Discuss events which are simultaneous in one frame?

In summary, the idea of simultaneity is often discussed in relation to the theory of relativity, specifically in the context of how events can be perceived differently by observers in different frames of reference. There are two main types of simultaneity - reception simultaneity, where two events are perceived as happening at the same time by an observer, and transmission simultaneity, where the events were actually released at the same time but may not be perceived as simultaneous due to the time it takes for light to travel. The concept of simultaneity is important in understanding how time is perceived in different frames of reference, and it is often discussed in relation to the Lorentz invariance of the fundamental laws of physics.
  • #36
JesseM said:
Can you specify what you're talking about here?

I knew about the FTL limitations and the inability to violate causality (outside of the theoretical tachyon). I just didn't want someone to assume that I thought that causality was being violated and that I was therefore trying to invalidate relativity.

I am also aware that you can't see causally related events simultaneously without something going FTL.

Writing something that could be interpreted as implying that I wasn't aware of this FTL issue did get you to respond though. I note that you tend to respond very quickly when you can easily find an error. But not when it seems you can't, such as in the post before the one you are responding to here (eg #33). I am trying my best not to impugn motives, but it would help me to resist the temptation if you could respond to that one now.

In addition, while we both accept that you can't see causally related events simultaneoulsy, can we both accept that we can observe causally related events from an inertial frame such that the events are skewed (ie the events may not be simultaneous in the frame in which they occur, but the relationship between them is skewed, similarly to the relationship between the two clocks on the table are skewed)?

Can you then see if you can understand what I said here?

neopolitan said:
In the frame not at rest relative to the table, there is an event "now" in which a past event and future event are observed simultaneously. To me that means the past event is brought forward to the future (now is in the future relative to the past) and the _future_ event is brought back to the past (now is in the past relative to the future).

Note that I not claiming the ability to see the cue ball both unwhacked and in the pocket, just what we have talked about before - the clock at one end of the table saying 2s and the clock at the other end saying 10s. These clocks are not causally related.

cheers,

neopolitan
 
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  • #37
neopolitan said:
I am also aware that you can't see causally related events simultaneously without something going FTL.
Sorry if I'm telling you something you already know, but you said "What apparently can happen is that from well selected frame, you can see the cue ball being whacked and the cue ball sitting in the pocket simultaneously"--are the ball being whacked and the ball ending up in the pocket not causally related? Did you mean to write "what apparently can't happen"?
neopolitan said:
Writing something that could be interpreted as implying that I wasn't aware of this FTL issue did get you to respond though. I note that you tend to respond very quickly when you can easily find an error. But not when it seems you can't, such as in the post before the one you are responding to here (eg #33). I am trying my best not to impugn motives, but it would help me to resist the temptation if you could respond to that one now.
Please, please, please, stop with the paranoid fantasies about my secret motives. Making rather unsubtle hints about my motives and then saying "I am trying my best not to impugn motives" is not somehow more acceptable to me, if anything it is even more irritating. And the posts I respond quickly to are just the ones I can immediately think of an obvious response to--could be because there's an error, could be just to answer a question that has a simple answer, could just be to confirm that what someone is thinking is correct. I tend to respond slower to posts where I'm not sure what the person is trying to say and I don't know what I can say or ask to bring things into focus (as has been true with many of your posts on this thread). Sometimes I respond slower when I know basically what I'd like to say but I know the response will be rather involved, so it's something I put off. Sometimes it's just random laziness or getting distracted by something else in my life. But it's not like I owe you my time here--and I've spent quite a lot of accumulated time so far responding to your posts and emails--and this demanding, impatient attitude (it's only been a day since you posted the post you're now complaining I haven't gotten to yet!) is really not the best way to convince me to continue.
neopolitan said:
In addition, while we both accept that you can't see causally related events simultaneoulsy, can we both accept that we can observe causally related events from an inertial frame such that the events are skewed (ie the events may not be simultaneous in the frame in which they occur, but the relationship between them is skewed, similarly to the relationship between the two clocks on the table are skewed)?
I don't know what you mean by "the relationship between them is skewed". Every frame agrees on which of three causally related events happens first, middle, and last, the only difference is in the amount of time between them (but that's basically just time dilation).
neopolitan said:
Can you then see if you can understand what I said here?
In the frame not at rest relative to the table, there is an event "now" in which a past event and future event are observed simultaneously. To me that means the past event is brought forward to the future (now is in the future relative to the past) and the _future_ event is brought back to the past (now is in the past relative to the future).
I can't really see what you mean here either, the language is again pretty vague. 'There is an "event" now (what event? now in what frame?) in which a past event (what past event? past of what, and in what frame?) and future event (what future event? future of what, and in what frame?) are observed simultaneously (simultaneously in what frame?)' Can you give some kind of specific example--preferably a numerical example--and be consistently clear about what frame every statement you're making is supposed to refer to?
neopolitan said:
Note that I not claiming the ability to see the cue ball both unwhacked and in the pocket, just what we have talked about before - the clock at one end of the table saying 2s and the clock at the other end saying 10s. These clocks are not causally related.
If you're talking only about events which are not causally related, what point were you making about the cue ball?
 
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  • #38
neopolitan said:
If I have it right, DaleSpam is saying is that we cannot skew spacetime enough to make three related events simultaneous in any frame:

event one = where and when the cue ball is just as I whack it
event two = where and when the cue ball is between event one and event three
event three = where and when the cue ball is just after it falls into the pocket.

Relative to event three, events one and two are in the past. Relative to event one, events two and three are in the future. Relative to event two, event one is in the past and event three is in the future - irrespective of which frame you observe it from. So, according to DaleSpam, you can't do is choose a frame such that me whacking the cue ball comes before the cue ball sitting the pocket.

Is this correct?
Correct, that is exactly what I was saying.

In SR this is called "timelike". When two events have timelike separation one will be inside the future light cone of the other. The inside of the future light cone contains only future events (but not all future events). Because the light cone is preserved in all frames, it should come as no surprise that events with timelike separation in one frame will have timelike separation in all frames and therefore the temporal ordering of timelike separated events is preserved in all frames.

neopolitan said:
If we place two synchronised clocks on the table, one next to the start position of the cue ball and the other next to the pocket and then observe from another frame in which the table is not at rest, then this is equivalent to our rocket scenario.
No, it is not equivalent to the rocket scenario.

In the rocket scenario the event that the nose clock reads 12:00 is not in the future nor past light cones of the event that the tail clock reads 12:00. In SR this is called "spacelike". Again, because the light cone is preserved in all frames it should come as no surprise that events with spacelike separation in one frame will have spacelike separation in all frames. However, what is surprising is that due to the relativity of simultaneity the temporal ordering of spacelike separated events is not preserved in all frames. This can occur because outside of the light cones are future, past, and simultaneous events .
 
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  • #39
JesseM and neopolitan,

Here is a pair of spacetime diagrams depicting the situation in various frames. The primed frame is the rest frame of the rocket and the unprimed frame is a frame where the rocket is moving at .6c (aka "observer frame"). In each frame the rocket's position at a specific time is shown with the solid outline. The dashed outline is the transform of the other frame's solid outline.

The argument here boils down to one of you looking at one frame and saying "the nose is further up the diagram" and the other looking at the other frame and saying "the nose is further down the diagram". Otherwise the argument is purely semantics.
 

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  • #40
neopolitan said:
I was thinking of something a little different, that option two would involve looking at the readings of the clocks and comparing them simply (so the order they appear in the B frame). That would be a blend of three frames and it didn't make sense to me. What you are saying now makes more sense to me, I think.

