Can time run backwards in an accelerating frame?

In summary, the conversation is discussing the concept of time dilation in the context of the twin paradox. The Earthbound twin experiences slower time during each leg of the journey, but catches up with the traveling twin due to the leap forward in time during the turnaround. The same applies for a distant observer, but to a greater extent. The main question is whether there is a single correct way to track time in both frames, and the answer is that time tracking is frame independent. Conventions such as Einstein synchronization may be used, but are not physically significant. The relativity of simultaneity is also discussed, with the suggestion that it is the same as the conventionality of simultaneity. However, this conflicts with previous studies and the conversation
  • #36
Dale said:
Aging is the increase of proper time, particularly of an object whose internal state is a function of proper time.
That's precisely the further specification required!
 
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  • #37
Ibix said:
As PeroK says, your phrasing is sloppy. I think you mean "moving inertially at 0.6c with respect to the earthbound twin".
Of course that's what I mean. In the scenario that I set up and have been using this whole time, I said that one twin is considered to be stationary and the other is considered to be traveling (since he's doing the accelerating), for the sake of conciseness.
Ibix said:
Assuming that, sure. He could choose to work in the frame where both he and the stay at home are moving at equal speeds in opposite directions... His clock readings don't match with his coordinate time, but that's fine.
But then he'd be pretending to be in a frame other than the one he's actually in, which wouldn't be truthful, would it? Everything I've ever read about SR says that each observer should conclude that they're stationary and that anything that moves relative to them is actually moving. If someone wants to know what's true in their own frame, isn't that what they need to do?
 
  • #38
Gumby The Green said:
All I care about is what's true in the Traveler's frame.
The issue then is to clearly specify what you mean by the Traveler’s frame. There is no standard meaning since the Traveler is non-inertial. Such a frame has few requirements, but those few include the requirement that it be invertible.

I recommend the Dolby and Gull definition, but it is not mandatory. What is mandatory is some valid definition of exactly what coordinates you mean by “the Traveler’s frame”
 
  • #39
Gumby The Green said:
Of course that's what I mean. In the scenario that I set up and have been using this whole time, I said that one twin is considered to be stationary and the other is considered to be traveling (since he's doing the accelerating), for the sake of conciseness.
Note that you can do the twin paradox without acceleration:

https://en.wikipedia.org/wiki/Twin_paradox#Role_of_acceleration

The eventual differential ageing, therefore, is entirely due to the geometry of spacetime and is effectively the triangle inequality in Minkowski spacetime.
 
  • #40
PeroK said:
Note that you can do the twin paradox without acceleration
I've looked at that. While the relay race scenario might model the earthbound (inertial) twin's perspective fine, it implies a discontinuity in the traveling (non-inertial) twin's perspective at the turnaround, i.e., it implies that his sibling's age jumps instantly, which doesn't make physical sense and isn't what an actual traveling twin would record. So acceleration is still needed to bridge the two frames in order to realistically model the traveler's perspective, which, again, is the only perspective I care about here.
 
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  • #41
Gumby The Green said:
I've looked at that. While the relay race scenario might model the earthbound (inertial) twin's perspective fine, it implies a discontinuity in the traveling (non-inertial) twin's perspective at the turnaround, i.e., it implies that his sibling's age jumps instantly, which doesn't make physical sense. So acceleration is still needed to bridge the two frames in order to realistically model the traveler's perspective, which, again, is the only perspective I care about here.
Here's a question. Just suppose there is a traveller turning round now and, according to him, your time is running backwards. How do you feel as a result of this?
 
  • #42
PeroK said:
Note that you can do the twin paradox without acceleration
Gumby The Green said:
acceleration is still needed to bridge the two frames in order to realistically model the traveler's perspective
Alternatively, we could say that the relay race scenario models the perspective of a traveling twin with infinite acceleration.
 
  • #43
PeroK said:
Just suppose there is a traveller turning round now and, according to him, your time is running backwards. How do you feel as a result of this?
Probably about the same as I would feel if he said that my time was running slower than his—especially since I'd be under the impression that his was running slower than mine, which is intuitively quite strange. But then I'd remember that it isn't that strange since time proceeds in different directions in different reference frames, which is analogous to two people walking at the same speed in different directions and each believing that the other is walking slower. In the case of time inversion, I'd think to myself that if one's time can rotate relative to someone else's like that, perhaps in some cases, it can simply continue rotating until it has a component in the opposite direction.

