Confirm: Smooth Twin Paradox Intuition

[epovo]

I'd like someone to confirm whether I am on the right track here.
Most formulations of the twin paradox involve a sharp turn-around with infinite acceleration. I suppose that there is an SR-only description of a non-infinite acceleration - a kind of 'smooth' version of the twin paradox. But my maths are a bit rusty to work that out. I suppose that as a→∞, we will get the sharp twin paradox.
If we invoke GR to let our traveling twin to declare himself at rest, then during the acceleration phase a uniform gravitational field fills the universe. During that phase our traveling twin experiences gravitation and his clock starts ticking more slowly than his far-away, free falling twin. The more far-away the stay-at-home twin is, the larger his gravitational potential is and the fastest his clock will tick with respect to the traveling twin. I feel confident that (if I were able to work out the math, which sadly I am not), the equations for the GR solution and the SR solution would yield the same result.
Is my intuition correct?

 

[PAllen]

Yes, what you say is generally correct.

 

[Bill_K]

Funny, I would not have said that. I would have answered that the effect of smooth acceleration can and must be understood in terms of special relativity. It has nothing to do with GR, and trying to explain it by imagining that the universe has been filled with a uniform gravitational field is not only incorrect, but also actually harmful, if one's goal is to gain better understanding. So there.

 

[pervect]

I remember a quote from Einstein on the topic:

http://www.bartleby.com/173/20.html

We can now appreciate why that argument is not convincing, which we brought forward against the general principle of relativity at the end of Section XVIII. It is certainly true that the observer in the railway carriage experiences a jerk forwards as a result of the application of the brake, and that he recognises in this the non-uniformity of motion (retardation) of the carriage. But he is compelled by nobody to refer this jerk to a “real” acceleration (retardation) of the carriage. He might also interpret his experience thus: “My body of reference (the carriage) remains permanently at rest. With reference to it, however, there exists (during the period of application of the brakes) a gravitational field which is directed forwards and which is variable with respect to time. Under the influence of this field, the embankment together with the Earth moves non-uniformly in such a manner that their original velocity in the backwards direction is continuously reduced.”

So I wouldn't say the OP is wrong.

I have some sympathy for Bill K's point of view, though. I think a more careful way of describing what happens is that we are identifying the "gravitational field" with the Christoffel symbols. And while we appear to be describing gravity as arising due to a change of motion, more fundamentally it's a change in coordinates that causes the Christoffel symbols to appear.

Furthermore, the fact that they "apppear" and "disappear" in this manner is possible only because the Christoffel symbols aren't tensors. If they were tensors, a change in coordinates wouldn't have this effect.

Unfortunately, I think the more careful explanation is going to be opaque to a large section of the readership.

 

[phinds]

I'd like someone to confirm whether I am on the right track here.
Most formulations of the twin paradox involve a sharp turn-around with infinite acceleration. I suppose that there is an SR-only description of a non-infinite acceleration - a kind of 'smooth' version of the twin paradox. But my maths are a bit rusty to work that out. I suppose that as a→∞, we will get the sharp twin paradox.
If we invoke GR to let our traveling twin to declare himself at rest, then during the acceleration phase a uniform gravitational field fills the universe. During that phase our traveling twin experiences gravitation and his clock starts ticking more slowly than his far-away, free falling twin. The more far-away the stay-at-home twin is, the larger his gravitational potential is and the fastest his clock will tick with respect to the traveling twin. I feel confident that (if I were able to work out the math, which sadly I am not), the equations for the GR solution and the SR solution would yield the same result.
Is my intuition correct?

You're over complicating things for a "smooth twin paradox". Try this:

https://www.physicsforums.com/showthread.php?t=551083&highlight=twin+paradox

 

[Bill_K]

I remember a quote from Einstein on the topic:

Same difference - if Einstein said that the twin paradox results from a universe-filling gravitational field - well he's wrong too!

Unfortunately, I think the more careful explanation is going to be opaque to a large section of the readership.

Hm, that attitude probably explains some of the things that come out of Brian Greene's mouth!

I think a more careful way of describing what happens is that we are identifying the "gravitational field" with the Christoffel symbols.

