Whose Clock Slows Down in Relativity?

[yuiop]

In Euclidean geometry, you can compare lengths of any curves on the plane. In SR you can compare invariant interval along any timelike or spacelike paths. In your 'heuristic Euclidean metric' any conclusion you draw for curves that don't begin and end at the same points are likely wrong. This is an opportunity for major confusion, despite the 'teaching opportunity you describe.

I disagree that you can compare invariant intervals along any timelike paths. For example if we have one clock (A) that remains at rest in frame S and another clock (B) that moves at 0.8c relative to S, then according to an observer in S, less proper time has elapsed for clock B at any coordinate time t. Now if we switch to frame the rest frame of clock B, (S') then an observer at rest in frame S' will say less proper time has elapsed for clock A relative to clock B at any coordinate time t'. In other words asking "whose clock is really slower?" for two clocks that are not initially and finally co-located is meaningless even in the purely SR/Minkowski context. While B is still going away from A, which clock would end up with the least elapsed proper time when they come back together again, depends entirely upon which clock turns around and accelerates towards the other and without a crystal ball to predict the future, we do not know which clock that will be until they are actually alongside each other again. This is a very important point is the context of the main question originally posed in this thread.

 

[yuiop]

However, let's assume that there exist instances whereby we desire calculations that require no transformations between the 2 systems ... eg the one yuiop desires. OK, so we ignore the fact that time axes are imaginary, and inquire as to the length of the spacetime interval, per the stationary POV. We then attain this metric ... (ct)²+(vt)².

I am not sure why you are fixated on time axes being imaginary, when we can do the calculations for timelike paths without any imaginary or complex values. Sure, imaginary quantities can turn up when considering spacelike intervals, but if we restrict ourselves to the subject of this thread (clock rates) then this does not apply. I am also not sure why you think it is impossible to do transformations to different POVs. If we take two paths in a given reference frame where the path with the least proper time has the greatest Euclidean 4d length (ct)²+(vt)² and use the regular Lorentz transformation to obtain the POV of an observer at rest in a different reference frame, then it is still true that the path with the least proper time has the greatest Euclidean 4d length.

Now, we inquire as to the physical meaning of (ct)²+(vt)². The soln would represent the distance traversed thru euclidean 4-space, per only the stationary POV. Next we ask, of what benefit might this be? I can think of none, personally. Yuiop has pointed out that it may make the determination as to who aged the least (vs most) more convenient.

It is the POV of a single inertial observer, but as mentioned above we can transform to the POV of any inertial observer and the inequality still holds. Let us say we have clocks 1 and 2 with paths (\Delta t1,\Delta x1) and (\Delta t2,\Delta x2) respectively. According to any inertial observer, if the inequality (\Delta t2)^2+(\Delta x2)^2 > (\Delta t1)^2+(\Delta x1)^2 then it always holds that the inequality \Delta \tau 2 > \Delta \tau 1 is also true (given the 3 conditionals* that I gave earlier). Although the Euclidean 4d calculation is qualitative rather than quantitative (as PAllen pointed out) it is instantly visually obvious which is the path with the least proper time because any student already knows that the combined lengths of any two sides of a triangle is greater than the length of the third side and that the shortest spatial distance between any two points is a straight line. Here they can use their Euclidean intuition and then apply the relativistic "twist".

Are there any disadvantages ... The magnitude of this 4d distance vector would pertain only to the stationary POV, it cannot be measured, and can only be predicted. There would exist no way of confirming the prediction correct, nor would we ever expect it to. It surely does not represent clock readings of either system. We'd now have a new (extra) description of the spacetime interval, a frame dependent solution, which would differ from the related invariant soln of Minkowski. Lengths of worldlines would no longer (numerically) equal the proper time experienced in all cases, and the meaning of a worldline length would likely become confused.

Sure we would a different name for Euclidean 4d length other that worldline length or spacetime interval to avoid confusion. I disagree that it cannot be measured. Many so called measurements are not direct measurements but rather are calculations, e.g. velocity is not always a direct measurement but a calculation of distance versus time and kinetic energy etc. I agree that the Euclidean 4d distance is not frame invariant, but as an logical (true/false) inequality it is invariant.

It seems inappropriate IMO, that within the same theory we might consider axes as imaginary in some circumstances and real in others. The imaginary axes are required for transformation (ie rotation) between the 2 systems, given the 2 postulates true.

Again, I think you need to elucidate on why you think time is imaginary and why imaginary axes are required for transformation.

*As for the 3 conditionals I gave earlier, these are not over and above the conditionals required by SR for real clocks. (By "real clocks" I mean clocks with non zero real rest mass rather than imaginary or complex rest mass) These conditionals are required by SR too.

 

[PAllen]

I disagree that you can compare invariant intervals along any timelike paths. For example if we have one clock (A) that remains at rest in frame S and another clock (B) that moves at 0.8c relative to S, then according to an observer in S, less proper time has elapsed for clock B at any coordinate time t. Now if we switch to frame the rest frame of clock B, (S') then an observer at rest in frame S' will say less proper time has elapsed for clock A relative to clock B at any coordinate time t'. In other words asking "whose clock is really slower?" for two clocks that are not initially and finally co-located is meaningless even in the purely SR/Minkowski context. While B is still going away from A, which clock would end up with the least elapsed proper time when they come back together again, depends entirely upon which clock turns around and accelerates towards the other and without a crystal ball to predict the future, we do not know which clock that will be until they are actually alongside each other again. This is a very important point is the context of the main question originally posed in this thread.

Invariant means all observers agree on it, independent also of coordinate system. Otherwise it isn't invariant. All observer's agree on the interval along a world line: that a give particle decays or not, that a person dies of old age or not, what a clock reads along a world line. They may perceive that a clock on some world line goes a different rate than their own, they will radically disagree on which points of different world lines are 'simultaneous', but there is never any disagreement on what the clock on some world line does between two physical events. This is fundamental to both SR and GR.

Colocation is irrelevent. Suppose you are traveling at .99c from start s1 to s2. You send me a picture of yourself at s1, and send me another picture when you get to s2. However long it takes me to get these signals, I will agree on how much aging you will experience (as long as I know the distance from s1 to s1 as I would measure it, and your speed). You will experience the aging I compute. Further, if you know my start and end points as you would measure them, and my speed as you measure it, and compute the invariant interval along my path in these coordinates, you will get the same longer age that I experience. The explanation of why you agree on my invariant interval while still seeing my clock going slow is that you radically disagree on simultaneity. The event on my worldline I say is simultaneous to your passing s1 is not at all what you would say.

