# Graphical Derivation of the CADO Equation

I'm not sure what you mean. The whole purpose of the ground DVR recording was to show that letting the information travel a known distance at the speed of light does not change the end result.
I never claimed that it would. My point is: what does Charlie measure about his time at stand-still, now that he has been travelling again and received the information about what Alice's clock was doing at the time.

The time information received by the ground DVR (t=40 x=0) matches Charlie's calculation.
Yes, obviously. But what does Charlie now measure "really" happened back there, now that he's moved on?

So that's why I am not sure what you mean. Perhaps you mean that after Charlie jumps on the train, and he waits for the time information to arrive to himself as he rides on the train, then the information would be t=10, x=0 for the moment he landed on the train?
That's exactly what I meant. I don't see how what I said can be interpreted in another way? At what other point earlier than that does "the information about Alice's clock reach[es] him?" I know the second half of the scenario is your blind spot, so I'm glad you managed to remember it this time!

Yes, that is true, and that is what the train DVR records. But that pertains to the moment he landed on the train, not the moment he landed on the ground.
And what would you call that, the fact that Charlie changing frames changed what DVR is relevant to him, and his conclusions about Alice's clock? I'm still open to a better word that "overwrite".

Yes, I agree. The t=40 x=0 calculation is specifically for the moment he is on the ground. Once he has jumped back on the train, his calculation produces t=10, x=0 which is the only calculation that applies to him then. That is the whole point isn't it? First it was t=40 then it was t=10.
The calculations, yes. But Alice's clock never ticked backwards; that was the whole point.

1. If someone asks Charlie what time (t) it was at x=0 just before he jumped off the train, he would have to answer t=10.

2. If someone asks Charlie what time (t) it was at x=0 just after he jumped off the train, he would have to answer t=40.

3. If someone asks Charlie what time (t) it was at x=0 just after he jumped back on the train, he would have to answer t=10.

Even if the answer to #3 "overwrites" the answer to #2, those are still the answers he has to give to those questions.
Yes, obviously. But you've dodged the whole point: is Charlie now justified in saying Alice travelled back in time? My answer: no.

I honestly don't see the conflict. Sorry you find that so offensive.
No, I don't find ignorance offensive.

If I could see it, I would say so.
Go back a couple of sentence in your own post just now. You have, in order:
1) t=10
2) t=40
3) t=10

Any object or clock going from t=40 to t=10 is travelling back in time. However, everybody knows that currently, we haven't got any evidence travelling back in time is possible. So, it can't be travelling back in time.

Thus, Charlie must not be able to draw the conclusion that Alice's clock ticked backwards, even though the (correct) calculations suggest it. How do you address this?

I've been doing SR calculations a long time,
Same here.

so maybe I'm just used to it being this way.
Well, I've learned that the best way to really get to know a theory like SR, is to play with it, and address any questionable answer you get. For example, obviously the travelling backwards in time cannot be real, so I try to answer that question. I feel sorry for you that you don't experience such journeys anymore.

I guess it does seem strange sometimes, but I just accept that's how it is.
I on the other hand have an internal drive to understand. Any questionable answer in a theory so obviously self-consistent and worked-out as SR must have a resolution, so I go looking for it. I really do feel sorry for you that you've lost that curiosity.

But you say none of that happens, even though Charlie's answers to the 3 questions are t=10, t=40, t=10 respectively, right?
Do you think travelling backwards in time is possible through this method?

So you have already explained to yourself that none of that happens. So why is it still a problem?
Erm, you've just stated that Charlie calculates t=10, then t=40, and then t=10, and that those calculations are correct, but in the very next line you say that's not possible. Don't you see the problem?

If travelling backwards in time isn't possible, then how it t=10, followed by t=40, and then followed by t=10 possible?

Well if the word "overwrites" prevents all of those problems, then I guess it is a good word to use.
I'm not sure if it does; it doesn't "feel" right, but I can't come up with a better one at the moment. "Invalidates" would be better, but that sounds like the old calculation was wrong in some way. "Supersedes" perhaps?

I didn't really realise there were all those problem to begin with, so I didn't realise how important it was to use the right words.
In science, it's actually quite important to use the right words.

Measured reality is what is on the two DVR recordings.
Right, and neither of those show Alice travelling backwards in time at any moment.

They corroborate Charlie's answers to the 3 questions t=10, t=40, and t=10 respectively.
Correct.

That's all there is to this, right?
You keep ignoring the second half. How can t=10 be followed by t=40 be followed by t=10, if travelling backwards in time isn't allowed?