No need to invoke anything like Christoffel symbols. The stated idea was to "let our traveling twin to declare himself at rest". All this means is, at each point of his world line, take his instantaneous rest frame and draw the surface τ = const. In a loose sense these serve to represent his proper time τ at points off his world line, and in particular at points on Earth's world line. it's like using a Rindler coordinate. But it enables a point-by point comparison of his proper time with Earth's time.

 

[thenewmans]

I’ll take a crack at it. I’m not great with GR. But SR I got. GR covers the special case of SR. So, yeah, you should get the same answer. But I don’t think it’s the way you’re thinking. It’s not the acceleration. The twin paradox can be demonstrated without any acceleration if you use a third traveler.

Let’s say the triplets are Alison, Brenda and Cathy and Alpha Centauri is 2 lightyears away. Alison travels out past Alpha Centauri and waits a couple years to return. Next Brenda flies out. To Cathy knows enough to take into account the time it takes for light to travel when figuring out how fast Brenda is traveling. Given that, it looks to Cathy that Brenda is traveling at 80% C. To Cathy it should take Brenda 2.5 years to get to Alpha Centauri. To Brenda, the trip takes 1.2 years. That’s because Alpha Centauri appears closer than it does to Cathy. Just as Brenda gets there, Alison passes her going the opposite direction. To Cathy, Alison is traveling at 80% C. Alison has already gone through her acceleration by the time the pass happens. To Cathy, the total travel time is 5 years. But when Alison and Brenda get a chance to total it up, it’s 2.4 years. No acceleration was needed for this time dilation or the turnaround.

Let’s take the moment Alison passes Brenda. To Cathy, that happens at 2.5 years. To Brenda, only 9 months have passed on earth. To Alison, it appears 4 years and 3 months have passed on Earth since Brenda left it. So what happened to the 3.5 years in between? That’s usually explained by the turnaround. But there was no turnaround. Therefore, that 3.5 year gap is not explained by the acceleration in the turnaround.

It’s explained by the shift in the lines of simultaneity. If you ask me, the trick to understanding relativity is understanding simultaneity. In gravity or acceleration, time is curved making it harder to calculate the effect they have on things happening at the same time. SR makes all that a lot easier since it’s relatively flat.

http://en.wikipedia.org/wiki/Relativity_of_simultaneity

But let’s say we were to try to explain it all with acceleration. It that case, the traveler would have to spend 3.5 years turning around. I’m not about to guess at the force or Gs it would take to go from -80% C to 80% C in 3.5 years. But I’m sure it’s a lot and there would be plenty of time dilation. But it’s not infinite and some time would pass. But let’s say time froze for the traveler. That would do it.

Finally, let’s say the turnaround happens in a day (from the Earth point of view). To the traveler, time outside passes very quickly. But never more than a day can pass outside. To the traveler, the turnaround happens in a fraction of a second.

Sorry. I think I just got carried away with my answer.

 

[epovo]

Thank you all. I understand the 'sharp' twin paradox in SR, it causes no problems for me. I was just wondering if the smooth twin paradox can be also solved with SR (I suspected yes, and now I understand that it can indeed be done). Then I wondered what it would be like for the traveling twin if he considered himself at rest all the time. During the turn-around time, he would feel a strong gravitation that would push him to against the wall (or the floor). His twin (millions of miles up in the gravitational field) would behave like a ball that a child throws up, stops and comes down again. When the acceleration disappears, gone is the gravitational field as well.
Of course, this gravitational field filling the whole universe instantaneously has no physical existence, I see it rather as a mathematical consequence of the equivalence principle.
The fact that GR and SR give the same result is (for me) an astonishing proof of the internal consistency of relativity.
I am no physicist and my maths are poor, so this talk of Rindler coordinates and Christoffel symbols really went over my head. :( sorry.

 

[Bill_K]

Thank you all. I understand the 'sharp' twin paradox in SR, it causes no problems for me. I was just wondering if the smooth twin paradox can be also solved with SR (I suspected yes, and now I understand that it can indeed be done). Then I wondered what it would be like for the traveling twin if he considered himself at rest all the time. During the turn-around time, he would feel a strong gravitation that would push him to against the wall (or the floor). His twin (millions of miles up in the gravitational field) would behave like a ball that a child throws up, stops and comes down again. When the acceleration disappears, gone is the gravitational field as well.
Of course, this gravitational field filling the whole universe instantaneously has no physical existence, I see it rather as a mathematical consequence of the equivalence principle.
The fact that GR and SR give the same result is (for me) an astonishing proof of the internal consistency of relativity.
I am no physicist and my maths are poor, so this talk of Rindler coordinates and Christoffel symbols really went over my head. :( sorry.

