Is Event Sequencing Relative in the Theory of Manifolds?

[yossell]

This is correct for inertial reference frames, but not true in general for arbitrary reference frames. Consider a ship moving toward Earth at 0.8c that decelerates to come to rest with Earth at a distance of 10 light years. Using the standard SR simultaneity convention, wrt the ship observer, people who were dead on Earth simultaneous with the ship beginning deceleration are alive on Earth simultaneous with the ship stopping its deceleration.

Yep, dead people can rise from the grave in SR.

Ok, thanks, good point - I'll be more careful to write `inertial frame' instead of just `frame' in future.

Having said that, and I ask out of interest rather than feistiness, and because this would be a very cool example, what's the more general definition of a frame? Accelerating observers who construct frames using that simultaneity convention can't define 1-1 continuous mappings from R^4 to all of Minkowski space-time; does a frame drop the idea that it's a global mapping? (that's what happens in GR). I'm trying to think of a coordinate patch which connects the rocket at the beginning of deceleration and the Earth when the people are dead - for temporal comparison - and also connects the rocket at the end of deceleration with the Earth when the people are alive. I can't quite yet convince myself that's it's possible, but I don't want to spend too much time if you had another idea in mind.

 

[starthaus]

This is correct for inertial reference frames, but not true in general for arbitrary reference frames. Consider a ship moving toward Earth at 0.8c that decelerates to come to rest with Earth at a distance of 10 light years. Using the standard SR simultaneity convention, wrt the ship observer, people who were dead on Earth simultaneous with the ship beginning deceleration are alive on Earth simultaneous with the ship stopping its deceleration.

Yep, dead people can rise from the grave in SR.

Do you have a reference for that? My calculations show that your claim is false. The line of simultaneity keeps rotating wrt the x-axis in the Minkowski diagram as the speed varies but it never crosses over itself, even at the inflection points.

 

[Rasalhague]

Do you have a reference for that? My calculations show that your claim is false. The line of simultaneity keeps rotating wrt the x-axis in the Minkowski diagram as the speed varies but it never crosses over itself, even at the inflection points.

I get 10*Sinh[ArcTanh[.8]] = 10*0.8/Sqrt[1 - 0.8^2] = 13 + 1/3 years. So, unless I'm mistaken, people dead for less than this time, in the rest frame of the ship before it decelerates, will be alive still in the rest frame of the ship after it decelerates, i.e. in the rest frame of the earth.

EDIT: Correction: that should be 10*Sinh[ArcTanh[.8]]/Cosh[ArcTanh[.8]] = 10*Tanh[ArcTanh[0.8]] = 10*Sqrt[1 - 0.8^2]*0.8/Sqrt[1 - 0.8^2] = 10*0.8 = 8 years. I forgot to first convert the distance of 10 light years to 6, as it is in the rest frame of the ship before it decelerates. (Thanks for pointing that out, Al68.)

 

[yossell]

Do you have a reference for that? My calculations show that your claim is false. The line of simultaneity keeps rotating wrt the x-axis in the Minkowski diagram as the speed varies but it never crosses over itself, even at the inflection points.

What are your calculations? I have a counter-argument:

Restricting all de- and acceleration to the x-axis, to the 2-d case, consider a curving world-line w, and consider any two lines of simultaneity l1 and l2 of w that intersect w at p1 and p2. If these lines are not parallel, then l1 and l2 must at some point intersect and cross, say at point z. Thus lines of simultaneity of an accelerating object must intersect and cross.

 

[starthaus]

What are your calculations? I have a counter-argument:

Restricting all de- and acceleration to the x-axis, to the 2-d case, consider a curving world-line w, and consider any two lines of simultaneity l1 and l2 of w that intersect w at p1 and p2. If these lines are not parallel, then l1 and l2 must at some point intersect and cross, say at point z.

What if they intersect to the right of the wordline? (i..e. their extensions intersect)

Thus lines of simultaneity of an accelerating object must intersect and cross.

I want to see Al68's reference.

 

[yossell]

What if they intersect to the right of the wordline? (i..e. their extensions intersect)

I don't know what you're asking for here. But by changing the direction of acceleration, you can get an intersection on either `side' of the world line.