It's not intentional. I honestly don't understand what else there is to this besides Charlie's answers to the 3 questions t=10, t=40, and t=10 respectively.
Charlie's measurements also show that Alice at no point in time travelled backwards in time. However, t=10 followed by t=40 followed by t=10 suggests otherwise. It's really weird... You were just a few posts ago making all kinds of silly statements like me not understand that 40>10, but here you are, doing that exact thing. If time (t) starts at 10, then takes a value of 40, and then a value of 10, how did time not decreasing in that last step? And since it's the time on a clock; how is that not travelling backwards in time?

Really, you say there's a clock going from t=40 to t=10, but it's not ticking (or jumping) backwards?

I thought I already did that. On the Minkowski diagram, there are two different planes of simultaneity for the two frames. One intersects the t axis at t=40 x=0 and the other intersects the t axis at t=10 x=0. When Charlie jumps from the ground to the train, his plane of simultaneity changes from the former to the latter.
No, I didn't ask about Charlie or his calculations. Show on Alice's worldline where she does it. Show me where her worldline changes direction, and starts flowing downwards.

I honestly don't know what else you think there is to this.
Nothing more that I've already explained to you multiple times. It's just that you often ignoring half of it.

Yes that is true! But that is because I thought the only moments we were concerned with were the moment before he jumped off the train, the moment after he jumped off the train, and the moment after he jumped back on the train. It would be 60 years after that (in Charlie's own time) when he would calculate t=40 x=0 again.
I probably should've picked t=10 to avoid confusion. But I'm glad you agree I was correct in my original statement.

Don't worry, the sore thumb is there whether Charlie jumps off the train or on the train. You just have to look at the right location on the x axis to find the sore thumb in both cases. There is no difference jumping off or on a train, it is just changing frames either way.
Charlie can jump off and on the train just fine. But time ticks forward, but not backwards (as far as we know). I don't see how you think the former is some big issue, but dismiss the later? This feels like you are backpeddling on your original statement?

The train assistant never leaves the train frame, so it is t=10 x=0 during the whole jumping routine.
So no Alice travelling backwards in time.

But Charlie and the assistant are not in the same frame when Charlie is on the ground.
And I never claimed otherwise.

He could have an assistant on the ground tell him it is t=40 in the moment he is on the ground, and he would have to trust the ground assistant, not the train assistant.
True, and that assistant also says that Alice never travelled backwards in time.

The train assistant doesn't have Alice traveling back in time.
True. In fact, nobody has. Even Charlie's Minkowski diagram clearly shows this.

But Charlie's answers to the 3 questions are still t=10, t=40, and t=10 respectively, because of his jumping routine.
True, but Charlie's Minkowski diagram also shows Alice not travelling backwards in time. Are you seeing the conflict?
Here is something interesting I came up with over the weekend.

This spreadsheet shows the traveling twin's perspective. The top half is the outbound leg of the trip, and the bottom half is the inbound leg. In the middle you can see where his t=10 calculation suddenly becomes t=70 after the turnaround point.

The far right hand column shows what the traveling twin actually sees with his eyes, as he looks at the stay-home twin's clock. For example, on the outbound leg, he sees 5.09 on the distant clock at the same time that his own clock displays t'=19. Also, on the inbound leg, he sees 72.53 on the distant clock at the same time that his own clock displays t'=38.

I chose to use those examples because they happen to be close to some values for t that the traveling twin had calculated earlier, (please refer to the blue boxes & arrows). What is interesting is that this provides visual confirmation to the traveling twin that his earlier calculations were correct, as follows:

Back when the traveling twin's own clock displayed t'=10, he calculated t=5 for the stay-home twin's clock. And at that time, the distance to the stay-home twins clock was -x'=8.66 so it follows that he should actually see the t=5 after 8.66 more time has elapsed, which would be at t'=10+8.66=18.66 but the closest thing on the chart is t'=19 and he sees 5.09 at that time (see the red boxes & arrows on the top half of the chart). That is a confirmation that his calculations for t are correct on the outbound half of the trip.

And back when the traveling twin's own clock displayed t'=25, he calculated t=72.5 for the stay-home twin's clock. And at that time, the distance to the stay-home twins clock was -x'=12.99 so it follows that he should actually see the t=72.5 after 12.99 more time has elapsed, which would be at t'=25+12.99=37.99 but the closest thing on the chart is t'=38 and he sees 72.53 at that time (see the red boxes & arrows on the bottom half of the chart). That is a confirmation that his calculations for t are correct on the inbound half of the trip.

