## "Final Solution" to "Ralf's Problem"

Discussions on classical and modern physics, quantum mechanics, particle physics, thermodynamics, general and special relativity, etc.

### "Final Solution" to "Ralf's Problem"

Let's do a variant of the 'twin-paradox' with some uncertainty sneaked in. Assume Alice and Bob flew past each other inertially at a relative speed of 0.6c, having set their clocks to zero at fly-by. Say Alice and Bob have agreed beforehand that each will toss a true coin when his/her clock reads 4 years exactly.

If the coin comes up tails, the tosser just flies on. If Alice gets heads, she will 'stop' relative to Bob. If Bob gets heads, he will set off after Alice at the same speed as her original fly-by speed (essentially, Bob 'stops' relative to Alice). Each sends a light signal to advise the other party of the respective toss result and action taken at the event.

One of four possible scenarios can play out:
1. Both tails: they just coast away from each other at 0.6c relative, forever...
2. Alice heads and Bob tails: we have the standard half-twin-scenario, as Ralf described, with both ending up at rest in Bob's inertial frame.
3. Bob heads and Alice tails: we have the reverse of 2, with both ending up at rest in Alice's inertial frame.
4. Both heads: they will approach each other at 0.6c and have a second fly-by, just like in the standard full-twin-scenario.

The reason for the variant is to make it clear that there is no preferred reference frame and both participants are treated equally from the start. There is no way to predict, at the beginning of the scenario, the final outcome of the age comparison .

Now, since this thread is mainly for Ralf's benefit, and since he is good with Minkowski diagrams, I want to request Ralf to prepare the 4 Minkowski's (or roll them into two, but no less, for clarity). We will probably all learn at least something from this scenario. You may be surprised how different some of them look, compared to the usual twin visualization.

BurtJordaan
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### Re: "Final Solution" to "Ralf's Problem"

I would object to the asymmetric language here.

BurtJordaan » January 17th, 2018, 11:50 pm wrote:If the coin comes up tails, the tosser just flies on. If Alice gets heads, she will 'stop' relative to Bob. If Bob gets heads, he will set off after Alice at the same speed as her original fly-by speed (essentially, Bob 'stops' relative to Alice). Each sends a light signal to advise the other party of the respective toss result and action taken at the event.

Instead I would suggest the following

If the coin comes up tails, the tosser just flies on. If coin comes up heads, the tosser will alter their velocity to match the velocity the other person had when they passed each other. Each sends a light signal to advise the other party of the respective toss result and action taken.

The point is that we have absolutely no warrant to consider one at rest and the other moving. The problem is entirely symmetrical in EVERY way.

mitchellmckain
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### Re: "Final Solution" to "Ralf's Problem"

Beautiful challenge, I accept it wholeheartedly. You might want to take a gander at my last post on ralfativity 2.0. I was able to work out the relativistic math to prove in one scenario so far, the way age difference accumulates while Bob is waiting for info from Alice. I'm going to use that math for your challenge.
ralfcis
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### Re: "Final Solution" to "Ralf's Problem"

mitchellmckain » 18 Jan 2018, 10:15 wrote: The problem is entirely symmetrical in EVERY way.

Yea, your scenario statement maybe a better way of wording it, but the two mean exactly the same thing, AFAICS. The problem, as intended, is however not quite symmetrical - each of the four possibilities have a different final outcome.

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### Re: "Final Solution" to "Ralf's Problem"

ralfcis » 18 Jan 2018, 14:32 wrote:You might want to take a gander at my last post on ralfativity 2.0. I was able to work out the relativistic math to prove in one scenario so far, the way age difference accumulates while Bob is waiting for info from Alice. I'm going to use that math for your challenge.

I did have a look and I can see that it has lead you to a fatally flawed conclusion. I'm respectfully requesting you to just give us standard Minkowski's, without fancy interpretations. Once we have all 4 scenarios in a clear and simple fashion, then let's discuss interpretations. This thread has been created to help you see and understand the flaws in your argument, but then you have to let it go along it's intended path.