While both A and B will read the clocks so that the nose reads less than the tail, if both read the clocks simultaneously in their own frames, you are saying that it can be so that the tail reading as observed by B (10s in your example) will be observed by A later than A observes the nose reading that B reads (2s in your example) - if the readings are simultaneous to B. Simultaneous readings taken in the A frame could be something like 6s on the tail and 2s on the nose. Is that what you mean?
Did you mean to write "C" rather than "B" here? B was the rocket's frame, in that frame the nose doesn't read less than the tail, assuming the two clocks are synchronized in their own rest frame. But in the C frame, yes, the tail clock could read 10 s simultaneously with the nose clock reading 2 seconds, so if we go back to the observer's frame A where the tail clock reads 8 s simultaneously with the nose clock reading 2 seconds (using the numbers from my example near the end of post #28), then the tail reading 10 s is still in the future.
neopolitan said:
JesseM said:
Again, the only sensible way I can interpret your claim about the nose clock being further in the future is to follow a procedure just like this, except in place of frame C, use the rest frame of the rocket which I called frame B, pick clock readings which are simultaneous in B, and see which happens further in the future in the observer's frame A (in this case it will be the reading on the nose clock). I'm just saying there's nothing special about frame B, you could equally well use C and conclude that the tail clock is the one that's further in the future.
Here is, possibly, the crux of our misunderstanding.

I see there being something special about the frame B in that both the items which are being observed share that frame. While the actual synchronisation is not overly important as you pointed out, the fact that the clocks are in phase and at rest relative to each other is important (in phase time-wise, not necessarily timekeeping-wise, since clocks can run slow for mechanical reasons). This, I think, make a simultaneous reading in this frame different to a simultaneous reading made in another frame.
OK, but "special" and "different" in what sense? Also, one of the reasons I brought up frame C was because it sounded a little like you were saying the whole thing about the nose being further in the future for the outside observer only required us to think about the outside observer's frame A, not to even bother with thinking about how things work in other frames, even the rocket's rest frame. I was just making the point that your argument does require us to consider another frame, although I agree that it seems more "elegant" or more intuitive to use the rocket's rest frame as the other one rather than another frame like C. But saying it's more elegant/intuitive doesn't mean it's somehow fundamentally more correct in terms of actual physics, and I wasn't sure if you were suggesting that the statement "the nose is further in the future" was supposed to be something more fundamental than just a comparison of simultaneity in two different reference frames...I was thinking specifically of your comment in post #25:
Just try to apply the same logic to the rocket and the two clocks. Relative to an observer not at rest relative to the rocket, the clock on the nose travels into the observer's future faster than the clock on the tail. The clock on the tail travels into the observer's future faster than the observer.

The observer also moves into the clocks' future faster than the clocks do.

This is where it gets less like semantics and more like something interesting ... can you model that? Not just wave it away, not just say "that's just relativity", not just show the mathematics on what must happen, but describe a model in which that is possible.
Your comments about it being "less like semantics and more like something interesting", and more than "that's just relativity", made it sound like you thought there was something more to this than just a comparison of simultaneity in frame A and frame B, something more like new physics of some kind. So that's one of the reasons I brought up the frame C, to show that which clock is "more in the future" for the observer A really does crucially depend on your choice of the second frame to use, even if frame B may be the most natural choice in this situation...but we are in no way forced to use frame B by nature here.

Also, in some later posts you seemed to deny that your comments about the nose being further in the future for A depended on using a second frame at all...like your comments in post #27 about not being interested in the perspective of a "dorky physics guy" on the rocket, only in the perspective of the outside observer, and then especially your comment in post #29 where you said:
I only ever talked about one observer. I never invited a second one (on the rocket) and certainly not a third (alternatively one relative to which the rocket is moving backwards or one relative to which the rocket is moving forwards but twice as fast as the first observer perceives).
So, are you now in agreement that your comment about the nose clock being further in the future for the outside observer A depends crucially on looking at the rocket frame B and picking events which are simultaneous in that frame, then seeing the order of the same events in frame A?

If we can agree on that, then I guess we really need to go back to the question of whether you're just trying to talk about how to conceptualize the different definitions of simultaneity in the rocket's frame vs. the outside observer's frame, or if your comment that this is "less like semantics and more like something interesting", and your question about whether we can "model" that, suggests you're talking about something more, perhaps a new physical model. If the former, then sure, I think this is an OK way to conceptualize the relation between the two frames' definitions of simultaneity, but if the latter, then I'm still having trouble understand what you're trying to get at, and I'm not sure what I can ask you to get you to state it in a way that makes more sense to me. If this is a case, maybe a start would be for you to try to address my questions from post #28:
This is where it gets less like semantics and more like something interesting ... can you model that? Not just wave it away, not just say "that's just relativity", not just show the mathematics on what must happen, but describe a model in which that is possible.

This also may be the point at which I get stomped on, so if you feel like coming back with "can you?" then I will have to politely decline.
What is the difference between a "model" and just showing the "mathematics on what must happen" according to relativity? In physics when I hear the word "model" I just interpret it to mean a mathematical model, do you mean something different? And when you say "describe a model in which that is possible", what did you mean by "that" if you weren't referring back to your earlier picture involving one guy moving into the future faster than the other? Describe a model in which what is possible?
neopolitan said:
Any other (non-rest) frame will observe a skewing of spacetime in the B frame where the clocks are at rest - which makes a difference. Doesn't it?
It makes a difference in being harder to conceptualize intuitively, but if you're talking about some other kind of "difference", then what kind of difference do you mean?
neopolitan said:
What I think you are effectively doing by introducing a third observer is comparing the extent of skewing, which is valid enough on its own terms, but not really part of what I was getting at. Still I think we agree on what happens with third observers, can we go back to only one observer (flesh and blood) and two clocks on rocket with forward motion relative to the observer (ie nose first)?
Again, when you say you want only one observer, it worries me a little because of my concern that you might be suggesting that your comments about the nose clock being further in the future do not require us to think about two different frames, both the outside observer's frame and the rocket's frame, but can somehow be understood purely in terms of the observer's frame. If you agree we need to think about two different frames to make sense of your comments, then I don't see why you don't want to consider two different observers, since as I was saying in post #30, talking about particular observers is usually just understood as a kind of shorthand for talking about what's going on in particular frames.
 
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  • #41
DaleSpam said:
The argument here boils down to one of you looking at one frame and saying "the nose is further up the diagram" and the other looking at the other frame and saying "the nose is further down the diagram". Otherwise the argument is purely semantics.
I appreciate you drawing the diagrams, and the argument was about that at one point, but I think it's moved on a little. In post #26 I think I figured out what neopolitan meant about the nose being further in the future (as illustrated in your second diagram where at a single instant in the rocket frame, the nose is crossing a later line of simultaneity from the outside observer's frame than the line of simultaneity the tail is crossing), but I have some questions about some of neopolitan's other comments as explained in the post immediately above, and that's what the more recent discussion has been oriented around.
 
  • #42
JesseM said:
Did you mean to write "C" rather than "B" here? B was the rocket's frame, in that frame the nose doesn't read less than the tail, assuming the two clocks are synchronized in their own rest frame. But in the C frame, yes, the tail clock could read 10 s simultaneously with the nose clock reading 2 seconds, so if we go back to the observer's frame A where the tail clock reads 8 s simultaneously with the nose clock reading 2 seconds (using the numbers from my example near the end of post #28), then the tail reading 10 s is still in the future.