Sure, epovo, but acceleration is not the same as gravity! That is not what the equivalence principle says. Tie a ball to a string and swing it around over your head. It accelerates inward. Would you say that's because there's an inward gravitational field, and general relativity is required?

 

[epovo]

@thenewmans: I am not sure I followed your story of the triplets. I am aware that you don't need acceleration to explain the twin paradox. It's the twin changing from one inertial frame to another that does it - therefore changing his plane of simultaneity.

The GR explanation is completely different. It does NOT rely on the concept of planes of simultaneity. The explanation is wholly due to the effect of gravitation on the passage of time. Even if the acceleration lasts for a single second, during that second the stay-at-home twin can age years (as measured by the accelerating twin) - it's proportional to the distance that separates them (in other words, to the gravitational potential of the stay-at-home twin)

 

[ghwellsjr]

I'd like someone to confirm whether I am on the right track here.
Most formulations of the twin paradox involve a sharp turn-around with infinite acceleration. I suppose that there is an SR-only description of a non-infinite acceleration - a kind of 'smooth' version of the twin paradox.

There certainly is and it's the one Einstein proposed at the end of section 4 of his 1905 paper introducing Special Relativity. There you have the traveler moving in a circle which is a smooth acceleration. You just determine his constant speed and plug that into the Lorentz Factor to calculate a Time Dilation and knowing how long his trip took in the Inertial Reference Frame of the other twin and you calculate their relative aging. It's real simple, don't you think? One frame is all it ever takes to analyze any scenario and if gravity is negligible (or ignored), we don't have to resort to General Relativity, Special Relativity will do just fine.

 

[tom.stoer]

I think we should have a look at the math.

Time dilation can be derived from the rather general proper time formula for an arbitrary (light-like) curve C and a spacetime metric g.

$$\tau[C] = \int_C d\tau = \int_C \sqrt{g_{\mu\nu}\,dx^\mu\,dx^\nu}$$

This formula is valid in GR.

SR follows when restricting to a flat metric, e.g diag(+1,-1,-1,-1). Then the only effect for time dilation is due to the curve C, i.e. due to velocity v along C.

One can rewrite the integral as

$$\tau[C] = \int_C d\tau = \int_{t_a}^{t_b} dt \sqrt{1-\vec{v}^2(t)}$$

where one specific reference frame with coordinates (t,x) is used; t is the time coordinate and v(t) is the velocity along C in this reference frame w.r.t. to the coordinates x(t). The reference frame is an inertial frame, whereas the object moving along the curve C need not define an inertial frame.

Time dilation can be derived via comparing two curves C_i for two different observers i=1,2 with intersecting world lines. Using two curves C_i you will find two proper times τ_i

$$\tau[C_i] = \int_{C_i} d\tau = \int_{t_a}^{t_b} dt \sqrt{1-\vec{v}_i^2(t)}$$

The two curves are defined such that the they intersect at coordinate time t= t_a and t_b. At t_b the two observers i=1,2 can compare their proper times τ_i.

Using this formula one can calculate time dilation for two observers along arbitrary, time-like, intersecting curves with velocities v(t) along the curves. It is remarkable that only v₂ is required to find τ.

 

[epovo]

Yes, ghwellsjr. The circling example is very good and simple. However my aim is to reconcile the SR description with the GR description. For the circling twin, it's the stay-put twin who does the circling -within a gravitational field. For him to calculate how his twin ages I suppose the maths are difficult.

 

[tom.stoer]

Yes, ghwellsjr. The circling example is very good and simple. However my aim is to reconcile the SR description with the GR description. For the circling twin, it's the stay-put twin who does the circling -within a gravitational field. For him to calculate how his twin ages I suppose the maths are difficult.

I think you need nothing else but the above mentioned formula.

I don't think that it's a good idea to introduce a "gravitational field" to describe the accelaration of the circling twin; it's certainly very complicated (just have a look at the Rinder coordinates for linear, constant accelaration), and it's not necessary.

