"Final Solution" to "Ralf's Problem"

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

Postby BurtJordaan on January 20th, 2018, 1:44 am 

JMP1958 » 19 Jan 2018, 22:51 wrote:That being said, here are the possible outcomes according to Bob's starting rest frame.

Superbly, simply done and discussed, with no superfluous lines, no interpretations, just what I was looking for (for a reason). Thanks JMP.

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 ...

Since 3.2 years is not an event of interest, I would prefer to leave it out of the discussion. It is reciprocal time dilation and has no bearing on the results. I would have stuck to the 'real' events, when something is sent or arrives, locally. And they are all there, clear and readable from the the chart.

Just to avoid miscommunication, I think you have my scenarios 2 and 3 swapped, but that's irrelevant. I just want to write the results of each case here for reference.

1. Both tails. No difference in aging can be determined.

2. Only Alice heads. It is pretty clear from the dark blue and bright blue lines that Alice has aged one year less than Bob by the time they could have compared clocks at a distance and post-process

3. Only Bob heads: From JMP's second chart, it is clear from the vertical green and red lines that the roles are reversed.

4. Both heads. They have both aged 8 yrs when the red and blue lines cross at their second flyby.

While this is checked for errors by the community, I will start to write my interpretation of when what aging can be said (or not said) to have happened.
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Re: "Final Solution" to "Ralf's Problem"

Postby BurtJordaan on January 20th, 2018, 2:16 am 

mitchellmckain » 19 Jan 2018, 22:53 wrote:Perhaps this reveals our habit of thinking in such a manner when experience tells us we can get away with it.

Yup, since we know that in flat spacetime, spatially separated clocks that are stationary relative to each other, show no elapsed proper time differences - and we can synchronize them in their inertial frame. But we know they will not be synchronized in all frames. So their time differences are not frame independent.
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Re: "Final Solution" to "Ralf's Problem"

Postby mitchellmckain on January 20th, 2018, 2:37 am 

mitchellmckain » January 18th, 2018, 8:34 pm 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.


I was checking my calculation because JMP1958 got a different answer of 6 years. I have concluded that 6 years is correct. 2.4 ly is the distance traveled in the 4 years before accelerating to the other inertial frame, where it is the uncontracted distance of 3 ly, and that is where he is waiting for the message if the coin toss came up tails. Thus the correct calculation is 4+3-1 = 6 years.
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Re: "Final Solution" to "Ralf's Problem"

Postby BurtJordaan on January 20th, 2018, 3:38 am 

ralfcis » 20 Jan 2018, 05:44 wrote: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.)

Ralf, I think you should save your concentration for simply looking at JMP's two charts. Magnificently simple and accurate. No math needed. And then reconsider your interpretations. The answer lurks in there, but not everyone spots it quickly.
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Re: "Final Solution" to "Ralf's Problem"

Postby ralfcis on January 20th, 2018, 7:28 am 

Yes thanks to jmp for clarifying what you wanted. I thought you were looking foe a reverse perspective analysis of Alice stopping, then a reverse direction (turnaround) and possibly a clock handoff turnaround as a follow up. If I have mistakes they would appear in those scenarios. So I'll just do the same scenarios as jmp for now.
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Re: "Final Solution" to "Ralf's Problem"

Postby ralfcis on January 20th, 2018, 8:06 am 

But first I will continue where I left off, Bob's post processing re-evaluation of the per yr accumulated age difference between him and Alice after he receives her pink signal. Here's the STD again:

6cj6.jpg


Up until t'=4, There was no age difference but until t=8 Bob was still receiving Alice's doppler ratio as 1/2 indicating he was going at .6c relative velocity to her. He would chart his equivalent yrs between Alice's lines of present and conclude no age difference between the two up until t=8. But Alice's info would force Bob to re-calculate Alice's lines of present after t=3.2 (equivalent Bob yr 4). The wider gap between his blue lines now indicates his equivalent yr had changed to a Alice full yr (no time dilation in 0 relative velocity). So at t'=5, the first circle shows a .2yr Bob yr age difference between him and Alice. To undilate that into Alice yrs you multiply that by Y and get Alice has aged .25 yrs less that Bob up to this point.

Similarly, the next year's accumulated difference, represented by the next circle, is a total of .-4 Bob yrs or .-5 Alice yrs which is another .2 yr loss over the last yr. The same goes for the next yr, an accumulated total loss of .6 (.75 Alice yrs) or another .2yr loss per year. Same for the next yr with an accumulated loss of .8 (1 Alice yr) or another .2yr loss per yr.