Looking over it again, yes, I meant C rather than B. It think it is reasonably clear that I was referring to another frame which was not at rest relative to the rocket. I am not 100% sure that we are there yet with what I meant about "the nose being more in the future as compared to the tail" - noting that I was only referring to a frame where two synchonised clocks were at rest and one observer who was not at rest relative to the clocks such that one "nose" clock is ahead of the other "tail" clock in terms of their relative motion according to the observer. In that limited scenario, do you agree?

To try to clarify again, in a now moment in the observer's frame (all now moments are relative, since "now" changes all the time), the observer may observe the tail clock reading 10s and the nose clock reading 2s. IF the clocks are synchonised relative to their rest frame - noting that the observer can work this out from the relative velocity of the clocks and their apparent separation from each other - THEN the observer can further deduce that the nose clock he sees "now" is a younger version of the nose clock and an older version of the tail clock (the observed nose clock manifests earlier in the clocks's rest frame than the observed tail clock - in our example 8s earlier). The nose clock, if you like, has reached the observer's "now" before the tail clock has.

I am sorry to have to do this, but I hope I can justify it. Let's introduce a third clock - on the rocket, in the midpoint between the nose and the tail. That clock will read a midpoint value. Without thinking too deeply about the specifics, I suspect it is 6s (midway between 10s and 2s) but the acutal reading is immaterial - what is important is that it is more than 2s and less than 10s.

If the observer not at rest relative to the clocks observes a reading of 6s on the midpoint clock, 2s on the nose clock and 10s on the tail clock - and knows from his deductions that in their own rest frame the clocks are synchonised then he can say, taking the midpoint clock as his reference, that the nose clock he "should" (see since the clocks are synchronised) is in the future and the tail clock he "should" see is in the past. Whose past and whose future? the past and future of the observer.

What that observer sees, as you point out (I think), is a past version of the nose clock, relative to the midpoint clock, and a future version of the tail clock, relative to the midpoint clock.

You don't really need the third clock, since the same logic applies with only two points, but hopefully the temporary introduction of a third clock makes it easier to understand.

DaleSpam is most probably right, we are probably arguing over semantics.

More later, I must attend a meeting.

cheers,

neopolitan
 
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  • #43
JesseM said:
<snip>
Also, in some later posts you seemed to deny that your comments about the nose being further in the future for A depended on using a second frame at all...like your comments in post #27 about not being interested in the perspective of a "dorky physics guy" on the rocket, only in the perspective of the outside observer,
<snip>

Fortunately I like my sense of humor, so to have an example of it reflected back at me gave me a good laugh.

Then I read a bit closer.

Please please please reread the second paragraph of post #27.

There was only one observer and that observer was not on the rocket, not at rest relative to the rocket or the clocks. The "dorky physics guy" was the observer, the only observer, the observer who I specifically stated was not on the rocket. There was no other observer on the rocket who could observe the "dorky physics guy", which was the whole point.

It is as if I have said "there is only one observer, an observer who is not at rest relative to the rocket or the clocks and we calculate things like this and come out with this answer" then you have effectively said "okay, so I take your observer and put him on the rocket, bring in another observer to replace yours, do my calculations from a different perspective than the one you have used and - look! - your results are wrong".

I am not saying you have done this deliberately, intentionally or with malice. It's an "as if". Can you see how I arrived at it?

cheers,

neopolitan
 
  • #44
Jesse,

I think that most of your other questions in #40 are possibly answered by the contents of #42 where I use the midpoint clock to explain what I mean by "the nose clock is in the future".

The only that remains open is about why the rocket's frame is "special". Well, it is not special other than it is the one that the "outsider observer" is observing. The two items being observed share that frame. In your way of thinking there is a virtual observer at rest in the rest frame of the rocket and the clocks and there is an "outside observer". The virtual observer, observer B, is special because this observer is the only one of the three you discussed before who is at rest in the B frame, along with the rocket and the clocks being observed, neither A nor C are at rest in the B frame.

Observer B is the only observer who can observe the clocks to be synchronised (after taking into account travel times, if he is not at the midpoint between the clocks - since the observer is virtual we can even nominate the midpoint between the clocks as her virtual location to make it easier for ourselves). Is this not special or different?

cheers,

neopolitan
 
  • #45
Hi Neopolitan,

Say I place two mickey mouse clocks next to each other on a table. Say the clocks are syncronised with each and nothing is moving except for the hands of the clocks advancing at the same rate as far as I am concerned. Now if I advance the hands of the right hand clock by 2 hours, is the right hand clock 2 hours in the future of the left hand clock in the meaning of "future" that you are using? Or is it just that I advanced the hands of the right hand clock and there is nothing fundamental about the time displayed. For instance the atoms of the two clock cases would have (near enough) the same "age".
 
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  • #46
neopolitan said:
Fortunately I like my sense of humor, so to have an example of it reflected back at me gave me a good laugh.

Then I read a bit closer.

Please please please reread the second paragraph of post #27.

There was only one observer and that observer was not on the rocket, not at rest relative to the rocket or the clocks. The "dorky physics guy" was the observer, the only observer, the observer who I specifically stated was not on the rocket. There was no other observer on the rocket who could observe the "dorky physics guy", which was the whole point.
I understood that you only wanted to talk about one observer outside the rocket--that was why I described the paragraph as "Also, in some later posts you seemed to deny that your comments about the nose being further in the future for A depended on using a second frame at all...like your comments in post #27 about not being interested in the perspective of a "dorky physics guy" on the rocket, only in the perspective of the outside observer". I thought your point was to denigrate the notion of a second observer on the rocket by calling him a "dorky physics guy"...what you said was:
We have a rocket with two clocks and we have an observer who is not at rest relative to the rocket. If I want to know what it observed I expect to hear "I see a rocket in motion with two clocks on it, one on each end" not "I see some dorky physics guy observing me".
In this paragraph, I thought you were saying that if you asked the observer outside what he saw, you wanted him to just see a rocket, not a dorky physics guy on board the rocket observing him back. But I guess what you're telling me is that I misunderstood, and you actually meant that if you asked what was observed by anyone, you only wanted to hear the observations of the guy outside, you didn't want to hear the observations of someone on the rocket who looks outside and sees a "dorky physics guy" (the outside observer) watching him. However, you can see that although I misunderstood who the dorky physics guy was supposed to be, I understood your main point just fine--that you didn't want to have a second observer on the rocket, just the guy outside the rocket. So this is a "difference that makes no difference".

Anyway, as I said in my last post, I don't understand why you're adamant that there be only one observer--do you agree that in relativity we basically just talk about "observers" as shorthand for different frames? Do you agree with my point that it's impossible to understand the meaning of "the nose clock is farther in the future" if we talk solely about the frame of the outside observer, that this is really a comparison of the definitions of simultaneity in the outside observer's frame and in the rocket's frame? If you agree that we have to talk about the rocket's frame in order to pick two clock readings which are simultaneous in that frame (since again, your statement about which clock is further in the future shouldn't depend on assuming the clocks are actually properly synchronized in their own rest frame), then why not just talk about the observations of the two clocks by someone on the rocket as shorthand for statements about simultaneous readings in the rocket's frame?
neopolitan said:
It is as if I have said "there is only one observer, an observer who is not at rest relative to the rocket or the clocks and we calculate things like this and come out with this answer" then you have effectively said "okay, so I take your observer and put him on the rocket, bring in another observer to replace yours, do my calculations from a different perspective than the one you have used and - look! - your results are wrong".

I am not saying you have done this deliberately, intentionally or with malice. It's an "as if". Can you see how I arrived at it?
I think it was you who was not reading my own words carefully enough here. Although I misunderstood who the "dorky physics guy" was supposed to represent, I made it clear that I understood that your point in that paragraph was that you didn't want to talk about any observations made on the rocket, only about the observations of the observer outside.
 
