# Graphical Derivation of the CADO Equation

Discussion in 'Physics & Math' started by Mike_Fontenot, Sep 5, 2018.

1. ### NotEinsteinValued Senior Member

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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.

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

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!

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".

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

Yes, obviously. But you've dodged the whole point: is Charlie now justified in saying Alice travelled back in time? My answer: no.

No, I don't find ignorance offensive.

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?

Same here.

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.

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

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?

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?

In science, it's actually quite important to use the right words.

Right, and neither of those show Alice travelling backwards in time at any moment.

Correct.

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?

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?

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.

Nothing more that I've already explained to you multiple times. It's just that you often ignoring half of it.

3. ### NotEinsteinValued Senior Member

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I probably should've picked t=10 to avoid confusion. But I'm glad you agree I was correct in my original statement.

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?

So no Alice travelling backwards in time.

And I never claimed otherwise.

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

True. In fact, nobody has. Even Charlie's Minkowski diagram clearly shows this.

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

5. ### Neddy BateValued Senior Member

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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 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.

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.

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.

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.

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.

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

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

You should tell me.

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.

Last edited: Oct 26, 2018

7. ### Neddy BateValued Senior Member

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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.

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

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.

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.

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.

Last edited: Oct 26, 2018
8. ### Neddy BateValued Senior Member

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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.

9. ### NotEinsteinValued Senior Member

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1,986
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.

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

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).

Yep, length contraction and time dilation are funny things, but they are measurably true. Clocks ticking backwards though...

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

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?

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"!

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.

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

Good. Seems I was right all along.

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

Nope, that's just you imagining that.

Same here, and I posted that in this thread many, many pages ago.

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?

They really don't, though.

Please stop feigning ignorance; you've used my "overwrites"-solution in your very post.

I'm sorry, I don't know what you think. I'm not psychic.

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.

Yep, obviously.

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.

10. ### NotEinsteinValued Senior Member

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So travelling back in time is possible, according to you? Please go write that paper, and collect your Nobel Prize!

You tell me; I've been asking you that question for pages now. (You already have my answer.)

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.

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.

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.

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...?

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.)

11. ### Mike_FontenotRegistered Senior Member

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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.

12. ### Mike_FontenotRegistered Senior Member

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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.

Neddy Bate likes this.
13. ### NotEinsteinValued Senior Member

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It's probably better to discuss that with Neddy Bate, since he's the one that brought it up.

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?

14. ### Mike_FontenotRegistered Senior Member

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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."

15. ### NotEinsteinValued Senior Member

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Yes, all three of us are in agreement here.

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?

16. ### Mike_FontenotRegistered Senior Member

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

17. ### NotEinsteinValued Senior Member

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So you are unwilling to answer the issue raised?

18. ### Neddy BateValued Senior Member

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

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.

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.

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.

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.

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

Again, this is old news.

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

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.

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.

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.

It is Charlie's simultaneity changing.

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

Charlie would say it does, not Alice.

19. ### Neddy BateValued Senior Member

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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.

Last edited: Nov 2, 2018
20. ### Neddy BateValued Senior Member

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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.

Last edited: Nov 2, 2018
21. ### Q-reeusBannedValued Senior Member

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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'.

22. ### NotEinsteinValued Senior Member

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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?

And Mike_Fontenot appears to hold this position too:

23. ### Q-reeusBannedValued Senior Member

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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.