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### Re: "Final Solution" to "Ralf's Problem"

Here are the 1st 2. I had to change it to when each hit 1 yr they'd flip the coin so the STD could fit on one page. The pink line is Alice's signal to Bob and the yellow is Bob's coin flip result to Alice. Bob's yrs are in blue on the t-axis and Bob's equivalent dilated yrs are in red on the t-axis. The red lines of present from Alice's perspective align with Bob's dilated yrs to show no age difference if Alice doesn't change velocity. When she does at t'=2, She will end up aging a half yr less than Bob when her 2nd pink signal reaches him. I have also included my method for determining how much she loses in the 1.5 yrs it takes for her signal to reach Bob. Bob spends 3 equivalent yrs thinking he's aging the same as Alice but when her signal reaches him, he finds she has lost .2 of his yrs (.25 of hers) the first equivalent yr, the same for the 2nd and 0 in the third which tally matches what relativity says Alice loses at the end of the spacetime path.

ralfcis
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### Re: "Final Solution" to "Ralf's Problem"

Things don't tie up between text and graphic. Did you mean you have halved the scale so that they both flip a coin at 2 years on their own clocks? Because that's when you show Alice to have "stopped" relative to Bob. Then the pink lines at 1 yr is superfluous? And you need a yellow signal at Bob 2 yrs?

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### Re: "Final Solution" to "Ralf's Problem"

They both flip when their clocks hit 1 and then they must transmit the result to each other. Alice gets the result from bob when she's 2. It's a trick I use to keep c constant for all frames without using length contraction for the distance Alice travels. I guess I forget how relativity would draw it. The results should be the same, Alice shoulld get the result of Bob's coin flip at t=1 for her t'=2 no?
ralfcis
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### Re: "Final Solution" to "Ralf's Problem"

Changing the problem without stating the new problem looks to me like throwing up a smokescreen of confusion. Why aren't you sticking to the original problem?

The velocity is .6c which gives a dilation factor of .8, and the coin flips are performed at 4 years proper time.

The good thing about the problem is that there are only two inertial frames involved. The difficult thing is that what happens depends on coin tosses and it is a problem we face right away.

spoiler (highlight/select to see)
---------------------------

In each person's own inertial frame the other person does their coin flip and sends their message after 5 years when they are 3ly away. Assuming their own coin toss came up heads, this is therefore one year after they have done their own coin flip and sent their own message. If their own coin toss came up tails, however, then after sending their message, they accelerate to the other person's inertial frame where the other person did their coin toss one year earlier rather one year later, and thus the other person's message is already enroute.

Thus if their coin toss is heads, then they will receive the message from the other person on year 8. If their coin toss is tails, then they will receive the message from the other person on year 4+2.4-1=5.4 according to their own clock.

---------------------------
This still leaves the diagrams up to ralfis.

mitchellmckain
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### Re: "Final Solution" to "Ralf's Problem"

"I had to change it to when each hit 1 yr they'd flip the coin so the STD could fit on one page."

Yup an absolutely huge smokescreen of confusion particularly if you don't read any of the text. The farther out you put the coin flip, the longer it takes for the result of that coin flip to reach the other party. What on earth could be so significant of yr 4 over yr 1? I think you're confusing the year Alice changes the velocity in previous examples with the year the coin is flipped.

Yes the 1st pink line is superfluous, I was going to put all 4 scenarios on 1 page but it got really busy. Bob still has to be informed not to take any action as a result of Alice's flip.
Last edited by ralfcis on January 18th, 2018, 10:32 pm, edited 1 time in total.
ralfcis
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### Re: "Final Solution" to "Ralf's Problem"

ralfcis » January 18th, 2018, 9:07 pm wrote:"I had to change it to when each hit 1 yr they'd flip the coin so the STD could fit on one page."

Yup an absolutely huge smokescreen of confusion particularly if you don't read any of the text. The farther out you put the coin flip, the longer it takes for the result of that coin flip to reach the other party. What on earth could be so significant of yr 4 over yr 1? I think you're confusing the year Alice changes the velocity with the year the coin is flipped.