  • #47
Now say I get 3 lumps of radioactive material that decay in a consistant and predictable manner at the same rate. The half life of the material is defined as the time it takes for half the remaining atoms of isotope1 to convert to isotope2. These lumps provide a fundamental clock that cannot be arbitarily advanced, retarded or syncronised with respect to each other when they are at rest wrt each other.

I place one lump at the back of the rocket and one at the nose while the rocket is at rest wrt me. I keep one lump for reference and there are also the 3 conventional clocks of your original experiment (one alongside each lump). The rockets accelerates away for a time until coming to a final cruising speed where the rear and the nose of the rocket are going at the same speed relative to us. We note that the rocket has length contracted and conclude that the rear of the rocket must have been going faster than the nose of the rocket in order to "catch up" a little bit. It is reasonable to assume the rear clock and the rear isotope lump have time dilated the most out of all the clocks. In that sense the isotope lump at the nose of the rocket will have decayed more than the lump at the rear and so is in a fundamental sense in the future of the rear clock (but also in the past of reference lump that remained with us).

Now if some robot on the rocket is programmed to syncronise the clocks at the front and rear it will arbitarily retard the front clock by 8 seconds or advance the rear clock 8 seconds. They are both equally valid methods but the choice of adjusting the rear clock or the front clock to achieve syncronisation is arbitary and artificial. Does that make any sense?

[EDIT] Notice that after syncronisation we would be able to tell which clock was advanced or retaded by comparing the conventional clcocks to the radioactive lumps sitting next to them.
 
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  • #48
neopolitan said:
Looking over it again, yes, I meant C rather than B. It think it is reasonably clear that I was referring to another frame which was not at rest relative to the rocket. I am not 100% sure that we are there yet with what I meant about "the nose being more in the future as compared to the tail" - noting that I was only referring to a frame where two synchonised clocks were at rest and one observer who was not at rest relative to the clocks such that one "nose" clock is ahead of the other "tail" clock in terms of their relative motion according to the observer. In that limited scenario, do you agree?
I don't think the statement about the nose clock being further in the future can make sense except as a comparison of simultaneity in two frames, the outside observer's frame and the rocket's rest frame. The idea as I understood it is to take two readings on the nose and tail clock which are simultaneous in the rocket's frame, then look at when these same two readings occur in the outside observer's frame, and note that the nose reading happens further in the future than the tail reading in the observer's frame. Do you think it is possible to explain your idea of the nose being further in the future without referring to simultaneity in the rocket's rest frame, and also without necessarily assuming the clocks themselves have been properly synchronized in the rocket's rest frame?
neopolitan said:
To try to clarify again, in a now moment in the observer's frame (all now moments are relative, since "now" changes all the time), the observer may observe the tail clock reading 10s and the nose clock reading 2s. IF the clocks are synchonised relative to their rest frame - noting that the observer can work this out from the relative velocity of the clocks and their apparent separation from each other - THEN the observer can further deduce that the nose clock he sees "now" is a younger version of the nose clock and an older version of the tail clock (the observed nose clock manifests earlier in the clocks's rest frame than the observed tail clock - in our example 8s earlier). The nose clock, if you like, has reached the observer's "now" before the tail clock has.
Well, if you talk about which clock "reaches the observer's now first", this sounds more like viewing the outside observer's plane of simultaneity (his 'now') from the perspective of the rocket rest frame--as illustrated in DaleSpam's second diagram, the observer's planes of simultaneity are tilted in the rocket's frame so that the the nose will hit a given plane before the tail (here the 'before' refers to time in the rocket's own frame). So again, considering both the outside observer's frame and the rocket's frame seems critical here, which you seemed to at leas partially acknowledge when you said above "IF the clocks are synchonised relative to their rest frame..."
neopolitan said:
I am sorry to have to do this, but I hope I can justify it. Let's introduce a third clock - on the rocket, in the midpoint between the nose and the tail. That clock will read a midpoint value. Without thinking too deeply about the specifics, I suspect it is 6s (midway between 10s and 2s) but the acutal reading is immaterial - what is important is that it is more than 2s and less than 10s.
That's right, it'd be 6s. If two clocks are synchronized in their own frame, then in a frame where they're moving at speed v they'll be out-of-sync by vx/c^2, where x is the distance between them in their own rest frame. The clock in the middle is the same distance from the clock on the nose as it is from the clock on the tail, so it must be out-of-sync with each by the same amount (ahead of one and behind the other).
neopolitan said:
If the observer not at rest relative to the clocks observes a reading of 6s on the midpoint clock, 2s on the nose clock and 10s on the tail clock - and knows from his deductions that in their own rest frame the clocks are synchonised then he can say, taking the midpoint clock as his reference, that the nose clock he "should" (see since the clocks are synchronised) is in the future and the tail clock he "should" see is in the past. Whose past and whose future? the past and future of the observer.
OK, if he takes the midpoint clock as a reference for what he "should" see (though I hope you agree he could equally well take another clock for his reference) then he'll only see the nose clock give the same reading in the future, and he's already seen the tail clock show this reading in the past. I understand, and this is equivalent to the interpretation of you're comments that I've been talking about since post #26 (again, my interpretation is just that you pick simultaneous readings in the rocket's rest frame--in this case each of the three clocks reading 6 s--and then look at the order of these same readings in the observer's frame, noting that the nose reaches its reading at a time more 'in the future' for the observer than the tail reaches its own reading, and likewise the middle clock reaches its reading at a time midway between the other two in the observer's frame).
JesseM said:
DaleSpam is most probably right, we are probably arguing over semantics.
If you're just talking about the best way to conceptualize the relation between simultaneity in the observer's frame and the rocket's frame, then I'm not really arguing with you at all, I've said since post #26 that I think "the nose clock is further in the future" can be interpreted in a reasonable way. My issues were the ones I mentioned in post #40--that sometimes you seemed to suggest we didn't have to think about the rocket frame at all, and also that you wrote this paragraph which suggested you might be hinting at something more than just a way of conceptualizing simultaneity in relativity:
Just try to apply the same logic to the rocket and the two clocks. Relative to an observer not at rest relative to the rocket, the clock on the nose travels into the observer's future faster than the clock on the tail. The clock on the tail travels into the observer's future faster than the observer.

The observer also moves into the clocks' future faster than the clocks do.

This is where it gets less like semantics and more like something interesting ... can you model that? Not just wave it away, not just say "that's just relativity", not just show the mathematics on what must happen, but describe a model in which that is possible.
If you didn't mean to suggest here that what you were talking about was anything more than a way of conceptualizing simultaneity in SR, then just say so and my mind will be put at ease that you're not making any claims I would need to argue with.
 
  • #49
I can partly justify the nose isotope lump being ahead of the rear lump in a fundamental sense by getting the onboard robots to slowly transport both lumps to the centre of the ship so that they are right next to each other. We would notice that the lump originally at the nose has decayed more than the isotope lump that was originally at the rear. Since both the isotope lump "clocks" are essentially at the same location any observer would agree with the comparison at that point.
 
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  • #50
neopolitan said:
Jesse,

I think that most of your other questions in #40 are possibly answered by the contents of #42 where I use the midpoint clock to explain what I mean by "the nose clock is in the future".

The only that remains open is about why the rocket's frame is "special". Well, it is not special other than it is the one that the "outsider observer" is observing. The two items being observed share that frame. In your way of thinking there is a virtual observer at rest in the rest frame of the rocket and the clocks and there is an "outside observer". The virtual observer, observer B, is special because this observer is the only one of the three you discussed before who is at rest in the B frame, along with the rocket and the clocks being observed, neither A nor C are at rest in the B frame.