Choose a frame, let rocket and person begin stationary in this frame, with the human remaining and having been stationary in this frame. Human having just died at (0, 0, 0, 0), while the rocket is at (0, X, 0, 0) (coordinates of the form (t, x, y, z).

Let rocket quickly accelerate away in the positive x direction, reaching speed v at time t (in stationary frame) in positive x direction and then maintaining v. Its lines of simultaneity are now tilted, and simply by letting X and v be big enough and t small enough, arbitrary events of coordinate (-T, 0, 0, 0) intersect its simultaneity line. For some of these events, the human is not a corpse.

 

[starthaus]

I don't know what you're asking for here. But by changing the direction of acceleration, you can get an intersection on either `side' of the world line.

Choose a frame, let rocket and person begin stationary in this frame, with the human remaining and having been stationary in this frame. Human having just died at (0, 0, 0, 0), while the rocket is at (0, X, 0, 0) (coordinates of the form (t, x, y, z).

Let rocket quickly accelerate away in the positive x direction, reaching speed v at time t (in stationary frame) in positive x direction and then maintaining v. Its lines of simultaneity are now tilted,

Yes, they start at 0 degrees wrt x and they tilt continously (with an asymptote angle of \pi/4 )as the rocket accelerates. Yet, starting point of each line of simultaneity also moves since the rocket is tracing the wordline.

and simply by letting X and v be big enough and t small enough, arbitrary events of coordinate (-T, 0, 0, 0) intersect its simultaneity line. For some of these events, the human is not a corpse.

Try proving this mathematically.

 

[yossell]

Yes, they start at 0 degrees wrt x and they tilt continously (with an asymptote angle of )as the rocket accelerates. Yet, starting point of each line of simultaneity also moves since the rocket is tracing the wordline.

But although the line of simultaneity has moved on, the lines are indefinitely extended in the x and -x directions. The equations mx + c and m'x + c' always have a solution if m and m' are distinct, no matter how big c - c' is.

Try proving this mathematically

Ok - I'll give it a shot - but what will you allow me? Can I work in natural units where lightrays equation is t = x and t = -x, and where objects with velocity v has a simultaneity line of 1/v? Or does this need to be established too?

 

[Al68]

Do you have a reference for that?

Sure: http://www.bartleby.com/173/11.html. That's just a reference for the lorentz transformations. I assume you know how to use them.

Assuming that the ship starts a (short duration) decel simultaneous with the 2020 Presidential election in the inertial frame of the ship prior to decel, it will be 2012 on Earth simultaneous with the decel in the inertial frame of the ship after decel.

Notice that Earth's clock "jumps backward" in my example for the same reason that Earth's clock "jumps ahead" in the standard twins paradox resolutions. It's just an artifact of using the SR simultaneity convention to assign coordinates to distant events.

 

[yossell]

starthaus, I know you're the hard man of forum and that you run a tight ship, but won't you take a look at this diagram? Red dotted line a null beam, green arrow path of rocket. It starts stationary and then gently accelerates to a velocity where the slanty line is its line of simultaneity. As X is general, it's easy to find an X where it can gently accelerate to a v and a point where the line of simultaneity hits (-T 0 0 0)

 

[bcrowell]

In SR, for any two events A and B that are time like separated, if A is earlier than B in one frame, it is earlier than B in all frames.

This is correct for inertial reference frames, but not true in general for arbitrary reference frames. Consider a ship moving toward Earth at 0.8c that decelerates to come to rest with Earth at a distance of 10 light years. Using the standard SR simultaneity convention, wrt the ship observer, people who were dead on Earth simultaneous with the ship beginning deceleration are alive on Earth simultaneous with the ship stopping its deceleration.

Yep, dead people can rise from the grave in SR.

The reason we care about frame-independence of time ordering in SR is that it's necessary for causality. Your example refers to simultaneity between events that are spacelike separated, so it doesn't have any implications for causality. Depending on how you interpret yossell's statement of the principle, I'm not even sure that your example is a counterexample.

Here is a somewhat different statement of the principle that may be more clearcut. Let A and B be the end-points of a curve S in spacetime, such that the curve's orientation is always in the positive timelike direction as we traverse it from A to B. If the spacetime is flat, then this property of S is coordinate-independent.

