# Every event is simultaneous with you at some time: proof?

Staff Emeritus
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## Main Question or Discussion Point

Theorem: In Minkowski space, let ##r(\tau)## define a timelike, ##C^2## world-line W parametrized by proper time for ##\tau## covering the whole real line. Here ##r## is the radius four-vector extending from the origin O to a given point. Then there exists a unique time ##\tau## such that ##r(\tau)## is orthogonal to W at the point defined by ##\tau##.

******* This version is false. See my attempt at revision below. ********

In other words, if you're an observer, every event is simultaneous with you at some uniquely defined time.

How would you go about proving this? I would prefer a proof that avoids invoking any coordinates.

At first I expected this to be trivial, but it seems a little tricky because it seems like you need a topological component to the argument. For example, I had been thinking to argue something like this. Let ##v(\tau)## be the unit tangent vector to W, and let ##f(\tau)=r(\tau)\cdot v(\tau)##. Then it seems obvious, in Minkowski space, that ##f## starts out negative and ends up positive, and since it's continuous it must be zero somewhere. But the fact that ##f## starts negative and ends positive is something that can be false if, for example, you cut out all of Minkowski space for ##t\ge 0##.

I want to avoid making assumptions in the proof that seem reasonable but that require explicit justification to rule out counterexamples such as a flat spacetime with a nonstandard topology.

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PAllen
2019 Award
I'm confused. There are trivial counter-examples for a non-inertial world line in perfectly normal Minkowski space. Just consider a world line with uniform proper acceleration, e.g. to the left, starting from velocity c to the right at minus infinity, going to c to the left at plus infinity. Then, there is an event on the Rindler horizon to the right, such that two or more different events on the accelerating world line would be simultaneous to this one event, by this definition.

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Dale
Mentor
Then there exists a unique time τ\tau such that r(τ)r(\tau) is orthogonal to W at the point defined by τ\tau.
What is W here?

Erland
??? This must be wrong. For example, the birth of my mother cannot be simultaneous with any event experienced by me. All these events must be later than the birth of my mother, for all observers.

PeterDonis
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2019 Award
the birth of my mother cannot be simultaneous with any event experienced by me
But your ##\tau## doesn't cover the whole real line; it only covers a finite range of it (you only experience a finite range of proper times). The OP is talking about idealized "observers" that exist forever, infinitely in the past and infinitely in the future.

PAllen
2019 Award
??? This must be wrong. For example, the birth of my mother cannot be simultaneous with any event experienced by me. All these events must be later than the birth of my mother, for all observers.
You missed that the world line must extend from -infinity to +infinity of proper time on the world line. I think there may be a problem with the proposition (or with my understanding of it), but that is certainly not the problem.

Staff Emeritus
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What is W here?
Oops, W was meant to be the world-line. I've edited the OP. Thanks!

PAllen
2019 Award
Oops, W was meant to be the world-line. I've edited the OP. Thanks!
That clarifies it, and my objection stands - the uniqueness part of the proposition is false. The existence part is true.

Staff Emeritus
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I'm confused. There are trivial counter-examples for a non-inertial world line in perfectly normal Minkowski space. Just consider a world line with uniform proper acceleration, e.g. to the left, starting from velocity c to the right at minus infinity, going to c to the left at plus infinity. Then, there is an event on the Rindler horizon to the right, such that two or more different events on the accelerating world line would be simultaneous to this one event, by this definition.
I don't understand your counterexample. Let's say the world-line W is ##r(\tau)=(t,x)=(\sinh\tau,-\cosh\tau)## in Minkowski coordinates. Then O=(0,0) is on the Rindler horizon, and ##r(0)=(0,-1)## is orthogonal to W at ##\tau=0##. I don't see how you get another point on W that has this property...??

PeterDonis
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2019 Award
I don't see how you get another point that has this property...??
Every point on the worldline has this property; that is, the line ##r(\tau)## from the origin ##(0, 0)## to every point of the form ##(\sinh \tau, - \cosh \tau)##, for all real values of ##\tau##, is orthogonal to the tangent vector to the worldline at that point. That's a fundamental property of every worldline in a Rindler congruence.

bcrowell
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Every point on the worldline has this property; that is, the line ##r(\tau)## from the origin ##(0, 0)## to every point of the form ##(\sinh \tau, - \cosh \tau)##, for all real values of ##\tau##, is orthogonal to the tangent vector to the worldline at that point. That's a fundamental property of every worldline in a Rindler congruence.
Aha! Very interesting. That invalidates what I thought was a nice simple proof of the 1+1-dimensional Herglotz-Noether theorem ( http://www.lightandmatter.com/sr/ , section 9.5.3).

So maybe this is what I want:

Theorem: In Minkowski space, let W be a timelike, ##C^2## world-line parametrized by proper time ##\tau## for ##\tau## covering the whole real line. Then (1) for any point P, there exists a point Q on W such that line PQ is orthogonal to W. Furthermore, (2) there exists a neighborhood A of W such that if P is chosen from within A, then Q is unique.

