Discuss events which are simultaneous in one frame?

[neopolitan]

I have scanned the simultaneity related posts and cannot find a succinct answer to a question I have - I accept that the answer is probably buried in one or more of them.

My question is related to the common claim that you somehow lose simultaneity in relativity, or that the idea of simultaneity is somehow meaningless.

What, precisely, are we referring to when we discuss events which are simultaneous in one frame?

I give you the two options which I think we could be talking about:

Reception simultaneity - photons from two events reach the observer together (this is a third event if you like, one in which the observer and the two photons are collocated in both time and space).

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.

Is the simultaneity we talk about one of these, or something different?

Am I mistaken about the concept that "simultaneity is lost" and/or "simultaneity is meaningless in relativity"?

I do have a follow on question, but it may be moot if the answers I get to this clarify something else for me.

cheers,

neopolitan

 

[JesseM]

I have scanned the simultaneity related posts and cannot find a succinct answer to a question I have - I accept that the answer is probably buried in one or more of them.

My question is related to the common claim that you somehow lose simultaneity in relativity, or that the idea of simultaneity is somehow meaningless.

What, precisely, are we referring to when we discuss events which are simultaneous in one frame?

I give you the two options which I think we could be talking about:

Reception simultaneity - photons from two events reach the observer together (this is a third event if you like, one in which the observer and the two photons are collocated in both time and space).

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.

Is the simultaneity we talk about one of these, or something different?

It's the second one--if an observer assumes that both signals move at c relative to himself, and subtracts the calculated time for the light to traverse the distance between the point of emission and himself (as measured by rulers at rest in his frame) from the time the light actually reaches him (as measured by his own clock), giving the calculated time the light was actually transmitted in his frame, then two events are simultaneous in his frame if the calculated transmission time for each is identical. A functionally identical way of defining the time of events in a given frame is for each observer to have an array of clocks at rest relative to themselves, attached to rulers also at rest relative to them, and with the clocks "synchronized" using the assumption that light travels at the same speed in all directions in their frame--for example, one could synchronize two clocks by setting off a flash at their exact midpoint, and making sure they read the same time when the light from the flash reaches them (this is one version of the 'Einstein clock synchronization convention'). Then to find the time of a given event, I just look at the reading on the clock right next to it as the event happened, so I'm defining times of events purely in terms of local measurements and don't have to worry about light delays; this will give the same answer as the earlier method, and two events will be simultaneous if the clock that was next to one event when it happened showed the same time as the clock that was next to the other event when it happened.

However, if different observers in relative motion each define simultaneity in this way, making the assumption that light travels at the same speed in all directions in their own frame, then a consequence of this is that they will disagree about whether two given events happened simultaneously or not. For example, suppose I am on board a rocket which is moving relative to you from left to right, and in my rest frame I synchronize clocks on either end of the rocket by setting off a flash at the midpoint of the rocket and making sure each clock reads the same time when the light from the flash reaches them. But in your frame the clock on the right side is moving away from the position on your ruler where the flash was set off, while the clock on the left side is moving towards it, so if you assume both light signals move at the same speed in your frame, you must conclude the light reached the left clock before the right clock; thus, in your frame my two clocks are out-of-sync and the events of the light hitting each one are non-simultaneous.

Why is the assumption made that each observer should assume light moves at the same speed in each direction, then? Basically the reason is that all the fundamental laws of physics seem to have the property of "Lorentz invariance" meaning that if different observers construct their coordinate systems in this way, the basic equations representing the laws of physics will be the same when expressed in these different coordinate systems. If the laws of physics weren't Lorentz-invariant, but instead were invariant under the Galilei transformation where different coordinate systems have no disagreements about simultaneity, then there would be a physically preferred definition of simultaneity and we wouldn't be using coordinate systems where light moves at the same speed in all directions in every frame.

If you're interested, I drew up some diagrams of two ruler-clock arrays sliding next to each other in this thread, showing how in each array's rest frame the clocks of the other array are out-of-sync, and how this is crucial to understanding how it can be that in each array's rest frame the clocks of the other array are running slower and the distance between markings on the other ruler are shrunk.

 

[neopolitan]

But doesn't assume privileged information?

Either you give the location of each event to both of the observers, each in terms of their own frames (in which case they can both work out that the transmissions were simultaneous) or you give neither information about location, and there is no longer enough information to detemine whether any simultaneity was involved.

Of course simultaneity will be lost if one frame is given privileged status.

Think of the synchonisation convention that you referred to. An observer at rest relative to the two clocks at their midpoint will observe the clocks as synchronised. This has to do with location. If that observer gets up, takes a step towards one or other of the clocks and stops again (returning to rest), the clocks will no longer be directly observed as synchronised. They are still synchronised though, irrespective of the direct observations made.

If an observer in a frame which is not at rest relative to the clocks makes an observation at the midpoint between the clocks, the clocks will be seen as synchronised - at that point. This must be the case since photons released from each clock meet each other at that midpoint (the photons not being affected by the inertia of any observer). The photons will hit the eye of the observer together.

That observer in a frame which is not at rest relative to the clocks will directly observe the same apparent lack of simultaneity as an observer at rest relative to the clocks does if they are in the same location (by this I mean if they are physically collocated, not that they agree about where they are). Both will be able to use the information they have to hand to work out that the clocks are simultaneous - with each other.

If I used your example of a rocket with lights flashing on each end, I could actually work out why the disagreement occurs, and timing of the flashes would make sense to both observers. (The maths is essentially the same as I used in an earlier strand to point out that there is no twins paradox, just a poorly framed scenario.)

Is the simultaneity that you are talking about? or is it a simultaneity that is "lost" on another level - that is, both observers can work out that the clocks are synchronised, but there is no simultaneity between the observations, direct or inferred, because the when of each tick of the clocks is not agreed?

To try to clarify what I mean here, at the instant when both observers are collocated at the midpoint between two clocks, the one who is at rest relative to the clocks will say that the ticks observed happened together a period of t1 ago, while the one in motion relative to the clocks will say that the ticks observed happened together a period of t2 ago. t1 does not equal t2.

cheers,

neopolitan

 

[Xantos]

However, if different observers in relative motion each define simultaneity in this way, making the assumption that light travels at the same speed in all directions in their own frame, then a consequence of this is that they will disagree about whether two given events happened simultaneously or not. For example, suppose I am on board a rocket which is moving relative to you from left to right, and in my rest frame I synchronize clocks on either end of the rocket by setting off a flash at the midpoint of the rocket and making sure each clock reads the same time when the light from the flash reaches them. But in your frame the clock on the right side is moving away from the position on your ruler where the flash was set off, while the clock on the left side is moving towards it, so if you assume both light signals move at the same speed in your frame, you must conclude the light reached the left clock before the right clock; thus, in your frame my two clocks are out-of-sync and the events of the light hitting each one are non-simultaneous.

This simply cannot be true. If you assume that the speed of light is constant for everyone meaning information moves at constant C and that laws of physics are in all frames the same then I (on the spaceship) and you (observer) would see the same time dilation (laser beam takes the same amount of time to get to the front clock).

 

[JesseM]

This simply cannot be true. If you assume that the speed of light is constant for everyone meaning information moves at constant C and that laws of physics are in all frames the same then I (on the spaceship) and you (observer) would see the same time dilation (laser beam takes the same amount of time to get to the front clock).

Not true. Suppose the ship is 24 light-seconds long in my frame, moving at 0.5c to the right, and at time t=0 in my frame the flash is set off at position x=12 ls, and at that moment the left end is at position x=0 ls while the right end is at x=24 ls. In this case, at t=8 seconds the light heading in the left direction will have moved 8 ls to the left of x=12 ls, so it'll be at x=12-8=4 ls. Meanwhile, since the left end of the ship started at x=0 ls and is moving to the right at 0.5c, after 8 seconds it'll have moved 8*0.5 = 4 ls to the right, so it'll be at x=4 ls too; thus, the light must hit the left clock at time t=8 ls.

Meanwhile, at t=24 s, the light moving to the right has moved 24 ls to the right of x=12 ls, so it'll be at x=36 ls. And the right end of the ship started at x=24 ls, and since it's moving to the right at 0.5c, 24 s later it will have moved 24*0.5 = 12 ls to the right of this position, so it'll be at x= 24 + 12 = 36 ls as well. So, this must be the time the light catches up to the right clock, at time t=24 s, a full 16 s after it caught up to the left clock.

Of course, this is just in my frame where the ship is moving. In the ship's own rest frame, the two clocks are at rest at equal distances from the position the flash was set off, so naturally if we assume both signals move at the same speed in this frame we'll conclude the light must have hit both ends at the same time. This is what is meant by the relativity of simultaneity--different frames disagree on whether a pair of events at different locations happened "at the same time" or "at different times".

 

[Xantos]

Not true. Suppose the ship is 24 light-seconds long in my frame, moving at 0.5c to the right, and at time t=0 in my frame the flash is set off at position x=12 ls, and at that moment the left end is at position x=0 ls while the right end is at x=24 ls. In this case, at t=8 seconds the light heading in the left direction will have moved 8 ls to the left of x=12 ls, so it'll be at x=12-8=4 ls. Meanwhile, since the left end of the ship started at x=0 ls and is moving to the right at 0.5c, after 8 seconds it'll have moved 8*0.5 = 4 ls to the right, so it'll be at x=4 ls too; thus, the light must hit the left clock at time t=8 ls.

Meanwhile, at t=24 s, the light moving to the right has moved 24 ls to the right of x=12 ls, so it'll be at x=36 ls. And the right end of the ship started at x=24 ls, and since it's moving to the right at 0.5c, 24 s later it will have moved 24*0.5 = 12 ls to the right of this position, so it'll be at x= 24 + 12 = 36 ls as well. So, this must be the time the light catches up to the right clock, at time t=24 s, a full 16 s after it caught up to the left clock.

