An event being in the past of a timelike separated one

[sophiatev]

TL;DR Summary

Confused as to why an event A that is timelike separated from some event P, and is in P's future, is always in P's future in all reference frames.

In Hartle's book Gravity: An Introduction to Einstein's General Relativity he asserts that "It does not make sense in general to say that one event is later than another. An event can be later than another spacelike separated event in one inertial frame and earlier in another. But it does make sense to say which is the earlier of two timelike separated events. That's because events to the future of P are inside its future light cone, and the inside and outside of a lightcone are properties of the geometry of spacetime - the same in all frames" (page 60, Ch. 4.4).

Let us consider the lightcone of some event P, and suppose that another event A lies inside its future lightcone. No matter what reference frame we are in, A will always be inside P's lightcone. This is because A being inside P's lightcone means that the spacetime distance between A and P is less than 0, and distances are invariant. So no matter what reference frame we are in, or put another way, no matter what coordinate system we use, the distance between A and P is always less than 0, and thus A is always inside P's lightcone.

I am however confused by Hartle's assertion that A will remain inside the future lightcone of P in all reference frames. Let us consider a coordinate system that differs from our original coordinate system via the following transformation:
t'= -t
x' = x
y' = y
z' = z
Suppose in the original reference frame, P has coordinates (0, 0, 0, 0) and A has coordinates (3, 0, 0, 0). In the new reference frame P still has coordinates (0, 0, 0, 0) but now A has coordinates (-3, 0, 0, 0). This makes it seem like A is now in P's past, whereas in our original frame A was in P's future. Put another way, in our original frame A was in P's future lightcone, whereas now it is in P's past lightcone. Where am I going wrong?

 

[PeterDonis]

Let us consider the lightcone of some event P, and suppose that another event A lies inside its future lightcone. No matter what reference frame we are in, A will always be inside P's lightcone.

That is correct. Light cones are part of the invariant geometric structure of spacetime, so they don't change when you change reference frames.

I am however confused by Hartle's assertion that A will remain inside the future lightcone of P in all reference frames.

That's because you have picked a transformation in your example that reverses the sign of the $t$ coordinate. Such a transformation reverses how the $t$ coordinate represents past vs. future: now larger values of $t$ are further in the past, not further in the future. So A is still in the future light cone of P; but now this fact is represented by A having a negative value of $t$ instead of a positive one.

In other words, you can't change which half of the light cone is the "future" half by changing coordinates; the "future" vs. "past" distinction is an invariant geometric property of the spacetime.

 

[PAllen]

The assignment of lightcones as to future or past is meant to be an invariant feature of a spacetime. That is, given an arbitrary spacetime, there might not even be a globally consistent orientation - you might find that extending a local choice you reach another point where one path suggests one light cone a the future, while another path suggests the other as the future. However physically plausible spacetimes are typically considered those for which a globally consistent time orientation is possible - these are called time orientable manifolds.
Then, given a time orientable manifold, you choose one of the two orientations as to what is future versus past. This is then considered coordinate invariant. Then, in your example, you say that in t’ coordinates, the negative coordinate direction is the future, in order to preserve the global choice of orientation.

 

[sophiatev]

That's because you have picked a transformation in your example that reverses the sign of the ttt coordinate. Such a transformation reverses how the ttt coordinate represents past vs. future: now larger values of ttt are further in the past, not further in the future.

I understand that in my new coordinates, larger values of t' mean larger values of -t, and so larger values of t' mean going further into the past in the original reference frame. But is it not the case that the time that an event happens at in a given reference frame is defined by the t coordinate in that reference frame? Based on this, shouldn't event A happen in the past in the second reference frame (even though it happens in the future in the first), since its time coordinate is negative?

 

[PAllen]

I understand that in my new coordinates, larger values of t' mean larger values of -t, and so larger values of t' mean going further into the past in the original reference frame. But is it not the case that the time that an event happens at in a given reference frame is defined by the t coordinate in that reference frame? Based on this, shouldn't event A happen in the past in the second reference frame (even though it happens in the future in the first), since its time coordinate is negative?

No, coordinates are labels people assign to events. Causal future is a physical, a invariant relation of model of reality. It is true that the time coordinate defines what time value is assigned to events. But it does not change what is causal future and past. So, as noted twice, the negative direction in t’ coordinates must be taken to be the future direction as long at the positive direction of t is taken to be the future

 

[PeterDonis]

is it not the case that the time that an event happens at in a given reference frame is defined by the t coordinate in that reference frame?

Yes, but "the time that an event happens at in a given reference frame" doesn't mean what you think it means. That "time" does not tell you which events are to the future or past of which other events.