Let's make this last point more concrete. Let's say I blow up an H-bomb at t1 and t2 on earth. When I get your signals sent from s1 and s2, as I inerpret where s1 and s2 are, factoring in light delay, I find (miraculous coincidence) the you sent your singal from s1 at t1, and from s2 at t2. I compute both intervals and find you aged much less, consistent with the pictures you sent.

You do the same thing. Only you find that t1 and t2 are not remotely simultaneous with when you were at s1 and at s2. However, taking into account when they occurred as you see them, and where I was at t1 and t2, as you see it, would would compute the same age difference for me as I actually experienced.

 

[GrayGhost]

I am not sure why you are fixated on time axes being imaginary, when we can do the calculations for timelike paths without any imaginary or complex values. Sure, imaginary quantities can turn up when considering spacelike intervals, but if we restrict ourselves to the subject of this thread (clock rates) then this does not apply.

Hmmm. Well, we've been discussing the length of the spacetime interval as defined under the Minkowski model. OK then, are you suggesting here that imaginaries do not apply wrt clock rates?

Seems to me that you might not understand the meaning of complex systems. I do realize that Einstein did not use imaginaries in his OEMB. But Minkowski recognized that the Einstein model was equivalent to a relative rotation if the time axes were assumed orthogonal wrt 3-space. Remember, we begin with a spacetime illustration with imaginary time axes. Minkowski did not designate these as such for no good reason. They in fact were necessary, if Einstein's kinematic scenario is to be modeled consistent with his LT solns.

Where ict' = s, we start with this vector equation ...

(ict')² = (ict)²+(vt)²​

And we end up with this vector equation ...

(ct')² = (ct)²-(vt)²​

which reduces to ...

t' = t(1-v²/c²)^1/2​

If you think that the 2nd equation above exists regardless as to whether the i's exist in 1st eqn or not, you are mistaken. The 2nd eqn exists only after applying the Pathorean theorem, which itself requires the squaring of imaginary vectors (and where i² = -1) ... physically, the ict-axis (or likewise the ict'-axis) is rotated 90 deg into a real 3-space plane. It is this rotation that allows the LTs to result in the precise way they did.

That said, it is impossible to obtain this ...

(ct')² = (ct)²+(vt)²​

if starting from this ...

(ict')² = (ict)²+(vt)²​

On the other hand, one can obtain this last metric if one ignores the fact that the ict time axis is imaginary (as you wish to), because then no rotations/transformations are done. That is, ict' would not then be indicative of either the moving (or even the stationary) frame's measure of space or time. The resultant value of ict' no longer represents the time of anything real, and would represent a distance thru spacetime that cannot be measured or verified by anyone. I realize what you are trying to do yuiop, however if done, I see more losses than gains, IMO.

GrayGhost

 

[bobc2]

Hmmm. OK then, so you claim here that imaginaries do not apply wrt clock rates.

Seems to me that you may not understand the meaning of complex systems. Remember, we begin with a spacetime illustration with imaginary time axes. Minkowski did not designate these as such for no good reason. They in fact were necessary, if Einstein's kinematic scenario is to be modeled consistent with his LT solns.

GrayGhost, I always am impressed with your posts and do not wish to argue against the very good presentation you've made on this. But here is the derivation I've done before (sorry I don't remember the link). This is just to show there is an alternative derivation that begins with a 4-dimensional spacetime--just to show it is possible to develop the Minkowski metric without resorting to an imaginary axis. You can always put one in, just because any parameter or variable shown as a negative squared quantity, i.e., -X^2, can be represented with an imaginary number arbitralily inserted, i.e., -X^2 = (iX)^2. After developing the metric as shown below, you can of course do this in order to have an imaginary axis.

I derived the metric, then substituted in the ct without ever resorting to the imaginary i to obtain the Lorentz transformation for the double-bar time.

In this derivation we just use spatial coordinates throughout--just the spatial length of legs and hypotenuse of a right triangle. You can take any right triangle in normal X-Y space and then use the Pythagorean theorem--then solve for the length of one leg-- followed by a substitution of the imaginary number i in the negative hypotenuse squared term.

Some people object to this derivation claiming that the original diagram is arbitrarily contrived. But this is not the case. It is required if the speed of light is to be the same for all observers. Furthermore it is very general, because for any two observers moving relative to each other, you can always find a rest system that has the two movers going in opposite directions at the same speed.

 

[moonman239]

Here's an idea: discuss these ideas with Stephen Hawking.

 

[Dale]

Hmmm. Well, we've been discussing the length of the spacetime interval as defined under the Minkowski model. OK then, are you suggesting here that imaginaries do not apply wrt clock rates?

Seems to me that you might not understand the meaning of complex systems. I do realize that Einstein did not use imaginaries in his OEMB.

Again, see my post 91. The use of ict is not necessary and is rarely used any more.

 

[yuiop]

Colocation is irrelevent. Suppose you are traveling at .99c from start s1 to s2. You send me a picture of yourself at s1, and send me another picture when you get to s2. However long it takes me to get these signals, I will agree on how much aging you will experience (as long as I know the distance from s1 to s1 as I would measure it, and your speed). You will experience the aging I compute. Further, if you know my start and end points as you would measure them, and my speed as you measure it, and compute the invariant interval along my path in these coordinates, you will get the same longer age that I experience. The explanation of why you agree on my invariant interval while still seeing my clock going slow is that you radically disagree on simultaneity. The event on my worldline I say is simultaneous to your passing s1 is not at all what you would say.

This avoids the subject of the OP "which clock is really slower?". Until the clocks are co-located again, you can not in any meaningful (invariant) way say which clock is ticking slower. This is true, whether using Euclidean 4d or Minkowski 4d calculations.

 

[PAllen]

This avoids the subject of the OP "which clock is really slower?". Until the clocks are co-located again, you can not in any meaningful (invariant) way say which clock is ticking slower. This is true, whether using Euclidean 4d or Minkowski 4d calculations.