Of course, this is just in my frame where the ship is moving. In the ship's own rest frame, the two clocks are at rest at equal distances from the position the flash was set off, so naturally if we assume both signals move at the same speed in this frame we'll conclude the light must have hit both ends at the same time. This is what is meant by the relativity of simultaneity--different frames disagree on whether a pair of events at different locations happened "at the same time" or "at different times".

Very nice explained, but i still don't agree. I cannot and will not accept the modern explanation of relativity because it's wrong and full of illogical assumptions.

Here, look at this link:
http://www.phys.unsw.edu.au/einsteinlight/jw/module3_weird_logic.htm"

This animation is clearly evidence that light is just like sound only that it has a different base. Sound is also constant and it is the basic physical phenomenon that gives us information about sound of the "flying block". Light is also constant and it is th basic physical phenomenon that gives us information about what it is around us. If you went to do experiments with sound and sound barrier you'd also find the same time dilation as you find it with light. So if you study sound and its properties you indirectly study the behaviour of light and infromation that it sends to you. Light barrier exists and it manifests in Cherenkov radiation and its blueish glow. The only thing we have to figure out is how to break it.

 

[JesseM]

But doesn't assume privileged information?

Either you give the location of each event to both of the observers, each in terms of their own frames (in which case they can both work out that the transmissions were simultaneous) or you give neither information about location, and there is no longer enough information to detemine whether any simultaneity was involved.

I don't understand what you mean by "give" the location. Each observer has their own ruler at rest relative to themselves--you can imagine the two observers' rulers moving alongside one another at arbitrarily small separation, as in my illustration--so to figure out the position of a given event in each frame, you just have to look at which marking on each ruler was right next to the event as it happened.

Think of the synchonisation convention that you referred to. An observer at rest relative to the two clocks at their midpoint will observe the clocks as synchronised. This has to do with location. If that observer gets up, takes a step towards one or other of the clocks and stops again (returning to rest), the clocks will no longer be directly observed as synchronised. They are still synchronised though, irrespective of the direct observations made.

Again, synchronization has nothing to do with when observers see events, I thought I made that clear. An observer defines the time of an event using rulers and clocks at rest relative to themselves--they can either use their ruler to determine the distance of the event and subtract the time the light would be calculated to reach them from the time they actually see it to determine the "actual" time of the event in their frame (the first method I discussed), or they can presynchronize clocks at different locations using light signals, and then define the time of the event in terms of the reading on a local clock from their system that was right next to the event as it happened (the second method I discussed, which will give exactly the same answer as the first).

If an observer in a frame which is not at rest relative to the clocks makes an observation at the midpoint between the clocks, the clocks will be seen as synchronised - at that point.

They'll be seen as showing the same time when he looks through his telescope at each one at that moment, but this has nothing to do with what it means for an observer to say the clocks are "synchronized" in his frame, it's completely irrelevant.

That observer in a frame which is not at rest relative to the clocks will directly observe the same apparent lack of simultaneity as an observer at rest relative to the clocks does if they are in the same location (by this I mean if they are physically collocated, not that they agree about where they are). Both will be able to use the information they have to hand to work out that the clocks are simultaneous - with each other.

This paragraph seems to be operating under the assumption that simultaneity has something to do with whether or not a given observer sees two clocks showing the same time at a given instant, so all I can tell you is that your assumption is wrong, simultaneity has nothing to do with that. It is determined by either the first method where you subtract off the time the signal took to reach you from the time you see the event, or by the second method where you have an array of clocks that are synchronized using the Einstein synchronization convention (itself based on the assumption that light moves at the same speed in all directions in your frame), and you use only local measurements from clocks that were right next to each event as it happened.

If I used your example of a rocket with lights flashing on each end, I could actually work out why the disagreement occurs, and timing of the flashes would make sense to both observers. (The maths is essentially the same as I used in an earlier strand to point out that there is no twins paradox, just a poorly framed scenario.)

It would make sense to both observers, but the two observers would disagree about whether the two events happened simultaneously or not.

Is the simultaneity that you are talking about? or is it a simultaneity that is "lost" on another level - that is, both observers can work out that the clocks are synchronised, but there is no simultaneity between the observations, direct or inferred, because the when of each tick of the clocks is not agreed?

I don't understand this paragraph at all. What does "simultaneity between the observations" mean, and what does "the when of each tick of the clock" mean? What specific method are you imagining the observers use to assign time-coordinates to events? Maybe it would help if you gave a numerical example. Do you agree that if the observers use either of the two equivalent methods I discussed above, then if one observer finds that two events happened at the same time-coordinate in his frame, the other observer will find they happened at different time-coordinates in his own frame?

To try to clarify what I mean here, at the instant when both observers are collocated at the midpoint between two clocks, the one who is at rest relative to the clocks will say that the ticks observed happened together a period of t1 ago, while the one in motion relative to the clocks will say that the ticks observed happened together a period of t2 ago. t1 does not equal t2.

No, the one in motion will say that the ticks happened at two different times, because on his ruler they happened next to markings which are at unequal distances from him, so the only way he can account for the fact that they both signals reached him at the same time while still assuming they traveled at the same speed is to conclude they happened at different times.

Imagine we have two observers A and A' whose rulers are sliding next to each other at 0.6c, with A sitting at the x=0 ls mark on his ruler and A' sitting on the x'=0 mark on his own ruler. Let's suppose two events happened simultaneously in A's frame, so at t=0 seconds, event L happened next to the x=-10 ls mark on his ruler, and event R happened next to the x=+10 ls mark on his ruler. Meanwhile, at this time observer A', sitting at x'=0 ls on his own ruler, is next to the x=-6 ls mark on the ruler of A. So in the frame of A, A' is 4 ls from L and 16 ls from R; but since the markings on the ruler of A' seem to be shrunk by a factor of 0.8 in the frame of A, that must mean event L happens next to the mark x'=4/0.8 = 5 ls on the ruler of A', and the event R happens next to the mark x'=16/0.8 = 20 ls on the ruler of A'. So, we see that according to the ruler used by A the events happened at equal distances from himself, and according to the ruler used by A' they happened at unequal distances from himself.

Now since A' is moving at 0.6c in the frame of A, in 10 seconds he moves 6 ls to the right; so since A' was at x=-6 ls at t=0 s in the frame of A, at t=10 s A' will have reached the position x=0 ls, where A is too. And since the events L and R each happened 10 ls away from A at t=0 s, then at t=10 s both must be reaching the position of A as well. So, both observers see the signals from both events reach them at the same moment, when their positions coincide. If A assumes each signal moved at the same speed, then since they happened at equal distances from him according to his ruler, this is consistent with the notion that the two events must have happened at the same time in his frame. But since A' measured the two events to have occurred at different distances from himself (according to his ruler, L happened 5 ls away and R happened 20 ls away), if he also assumes the signals traveled at the same speed, he must conclude the events happened at different times in his frame (L must have happened 5 seconds before the signal reached his eyes, while R must have happened 20 seconds before the signal reached him).

 

[JesseM]

This animation is clearly evidence that light is just like sound only that it has a different base. Sound is also constant and it is the basic physical phenomenon that gives us information about sound of the "flying block".

Sound waves only moves at the same speed s in all directions if you're in the rest frame of the medium (air) which the sound waves are vibrations in. If someone is moving at a speed v relative to this medium s, then in their own rest frame they'll measure sound waves to travel at s+v in one direction and s-v in the other, according to classical physics. Before relativity physicists used to imagine that light waves actually were vibrations in a medium called the luminiferous aether which was supposed to fill all of space, which led them to conclude that light would only be measured to move at the same speed in all directions when the Earth was at rest relative to this fluid, so that they could determine the rest frame of the aether by checking for differences in the speed of light in different directions when the Earth was at different points in its orbit. This was the famous Michelson-Morley experiment which gave the surprising result that no matter what point in the Earth's orbit the experiment was done, light always seemed to have the same speed in all directions. The failure of these types of aether experiments was one of the major inspirations for Einstein's theory of relativity, which postulated that the laws of physics were such that all observers could measure the speed of light to be the same in all directions, and that if they constructed their coordinate systems under this assumption they would find that all the fundamental laws of physics had the property of obeying the same equations in each coordinate system (which can only be true for laws that have a mathematical property known as 'Lorentz-invariance', which all the known fundamental laws seem to have).

Light barrier exists and it manifests in Cherenkov radiation and its blueish glow. The only thing we have to figure out is how to break it.

Cherenkov radiation is only observed when the speed of light is artificially slowed in a medium; the 'c' in the equations of relativity is always understood to be the speed of light in a vacuum, and no one has observed Cherenkov radiation in a vacuum.

 

[neopolitan]

No, the one in motion will say that the ticks happened at two different times, because on his ruler they happened next to markings which are at unequal distances from him, so the only way he can account for the fact that they both signals reached him at the same time while still assuming they traveled at the same speed is to conclude they happened at different times.

Ok, taking this to be the the situation where the observer at rest and the observer in motion are collocated (the only time when the observer in motion will receive signals at the same time), I agree. Since the photons reach him at the same time as he is equidistant from the clocks, and they are in relative motion, he must assume one of the following: it is he who is in motion rather than the clocks, the speed of light (in a vaccuum) is not a constant, the ticks actually were not simultaneous in his frame, or something funny is happening with length contraction based on his location relative to the clocks.

I am willing to reject the last one there, since it implies that the universe cares where the observer is. I am willing to reject the first, since our scenario doesn't specify which is in motion. I at least only talked about relative motion.

That leaves us with two options. Personally I am not inclined to suggest that speed of light in not a constant, I actually think it is but a better reason than "it is a postulate" or "Michelson and Morley didn't detect an aether wind". I'd be interested to hear what your justification is, if it is discussable (mine is possibly not discussable in this forum).

So I am left with one option, ticks are not actually simultaneous in the frame of the observer who is in motion relative to the clocks. I agree.

Is it permitted to go further than that and discuss how that can be the case? Is it an illusion? Is it because world line for each is skewed with respect to each other (and again is this something real or just a "lie to children")?