Observer B is the only observer who can observe the clocks to be synchronised (after taking into account travel times, if he is not at the midpoint between the clocks - since the observer is virtual we can even nominate the midpoint between the clocks as her virtual location to make it easier for ourselves). Is this not special or different?
Observer B is the only one who observes the clocks to be synchronized if they are indeed synchronized in their own rest frame--it would be possible, although unconventional, to set the clocks so that they were synchronized in some other frame, like the outside observer's frame. And didn't you agree with me earlier that the question of which clock is "further in the future" shouldn't depend on how they are synchronized?

Anyway, whether the rocket's frame is special really depends what you mean by "special", as I already said it is certainly the most simple and elegant choice if you're interested in comparing the outside observer's definition of simultaneity with some other frame's definition of simultaneity, but it isn't physically more special than some other frame like frame C in the sense that statements made based on using the rocket's frame (like, 'the nose clock is further in the outside observer's future') will be more "objectively true" in some sense than statements made based on using frame C (like 'the tail clock is further in the outside observer's future'). So once again we come back to the question at the end of my most recent post about whether you're just talking about a nice way to conceptualize the relation between two frames' definitions of simultaneity, or whether your "describe a model in which that is possible" was meant to suggest some kind of new truths about physics beyond what SR tells us.
 
  • #51
kev said:
Hi Neopolitan,

Say I place two mickey mouse clocks next to each other on a table. Say the clocks are syncronised with each and nothing is moving except for the hands of the clocks advancing at the same rate as far as I am concerned. Now if I advance the hands of the right hand clock by 2 hours, is the right hand clock 2 hours in the future of the left hand clock in the meaning of "future" that you are using? Or is it just that I advanced the hands of the right hand clock and there is nothing fundamental about the time displayed. For instance the atoms of the two clock cases would have (near enough) the same "age".

The clocks are desynchronised as soon as you move the hands. It's not what I meant at all.

cheers,

neopolitan
 
  • #52
Jesse,

I will get back to you tomorrow. For the moment, I was adamant that there be only one observer so you knew which equations mattered. We have had so much trouble in other exchanges by your swapping observers.

One observer, only one observer, and we should get past that.

cheers,

neopolitan
 
  • #53
kev said:
I can partly justify the nose isotope lump being ahead of the rear lump in a fundamental sense by getting the onboard robots to slowly transport both lumps to the centre of the ship so that they are right next to each other. We would notice that the lump originally at the nose has decayed more than the isotope lump that was originally at the rear. Since both the isotope lump "clocks" are essentially at the same location any observer would agree with the comparison at that point.

Um, where are "we"? And in any case, I think when you bring them together you should find that they have decayed equally.

I am ready to be proved wrong on this though.

cheers,

neopolitan
 
  • #54
neopolitan said:
Um, where are "we"? And in any case, I think when you bring them together you should find that they have decayed equally.

I am ready to be proved wrong on this though.

cheers,

neopolitan

"We" are in the original frame before the rocket accelerated away from "us". We are not onboard the rocket and have been stationary and never experienced any acceleration for the duration of the experiment. We count as one observer because we are next to each other and at rest with respect to each other. Sorry, I should have said "I" or "you" rather than "we" to stay in line with your condition of only one observer, but I get lonely sometimes :P The robots are programmed not to have an opinion and so do not count as observers ;)

When the radiactive lumps are being brought together by the inpartial robots, the rear lump is moving even faster (relative to the one and only sentient observer) than the nose lump that is transported "backwards" from the nose to the centre and so the rear lump experiences even more time dilation relative to the front lump from the point of view of the only observer.


When I get time I will do the calculations using the equations for the proper time experienced by accelerated and transported clocks.

[EDIT] Without doing the formal calculations we can note that when the rocket is cruising at constant velocity and when the convential rear and nose clocks have been syncronised that they will show the same time when brought together at the centre of the rocket. Since the radioctive lumps were not syncronised prior to bringing them together they will not show the same time when bringing them together.
 
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  • #55
kev said:
[EDIT] Without doing the formal calculations we can note that when the rocket is cruising at constant velocity and when the convential rear and nose clocks have been syncronised that they will show the same time when brought together at the centre of the rocket. Since the radioctive lumps were not syncronised prior to bringing them together they will not show the same time when bringing them together.

I should let someone else respond, but I make two comments anyway.

Comment one: Are we not comparing the extent to which each radiactive lump has decayed? I would have thought that this implies a synchronisation at a reference event after which the decay of each lump is measured.

Comment two: I assume that, in the rest frame of the rocket, the robots move at the same speed from the nose and tail towards the middle. This means that in the rocket frame there will be a very small amount of time dilation due to the relative motion of the robots (very small because you stipulated "gradually" as the magnitude of their velocities) and that time dilation would affect each robot and lump equally. The lumps will be simultaneous at the midpoint and will have undergone the same amount of time dilation, and, if my comment one is right, still synchonised (in the rocket's frame). Since the lumps will now be collocated, and collocation applies to all frames, I believe that they will then be synchonised in all frames.

Again, I may be wrong and stand ready to be corrected :)

cheers,

neopolitan
 
  • #56
neopolitan said:
I should let someone else respond, but I make two comments anyway.

Comment one: Are we not comparing the extent to which each radiactive lump has decayed? I would have thought that this implies a synchronisation at a reference event after which the decay of each lump is measured.

We do a crude form of syncronisation by starting with one large lump of radioactive material that consists of billions of atoms and is essentially homogenous and then divide the large lump into 3 equal portions to form 3 crude clocks.

neopolitan said:
Comment two: I assume that, in the rest frame of the rocket, the robots move at the same speed from the nose and tail towards the middle. This means that in the rocket frame there will be a very small amount of time dilation due to the relative motion of the robots (very small because you stipulated "gradually" as the magnitude of their velocities) and that time dilation would affect each robot and lump equally. The lumps will be simultaneous at the midpoint and will have undergone the same amount of time dilation, and, if my comment one is right, still synchonised (in the rocket's frame). Since the lumps will now be collocated, and collocation applies to all frames, I believe that they will then be synchonised in all frames.

Again, I may be wrong and stand ready to be corrected :)

cheers,

neopolitan

The two robots will have undergone the same amount of time dilation in the rest frame of the rocket relative to clocks on the rocket that remain at rest with rocket. However from the point of view of our single observer not onboard the rocket the two robots will not have experienced the same amount of time dilation.

I added an edit to the end of my last post (#47) that you may have missed.

--> Without doing the formal calculations we can note that when the rocket is cruising at constant velocity and when the convential rear and nose clocks have been syncronised that they will show the same time when brought together at the centre of the rocket. Since the radioctive lumps were not syncronised prior to bringing them together they will not show the same time when bringing them together. <--

Does that additional comment help any?
 
  • #57
kev,

If you check my previous post, you will see that your edit was the topic of my reply - I quoted your edit and nothing else.

I would appreciate a third opinion on this. I do still think that, when collocated, the lumps will have decayed equally - in all frames.

cheers,

neopolitan

MY EDIT - Here is an explanation for why. Perhaps it works, perhaps not. The time dilation for both lumps will be the same, relative to an outside observer - so long as the lumps are at rest in the rest frame they share with the rocket. There will just be an offset between due to their not being collocated. The reduction of that offset, to zero, can be accounted for by the different rates of time dilation as observed by the outside observer when they are not at rest in what was otherwise their shared rest frame (and, for the duration, only the rocket's rest frame).