But that wasn't the point of our discussion. Our discussion was around the limitations of trying to explain features of the Minkowski metric using a Euclidean metric. Using each metric where it really applies, you compute invariants that are coordinate independent (and observer independent, where applicable). Using the Euclidean metric in SR you have to be aware that nothing you compute with it is invariant, and you can't use it to compare anything across two arbitrary timelike, physically plausible paths. Meanwhile, the Minkowski metric can be used to compare intervals across arbitrary paths, just like the Euclidean could if it were really applicable.

My example is on point for this. I experience that the time between my bomb detonations is 10 years, and compute and observe that you only age 1 year traveling between s1 and s1 (which are say 8 light years apart, according to me). You experience 1 year going from s1 to s1, and compute that I age 10 years between between my H-bomb detonations. This is what one expect a metric to accomplish, and thus using the Euclidean metric in SR will lead to many confusions and false conclusions.

 

[GrayGhost]

accidental premature post. This post was reposted here ...

https://www.physicsforums.com/showpost.php?p=3113986&postcount=112"

GrayGhost

 

[John232]

Why does one's clock really have to be the one that is slower? Couldn't they both really be slower at the same time from the perspective of each observer?

Without a preferred frame of reference you would think it would force this situation for two observers that have always traveled at a constant speed relative to each other. You couldn't ever say this object is the one that has the true velocity so then how could you say that the others clock is the one that is really slower. Without an absolute frame of reference they both have to be slower at the same time...

 

[GrayGhost]

DaleSpam,

Wrt you prior post 91 ... I had assumed that the metric was the result "after having had multiplied by i". That's why I was unaware of any (++++) metric, given I understood it in that way. So when you said the 1st metric is no longer used, you mean that no axes are assumed imaginary in the very first place anymore (on a spacetime diagram), yes?

BobC2,

Indeed, spacetime diagrams show the relative spacetime orientations between observers moving relatively, regardless as to whether the i is used. Yours (of course) points that out nicely. Einstein used no imaginaries in OEMB, so their use by Minkowski was not necessary although mathematically appropriate.

yuiop,

Thanx for your patience here. While I was correct in that you cannot get from (ict)²+(vt)² to (ct)²+(vt)² (unless you discard the i's), you indeed CAN get to -(ct)²+(vt)² or (ct)²-(vt)² w/o invoking the use of imaginaries. Obviously, since OEMB used no imaginaries. So Minkowski's is nothing more than a different process to do the very same thing...

That said, no imaginaries need be used, as you said. I still contend that the length (ct)²+(vt)² (per the stationary POV) may well hinder more than aid in expediting understanding of relative aging. Those related points I mentioned prior still stand, IMO. It'd be interesting to see how it turns out if used in classroom. I'm with PAllen on that point though ... it's probably better to adapt to Minkowski's metric than to add complexity using euclidean perspectives with stipulations. Yet at the same time, any inertial perspective in SR is euclidean, so.

GrayGhost

 

[PAllen]

Why does one's clock really have to be the one that is slower? Couldn't they both really be slower at the same time from the perspective of each observer?

Without a preferred frame of reference you would think it would force this situation for two observers that have always traveled at a constant speed relative to each other. You couldn't ever say this object is the one that has the true velocity so then how could you say that the others clock is the one that is really slower. Without an absolute frame of reference they both have to be slower at the same time...

Correct. You cannot say which clock is slower, each observes the other clock is slower, and neither is more right than the other. However, each agrees on how much 'physical time' elapses between any two given physical events on any worldline (spacetime path of some observer). They differ on their explanations of this, the key difference being different perceptions of simultaneity.

 

[GrayGhost]

bobc2,

I'm curious, how would you apply your method of determining the spacetime interval under the situ per the attached figure, which is not a Loedel figure?

GrayGhost

 

[GrayGhost]

Correct. You cannot say which clock is slower, each observes the other clock is slower, and neither is more right than the other. However, each agrees on how much 'physical time' elapses between any two given physical events on any worldline (spacetime path of some observer). They differ on their explanations of this, the key difference being different perceptions of simultaneity.

While your response is well stated IMO, and cuts to the chase, I suppose there may be debate over whether moving clocks tick slower "per the inertial observer". I suppose it comes down to "slower wrt what".

GrayGhost

 

[yuiop]

Note, it is not so trivial to banish forward and back in time.

I thought it might be interesting to demonstrate a problem that arises if we allow a physical object to travel back in time. Let us say we have two clocks at the origin of an inertial reference frame (S). One clock (A) travels directly to (x,t) coordinates (1,2) at half the speed of light. The other (B) remains at rest in S for 2.5 time units arriving at (0,2.5) and then travels back in time to meet clock A at coords (1,2). Now at coordinate time t=2.25 in S we can see there are two copies of clock B, one at coords (0,2.25) and the other at (0.5,2.25). The same clock exists in two places at the same time and there is possibly a violation of conservation of energy, because we have duplicated the clock. An additional complication is that it is possible to find an observer at rest wrt to a different frame (S') who does not ever see the second copy of clock B, so the cloned version of clock B does not have an observer independent existence.

 

[PAllen]

I thought it might be interesting to demonstrate a problem that arises if we allow a physical object to travel back in time. Let us say we have two clocks at the origin of an inertial reference frame (S). One clock (A) travels directly to (x,t) coordinates (1,2) at half the speed of light. The other (B) remains at rest in S for 3 time units arriving at (0,3) and then travels back in time to meet clock A at coords (1,2). Now at coordinate time t=2.5 in S we can see there are two copies of clock B, one at coords (0,2.5) and the other at (0.5,2.5). The same clock exists in two places at the same time and there is possibly a violation of conservation of energy, because we have duplicated the clock. An additional complication is that it is possible to find an observer at rest wrt to a different frame (S') who does not ever see the second copy of clock A, so the cloned version of clock A does not have an observer independent existence.

I believe this comment of mine was in relation to spacelike paths. And the original comment about timelike curves (that go forward and back in time) was to highlight the confusion that might be possible by thinking in Euclidean terms about spacetime - in Euclidean terms there is no special coordinate associated with a different signature sign, and coordinate paths can freely move back and forth with respect to any coordinate. Introducing a Euclidean metric encourages such confusion.