In both cases, I'd be inclined to ask him why he believes that, but in the case of the traveler who's accelerating away and thinks my time is running in reverse, I wouldn't even be able to ask him the question since none of my signals can reach him (since I'm behind his Rindler horizon). After realizing that, I'd realize that he can't actually observe my time moving backwards (just as time dilation can't be directly observed in real time), so I'd know that he was merely inferring it from his understanding of physics and his knowledge of his proper acceleration relative to me and his distance from me.
 
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  • #44
Gumby The Green said:
In both cases, I'd be inclined to ask him why he believes that, but in the case of the traveler who's accelerating away and thinks my time is running in reverse, I wouldn't even be able to ask the question since none of my signals can reach him (since I'm behind his Rindler horizon). After realizing that, I'd realize that he can't actually observe my time moving backwards (just as time dilation can't be directly observed in real time), so I'd know that he was merely inferring it from his understanding of physics and his knowledge of his proper acceleration relative to me and his distance from me.
That reads like a fairly good summary of why your conclusions are physically meaningless!
 
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  • #45
Gumby The Green said:
But then he'd be pretending to be in a frame other than the one he's actually in, which wouldn't be truthful, would it?
Everything is in every frame. You don't find rivers only on maps where they run straight up and down the page, do you? Can you only use maps when you are facing north?
Gumby The Green said:
Everything I've ever read about SR says that each observer should conclude that they're stationary and that anything that moves relative to them is actually moving.
That's not correct. The point of the principle of relativity is that you cannot determine whether or not you are moving - the question is meaningless in fact. So you may always treat yourself as at rest (even when you accelerate, if you are willing to pay the price in mathematical complexity when you do). But you can choose otherwise (we frequently do - "I was doing 40mph") and that's fine too.
Gumby The Green said:
If someone wants to know what's true in their own frame, isn't that what they need to do?
If they want to use their own inertial rest frame it is tautologically true that they are at rest, yes, but that choice is no righter than any other. "What time it is now over there" will depend on that choice.

In a purely inertial case you mightas well adopt your rest frame's simultaneity condition. But you don't have to. And in a non-inertial case there isn't an obvious choice of simultaneity condition.
 
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  • #46
PeroK said:
That reads like a fairly good summary of why your conclusions are physically meaningless!
I think I'm dropping the word "physical" because it doesn't seem to have a clear standard definition (if it does, I don't know what it is, and I suspect that trying to pin it down would mire us in a philosophical debate) and thus seems to be causing most of the misunderstandings here. What I care about is what answers you get when you assume that the traveler is stationary throughout the journey (that's what I mean when I say "what's true in the traveler's frame/perspective"—perhaps "perspective" is a better word).
 
  • #47
Gumby The Green said:
What I care about is what answers you get when you assume that the traveler is stationary throughout the journey (that's what I mean when I say "what's true in the traveler's frame/perspective").
Again, the central issue is that there is no one unique way to construct this frame, but any way you construct it must satisfy the standard requirements.
 
  • #48
Gumby The Green said:
What I care about is what answers you get when you assume that the traveler is stationary throughout the journey (that's what I mean when I say "what's true in the traveler's frame/perspective").
That's not an assumption. That's a choice of reference frame where one particular object remains at the spatial origin.

If we look at an elementary mechanics problem with a moving train, say, in which someone is tossing a coin. We can choose to analyse that from the train frame or from the ground frame. These are not conflicting assumptions; these are simply two ways to look at the same scenario.

Also, something which in your words is "true in the traveller's frame" in simply some coordinate dependent statement. For example, if we say that object A moves with velocity ##v##. That's a coordinate dependent statement. It's a true statement in that everyone agees that object A moves with velocity ##v## in a certain reference frame. But, there is no sense in which the object A has absolute velocity ##v##.

The principle of relativity tells us that we can do physics in any (inertial) reference frame we choose. There is no right and wrong.

You are in deep looking at the Rindler horizon without really understanding the basics of reference frames and what physics is all about.
 
  • #49
Gumby The Green said:
What I care about is what answers you get when you assume that the traveler is stationary throughout the journey (that's what I mean when I say "what's true in the traveler's frame/perspective"—perhaps "perspective" is a better word).
Depends how you construct the coordinates. As long as surfaces of constant time are spacelike and non-crossing, the coordinate system is valid and you can get a great many different answers. None of them will involve any clocks going backwards, though. The non-crossing condition enforces that.
 
  • #50
PeroK said:
That's not an assumption. That's a choice of reference frame where one particular object remains at the spatial origin.
I realize that. "Assume X" can mean "treat X as true".
PeroK said:
These are not conflicting assumptions; these are simply two ways to look at the same scenario...