 

[epovo]

Yes, tom.stoer. Those formulas are all I need. They allow me to compute the proper time of any worldline between events A and B which have ta and tb time coordinates in an inertial reference frame. Cool. But then I think how things look to someone who is not in an inertial reference frame. Let's take the circling twin (B) who considers himself at rest. He sees his twin (A) circling and he tries to figure out why A ends up older than himself. To do that he has to postulate a gravitational field, in which A does the circling. When A is climbing up that circle, he is gaining gravitational potential and when he's climbing down, he's losing it. At the top, A's clock ticks faster than B's. But there is also the Lorentz time dilation to consider, which has the opposite effect. The gravitational effect needs to trump that the Lorentz time dilation to explain why A is older after completing the circle (if B chooses to ignore the gravitational effect, he would expect A to be younger).

 

[Myslius]

I don't think acceleration has anything to do with twin paradox, it's perfectly fine to approximate as instantaneous turn around. If it's hard to imagine a body turning around so quickly, you can imagine some particles turning around quickly. Particles at LHC are moving quite fast, with Lorentz factor of 7500 or similar, and they are turning around quite fast with respect to us. So a twin turning around without approximation could be like half of the circle in LHC, where he spends a second (not years) for turn around (according to non moving twin on earth), and even less for moving twin. Also, turn around has nothing to do with the distance traveled, and time dilation does. So if twin travel time is longer, in turn-around he experiences same G forces, but time difference will be different depending on travel time.
Here's some calculation, twin at Earth A, and twin on ship B,
if B were traveling at 0.8c for 10 years forward and back according to A's clock, time difference would be 4 years. If he did the same for 20 years - 8 years. But B experiences same G forces due to acceleration and due to turn-around.

 

[epovo]

Yes, Myslius. Twin B experiences the same G forces, whether he's been traveling for 10 or 20 years. That bothered me a lot when I started thinking about this. But think that under GR rules, he can consider himself at rest - and then figure out why it is that his stay-at-home twin has aged twice as much when he takes the 20-year trip than when he took the 10 year trip. Until I realized that it is the gravitational potential of the stay-at-home twin that makes the difference.To reiterate: I understand the twin paradox in SR, both the sharp and the smooth variety. I am just trying to reconcile it with the description of our traveling twin, who stubbornly insists that he doesn't move at all. We need GR for that and we need that spooky gravitational field which, I admit, it's just a mathematical tool rather than a *real* gravitational field that instantaneously appears in the whole universe and has no effect on anyone except him.

 

[Myslius]

Something like this:

If I understood you correctly, twin B should also have gravitational potential something like mg(1 - v/c) * R (Newtonian approximation), just smaller one. Anyway, I would like to see twin's paradox solution from GR perspective.

 

[tom.stoer]

I think I understand what epovo has in mind:
- find a time-dependent metric (*) in which the second twin moving w.r.t. to the Minkowski metric is at rest
- calculate the proper times using this metric

(*) this is analogous to the Rindler metric for linear, constant acceleration

 

[PAllen]

I think I understand what epovo has in mind:
- find a time-dependent metric (*) in which the second twin moving w.r.t. to the Minkowski metric is at rest
- calculate the proper times using this metric

(*) this is analogous to the Rindler metric for linear, constant acceleration

Right, and this is the sense in which it has validity. I always prefer to say the proper explanatory direction is to do this, using 'GR techniques' in SR; then justify what this predicts about behavior of time in situations with gravity. Thus, instead of using GR based arguments to justify non-gravity situations, it is more logical to me to use SR treatment of acceleration to see why gravity must have temporal effects.

 

[epovo]

Yes, tom.stoer. That's what I mean.
My apologies for this talk about gravitational fields to refer to non-Minkowski metrics.

 

[ghwellsjr]

Yes, ghwellsjr. The circling example is very good and simple. However my aim is to reconcile the SR description with the GR description. For the circling twin, it's the stay-put twin who does the circling -within a gravitational field. For him to calculate how his twin ages I suppose the maths are difficult.

If you're trying to explain a twin paradox scenario in terms of only gravity, then having the stay-put twin circling will be a combination of time dilation due to speed and gravity, won't it?

