & = \arcsin \left( \beta \sin \alpha \right) \\
\alpha_o & = \alpha - \arcsin \left( \beta \sin \alpha \right) \end{align}$$
Which is the observed angle equation I posted as well.

Thank you for your attention, and I hope some find this interesting.

 

[Mister T]

The observer observes an object at the observed position which has since moved to the current position

How is the "current position" established? Somehow the observer would have to know where the object is "now". And that notion of what's happening somewhere else right now is relative.

Please understand that I'm not saying anything negative about your derivation. I'm simply pointing out that in order to arrive at an understanding of what you see now you are analyzing what was happening when the light you're seeing left the the thing you're looking at. This involves knowledge of distant position and time coordinates, which are the very same coordinates that appear, for example, in the expression ##(c\Delta t)^2-(\Delta x)^2##.

 

[Arkalius]

How is the "current position" established? Somehow the observer would have to know where the object is "now". And that notion of what's happening somewhere else right now is relative.

Well, the equations could be used in the reverse. An observer can look at where an object appears to be and its observed velocity and work back to where it actually is. However, this really isn't so much about making calculations as the observer in a particular scenario, but rather a way of understanding how our view of reality actually distorts during relativistic motion. If you begin to move at relativistic speeds, your view of reality shows a kind of zoom-out effect, with things stretching away in front of you (or rather the opposite if you look behind). I'd seen depictions of this effect and wanted to better understand the mathematical basis for it, so here we are.

 

[Mister T]

An observer can look at where an object appears to be and its observed velocity and work back to where it actually is.

Right. That's what you're doing in your derivation and there's nothing wrong with doing that. All I'm saying is that the notion of "where it actually is" means you have to assign a time coordinate to that distant "actually is" event. And when you say "is" you're implying that it's the same time at your location as it is at that distant location. There is no universally valid way to do that.

 

[Bartolomeo]

An observer can look at where an object appears to be and its observed velocity and work back to where it actually is.

Aberration of light
https://en.wikipedia.org/wiki/Aberration_of_light