If all the inbound and outbound calculations are correct, then it follows that the t=10 calculation just before the turnaround, and the t=70 calculation just after the turnaround are both correct. The traveling twin could even visually confirm the t=70 in this same manner at his own time t'=20+17.32=37.32, but it just does not happen to be on the chart. The closest is t'=37 and he sees 68.8 at that time.
Great demonstration, but I think everybody in this thread already agreed on this, but I guess it's good to have it spelled out explicitly. Now please do the same for a calculation on the outbound leg, but with the observation on the inbound leg, because that's what we were talking about.

My point is: what does Charlie measure about his time at stand-still, now that he has been travelling again and received the information about what Alice's clock was doing at the time.

But what does Charlie now measure "really" happened back there, now that he's moved on?

It seems to me that Charlie has no choice but to say that the time "really" was t=40 in ground frame at the moment when he was on the ground. What would you say is the answer to your own question?

And what would you call that, the fact that Charlie changing frames changed what DVR is relevant to him, and his conclusions about Alice's clock? I'm still open to a better word that "overwrite".

I would call it frame-dependence. Certain things are frame-dependent. In this case, what we are really discussing is that simultaneity is frame-dependent. Charlie's moment on the ground is simultaneous with t=40 x=0, whereas Charlie's moment on the train (either just before or just after he jumps off) is simultaneous with t=10 x=0. He can only be in one frame at a time, and so only one DVR can be relevant to him at any moment.

There are other things which are frame-dependent also, things which we have not even discussed yet: For example, the distance between Charlie and Alice is 34.64 when he is on the ground, but the distance between Charlie and Alice is 17.32 when he is on the train.

I think we can agree that the speed of light is the fastest possible speed in any given inertial frame, and yet Charlie has no choice but to say that Alice's location relative to himself changes by 17.32 light years in the infinitesimal time it takes him to jump on/off the train.

This is just another one of those strange things that comes along with changing reference frames. An accelerating reference frame is not inertial, and so it has different rules than inertial frames.

The calculations, yes. But Alice's clock never ticked backwards; that was the whole point.

is Charlie now justified in saying Alice travelled back in time? My answer: no.

I would answer that Charlie must conclude that the time in Alice's location was t=40 x=0 in the moment he was on the ground, and that Charlie also must conclude that the time in Alice's location was t=10 x=0 in the moment he jumped back on the train. On the face of it, that looks like Alice must go back in time, just as her changing from 34.64 to 17.32 light years distance makes it look like she travels faster than light. But this is not one inertial frame we are talking about here, it is an acceleration, or changing of inertial frames.

Go back a couple of sentence in your own post just now. You have, in order:
1) t=10
2) t=40
3) t=10

Any object or clock going from t=40 to t=10 is travelling back in time. However, everybody knows that currently, we haven't got any evidence travelling back in time is possible. So, it can't be travelling back in time.

Thus, Charlie must not be able to draw the conclusion that Alice's clock ticked backwards, even though the (correct) calculations suggest it. How do you address this?

You are the one who says traveling back in time is impossible, so you have to address this, not me. Your answer seems to be that the t=10 overwrites the t=40 in a way that makes it not really going back in time. Similarly, I suppose you could say that the distance 17.32 overwrites the distance 34.64 in a way that makes it not really moving at superluminal speed. I think that is as good an explanation as we are going to get, actually.

Well, I've learned that the best way to really get to know a theory like SR, is to play with it, and address any questionable answer you get. For example, obviously the travelling backwards in time cannot be real, so I try to answer that question. I feel sorry for you that you don't experience such journeys anymore.

I on the other hand have an internal drive to understand. Any questionable answer in a theory so obviously self-consistent and worked-out as SR must have a resolution, so I go looking for it. I really do feel sorry for you that you've lost that curiosity.

Don't feel bad for me. I am quite content just stating that accelerated reference frames have some strange rules which are quite different from the rules of inertial frames.

I feel bad for you actually, because you are probably trying to imagine the gigantic "pseudo forces" that caused the distance between Alice and Charlie to change by 17.32 light years in one moment.

Do you think travelling backwards in time is possible through this method?

When I first discovered these strange effects of jumping on/off trains, I tried a number of thought experiments to see if a train hopper and his various assistants could ever get some information from the future. For example, if a lottery number was being drawn at the back of the train, was there any way to get the information of the winning number to the front of the train in time to have someone buy a winning ticket using that inside information? I could not come up with any way to do so.

Erm, you've just stated that Charlie calculates t=10, then t=40, and then t=10, and that those calculations are correct, but in the very next line you say that's not possible. Don't you see the problem?