To see why not, suppose an observer's worldline passes through both event P and event A. We will suppose that this observer, whom we will call O, is at rest in the original reference frame, so his worldline is just the $t$ axis. That means O's clock reads 0 at event P and 3 at event A.

In the new frame, the value of $t'$ is minus 3 at event A instead of 3; but observer O's clock still reads 3 at event A, because observers' clock readings are frame invariant; they don't change when you change frames. So as far as observer O is concerned, event A is still to the future of event P, because his "future" and "past" are shown by his clock readings--larger clock readings are to the future of smaller ones. And since "future" and "past" are defined physically, by what observers observe, not by coordinate values, event A is still to the future of event P. The new coordinates are just weird in that, as observer O's clock readings increase, the $t'$ coordinate values decrease. But that's clearly an issue with the $t'$ coordinates, not with any physical change in the future-past relationship of events P and A.

 

[PeterDonis]

That "time" does not tell you which events are to the future or past of which other events.

It's worth noting that virtually all textbooks on SR do not explicitly discuss this issue, because they always pick frames in which the increasing direction of the time coordinate is the same as the increasing direction of observers' clock readings. So there is never a mismatch between the two. But that doesn't mean the issue isn't there; it just means the textbooks avoid it.

 

[sophiatev]

because observers' clock readings are frame invariant; they don't change when you change frames

Can you elaborate on why this is true? Doesn't time dilation imply that observers' clock readings in different reference frames would register different time deltas between events, and thus that the readings changed?

 

[Dale]

Summary:: Confused as to why an event A that is timelike separated from some event P, and is in P's future, is always in P's future in all reference frames.

Let us consider a coordinate system that differs from our original coordinate system via the following transformation:

In addition to @PeterDonis comment, I would also point out that this transform is not in the Poincare group so you wouldn’t expect it to be a physical symmetry.

 

[Dale]

Based on this, shouldn't event A happen in the past in the second reference frame (even though it happens in the future in the first), since its time coordinate is negative?

If you consider a specific reference event, eg the origin, then surfaces of constant spacetime interval form hyperboloids. Spacelike intervals form hyperboloids of one sheet, so you can smoothly transform any into the other with differential transformations. In contrast, timelike intervals form hyperboloids of two sheets. One sheet is the future and one sheet is the past events. There is no smooth differential transforms that can go from one sheet to the other.

 

[PAllen]

Can you elaborate on why this is true? Doesn't time dilation imply that observers' clock readings in different reference frames would register different time deltas between events, and thus that the readings changed?

Absolutely not. Time dilation refers to a comparison between one clock moving relative to two separated synchronized clocks at mutual rest. All observers agree on all clock readings. The time difference between the two clocks at mutual rest between crossings of the other clock is larger than the time difference between the readings on the third clock at crossing points. All observers agree on all of these facts.

 

[PeterDonis]

Can you elaborate on why this is true?

Physically, it's because observer O's clock readings are direct physical observables, and direct physical observables are always frame invariant.

Geometrically, it's because observers' clock readings are the invariant signed arc lengths along their worldlines. Observer O's worldline between event P and event A has an invariant signed arc length, which is the difference between O's clock readings at those two events. Arc lengths of curves are always invariant.

Doesn't time dilation imply that observers' clock readings in different reference frames would register different time deltas between events

No. Suppose we do a Lorentz transform from your original frame to a frame that is moving in the $x$ direction at speed $v$ (without any sign changes). We take event P to be the common spacetime origin of both frames. In this new frame, observer O is moving in the minus $x$ direction at speed $v$. If you work out the math, you will see that the time coordinate of event A in the new frame will be larger than $3$, but if you compute the spacetime interval between events P and A in the new frame, it will still be $3$. In other words, observer O's proper time (which is another name for the elapsed time shown on his clock) from event P to event A is the same in any frame; but the coordinate time difference from event P to event A in any frame in which observer O is not at rest will be larger than it is in O's rest frame.

In the new frame, there will be some other observer, whom we can call F, who is at rest and whose worldline passes through event P. This observer's elapsed clock time from event P to the coordinate time in the new frame of event A will be larger than $3$. But observer F's worldline does not pass through event A; he can't directly measure his own clock time at event A because he is never at it. He can only calculate the coordinate time in his rest frame of event A. And this calculation, by itself, does not tell observer F what observer O's clock reading is at event A.

 

[sophiatev]

Geometrically, it's because observers' clock readings are the invariant signed arc lengths along their worldlines.

