It is important that this motion,

if you accelerate for a finite period of time, is homogenous.

It is this moment of time when you switch on eternal acceleration and

switch it off, is of course different from any other moment on this world line.

And that moment, any moment of time on this part of the world line,

is different from any moment here, and different from any moment here.

So, this motion isn't homogenous,

and that is the reason we do not consider this kind of motion.

Because in this case we do not expect to get such a nice looking metric.

We expect to get such a nice looking stationary metric,

only if we consider homogenous eternal acceleration.

That is the reason we consider this a physical situation,

which allows us to draw some interesting conclusions.

That's the reason we consider this.

So, now what is important, that if you eternally accelerate,

if your acceleration is constant, let me remind you that

here you decelerate starting from the speed of light, you eternally decelerate.

Then here you accelerate actually,

because three-dimensional acceleration is always directed along this line.

But what is important that if someone decides at this moment of time, t = 0,

emit a light ray which tries to catch up with you, this light

ray is always parallel to this asymptote and never intersects with the hyperbola.

It means that the light ray emitted from any point behind the asymptote,

never catches up with the eternally accelerated observer.

But it does catch up with someone who stopped the acceleration

at finite moment of time.

Of course, because if you stop accelerating at finite moment of time,

you continue your motion along the line tangential

to the hyperbola because it has angle less than 45 degrees with respect to this line.

It is not parallel to this line, it intersects with it.

So, only in case if you externally accelerate,

you have this particularity that the light ray doesn't catch up with you.

How does this fact reveal itself in this formula?

It reveals itself as follows.

Suppose you have a motion with constant velocity along this direction.

Then it means that dx1/dt, in our annotation, is equal to c which is 1.

As a result, ds squared = 0.

But if ds squared = 0,

here we obtained that d rho/d tau = rho,

which is not necessarily equal to 1.

It is equal to 1 only if rho equals to 1.

So now we obtained the strangest situation, that in non-inertial reference

frame, we obtained that the speed of light does depend on the coordinate.

And that has to do with these particularities that I

just described to you.

It becomes even 0 when rho goes to 0,

when the hyperbolas degenerate to these straight lines.

So if light ray is emitted in the very vicinity of this asymptote and

tries to catch up with someone who accelerates, it takes very long time for

this light ray to catch up with the acceleration.

That reveals itself through this formula.

And this degeneracy of the metric is the price we have to pay

to consider such unphysical situation of eternal acceleration,

which has this property that light rays can not catch up with it.

So, this is the first thing I wanted to stress.

And another interesting observation that one can make here is that, for

example, consider an observer which is just stationary in the Minkowskian metric.

Its world line in this coordinate looks like this.

Of course it's world line in this coordinate looks as follows.

But it is important to stress that for this particle,

which is just stationary in original Minkowski spacetime,

it takes infinite time, tau, to get from here to here.

In fact, this line corresponds to tau =- infinity,

and this line corresponds to tau = + infinity.

In new coordinate system it takes infinite time, tau, to get from here to here.

So if we for example try to draw this coordinate system in the rectangular

fashion, so it means that we can see the, let me stress that we can see the rho > 0.

So we can see the lines of constant rho like this.

These are just lines of constant rho.