My conceptualisation is that the rocket is skewed a little in spacetime by virtue of its motion such that the forward end is a little in the future, relative to the rear end. Relative the rocket, of course, it is not in motion so the clocks at each end are simultaneous. Relative to another observer, not at rest relative to the rocket, the clocks are not simultaneous. Such an observer will see the front end first (so see the tick from that one first) and then the rear end (so the tick from this one will be seen second) - of course traveling times for the photons have to be accounted for, if the separation between a departing rocket's nose and tail are sufficiently large or the speed is sufficiently low, then the nose may be seen second but still sooner than it may otherwise would have been expected.

Is this way off?

cheers,

neopolitan

Note to moderators and other similar types: I have qualified this with "my conceptualisation", I am not claiming any more than that.

 

[JesseM]

Ok, taking this to the the situation where the observer at rest and the observer in motion are collocated (the only time when the observer in motion will receive signals at the same time), I agree. Since the photons reach him at the same time as he is equidistant from the clocks, and they are in relative motion, he must assume one of the following: it is he who is in motion rather than the clocks, the speed of light (in a vaccuum) is not a constant, the ticks actually were not simultaneous in his frame, or something funny is happening with length contraction based on his location relative to the clocks.

I am willing to reject the last one there, since it implies that the universe cares where the observer is. I am willing to reject the first, since our scenario doesn't specify which is in motion. I at least only talked about relative motion.

That leaves us with two options. Personally I am not inclined to suggest that speed of light in not a constant, I actually think it is but a better reason than "it is a postulate" or "Michelson and Morley didn't detect an aether wind". I'd be interested to hear what your justification is, if it is discussable (mine is possibly not discussable in this forum).

You're talking as if there is some objective truth about coordinate-dependent facts like whether the speed of light is the same in all directions, or whether two events at different locations are "really" simultaneous. But I don't see how experimental physics can give you objective truths about which coordinate system you "should" use. All that experiment can do is tell you that if observers construct their coordinate systems in some particular way, then the laws of physics expressed in that coordinate system will obey certain equations. The only physical claim of relativity is that if inertial observers construct their coordinate systems in the way described by Einstein, including the arbitrary axiom that the coordinate speed of light signals should always be c, then the equations for the fundamental laws of physics will look the same in all the coordinate systems constructed this way. This is a physical claim that could potentially be falsified by experiment. But even if it's true, that doesn't somehow prevent you from constructing the coordinate systems of inertial observers using a different rule, perhaps one under which the coordinate speed of light will be different in different directions, or one in which all inertial observers will have the same definition of simultaneity; it's just that when expressed in these coordinate systems, the equations describing the fundamental laws of physics will vary from one inertial coordinate system to another. But as long as you find the correct equations for whatever screwy coordinate system you use, you'll still get all the same predictions about coordinate-invariant physical facts (like what two clocks will read when they pass next to one another) as you would if you were using the standard inertial coordinate systems of SR.

So I am left with one option, ticks are not actually simultaneous in the frame of the observer who is in motion relative to the clocks. I agree.

Is it permitted to go further than that and discuss how that can be the case? Is it an illusion? Is it because world line for each is skewed with respect to each other (and again is this something real or just a "lie to children")?

Well, like I said above, I don't think that any coordinate-dependent statement is either "truth" or "illusion", it's just specific to your choice of coordinate system; the inertial coordinate systems related by the Lorentz transform are just the most "elegant" since the laws of physics have the same form in all these coordinate systems. But in terms of these coordinate systems, if you draw the lines of constant-coordinate-position and the lines of constant-coordinate-time (lines of constant time are also called 'surfaces of simultaneity') for two different frames in the same minkowski diagram, you do see that the constant-t lines defining a set of simultaneous events for each coordinate system are skewed relative to one another.

My conceptualisation is that the rocket is skewed a little in spacetime by virtue of its motion such that the forward end is a little in the future, relative to the rear end.

That's a decent way of thinking about it, although you actually have it backwards--as seen in a frame where the rocket is moving, at any given moment in this frame, the reading on the clock at the back end will be ahead of the reading on the clock at the front end at the same moment...the amount it's ahead is vx/c^2, where v is the speed of the rocket in this frame, and x is the separation between the two clocks in the rocket's own rest frame (so the separation between the clocks is actually smaller in the frame where it's moving)

Relative the rocket, of course, it is not in motion so the clocks at each end are simultaneous.

Yes, if the clocks are synchronized in the rocket's rest frame, then they both show the same reading simultaneously (at the same time coordinate) in the rocket's rest frame.

Relative to another observer, not at rest relative to the rocket, the clocks are not simultaneous. Such an observer will see the front end first (so see the tick from that one first) and then the rear end (so the tick from this one will be seen second) - of course traveling times for the photons have to be accounted for, if the separation between a departing rocket's nose and tail are sufficiently large or the speed is sufficiently low, then the nose may be seen second but still sooner than it may otherwise would have been expected.

Which clock an observer would see as showing the more advanced time could depend on whether they were in front of the rocket (so it was coming towards them) or behind it (so it was moving away), I think. But if you have two observers, one ahead and one behind, and both are at rest with respect to each other (so they share the same rest frame) and measure the rocket moving at speed v relative to themselves, then although they will see different things, as long as they compensate for signal delays when they calculate what time-coordinate a given tick actually happened at in their frame, then they will both come to the conclusion that the back clock's reading is ahead of the front clock's reading by the same amount (the vx/c^2 I mentioned earlier).

 

[neopolitan]

That's a decent way of thinking about it, although you actually have it backwards--as seen in a frame where the rocket is moving, at any given moment in this frame, the reading on the clock at the back end will be ahead of the reading on the clock at the front end at the same moment...the amount it's ahead is vx/c^2, where v is the speed of the rocket in this frame, and x is the separation between the two clocks in the rocket's own rest frame (so the separation between the clocks is actually smaller in the frame where it's moving).

We always seem to get stuck at these little points.

The nose is in the future, the tail in the past. Therefore you see the tail before the nose, because it's been there longer. Yes, the clock on the tail will appear to be ahead of the clock on the nose from a frame which is not at rest relative to the rocket, if the clocks are simultaneous in the rocket's frame.

Think about clocks which are simultaneous in the rocket's frame both saying, for instance, 1 minute past Jesse o'clock. If I am in a frame which is not at rest relative to the rocket, then (taking into account the time taken for the photons to travel), I will see 1 minute past Jesse on the tail first, because I see things in this order "the past, now, the future", rather "the future, now, the past". I have to wait a little until I catch up to the nose of the rocket (time wise) before I see 1 minute past Jesse o'clock on the nose.

So, as you say, yes - the clock on the tail will run ahead of the clock on the nose as observed by an observer not at rest relative to the rocket. And like I said, the clock on the nose is a little in the future relative to the one on the tail.

Do you see what I mean?

cheers,

neopolitan

 

[neopolitan]

You're talking as if there is some objective truth about coordinate-dependent facts like whether the speed of light is the same in all directions, or whether two events at different locations are "really" simultaneous.

Glad you have an "or" there.

There is no real simultaneous, just "simultaneous in a frame".

Are you saying that the speed of light in a vacuum isn't the same in all directions? Let's get more specific, do you think there is any way to measure the speed of light in a vacuum so that it is not a constant?

If so, I'd be interested to hear it from you. If not, what's the real issue here?

Note that to the best of my knowledge I didn't suggest any screwy coordinate system designed to make the speed of light different in different directions, from what I can tell that was your own personal strawman.

cheers,

neopolitan

 

[PatPwnt]

That is true... there is no absolute 0 velocity. You are always moving with respect to something, but at rest with yourself. Your simulaneity will be different from any other observers not at rest with you. Simultaneous is meaningless when you have different time frames.

The speed of light in a vacuum is the same in all directions no matter what frame you are in. This is why it is different from sound. If I am traveling at the speed of sound in air, then the velocity of air will be 0 to me in the direction I am going. This is not the case with light. Even if I go .99 the speed of light, light will still speed away from my perspective at c.

 

[JesseM]

Glad you have an "or" there.

There is no real simultaneous, just "simultaneous in a frame".

Are you saying that the speed of light in a vacuum isn't the same in all directions? Let's get more specific, do you think there is any way to measure the speed of light in a vacuum so that it is not a constant?

To use your phrasing above, there is no speed (of anything, not just light), just "speed in a frame". If I were to first synchronize my array of clocks using the Einstein method, but then move each clock forward or backwards based on their distance from me (for instance, a clock one meter to my right could be set one second forward, two meters to my right two seconds forward, one meter to my left one second backward, etc.), then this array of clocks with the altered times would define a new coordinate system with a different definition of simultaneity than my SR rest frame, and in this new coordinate system the one-way speed of light wouldn't be the same in both directions.

But I should point out that as long as the time-coordinate at a single location in my frame moves forward at the same rate as a normal physical clock at rest at that location, and distance in my frame is measured using rulers at rest in my frame, then the average two-way speed of light will always be c even if the one-way speed of light is different than c thanks to an odd clock synchronization scheme. Measuring the average two-way speed of light just requires a single clock (I can send a light signal away from the clock, bounce if off a mirror at a known distance, and divide the total distance it travels to the mirror and back to the clock by the difference in readings on the clock between the time the signal left and the time it returned), so synchronization schemes don't matter here. Even in this case, though, if I wanted to be contrarian I could define my coordinate system's time coordinate in such a way that a normal clock at rest relative to me was not ticking at the same rate coordinate time was advancing (or even where the coordinate time was such that the rate of ticking of such a clock is varying), and then the two-way speed of light in my coordinate system might not be c either.

Note that to the best of my knowledge I didn't suggest any screwy coordinate system designed to make the speed of light different in different directions, from what I can tell that was your own personal strawman.

