More questions about relative speeds.

Something must be wrong with my reasoning in the following thought experiment. Can someone, please, point out my error.

This thought experiment involves an empty cosmos which contains three space craft. Two of these, piloted by those intrepid adventurers Alice and Bob, are twenty light seconds apart and maintaining that distance rigorously. Each has a very accurate clock, and these clocks are synchronised. (For simplicity we will say that each reads “00.00” at an appropriate point in the thought experiment) The third craft, piloted by the less well known, but no less adventurous Charlie, is in motion relative to the other two. Relativity tells us that we cannot know which craft is moving and which, if any, is stationary. All we can say is that Alice and Bob are stationary relative to each other and that Charlie is in motion relative to the other two, or they are in motion relative to Charlie.

The distance between Alice and Charlie is closing at “0.8c”. It so happens that as Alice and Charlie pass each other their clocks are synchronised. (Charlie’s clock also reads “00.00”). It seems reasonable, therefore, to assume that Charlie’s clock must also be synchronised with that of Bob, but, of course, synchronicity is also subject to the rules of relativity, and therefore, events that are perceived as synchronous in one frame of reference may not be perceived as synchronous in another.

First, let’s consider this scenario from the point of view of Alice. Her perception is that she is stationary and that Charlie is approaching, and then passing her at 0.8c. According to her clock it takes twenty-five seconds for Charlie to reach Bob. Alice’s and Bob’s clocks remain synchronised, so they both read “00.25”. As Charlie is the one we chose to regard as being in motion at 0.8c, his clock, when he passes Bob, will read “00.15”, because, in his frame of reference it will have taken him only fifteen seconds to travel from Alice to Bob. Thus, relative to Alice and Bob, Charlie is now ten seconds younger than he would have been had he not travelled at 0.8c, relative to Alice and Bob for fifteen seconds (in his frame of reference).

Look at the scenario again, but this time from a F of R in which Charlie perceives himself as being stationary. We saw that Alice and Bob were maintaining a distance of twenty light seconds between their craft. The relative speed at which the gap between Alice and Charlie is closing remains 0.8c. Because we are now regarding Charlie as being stationary, it seems logical, to conclude that Charlie will now measure the time from when Alice passes him, to when Bob passes him as being twenty-five seconds, while Alice and Bob will measure it as being only fifteen seconds, because they have been travelling at 0.8c. In this F of R Alice and Bob will end up being ten seconds younger, relative to Charlie. How can this be? After all, the whole point of looking at this relativistically is to establish that any motion between the various elements in the scenario must be regarded as relative motion. Surely, it should make no difference who is thought of as being in motion; the outcome should be the same; yet, clearly, it is not.

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 Mentor You are neglecting both the relativity of simultaneity and length contraction. I will explain in more detail later.
 Thanks daleSpam, I shall look forward to your explanation. I was sure there was a bloomer there somewhere, but I seem to have ended up going round in circles looking for it. Not an unusual situation for me! There is a second part to this as well, but hopefully your explanation will be able to lay that to rest before I have to admit to it.

Mentor

More questions about relative speeds.

 Quote by Endervhar Each has a very accurate clock, and these clocks are synchronised.
For clarity I will denote Alice's/Bob's frame as the unprimed frame (t,x,y,z) and Charlie's frame as the primed frame (t',x',y',z').

First, remember the relativity of simultaneity. Whenever you say "these clocks are synchronized" you must also specify in which frame they are synchronized. I assume that you meant that they were synchronized in the unprimed frame. This will become important soon.

 Quote by Endervhar First, let’s consider this scenario from the point of view of Alice. Her perception is that she is stationary and that Charlie is approaching, and then passing her at 0.8c. According to her clock it takes twenty-five seconds for Charlie to reach Bob. Alice’s and Bob’s clocks remain synchronised, so they both read “00.25”. As Charlie is the one we chose to regard as being in motion at 0.8c, his clock, when he passes Bob, will read “00.15”, because, in his frame of reference it will have taken him only fifteen seconds to travel from Alice to Bob. Thus, relative to Alice and Bob, Charlie is now ten seconds younger than he would have been had he not travelled at 0.8c, relative to Alice and Bob for fifteen seconds (in his frame of reference).
So far so good.