Of course, I assume you know that GR allows timelike curves that go forward and back in time, and that, so far, no one has clearly demonstrated they can only occur in 'impossible' situations. I have a strong bias against them, but until someone demonstrates a clear impossibility argument, I admit my view is a philosophic bias.

 

[yuiop]

Of course, I assume you know that GR allows timelike curves that go forward and back in time, and that, so far, no one has clearly demonstrated they can only occur in 'impossible' situations. I have a strong bias against them, but until someone demonstrates a clear impossibility argument, I admit my view is a philosophic bias.

Me too ...

 

[GrayGhost]

I thought it might be interesting to demonstrate a problem that arises if we allow a physical object to travel back in time. Let us say we have two clocks at the origin of an inertial reference frame (S). One clock (A) travels directly to (x,t) coordinates (1,2) at half the speed of light. The other (B) remains at rest in S for 3 time units arriving at (0,3) and then travels back in time to meet clock A at coords (1,2). Now at coordinate time t=2.5 in S we can see there are two copies of clock B, one at coords (0,2.5) and the other at (0.5,2.5). The same clock exists in two places at the same time and there is possibly a violation of conservation of energy, because we have duplicated the clock. An additional complication is that it is possible to find an observer at rest wrt to a different frame (S') who does not ever see the second copy of clock A, so the cloned version of clock A does not have an observer independent existence.

yuiop,

I'd throw a few comments your way, but I fear DaleSpam would cry metaphysical foul. I'll just sit back and process the exchange, which I find "interesting enough".

EDIT: Looks like PAllen's response beat mine. Indeed, no one can travel back in time (per himself) per the theory.

GrayGhost

 

[GrayGhost]

Of course, I assume you know that GR allows timelike curves that go forward and back in time, and that, so far, no one has clearly demonstrated they can only occur in 'impossible' situations. I have a strong bias against them, but until someone demonstrates a clear impossibility argument, I admit my view is a philosophic bias.

Outside of falling into a rotating black hole, how else could you travel thru time if c is a speed limit? If one falls into a black hole, isn't it true that one would never survive the gravitational gradient during the free fall? What situ in GR would allow for a successful "time travel" in theory?

GrayGhost

 

[Dale]

What situ in GR would allow for a successful "time travel" in theory?

Wormholes, a rotating universe, etc. They are called "closed timelike curves".

 

[GrayGhost]

Wormholes, a rotating universe, etc. They are called "closed timelike curves".

Outside of the situ I mentioned, ie a rotating singularity, what would form a wormhole per GR?

When you say the "rotating universe", are you talking Goedel's theory?

GrayGhost

 

[Dale]

When you say the "rotating universe", are you talking Goedel's theory?

Yes.

 

[bobc2]

Yes.

I believe Godel found closed timelike curves for his initial static without expansion model. I may be mistaken, but I think he later did solutions for the rotating and expanding model without finding closed timelike curves. However, this motivated many many other models and solutions that included black holes, wormholes, etc. I'll double check my literature to see if I've remembered this correctly.

 

[bobc2]

bobc2,

I'm curious, how would you apply your method of determining the spacetime interval under the situ per the attached figure, which is not a Loedel figure?

GrayGhost

GrayGhost, I recognize the sketch without confusion. However, the t' and t'' coordinates are certainly not symmetric with respect to the x-t coordinate system (this sketch has the t' observer and the t'' observer moving at different speeds in opposite directions with respect to x-t). The diagram would have been symmetric if the t' and t'' observers had been moving at the same speeds in opposite directions (you can always find a rest system for which the speeds are the same).

The x' seems to be a valid instantaneous 3-D cross-section for the observer moving along t'. The x'' is likewise a valid instantaneous 3-D cross-section for the observer moving along the t'' axis. However, the x'' coordinate axis direction should be pointing down and to the right (what is indicated in the sketch is the negative direction along x''. I'm at work still but will get back with you with completed sketches this evening.

 

[bobc2]

bobc2,

I'm curious, how would you apply your method of determining the spacetime interval under the situ per the attached figure, which is not a Loedel figure?

GrayGhost

Here is a comparison of your diagram with the coordinates explicitly represented. On the right I've created another diagram that keeps your x'', t'' coordinates (red), but then I've put in new blue coordinates so as to present a Loedel figure (a symmetric space-time diagram).

I could have picked a rest system that kept both red and blue velocities the same as you had them originally, but also structured the diagram as Loedel. I would have had to first estimate each speed (for red and blue in your original diagram), then computed the difference between the speeds. The new symmetric rest system would then show red and blue moving away from each other at the same relative speed but the speeds relative to the rest system would be one half of the speed of blue relative to red (or red relative to blue). The reason for the symmetric presentation is that now the line lengths on both blue and red scale the same for actual distances and times.

Having completed the new diagrams, I understand why you showed it the way you did--I have added many more lines to the picture, and you were trying to keep it simple.

 

[GrayGhost]

Having completed the new diagrams, I understand why you showed it the way you did--I have added many more lines to the picture, and you were trying to keep it simple.

I mentioned that it was not a Loedel figure, upfront. The reason I drafted the diagram as such, was to see whether you could apply your method of determining the spacetime interval (via right triangles) on a diagram which was not tactically symmetric. The fact is, it doesn't ... although this is not to say that the Loedel figure is not useful.

My position is that the Loedel figure determines the spacetime interval length (via your method) by the "luck of symmetry". Therefore, it's lacking in its ability to explain the total picture of the invariant spacetime interval length. One might ask ... why does your method not work if the Loedel symmetry is not strategically selected?

GrayGhost

 

[bobc2]

I mentioned that it was not a Loedel figure, upfront. The reason I drafted the diagram as such, was to see whether you could apply your method of determining the spacetime interval (via right triangles) on a diagram which was not tactically symmetric. The fact is, it doesn't ... although this is not to say that the Loedel figure is not useful.

Sorry I totally missed the point you were inquiring on (as so often happens with me on these posts). I could draw a right triangle on your coordinates, but the legs of the triangle would not be scaled the same for true distance measurements. Now that I see what you are driving at, of course your triangle did not turn out to be a right triangle when you used a line parallel to the x' axis in a purposely failed attempt to form one leg of a right triangle--that can only happen with a symmetric (Loedel) diagram. So, you were purposefully making the point that a right triangle would not result in this case (the whole point of what you were showing--the point that went over my head--I just assumed you did not understand how to make a symmetric diagram--again, sorry about that).