It wasn't a strawman because I never accused you of doing so, I was just making a point that the speed of light is coordinate-dependent, and that it's only constant if inertial observers construct their coordinate systems in the way Einstein suggested. Again, this choice of coordinate systems is the most "elegant" since it ensures that each observer will have the same equations for the laws of physics, but I don't think that coordinate-dependent statements in these systems are somehow more "objectively true" than coordinate-dependent statements in more "screwy" coordinate systems.

 

[neopolitan]

Jesse

I knew once I had posted #12, that I should not have.

#11 is the post I really want a reply to.

Any chance of that?

 

[JesseM]

The nose is in the future, the tail in the past. Therefore you see the tail before the nose, because it's been there longer.

I don't know what you mean here--been where longer? Yes, in your frame the tail clock will hit a given time before the nose clock hits it, but I don't see the connection between this and "the nose is in the future, the tail in the past".

Think about clocks which are simultaneous in the rocket's frame both saying, for instance, 1 minute past Jesse o'clock. If I am in a frame which is not at rest relative to the rocket, then (taking into account the time taken for the photons to travel), I will see 1 minute past Jesse on the tail first, because I see things in this order "the past, now, the future", rather "the future, now, the past".

Don't get what you mean by "order"--if you instead saw a given time on the head first, are you saying that would mean you'd be seeing things in the order "the future, now, the past"? Regardless of which clock is ahead, you'll always see any given clock show earlier readings before later ones, so doesn't that mean you're always seeing things in the order "past, now, future"? Or are you talking about the spatial "order" when you look at different clocks on the ship from back to front, rather than the temporal order of readings on any given clock? Or something else entirely?

So, as you say, yes - the clock on the tail will run ahead of the clock on the nose as observed by an observer not at rest relative to the rocket. And like I said, the clock on the nose is a little in the future relative to the one on the tail.

Do you see what I mean?

I don't at all see by what you mean by "X is in the future relative to Y"--I can't think how this phrase would make sense as anything other than idea that X showed a later date than Y, rather than an earlier one--but if you agree the clock on the tail is ahead at of the one at the nose at a given moment in the observer's frame, I suppose it doesn't really matter how you choose to conceptualize this.

 

[neopolitan]

While my description at #11 was in part flippant, I am still keen to hear if anyone else can understand what I mean.

It could just be that my description was unclear. It could be that I am conceptualising things a different and difficult way. It could be that you have a fundamental misunderstanding of how times works. Or it could be no more than some sort of willful (albeit most likely subconscious) incomprehension on your part, Jesse, since at the end of your post, you seem to grudgingly accept that what I have to say has some sort of twisted validity anyway. I am not totally sure where the problem lies.

So ... let's see if anyone else reading this thread can work out what I mean before I have another hack at producing an explanation which might meet with your official approval.

 

[Dale]

I know what you mean, but I don't think it is an important argument. I can say "New York is an hour earlier than Chicago" meaning something like the sun rises in NY an hour before it rises in Chicago, or I can say "New York is an hour later than Chicago" meaning that the clock in NY shows 08:00 when the clock in Chicago shows 09:00. I don't think your post 11 is saying anything more important than that. It isn't a confusion about the math or physics, just an ambiguity in the english. Not worth arguing.

There is no real simultaneous, just "simultaneous in a frame".

I think this is the key. The point is that simultaneity is an artificial construct arising from the definition of a coordinate system, not something objectively real in its own right. Fundamentally it appears that the universe doesn't care about simultaneity, only about causality. Two simultaneous events cannot be causally connected, so what does it matter if one happened before the other? On the other hand, a cause should always come before an effect, and this is exactly what we see in relativity. A cause will preceed the effect in all reference frames, and for the rest it doesn't really matter.

 

[neopolitan]

Dale,

It could be just the English, but the way I see it the nose end of the rocket will be more in the future, due to its being skewed in spacetime by virtue of its speed (relative to an observer who is not at rest relative to the rocket).

Possibly Jesse is confused because I used the word "relative" in an attempt to indicate that the nose is more in the future than the tail is (so in the future relative to the tail). I can see how this could be confusing.

But his understanding seems to be the reverse - that the nose is more in the past (because the nose clock reads an earlier time than one at the tail).

For me, this is just plain wrong, but it could be a matter of perspective. The question then is whose perspective is more valid. I feel that Jesse's perspective almost presupposes absolute time, linked to clocks.

Clocks don't tell you "when" you are any more than odometers tell you where you are. They just tell you how much time has elapsed. He seems to think that when you are in time is related to what your clock says.

What I am saying instead is that while the rocket is in motion, the nose will reach the instant when it is observed before the tail reaches that same instant (relative to an observer who is not at rest relative to the rocket). It therefore could be said to be "in the future" - although of course when it is observed it is "in the now".

cheers,

neopolitan

 

[JesseM]

Or it could be no more than some sort of willful (albeit most likely subconscious) incomprehension on your part, Jesse, since at the end of your post, you seem to grudgingly accept that what I have to say has some sort of twisted validity anyway. I am not totally sure where the problem lies.

How did I grudgingly accept it? I just said that ultimately, how you conceptualize it doesn't matter as long as you get the right answers. But I still don't understand how you're conceptualizing it. Why does older = further in the past and younger = further in the future for you? Can you try to explain again?

 

[Dale]

Neopolitan, you are arguing over really unimportant things now compared to your initial question.

There are two important points in this thread. 1) in SR all observers are intelligent, meaning that they correct for the light propagation delay to figure out when an event really occurred not merely when it was observed. 2) when they do that they find that they disagree on wether or not two distant events really occurred simultaneously.

I'm not going to engage you in an unimportant debate about semantics.

 

[JesseM]

I know what you mean, but I don't think it is an important argument. I can say "New York is an hour earlier than Chicago" meaning something like the sun rises in NY an hour before it rises in Chicago, or I can say "New York is an hour later than Chicago" meaning that the clock in NY shows 08:00 when the clock in Chicago shows 09:00. I don't think your post 11 is saying anything more important than that. It isn't a confusion about the math or physics, just an ambiguity in the english. Not worth arguing.

OK, this kind of helps me see what neopolitan might have been talking about actually, although "New York is an hour later" (i.e., it's later in realtime to reach the same clock time) is a little different from "New York is an hour in the future", the latter just seems like really confusing phrasing if New York actually shows an earlier time than Chicago. Neopolitan, is this basically what you meant? If so, would you agree it's just a matter of ambiguity in language, that neither version is "more correct" than the other?

 

[JesseM]

But his understanding seems to be the reverse - that the nose is more in the past (because the nose clock reads an earlier time than one at the tail).

For me, this is just plain wrong, but it could be a matter of perspective. The question then is whose perspective is more valid. I feel that Jesse's perspective almost presupposes absolute time, linked to clocks.

Clocks don't tell you "when" you are any more than odometers tell you where you are. They just tell you how much time has elapsed. He seems to think that when you are in time is related to what your clock says.

I'm not presupposing absolute time, I'm comparing the opinions about simultaneity of two different frames. In the ship's own frame, both clocks show the same reading at the same time, i.e. simultaneously. In that frame (call it frame A), the event of the tail clock reading 4 seconds would be one second in the future of the event of the nose clock reading 3 seconds. So, in the frame where the ship is moving forwards (call it frame B), if the event of the tail clock reading 4 seconds and the nose clock reading 3 seconds are simultaneous, then in this frame one can observe, in a single moment, a reading on the tail clock that is "in the future" of the reading on the nose clock as understood in frame A. That's all I meant! One could find another frame (call it frame C) in which the rocket is moving backwards, and in this frame the event of the tail clock reading 4 seconds might happen earlier than the event of the nose clock reading 3 seconds. So, going back to frame B, in frame B one can observe, in a single moment, a reading on the tail clock that is "in the past" of the reading on the nose clock as understood in frame C. There's obviously no absolute truth about whether one event is "really" in the future or the past of another event (unless one event lies in the other event's future light cone, in which case all frames agree on the order), all we can do is talk about the opinions of different frames, and perhaps relate them to one another as I do above.

What I am saying instead is that while the rocket is in motion, the nose will reach the instant when it is observed before the tail reaches that same instant (relative to an observer who is not at rest relative to the rocket). It therefore could be said to be "in the future" - although of course when it is observed it is "in the now".

What do you mean by "reaching an instant"? By instant do you mean something different than just a given reading on each clock? After all, if we pick any give reading--say, 3 seconds--then in the frame B where the rocket is moving forward, the tail will get to that reading first, not the nose, so I don't see what you mean by "the nose will reach the instant when it is observed before the tail reaches that same instant".

 

[neopolitan]

re DaleSpam and arguments about semantics - it is possible that what I am thinking about it not merely semantics. I don't know what is inside your head and, I hope, you don't know what is inside mine. I am trying to gain a better understanding of whether what Jesse and I are saying are the same thing or not.

re JesseM's reply to DaleSpam - it could be an ambiguity of language, but I have the benefit of being biligual (English is my mothertongue though) and know that sometimes limitations in language make it impossible to convey certain concepts. It is possible that what I am trying to express is being converted in your mind to precisely what you originally thought, but isn't what I meant. It seems it is just you, me and Dale - so I will have another crack at explaining.

 

[neopolitan]

What do you mean by "reaching an instant"? By instant do you mean something different than just a given reading on each clock? After all, if we pick any give reading--say, 3 seconds--then in the frame B where the rocket is moving forward, the tail will get to that reading first, not the nose, so I don't see what you mean by "the nose will reach the instant when it is observed before the tail reaches that same instant".

An instant I mean as a "line of simultaneity", the events which share the same value of t in the frame to which it pertains.

So I am thinking of the rocket's frame, in which the rocket is not moving, and an observer's frame, in which the rocket is moving. An instant in the observer's frame is the set of events which share the same value of t. This "observer's instant" is not an instant in the rocket's frame, but a set of events which we can work out using lorentz transformations.

An instant in the rocket's frame, in which clocks at the nose and the tail read the same, is similarly not an instant in the observer's frame.