It is just a matter of scale. Change the labeling on the graph and it is the same graph. So no, I don't see any good reason for changing the problem. But in the original problem they do not flip their coin and change their velocity at different times -- thus you demonstrate the confusion I was suspicious concerning.

mitchellmckain
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### Re: "Final Solution" to "Ralf's Problem"

I didn't read your interpretation of the problem because this just isn't your show. Jorrie, what did you intend, flip then action or flip and action simultaneous? I don't care which, I'm not part of Mitch's conspiracy suspicions.
ralfcis
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### Re: "Final Solution" to "Ralf's Problem"

Flip and act together, i.e. see them as one event for each person. Why would you wait a year to execute? I think your "not fitting on the page" is caused by the error of using flip and act as two events. {AFAICS, when the scenarios are drawn correctly, they fit on your standard page.}

I agree with Mitch that the original values would have been better, because it would have made comparisons to previous efforts easier. But the flips and acts must be at the same proper time for each of them.
Last edited by BurtJordaan on January 19th, 2018, 2:05 am, edited 2 times in total.
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### Re: "Final Solution" to "Ralf's Problem"

mitchellmckain » 19 Jan 2018, 03:34 wrote:In each person's own inertial frame the other person does their coin flip and sends their message after 5 years when they are 3ly away. Assuming their own coin toss came up heads, this is therefore one year after they have done their own coin flip and sent their own message. If their own coin toss came up tails, however, then after sending their message, they accelerate to the other person's inertial frame where the other person did their coin toss one year earlier rather one year later, and thus the other person's message is already enroute.

Thus if their coin toss is heads, then they will receive the message from the other person on year 8. If their coin toss is tails, then they will receive the message from the other person on year 4+2.4-1=5.4 according to their own clock.

Mitch, I think these calcs clutter the scenario a little. I would suggest that he just draw the diagrams. Ralf should first get scenarios 1 and 2 right, where only Alice can possibly get heads. It is essentially the same as this post, minus some unnecessary and cluttering lines (e.g. Alice never 'returns' in these scenarios). The only lines required are the world lines of each, marked with their proper time and the signals to communicate the toss. No multiple lines of simultaneity and no presumed 'age difference' ellipses, because they may just steer the 'tutorial' off course.

Scenarios 3 and 4 should then follow easily. Remember that the idea is to cure Ralf of his mistaken view that the "real aging difference" only happens over the latter part of the scenario.
Last edited by BurtJordaan on January 19th, 2018, 1:25 am, edited 1 time in total.
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### Re: "Final Solution" to "Ralf's Problem"

BurtJordaan » January 19th, 2018, 12:06 am wrote:Mitch, I think these calcs clutter the scenario a little.

It is true that these are answers to questions you did not ask. They were the only obvious thing to calculate. And they do demonstrate the symmetry between the two people. But of course, I left the diagrams to ralfis as you wanted.

BurtJordaan » January 19th, 2018, 12:06 am wrote:Remember that the idea is to cure Ralf of his mistaken view that the "real aging difference" only happens over the latter part of the scenario.

I understood the relevance immediately. After all, it is what I kept doing in these scenarios. Show him that if you change which inertial frame you end up in then time dilation applies to a different person.

mitchellmckain
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### Re: "Final Solution" to "Ralf's Problem"

No problem, I always appreciate your inputs.

I found the most difficult insight to convey to Ralf is that velocity time dilation is a relative thing and aging is an absolute thing. As you have noticed, his latest effort is trying to 'prove' that the difference in aging happens gradually, but only while Alice and Bob are static in the same inertial frame again (after only she has rolled heads). Patently wrong.

He has difficulty swallowing that only a spatially closed path can provide demonstrable proper aging differences. As you know, we make a concession for when the path is not closed, but the participants end up static relative to each other - and enough time is allowed for both-way confirmation to take place of the fact.
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### Re: "Final Solution" to "Ralf's Problem"

Why would i wait a year?

Look at scenario 4:

One of four possible scenarios can play out:
1. Both tails: they just coast away from each other at 0.6c relative, forever...
2. Alice heads and Bob tails: we have the standard half-twin-scenario, as Ralf described, with both ending up at rest in Bob's inertial frame.
3. Bob heads and Alice tails: we have the reverse of 2, with both ending up at rest in Alice's inertial frame.
4. Both heads: they will approach each other at 0.6c and have a second fly-by, just like in the standard full-twin-scenario.