My position is that the Loedel figure determines the spacetime interval length (via your method) by the "luck of symmetry". Therefore, it's lacking in its ability to explain the total picture of the invariant spacetime interval length. One might ask ... why does your method not work if the Loedel symmetry is not strategically selected?

That seems like a reasonable argument, but I really have to disagree with that assessment, because this symmetry is definitely not by luck. That is because no matter what set of speeds with respect to a rest system you wish to give me, I can always represent it with a symmetric diagram. I can always find a rest system that has each observer moving in opposite directions with the same speed.

Or, if you give me an observer moving with some velocity with respect to a rest system, I will choose a new rest system such that you would then show two observers moving in opposite directions, each with one half the speed of the original speed.

Therefore, this is a very general result of special relativity, one that gives the genreral expression--the Minkowski metric.

And by the way, you can solve many tricky problems using a symmetric diagram, such as the pole-in-the-barn thought experiment, and also demonstrate situations like the Penrose Andromeda Galaxy Paradox--to say nothing about questions like, "Whose clock is slower?"

 

[GrayGhost]

BobC2,

Ahhh, but no matter how you slice it, your method does not work unless you force the Loedel symmetry. The Minkowski model (using imaginary time) works everytime no matter how its presented.

The reason Minkowski's model is-not-so-restricted resides in the fact that ict' (which is the spacetime interval length s) is derived from real spatial axes orthogonal to the direction of motion (eg y=y' =ct' or z=z' =ct').

GrayGhost

 

[bobc2]

BobC2,

Ahhh, but no matter how you slice it, your method does not work unless you force the Loedel symmetry. The Minkowski model (using imaginary time) works everytime no matter how its presented.

The reason Minkowski's model is-not-so-restricted resides in the fact that ict' (which is the spacetime interval length s) is derived from real spatial axes orthogonal to the direction of motion (eg y=y' =ct' or z=z' =ct').

GrayGhost

Good job, GrayGhost! You win. It's been a stimulating discussion.

 

[GrayGhost]

I should also have added, for the benefit of others, that ...

after determining y=y'=ct', subsequently multiplying by i serves to rotate this resultant ct' distance vector 90 degrees (away from y') into what is the ict'-axis on the Minkowski illustration. That is, time is orthogonal to real space.

GrayGhost

 

[bobc2]

I should also have added, for the benefit of others, that ...

after determining y=y'=ct', subsequently multiplying by i serves to rotate this resultant ct' distance vector 90 degrees (away from y') into what is the ict'-axis on the Minkowski illustration. That is, time is orthogonal to real space.

GrayGhost

Could you illustrate that graphically so we can be clear which axes you're talking about?

 

[Dale]

I should also have added, for the benefit of others, that ...

after determining y=y'=ct', subsequently multiplying by i serves to rotate this resultant ct' distance vector 90 degrees (away from y') into what is the ict'-axis on the Minkowski illustration. That is, time is orthogonal to real space.

The orthogonality of two vectors depends on the metric. The i has nothing to do with it.

 

[bobc2]

The orthogonality of two vectors depends on the metric. The i has nothing to do with it.

GrayGhost, I'm afraid DaleSpam really knows what he is talking about here--he really is right about that.

 

[GrayGhost]

DaleSpam,

I'm not suggesting that real axes cannot be orthogonal wrt one another, nor that metrics which require orthogonal axes produce non-orthogonal axes instead. I'm merely pointing out that mathematically, and as per the Minkowski model, the multiplication by i corresponds to a 90 deg rotation of the distance vector within the coordinate system, which is a complex system in Minkowski's model.

BobC2,

I fully recognize that DaleSpam knows his stuff. He's clearly an expert, and there are a number of other forum responders here that are very good. However, he's suggesting I said something there that I didn't. In mathematics, the multiplication by i "means something". That "something" applies to the Minkowski model.

GrayGhost

 

[GrayGhost]

Could you illustrate that graphically so we can be clear which axes you're talking about?

bobc2,

Might take a little time, but I can do that, yes. Before doing so, and before others claim otherwise, I should point out that this does not mean that imaginaries must be used. Again, OEMB didn't use imaginaries. I'm merely pointing out the meaning of the Minkowski model.

GrayGhost

 

[Dale]

I'm not suggesting that real axes cannot be orthogonal wrt one another, nor that metrics which require orthogonal axes produce non-orthogonal axes instead. I'm merely pointing out that mathematically, and as per the Minkowski model, the multiplication by i corresponds to a 90 deg rotation of the distance vector within the coordinate system, which is a complex system in Minkowski's model.

This is not correct either. The multiplication by i makes the time axis timelike, not orthogonal. Consider an orthogonal metric:

g=<br /> \left(<br /> \begin{array}{cccc}<br /> 1 &amp; 0 &amp; 0 &amp; 0 \\<br /> 0 &amp; 1 &amp; 0 &amp; 0 \\<br /> 0 &amp; 0 &amp; 1 &amp; 0 \\<br /> 0 &amp; 0 &amp; 0 &amp; 1<br /> \end{array}<br /> \right)

and a non-orthogonal metric:
h=<br /> \left(<br /> \begin{array}{cccc}<br /> 1 &amp; 1 &amp; 0 &amp; 0 \\<br /> 1 &amp; 1 &amp; 0 &amp; 0 \\<br /> 0 &amp; 0 &amp; 1 &amp; 0 \\<br /> 0 &amp; 0 &amp; 0 &amp; 1<br /> \end{array}<br /> \right)

with the basis vectors:
q=(1,0,0,0) and r=(i,0,0,0) and s=(0,1,0,0)

Then:
g_{\mu\nu}q^{\mu}s^{\nu}=0 and g_{\mu\nu}r^{\mu}s^{\nu}=0
while
h_{\mu\nu}q^{\mu}s^{\nu}\neq 0 and h_{\mu\nu}r^{\mu}s^{\nu}\neq 0

So the multiplication by i does not have anything to do with orthogonality. It does not make two orthogonal vectors non-orthogonal and it does not make two non-orthogonal vectors orthogonal. What it does is to make the one axis timelike, meaning that
g_{\mu\nu}q^{\mu}q^{\nu}=1 but g_{\mu\nu}r^{\mu}r^{\nu}=-1

This is a different concept than orthogonality.