A related effect must take place if a moving pair of clocks are slowed, so that they are and remain synchonised in their rest frame and become synchonised in a frame in which they were not at rest before deceleration but are at rest after deceleration. I am not saying it is the deceleration that does it.
 
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  • #58
Back to topic - Simultaneity

I would like to return to the original core topic, simultaneity.

Jesse confirmed that what we call simultaneity is "transmission simultaneity". Here is my definition:

neopolitan said:
Transmission simultaneity - photons from two events are released simultaneously, such that if the sources were equidistant (and remain equidistant - in other words the observer is at rest), the photons would reach the observer at rest together. Under most circumstances however, the photons will not reach the observer simultaneously and knowledge of where the photons were released is required to know that their release was in fact simultaneous.

If this is the case, and I have no doubt that it is, then is there validity in conceptualising an "instant", or a "surface of simultaneity"? Such an "instant" would comprise of an "event space" in which, relative to an inertial observer (at rest in that observer's own frame), all events are simultaneous. I could pick any instant, for example the instant when I absorbed the first photon from the sun to ever hit my retina (we could argue endlessly about how long that absorption process takes and the quantum uncertainty about when precisely the photon was absorbed, but the idea is to pick an instant, so we pick an instant in which the probability that the photon has just been absorbed is maximal), and label that t=0. Relative to my rest frame, there would be an event space which was the set of events (x,y,x,0) where x, y and z are unbounded. That event space would constitute an instance or a surface of simultaneity.

Are there any conceptual problems with that?

To prevent diversions, I state explicitly that I am aware that relative to other observers who might not share my rest frame the event frame I just define is not necessarily an instant or a surface of simultaneity. I will get to that later, if there are no real conceptual problems with what I am proposing.

cheers,

neopolitan
 
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  • #59
neopolitan said:
I would like to return to the original core topic, simultaneity.

Jesse confirmed that what we call simultaneity is "transmission simultaneity". Here is my definition:



If this is the case, and I have no doubt that it is, then is there validity in conceptualising an "instant", or a "surface of simultaneity"? Such an "instant" would comprise of an "event space" in which, relative to an inertial observer (at rest in that observer's own frame), all events are simultaneous. I could pick any instant, for example the instant when I absorbed the first photon from the sun to ever hit my retina (we could argue endlessly about how long that absorption process takes and the quantum uncertainty about when precisely the photon was absorbed, but the idea is to pick an instant, so we pick an instant in which the probability that the photon has just been absorbed is maximal), and label that t=0. Relative to my rest frame, there would be an event space which was the set of events (x,y,x,0) where x, y and z are unbounded. That event space would constitute an instance or a surface of simultaneity.

Are there any conceptual problems with that?

To prevent diversions, I state explicitly that I am aware that relative to other observers who might not share my rest frame the event frame I just define is not necessarily an instant or a surface of simultaneity. I will get to that later, if there are no real conceptual problems with what I am proposing.

cheers,

neopolitan

What you are describing here is pretty standard. See this link http://casa.colorado.edu/~ajsh/sr/simultaneous.html. Ceruleans "hypersurface of simultaneity" is the tilted blue plane as seen by Vermilion whos own hypersurface of simultaneity is the red plane orthogonal to the time axis of his coordinate system.
 
  • #60
Thanks kev,

I'll wait a bit to see if anyone thinks it is not standard.

cheers,

neopolitan
 
  • #61
neopolitan said:
If this is the case, and I have no doubt that it is, then is there validity in conceptualising an "instant", or a "surface of simultaneity"? Such an "instant" would comprise of an "event space" in which, relative to an inertial observer (at rest in that observer's own frame), all events are simultaneous. I could pick any instant, for example the instant when I absorbed the first photon from the sun to ever hit my retina (we could argue endlessly about how long that absorption process takes and the quantum uncertainty about when precisely the photon was absorbed, but the idea is to pick an instant, so we pick an instant in which the probability that the photon has just been absorbed is maximal), and label that t=0. Relative to my rest frame, there would be an event space which was the set of events (x,y,x,0) where x, y and z are unbounded. That event space would constitute an instance or a surface of simultaneity.
All of the above is correct. An observer has, for each event, a surface of simultaneity. In special relativity (i.e. ignoring gravity) the surface is a 3-dimensional "plane" in 4-dimensional spacetime. For a single observer, all the planes of simultaneity (for different events) stack up in parallel. But, as you suggest, different observers have different planes of simultaneity. This is standard, mainstream special relativity.

neopolitan said:
Comment two: I assume that, in the rest frame of the rocket, the robots move at the same speed from the nose and tail towards the middle. This means that in the rocket frame there will be a very small amount of time dilation due to the relative motion of the robots (very small because you stipulated "gradually" as the magnitude of their velocities) and that time dilation would affect each robot and lump equally. The lumps will be simultaneous at the midpoint and will have undergone the same amount of time dilation, and, if my comment one is right, still synchonised (in the rocket's frame). Since the lumps will now be collocated, and collocation applies to all frames, I believe that they will then be synchonised in all frames.
If I understand you correctly, you are claiming that, if two separated clocks on the rocket are synchronised according to an outside observer, when you slowly move the clocks together they should remain synchronised according to that same observer.

This may seem reasonable on the grounds that both clocks undergo the same time dilation. This would be true relative to the rocket, but it's not true relative to the outside observer; from the observer's point of view, one clock experiences more dilation than the rocket and the other less.

You might then argue that any change of dilation can be ignored if the clocks move slowly enough. However, this ignores the fact that the slower the clocks move (relative to the rocket), the longer it will take to bring them together. This lengthening of time taken increases the effect of the dilation. In fact the decrease in dilation-change and increase in duration tend to cancel each other out, and no matter how slow the clocks are moved, the there is a change between the clocks that will not go away, relative to the outside observer.

If you really wanted I could prove all this mathematically, but it would take a page or two of calculation.

Note that synchronising clocks by slowly moving them apart is referred in the literature as "slow clock transport" or "ultra slow clock transport". It can be proved that synchronisation by slow clock transport is exactly the same as Einstein synchronisation. For example, see this post.
 
  • #62
DrGreg said:
If I understand you correctly, you are claiming that, if two separated clocks on the rocket are synchronised according to an outside observer, when you slowly move the clocks together they should remain synchronised according to that same observer.

This may seem reasonable on the grounds that both clocks undergo the same time dilation. This would be true relative to the rocket, but it's not true relative to the outside observer; from the observer's point of view, one clock experiences more dilation than the rocket and the other less.

You might then argue that any change of dilation can be ignored if the clocks move slowly enough. However, this ignores the fact that the slower the clocks move (relative to the rocket), the longer it will take to bring them together. This lengthening of time taken increases the effect of the dilation. In fact the decrease in dilation-change and increase in duration tend to cancel each other out, and no matter how slow the clocks are moved, the there is a change between the clocks that will not go away, relative to the outside observer.

What I was trying to say is that, within the rocket's frame, the clocks that are at first synchronous, at rest and separated, will in the scenario presented arrive at the midpoint having undergone the same amount of time dilation - within rocket's frame - and therefore be synchronous, at rest and collocated.

Once collocated, their being synchonrous should be frame independent.

I was further trying to say, perhaps not sufficiently clearly, that the unequal time dilation effects observed in another frame should explain how the clocks end up being synchronous in that other frame when they weren't initially.

