Whose Clock Slows Down in Relativity?

[Mike_Fontenot]

No one in this forum can adequately answer the question because no one has a scientific definition of the word "real".

Einstein formulated and made productive use of a specific definition of "real" in his EPR paper. I did the same in my paper. Our two definitions were different, but they both served a useful purpose.

Mike Fontenot

 

[yuiop]

I also think the statement is false anyway. Consider the path (using (t,x)) from (0,0) to (2,.1) to (0,1). This is 'longer' than from (0,0) to (0,1) using your alternate metric, and *also* longer using the Minkowski metric.

Did you mean a clock traveling (0,0) to (2,0.1) to (0,1)?
The path from (2,0.1) to (0,1) or even (2,1) to (0,1) is unrealistic as it implies a particle traveling backwards in time using (t,x) notation. I perhaps should have added that all particles are restricted to traveling at the speed of light or less but that usually goes without saying. Could you clarify your intention?

 

[PAllen]

Did you mean a clock traveling (0,0) to (2,0.1) to (0,1)?
The path from (2,0.1) to (0,1) or even (2,1) to (0,1) is unrealistic as it implies a particle traveling backwards in time using (t,x) notation. I perhaps should have added that all particles are restricted to traveling at the speed of light or less but that usually goes without saying. Could you clarify your intention?

I was talking geometry, not particle paths. Also getting at the feature that with Euclidean metric you can make strong global statements about extremal properties of geodesics. With SR or GR you cannot. Extremal properties need qualification and with GR are only local at best. Note, geometrically, is there anything wrong with going forward and backward in 'x' along a path? If you are choosing to introduce a Euclidean metric, why should going forward and backward in 't' be any different?

When one discusses Euclidean space, one uses one metric, and pure and simple. You seemed to say using one metric in SR leads to circular definitions. I find this a strange statement which I disagree with.

 

[yuiop]

I was talking geometry, not particle paths. Also getting at the feature that with Euclidean metric you can make strong global statements about extremal properties of geodesics. With SR or GR you cannot. Extremal properties need qualification and with GR are only local at best. Note, geometrically, is there anything wrong with going forward and backward in 'x' along a path? If you are choosing to introduce a Euclidean metric, why should going forward and backward in 't' be any different?

When one discusses Euclidean space, one uses one metric, and pure and simple. You seemed to say using one metric in SR leads to circular definitions. I find this a strange statement which I disagree with.

I agree if we allow particles to exceed the speed of light or travel backwards in time, then yes you can construct "longer paths" through space-time with greater elapsed proper times, but if we restrict ourselves to measurements or real particles then this problem does not occur. Perhaps another of expressing what I was getting at, is that if (and only if) we have two clocks that are initially and finally co-located, then the the clock that has traveled the furthest distance through coordinate space will have accumulated the least elapsed proper time.

In the context of the twins paradox we might explain it like this. Calculate the proper time that elapses for each segment of the each twins journey. Add up the elapsed proper times of the segments. The twin that has the least total of proper times, will as a general rule, be the twin that has experienced the least proper time. This is a true and correct statement, but it is not very helpful to anyone. It amounts to this. Question: Which twin experiences the least proper time? Answer: The twin that experiences the least proper time.

Perhaps the problem lies in semantics. The Minkowski norm of the path is the invariant proper time interval, while I am talking about the Euclidean norm where time is on the same footing (with the same sign) as space. The path with the longest Euclidean norm has the shortest Minkowski norm for paths constrained to the future light cone. Identifying the path with the shortest proper time on a spacetime diagram, can be done at a glance without any calculations using this method and *might* be a helpful aid to beginners. If you think this confusing to beginners then I respect your right to your opinion.

 

[PAllen]

I agree if we allow particles to exceed the speed of light or travel backwards in time, then yes you can construct "longer paths" through space-time with greater elapsed proper times, but if we restrict ourselves to measurements or real particles then this problem does not occur. Perhaps another of expressing what I was getting at, is that if (and only if) we have two clocks that are initially and finally co-located, then the the clock that has traveled the furthest distance through coordinate space will have accumulated the least elapsed proper time.

In the context of the twins paradox we might explain it like this. Calculate the proper time that elapses for each segment of the each twins journey. Add up the elapsed proper times of the segments. The twin that has the least total of proper times, will as a general rule, be the twin that has experienced the least proper time. This is a true and correct statement, but it is not very helpful to anyone. It amounts to this. Question: Which twin experiences the least proper time? Answer: The twin that experiences the least proper time.

If it helps your thinking to have two metrics, so be it. I don't see either the necessity or physical validity of it. When you say: "then the the clock that has traveled the furthest distance through coordinate space will have accumulated the least elapsed proper time" you are introducing a Euclidean metric in an SR problem; for me, this is physically invalid.

 

[Mentz114]

For example:

"It is by this definition that I mean that the longest path through spacetime (with time substituted for the y dimension) is the hypotenuse and it the longest path that experiences the least proper time. Saying the path with the shortest proper time interval is the path that experiences the least proper time is a circular definition."