 

[bobc2]

This is not correct either. The multiplication by i makes the time axis timelike, not orthogonal. Consider an orthogonal metric:

g=<br /> \left(<br /> \begin{array}{cccc}<br /> 1 &amp; 0 &amp; 0 &amp; 0 \\<br /> 0 &amp; 1 &amp; 0 &amp; 0 \\<br /> 0 &amp; 0 &amp; 1 &amp; 0 \\<br /> 0 &amp; 0 &amp; 0 &amp; 1<br /> \end{array}<br /> \right)

and a non-orthogonal metric:
h=<br /> \left(<br /> \begin{array}{cccc}<br /> 1 &amp; 1 &amp; 0 &amp; 0 \\<br /> 1 &amp; 1 &amp; 0 &amp; 0 \\<br /> 0 &amp; 0 &amp; 1 &amp; 0 \\<br /> 0 &amp; 0 &amp; 0 &amp; 1<br /> \end{array}<br /> \right)

with the basis vectors:
q=(1,0,0,0) and r=(i,0,0,0) and s=(0,1,0,0)

Then:
g_{\mu\nu}q^{\mu}s^{\nu}=0 and g_{\mu\nu}r^{\mu}s^{\nu}=0
while
h_{\mu\nu}q^{\mu}s^{\nu}\neq 0 and h_{\mu\nu}r^{\mu}s^{\nu}\neq 0

So the multiplication by i does not have anything to do with orthogonality. It does not make two orthogonal vectors non-orthogonal and it does not make two non-orthogonal vectors orthogonal. What it does is to make the one axis timelike, meaning that
g_{\mu\nu}q^{\mu}q^{\nu}=1 but g_{\mu\nu}r^{\mu}r^{\nu}=-1

This is a different concept than orthogonality.

GrayGhost,

I have the impression from some of your posts that you are probably not familiar with tensor analysis--metrics, and terms like covariant, contravariant and raising and lowering indices. DaleSpam has made his point quite efficiently and accurately, but it may not have done much for you if you do not know tensor analysis.

I assume you know some vector analysis and scalar products. If so we could do a short tutorial on covariant and contravariant vector components, and we could discuss the symmetric special relativity diagram I've been using in that context. But, again, I would probably get way off point for this thread. But I was trying to think of some way to help you reconcile your thoughts to leave you with some satisfaction.

 

[GrayGhost]

BobC2,

What makes axes orthogonal as I understand it, is that they are either defined as such or a metric requires it. It's not the .However, from what I've studied on complex systems, the multiplication by i rotates a vector by 90 deg in a complex system. Consider a real light ray's pathlength from origin thru (say) the +x+y quandrant over time t. It exists as a length ct in the real xy plane. Mulitply ct by i, and that vector then rotates 90 deg from real space, and becomes colinear with Minkowski's ict axis (not ict'). The length of that ray is in fact the length of the time interval, because time t = ct per Minkowski. Granted, no i is really necessary, but imaginaries were (no doubt) more commonly used back then. So I don't see why DaleSpam objected to it, but I suspect it had to do with my wording "that could have been stated better" in a prior post here.

Folks always ask why s < ict, given ict' is the hypthenuse in a standard Minkowski illustration. Usually the Minkowski metric is simply stated and restated, but it does not seem often that anyone answers this to asker satisfaction. This also has to do with why Loedel figures do not work for non-symmetric presentations of the moving worldlines. Wrt that stated scenario in the prior para, assume the emitted light (from origin) an expanding lightsphere. Another system x',y',z' moves at v along +x, and is momentarily colocated with the stationary system when the EM is emitted. Wrt the ray that travels direct along +y' of the moving system, ct' is the distance the lightsphere expands along the y' (or z') axis. Because there are no contractions wrt axes orthogonal to the axis of motion, y = y' = ct' ... where y = sqrt( (ct)²-(vt)²) ), as viewed in the real xy plane over time. Multiply ct' by i, then it rotates 90 deg from the y'-axis and becomes colinear with ict'. So although it appears as a hypothenuse in a euclidean (stationary) system, it has the length of a shorter leg of the right triangle (since ct' = y, where y < ct).

In answer to your question Bob, I am not proficient in GR tensor math. Don't need it for SR, but I wouldn't mind studying it eventually for GR purposes. I am not degreed in math, but I have taken trig, analytical geometry, calculus, some matrix math, vector math, physics, quantum mechanics, the ole fields and waves, etc., at university. It has been many many years though, and I must admit that I haven't used much of it since. Thanx for the tensor tutorial offer, as I may well come back and ask for that when time permits, assuming you're still willing and free enough then. Very much appreciated !

GrayGhost

 

[Dale]

Consider a real light ray's pathlength from origin thru (say) the +x+y quandrant over time t. It exists as a length ct in the real xy plane.

Here is what I am objecting to, this is not correct. It can be projected as a length ct onto the Euclidean xyz space, but it already exists as a null-length path in the Minkowski txyz spacetime.

Mulitply ct by i, and that vector then rotates 90 deg from real space, and becomes colinear with Minkowski's ict axis (not ict').

The t axis is already 90 degrees from the x, y, and z axes.

If you want to add a 5th axis (3 real spatial axes, 1 real time axis, 1 imaginary time axis) then you can indeed claim that it rotates 90 degrees in a plane which is already orthogonal to xyz space. I.e. it starts out orthogonal to the x, y, and z axes and parallel to the t axis and after the rotation it is still orthogonal to x y and z, but is now also orthogonal to t. This is NOT Minkowski's approach AFAIK.

If you do not want to add a 5th axis then there is no rotation involved and the multiplication by i only serves to identify the signature of the metric.

So I don't see why DaleSpam objected to it, but I suspect it had to do with my wording "that could have been stated better" in a prior post here.

It is not about your wording. I am trying to teach you something here. You don't seem to understand the difference between the orthogonality of two vectors and the signature of a metric. The purpose of i in the ict convention is not to make anything orthogonal to anything else (they are already orthogonal); the purpose is to make the signature (-+++).

Do you understand how orthogonality is defined in a metric space? Do you understand what is meant by the signature of a metric? Do you see from the math above how multiplying by i does not change any orthogonality relationships (i.e. no rotation)? Do you see from the math above how multiplying by i does change the signature?

