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#55
Nov2112, 10:29 AM

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#56
Nov2112, 11:10 AM

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If you don't want to make the graphs correctly, I will. And I will also explain the scenario without the use of graphs. By the way, your graphs are not Minkowski diagrams, they are simply conventional position versus time graphs. And I'm not saying that simply because you are interchanging the time versus distance axes that is more common for a Minkowski diagram. 


#57
Nov2112, 11:34 AM

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#58
Nov2112, 01:18 PM

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Minkowski diagrams have at least two sets of axes to show how each event has two sets of coordinates for two different reference frames. All of the graphs that greswd presented have only one set of axes corresponding to the frame in which Charles remains at rest and in which Adam and Bob start out at rest and end up at rest.
People were drawing position versus time graphs long before Minkowski or Einstein or Lorentz or even Maxwell. I don't think Minkowski gets backwards credit for all those graphs just because they only have one set of axes. 


#59
Nov2112, 03:21 PM

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#60
Nov2112, 03:51 PM

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If someone draws a graph of position versus time, does that automatically make it a Minkowski diagram? Would you call greswd's graphs on this thread Minkowski diagram's? 


#61
Nov2112, 04:15 PM

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#62
Nov2112, 04:36 PM

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So then if Newton drew a distanceversustime diagram but he did not show that a traveler's clock was running slower than the coordinate time, then it would not be a Minkowski diagram, correct? Or more specifically, if the diagram shows somehow that the traveler's clock is indicating a slower time than the coordinate time, then that makes it Minkowski? In other words, it doesn't have to explicitly use a second set of axes to show the slower time, it can just do it as points spaced further apart than the coordinate spacing, correct?



#63
Nov2112, 09:17 PM

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Technically, all displacementtime graphs look the same with reference to one particular frame.
The distinguishing factor is the transformation from one frame to another. Now we've learnt about two transformations, Galilean and Minkowskian. Of course, we may come up with others, but they may not make physical sense. When transforming between inertial frames, all transformations have to use the worldline as the time axis, and ensure that relative velocity between both frames is the same. 


#64
Nov2312, 11:02 AM

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Minkowski diagrams replace the traditional time axis with ct (distance).
This results from transforming the equality for the invariant interval into a 4D expression, via t'=ict. The benefit is twofold. Unless one of the coordinates t or x is scaled, you could never graph it to scale! It reveals what's really being compared. The object speed vs light speed, i.e. vt/ct=v/c. It's the only variable in the gamma expression, which is the only factor distinguishing SR from prerelativity physics. 


#65
Dec412, 02:37 AM

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Similarly, I haven't shown dots along Charles's path (blue) to show where he receives signals from Adam (black): Finally, I suggested that you overlay these two graphs to get a correct third graph that shows everything, including Adam's perspective: Note that we can now see Adam's time dilation. Since his speed is 0.6c, gamma is 1.25 and his tick marks are spaced at 1.25 of the coordinate grid. This illustrates that his Proper Time is equal to gamma multiplied by the coordinate time. You can also see that he sends a signal to Charles at every tick but during the outbound portion of his trip, he receives the signals from Charles every other month corresponding to a redshift Doppler factor of 0.5. Charles is also sending out a signal every month to Adam but since he is stationary in this frame, his Proper Time is coincident with the coordinate time. Still, you can see that he receives signals from Adam every other month at the beginning. His redshift Doppler factor during this time is also 0.5. Charles continues to see Adam' clock running at 1/2 the rate of his own until he sees Adam turn around when Adam's clock reaches 12 months. This occurs when Charles's clock is at 24 months. From then on, he sees Adam's clock running at twice the rate of his own for a blueshift Doppler factor of 2 so that in 6 more months of his own time, he sees Adam's clock adanvce by 12 months. At the end, he has seen Adam's clock advance 12 months in slow motion and 12 months in fast motion for a total of 24 months. Meanwhile, Adam has been watching Charles's clock advance at 1/2 his own rate (redshift Doppler) so that after one year of his own time, Charles has advanced by 6 months. When he turns around, he sees Charles's clock advancing at double the rate of his own (blueshift Doppler) so that in the one year that it takes him to return, he sees Charles advance by 2 years for a total of 30 months. 