If travelling backwards in time isn't possible, then how it t=10, followed by t=40, and then followed by t=10 possible?

Wait, where did I say that's not possible? I think t=40 x=0 is followed by t=10 x=0 when Charlie jumps on the train.

I'm not sure if it does; it doesn't "feel" right, but I can't come up with a better one at the moment. "Invalidates" would be better, but that sounds like the old calculation was wrong in some way. "Supersedes" perhaps?

I just go with the idea that some things are frame-dependent, and that accelerated frames have different rules than inertial frames.

You keep ignoring the second half. How can t=10 be followed by t=40 be followed by t=10, if travelling backwards in time isn't allowed?

You should tell me, since you are the one saying that time going backwards isn't allowed.

If time (t) starts at 10, then takes a value of 40, and then a value of 10, how did time not decreasing in that last step? And since it's the time on a clock; how is that not travelling backwards in time?

Really, you say there's a clock going from t=40 to t=10, but it's not ticking (or jumping) backwards?

You should tell me.

No, I didn't ask about Charlie or his calculations. Show on Alice's worldline where she does it. Show me where her worldline changes direction, and starts flowing downwards.

That doesn't happen, obviously. But when Charlie changes frames, his plane of simultaneity changes, and that is apparent on the Minkowski diagram. Time at x=0 can go from t=40 to t=10 when Charlie does this.

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Charlie can jump off and on the train just fine. But time ticks forward, but not backwards (as far as we know). I don't see how you think the former is some big issue, but dismiss the later? This feels like you are backpeddling on your original statement?

When Charlie says that t=10 x=0 changes to t=40 x=0 at the moment Charlie jumps off the train, we could consider that to be time going forward, because 10<40. However, there are other places on the x axis that we could have considered besides x=0. If we had considered x=69.28 we would have found that Charlie says that t=70 x=69.28 changes to t=40 x=69.28 at the moment Charlie jumps off the train, and we could consider that to be time going backward, because 70>40.

Every jump, whether it is on or off the train, has these same effects. Time goes forward over there, backward over there, etc.

So no Alice travelling backwards in time.

What about when t=40 is followed by t=10 in the case of Charlie jumping from the ground to the train?

True, and that assistant also says that Alice never travelled backwards in time.

In fact, nobody has [Alice going back in time]. Even Charlie's Minkowski diagram clearly shows this.

The Minkowski diagram shows two different planes of simultaneity, and when Charlie jumps on the train it goes from t=40 x=0 to t=10 x=0. You can call that whatever you like.

True, but Charlie's Minkowski diagram also shows Alice not travelling backwards in time. Are you seeing the conflict?

I guess I see the conflict that you have set up by making a rule that time cannot go backward, even in the case of a distant location and an accelerating reference frame. The conflict arises because we clearly have t=40 being followed by t=10.

Great demonstration, but I think everybody in this thread already agreed on this, but I guess it's good to have it spelled out explicitly. Now please do the same for a calculation on the outbound leg, but with the observation on the inbound leg, because that's what we were talking about.

Thank you. I agree that we all would have agreed on this anyway, but better to have it spelled out explicitly.

What I show in the spreadsheet is the standard twin scenario, with one outbound leg and one inbound leg. In row 22 of the spreadsheet we see the moment just before the turnaround point where the traveling twin's own time is t'=20 and the stay home twin's time is t=10. In row 24 of the spreadsheet we see the moment just after the turnaround point where the traveling twin's own time is t'=20 and the stay home twin's time is t=70.

What you and I have been talking about is a different scenario where the traveler stops completely, rather than reversing direction. Nothing like that is not shown on the spreadsheet. However, something similar can be seen on the spreadsheet if we imagine that the traveling twin turns around once (i.e. row 24 where t=70) and then quickly turns around again, returning to row 22 where t=10 again. That would be similar to what you and I are discussing in that the t=10 would come back again, even though the later time of t=70 had already happened just after the first turnaround.

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Great demonstration, but I think everybody in this thread already agreed on this, but I guess it's good to have it spelled out explicitly. Now please do the same for a calculation on the outbound leg, but with the observation on the inbound leg, because that's what we were talking about.

Here is the revised spreadsheet showing Charlie riding the train from his own time t'=0 to t'=20, then jumping off & right back on the train again.

Because x'=-17.32 is a train frame measurement, I had to keep gamma=2 even though v=0.00 at the moment Charlie is on the ground. If I had used the CADO equation in column E, instead of the Lorentz equation, I could have avoided that issue. So the CADO equation does have its usefulness.