 

[GrayGhost]

DaleSpam,

Here is what I am objecting to, this is not correct. It can be projected as a length ct onto the Euclidean xyz space, but it already exists as a null-length path in the Minkowski txyz spacetime. The t axis is already 90 degrees from the x, y, and z axes.

I wasn't suggesting in my last post here that the ict (or ict') axis did not exist before the distance vector rotation. I said the complex system exists, and the multiplication by i rotates the ct distance vector by 90 deg from real space into the ict axis. I know what you're saying here ... the i does not make the ict orthogonal wrt 3-space, the metric does.

IMO, starting with ct of the real xy plane seems satisfactory, because that's what we measure. I mean, the nature of the light ray is what it all comes from. From the overall POV of the Minkowski 4-space, we (of course) know that ct only a projection from a higher 4-space POV. I don't disagree there.

If you want to add a 5th axis (3 real spatial axes, 1 real time axis, 1 imaginary time axis) then you can indeed claim that it rotates 90 degrees in a plane which is already orthogonal to xyz space. I.e. it starts out orthogonal to the x, y, and z axes and parallel to the t axis and after the rotation it is still orthogonal to x y and z, but is now also orthogonal to t. This is NOT Minkowski's approach AFAIK.

Hmm. Well, it just seems to me that one can define any number of axes orthogoinal wrt each other, if they wish. If it's imaginary, it alters the metric negatively for that dimenson.

But wrt Minkowski, I venture he set up the LTs as a system of linear equations, obtained the matrix of eigenvalues, and the inner products were all 0 ... so ict was orthogonal. No? ... I suspect he saw the similarity between this matrix and the matrix indictative of euler rotations.

If you do not want to add a 5th axis then there is no rotation involved and the multiplication by i only serves to identify the signature of the metric.

Understood.

It is not about your wording. I am trying to teach you something here.

It's about both, and I do appreciate it. It's easy in relativity discussions to word things hastely, or to develop poor habits wrt wording. Need to be more careful. Also, I'm not pretending to remember everything I learned at university long ago. I forgot most of it, but it does come back some if I dig into.

You don't seem to understand the difference between the orthogonality of two vectors and the signature of a metric. The purpose of i in the ict convention is not to make anything orthogonal to anything else (they are already orthogonal); the purpose is to make the signature (-+++).

I think I understand those better now, thanx.

So wrt the rest of your post ...

Do you understand how orthogonality is defined in a metric space?

I'd say ... inner products are all zero.

Do you understand what is meant by the signature of a metric?

I'd say ... Wrt SR, a diagonal matrix of either +1 or -1. The sign indicates whether the eigenvalues are added or subtracted, one of time and 3 of space.

Do you see from the math above how multiplying by i does not change any orthogonality relationships (i.e. no rotation)?

yes. Mulitplying by i only rotates the vector by 90 deg in a complex system, where the imaginary axis has already been defined as orthogonal to 3-space.

Do you see from the math above how multiplying by i does change the signature?

Yes, because we're dealing with quadratics, and since (i)² = -1, that dimension is reversed in polarity thereby effecting the equation on the whole.

GrayGhost

 

[GrayGhost]

Consider a real light ray's pathlength from origin thru (say) the +x+y quandrant over time t. It exists as a length ct in the real xy plane. Mulitply ct by i, and that (distance) vector then rotates 90 deg from real space, and becomes colinear with Minkowski's ict axis (not ict'). The length of that ray is in fact the length of the time interval, because time t = ct per Minkowski.

Here is what I am objecting to, this is not correct. It can be projected as a length ct onto the Euclidean xyz space, but it already exists as a null-length path in the Minkowski txyz spacetime.

DaleSpam,

I don't see the 90 deg ct vector rotation "from real 3-space into the ict-axis" as a representation of the lightray traveling thru 4-space (which is a zero pathlength). I see it as the required traversal of the stationary observer thru 4-space, wrt the photon's pathlength in real 3-space as the reference ... given Minkowski set t = ict.

Yet, I'm curious as to how you might respond to this ...

Q) How does a null (zero) pathlength in spacetime produce a non-null projection into real euclidean 3-space?​

I know it does, but I'm curious as to how you'd explain that in layman's terms.

GrayGhost

 

[Dale]

Yet, I'm curious as to how you might respond to this ...

Q) How does a null (zero) pathlength in spacetime produce a non-null projection into real euclidean 3-space?​

I know it does, but I'm curious as to how you'd explain that in layman's terms.

From Euclidean geometry you should already be familiar with the idea that a projection of a line segment will have a different length than a line segment. For instance the projection of a hypotenuse onto the x-axis is h cos(theta). The concept is similar in Minkowski geometry except that projections may be longer than the segment itself.

 

[GrayGhost]

From Euclidean geometry you should already be familiar with the idea that a projection of a line segment will have a different length than a line segment. For instance the projection of a hypotenuse onto the x-axis is h cos(theta). The concept is similar in Minkowski geometry except that projections may be longer than the segment itself.

Yes, however I was interested in how you would answer that for the lightpath ...

Q) How does a null (zero) length produce a non-null projection unto 3-space, in layman's terms?​

GrayGhost

 

[bobc2]

Yes, however I was interested in how you would answer that for the lightpath ...

Q) How does a null (zero) length produce a non-null projection unto 3-space, in layman's terms?​

GrayGhost

GrayGhost, you set up a laser beam, pointed from one end of the room to the other along your x axis, then observe the projection.

Actually, you are wanting the projection of a single photon world line to project on your x axis. That would be difficult to do, even aside from quantum mechanical issues, because the sequence of photon positions along your x-axis occurs so fast. After all, you are moving along your own 4th dimension (X4) at 186,000 miles every second, so you are traveling enormous distances in a fraction of a second while trying to observe photon movement of just 20 ft or so.

So it is not unreasonable to produce a steady stream of photons in order to generate the picture you need to infer what is going on with the photon world lines and their projection onto your X axis.

By the way, I was a little confused about the process you were trying to describe when multiplying by the imaginary i. It sounds like you are trying to apply an operator (as opposed to doing coordinate transformations).