#66
Jan1013, 05:10 PM

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Sorry for taking such a long hiatus. I was very busy with other matters. Anyway, I understand the diagrams well.
With regards with my rejected 3rd diagram. I found this from UNSW that looks very similar. http://www.phys.unsw.edu.au/einstein...in_paradox.htm 


#67
Jan1013, 06:32 PM

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They did a better job on their first diagram compared to the one shown in post #34 from wikipedia but it's still not perfect. Here's what it should look like: They show Joe 7 after he receives the signal from Jane 3 but he should be slightly before. Note also how they show Jane 5 receiving the signal from Joe 6 at the same time but she should be slightly earlier. In fact these two situations should show the same relationship because it is a reciprocal Doppler shift. And their second diagram is wrong, not to mention ridiculous. They should show Joe 2 on the bottom diagonal and they should show Joe 6 on the top diagonal. Here is the diagram for the frame in which Jane is at rest during the outbound portion of her trip: And here is the diagram for the frame in which Jane is at rest during the inbound portion of her trip: What they were trying to do is combine the bottom part of the outbound portion of the trip with the top part of the inbound portion of the trip while showing in dotted lines the signals coming from Joe. They did a fairly good job of that but why don't they correctly show the signals going to Joe from Jane? They do show all five of Jane's signals but if they had shown Joe 6 at the correct location he would have received the signal from Jane 2 after the correct position for Joe 6 and it should be coming between Joe 4 and Joe 5 as they indicate in their first diagram. It is impossible to combine the two parts of the outbound and inbound portions of Jane's two rest frames into one like this. If you're going to do it correctly, you need a much more complicated diagram. You need to show the correct Doppler signals for both twins throughout the diagram, just like all three of my diagrams show. It's so easy to do in an Inertial Reference Frame, why do you feel the need to do it in a noninertial frame? 


#68
Jan1013, 08:05 PM

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Is this all true if it were two twins and an older unrelated person?



#71
Jan1113, 06:27 AM

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#72
Jan1113, 01:48 PM

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I said their first diagram is close to being right but their second diagram is wrong. If they had used their second diagram to show just the messages coming from Joe to Jane they would have made a diagram that was more like the one DrGreg made in post #39. Note that he is only showing the messages going from the inertial twin to the traveling twin. That part of the diagram, as I already stated, is fairly good. But the part that is completely wrong is where they also try to show the messages going from Jane to Joe. If you look at their first diagram, you can see that Joe receives these messages at two different rates. The first three messages take over two years each for him to receive, then in his last year he receives all the rest of them. They show this pretty close to being right and it's important that a diagram show that the inertial twin receives half of the messages at a slow rate and half of the messages at a fast rate and it's important to show that the time interval over which he receives those message is not evenly spaced. He spends way more of his time receiving the low rate messages and only a short time near the end receiving the high rate messages. It's also important that a diagram show that the traveling twin spends exactly half her time receiving the low rate messages during the outbound portion of her trip and the other half of her time receiving the high rate messages during the inbound portion of her trip. They do a good job of showing this aspect in both diagrams. However, if you look at their second diagram, you see that Joe does not receive any messages until half way through the diagram at which point he receives all the messages from Jane equally spaced in time. It's the correct spacing in time for the last messages but not for the first three. It's faster than it should be. This is wrong and it's a ridiculous concept to try to show on a combined diagram like this. In fact, I have no idea how to correctly show Joe receiving the messages from Jane at the correct rates and to show the transmission of the signals traveling at c between the two twins. I'm not saying it can't be done, just that I don't know how to do it. And again, I ask you, why do you feel compelled to combine portions of two perfectly good Inertial Reference Frames into one ridiculous monstrosity? Why not just show everything in each one of the Inertial Reference Frames like I did in post #67? 


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