When he says 'longest' he is using a Euclidean metric; when he says proper time, he is using the Minkowski metric. I find it confusing to use two metrics at the same time.

Then, I gave a specific path such that the longer Euclidean path was also the longer proper time, contradicting his statement.

OK, I missed that somehow.

 

[Dale]

Einstein formulated and made productive use of a specific definition of "real" in his EPR paper.

He actually avoided making a definition of "real", but he did give a definition that he considered a sufficient condition. That was probably a wise choice.

 

[GrayGhost]

yuiop,

I understand your contention. The issue as to whether s is "the longest vs shortest path thru spacetime" is an old one, one that most folks debate at some point or another. This is not an easy topic to convey if addressed properly, so please allow me just a wee bit of time to respond in a good, short, and concise way that lays this to rest ... ie. do it right and do it justice. Hopefully, that'll be sometime this evening.

GrayGhost

 

[GrayGhost]

yuiop,

OK then. Let’s begin as you did, with a standard spacetime diagram and the use of Pathagorus’ theorem, where ict’ is the worldline vector of the moving observer (and hence becomes the hypthenuse) …

(ict’)² = (ict)² + (x² + y² + z²)

Minkowski designates ict’ as the spacetime interval, and so ict’ = s, where s is the length the moving observer travels thru the 4-space (or spacetime) between some 2 events, and he resides at both events …

s² = (ict)² + (x² + y² + z²)

OK, so here’s the big question … Which spacetime pathlength is longer, s or ict? By pathagorean’s theorem, it looks fairly simple … the longer path is the hypothenuse, which is s. In euclidean space, the euclidean metric is used. However as PAllen stated, the metric here is not euclidean, it’s Minikowski’s.

To ask “which pathlength is the longer”, is to ask this … is s < ict? It’s pretty much that simple. We know what ict is, and we know what vt is, and so we need only calculate s.

s² = (ict)² + (x² + y² + z²)
s² = -(ct)² + (x² + y² + z²)
s² = -(ct)² + (vt)²
s² = (vt)² - (ct)²
s² = t² (v²-c²)
s² = c²t² (v²/c²-1)
s = ct (v²/c²-1)^1/2 ... and since i²=-1
s = ict (1-v²/c²)^1/2

which requires that the length of the spacetime interval s be a Lorentz contracted ict, were s = ict’. Therefore, s < ict. Bottom line, the shorter pathlength thru the continuum is s, even though it’s the hypothenuse on a standard Minkowski spacetime diagram.

It’s the use of complex spatial axes (for time) that changes the metric from Euclidean to Minkowski’s. It has to do with the 90 degree counterclockwise rotation of a vector when multiplied by i, or a 180 deg rotation when multiplied by i².

In fact, consider the original formula for the spacetime interval …

s² = (ict)² + (x² + y² + z²)

or, since s = ict’ …

(ict’)² = (ict)² + (x² + y² + z²)

Time must always be orthogonal to space. Yet, there are 2 time axes here, each represented as a complex spatial axis. That is to say, we have 2 axes that are each orthogonal wrt space, “in the same triangle”. So as you can see, there’s more to this equation than casually meets the eye.

That all said, it would be improper in the case of relativity to say that he who travels the longer worldline path experiences the lesser time. Distance thru spacetime is nothing more than the accrued proper-time over a defined interval as reference, and the rate of proper-time is the very same for each and all. He who resides at both events always travels the shorter path, and thus ages the least. If both observers reside at both events, then the one who travels the shorter path ages the least.

Do you agree?

GrayGhost

 

[GrayGhost]

Mike Fontenot,

Wrt your brother/sister acceleration scenario data in ...

http://home.comcast.net/~mlfasf/"​

I haven't verified your data per se, however from basic concepts wrt an accelerated POV, I don't see anything that rubs me the wrong way. Clearly, periods of acceleration and/or deceleration can produce wild clock rates of distant inertial clocks per the borther's POV, so the way the sister's age bounces around do not really look suspicious or anything. And, the inertial periods were fine.

The only data point that I wondered about, just shooting from the cuff, was when the traveler became 26 and his sister 17 ... which occurs on his return leg after completing his +1g burn period away from Earth again. That said, my question is this ... (if you have your data at hand), what are the ages of both at the initial turnabout point when he drops momentarily back into the Earth frame, and (if you know) what was their separation then?

GrayGhost

 

[yuiop]

That all said, it would be improper in the case of relativity to say that he who travels the longer worldline path experiences the lesser time. Distance thru spacetime is nothing more than the accrued proper-time over a defined interval as reference, and the rate of proper-time is the very same for each and all. He who resides at both events always travels the shorter path, and thus ages the least. If both observers reside at both events, then the one who travels the shorter path ages the least.

Do you agree?