It is not a case of the problem "going away".

cheers,

neopolitan
 
  • #63
neopolitan said:
What I was trying to say is that, within the rocket's frame, the clocks that are at first synchronous, at rest and separated, will in the scenario presented arrive at the midpoint having undergone the same amount of time dilation - within rocket's frame - and therefore be synchronous, at rest and collocated.
You started out this thread stating you wanted to stick to one observer (Let's call him Fred) to avoid confusion. Fred is an observer that is not onboard the rocket but was at one point at rest with rocket before the rocket accelerated. I have inserted red text in square brakets to make clear which measurements are Fred's observations. Consciously or unconsiously you now talking in terms of another observer (call him Barney?) who's observations I have inserted in square brakets and blue text in your statements.

"What I was trying to say is that, within the rocket's frame [Barney's frame], the clocks that are at first synchronous [According to the observer onboard the rocket], at rest and separated, will in the scenario presented arrive at the midpoint having undergone the same amount of time dilation [As measured by Barney] - within rocket's frame - and therefore be synchronous, at rest and collocated."


neopolitan said:
Once collocated, their being synchonrous should be frame independent.
Correct

neopolitan said:
I was further trying to say, perhaps not sufficiently clearly, that the unequal time dilation effects observed in another frame should explain how the clocks end up being synchronous in that other frame when they weren't initially.

It is not a case of the problem "going away".

cheers,

neopolitan

"I was further trying to say, perhaps not sufficiently clearly, that the unequal time dilation effects observed in another frame [By Fred] should explain how the clocks end up being synchronous in that other frame [In Fred's frame] when they weren't initially [According to Fred].


Ok, you seem to ready to talk in terms of two observers.

We are agreed that two clocks that are syncronised [according to Barney] will still be syncronised when they are transported to the centre of the rocket. In the example I gave the two tamper proof radioactive decay clocks are not syncronised after the acceleration phase of the rocket [as far as Barney is concerned] and they will still not be syncronised when transported to the centre so as to be co-located (as far as any observer is concerned). The tamper proof clock that was at the nose will be "in the future" (to use your expression) of the tamper proof clock that was at the rear.
 
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  • #64
kev said:
You started out this thread stating you wanted to stick to one observer (Let's call him Fred) to avoid confusion. Fred is an observer that is not onboard the rocket but was at one point at rest with rocket before the rocket accelerated. I have inserted red text in square brakets to make clear which measurements are Fred's observations. Consciously or unconsiously you now talking in terms of another observer (call him Barney?) who's observations I have inserted in square brakets and blue text in your statements.

"What I was trying to say is that, within the rocket's frame [Barney's frame], the clocks that are at first synchronous [According to the observer onboard the rocket], at rest and separated, will in the scenario presented arrive at the midpoint having undergone the same amount of time dilation [As measured by Barney] - within rocket's frame - and therefore be synchronous, at rest and collocated."



Correct



"I was further trying to say, perhaps not sufficiently clearly, that the unequal time dilation effects observed in another frame [By Fred] should explain how the clocks end up being synchronous in that other frame [In Fred's frame] when they weren't initially [According to Fred].


Ok, you seem to ready to talk in terms of two observers.

We are agreed that two clocks that are syncronised [according to Barney] will still be syncronised when they are transported to the centre of the rocket. In the example I gave the two tamper proof radioactive decay clocks are not syncronised after the acceleration phase of the rocket [as far as Barney is concerned] and they will still not be syncronised when transported to the centre so as to be co-located (as far as any observer is concerned). The tamper proof clock that was at the nose will be "in the future" (to use your expression) of the tamper proof clock that was at the rear.

Sorry for intervening but I think you need to pay attention to what neopolitan just told you:

neopolitan said:
What I was trying to say is that, within the rocket's frame, the clocks that are at first synchronous, at rest and separated, will in the scenario presented arrive at the midpoint having undergone the same amount of time dilation - within rocket's frame - and therefore be synchronous, at rest and collocated.

Once collocated, their being synchonrous should be frame independent.

The fact that the two clocks move at different speeds from the point of view of an observer external to the rocket is purely irrelevant. The fact that the clocks are synchronized and collocated takes precedence.
 
  • #65
My strict regime of one and only one observer was an attempt to avoid a situation as happened when I talked about the dorky physics guy (see my earlier post and a later reply from JesseM) in which I wanted to look at things from one specific perspective, where one specific set of equations applied, and JesseM wanted to look at things from another perspective, where another specific set of equations applied.

In my reply to you, it is not possible or sensible to only consider one observer, since we are inherently, as you point out, discussing two points of view.

What I want to know is how we can have two clocks, which are both collocated and synchronous in one frame (Barney's) and collocated and asychonous in another frame (Fred's) given that we both agree that being synchronous when collocated is frame independent.

Is this not paradoxical?

You may understand that it is difficult to accept this as objectively true.

cheers,

neopolitan
 
  • #66
DrGreg said:
If I understand you correctly, you are claiming that, if two separated clocks on the rocket are synchronised according to an outside observer, when you slowly move the clocks together they should remain synchronised according to that same observer.

.

I think (IMHO) that neopolitan talks about two clocks synchronized from the perspective of an on-board observer , i.e. inside the rocket.
This is why, he claims , correctly (IMHO) that once the two clocks are "slow transported" to the center of the rocket, they will be still synchronized (by the very definition of slow clock transport).This is regardles as to how long the transport took.

He further claims that, since the clocks are now collocated , they are synchronized in any frame of reference.

If this is what neopolitan is saying, I think (again, IMHO) that he is correct.
 
  • #67
I'm away from home but I've been using my friend's laptop a bit this afternoon, so I thought I'd give a quick reply to this:
neopolitan said:
My strict regime of one and only one observer was an attempt to avoid a situation as happened when I talked about the dorky physics guy (see my earlier post and a later reply from JesseM) in which I wanted to look at things from one specific perspective, where one specific set of equations applied, and JesseM wanted to look at things from another perspective, where another specific set of equations applied.
But as I said, I don't think it's possible to make sense of your claim that the nose is "more in the future" in the outside observer's frame without making some reference to the rocket's own rest frame--the idea is to pick to clock readings which are simultaneous in the rocket's frame, then switch to the outside observer's frame, and note that the reading on the nose happens at a later time in this frame than the reading on the tail. But even if we do have two frames, I suppose it might be a bit easier to have just one "observer" who measures the rocket moving, that way we can refer to "the observer's frame" and "the rocket's frame" instead of "the frame of the outside observer who sees the rocket moving" and "the frame of the observer on board the rocket".

Do you disagree with the basic point that two frames are needed? If so, how would you explain the notion that the nose clock is "further in the future" without making reference to the rocket frame, and without assuming the clocks have actually been synchronized in the rocket frame?
neopolitan said:
What I want to know is how we can have two clocks, which are both collocated and synchronous in one frame (Barney's) and collocated and asychonous in another frame (Fred's) given that we both agree that being synchronous when collocated is frame independent.
We don't (referring to the bolded part). When they're collocated, both frames agree they show the same time, assuming they were brought together at equal speeds in the frame where they were synchronized when located at opposite ends of the rocket (the rocket's rest frame). In this case, that means that in the frame where the rocket is moving forward, the clock that is being pushed in the same direction that the rocket is moving (from tail to middle) will have a slightly higher speed than the clock that is being pushed opposite the rocket's direction of motion (from nose to middle), so the clock that's pushed from tail to middle will tick slightly slower as it's brought together with the clock that's pushed from nose to middle, which means even though the clock at the tail was originally ahead of the clock at the nose, the time difference decreases as they're brought together until they show the same time at at the middle (and also show the same time as a clock that was fixed at the middle and previously synchronized with the other two in the rocket's frame, before they were moved).