The principle of relativity tells us that we can do physics in any (inertial) reference frame we choose...
I understand all of that; I just haven't been using all the right words apparently.
 
  • #51
Dale said:
The issue then is to clearly specify what you mean by the Traveler’s frame...

I recommend the Dolby and Gull definition, but it is not mandatory.
I'll be able to look at the radar coordinates later today. In the meantime, can't I just say that the traveler's perspective is comprised of five different frames, two of which are inertial (Minkowski coordinates), three of which involve constant proper acceleration (Rindler coordinates), and all of which treat the traveling twin as stationary, stitched together so that the final position and velocity of the Earth in one frame is the initial position and velocity of the Earth in the next? Isn't this basically how the traveler's perspective in the twin paradox is usually analyzed (minus the accelerating frames at the beginning and end)? Would you say that this works with the basic twin paradox—but not with my variation of it—because it doesn't involve any time inversion?
 
  • #52
Gumby The Green said:
What I care about is what answers you get when you assume that the traveler is stationary throughout the journey (that's what I mean when I say "what's true in the traveler's frame/perspective"—perhaps "perspective" is a better word).
The traveller’s perspective is going to be based on what they sees/experience, so let’s set up a strobe light on Earth that flashes once every second. Every time that the traveller sees a flash, they can say “OK, that’s another tick of the Earth twin’s clock” and they can look at their clock to see how much time has passed since the previous flash arrived.
On the outbound leg, the flashes arrive more than one second apart according to the traveler’s clock (which is measuring the traveler’s proper time). On the inbound leg, the flashes arrive less than one second apart according to the traveler’s clock. The total number of flashes received during the entire journey is greater than the number of seconds measured by the traveller’s clock during the journey - that is, the Earth clock ticked more often during the journey than the traveling clock so the Earth twin experienced more time and aged more between the departure and the return.

Note that there is no frame involved here. We’re just stating facts about the the time on the traveller’s clock when a flash of light reaches it. Everyone everywhere will agree about these statements of fact, no matter their state of motion and no matter what their clocks might be doing.

(You might want to take a moment to work out the Earth twin’s experience/perspective, assuming that the spaceship is also equipped with its own strobe light. The key difference will be when the flashes start arriving more rapidly - for the traveller it is halfway between departure and return, but for the Earth twin it will be closer to the return).
 
  • #53
Let's go back to your initial question:

Gumby The Green said:
Summary:: In the twin paradox, if the traveling twin keeps track of the proper time of a stationary observer who's farther away than the earthbound twin, how can he avoid concluding that the observer's time ran backwards at some points during the journey?
The proper time of an object never runs backwards, no matter what coordinates you choose. This is shown by considering a continuous emission of radio signals communicating the objects proper time. All the arbitrary changes in coordinate time that you make have no bearing on the matter. You can chop and change your simultaneity conventions as often as you please: that does not and cannot make an object's proper time run backwards.

Time to move on!
 
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  • #54
Gumby The Green said:
can't I just say that the traveler's perspective is comprised of five different frames, two of which are inertial (Minkowski coordinates), three of which involve constant proper acceleration (Rindler coordinates), and all of which treat the traveling twin as stationary, stitched together so that the final position and velocity of the Earth in one frame is the initial position and velocity of the Earth in the next?
I don’t believe so. The tricky part is that “stitching together” part. I don’t believe that there is a way to do that which satisfies the requirements.

Gumby The Green said:
Isn't this basically how the twin paradox is usually analyzed (minus the accelerating frames at the beginning and end)? Would you say that this works with the basic twin paradox—but not with my variation of it—because it doesn't involve any time inversion
I have never seen anyone actually do the analysis you propose. People worrying about the twin’s paradox almost universally refer to “the traveler’s frame” without ever even acknowledging the fact that there is no standard definition of such a frame
 
  • #55
Sagittarius A-Star said:
It is physically impossible, that the traveler can receive light from behind the Rindler horizon.
That's part of my point and I should've stated it more clearly: If the traveler, while accelerating away, can't even receive signals that the distant observer sends while behind the Rindler horizon, how can we say that he'll receive those signals in any particular order? Now the counterpoint would probably be, "But if he stops accelerating, then he'll receive all of those signals in forward order", to which I would respond, "That's because at that point, he's back in a frame whose time is not inverted relative to hers. The time inversion is relative, not absolute!"
 