 Quote by Endervhar Look at the scenario again, but this time from a F of R in which Charlie perceives himself as being stationary. We saw that Alice and Bob were maintaining a distance of twenty light seconds between their craft.
Here you have forgotten length contraction. Although the distance is 20 ls in the unprimed frame, in the primed frame it is only 12 ls.

 Quote by Endervhar The relative speed at which the gap between Alice and Charlie is closing remains 0.8c.
Correct.

 Quote by Endervhar Because we are now regarding Charlie as being stationary, it seems logical, to conclude that Charlie will now measure the time from when Alice passes him, to when Bob passes him as being twenty-five seconds,
Since the distance is 12 ls and the speed is .8 c Charlie measures the time to be 15 s, which agrees with the analysis in the unprimed frame.

 Quote by Endervhar while Alice and Bob will measure it as being only fifteen seconds, because they have been travelling at 0.8c. In this F of R Alice and Bob will end up being ten seconds younger, relative to Charlie.
Here you have forgotten the relativity of simultaneity. At t'=0 both Charlie's and Alice's clocks read 0, but Bob's clock is not synchronized with Alice's in the primed frame and it already reads 16 s. Due to time dilation, during the 15 s that it takes for Bob to reach Charlie in the primed frame, Bob's clock only advances 9 s. So the time on Bob's clock when he reaches Charlie is 25 s, which also agrees with the analysis in the unprimed frame.

 Quote by Endervhar Surely, it should make no difference who is thought of as being in motion; the outcome should be the same
Correct. I can walk you through the math if you wish.

 Thanks again, DaleSpam; part of the downside of getting ancient is the inability to retain details, so I shall have to work through this with "pencil and paper" to make sure I have grasped it. Having done that, if I still feel brave enough, I will take you up on your offer of a walk through the maths, but only if you remember that I am not a mathematician. :)

 Quote by DaleSpam Here you have forgotten length contraction. Although the distance is 20 ls in the unprimed frame, in the primed frame it is only 12 ls.
.

Regarding Charlie as stationary. As Alice passes Charlie, does Charlie measure the distance to Bob as 20 ls?

If so, should Charlie measure the time taken by Bob to cover that distance as 20s?

As we are regarding Bob as moving, would it not be Bob who would measure the time as 15s, and see the distance contract to 12 ls?

If so, would this not place the length contraction in the unprimed frame?

Recognitions:
Science Advisor
 Quote by Endervhar Regarding Charlie as stationary. As Alice passes Charlie, does Charlie measure the distance to Bob as 20 ls?
If Alice and Bob are 20 ls apart in the frame where they are at rest, and in Charlie's rest frame they are moving at 0.8c, then in Charlie's rest frame the distance between Alice and Bob is a constant 12 ls, because the length contraction factor is $$\sqrt{1 - 0.8c^2/c^2}$$ = 0.6
 Quote by Endervhar As we are regarding Bob as moving, would it not be Bob who would measure the time as 15s
No, because in Charlie's rest frame Bob's clock is out-of-sync with Alice's...in general if two clocks are synchronized and a distance L apart in their own rest frame, then in another frame which sees them moving at speed v, at any given moment the two clocks will be out-of-sync by vL/c^2, with the clock in the rear being ahead of the front clock by this amount. So in this case, at the moment Alice passes Charlie and her clock reads 0 seconds, Bob's clock already reads (0.8c)(20 ls)/c^2 = 16 seconds at that moment (in Charlie's frame). At this moment Bob is 12 ls away and moving towards Charlie at 0.8c, so in Charlie's frame it takes Bob 12/0.8 = 15 seconds to reach Charlie's position. Because of time dilation, Bob's clock is running slow by a factor of 0.6 in Charlie's frame, so in these 15 seconds Bob's clock only ticks forward by 15*0.6 = 9 seconds, so by the time Bob reaches Charlie's position Bob's clock reads 16 + 9 = 25 seconds.
 Quote by Endervhar , and see the distance contract to 12 ls?
But didn't you say originally that the distance in Bob's rest frame was 20 ls? You have to pick a single well-defined scenario where there is a single well-defined distance in each frame...I think you may be confusing yourself with the whole "regarding Charlie as stationary" and "regarding Bob as moving" stuff, in relativity you don't pick a single frame to label as "stationary", each frame has its own definition of stationary and all frames are equally valid. In Charlie's rest frame Charlie is stationary, and in Bob's rest frame Bob is stationary, there's no extra step where we declare that one of them is "really" stationary.