If this version is right, then I think my proof of H-N in 1+1 dimensions works, but the theorem has to be weakened in a similar way to the weakening of the theorem above.

PeterDonis
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If this version is right
I think it is, since there is no constraint on how small the neighborhood can be. In fact, I believe the construction of Fermi normal coordinates depends on this theorem being true in a general spacetime, not just Minkowski spacetime; the size of the neighborhood is constrained in the general case by both the path curvature of the worldline (which is also a constraint in the Minkowski case) and the curvature of spacetime near the worldline (which of course is not present in the Minkowski case).

Staff Emeritus
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So for H-N in 1+1, I think what this means is that a rigid motion is unique given the motion of one point on the object, but existence is not guaranteed unless the object is small enough and the proper acceleration is small enough. Given the size L and the proper acceleration a, I think we need something like ##L\lesssim c^2/a##. Basically you can't extend a rigid object beyond its own Rindler horizon.

Dale
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This seems like a plausible theorem, but difficult to prove. The proof will certainly have to involve the ##C^2## property, since a worldline with a sharp bend is a counter example.

PAllen
2019 Award
This seems like a plausible theorem, but difficult to prove. The proof will certainly have to involve the ##C^2## property, since a worldline with a sharp bend is a counter example.
Yes, I think C^2 is necessary and sufficient for a nonempty neighborhood. A Peter pointed out, the place to look in the literature for a fully rigorous proof would mathematical relativity treatments of Fermi-Normal coordinates. Physicists will take shortcuts (though, I guess, Hawking and Ellis was a rare counterexample).

PAllen
2019 Award
Yes, I think C^2 is necessary and sufficient for a nonempty neighborhood. A Peter pointed out, the place to look in the literature for a fully rigorous proof would mathematical relativity treatments of Fermi-Normal coordinates. Physicists will take shortcuts (though, I guess, Hawking and Ellis was a rare counterexample).
Note that physically, this is very plausible. As acceleration approaches infinite, maximum Born rigid object size approaches zero.

DrGreg
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Theorem: In Minkowski space, let W be a timelike, ##C^2## world-line parametrized by proper time ##\tau## for ##\tau## covering the whole real line. Then (1) for any point P, there exists a point Q on W such that line PQ is orthogonal to W. Furthermore, (2) there exists a neighborhood A of W such that if P is chosen from within A, then Q is unique.
(1) still isn't true. Use the example of posts #11 & #12, and take P to be any point with $|x| \leq |t|$, i.e. on or behind the Rindler horizon (and its mirror image), e.g. (1,0).

I believe (2) is true, but I haven't proved it.

bcrowell
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(1) still isn't true. Use the example of posts #11 & #12, and take P to be any point with $|x| \leq |t|$, i.e. on or behind the Rindler horizon (and its mirror image), e.g. (1,0).

I believe (2) is true, but I haven't proved it.
Yep. I guess I'd better stop writing revisions labeled optimistically as "theorem" :-)

Conjecture: In Minkowski space, let W be a timelike, ##C^2## world-line parametrized by proper time ##\tau## for ##\tau## covering the whole real line. Then there exists a neighborhood A of W such that for any point P in A, there exists a unique point Q on W such that line PQ is orthogonal to W.

This version seems to me like it really must be weak enough to hold. If we zoom in close enough on any part of a ##C^2##, timelike world-line, then it looks like a curve with constant proper acceleration, so (assuming ##a=1##, which is clearly inessential) we can make it look just like the point ##(0,-1)## on our canonical example ##(t,x)=(\sinh\tau,-\cosh\tau)##. We know that the conjecture holds in a neighborhood of this point, since the Rindler horizon lies at a finite distance from it.

But I've been wrong so many times that I'm probably wrong again now :-)

I originally thought that this was going to require some global knowledge of Minkowski space such as topology or causal structure, but it looks like the version of the conjecture that is actually true is much weaker and purely local, which makes proving it look much more straightforward.

Nugatory
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Note that physically, this is very plausible. As acceleration approaches infinite, maximum Born rigid object size approaches zero.
@Dadface may find this interesting - it's part of what I was getting at when I suggested that he look into Born rigid motion in the Bell paradox thread.

vanhees71
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2019 Award
I've found a very clear proof of the Herglotz-Noether theorem in Chapter 8 of

Trautman, A., F. A. E. Pirani, and H. Bondi. "Lectures on general relativity. Brandeis 1964 Summer Institute on Theoretical Physics. Vol. 1." (1965).

It's investigating the "rigid motion" more generally in any GR spacetime, but gives also the flat-space results as an example. The proof is elementary in the sense that it just uses standard Ricci notation for the non-expert theoretical physicist. I guess, when I find the time, I'll study this proof specializing to the simpler case of flat Minkowski space right away and then put it in my SRT writeup, but that will take some time for sure.