I gave a numerical example showing that slow transport should always cause the clock that gets slowly transported to read the same time as local untransported clocks at rest and synchronized in the frame where its speed is arbitrarily close to zero (in the above example, this means if you had a row of synchronized clocks on board the rocket, then during the process of transporting two clocks at either end to the middle, they would always read the same time as whichever of the other clocks in the row they were passing next to), even if you analyze the sitation from a frame where the transported clock's speed is not close to zero and the untransported clocks are not in sync, in post #37 of this thread, and a more general proof at the end of post #41.
 
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  • #68
1effect said:
I think (IMHO) that neopolitan talks about two clocks synchronized from the perspective of an on-board observer , i.e. inside the rocket.
This is why, he claims , correctly (IMHO) that once the two clocks are "slow transported" to the center of the rocket, they will be still synchronized (by the very definition of slow clock transport).This is regardles as to how long the transport took.

He further claims that, since the clocks are now collocated , they are synchronized in any frame of reference.

If this is what neopolitan is saying, I think (again, IMHO) that he is correct.

You are correct and so is neopolitan in this particular aspect and I never disagreed with this view. In the last paragraph of post #63 I showed I also agreed with this aspect in this sentence --> "We are agreed that two clocks that are syncronised [according to Barney] will still be syncronised when they are transported to the centre of the rocket."


neopolitan said:
...

What I want to know is how we can have two clocks, which are both collocated and synchronous in one frame (Barney's) and collocated and asychonous in another frame (Fred's) given that we both agree that being synchronous when collocated is frame independent.

Is this not paradoxical?

You may understand that it is difficult to accept this as objectively true.

cheers,

neopolitan

I never said that two clocks that are co-located and syncronous in one frame would not be syncronous in another frame. What I said in post #63 was "In the example I gave the two tamper proof radioactive decay clocks are not syncronised after the acceleration phase of the rocket [as far as Barney is concerned] and they will still not be syncronised when transported to the centre so as to be co-located (as far as any observer is concerned)" which is not the same thing.

I also said in post #54 I said "...that when the rocket is cruising at constant velocity and when the convential rear and nose clocks have been syncronised that they will show the same time when brought together at the centre of the rocket." and in post #49 I said "Since both the isotope lump "clocks" are essentially at the same location any observer would agree with the comparison at that point."

I hope you can agree I never said (as far as I am aware) anything that amounts to your suggestion that I implied two clocks, which are both collocated and synchronous in one frame (Barney's) could be collocated and asychonous in another frame (Fred's).

To summerise:

Two clocks that are co-located and syncronised in any observers frame will be syncronised in any other observer's frame.

Two clocks that are co-located but not syncronised in any observer's frame will not be syncronised in any other observer's frame.

[EDIT] Can either of you show me where I said anything that contradicted that basic concept? If I did it is a typo and I will correct it.
 
Last edited:
  • #69
JesseM

My comment was based in so much context that I don't think I can respond any better than "please see the ealier posts". I should not have tried to defend against a comment made referring to something I had written, but presented out of context. I've explained myself before. Please find the relevant posts which appear earlier in the thread.

Equally, Barney and Fred are not mine. They are kev's creations from the very start (along with the decaying lumps and disinterested robots). I will continue to try to work out if kev was claiming what I thought he was, if he had a typo or if I misread what he wrote.

cheers,

neopolitan
 
  • #70
kev said:
I never said that two clocks that are co-located and syncronous in one frame would not be syncronous in another frame. What I said in post #63 was "In the example I gave the two tamper proof radioactive decay clocks are not syncronised after the acceleration phase of the rocket [as far as Barney is concerned] and they will still not be syncronised when transported to the centre so as to be co-located (as far as any observer is concerned)" which is not the same thing.

I also said in post #54 I said "...that when the rocket is cruising at constant velocity and when the convential rear and nose clocks have been syncronised that they will show the same time when brought together at the centre of the rocket." and in post #49 I said "Since both the isotope lump "clocks" are essentially at the same location any observer would agree with the comparison at that point."

I hope you can agree I never said (as far as I am aware) anything that amounts to your suggestion that I implied two clocks, which are both collocated and synchronous in one frame (Barney's) could be collocated and asychonous in another frame (Fred's).

To summerise:

Two clocks that are co-located and syncronised in any observers frame will be syncronised in any other observer's frame.

Two clocks that are co-located but not syncronised in any observer's frame will not be syncronised in any other observer's frame.

[EDIT] Can either of you show me where I said anything that contradicted that basic concept? If I did it is a typo and I will correct it.

kev,

In #54 you said
kev said:
[EDIT] Without doing the formal calculations we can note that when the rocket is cruising at constant velocity and when the convential rear and nose clocks have been syncronised that they will show the same time when brought together at the centre of the rocket. Since the radioctive lumps were not syncronised prior to bringing them together they will not show the same time when bringing them together.

Then in #55 I said
neopolitan said:
Comment one: Are we not comparing the extent to which each radiactive lump has decayed? I would have thought that this implies a synchronisation at a reference event after which the decay of each lump is measured.

Comment two: I assume that, in the rest frame of the rocket, the robots move at the same speed from the nose and tail towards the middle. This means that in the rocket frame there will be a very small amount of time dilation due to the relative motion of the robots (very small because you stipulated "gradually" as the magnitude of their velocities) and that time dilation would affect each robot and lump equally. The lumps will be simultaneous at the midpoint and will have undergone the same amount of time dilation, and, if my comment one is right, still synchonised (in the rocket's frame). Since the lumps will now be collocated, and collocation applies to all frames, I believe that they will then be synchonised in all frames.

Then in #56 you said
kev said:
The two robots will have undergone the same amount of time dilation in the rest frame of the rocket relative to clocks on the rocket that remain at rest with rocket. However from the point of view of our single observer not onboard the rocket the two robots will not have experienced the same amount of time dilation.

Then in #62 (responding to Dr Greg) I said
neopolitan said:
What I was trying to say is that, within the rocket's frame, the clocks that are at first synchronous, at rest and separated, will in the scenario presented arrive at the midpoint having undergone the same amount of time dilation - within rocket's frame - and therefore be synchronous, at rest and collocated.

Once collocated, their being synchonrous should be frame independent.

I was further trying to say, perhaps not sufficiently clearly, that the unequal time dilation effects observed in another frame should explain how the clocks end up being synchronous in that other frame when they weren't initially.

As far as I can tell it was not until #63 that you introduced tamper proof clocks which you claim would lose synchronicity after acceleration - specifically, you claim that the clocks lose synchronicity in the rest frame due to acceleration:

kev said:
We are agreed that two clocks that are syncronised [according to Barney] will still be syncronised when they are transported to the centre of the rocket. In the example I gave the two tamper proof radioactive decay clocks are not syncronised after the acceleration phase of the rocket [as far as Barney is concerned] and they will still not be syncronised when transported to the centre so as to be co-located (as far as any observer is concerned). The tamper proof clock that was at the nose will be "in the future" (to use your expression) of the tamper proof clock that was at the rear.

If, for any reason, the clocks are not synchronised in their shared rest frame (the rocket's frame, or Barney's frame), then moving them to the centre of the rocket in the manner described will not make them synchronised. However, the extent to which they are not synchronised will not be affected either.

The question I must have then is, does acceleration of a rest frame - relative to an outside observer - cause synchronised clocks in that rest frame to lose synchronisation?

I doubt that it does, since each clock will be accelerated equally (since otherwise the clocks don't share a rest frame). All effects will be equal and synchronisation in that rest frame will be maintained.

Yet again, I stand ready to be corrected.

cheers,

neopolitan
 

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