 

[starthaus]

starthaus, I know you're the hard man of forum and that you run a tight ship,

Yes, I do :-)

but won't you take a look at this diagram? Red dotted line a null beam, green arrow path of rocket. It starts stationary and then gently accelerates to a velocity where the slanty line is its line of simultaneity. As X is general, it's easy to find an X where it can gently accelerate to a v and a point where the line of simultaneity hits (-T 0 0 0)

bcrowell beat me to the disproof but I can disprove your statement mathematically (Al68 offered nothing but some handwaving). Are you interested in the mathematical disprroof?

 

[yossell]

bcrowell beat me to it but I can disprove your statement mathematically (Al68 offered nothing but some handwaving). Are you interested in the mathematical disprroof?

I'd be interested, but I now think we must be talking at cross-purposes. I (and I think Al68), are considering examples where an object, by accelerating or decelerating, changes inertial frames in a way which affects the T-coordinate of distant events - bcrowell is talking about the property of particular CURVES in spacetime, and the question of whether they are timelike or not is not a coordinate dependent matter. This latter I think is true, and am not arguing against it.

 

[Rasalhague]

In #33, I made a calculation based on my interpretation of Al68's resurrection statement. Is this what you had in mind, Al68? Are other people here interpreting it differently? Did I get the calculation right?

 

[Al68]

The reason we care about frame-independence of time ordering in SR is that it's necessary for causality. Your example refers to simultaneity between events that are spacelike separated, so it doesn't have any implications for causality. Depending on how you interpret yossell's statement of the principle, I'm not even sure that your example is a counterexample.

You're right, it's not a counterexample (to causality), yossel clearly meant to refer strictly to inertial reference frames, not an arbitrary coordinate system. And you're right that there are no implications for causality. Assigning coordinates to distant events with the SR simultaneity convention doesn't cause anything.

 

[Al68]

In #33, I made a calculation based on my interpretation of Al68's resurrection statement. Is this what you had in mind, Al68? Are other people here interpreting it differently? Did I get the calculation right?

Yeah your interpretation is right, but I think you missed that the coordinate distance between the ship and Earth is only 6 ly in the ship's frame prior to decel. So it should be 8 yrs instead of 13.33 yrs.

 

[Al68]

...Al68 offered nothing but some handwaving...

LOL. You didn't ask for a proof. You asked for the reference I used. I provided it.

Edit: Rasalhague shows the math in the next post, but I assumed an experimental physicist like yourself wouldn't need it to be shown.

 

[Rasalhague]

Yeah your interpretation is right, but I think you missed that the coordinate distance between the ship and Earth is only 6 ly in the ship's frame prior to decel. So it should be 8 yrs instead of 13.33 yrs.

Oh yeah, of course! 10*Sqrt[1-0.8^2]*0.8/Sqrt[1-0.8^2] = 10*0.8 = 8. Sorry deads! At least we get some back, and with a bigger boost as many as we want...

 

[Al68]

Oh yeah, of course! 10*Sqrt[1-0.8^2]*0.8/Sqrt[1-0.8^2] = 10*0.8 = 8. Sorry deads! At least we get some back, and with a bigger boost as many as we want...

Yep. Too bad the ship's observer can never actually observe them alive again. They're alive "now" according to the SR simultaneity convention, but no "they're alive" signal will ever reach the ship "later" regardless of its subsequent motion. In fact, the faster the ship tries to reach earth, the quicker they will die off again.

 

[starthaus]

LOL. You didn't ask for a proof. You asked for the reference I used. I provided it.

I asked you for a valid reference to your claim that acceleration can produce event order reversal. So, you provided no reference. Not only that, you got the wrong answer.

 

[JesseM]

Do you have a reference for that? My calculations show that your claim is false. The line of simultaneity keeps rotating wrt the x-axis in the Minkowski diagram as the speed varies but it never crosses over itself, even at the inflection points.