If Alice gets heads you say she stops immediately without checking whether Bob also has heads which would mean she turns around which is a completely different action. Both of you don't see that? The same is true if Bob gets heads and doesn't wait to see what Alice flipped. She might have also flipped heads and is stopped instead of moving on while Bob is approaching her at .6c. Did you want that as a 5th scenario? What possible purpose does the coin flip have. Just say at Alice's year 4 she either stops or keeps going or returns to Bob. At Bob's year 4, he stops wrt Alice if she keeps going. Explore the 4 or 5 or 6 scenarios please. Yeah, the 6th one could be both heads and both Bob and Alice go to .33 c wrt earth which is .6c wrt each other so please stop with the condescending attitudes like I got your truth table wrong. And wait to draw your conclusions until after this exercise is done and you have some proof of what you're saying.
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### Re: "Final Solution" to "Ralf's Problem"

ralfcis » 19 Jan 2018, 12:57 wrote:Why would i wait a year?

That's what your Minkowsi said. Alice apparently tossed at 1 yr, but only stopped at 2 yrs. You have not really answered my query on that.

If Alice gets heads you say she stops immediately without checking whether Bob also has heads which would mean she turns around which is a completely different action. Both of you don't see that?

Please read the whole scenario description again. Nobody turns around anywhere. You apparently understand it incorrectly, not us.

I wrote: "If Alice gets heads, she will 'stop' relative to Bob. If Bob gets heads, he will set off after Alice at the same speed as her original fly-by speed (essentially, Bob 'stops' relative to Alice)". Mitch has put it even more clearly: "If coin comes up heads, the tosser will alter their velocity to match the velocity the other person had when they passed each other." Both mean the same thing - no turnarounds.

And yes, they do their delta-V's without any knowledge of the other parties toss. That's the whole idea behind the scenario.

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### Re: "Final Solution" to "Ralf's Problem"

Anyways as long as I'm waiting for clarification of what you want me to do, the question did arise of whether I am interpreting the light signals after the coin flip correctly. Now that is a real issue that made me kind of nervous. The light signal from Bob to Alice has to cover a greater distance from Bob's perspective. Yes I know relativity invokes length contraction from Alice's perspective so the light from Bob is traversing a contracted length.But just looking at the STD and having proper length markers everywhere in space, when Alice comes close to one of these markers, she is aware of proper distance from Bob. The exact moment between approach and separation to and from one of these markers is an instantaneous stop wrt one of these markers because if you're between approaching and leaving something you're instantaneous relative velocity is 0, no?

Further the STD shows Alice's line of present at t'= 1.25 intersects when Bob sends out his light signal at t=1. So presently from her perspective and reading her distance marker and her own clock, she has a 1.25 yr and a .9375 ly head start on that light signal. That signal will meet Alice when she travels another .75 ys her time after it has traveled 1.5yr proper distance which will be on the next proper distance marker she encounters. So the light in her time is basically moving 2c. Is this not an equivalent way to explaining the constancy of the speed of light for each frame without invoking length contraction? Ok, go ahead and shower me with abuse for even thinking such blasphemy but technically, isn't it true?
ralfcis
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### Re: "Final Solution" to "Ralf's Problem"

You wrote this:

"4. Both heads: they will approach each other at 0.6c and have a second fly-by, just like in the standard full-twin-scenario."

This doesn't mean turnaround? What does it mean then?

Alice stopped at 2 because that's when Bob's signal of his coin flip result from when he turns 1 reaches her so that she can take action based on his flip. This is really turning out to be no fun at all and we haven't even started yet.
ralfcis
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### Re: "Final Solution" to "Ralf's Problem"

No, you completely overthink the problem. It is really simple. Scenario 1 is trivial and 2 is just the stock standard Alice 'stopping' permanently relative to Bob, which you have already given in the other thread.

Try to get these right before we move on to 4 and 5.

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### Re: "Final Solution" to "Ralf's Problem"

If 'overthinking" means applying simple logic then guilty as charged. Why don't you just answer the question because what you said in 4 has implications for 1 and 2. Does 4 mean turnaround or not? Just answer a simple question for once.
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### Re: "Final Solution" to "Ralf's Problem"

No, Alice stops at 4 years her time. Bob accelerates at 4 years his time. They have (and need) no clue as to what the other one did by then. Take into account that there are other scenarios as well.

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### Re: "Final Solution" to "Ralf's Problem"

If you want to use less than 4 years, do so, but I can see no reason for it.

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### Re: "Final Solution" to "Ralf's Problem"

ralfcis » 19 Jan 2018, 14:19 wrote:Why don't you just answer the question because what you said in 4 has implications for 1 and 2.

