Lorentz Invariant Paths in Spacetime

[snoopies622]

Last night I was pleasantly surprised to discover that, given a particle trajectory

$x^2 - c^2t^2 = a^2$

when viewed through a Lorentz transformation

$x' = \gamma (x-vt)$
$t' = \gamma (t - vx/c^2)$

produces exactly the same shape

$x'^2 - c^2t'^2 = a^2$.

I suppose this is equivalent to the way a circle of radius $a$ looks the same after an ordinary coordinate system rotation. My question is simply, are there shapes besides this hyperbola with this Lorentz invariant property? Thanks.

 

[PeterDonis]

I suppose this is equivalent to the way a circle of radius aa looks the same after an ordinary coordinate system rotation.

It's analogous, yes. A Lorentz boost is a hyperbolic "rotation".

are there shapes besides this hyperbola with this Lorentz invariant property?

What, exactly, is the property you think is Lorentz invariant here?

 

[snoopies622]

I would call the trajectory $x^2 - c^2t^2 = a^2$ Lorentz invariant because after being subjected to a Lorentz transformation it still looks exactly the same.

This is helpful for me to see how this path represents uniform proper acceleration, since for any instantaneous co-moving reference frame, the rightward acceleration at t=0 (the moment the particle has velocity zero) must be the same as well. $c^2/a$ in this case I think.

 

[PeterDonis]

I would call the trajectory $x^2 - c^2t^2 = a^2$ Lorentz invariant because after being subjected to a Lorentz transformation it still looks exactly the same.

What does "looks exactly the same" mean? I know it seems obvious to you, but it's not; you need to carefully state, in precise language, what you mean. Otherwise there is no way to answer the question you posed in the OP, because the question itself is not well defined.

 

[PeterDonis]

This is helpful for me to see how this path represents uniform proper acceleration, since for any instantaneous co-moving reference frame, the rightward acceleration at t=0 (the moment the particle has velocity zero) must be the same as well. $c^2/a$ in this case I think.

As you've written it, yes, $c^2 / a$ is the proper acceleration. Most treatments use $a$ for the proper acceleration, in which case the equation would be $x^2 - c^2 t^2 = c^4 / a^2$.

 

[snoopies622]

What does "looks exactly the same" mean?

Well, $x^2 - c^2t^2 = a^2$ and $x'^2 - c^2t'^2 = a^2$ are the same equation, the only difference is the change from one coordinate system (x,t) to the other (x',t').

 

[PeterDonis]

Well, $x^2 - c^2t^2 = a^2$ and $x'^2 - c^2t'^2 = a^2$ are the same equation

Ok, then your question could be rephrased as, what equations relating the coordinates are left invariant by Lorentz transformations. However, you should note that this is not the same question as the one in the title of this thread, what paths are left invariant by Lorentz transformations.

The answer to the "what equations" question is that only the hyperbola equations are left invariant by Lorentz transformations. The analogy with circles and ordinary rotations holds here (the circle equations are the only equations left invariant by Lorentz transformations).

However, any path through spacetime is left invariant by Lorentz transformations; Lorentz transformations don't change the geometry of spacetime, or of curves within it, they only change how that geometry and those curves are described by the coordinates. Again, the analogy with ordinary rotations is instructive: an ordinary rotation doesn't change the geometry of a plane or of curves on it, it only changes how those curves are described by the coordinates.