It depends on how fast the acceleration is, i.e. how long it takes the ship to go from 0.8c to rest in the Earth frame. For example, suppose the ship is traveling towards Earth at 0.8c until it is 10 light years from Earth at t=0 years in the Earth frame (so if the Earth is at x=0 light years, the ship could be at x=-10 light years), then it accelerates for a year until it is at rest at t=1 year the Earth frame. In that case, in the ship's inertial rest frame at the moment it begins to accelerate, the moment of its beginning to accelerate is simultaneous with an event that occurs on Earth at t=8 years in the Earth frame (since in the Earth's frame we have dx=10 and dt=8 between these events, meaning in the ship's frame dt'=gamma*(dt - v*dx/c^2)=0), but in the ship's inertial rest frame at the moment it stops accelerating, the moment it stops accelerating is simultaneous with an event that occurs on Earth at t=1 year in the Earth frame.

 

[starthaus]

It depends on how fast the acceleration is, i.e. how long it takes the ship to go from 0.8c to rest in the Earth frame. For example, suppose the ship is traveling towards Earth at 0.8c until it is 10 light years from Earth at t=0 years in the Earth frame (so if the Earth is at x=0 light years, the ship could be at x=-10 light years), then it accelerates for a year until it is at rest at t=1 year the Earth frame. In that case, in the ship's inertial rest frame at the moment it begins to accelerate, the moment of its beginning to accelerate is simultaneous with an event that occurs on Earth at t=8 years in the Earth frame (since in the Earth's frame we have dx=10 and dt=8 between these events, meaning in the ship's frame dt'=gamma*(dt - v*dx/c^2)=0), but in the ship's inertial rest frame at the moment it stops accelerating, the moment it stops accelerating is simultaneous with an event that occurs on Earth at t=1 year in the Earth frame.

I don't think that the above is correct. The issue in discussion was whether acceleration can invert the ordering of events. I will repeat the proof I gave for inertial frames and I will generalize to accelerated frames:

dt'=\gamma(\tau-vx/c^2)

dt'=\gamma (d\tau-vdx/c^2)

dt'=\gamma d\tau (1-v/c^2*\frac{dx}{d\tau})

Since v \frac{dx}{d\tau}<c^2 it follows that dt' and d\tau always have the same sign. If you are ok with the above, I can post the generalization to accelerated frames.

 

[JesseM]

I don't think that the above is correct. The issue in discussion was whether acceleration can invert the ordering of events. I will repeat the proof I gave for inertial frames and I will generalize to accelerated frames:

dt'=\gamma(\tau-vx/c^2)

dt'=\gamma (d\tau-vdx/c^2)

dt'=\gamma d\tau (1-v/c^2*\frac{dx}{d\tau})

How would you define the non-inertial frame for a non-inertial observer? There's no single "correct" way, but I think the most common type of non-inertial frame used by physicists in SR would be one that has the following properties:

1. For point on the non-inertial observer's worldline, the coordinate time is just equal to the observer's proper time at that point

2. At any point on the observer's worldline, a line of simultaneity in the non-inertial coordinate system which goes through that point would be identical to a line of simultaneity in the inertial frame where the observer has an instantaneous velocity of zero at that point

3. For two events which lie on a single line of simultaneity in the non-inertial coordinate system, the coordinate distance between them should be the same as the coordinate distance in the inertial frame which also has a line of simultaneity going through both events

4. The non-inertial observer has a coordinate position that doesn't change with coordinate time in the non-inertial system

If you define the non-inertial coordinate system in this way, then (3) implies the system's judgments about simultaneity always agree with those of the observer's instantaneous inertial rest frame, so this was the basis for my comments above (although I think it would actually be more common to just say the coordinate system "ends" at the point where lines of simultaneity would cross over one another, as with Rindler coordinates which don't extend past the Rindler horizon). I think Al68 was also assuming this sort of coordinate system when he commented about the dead rising from the grave in post #30. If you're assuming a different type of non-inertial coordinate system, you'll have to explain how it's defined for an observer with a non-inertial worldline that has a known parametrization x(t) relative to some inertial frame (for an observer experiencing constant proper acceleration a, their x(t) in an inertial frame where they started at rest would be x(t) = (c^2/a) (sqrt[1 + (at/c)^2] - 1) according to the http://math.ucr.edu/home/baez/physics/Relativity/SR/rocket.html ).

 

[starthaus]

How would you define the non-inertial frame for a non-inertial observer?

I'll show you next. First off, do you agree that inertial motion cannot reverse the order of events between frames? Yes or no?