  • #56
Gumby The Green said:
I understand all of that; I just haven't been using all the right words apparently.
One way to look at it is this. When you are moving inertially your worldline is straight. You can draw a unique grid on all of spacetime just by drawing lines orthogonal to or parallel to your worldline. But there is no suchgl grid in reality, any more than there are longitude and latitude lines marked on the Earth. So you aren't obliged to use the orthogonal grid you drew - you can just draw a different grid, possibly one that isn't even drawn with the parallel/perpendicular construction. It's usually unnecessarily complicated to do it, which is why people don't, but you can do it.

Once you accelerate, you can't just draw straight lines orthogonal to your worldline any more because they cross and you end up assigning the same event multiple coordinates. So we have to used curved lines. There is only one way to draw a straight line, but "draw a curved line" is a much less tightly specified instruction. And that's, basically, why there's no single answer to "what is the traveller's perspective" - it depends what curves you choose to draw. You need to add other constraints to get a unique answer, such as saying "I'll use Dolby and Gull's radar coordinates", to get an answer, but those constraints are your choice.
 
  • #57
Gumby The Green said:
"That's because at that point, he's back in a frame whose time is not inverted relative to hers. The time inversion is relative, not absolute!"
The problem if you try this is that your frames either overlap or are discontinuous - and we're back looking at a map with two copies of your house wondering if that means you've really got two houses.
 
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  • #58
PeroK said:
The proper time of an object never runs backwards, no matter what coordinates you choose.
I think I've made it clear that I'm not talking about anyone's proper time running backwards. I'm only saying that the time of an inertial observer, as inferred from a non-inertial frame (i.e., a coordinate system or combination of them in which a non-inertial traveler is treated as stationary), can run backwards.
 
  • #59
Gumby The Green said:
I'm only saying that the time of an inertial observer, as inferred from a non-inertial frame (i.e., a coordinate system or combination of them in which a non-inertial traveler is treated as stationary), can run backwards.
It cannot, unless coordinate lines cross. If they cross then you don't have a coherent perspective - you have one where you claim that you have two houses because your house appears twice on your (bad) map.
 
  • #60
Dale said:
I don’t believe so. The tricky part is that “stitching together” part. I don’t believe that there is a way to do that which satisfies the requirements.
If we take a basic acceleration scenario we can see the problem immediately. In some IRF at time ##t = 0## we have:

Object ##A## at position ##x = -L##; object ##R## at ##x = 0## and object ##B## at ##x = L##.

Now ##R## is a rocket that accelerates very quickly to speed ##v## towards ##B##. We can assume that the distance traveled and the time taken to accelerate are small compared with ##L##. I.e. the rocket is still near the origin of the original frame.

Now, the new rest frame for ##R## at time ##t' = 0## has the time at ##A## of ##+\gamma v L## and the time at ##B## at ##-\gamma vL## (using the standard Einstein sync convention).

This, in the OP's terms, is ##B##'s time running backwards from the original ##t' = t = 0## when the acceleration began. Moreover, as the rocket moves inertially, the time at ##B## (in the rocket frame) advances to ##t' = 0## again. We end up with two events that have ##t' = 0## in the rocket's changing inertial frames.

This is not a paradox because a) we can treat the coordinate time at ##B## going backwards as a change like the autumn clock changes; and b) we can use a different simultaneity convention to ensure that the coordinate time at ##B## never goes backwards.

In any case, it's only a technical problem with coordinates; and nothing more.
 
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  • #61
PS in the above scenario if we imagine a clock at rest in the rocket's frame after acceleration, Einstein sync'd with the rocket clock, and colocated with ##B## then that clock shows a different local time from the clock at ##B##. That clock does not show ##B##'s proper time. That's the critical point.

Instead of ##B##'s time having run backwards, ##B##'s clock is ahead of the hypotehtical "rocket" clock at ##B##.
 
  • #62
Ibix said:
Once you accelerate, you can't just draw straight lines orthogonal to your worldline any more... So we have to use curved lines... there's no single answer to "what is the traveller's perspective" - it depends what curves you choose to draw. You need to add other constraints to get a unique answer, such as saying "I'll use Dolby and Gull's radar coordinates"
That's why I proposed using Rindler coordinates during the acceleration phases of the journey (but those aren't defined behind the horizon where the distant stationary observer resides). Are (and Dale) you saying that there are other coordinates that would always 1) treat the traveler as stationary, 2) be defined where the distant observer resides, and 3) not show the distant observer's time as running backwards? If so, would the radar coordinates be an example?
 