Coin tosses are random, independent events. Scenario 4 has zero impact on any other scenario. But you have to read it with 1,2,3 in order to comprehend 4. Is that so unreasonable?

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### Re: "Final Solution" to "Ralf's Problem"

ralfcis » January 19th, 2018, 5:19 am wrote:If 'overthinking" means applying simple logic then guilty as charged. Why don't you just answer the question because what you said in 4 has implications for 1 and 2. Does 4 mean turnaround or not? Just answer a simple question for once.

In this case by "overthinking" he means that you are adding a condition that is not in the original scenario: Namely, that either Alice's or Bob's actions are dependent on what the other has done.

Four years by their own clock, Bob and Alice each flip a coin. Depending on the result, they do 1 of 2 things: 1. Nothing
2. Alter their velocity so that it becomes zero with respect to the velocity that the other had at the onset of the scenario (In other words Bob does not take into consideration any velocity change made by Alice due to her coin flip, nor does Alice do so for Bob's coin flips. The coin flips of either party has no effect on the other)

That being said, here are the possible outcomes according to Bob's starting rest frame.

The dark blue line is Bob's starting world-line. Upon reaching 4 years, he either continues along the Dark blue line or changes velocity to follow the red line.
Alice starts along the green line, and when her clock reads 4 years, she either continues along the green or alters velocity to follow the light blue line.
Each of them can send a signal to the other at the Four year mark or their journeys( the yellow lines).

Thus we have four possible outcomes depending on the coin flips.
1. Both Alice and Bob do nothing, in which case they continue to separate.
The light signal sent by Bob when his clock reads 4 years and Alice's clock read 3.2 years will reach Alice when Bob's clock reads 10 yr and Alice's clock reads 8 yr.
The light signal sent by Alice is sent when Bob's clock reads 5 yr and Alice's reads 4 yr, and arrives when Bob's clock reads 8 yr and Alice's reads 6.4 yrs.

2. Bob changes velocity and Alice does nothing. Bob has matched velocities with Alice and they maintain a constant distance.
The light signal from Bob still leaves when his clock reads 4 yr and Alice's reads 3.2 yr, and it still arrives when Alice's clock reads 8 yr, but now Bob's clock read ~8.8 yr when it reaches Alice.
The light signal from Alice leaves when Bob's clock reads 4.8 yrs and Alice's reads 4 yr, and Arrives when Bob's clock reads 6 years and Alice's reads 5.2 years.

3. Bob does nothing and Alice changes velocity. Once again maintain a constant distance afterward.
The light signal from Bob leaves when his clock reads 4 yr and Alice's clock reads 3.2 yr and it arrives at Alice when his clock read 7 yrs and Alice's reads 6yrs.
The light signal from Alice leaves when Bob's clock reads 5 yrs and hers reads 4 yrs, and arrives when His reads 8 years and hers reads 7 yrs.

4. Both Bob and Alice change velocities and they end up passing each other again.
The light from Bob leaves when his clock reads 4 yr and Her clock read 3.2 yr and arrives when his clock reads 6 yr and his reads 7 yr.
The light from Alice leaves when his clock reads 5 and hers reads 4 and arrive when his clock reads 6 yrs and hers reads ~5.5 years.
In addition, Bob and Alice will meet when both of their clocks read 8 yrs.

From Alice's original rest frame, the ST diagram looks like this:

The events for this rest frame can be described by taking above description from Bob's Frame and replacing " Bob" with "Alice" and "Alice" with "Bob".

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### Re: "Final Solution" to "Ralf's Problem"

BurtJordaan » January 19th, 2018, 3:08 am wrote:He has difficulty swallowing that only a spatially closed path can provide demonstrable proper aging differences. As you know, we make a concession for when the path is not closed, but the participants end up static relative to each other - and enough time is allowed for both-way confirmation to take place of the fact.

We also tend to treat things as an instantaneous present when they end up in the same inertial frame. You are correct that this isn't entirely warranted and a closed path is preferable. Perhaps this reveals our habit of thinking in such a manner when experience tells us we can get away with it.