But the advantage of doing it this way is that we can use t'=20.00 and x'=-17.32 to calculate that the time when Charlie would actually see the t=10 with his eyes would be at his own time t'=37.32, and the chart confirms that.

It seems to me that Charlie has no choice but to say that the time "really" was t=40 in ground frame at the moment when he was on the ground. What would you say is the answer to your own question?
I say that Charlie cannot come to the conclusion that Alice travelled back in time, and thus that any such claim (direct or indirect) must be incorrect.

I would call it frame-dependence. Certain things are frame-dependent. In this case, what we are really discussing is that simultaneity is frame-dependent.
Right, so Charlie cannot conclude that Alice's clock went backwards, because he's comparing a frame-dependent quantity measured in two separate frames.

Charlie's moment on the ground is simultaneous with t=40 x=0, whereas Charlie's moment on the train (either just before or just after he jumps off) is simultaneous with t=10 x=0. He can only be in one frame at a time, and so only one DVR can be relevant to him at any moment.
And thus his initial calculation of t=40 becomes irrelevant when he switches frames. The new calculations "overwrites" the old one. He changes what time this variable "t" is referring to (i.e. it is frame-dependent, and is switched from one frame to another).

There are other things which are frame-dependent also, things which we have not even discussed yet: For example, the distance between Charlie and Alice is 34.64 when he is on the ground, but the distance between Charlie and Alice is 17.32 when he is on the train.

I think we can agree that the speed of light is the fastest possible speed in any given inertial frame, and yet Charlie has no choice but to say that Alice's location relative to himself changes by 17.32 light years in the infinitesimal time it takes him to jump on/off the train.

This is just another one of those strange things that comes along with changing reference frames. An accelerating reference frame is not inertial, and so it has different rules than inertial frames.
Yep, length contraction and time dilation are funny things, but they are measurably true. Clocks ticking backwards though...

I would answer that Charlie must conclude that the time in Alice's location was t=40 x=0 in the moment he was on the ground, and that Charlie also must conclude that the time in Alice's location was t=10 x=0 in the moment he jumped back on the train. On the face of it, that looks like Alice must go back in time, just as her changing from 34.64 to 17.32 light years distance makes it look like she travels faster than light. But this is not one inertial frame we are talking about here, it is an acceleration,
Are you arguing that the theory of special relativity doesn't hold when acceleration is involved?

or changing of inertial frames.
Yes, obviously. But please finish that thought: what does Charlie's changing of frame, causing a calculation to indicate a backwards ticking clock for Alice, physically mean for Alice?

You are the one who says traveling back in time is impossible,
If you say this effect is a case of travelling back in time, please write a paper about it, and collect your Nobel Prize. You claim to have disproven the "arrow of time"!

so you have to address this, not me.

But your resolution seems to be that this is indeed proof of travelling back in time? Again: publish a paper, and collect your Nobel Prize.

Your answer seems to be that the t=10 overwrites the t=40 in a way that makes it not really going back in time.

Similarly, I suppose you could say that the distance 17.32 overwrites the distance 34.64 in a way that makes it not really moving at superluminal speed.
That wouldn't be the same situation, because the change in distance can be measured.

I think that is as good an explanation as we are going to get, actually.
Good. Seems I was right all along.

Don't feel bad for me. I am quite content just stating that accelerated reference frames have some strange rules which are quite different from the rules of inertial frames.
Actually, they don't, once you figure them out. But whatever floats your boat.

I feel bad for you actually, because you are probably trying to imagine the gigantic "pseudo forces" that caused the distance between Alice and Charlie to change by 17.32 light years in one moment.
Nope, that's just you imagining that.

When I first discovered these strange effects of jumping on/off trains, I tried a number of thought experiments to see if a train hopper and his various assistants could ever get some information from the future. For example, if a lottery number was being drawn at the back of the train, was there any way to get the information of the winning number to the front of the train in time to have someone buy a winning ticket using that inside information? I could not come up with any way to do so.
Same here, and I posted that in this thread many, many pages ago.

Wait, where did I say that's not possible? I think t=40 x=0 is followed by t=10 x=0 when Charlie jumps on the train.
So you do think this is travelling back in time? As I said, write a paper about it, and collect your Nobel Prize.

I just go with the idea that some things are frame-dependent,
Travelling back in time is frame-dependent then?

and that accelerated frames have different rules than inertial frames.
They really don't, though.