If we pick any instant in the observer's frame, and look at the rocket, we will see that clock on the nose reads less than the clock on the tail. We agree about this.

What I am saying is that the nose reaches any given observer's instant before the tail.

Imagine that we sit in two separate time machines, machines that shunt us into the future at faster rate than normal life (like fun events seem to do). We each have a watch, and we synchronise them before we switch our machines on.

If my machine shunts me into the future twice as quickly than yours, which one of us will have more time on their watch? I put it to you that the one who reaches the future first will have less time on their watch (that means me).

I agree that if we were worried about who is able to say 5 minutes have elapsed on their watch first, then that will be you. If we both turn off our machines when five minutes have elapsed on our watches (inside the time machines), you will have to wait around a while for me to turn my machine off, and will be able to say that your watch read 5 minutes first. But ... I will have gone further into the future than you.

Now, this was just an explanation, I am not suggesting that such time machines are possible. 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.

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.

cheers,

neopolitan

 

[JesseM]

So I am thinking of the rocket's frame, in which the rocket is not moving, and an observer's frame, in which the rocket is moving. An instant in the observer's frame is the set of events which share the same value of t. This "observer's instant" is not an instant in the rocket's frame, but a set of events which we can work out using lorentz transformations.

An instant in the rocket's frame, in which clocks at the nose and the tail read the same, is similarly not an instant in the observer's frame.

If we pick any instant in the observer's frame, and look at the rocket, we will see that clock on the nose reads less than the clock on the tail. We agree about this.

What I am saying is that the nose reaches any given observer's instant before the tail.

But as you point out, different observers have different "instants"...whose are you talking about above? If we talk about a given instant (line of constant t) in the rocket frame, then as seen in the outside observer's frame, the clock on the tail will intersect with this line before the clock on the nose. Maybe you're taking the opposite perspective, and considering a line of constant t in the observer's frame, and then pointing out that, in the rocket's frame, the clock on the nose will reach a given point on this line before the clock on the tail? If so let me know, this would help clear up my confusion about what you're trying to say (and if it is what you're saying, would you agree either perspective is equally valid?)

Imagine that we sit in two separate time machines, machines that shunt us into the future at faster rate than normal life (like fun events seem to do). We each have a watch, and we synchronise them before we switch our machines on.

If my machine shunts me into the future twice as quickly than yours, which one of us will have more time on their watch? I put it to you that the one who reaches the future first will have less time on their watch (that means me).

I don't think that's a good analogy, because no matter what frame you pick, the rate at which each clock is advancing forward in that frame will be identical, it's not like one is being "shunted into the future" faster, just that one has a "head start" of sorts.

Now, this was just an explanation, I am not suggesting that such time machines are possible. 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.

Like I said, there's no difference in the rate no matter what frame you pick, so I would say "faster". But if my guess above is right, do you mean that in the rocket's frame at any given instant, the time on the nose-clock at that instant would be further in the outside observer's future than the time on the tail-clock at the same instant? In other words, if both clocks show a time of 3 seconds simultaneously in the rocket frame, we can note that in the outside observer's frame, the event of the nose clock showing 3 seconds happens further in the future than the event of the tail clock showing 3 seconds. So is that what you mean?

If so, I just want to note that even if we talk in this way, which clock is "further in the outside observer's future" depends on your choice of frame. We might take the frame of an observer who's moving relative to the first observer outside the rocket, but in the opposite direction as the rocket...in this new observer's frame, the time on the nose-clock at a given instant would be further in the first outside observer's past than the time on the tail-clock at the same instant.

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.

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.

As I said above, I think the problem with your analogy is that no matter which frame you choose both clocks tick at the same rate, and also that the question of which clock is further in the outside observer's future in the frame you're using will be different depending on the choice of frame (in the rocket's rest frame the nose clock will be, but in the frame of a third observer moving in the opposite direction relative to the outside observer the tail clock will be). But maybe I'm still not understanding what you're saying here, let me know.

 

[neopolitan]

But as you point out, different observers have different "instants"...whose are you talking about above? If we talk about a given instant (line of constant t) in the rocket frame, then as seen in the outside observer's frame, the clock on the tail will intersect with this line before the clock on the nose. Maybe you're taking the opposite perspective, and considering a line of constant t in the observer's frame, and then pointing out that, in the rocket's frame, the clock on the nose will reach a given point on this line before the clock on the tail? If so let me know, this would help clear up my confusion about what you're trying to say (and if it is what you're saying, would you agree either perspective is equally valid?)

We are discussing physics aren't we, not social work?

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".

Of course I am discussing the observer's instant. Why is that so hard for you to grasp?

I don't think that's a good analogy, because no matter what frame you pick, the rate at which each clock is advancing forward in that frame will be identical, it's not like one is being "shunted into the future" faster, just that one has a "head start" of sorts.

Like I said, there's no difference in the rate no matter what frame you pick, so I would say "faster". But if my guess above is right, do you mean that in the rocket's frame at any given instant, the time on the nose-clock at that instant would be further in the outside observer's future than the time on the tail-clock at the same instant? In other words, if both clocks show a time of 3 seconds simultaneously in the rocket frame, we can note that in the outside observer's frame, the event of the nose clock showing 3 seconds happens further in the future than the event of the tail clock showing 3 seconds. So is that what you mean?

Yes, that is what I mean. See how even if the analogy is not perfect (I knew it wasn't by the way) it did help you to understand? More perfect and it would be too complex, and you would have been running around point out other errors and uncertainties rather than concentrating on the issue I wished to convey.

If so, I just want to note that even if we talk in this way, which clock is "further in the outside observer's future" depends on your choice of frame. We might take the frame of an observer who's moving relative to the first observer outside the rocket, but in the opposite direction as the rocket...in this new observer's frame, the time on the nose-clock at a given instant would be further in the first outside observer's past than the time on the tail-clock at the same instant.

I guess I implied that you were willfully incomprehending so an implication here of stupidity on my part is fair enough. Your scenario is precisely the same as "what if the rocket was going backwards relative to the observer so that the tail was effectively the nose and the nose was effectively tail". Relabel nose and tail and we are back where we started, just as both of us expect. I am pretty sure that most of us understand that "nose" is the front bit and "tail" is the back bit. If you were thinking I had each part labeled permanently on the rocket so that the nose clock would behave the same way irrespective of the direction it traveled in relative to the observer then you did not read an earlier post properly.

Such an observer will see the front end first (so see the tick from that one first) and then the rear end (so the tick from this one will be seen second) - of course traveling times for the photons have to be accounted for, if the separation between a departing rocket's nose and tail are sufficiently large or the speed is sufficiently low, then the nose may be seen second but still sooner than it may otherwise would have been expected.

Perhaps this is too confusing for you, but if you think about it, I am discussing a rocket which is departing from the observer - a rocket which is going forwards since the nose is further away than the tail. If you want to change the scenario so that the tail is further away than the nose, fine, then I agree the tail will reach the future before the nose and the clock on the tail will read a lower elapsed time than the nose - if the rocket is in reverse relative to the observer.

As I said above, I think the problem with your analogy is that no matter which frame you choose both clocks tick at the same rate, and also that the question of which clock is further in the outside observer's future in the frame you're using will be different depending on the choice of frame (in the rocket's rest frame the nose clock will be, but in the frame of a third observer moving in the opposite direction relative to the outside observer the tail clock will be). But maybe I'm still not understanding what you're saying here, let me know.

Which analogy are you referring to? The explanation analogy with the time machines? If so, they have done their job. Forget the analogy and go back to trying to understand what I said originally.

You can try to describe the model now as requested, if you like.

cheers,

neopolitan

 

[JesseM]

We are discussing physics aren't we, not social work?

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".

Of course I am discussing the observer's instant. Why is that so hard for you to grasp?

If I've understood you correctly, you're discussing how the outside observer's instant is seen from the perspective of an observer on the rocket (i.e if we look at two simultaneous readings in the rocket-observer's frame, you're talking about which reading is in the future and which is in the past according to the outside observer's definition of simultaneity). I had been thinking you were talking about the perspective of an observer outside the rocket, and what time the two clocks show simultaneously in his frame. I think this is a pretty understandable confusion given that you didn't really spell out what you were talking about (DaleSpam also seems to have interpreted your comment as involving simultaneous readings in the rocket observer's frame rather than the rocket-observer's frame), so there's no need for condescending comments like "why is that so hard for you to grasp?"

Yes, that is what I mean. See how even if the analogy is not perfect (I knew it wasn't by the way) it did help you to understand?

No, it didn't. It was the previous comment about "instants" that helped me to understand (especially the comment 'An instant in the rocket's frame, in which clocks at the nose and the tail read the same, is similarly not an instant in the observer's frame', which clued me in that you might be talking about simultaneous readings in the rocket-observer's frame rather than the outside observer's frame), and you can see that I divined your meaning in the response to that earlier section. I hadn't even read the following paragraph when I fired off that response, and if your post had consisted only of the time travel analogy I don't think it would have helped me at all.

If so, I just want to note that even if we talk in this way, which clock is "further in the outside observer's future" depends on your choice of frame. We might take the frame of an observer who's moving relative to the first observer outside the rocket, but in the opposite direction as the rocket...in this new observer's frame, the time on the nose-clock at a given instant would be further in the first outside observer's past than the time on the tail-clock at the same instant.

guess I implied that you were willfully incomprehending so an implication here of stupidity on my part is fair enough.

I have to say neopolitan, I consider your constant attempts to mind-read my motives (inevitably in uncomplimentary ways) really disrespectful. I meant no implication of stupidity here, the fact that I "want to note" something doesn't even imply that I think you would disagree with it, and it certainly doesn't imply I think you're stupid. Anyway as seen below, I think you were actually leaping to incorrect conclusions about what I was saying in that comment.

Your scenario is precisely the same as "what if the rocket was going backwards relative to the observer so that the tail was effectively the nose and the nose was effectively tail".