  • #63
Gumby The Green said:
"That's because at that point, he's back in a frame whose time is not inverted relative to hers
But aren’t you only interested in “the traveler’s frame”. What does it matter if they are momentarily at rest in some other frame?
 
  • #64
PeroK said:
In any case, it's only a technical problem with coordinates; and nothing more.
Yes. But it is a technical problem that must be resolved for anything that claims to be “the traveler’s frame”
 
  • #65
Gumby The Green said:
Are (and Dale) you saying that there are other coordinates that would always 1) treat the traveler as stationary, 2) be defined where the distant observer resides, and 3) not show the distant observer's time as running backwards? If so, would the radar coordinates be an example?
Yes, and yes.
 
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  • #66
Dale said:
But aren’t you only interested in “the traveler’s frame”. What does it matter if they are momentarily at rest in some other frame?
Yes but (if I understand your question) what I'm saying is that when the traveler stops accelerating and then receives the backlog of signals from the distant observer in the forward order, that wouldn't prove to him that her time was never running backward in his frame; it would only prove that it's not doing so anymore.
 
  • #67
Gumby The Green said:
Yes but (if I understand your question) what I'm saying is that when the traveler stops accelerating and then receives the backlog of signals from the distant observer in the forward order, that wouldn't prove to him that her time was never running backward in his frame; it would only prove that it's not doing so anymore.
Because if time ran backwards, then you'd get, for example, two signals with "my time is ##t_0##". That's what time running backwards would mean. The time would go something like: ##t_0##, ##t_0 - 1##, ##t_0 - 2##. Then, running forwards again you would further get ##t_0 -1##, ##t_0##, ##t_0 + 1##.

You would receive signals ##t_0, t_0 - 1, t_0 - 2, t_0 -1, t_0, t_0 + 1##.

What you could get are signals where those were your coordinate times, calculated by the distant observer and communicated to you. In full, you could see messages something like:

"My clock reads ##0##, your coordinate time at my location is ##0##."

"My clock reads ##1##, your coordinate time at my location is ##-2.0##.

"My clock reads ##2##, your coordinate time at my location is ##-4.0##."

"My clock reads ##3##, your coordinate time at my location is ##-2.0##."

"My clock reads ##4##, your coordinate time at my location is ##0##.

My clock reads ##5##, your coordinate time at my location is ##+2.0##.

The distant observer could calculate your coordinate time at their location just as easily as you can. This exposes that although your coordinate time at their location is going backwards and forwards as you change your simultaneity convention, their proper time is behaving normally.
 
  • #68
Gumby The Green said:
when the traveler stops accelerating and then receives the backlog of signals from the distant observer in the forward order, that wouldn't prove to him that her time was never running backward in his frame; it would only prove that it's not doing so anymore
Yes, which is an indication that the question about time running backwards is not a physical question, it is a question about coordinates.

Nothing can be proven about coordinates from any physical measurement.
 
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  • #69
I've been making analogies with topographic maps quite a lot - thought it might be worth illustrating it a bit. Here's a map (public domain, taken from https://commons.wikimedia.org/wiki/File:Topographic_map_example.png) on which I've added a red line representing a simplified version of someone following the road through Lower Village and Stowe (it's in Vermont, if anyone's curious). There's only one corner in this trajectory, so it consists of just two straight line segments (just like the traveling twin's worldline in the idealised twin paradox).
1651342462681.png

Now, let's do what the OP wants to do. In the spacetime diagram, he takes the regions swept out by a perpendicular to each line segment (that is, the regions that are "next to" some part of the path), and glues them together so that the path is straightened out. Let's do that same thing on the map:
1651342850705.png

Note that in this version some parts of Cady Hill (left of the path) are missing while some parts of Stowe and Taber Hill (right of the path) are duplicated. A fair criticism of my demonstration of this approach is that the missing parts of Cady Hill arise because of the instantaneous nature of the corner. A more realistic smooth corner would include a distorted representation of those missing parts, but that's too hard to draw, and it doesn't change anything important about this argument, which is about what happens on the other side of the path. The duplication of Stowe and Taber Hill on the right of the path is unavoidable (well, to be precise you could corner slow enough not to duplicate those specific features, but there'd always be some feature on the right that was duplicated).

The question is: is this image the perspective of someone following that red route? Well, if you claim that it is, you claim that there is a perspective where Taber Hill exists in two places. So I'd say no.