 

[JesseM]

I'll show you next. First off, do you agree that inertial motion cannot reverse the order of events between frames? Yes or no?

Can you clarify what you mean by that? It is certainly true that for events with a space-like separation (|dx| > |c*dt| in any given inertial frame), you can find a pair of inertial frames which disagree on their order, are you denying that? If you don't deny it, why doesn't this qualify as an example of "inertial motion reversing the order of events between frames" as you define this phrase?

 

[yossell]

dt'=\gamma(\tau-vx/c^2)

dt'=\gamma (d\tau-vdx/c^2)

dt'=\gamma d\tau (1-v/c^2*\frac{dx}{d\tau})

Since v \frac{dx}{d\tau}<c^2 it follows that dt' and d\tau always have the same sign. If you are ok with the above, I can post the generalization to accelerated frames.

Your question is whether inertial motion can invert the order of events.

I do not think you have shown this claim in its full generality, but I think you have shown a weaker one, that inertial motion cannot invert the order of timelike events.

First, an aside: what's going on in step 1 to 2 - why the replacement of x and \tau with dx and d\tau? Just notational?

Secondly, I take the strategy to be this: dt' represents the difference in time between two events A and B in one inertial frame; d\tau and dx the difference in time and the difference in x-coordinate between these two events in another inertial frame. You then try to show that d\tau and dt' have the same sign, and therefore that any two frames must agree on the temporal order of the two events.

The trouble is that dx/d\tau is not necessarily a velocity of anything - it is *just* a ratio. The events may be spacelike separated. In which case, the ratio is not less than c. In such cases, your proof can be used to show that the temporal order can be inverted.

However, when attention is restricted to events which are timelike separated, then your proof seems to go through.

 

[starthaus]

Your question is whether inertial motion can invert the order of events.

More precisely, if order of timelike events (see the OP) is maintained in SR. We aren't talking about spacelike events.

I do not think you have shown this claim in its full generality, but I think you have shown a weaker one, that inertial motion cannot invert the order of timelike events.

...which is precisely what I have been talking about at posts 12 and 13. This answer is for both you and JesseM. In mathematical terms, as already shown twice, the proof is for \frac{dx}{d\tau}<1 (no "faster than light signalling"). I am addressing the OP, not the case of spacelike events.

First, an aside: what's going on in step 1 to 2 - why the replacement of x and \tau with dx and d\tau? Just notational?

No, this is standard differentiation.

Secondly, I take the strategy to be this: dt' represents the difference in time between two events A and B in one inertial frame; d\tau and dx the difference in time and the difference in x-coordinate between these two events in another inertial frame. You then try to show that d\tau and dt' have the same sign, and therefore that any two frames must agree on the temporal order of the two events.

yes

The trouble is that dx/d\tau is not necessarily a velocity of anything - it is *just* a ratio. The events may be spacelike separated.

No, see above.

However, when attention is restricted to events which are timelike separated, then your proof seems to go through.

It does not "seem", it does. It is just a repeat of the one at post 12. Same conditions, same outcome.

 

[yossell]

I still don't understand what you're calling differentiation or what you're doing from 1 - 2.

From line 2 onwards we're agreed that it shows that, for any timelike related (and I emphasise this because you've not been explicitly putting in this rider) events, all inertial observers agree about the events' temporal order ---- as I in fact wrote way back in post 5.

Great.

Now, I'm interested in the generalisation you've mentioned, to deal with Al68's point, that a rocket ship can, beginning at space time point A and in an inertial frame where events A and space time point B are simultaneous, travel to a spacetime point C and an inertial frame where C is simultaneous with an event that lies in B's past.

 

[starthaus]

I still don't understand what you're calling differentiation or what you're doing from 1 - 2.

Because this is a well known operation in calculus. It is important that you understand it for the next step, accelerated motion. How familiar are you with differential calculus?

 

[JesseM]

...which is precisely what I have been talking about at posts 12 and 13. This answer is for both you and JesseM. In mathematical terms, as already shown twice, the proof is for \frac{dx}{d\tau}<1 (no "faster than light signalling")

OK, so when you said "First off, do you agree that inertial motion cannot reverse the order of events between frames?" were you only talking about pairs of events for which |dx/dt| <= c? If so, I agree that different inertial frames won't disagree on their order.