PeterDonis said:
No, this is wrong. If you just rescale ##\tau## but don't change ##\rho##, then your observer is still at ##\rho_0 = 0.5## and his proper time is still ##\tau' / 2##. So, in these new coordinates, ##\Delta \rho## stays the same and ##\Delta \tau## gets multiplied by ##2##, so coordinate speed is half what it was in the old coordinates. And the "speed" for your observer by your definition is still twice the coordinate speed, so it also is half what it was in the old coordinates.
In other words, in these new coordinates, the coordinate speed will be ##\approx 0.6664## and the "speed" for your observer by your definition will be ##\approx 1.3328##.
Let me do it in detail, I change from ##(\tau, \rho)## to ##(\tau', \rho)##, therefore my new metric is ##ds^2=\frac{1}{4}\rho^2d\tau'^2 - d\rho^2##. And I'm still at ##\rho=0.5## so my proper time is related to ##\tau'## as follows:
$$ds^2=\frac{1}{4}\rho_0^2d\tau'^2 \Longrightarrow \Delta T = \frac{\rho_0}{2}\Delta \tau'$$
Now before the two event were separated by ##\Delta \tau = 0.55 \Longrightarrow \Delta \tau'=1.1##. Therefore the time between the two events is still ##\Delta T = \frac{0.5 \cdot 1.1}{2}=0.27## as before, I haven't change the way I define distance, so velocity is still ##2.7##.
However, as you say, coordinate velocity has changed from ##1.33## to ##0.66##, just because we have changed how we describe mathematically the real world.
Now if I rescale ##\rho'=2\rho## now the metric becomes ##ds^2=\frac{{\rho'}^{2}}{4}d\tau^2-\frac{1}{4}d\rho'^2##. Now of course I'm at ##\rho'_0=2\rho_0=1## and the proper time is
$$ds^2=\frac{{\rho'_0}^2}{4}d\tau^2 \Longrightarrow \Delta T = \frac{\rho'_0}{2}\Delta \tau = \frac{\Delta \tau}{2}$$. So again, the time between the two events is still ##0.27##. But now we have also modify how we measure distance, because at constant times
##-ds^2=\frac{1}{4}d\rho'^2 \Longrightarrow \Delta L = \frac{\Delta \rho'}{2}##. So the two event are separated by ##\Delta \rho' = 1.46##, which means a distance of ##\Delta L = 0.73##, the same as before, if I compute the velocity I get, again ##2.7##. And you're right now the coordinate velocity is in fact ##2.7## so both are equal, but I don't understand why you say that coordinate velocity stays the same while mine changes...
PeterDonis said:
I challenge you to prove this.
Well, then probably I'm wrong, I haven't proved rigorously, but my intuition was something like this:
The proper time between two points of my worldline is just the relativistic distance between them, this is a scalar quantity. I also define the physical distance as the relativistic distance between two points at the same time (because we have a scalar notion of time, there no problem there, and because the relativistic distance is invariant, we should have an invariant "physical" distance). Therefore the velocity as I compute it would be invariant. As I said, this was my reasoning and probably it's wrong. But surely, for the easy cases that I've done before, I still don't understand why the coordinate velocity is better.
PeterDonis said:
Everybody all along has been telling you that the things you are calculating are coordinate speeds, not physical speeds. Apparently you haven't been paying attention.
Well, I'm sorry, English is not my main language, and I really thought that everyone before message #17 was ok with my calculations.