You should tell me, since you are the one saying that time going backwards isn't allowed.
Please stop feigning ignorance; you've used my "overwrites"-solution in your very post.

You should tell me.
I'm sorry, I don't know what you think. I'm not psychic.

That doesn't happen, obviously.
So Alice doesn't travel back in time, great! So any conclusion that Charlie draws that Alice did, must be wrong. Which means that if Charlie says that Alice's clock shows t=10, then t=40, and then t=10, he is wrong.

But when Charlie changes frames, his plane of simultaneity changes, and that is apparent on the Minkowski diagram.
Yep, obviously.

Time at x=0 can go from t=40 to t=10 when Charlie does this.
Nope; please study more carefully what the plane of simultaneity is. You just said that Alice doesn't travel back in time, so Alice's time doesn't go from t=40 to t=10.

When Charlie says that t=10 x=0 changes to t=40 x=0 at the moment Charlie jumps off the train, we could consider that to be time going forward, because 10<40. However, there are other places on the x axis that we could have considered besides x=0. If we had considered x=69.28 we would have found that Charlie says that t=70 x=69.28 changes to t=40 x=69.28 at the moment Charlie jumps off the train, and we could consider that to be time going backward, because 70>40.

Every jump, whether it is on or off the train, has these same effects. Time goes forward over there, backward over there, etc.
So travelling back in time is possible, according to you? Please go write that paper, and collect your Nobel Prize!

What about when t=40 is followed by t=10 in the case of Charlie jumping from the ground to the train?
You tell me; I've been asking you that question for pages now. (You already have my answer.)

The Minkowski diagram shows two different planes of simultaneity, and when Charlie jumps on the train it goes from t=40 x=0 to t=10 x=0. You can call that whatever you like.
And what does Alice's worldline do? You know, the thing that solely determines whether Alice goes forward or backwards in time? You already stated in your previous post that it doesn't go backwards, so Alice doesn't travel back in time.

I guess I see the conflict that you have set up by making a rule that time cannot go backward,
Please point me to the worldline going backwards in the Minkowski diagram, because that would be an object going back in time.

And if you think this is time going backwards, blah blah Nobel Prize.

even in the case of a distant location and an accelerating reference frame. The conflict arises because we clearly have t=40 being followed by t=10.
Finally, you see the conflict! Now, judging by what you've been writing so far, it seems that you are going to resolve this conflict by saying that time goes backwards.

Blah blah Nobel Prize.

Thank you. I agree that we all would have agreed on this anyway, but better to have it spelled out explicitly.

What I show in the spreadsheet is the standard twin scenario, with one outbound leg and one inbound leg. In row 22 of the spreadsheet we see the moment just before the turnaround point where the traveling twin's own time is t'=20 and the stay home twin's time is t=10. In row 24 of the spreadsheet we see the moment just after the turnaround point where the traveling twin's own time is t'=20 and the stay home twin's time is t=70.

What you and I have been talking about is a different scenario where the traveler stops completely, rather than reversing direction. Nothing like that is not shown on the spreadsheet. However, something similar can be seen on the spreadsheet if we imagine that the traveling twin turns around once (i.e. row 24 where t=70) and then quickly turns around again, returning to row 22 where t=10 again. That would be similar to what you and I are discussing in that the t=10 would come back again, even though the later time of t=70 had already happened just after the first turnaround.
And...? Did you forget to finish that thought? What happens in the case where the validation of a calculation spans over a switch in frame? Is the calculation wrong, does it get overwritten by a new, updated calculation, or...?

Here is the revised spreadsheet showing Charlie riding the train from his own time t'=0 to t'=20, then jumping off & right back on the train again.

Because x'=-17.32 is a train frame measurement, I had to keep gamma=2 even though v=0.00 at the moment Charlie is on the ground. If I had used the CADO equation in column E, instead of the Lorentz equation, I could have avoided that issue. So the CADO equation does have its usefulness.

But the advantage of doing it this way is that we can use t'=20.00 and x'=-17.32 to calculate that the time when Charlie would actually see the t=10 with his eyes would be at his own time t'=37.32, and the chart confirms that.
So... you've completely ignored what I asked, and decided to point out something that we both agreed on from page 1. Care to actually address the point?

(For the record: I asked you to do such a calculation using what's now in E24, not E26.)

Charlie cannot conclude that Alice's clock went backwards, because he's comparing a frame-dependent quantity measured in two separate frames.

Charlie isn't using two separate frames. He is using his one single frame. It is called his CADO reference frame. The CADO frame is not an inertial frame, but it is a single frame. And in that frame, Alice has clearly gotten younger.