No, actually, it's a little more complicated. In your scenario, there have to be two observers whose frames we refer to--the first observer A who sees the rocket going forward, and a second B on board the rocket. Unless there's an error in my interpretation of your words, what you're saying is that if we look at readings on the two clocks which are simultaneous in the frame of B, then the reading on the nose clock has a greater time-coordinate in the frame of A than the time-coordinate of the reading on the tail clock in the frame of A. In my scenario, it's still true that the rocket is moving forward relative to A, but I'm introducing a third observer C who is moving in the opposite direction as the rocket in A's frame (so this observer C also sees the rocket moving forward, at an even greater speed than in A's frame), and saying that if we look at readings on the two clocks which are simultaneous in the frame of C, then the reading on the tail clock has a greater time-coordinate in the frame of A than the time-coordinate of the reading on the nose clock. This is an entirely different scenario from imagining that the outside observer A sees the rocket moving backwards, and then looking at readings on the two clocks which are simultaneous in the frame of an observer B on board the rocket, and noting that in this case if we look at two readings which are simultaneous in the frame of B, then the reading on the tail clock (which is now effectively the nose clock in A's frame) has a greater time-coordinate in the frame of A than the time coordinate of the reading on the nose clock (which is now effectively the tail clock in A's frame) in the frame of A.

Perhaps this is too confusing for you, but if you think about it, I am discussing a rocket which is departing from the observer - a rocket which is going forwards since the nose is further away than the tail.

Well, I'm sure you would agree with this, but for the rocket to be "going forward" that means the nose is further away than the tail when the rocket is moving away from the observer A, but it also means the nose is closer than the tail when the rocket is moving towards the observer A (and at some moment they will be equidistant as the rocket passes A). Whether the rocket is moving towards A or away from A makes no difference to our statements about simultaneity, as long as the rocket is moving forward in both cases.

If you want to change the scenario so that the tail is further away than the nose, fine, then I agree the tail will reach the future before the nose and the clock on the tail will read a lower elapsed time than the nose - if the rocket is in reverse relative to the observer.

Again, that wasn't what I was suggesting. I was still suggesting a scenario where the rocket was going forward relative to A (if the rocket was moving away from A, the nose would be further than the tail), but where instead of then looking at readings of the rocket's clocks which occur simultaneously in the frame of an observer B on board the rocket, we instead look at readings on the rocket's clocks which occur simultaneously in the frame of an observer C who is moving in the opposite direction as the rocket in A's frame, so that in C's frame the rocket is moving forward at an even greater speed than in A's frame. In this case the reading on the tail clock will be "further in the future" in A's frame than the reading on the nose clock (where again, the two readings we're talking about were chosen to be simultaneous in C's frame).

As an example, suppose the rocket is 10 light-seconds long in its own rest frame, and in A's frame its moving forward at 0.6c, so in A's frame the tail clock's reading is ahead of the nose clock's reading by 6 seconds. Now choose an observer C who sees the rocket moving forward at 0.8c, so in C's frame the tail clock's reading is ahead of the nose clock's reading by 8 seconds. Pick two readings which are simultaneous in C's frame, like the tail clock reading 10 seconds and the nose clock reading 2 seconds. In A's frame when the nose clock reads 2 seconds, the nose clock reads 8 seconds and won't read 10 seconds until a later time, so the event of the tail clock reading 10 seconds is "further in the future" in A's frame than the event of the nose clock reading 2 seconds. In contrast, if we picked two readings which were simultaneous in the frame of the observer B on board the rocket, like the tail clock reading 3 seconds and the nose clock reading 3 seconds, then we'd find that the event of the nose clock reading 3 seconds is "further in the future" in A's frame than the event of the tail clock reading 3 seconds.

Which analogy are you referring to? The explanation analogy with the time machines? If so, they have done their job. Forget the analogy and go back to trying to understand what I said originally.

You can try to describe the model now as requested, if you like.

OK, but then I don't really understand what you were asking with this earlier comment that I was responding to:

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]

Ok, sorry if I offend. I am merely a little frustrated.

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).

I just don't quite understand how you are helping by adding more and more observers.

I accept that it is confusing if you add more and more observers, but that with time and patience you can work out what each observes.

Can we dispense with the third observer at the very least. I am not sure what you want to prove with that observer. I admit that I misread it at first. But I understand less now what your intention is with the introduction of that observer than I did when I thought you wanted to somehow make the rocket appear to go backwards.

I am going to leave this for a while, because I do find it rather frustrating.

neopolitan

 

[JesseM]

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).

I just don't quite understand how you are helping by adding more and more observers.

I accept that it is confusing if you add more and more observers, but that with time and patience you can work out what each observes.

But in relativity talking about "observers" is basically just shorthand for talking about what's true in different frames, the actual presence or absence of biological humans at rest in a particular frame doesn't affect your actual problem. It seems to me that your argument depends critically on using the definition of simultaneity in the frame where the rocket is at rest as well as the definition in the frame where it's moving; as far as I can tell what you're saying is that if we take two simultaneous readings in the rocket's rest frame B, then of those two readings, the one on the nose will be further ahead in time in the frame where the rocket is moving A (the one where you want the 'observer' to be) then the one on the tail. Is this wrong?

If not, my point in introducing a third frame C was just to show that the nose-reading being further in the future than the tail-reading in the frame A of the "observer" depends critically on the fact that you picked two clock readings which were simultaneous in the rocket's rest frame B; if you instead picked simultaneous clock readings in another frame C moving in the opposite direction relative to A (still talking about the two clocks on board the rocket, and without changing the motion of the rocket), then out of these two readings, the one on the tail will be further ahead in time in A's frame than the one on the nose. Assuming you agree with this point, then that was my only reason for introducing the third frame C, we don't have to discuss it further.

 

[neopolitan]

Jesse,

Please read the whole thing before replying. Please also avoid adding complicating factors until we have clarified what we currently have. No more observers, no different rockets, no different trajectories, no different clocks. Thanks.

I have four questions (or five, depending on how you want to define "question", but two are really only one question with two options), which I have color coded red. I would appreciate you making the effort to answer them.

But in relativity talking about "observers" is basically just shorthand for talking about what's true in different frames, the actual presence or absence of biological humans at rest in a particular frame doesn't affect your actual problem. It seems to me that your argument depends critically on using the definition of simultaneity in the frame where the rocket is at rest as well as the definition in the frame where it's moving; as far as I can tell what you're saying is that if we take two simultaneous readings in the rocket's rest frame B, then of those two readings, the one on the nose will be further ahead in time in the frame where the rocket is moving A (the one where you want the 'observer' to be) then the one on the tail. Is this wrong?

I am back at work so there is no longer any weekend to be spoilt by my getting tetchy.

I am fully aware that flesh and blood observers are not required. However the mechanism of only nominating one observer was intended to get around the problem we seem to have with you being confused about which frame's perspective I was talking about - I mean the one with an observer, the only observer I ever stipulated.

Despite this, you seem to want to observe things from the rocket, where I never specified there would be an observer, just two clocks. All we know is that the clocks are set up to be synchronous in their own frame, as you are most likely aware. I don't require that you take simultaneous readings of the clocks.

Restating: there is only one observer, the one in reference to whom the rocket is moving forward - nose first (and I initially said departing, but it doesn't really matter if it is approaching, it just may be easier to visualise a departing rocket).

If that observer observes the clocks, the nose clock will read less than the tail clock (so if the tail reads 13:55 for instance, then the nose may read 13:00).

(Question One) Can we agree on this simple point? No more new observers until we have done that please.

(I cannot answer "Is this wrong?" directly because I am not certain what you mean by "the nose will be further ahead in time". The best I can do is rephrase in the hope that my rephrasing answers your question.)

If not, my point in introducing a third frame C was just to show that the nose-reading being further in the future than the tail-reading in the frame A of the "observer" depends critically on the fact that you picked two clock readings which were simultaneous in the rocket's rest frame B; if you instead picked simultaneous clock readings in another frame C moving in the opposite direction relative to A (still talking about the two clocks on board the rocket, and without changing the motion of the rocket), then out of these two readings, the one on the tail will be further ahead in time in A's frame than the one on the nose. Assuming you agree with this point, then that was my only reason for introducing the third frame C, we don't have to discuss it further.

If you agree on the simple point above, it would be worthwhile to try to explain what you are getting at here, because I can't see the relevance of it. I also seem to be lost, since you have written the following in different posts #23 and #25 respectively.

One could find another frame (call it frame C) in which the rocket is moving backwards, and in this frame the event of the tail clock reading 4 seconds might happen earlier than the event of the nose clock reading 3 seconds.

We might take the frame of an observer who's moving relative to the first observer outside the rocket, but in the opposite direction as the rocket...in this new observer's frame, the time on the nose-clock at a given instant would be further in the first outside observer's past than the time on the tail-clock at the same instant.

From what I can work out, you have two frames called C. One which has a new observer moving such that the rocket appears to be moving backwards, and one which has another new observer such that the rocket appears to be moving but faster than my single stipulated observer.

Since #25 is most recent, I now assume that you want to talk about the latter. (Question Two) Is that correct? In that case, I erred in post #27 because I was still referring to the former.

I do hope you can understand that it is getting a little crowded in our scenario with all these observers.

Anyway, dispensing with the observer introduced in #23, we have two relative velocities for the rocket, with the same direction and magnitudes such that:

relative velocity according to observer A (my "there can be only one" observer) < relative velocity according to C (as introduced in#25)

(Question Three) Is that correct?

Assuming this is indeed correct, then you want to take simultaneous readings of the clocks in the C frame. Then I am lost, I don't quite know what you want to do with those readings.

(Question Four) Do you want to take "the nose clock reads t1 and the tail clock reads t2" and see when those readings are observed by my observer (observer A) and compare the order in which these readings appear in the A frame?

Or do you want look at readings of the clocks taken by observer A which are simultaneous according to observer C, but not according simultaneous to observer A?