So far we've done the loss calculation from Alice yrs 4-5, 5-6, 6-7, and 7-8. We've already arrived at the correct tally yet there's another circle on my diagram, what's that all about? It shows the total loss remains the same at .8yrs so the extra per yrs loss is 0 which means this yr does not affect the tally. It would have been another .2yr loss but the pink signal cuts that .2yrs off before it can happen. The last blue slanted line that intersects Alice's pink line is forced to slam down hard horizontal by the pink line telling it to.

So we have a total that matches relativity but we get the added bonus of seeing how that total accumulated during the 3 yrs Alice's signal propagated when there was a relative velocity mismatch. While relativistic math does not prevent the same calculations to be made, relativistic theory does and prohibits them as being indeterminate.

Next up, Bob stops wrt Alice continuing at .6c. And finally Bob stopped and Alice stopped which results in Bob approaching Alice at .6c.
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Re: "Final Solution" to "Ralf's Problem"

Postby BurtJordaan on January 20th, 2018, 10:07 am 

ralfcis » 20 Jan 2018, 14:06 wrote:So we have a total that matches relativity but we get the added bonus of seeing how that total accumulated during the 3 yrs Alice's signal propagated when there was a relative velocity mismatch. While relativistic math does not prevent the same calculations to be made, relativistic theory does and prohibits them as being indeterminate.

If you have followed advice and studied JMP's charts carefully, you may have seen the fallacy of your claim jump out at you. In the absence of gravity, clocks at rest relative to each other do not accumulate any proper time differences. This is also what the math tells us.

Whatever the perceived clock rate differences have been along various parts of the spacetime paths, the integrated extent of the propertime (aging) differences only becomes defined when the spatial path is closed - either physically, or by means of light signals.

To post-process and then interpret it as if it accrued during the period stationary in a shared inertial frame is just silly. It is much more credible to post-process and interpret it as having accrued during the period when they were not in the same inertial frame, but made 'concrete' the moment the frame change occurred.

Even this view runs you into trouble, because things are only 'concrete' when observed in a frame-independent way. This only happens when the twins are co-located twice over time. The "stationary relative to each other" scheme is a special case that only works for the twins themselves, not for all observers.

This is the physics - your interpretation of it is, as has been said by many, physically seriously flawed.
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Re: "Final Solution" to "Ralf's Problem"

Postby BurtJordaan on January 20th, 2018, 10:41 am 

With Ralf's problem sorted, here is a variant for the quantum physics people to chew on.

Instead of coins to be flipped, say Alice and Bob could each take one of an entangled particle pair with them. At 4 years (exactly) on their respective clocks, they observe their respective particle's quantum state. Of course the result of the observations would be as random as the two coins, but there are now just two possible combinations: +- or -+. 'Jump frame' when it's a +, cruise on if it's a -.

Presumably, each party would immediately know what the other parties outcome was - the opposite. Would that change the twins interpretation?
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Re: "Final Solution" to "Ralf's Problem"

Postby ralfcis on January 20th, 2018, 10:58 am 

You know what, I'll just continue this discussion in the Personal Theories section because I can see you're just itching to close this thread down. Thanks for the opportunity to show only half of the scenarios you asked for. They all calculate the right answers and if you had let me continue I would have shown that any speed change would generate a communicable age difference between 2 participants. Physics and aging are analog not digital (I'm paraphrasing what you said in your own book).

I don't understand your point about the age difference only being set in concrete for all frames only when they meet physically. The communication of age difference would be just as valid universally for all participants in 0 velocity frames given enough time for the signals to propagate between them. You've done this before, add another previously undisclosed caveat and then terminated the discussion. Maybe see you in the land of word salad posts? I'm out.

P.S. As I've said before, the formula for relativity of simultaneity is vx/c. Therefore, mathematically either v or x can be zero. If x is zero it means the participants meet at any velocity, if v is zero it means they can be separated by any distance to achieve the same result for establishing an age difference as being co-located.

P.P.S And when you say "as has been said by many, physically seriously flawed", that's seriously misleading. A guy named Mitch who I've offended grievously does not constitute "many". A more accurate statement would be my views are supported by no one here and the opinions as to why are open for discussion.
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Re: "Final Solution" to "Ralf's Problem"

Postby BurtJordaan on January 20th, 2018, 12:49 pm 

Re P.P.S: I recall members like Faradave, JMP, Lincoln, probably more, that have tried to correct you, but to no avail...

You have been thoroughly beaten by JMP's charts, so sadly, we do not need your Minkowski's any more. What we also don't need here is someone that doggedly pushes a false interpretation, instead of trying to the understand the physics.

You have asked only one pertinent physics question:
I don't understand your point about the age difference only being set in concrete for all frames only when they meet physically.

Maybe someone here will attempt to answer this to your satisfaction, but I have tried for nearly a decade now, and have apparently failed.
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