Travelling back in time is frame-dependent then?

YES! Alice gets younger in Charlie's CADO reference frame. She doesn't get younger in her inertial frame. Or in any other inertial frame.

Charlie isn't using two separate frames. He is using his one single frame. It is called his CADO reference frame. The CADO frame is not an inertial frame, but it is a single frame. And in that frame, Alice has clearly gotten younger.
It's probably better to discuss that with Neddy Bate, since he's the one that brought it up.

YES! Alice gets younger in Charlie's CADO reference frame. She doesn't get younger in her inertial frame. Or in any other inertial frame.
So you too are arguing that travelling back in time is possible? Blah blah Nobel Prize for you too, then.
But perhaps you just worded it carelessly; what do you mean exactly by "gets"? As observed, measured, or calculated?

NotEinstein wrote:

"Travelling back in time is frame-dependent then?"

I (Mike Fontenot) wrote:

"YES! Alice gets younger in Charlie's CADO reference frame. She doesn't get younger in her inertial frame. Or in any other inertial frame."

Then NotEinstein wrote

"What do you mean exactly by "gets"? As observed, measured, or calculated?"

And my (Mike Fontenot's) response is

"Initially, Charlie calculates (using either the Lorentz equations, or the CADO equation, or even easier, the delta_CADO equation) that Alice has just gotten younger. But he can also eventually directly confirm (as Neddy Bate has so well shown) that his calculation was correct."

Initially, Charlie calculates (using either the Lorentz equations, or the CADO equation, or even easier, the delta_CADO equation) that Alice has just gotten younger.
Yes, all three of us are in agreement here.

But he can also eventually directly confirm (as Neddy Bate has so well shown) that his calculation was correct.
Charlie's own observations conflict with the calculation, as pointed out earlier by me, and backed up by Neddy Bate. But perhaps you can demonstrate how Charlie can make an observation actually confirming that calculation directly?

Charlie's own observations conflict with the calculation, as pointed out earlier by me, and backed up by Neddy Bate. But perhaps you can demonstrate how Charlie can make an observation actually confirming that calculation directly?

In my opinion, Neddy has already patiently and brilliantly shown that.

In my opinion, Neddy has already patiently and brilliantly shown that.
So you are unwilling to answer the issue raised?

I say that Charlie cannot come to the conclusion that Alice travelled back in time, and thus that any such claim (direct or indirect) must be incorrect.

I think you already agreed that Charlie's t=40 calculation is correct, and his subsequent t=10 calculation is also correct.

Right, so Charlie cannot conclude that Alice's clock went backwards, because he's comparing a frame-dependent quantity measured in two separate frames.

I think Charlie would be well aware that he changed frames. That is the only reason he would say that t=40 x=0 was followed by t=10 x=0 in the first place. Of course he could also say that the t=10 overwrites the t=40 but that does not mean the t=40 was wrong in any way.

And thus his initial calculation of t=40 becomes irrelevant when he switches frames. The new calculations "overwrites" the old one. He changes what time this variable "t" is referring to (i.e. it is frame-dependent, and is switched from one frame to another).

In both cases, the t variable represents the time in the ground frame at x=0. Even if t=10 overwrites t=40, that does not change the fact that t=40 came before t=10.

Are you arguing that the theory of special relativity doesn't hold when acceleration is involved?

No, not at all. It is special relativity which tells us that t=40 came before t=10 in the case of Charlie's acceleration changing his frame.

Yes, obviously. But please finish that thought: what does Charlie's changing of frame, causing a calculation to indicate a backwards ticking clock for Alice, physically mean for Alice?

The question is what it means for Charlie. Charlie is the one who says t=40 x=0 was followed by t=10 x=0. Alice does not accelerate so she never says anything like that.

If you say this effect is a case of travelling back in time, please write a paper about it, and collect your Nobel Prize. You claim to have disproven the "arrow of time"!

I don't think this is news to anyone who understand special relativity.

But your resolution seems to be that this is indeed proof of travelling back in time? Again: publish a paper, and collect your Nobel Prize.

Again, this is old news.

That wouldn't be the same situation, because the change in distance can be measured.

The change in time was also measured by the DVR recordings.

Good. Seems I was right all along.

It certainly helps to keep in mind that Charlie changed frames when he accelerated. Of course the t=40 x=0 pertained to when he was in the ground frame, and the t=10 x=0 pertained to when he was in the train frame.

Actually, they don't, once you figure them out. But whatever floats your boat.