In either case, I still can't see the relevance of the scenario.

I also don't quite know what you mean by

<snip> out of these two readings, the one on the tail will be further ahead in time in A's frame than the one on the nose.

If you mean that the clock on the tail will indicate that more time has elapsed than the clock on the nose indicates (for example tail clock time is 13:55 and nose clock time is 13:00), then I think I agree with you, but since I am not sure what measurements you want to take nor what "further ahead in time in A's frame" means to you, I can't be certain.

cheers,

neopolitan

 

[JesseM]

I am fully aware that flesh and blood observers are not required. However the mechanism of only nominating one observer was intended to get around the problem we seem to have with you being confused about which frame's perspective I was talking about - I mean the one with an observer, the only observer I ever stipulated.

Despite this, you seem to want to observe things from the rocket, where I never specified there would be an observer, just two clocks. All we know is that the clocks are set up to be synchronous in their own frame, as you are most likely aware. I don't require that you take simultaneous readings of the clocks.

I talked about the rocket frame simply because I can't understand how to interpret your claim about the nose clock being "further in the future" than the tail clock without referring to the rocket frame as well as the outside observer's frame. More on that below.

Restating: there is only one observer, the one in reference to whom the rocket is moving forward - nose first (and I initially said departing, but it doesn't really matter if it is approaching, it just may be easier to visualise a departing rocket).

If that observer observes the clocks, the nose clock will read less than the tail clock (so if the tail reads 13:55 for instance, then the nose may read 13:00).

(Question One) Can we agree on this simple point? No more new observers until we have done that please.

Sure, I have already made the point that in the observer's frame the tail clock is ahead of the nose clock in many posts before, this was the whole reason I was having trouble understanding your claim that the nose clock was "further in the future", remember?

(I cannot answer "Is this wrong?" directly because I am not certain what you mean by "the nose will be further ahead in time". The best I can do is rephrase in the hope that my rephrasing answers your question.)

OK. I said:

as far as I can tell what you're saying is that if we take two simultaneous readings in the rocket's rest frame B, then of those two readings, the one on the nose will be further ahead in time in the frame where the rocket is moving A (the one where you want the 'observer' to be) then the one on the tail. Is this wrong?

Take two simultaneous readings in the rocket frame, say the nose clock reading 13:55 and the tail clock reading 13:55. If we now switch to the frame of the outside observer, the event of the nose clock reading 13:55 actually happens at a later time-coordinate (further in the future) than the event of the tail clock reading 13:55--the tail clock reaches that time first in his frame, the nose clock reaches it later. So, this is how I interpreted your claim that the nose-clock was "further in the future" than the tail clock. If you think there is a way of making sense of that claim without referring to the rocket frame, then I still don't understand what you're saying.

If you agree on the simple point above, it would be worthwhile to try to explain what you are getting at here, because I can't see the relevance of it.

The relevance is:

1. The way I am interpreting your comment about the nose being "further in the future", it seems to depend on using both the outside observer's frame and the rocket frame's, since we're picking two simultaneous events on the rocket's clocks in the rocket's frame and noting that the event at the nose happens further in the future in the outside observer's frame

2. So, I just wanted to make the point that if we kept the outside observer and the rocket the same, but now used a different second frame in place of the rocket's frame, we could use the same argument to show that if we take two simultaneous events on the rocket's clocks in this new frame, then it could be that the event at the tail happens further in the future in the outside observer's frame.

Note that in all this, how the two physical clocks on the rocket are actually synchronized is pretty much irrelevant, we're just talking about what events on the clocks are simultaneous in a given frame. I'm assuming that your argument that the nose clock is "further in the future" doesn't depend on whether or not the two clocks on the rocket have actually been synchronized in the rocket's rest frame, does it? Even if in the rocket's rest frame they've been synchronized incorrectly and the event of the tail clock reading 13:55 is simultaneous with the event of the nose clock reading 19:22 in this frame, and in the outside observer's frame the clock at the nose has a greater reading than the clock at the tail as a result (as opposed to the clock at the tail having a greater reading as they would if the clocks were correctly synchronized in the rocket's frame), this wouldn't make any difference to your statement that the nose clock was further in the future, would it? If you say one clock is further in the future, it seems to me you're trying to say something a little more basic than just a statement about how the clocks have been set (for example, you wouldn't say clocks in the central time zone are further in the future than clocks in the easter time zone just because clocks in the eastern time zone are set one hour ahead, would you?)

I also seem to be lost, since you have written the following in different posts #23 and #25 respectively.

One could find another frame (call it frame C) in which the rocket is moving backwards, and in this frame the event of the tail clock reading 4 seconds might happen earlier than the event of the nose clock reading 3 seconds.

We might take the frame of an observer who's moving relative to the first observer outside the rocket, but in the opposite direction as the rocket...in this new observer's frame, the time on the nose-clock at a given instant would be further in the first outside observer's past than the time on the tail-clock at the same instant.

These two quotes aren't talking about the same thing at all. In the first quote, the context was that I hadn't yet developed a hypothesis about what you meant when you said the nose is more in the future than the tail--in an earlier post I had given a short explanation about why, when I thought about what it would mean to say one clock was more in the future, I would think it was natural to say the tail was more in the future, just because it shows a greater time in the observer's frame. Then in response you said "I feel that Jesse's perspective almost presupposes absolute time", so I explained that I wasn't saying the tail was further in the future in any absolute sense, and to illustrate this I pointed out that the question of which clock was further in the future (again according to my idea of the most natural interpretation of the phrase, which was the opposite of yours) would have the opposite answer if the outside observer was in a frame where the rocket was moving backwards...here's the full paragraph so you can review the context:

I'm not presupposing absolute time, I'm comparing the opinions about simultaneity of two different frames. In the ship's own frame, both clocks show the same reading at the same time, i.e. simultaneously. In that frame (call it frame A), the event of the tail clock reading 4 seconds would be one second in the future of the event of the nose clock reading 3 seconds. So, in the frame where the ship is moving forwards (call it frame B), if the event of the tail clock reading 4 seconds and the nose clock reading 3 seconds are simultaneous, then in this frame one can observe, in a single moment, a reading on the tail clock that is "in the future" of the reading on the nose clock as understood in frame A. That's all I meant! One could find another frame (call it frame C) in which the rocket is moving backwards, and in this frame the event of the tail clock reading 4 seconds might happen earlier than the event of the nose clock reading 3 seconds. So, going back to frame B, in frame B one can observe, in a single moment, a reading on the tail clock that is "in the past" of the reading on the nose clock as understood in frame C. There's obviously no absolute truth about whether one event is "really" in the future or the past of another event (unless one event lies in the other event's future light cone, in which case all frames agree on the order), all we can do is talk about the opinions of different frames, and perhaps relate them to one another as I do above.

In the second quote above the context was completely different. By that point I had developed a hypothesis about what you meant when you said it should be the nose that was farther in the future for the observer who sees the rocket moving forward. My hypothesis about your meaning, as I've explained earlier, involved taking two simultaneous events on the clocks in the rocket's frame, and then noting that of these two events, the event on the nose clock happens later in the outside observer's frame than the event on the tail clock. So again, I was making the point that you could keep the outside observer and the rocket the same, but now pick a different frame C to define simultaneous events on the two clocks on board the rocket, and if this frame C happens to be moving in the opposite direction as the rocket in the frame of the outside observer (which is why I said 'We might take the frame of an observer who's moving relative to the first observer outside the rocket, but in the opposite direction as the rocket'), then the result would be that the event on the tail clock happens later in the outside observer's frame than the event on the nose clock.

From what I can work out, you have two frames called C.

The second mention of a frame C was not meant to have any relation to the earlier mention of a frame C, sorry if using the same letters caused confusion, I don't think I remembered that I had used these labels earlier when I wrote the second quote.

Since #25 is most recent, I now assume that you want to talk about the latter. (Question Two) Is that correct? In that case, I erred in post #27 because I was still referring to the former.

Yes, that's correct, and now I can see that my reuse of the same letters was the cause of your confusion on that point, sorry about that.

relative velocity according to observer A (my "there can be only one" observer) < relative velocity according to C (as introduced in#25)

(Question Three) Is that correct?

Assuming you're talking about the velocity of the rocket in each frame, yes, that's correct. Another way of putting this is that in A's frame, an object at rest in C would be moving in the opposite direction as the rocket.

Assuming this is indeed correct, then you want to take simultaneous readings of the clocks in the C frame. Then I am lost, I don't quite know what you want to do with those readings.

I'm just doing something directly analogous with what I did in my guess about what you meant when you said the nose clock was more in the future. My guess was that you meant we take readings on the clocks which are simultaneous in the rocket's frame (for example, the tail clock reading 10 seconds and the nose clock reading 10 seconds), and then we see which event happens later in the observer's frame (here the nose clock will not read 10 seconds until after the tail clock has already read 10 seconds, because the tail clock is ahead in this frame), and whichever clock's reading happens later, that clock is "more in the future" for the outside observer. So I was just following almost the same procedure, except instead of picking two readings which are simultaneous in the rocket's frame, I was picking two readings on the rocket's clocks which are simultaneous in the frame C (like the tail clock reading 10 seconds and the nose clock reading 2 seconds in my example near the end of post #28), and noting that of these two readings, it's actually the tail clock reading that happens later in the observer's frame, so using the same meaning of "more in the future" it's now the tail clock that's more in the future for the outside observer.

(Question Four) Do you want to take "the nose clock reads t1 and the tail clock reads t2" and see when those readings are observed by my observer (observer A) and compare the order in which these readings appear in the A frame?

Or do you want look at readings of the clocks taken by observer A which are simultaneous according to observer C, but not according simultaneous to observer A?

I don't really see how these are distinct alternatives. I want to find two readings t1 and t2 on the nose and tail clock which are simultaneous in the frame of C but not in the frame of A, and look at the order in which these readings appear in the A frame. So I guess the answer is "all of the above"

In either case, I still can't see the relevance of the scenario.