How about a scenario where twin B is in a rocket undergoing constant circular acceleration a fixed distance around twin A who is inertial in space, no significant gravity anywhere. It would be easy to analyze the difference in aging per "orbit" according to SR. Then you could consider twin B to be at rest on an insignificantly small planet with sufficient mass to produce the same force of acceleration as the rocket did. Twin A is in orbit around the planet with the same period as Twin B determined from his Proper Time. Then you have the planet rotate at the same rate so that Twin B sees Twin A in synchronous orbit directly overhead and not moving. Now you're going to analyze everything from the non-inertial frame in which Twin B is stationary. Both twins can be considered stationary in this frame (remember, the planet is insignificantly small) but experiencing different gravitational potentials (I think that's the right term).

I have no idea how to do this analysis or even if the whole concept is legitimate but it would be interesting to see if some of the experts in GR on this forum could provide a solution or explain why it is not valid.

 

[tom.stoer]

Does anybody know about a metric solving this problem?

 

[Dale]

Does anybody know about a metric solving this problem?

Sure. Start with a standard inertial frame $(T,X,Y,Z)$ with metric $ds^2=-c^2 dT^2+dX^2+dY^2+dZ^2$ and consider a transform to a rotating grid $(t,x,y,z)$ as follows:
$t=T$
$x=X cos(\omega T) + Y sin(\omega T)$
$y=Y cos(\omega T) - X sin(\omega T)$
$z=Z$

Solving for $(T,X,Y,Z)$ and substituting into the metric we get:
$ds^2=(-c^2+(x^2+y^2)\omega^2)dt^2+2x \, \omega \, dy \, dt -2y \, \omega \, dx \, dt + dx^2 + dy^2 + dz^2$

 

[DrGreg]

Sure. Start with a standard inertial frame $(T,X,Y,Z)$ with metric $ds^2=-c^2 dT^2+dX^2+dY^2+dZ^2$ and consider a transform to a rotating grid $(t,x,y,z)$ as follows:
$t=T$
$x=X cos(\omega T) + Y sin(\omega T)$
$y=Y cos(\omega T) - X sin(\omega T)$
$z=Z$

Solving for $(T,X,Y,Z)$ and substituting into the metric we get:
$ds^2=(-c^2+(x^2+y^2)\omega^2)dt^2+2x \, \omega \, dy \, dt -2y \, \omega \, dx \, dt + dx^2 + dy^2 + dz^2$

When expressed in cylindrical polar coordinates, these are known as Born coordinates
$$ds^2 = -\left( c^2 - \omega^2 \, r^2 \right) \, dt^2 + 2 \, \omega \, r^2 \, dt \, d\phi + dz^2 + dr^2 + r^2 \, d\phi^2$$

- find a time-dependent metric

Actually the metric coefficients are time-independent but space-dependent, which accounts for the time dilation.

 

[epovo]

Wow!

 

[WannabeNewton]

When expressed in cylindrical polar coordinates, these are known as Born coordinates
$$ds^2 = -\left( c^2 - \omega^2 \, r^2 \right) \, dt^2 + 2 \, \omega \, r^2 \, dt \, d\phi + dz^2 + dr^2 + r^2 \, d\phi^2$$

The twin traveling the arc of the circle would be at rest in these coordinates correct? He/she would be following an orbit of $e_{0} = \frac{1}{\sqrt{c^{2} - \omega^{2}r^{2}}}\partial_{t}$ which, in Born coordinates, would just be a straight line in the "time direction" of said coordinate system yes?

 

[DrGreg]

The twin traveling the arc of the circle would be at rest in these coordinates correct? He/she would be following an orbit of $e_{0} = \frac{1}{\sqrt{c^{2} - \omega^{2}r^{2}}}\partial_{t}$ which, in Born coordinates, would just be a straight line in the "time direction" of said coordinate system yes?

Yes, the traveller is at rest, so $dr=dz=d\phi=0$ and the only relevant value is the coefficient of $dt$ which is $r$-dependent.

 

[WannabeNewton]

Yes, the traveller is at rest, so $dr=dz=d\phi=0$ and the only relevant value is the coefficient of $dt$ which is $r$-dependent.

And as for the twin who was at rest in the global inertial coordinates of Minkowski space-time: would he/she now be following an orbit (in Born coordinates) of something like $u = \frac{1}{\sqrt{1 - \omega^{2}r^{2}}}\partial_{t} - \frac{\omega r}{\sqrt{1 - \omega^{2}r^{2}}}\frac{1}{r}\partial_{\phi}$ i.e. the traveling twin who is now at rest in Born coordinates would see the "stay at home" twin moving along an arc of the same (allowable) radius but in the opposite azimuthal direction, as represented in Born coordinates?