If you have an X and iY pair of coordinates, you can define a phasor (an amplitude and a phase angle) by establishing a point in that complex plane. Now if you multiply the complex number representing that phasor by the imaginary, i, of course you rotate the phasor, and now you have a new phasor, i.e., a new point in that SAME complex plane. This is what operators do. You haven't created any new coordinates, you've just converted a phasor into a new phasor without doing anything to the coordinates.

So, with your imaginary example, you really implicitly started with a complex plane and a phasor having a zero phase angle (a point on the x axis). You then multiplied by i, rotating that phasor 90 degrees. But you certainly did not create an imaginary axis, and you certainly did not derive a Minkowski time axis. Go back to my original sketch of the pair of symmetric moving observers, because, using purely geometric principles I actually did derive the Minkowski metric. And I still don't understand why you can't recognize it as a very general derivation, since for any two observers in relative motion, you can always apply that analysis, i.e., it is definitely not a special case.

Also, I do not refer to your phasor as a vector, because they are not vectors under affine transformations (remember, you don't have a vector if the components do not transform like coordinates--I think there is a special case for which a complex phasor can be a vector).

Back to your original issue. I think you confuse the 4-dimensional objects with cross-section views computed using Lorentz transformations, which allows your chosen observer at rest to compute observations from the point of view of other observers. The photon is an external object with a world line in 4-dimensional space just as any other point object, i.e., electrons, quarks, etc. So, if you first establish the 4-D objects in the 4-D space, then start examining the various cross-section views of objects from various observer points of view, there may not be the confusion with questions like projecting a null world line to an x axis.

Thus, a 4-D photon does not have zero length as a 4-D object. Now, if you are at rest and you see another observer approaching the speed of light, knowing special relativity, you are thinking, "My gosh! That guy's X4 and X1 axes are rotating dangerously close together--much closer and his X4 and X1 will be colinear and his 3-D cross-section of the universe will be along his time axis--he will experience all past and future simultaneously--and yet time has not changed for him."

 

[GrayGhost]

Yes, however I was interested in how you would answer that for the lightpath ...

Q) How does a null (zero) length produce a non-null projection unto 3-space, in layman's terms?

GrayGhost, you set up a laser beam, pointed from one end of the room to the other along your x axis, then observe the projection.

Actually, you are wanting the projection of a single photon world line to project on your x axis. That would be difficult to do, even aside from quantum mechanical issues, because the sequence of photon positions along your x-axis occurs so fast. After all, you are moving along your own 4th dimension (X4) at 186,000 miles every second, so you are traveling enormous distances in a fraction of a second while trying to observe photon movement of just 20 ft or so.

It's seems you making a simple question complex here. The spacetime interval length for the photon is s=0. It's a null length, as DaleSpam mentioned. DaleSpam pointed out that one should envision projections to understand how a worldline's length in 4-space differs from its recorded length thru real 3-space. All well and true. However, my question was specific ...

Q) How is it that a zero length distance in 4-space projects a non-zero finite traversal thru real 3-space?​

I've got my idea here. I was curious as to how DaleSmapm would answer it, w/o just saying "it does so because the eqn requires it".

GrayGhost

 

[Dale]

Yes, however I was interested in how you would answer that for the lightpath ...

Q) How does a null (zero) length produce a non-null projection unto 3-space, in layman's terms?​

That is exactly what I answered in post 143.

 

[GrayGhost]

BobC2,

Nah, no confusion here. I'm very well aware for many years of differing POVs, the transformations, cross sectional considerations, relative simultaneity, effects of speed c motion, orientations within spacetime, Minkowski and Loedel figures, etc etc etc. My math is not the best, that I'll agree with.

Wrt the muliplication by i ... We're talking about distance vectors. Given the system complex, multiplication by i rotates it by 90 deg, and the magnitude remains unchanged, and so only its direction changes. It's all inherent in the geometric meaning of the Minkowski model, and it explains why s < ct.

Your prior response suggests that ... you still seem to think that when I say "rotate ct by 90 deg from the real xy plane" (via mulitplying by i) that I am suggesting this somehow creates the ict-axis. If you go back and reread my prior, I stated only that it becomes colinear with the already-existent ict-axis.

Wrt your Loedel figures ... Nothing against them, as they are useful. Your procedure to derive the spacetime interval requires right triangles. I pointed out that the Loedel figure works for you only because of the luck of symmetry ... which is why you specifically elect that symmetry. If the moving worldlines are not symmetric about the ficticious center observer POV, you have no right triangles, and your procedure cannot commence. I was merely driving the point home as to why it fails to work for all symmetries. It's the same reason that s < ct, even though ict' > ict in a standard Minkowski diagram.

Wrt the photonic worldlines ... I agree in that the photon has a worldline, as do material bodies. However, there is a difference ... the length of a worldline is defined between 2 events that reside upon the worldline. For the photon, it's length is always s = 0, a null length. If for any material entity its own s = 4 ls, it travels 4 ls thru 4-space and it experiences 4 sec proper duration. For the photon, its own s = 0, so it travels 0 ls thru 4-space and experiences 0 sec proper duration. This all takes us back to this ...

Q) How does a null length in 4-space project as a finite traversal thru real preceptable 3-space?​

IMO, I think your last para of your prior post touched on the answer to this question. Although a photon travels along its worldline at c per material observers, it does not travel at all thru 4-space. It cannot travel thru 4-space because it cannot experience any passage of proper time, since s=0. It simply exists at all locations of its propagational path (within the cosmos) at-once. Its worldline length is zero, but its projection unto real perceptable 3-space is not. The length of said projection is completely dependent upon the real events that create and destroy the photon (ie transform it), and this is why it projects a finite length. It's like an operation that results in an indeterminate mathematically, however said indeterminate is infact determined physically.

GrayGhost

 

[GrayGhost]

That is exactly what I answered in post 143.

I know. That's the traditional explanation, in general. I don't believe that explanation is enough to explain the projection of a lightpath onto perceptable 3-space. It definitely applies, but it just seems to fall short IMO.

GrayGhost

 

[Dale]

I know. That's the traditional explanation, in general. I don't believe that explanation is enough to explain the projection of a lightpath onto perceptable 3-space. It definitely applies, but it just seems to fall short IMO.

Then you will have to be a little more detailed about why you think it falls short. It seems perfectly adequate to me capturing both the similarities with Euclidean geometry as well as the essential difference due to the (-+++) signature. Anything more specific will require math.