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.

I also don't quite know what you mean by

<snip> out of these two readings, the one on the tail will be further ahead in time in A's frame than the one on the nose.

If you mean that the clock on the tail will indicate that more time has elapsed than the clock on the nose indicates (for example tail clock time is 13:55 and nose clock time is 13:00), then I think I agree with you, but since I am not sure what measurements you want to take nor what "further ahead in time in A's frame" means to you, I can't be certain.

Well, to see what I was referring two when I said "these two readings", look at the context:

if you instead picked simultaneous clock readings in another frame C moving in the opposite direction relative to A (still talking about the two clocks on board the rocket, and without changing the motion of the rocket), then out of these two readings, the one on the tail will be further ahead in time in A's frame than the one on the nose.

The "two readings" here refer to readings which are simultaneous in the frame of C; for example, in terms of my example near the end of post #28, the event of the tail clock reading 10 seconds happens simultaneously with the event of the nose clock reading 2 seconds, in the frame of C. And in this example, in the frame of the observer A, the event of the tail clock reading 10 seconds happens further ahead in time (at a later time-coordinate in A's frame) then the event of the nose clock reading 2 seconds, so that's what I mean when I say "out of these two readings, the one on the tail will be further ahead in time in A's frame than the one on the nose."

 

[neopolitan]

I think we have reached another good point. We both seem to have a better understanding of the other's point of view. I think we still disagree on some key points, but on most of the basics we seem to be in accord with each other. I hope you feel the same.

I don't really see how these are distinct alternatives. I want to find two readings t1 and t2 on the nose and tail clock which are simultaneous in the frame of C but not in the frame of A, and look at the order in which these readings appear in the A frame. So I guess the answer is "all of the above".

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?

For me this is quite obviously the case, since the skewing of the B frame is greater relative to C than it is relative to A.

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.

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?

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)?

cheers,

neopolitan

 

[neopolitan]

<snip> Fundamentally it appears that the universe doesn't care about simultaneity, only about causality. Two simultaneous events cannot be causally connected, so what does it matter if one happened before the other? On the other hand, a cause should always come before an effect, and this is exactly what we see in relativity. A cause will preceed the effect in all reference frames, and for the rest it doesn't really matter.

I was reviewing our thread and saw this.

I wonder what Jesse thinks of this, in reference to his third observer with a different perception of simultaneous.

Jesse was confused when I said things happen in this order "past, now, future" rather than "future, now, past". Is it easier if we call it "cause, process, result"? For example, a "cause" is me whacking a cue ball towards a pocket, the "process" is the cue ball having a rough approximation of inertial velocity across the table followed by the "result" which is cue ball in pocket (assuming my aim is true, I hit sufficiently hard and not too hard as to cause a rebound).

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?

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.

In the table's frame both clocks will remain synchronised for all events. Event one is relatively in the past since when event three happens the clocks may for example all read 10s as opposed to the 2s when event one happened - something that happened 8s ago.

We talked in earlier posts about being able to observe, simultaneoulsy in a frame which is not at rest with the rocket, the clock at the nose reading 2s and the clock at the tail reading 10s.

Let us then observe the table from such a frame in which where I whack the ball is "the nose" and the pocket is "the tail". I am not interested in the relative velocity of the cue ball, I can work that out myself thank you very much. I just want to consider my three events.

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 - in a frame which is not at rest with the table. From what DaleSpam says though, you can't select a frame where you see the cue ball in the pocket before you can see it is whacked.

This seems counterintuitive since I specified none of the following: the speed at which I hit the cue ball, the distance between the cue ball's start position and the pocket, or the relative velocity of the table in the frame from which it is observed (or the rest length of the rocket). I see no reason why I can't hit the cue ball so as to give it higher velocity, and do so when the clocks read 4s such that the ball is in the pocket at 10s. Then the observation from another frame will see the clock at the nose reading 2s (with the cue ball sitting there undisturbed) and the clock at the tail reading 10s (with the cue ball sitting there after being whacked).

This is where I think that the 2s cue ball at the nose is in the future - relative to where it "should" be since according to the observer it should be more in the past since cause should precede effect (result) - and the 10s cue ball at the tail is in the past - relative to where it "should" be. I think I can understand your perspective though since what we are observing, from a frame not at rest relative to the table, is a "past" event and a "future" event.

We tend to think of traveling to events rather than events traveling to us. For example in Sci-fi it is normally time travellers who travel to the past, rather than time-movers who bring to past to them. But in our example, I do think that we are considering something closer to the latter than the former.

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 past event is brought back to the past (now is in the past relative to the future).

To be honest, I don't expect many people to grasp this first time around. In any event, the (apparent) potential for violated causality may be a problem. It's all nice and simple when we just talk about clocks, not so easy when you add in chains of cause and effect.

Comments?

neopolitan

PS I had better admit to being a little naughty here. If you are looking for it, you can find in my scenario the reason why causality is most certainly not violated. I did want others to find it and point it out, but not so much that I am willing to risk being locked out by an overzealous moderator or give anti-relativity nuts something that may look like ammunition for their cause. Now I admit the reason is there, is anyone still willing to point it out?

 

[JesseM]

I was reviewing our thread and saw this.

I wonder what Jesse thinks of this, in reference to his third observer with a different perception of simultaneous.

Jesse was confused when I said things happen in this order "past, now, future" rather than "future, now, past". Is it easier if we call it "cause, process, result"? For example, a "cause" is me whacking a cue ball towards a pocket, the "process" is the cue ball having a rough approximation of inertial velocity across the table followed by the "result" which is cue ball in pocket (assuming my aim is true, I hit sufficiently hard and not too hard as to cause a rebound).

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.

Right, the only way these events could be simultaneous is if cue ball moved FTL. As long as there's no FTL in the universe, then all frames will agree on the order of causally-related events (which is why tachyons would cause serious problems for causality if they existed, see this recent thread).

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?

Right.

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.

In the table's frame both clocks will remain synchronised for all events. Event one is relatively in the past since when event three happens the clocks may for example all read 10s as opposed to the 2s when event one happened - something that happened 8s ago.

We talked in earlier posts about being able to observe, simultaneoulsy in a frame which is not at rest with the rocket, the clock at the nose reading 2s and the clock at the tail reading 10s.

Let us then observe the table from such a frame in which where I whack the ball is "the nose" and the pocket is "the tail". I am not interested in the relative velocity of the cue ball, I can work that out myself thank you very much. I just want to consider my three events.

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 - in a frame which is not at rest with the table. From what DaleSpam says though, you can't select a frame where you see the cue ball in the pocket before you can see it is whacked.

No, not only can you not see them in the wrong order, you can't see them simultaneously either. If events are simultaneous in one frame, then there is a spacelike separation between them, meaning that no matter what inertial coordinate system you use, if the spatial separation between them is $$\Delta x$$ and the temporal separation is $$\Delta t$$ in this coordinate system, then $$c^2 * \Delta t^2 - \Delta x^2 < 0$$...if you had $$c^2 * \Delta t^2 - \Delta x^2 > 0$$ then there'd be a timelike separation between them, and if $$c^2 * \Delta t^2 - \Delta x^2 = 0$$ then they have a lightlike separation. If events have a spacelike separation in one inertial frame's coordinates, they will have a spacelike separation in every inertial frame, and likewise with events that have a timelike or lightlike separation (in fact the value of $$c^2 * \Delta t^2 - \Delta x^2$$ is the same in every frame, this is the 'invariant spacetime interval'...for events with a timelike separation, it is just c^2 times the proper time elapsed on a clock that moves inertially between the events). For events with a spacelike separation, it is always possible to find an inertial frame where they are simultaneous, and also always possible to a pair of inertial frames that both say they were non-simultaneous but disagree on their order. Events with a timelike separation could be bridged by an object moving slower than light, events with a lightlike separation can only be bridged by something moving at exactly c, and events with a spacelike separation could only be bridged by a tachyon moving FTL, if such things existed.

This seems counterintuitive since I specified none of the following: the speed at which I hit the cue ball, the distance between the cue ball's start position and the pocket, or the relative velocity of the table in the frame from which it is observed (or the rest length of the rocket). I see no reason why I can't hit the cue ball so as to give it higher velocity, and do so when the clocks read 4s such that the ball is in the pocket at 10s. Then the observation from another frame will see the clock at the nose reading 2s (with the cue ball sitting there undisturbed) and the clock at the tail reading 10s (with the cue ball sitting there after being whacked).

This is where I think that the 2s cue ball at the nose is in the future - relative to where it "should" be since according to the observer it should be more in the past since cause should precede effect (result) - and the 10s cue ball at the tail is in the past - relative to where it "should" be. I think I can understand your perspective though since what we are observing, from a frame not at rest relative to the table, is a "past" event and a "future" event.

We tend to think of traveling to events rather than events traveling to us. For example in Sci-fi it is normally time travellers who travel to the past, rather than time-movers who bring to past to them. But in our example, I do think that we are considering something closer to the latter than the former.

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 past event is brought back to the past (now is in the past relative to the future).

To be honest, I don't expect many people to grasp this first time around. In any event, the (apparent) potential for violated causality may be a problem. It's all nice and simple when we just talk about clocks, not so easy when you add in chains of cause and effect.

Comments?

Well, tell me if I'm wrong, but it seems as if all your speculations here followed from the assumption that it would be possible for two causally-related events to be simultaneous in some frame (your comment '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'), when in fact that is not possible unless FTL exists, as I explained above (and if FTL exists, the problems this causes for causality in relativity are well-known).

PS I had better admit to being a little naughty here. If you are looking for it, you can find in my scenario the reason why causality is most certainly not violated. I did want others to find it and point it out, but not so much that I am willing to risk being locked out by an overzealous moderator or give anti-relativity nuts something that may look like ammunition for their cause. Now I admit the reason is there, is anyone still willing to point it out?

Can you specify what you're talking about here?