 

[Dale]

If I understood you correctly, twin B should also have gravitational potential something like mg(1 - v/c) * R (Newtonian approximation), just smaller one.

So in the coordinate system I posted above the time dilation formula can be derived as follows:

$-c^2 d\tau^2=(-c^2+(x^2+y^2)\omega^2)dt^2+2x \, \omega \, dy \, dt -2y \, \omega \, dx \, dt + dx^2 + dy^2 + dz^2$

$d\tau^2/dt^2=1-(x^2+y^2)\omega^2/c^2-2x/c^2 \, \omega \, dy/dt +2y/c^2 \, \omega \, dx/dt - (dx^2/dt^2 + dy^2/dt^2 + dz^2/dt^2)/c^2$

$d\tau^2/dt^2=1- \beta^2 - (x^2+y^2)\omega^2/c^2 + 2\omega/c^2 (-x \, dy/dt +y\, dx/dt)$

The first two terms in that expression are the standard velocity time dilation that you are used to in SR. The third term could be considered a gravitational time dilation as a function of the distance from the z axis. But the fourth term has no such interpretation that I am aware of. It is dependent on velocity, not speed, so it is not like the usual time dilation in a static gravitational field nor is it like the usual SR time dilation. You could consider it to be a gravitational time dilation due to the "gravity" of the Coriolis force, but that seems stretching the analogy a little too far for me.

 

[DrGreg]

And as for the twin who was at rest in the global inertial coordinates of Minkowski space-time: would he/she now be following an orbit (in Born coordinates) of something like $u = \frac{1}{\sqrt{1 - \omega^{2}r^{2}}}\partial_{t} - \frac{\omega r}{\sqrt{1 - \omega^{2}r^{2}}}\frac{1}{r}\partial_{\phi}$ i.e. the traveling twin who is now at rest in Born coordinates would see the "stay at home" twin moving along an arc of the same (allowable) radius but in the opposite azimuthal direction, as represented in Born coordinates?

I have to confess, on re-reading post #18, I'm not clear what the diagram there means. At first glance I had assumed both twins were at rest relative to the rotating disk, but in that case I don't understand what the "0 c" and "0.8c" would mean. Is B supposed to be moving "radially" (or rather "spirally"?)

The simplest scenario is simply to have one twin on the periphery of the disk and the other at its centre. Then both are at rest relative to the disk, so we just need to compare proper time at r=0 and r=r₀ (both for constant phi and z).

For a twin at rest in the inertial frame, but not at the disk centre, yes I think what you said is correct.

 

[WannabeNewton]

Oh ok, I didn't even catch that diagram. I was picturing something completely different: a "stay at home" twin and a second twin equipped with a rocket engine who would go out into space and at some point fire the rocket so as to traverse a semicircular arc and turn off the rocket to then head back the same path, towards the "stay at home" twin. But you are correct that the physical scenario in the diagram would be much simpler because we won't have to somehow "smoothly patch together" different metrics (i.e. for the transition from rocket off to rocket on and then back off) to describe the entire round trip if we wanted to go the metric tensor route.

In the scenario given in the diagram, it seems the traveling twin is, from the start, traveling in a circular path such that the "stay at home" twin is at the center of this circular path (which is the origin of the Born coordinates). The only problem I have with this is that the Born metric in Born coordinates is only defined for $0 < r < \frac{1}{\omega}$ so how would we even describe the worldline of the central twin in Born coordinates?

 

[Dale]

the Born metric in Born coordinates is only defined for $0 < r < \frac{1}{\omega}$

Hmm, that doesn't seem right to me. It should be $0<r<\infty$

 

[WannabeNewton]

Hmm, that doesn't seem right to me. It should be $0<r<\inf$

Those are indeed the bounds for the Minkowski metric in cylindrical coordinates but when we simultaneously boost to the momentarily comoving inertial frames of a circling observer at all events on his/her worldline, and construct the Born metric in Born coordinates, we have to restrict ourselves to the open subset $0 < r < \frac{1}{\omega}$ of Minkowski space-time so that the boost speed doesn't exceed the speed of light i.e. $v = \omega r < 1$ where $v$ is the boost speed.

 

[Dale]

There is no reason to do that. You just get that there are no timelike worldlines at rest outside that critical radius.