Staff Emeritus
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I think it ends up being straightforward to prove the uniqueness part of the weakened form of the conjecture in #19, which was:

In Minkowski space, let W be a timelike, ##C^2## world-line parametrized by proper time ##\tau## for ##\tau## covering the whole real line. Then there exists a neighborhood A of W such that for any point P in A, there exists a unique point Q on W such that line PQ is orthogonal to W.
Let ##r(\tau)## be the vector from P to the world-line, and let ##v(\tau)## and ##a(\tau)## be the corresponding velocity and acceleration vectors. Let ##f(\tau)=r\cdot v##. For uniqueness to fail, we would need ##f## to hit zero twice, and by Rolle's theorem, this would imply ##f'(\tau)=0## at some point ##\text{X} \in\text{W}##. We have ##f'(\tau)=1+r\cdot a##, so at X, ##r\cdot a=-1##. Pick Minkowski coordinates such that at X, ##r## lies along the x axis. In these coordinates, let ##\ell## be the spatial distance between X and P. Then ##a_x=1/\ell##. Since W is ##C^2##, we can put an upper bound on the acceleration if we restrict our attention to any finite part of W, so this puts a lower bound on ##\ell##. Therefore we can make an neighborhood A surrounding W out to some finite distance and be guaranteed uniqueness.

It would be interesting to see how this might connect to the pedagogy of the twin paradox. Many presentations try to attribute the difference in the twins' ages to the periods of acceleration that the traveling twin undergoes. I've never liked this approach, and I feel that it encourages misconceptions about frames of reference, such as the idea that a frame of reference is what you instantaneously "see." You can say that there's a discontinuity in the age of the earthbound twin as defined by the traveling twin, but since there is no theoretical bound on the acceleration of the traveling twin, it seems to me that the age of the earthbound twin isn't even necessarily single-valued.

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Staff Emeritus
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It would be interesting to understand how to put a rigorous quantitative lower bound on how far out one should be able to stretch the neighborhood A. MTW discusses this kind of thing in several places, but although they claim the result is something like ##\ell = 1/a##, I don't see where they ever prove this with a calculation, or even define carefully what they mean by the variables. You don't necessarily want ##\ell## to be proper distance, because the whole point here is kind of that we can't always define such a thing. It would be attractive to be able to write this as a rigorous inequality in terms of coordinate-independent quantities such as proper acceleration. All I did in #22 was to argue that in some coordinate system, the distance ##\ell=\Delta x## is bounded below, and that's all that needed in order to show that the neighborhood A exists, but that's different from giving a bound on the distance that would be meaningful without reference to some a priori unknown coordinate system.

In the argument I gave in #22, we have ##f(\tau_1)=0## and ##f(\tau_2)=0## for ##\tau_1<\tau_2##. Since we have ##r\cdot v =0## at these times, it is possible at these two times to define proper distances from the world-line to P. But when we apply Rolle's theorem we get a third time ##\tau^*## such that ##\tau_1<\tau^*<\tau_2##, and it's the value of the ##r## vector at *this* time that we get a bound on. I'm not immediately seeing any way to get any bound on anything like a minimum proper distance from P to the world-line, especially since such a proper distance is not guaranteed to be defined at all times. The times ##\tau_1## and ##\tau_2## are times at which the proper distance is stationary, and maybe it's possible to prove that they're local minima or something, but that doesn't necessarily mean much, since we're talking about locally minimizing something that is not even globally well defined.

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pervect
Staff Emeritus
I'd need to introduce coordinates to prove it, but the point as I see it as that if P and Q are displacement vectors in Minkowskii space, if P is timelike and P and Q are orthogonal , then Q must be space-like.

Given that a diagonal Minkowskii metric exits, then P can be represented as (a,0,0,0) (and a cannot be equal to zero), so if Q is (w,x,y,z) then ##P \cdot Q = -a w =0## which implies w=0

This is a coordinate independent way (though my proof isn't coordinate independent) of saying that given the timelike worldline of an "observer" in 4d space, there are 3d space-like subsspaces spanned by vectors orthogonal to the observer's worldline that we can call "surfaces of simultaneity".

I regard simultaneity as a convention. The Rindler case shows that while you can set up a coordinate / reference system that defines the time coordinate of every event, demanding that that the coordinate assignment is unique so that there is a 1:1 mapping between events and their coordinates is in some circumstances not globally possible. I really don't know what the best approach to teaching this is, by the time one makes things specific enough to illustrate the point, one has usually lost a large fraction of the audience who might benefit :(.

Another interesting sidenone is that the end results aren't so dissimilar if one has a more general manifold and not just Minkowskii space, though the route to get there is a bit trickier - definitely something one would prefer to avoid burdening a beginning SR student with. The concept of displacement vectors doesn't really apply to the manifold, as the displacement operators don't commute and can't form a vector space as is needed in an affine space. The route to getting similar results for a manifold ends up noting that you can define displacement vectors in the tangent space, and that you can construct maps from the tangent space to the manifold which are 1:1 in some limited region (usually one uses the exponential map to map from the tangent space to the manifold in the local convex region).

A related interesting question (that I don't really have an answer for) is how to introduce rotating frames of reference. Closed surfaces of simultaneity no longer exist, though it's possible to identify a "space" by considering a rotating coordinate system with a stationary metric (i.e. timelike Killing vectors), and creating what mathemeticians call a "quotient manifold".

PAllen