What does this second image actually show? Imagine slicing the image into a stack of narrow slivers:
1651343975189.png

Like this, you can see that each horizontal slice shows you an accurate map of the part of the world that lies on a line perpendicular to the path's current direction, so the glued-together version of these is just that: everything that's directly to the side of you at some point, glued together. That's why Taber Hill appears twice - because your perpendicular direction changed when you took the corner so it genuinely was directly abeam of you twice. But do the slices of the world that are currently next to you combine together to make an accurate picture of the world? Something that could reasonably be called a perspective? No, not really, because nobody's perspective has Taber Hill doubled up. So the lesson you should take away is that simply stitching together the bits of the world that are directly to your left or right does not yield a good map (unless you always travel in a straight line).

This is also true in relativity. Literally the only difference in the argument is that we use Lorentz notions of orthogonality when thinking about Minkowski space rather than Euclidean notions in Euclidean space. It is certainly true that the traveling twin can draw two (or more) lines orthogonal to his worldline that pass through one event (that line is spacelike, so he can't actually see the same event more than once). But the lesson you should take from this is not that "time runs backwards", but rather that naively combining the sets of events on the lines orthogonal to your worldline does not produce anything resembling anyone's perspective. The only exception is if your worldline is always purely inertial, and it's this fact that makes inertial frames so easy to use (and so easy to conflate with "an observer"), and makes it so easy to think that naively applying the same approach to more complicated circumstances will produce a helpful result.
 
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  • #70
Gumby The Green said:
What is that based on though?
The fact that what each observer sees in the Doppler shifted light signals they receive is a direct observable and is invariant. There's no calculation involved; there's no "correction" because of some choice of coordinates. There's just what each observer directly sees, and they directly see all other observers' clocks running forward.

Gumby The Green said:
If the signals don't accurately depict the exact speed of the distant clock's "flow"
No signal can possibly "accurately depict the exact speed of the distant clock's flow" if the distant clock is in motion relative to the observer.

Gumby The Green said:
how are you sure that they accurately depict its direction in the traveler's frame at all times
Because the ordering of the light signals between emitter and detector is invariant. That is a simple physical fact about how light propagates. All of the light signals are moving in the same direction and at the same speed.

Gumby The Green said:
(even while its behind the traveler's Rindler horizon)?
If the distant object is behind the traveler's Rindler horizon, the traveler can't see it at all; no light signals are received. And the "accelerating frame" you are implicitly using for the traveler won't even cover the region of spacetime behind the Rindler horizon; it only covers the portion of spacetime that the traveler can actually see, i.e., receive light signals from. So objects behind the traveler's Rindler horizon are simply not included at all in the entire scenario you are talking about.

Gumby The Green said:
I'm not making any claims about what's actually happening to the distant observer in their own frame. All I care about is what's true in the Traveler's frame.
There is no such thing as "what's actually happening to the distant observer in their own frame" as contrasted with "what's true in the Traveler's frame". Coordinates and frames don't tell you what's "true". They are conveniences for calculation. They are not physical things and they do not tel you physical things.

"What's true" is contained in invariants: things that are independent of any choice of coordinates. In other words, "what is true" must be the same in all frames. The time an observer reads on their own clock at a particular event, such as emitting a particular light signal, is an invariant. What an observer actually sees in a Doppler shifted light signal arriving from a distant object is an invariant. But "what time is it for a distant object at a given event for the traveler" is not an invariant. It depends on your choice of coordinates and is a convenience for calculation; it doesn't tell you anything about "what is true".

Unless and until you are able to properly grasp the above, you will continue to make mistakes and you will continue to say wrong things that we have to correct.

Gumby The Green said:
I'm not applying any correction to the Doppler shift.
Yes, you are: in order to calculate what you are calling "what's true in the traveler's frame", you have to apply corrections to the Doppler shifted information in the light signals the traveler receives. If you just take the Doppler shifted information as it is, you are working with invariants that aren't "in" any frame, they are just invariants. But you refuse to do that; you insist on talking about "what is true in the traveler's frame" as thought it had physical meaning. It doesn't.

Gumby The Green said:
I don't care about the Doppler shift or about anything the traveler sees.
And you should. That's the point. What the traveler actually sees is invariant. What you are calling "what's true in the traveler's frame" is not, and has no physical meaning. So you're focusing on the wrong thing.

Gumby The Green said:
My argument is all about what can be logically deduced at the end of the journey from the most basic principles.
You might think it is, but you are wrong. What you are doing is not based on any physical principles. It is based on incorrectly treating coordinate-dependent quantities as though they had physical meaning. They don't.
 
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