Things like the distance between Alice and Charlie changing by 17.32 light years in one moment do not happen in an inertial frame, but they do in an accelerating frame.

Nope, that's just you imagining that.

Earlier in the thread you talked about "pseudo forces" when I mentioned length contraction. So you must think that pseudo forces acted on the whole ground when Charlie jumps on/off the train.

So you do think this is travelling back in time? As I said, write a paper about it, and collect your Nobel Prize.

Travelling back in time is frame-dependent then?

It is Charlie's simultaneity changing.

So Alice doesn't travel back in time, great! So any conclusion that Charlie draws that Alice did, must be wrong. Which means that if Charlie says that Alice's clock shows t=10, then t=40, and then t=10, he is wrong.

What? I thought you agreed those calculations were correct. You must think SR is wrong then?

Nope; please study more carefully what the plane of simultaneity is. You just said that Alice doesn't travel back in time, so Alice's time doesn't go from t=40 to t=10.

Charlie would say it does, not Alice.

Care to actually address the point?

(For the record: I asked you to do such a calculation using what's now in E24, not E26.)

Alright let's look at row 24. Charlie is in the ground frame, and he sees the clock at x=0 displaying 5.36. The distance between him and that clock is 34.64. So, if Charlie stays there 34.64 more years, he would see the distant clock say 34.64+5.36=40.

Of course Charlie does not have to wait that long to make that conclusion. It is immediately evident from the 34.64 distance, and the time 5.36 which he sees with his eyes.

Case closed.

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In my opinion, Neddy has already patiently and brilliantly shown that.

Thank you, Mike. I would also like to say that the CADO equation would have made my most recent spreadsheet nicer, because then when Charlie was on the ground, the distance to Alice (34.64) would already have been in the data, instead of x'=-17.32. And then I could have let gamma go to zero when v went to zero, as it should be. So, congratulations on developing the CADO equation, I find it useful and interesting.

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An idle moment, a whimsical peek at where Physics & Maths has been randomly drifting, and ... what?! The CADO dead horse is still flogged?!
Mike Fontenot - are you a coward? Why did you avoid answering my call in #90 to answer my challenge even further back in #30? (minor correction re #90 - Andromeda galaxy is 'only' ~ 2.5 million ly distant, not 3+ billion ly). Naive extrapolation a la Brian Greene, vs 'reality' principle remains valid regardless. That is to say - to avoid stupid, unphysical inferences and conclusions from such - always check against the reliable, Doppler shift 'interpretation/analysis' of traveling-twin 'paradox'.

An idle moment, a whimsical peek at where Physics & Maths has been randomly drifting, and ... what?! The CADO dead horse is still flogged?!
Right now that's because Neddy Bate is defending the position that travelling back in time is real and happens all the time in SR.

What's your opinion on this? The situation: Alice is in an inertial frame, Charlie is somewhere out in space. Charlie changes speed while flying straight away from Alice. This causes his plane of simultaneity to change, and his calculation of Alice's age to go down (or up). Did Alice travel back in time?

You have, in order:
1) t=10
2) t=40
3) t=10

Any object or clock going from t=40 to t=10 is travelling back in time. However, everybody knows that currently, we haven't got any evidence travelling back in time is possible. So, it can't be travelling back in time.

You are the one who says traveling back in time is impossible, so you have to address this, not me.

And Mike_Fontenot appears to hold this position too:
YES! Alice gets younger in Charlie's CADO reference frame. She doesn't get younger in her inertial frame. Or in any other inertial frame.

Right now that's because Neddy Bate is defending the position that travelling back in time is real and happens all the time in SR.

What's your opinion on this? The situation: Alice is in an inertial frame, Charlie is somewhere out in space. Charlie changes speed while flying straight away from Alice. This causes his plane of simultaneity to change, and his calculation of Alice's age to go down (or up). Did Alice travel back in time?...
I believe we two are in agreement that actual backwards time travel, from the pov of either of two observers in arbitrary relative motion, never happens. Concluding otherwise via naive interpretation of instantaneous gradients on a space-time diagram is simply unwarranted extrapolation.
Sadly, not only Brian Greene but even Roger Penrose got to take the naive interpretation of such diagrams too seriously. About the only situation they can be considered to be accurately reflecting reality is for the case of uniform relative speed. Introduce any nonuniformity, and all you get from such, if trying to ascertain what's really happening 'now' over 'there', is paradox. Which never occurs in nature actually. Differentials and integrals are related but different animals. It's the closed path integral that unambiguously matters re 'reality'. An exception is the scenario given back in #30. Which though still falls under the case of uniform relative speed.