And the way length contraction works is that if you have two objects at rest relative to each other (or two ends of a single object), then the distance between them is maximized in the frame where they are at rest, while in any other frame the distance between them is smaller than in their rest frame.

Mentor
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 Quote by Endervhar . Regarding Charlie as stationary. As Alice passes Charlie, does Charlie measure the distance to Bob as 20 ls?
No. The distance between Alice and Bob is 20 ls according to Alice and Bob. But according to Charlie that distance is contracted to only 12 ls.

 Thanks, folks; I shall have to sleep on that, I think it is making sense, but there is still something niggling away in the back of my mind that I have to identify and deal with before I con let it go. Over 35 years ago, when doing S101 with the Open University, I had trouble with relativity. Circumstances made me pack it up then; now I have come back to it at 70+, I am determined to crack it, so I am grateful for all the help I can get.

Mentor
 Quote by Endervhar If so, would this not place the length contraction in the unprimed frame?
JesseM and DocAl essentially covered what I would have said, but I wanted to elaborate a bit on this comment. The designation of a frame as primed or unprimed is not important. What is important is that the length is maximum in the frame where it is at rest, in any frame where the length is moving it is divided by the Lorentz factor. The only length of interest in this problem is the length between Alice and Bob so it is 20 ls in Alice's frame where it is not moving and 12 ls in Charlie's frame where it is moving.

Btw, in general I do not like the length contraction and time dilation formulas. I recommend against their use and recommend in favor of the use of the Lorentz transforms instead. They will automatically simplify when applicable, and you will avoid using the simplified formulas in situations where they are not applicable.

 DaleSpam, I followed your "time dilation" link, apart from Sqrt. 1- v^2/c^2, the maths lost me completely, which might give you some idea of what you would be up against trying to walk me through any maths. :)
 It looks as though, a large part of my mistake stemmed from the fact that when I switched from the Alice/Bob I F to Charlie's I F, I made unjustified changes in my perception of what each would see. I regarded Charlie as stationary, as though this meant he was stationary in some absolute sense. Thus I interpreted Bob's perception in his I F on the basis of an arbitrary selection of a condition in Charlie's I F. Would I be right in thinking that although Alice's and Charlie's clocks might be in sync for an infinitesimal period of time, Charlie's and Bob's clocks would not be in sync at that same point, even though, in their I F, Alice & Bob's clocks would remain synchronised?

Recognitions:
Science Advisor
 Quote by Endervhar Would I be right in thinking that although Alice's and Charlie's clocks might be in sync for an infinitesimal period of time, Charlie's and Bob's clocks would not be in sync at that same point, even though, in their I F, Alice & Bob's clocks would remain synchronised?
Simultaneity depends on the choice of reference frame, there is no absolute answer to this question either (this is the "relativity of simultaneity", see here and here for discussion). In the inertial rest frame of Alice and Bob, at the moment Charlie's clock reads 0, Alice's clock reads 0 and Bob's clock also reads 0 at the same moment too. But in Charlie's inertial rest frame, at the moment that Charlie and Alice's clocks read 0, Bob's clock already reads 16 seconds as I mentioned earlier.

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