SR Thought: Spacetime Interval w/ Bounds of ##0<\alpha<1##

[benorin]

TL;DR

Just a note, wonder where the rabbit hole leads

Suppose the time interval in the Lab frame is a multiple of the time interval in the Rocket frame $\alpha \Delta t_L = \Delta t_R$, where $0 < \alpha < 1$ without loss of generality. Then the spacetime interval is

$\left( \Delta t_L\right) ^2-\left( \Delta x_L\right) ^2 = \left( \Delta t_R\right) ^2-\left( \Delta x_R\right) ^2\Rightarrow \left( \Delta t_L\right) ^2=\frac{\left( \Delta x_L\right) ^2 -\left( \Delta x_R\right) ^2}{1-\alpha^2}$

The thought ended abruptly there. I guess I just wanted to jot this down before I forgot, and hey, if you've got some good idea how to proceed from here: post it! Thanks!

Edit: something is off, like a minor error... but I can't put my finger on it. Should the bounds for be $1< \alpha < \infty$?

 

[Nugatory]

The spacetime interval between which pair of events? You haven’t specified that, and until you do there’s no way of checking your calculation.

Remember that an event is a point in spacetime, not time. “The moment when the lab clock reads $T$ and the rocket clock reads $T/\gamma$” is not an event. “The lab clock at this point in space reads $T$” is an event. “The rocket clock at that point in space reads $T/\gamma$” is a different event even if it happens at the same time.

So the first step is for you to write down the $x$ and $t$ coordinates of two events. You can use either the lab frame or the rocket frame to assign these coordinates, as long as you’re consistent and only use one frame or the other.

Once you’ve done that and you can calculate the spacetime interval between the two events, use the Lorentz transformations to find the coordinates of the two events in the other frame, and then check that you get the same interval using those coordinates.

 

[Ibix]

As Nugatory says, this depends on the pair of events you are considering. You can make $\alpha$ anything between $\pm\infty$ depending on that choice.

I would guess that you are thinking of $\alpha$ as the time dilation factor, which would imply $\alpha=\sqrt{1-v^2}$ (i.e. your $\alpha$ is the reciprocal of Lorentz' $\gamma$). If I've guessed correctly, then for your initial equation to hold you need $\Delta x_R=0$. In that case your final result can be rearranged to read $(\Delta x_L)^2=(v\Delta t_L)^2$, a restatement of the fact that the rocket moves at $v$ in the lab frame.

 

[benorin]

I was indeed as you suspected Ibix thinking along the lines of a rocket going with constant velocity through the Lab frame so that $\Delta x_R =0 \Rightarrow \alpha = \sqrt{1-\left( \frac{\Delta x_L}{\Delta t_L}\right) ^2}$ is what I derive from the OP. Now my question is doesn't $\frac{\Delta x_L}{\Delta t_L}:=v$? But then I also worked out the moving light-clock problem and suspect this should rather be $\frac{v}{c}$ to give the the expected Lorentz equation (or time dilation equation, whatever?). What am I missing?

 

[Ibix]

It's usually easier to work in units where $c=1$, since then you can drop factors of $c$ that pop up all over the place. I did that without commenting - apologies. Your $\alpha$ is $\sqrt{1-v^2/c^2}$ if you don't work in such units, yes.