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### Re: "Final Solution" to "Ralf's Problem"

Can we just cut to the chase here. What subset of all the possible permutations of coin flips would you like me to draw STD's for that will prove or disprove your eventual point that

"his latest effort is trying to 'prove' that the difference in aging happens gradually, but only while Alice and Bob are static in the same inertial frame again (after only she has rolled heads). Patently wrong."

Every time I'm caught doing physics instead of looking for a job I get a frying pan to the head. I know in relativity that doing the reverse scenario of Alice stopping where Bob stops will yield a different per year aging progression which is why relativity banned looking at that and only considers the end of the spacetime path as establishing a valid age difference. I don't know how many times I've stated this and only now are you taking notice. I'm saying my method should show no difference in per year age difference progression depending on who stops. So to save me some time can we just agree to look at that because I'm still totally confused as to what is the point of the coin flip especially after jmp's post. I can also do the full twin paradox with either turning around. Can we just do that huh? Isn't that the trap you were trying to lay for me? I can do whatever STD you want, just don't make me reach into your brain and guess what STD's you want to see.
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### Re: "Final Solution" to "Ralf's Problem"

Ok, let's start again with the simplest scenario just plain old .6c relative velocity between Bob and Alice. If you like symmetry, I could draw it as both Bob and Alice leaving the earth at .33c but I won't. I'll stick with relativity's convention of one being stationary and the other moving and when they reverse roles. However I have seen some texts with the earth as a background stationary frame and either Bob or Alice is attached to the stationary frame. This is wrong. When Bob is stationary with the earth, the reverse analysis should have both Bob and the earth moving away from a new stationary frame Alice creates in empty space. Here is the STD drawn in relativity style:

Bob's lines of present are blue horizontal and Alice's are slanted red. They are just indicators of reciprocal time dilation, depending on whose perspective you adopt. If you drew a .33c observer's line, he would see both Alice's and Bob's lines of present overlap and Bob and Alice's years at both ends of these lines would be the same number from the .33c perspective. I can't see how these lines have anything to do with age.When Bob is 4, Alice is not 3.2, she's only that from Bob's perspective using his line of present.

To avoid confusion when I just want to determine age difference, I add another set of labels on the t-axis of Bob's equivalent years to Alice. This sets a baseline for me to tell if Bob and Alice are no longer aging at the same rate. So long as they are at the same relative velocity, they are. This is not anti-relativity, it is not a convention, it's just a method to allow me to keep track of no age difference and age difference when it happens. Here is that STD:

Next we'll study the scenario of when Alice stops followed by when Bob stops.
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### Re: "Final Solution" to "Ralf's Problem"

Here is relativity's STD of Alice stopping at t'=4:

Before t'=4 there could have been no age difference between Bob and Alice because there was no difference in their relative velocity. However, those who misinterpret time dilation as age difference will say that Alice was either potentially aging 1/4 yr per yr less than Bob and Bob was also potentially aging 1/4 yr per yr less than Alice. No way for relativity to determine which is true so far as permanent age difference goes.

Bob during the transition delay (the 3 yrs it takes the pink line from Alice saying she has changed her relative velocity wrt Bob to 0c) still reads Alice's doppler ratio from Alice as 1/2 so he still thinks his relative velocity is .6c unchanged. So he still thinks his equivalent yrs are matching Alice's year for yr and no age difference is occurring.
Alice, on the other hand can see Bob's doppler ratio =1 immediately at the start of her transition. Bob will be able to see the same doppler ratio 3 yrs later when they will both agree that their relative velocity is now changed to 0. From then on they will have no further accumulated age difference because they both agree they have the same relative velocity.

But during the transition, relativity cannot make the call on how the age difference accumulated over those 3 yrs. It can only make the call when the pink signal reaches Bob and what was once a reading of reciprocal time dilation becomes a reading of age difference. Bob has aged 8 yrs, Alice has aged 7. My method agrees on the final answer but can also peer into those 3 yrs of indeterminacy to come up with a per yr age difference accumulation between the two. Jorrie believes that my method will be proven a fraud once I do the reverse analysis where Bob and the earth are considered moving and Alice is stationary. We'll see maybe tomorrow as I first have to show what yr by yr results my method calculates.

Here's a hint:

The circles represent accumulating yr by yr age difference but it takes a great deal of concentration on my part to keep relating the calculations to reality so I'll do that tomorrow. (I tend to make a lot of arithmetic mistakes.)
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