I actually had a question about this - why is this true? Hartle simply states as fact, without really explaining, that the proper time is given by
$dτ^2 = -ds^2/c^2$
and that this is what clocks measure.
My guess as to why this is true is
$$
ds^2 = -(cdt)^2 + dx^2 + dy^2 + dz^2 \rightarrow (-cdt)^2 = ds^2 - dx^2 - dy^2 - dz^2 \rightarrow c^2dt^2 = -ds^2 + dx^2 + dy^2 + dz^2 \rightarrow dt^2 = -ds^2/c^2 + dx^2/c^2 + dy^2/c^2 + dz^2/c^2.
$$
In the moving frame of the clock itself, am I correct in saying that $dx^2=dy^2=dz^2 = 0$ so the corresponding measurement of time is $-ds^2/c^2$, giving us $dτ^2 = -ds^2/c^2$?

 

[PeterDonis]

why is this true?

Um, because that's how the mathematical model is defined?

Any mathematical model in physics has to have some set of definitions for what quantities in the model correspond to what quantities in the real world that is being modeled. The statement that observers' clock readings are the invariant signed arc lengths along their worldlines is simply one of those definitions.

My guess as to why this is true

What you are showing here is not "why" it is true, it is simply a mathematical illustration of why choosing the definition the way I described makes sense. There is no "why it is true" for definitions beyond the simple fact that the definition was chosen that way because it works.

 

[Orodruin]

So let me point out the following:
As many other people before you, you seem to be assigning some sort of physical significance to the coordinate time t. This is not a good thing as it is a priori just a coordinate. There is nothing stopping you from using a coordinate system where either all or none of the coordinates are timelike (or even null). What is physical (and therefore invariant between coordinate systems) are clock readings.

 

[sophiatev]

Absolutely not. Time dilation refers to a comparison between one clock moving relative to two separated synchronized clocks at mutual rest. All observers agree on all clock readings. The time difference between the two clocks at mutual rest between crossings of the other clock is larger than the time difference between the readings on the third clock at crossing points. All observers agree on all of these facts.

I see, thank you, this was a clear and concise illustration and really helped.

 

[sophiatev]

Physically, it's because observer O's clock readings are direct physical observables, and direct physical observables are always frame invariant.

I think I understand now. Anyone at O's location looking at O's clock would see the same time reading on the clock. Clock readings aren't like velocities (i.e. if I'm in a train moving right relative to the ground I think someone outside is moving left, whereas they think I'm moving right). They are immutable aspects of our physical reality. And if that's the case, it makes sense why they are frame invariant. And so it makes sense to say that event A is to the future of event P if they are timelike separated, meaning we can physically move a clock to that location and show that it registers a higher reading. Is this also why it doesn't make sense to say that one spacelike separated event is later than another? Because for spacelike events, $Δx^2 + Δy^2 + Δz^2 > (cΔt)^2$, and so a clock would have to move faster than the speed of light to go from one to the other, which is impossible? So it's impossible to register a physical clock reading for both, and thus nonsensical to say one is later than the other?

Um, because that's how the mathematical model is defined?

Ah, in retrospect this is really obvious. I somehow didn't register that he was just stating a definition. Thank you.

 

[PeterDonis]

Anyone at O's location looking at O's clock would see the same time reading on the clock.

That's correct.

Is this also why it doesn't make sense to say that one spacelike separated event is later than another?

Because it is impossible for the same clock to be at both events if they are spacelike separated, yes.

 

[Mister T]

Doesn't time dilation imply that observers' clock readings in different reference frames would register different time deltas between events, and thus that the readings changed?

The readings on all instruments, clocks included, are frame invariant. All this means is that if the clock on my wall reads 2:00 pm all observers will agree that it reads 2:00 pm. And if that's when I start watching a 30-minute TV show then the clock will read 2:30 pm when it's over and all will agree. It's just that when they check this elapsed time with their own clocks they may not get 30 minutes.

 

[Nugatory]

All this means is that if the clock on my wall reads 2:00 pm all observers will agree that it reads 2:00 pm.

That's not quite the right way of phrasing it, though. Someone may "agree that it reads 2:00 PM", but unless we also specify that they and the clock are at the same place there's some ambiguity as to exactly what this means.

Maybe say "All observers will agree about what the clock reads at the moment that something happens at the same place as the clock."?

Of course @Mister T already understands this - I'm just trying to improve the wording.

 

[Mister T]

That's not quite the right way of phrasing it, though. Someone may "agree that it reads 2:00 PM", but unless we also specify that they and the clock are at the same place there's some ambiguity as to exactly what this means.

Good point. I recall authors writing that one can imagine a clock that produces a stamped card that reads, say, "2:00 pm". All observers will agree that the card reads "2:00 pm".