GrayGhost

I do not disagree with your definition of path length through Minkowski 4 space. I was simply pointing out that the particle that has the longest path length through Euclidean 3 space \sqrt{(x^2+y^2+z^2)} has the shortest path length through 4 space. There are conditionals on this observation. (1)That when comparing elapsed proper times of two particles that they are both present at the initial and final events. (2)That the particles are constrained to move at the speed of light or less. (3)That the particles always travel forward in coordinate time. If we take the path length in 4 space as:

\tau = \sqrt{(ct)^2 - (x^2+y^2+z^2)}

then we can rewrite this as:

\tau = \sqrt{(ct)^2 - (\sqrt{x^2+y^2+z^2})^2}

\rightarrow \tau = \sqrt{(ct)^2 - (vt)^2}

Now vt is the path length through 3 space and the particle with the greatest vt is the particle with the least proper time interval in 4 space, because vt is subtracted from the coordinate time component ct.

I disagree that we have to restrict ourselves to one set of coordinates as long as we make it clear what we are doing. In GR when we analyse what happens at the event horizon in Schwarzschild coordinates we can draw interesting conclusions by transforming to a different set of coordinates such as Kruskal-Szekeres coordinates.

In other words, if we draw the paths of particles on a time space diagram and measure the total path lengths with a ruler, then the paths with longest ruler measured paths lengths is always the path with least accumulated proper time if the constraints I mentioned above for realistic particles are adhered to. If you can show a valid counter example for realistic particles, then I will withdraw my assertion.

 

[Dale]

There are conditionals on this observation. (1)That when comparing elapsed proper times of two particles that they are both present at the initial and final events. (2)That the particles are constrained to move at the speed of light or less. (3)That the particles always travel forward in coordinate time.

There is one more condition. Consider the twin's from the reference frame where the traveling twin is initially at rest. In that frame they both travel the same distance through Euclidean 3 space, just one does it over a shorter period of time. So you are also limited to the condition that you are using an inertial frame where they traveled different (Euclidean 3-) spatial distances. Just because that condition is met in one inertial frame does not mean that it is met in all inertial frames.

 

[yuiop]

There is one more condition. Consider the twin's from the reference frame where the traveling twin is initially at rest. In that frame they both travel the same distance through Euclidean 3 space, just one does it over a shorter period of time. So you are also limited to the condition that you are using an inertial frame where they traveled different (Euclidean 3-) spatial distances. Just because that condition is met in one inertial frame does not mean that it is met in all inertial frames.

Good observation Dale. To circumvent the need for this additional condition I would like to revert to my original version, where the particle that travels the greatest distance through Euclidean 4 space \left(\sqrt{(ct)^2 + x^2 + y^2 +z^2}\right)/c has the shortest path length (proper time) in Minkowski 4 space \left( \sqrt{(ct)^2 - x^2 - y^2 - z^2}\right)/c. Does that seem reasonable with my original 3 conditions? I welcome any corrections to my terminology or conclusions here .

 

[PAllen]

Good observation Dale. To circumvent the need for this additional condition I would like to revert to my original version, where the particle that travels the greatest distance through Euclidean 4 space \left(\sqrt{(ct)^2 + x^2 + y^2 +z^2}\right)/c has the shortest path length (proper time) in Minkowski 4 space \left( \sqrt{(ct)^2 - x^2 - y^2 - z^2}\right)/c. Does that seem reasonable with my original 3 conditions? I welcome any corrections to my terminology or conclusions here .

I have no disagreement with this as stated. Pedagogical value is a matter of judgment and also differs by individual. My concern is that since only some of Euclidean intuitions carry over, rather than having to define the precise limits on this, just learn to use non-Euclidean metrics directly.

The intuition you want to encourage is 'longer on paper is shorter in time', but you have to add all those qualifiers:

- only between a given pair of point, not for comparing paths between different points
- only if all path segments are less the 45 degrees to the t axis (assuming c=1)
- only if you can not move back in t (while you can move back in x, and in Euclidean geometry there is no 'special coordinate' where moving back is prohibited).

I honestly don't see a net pedagogical benefit by the time you add all the qualifiers.

 

[Dale]

Does that seem reasonable with my original 3 conditions?

Seems reasonable to me. I think there is some value to being able to look at a spacetime diagram and know which is the shortest interval, and this could help. I agree with PAllen though, that it is best to learn to work with the metric directly. Maybe this would be a useful "stepping stone" towards that goal.

 

[Mike_Fontenot]

The only data point that I wondered about, just shooting from the cuff, was when the traveler became 26 and his sister 17
[...]
what are the ages of both at the initial turnabout point when he drops momentarily back into the Earth frame, and (if you know) what was their separation then?

Good (and perceptive) questions.

Starting at a separation L of 40 ly, when he (the traveler) is 26, she (according to him) is 17, she (according to her) is 47.9, and velocity beta = +0.774, then:

After 1 year at -1g, you get beta = 0, he is 27, she (according to both siblings) is 49.1, and L = 40.5 ly.

After 2 years at -1g, you get beta = -0.774, he is 28, she (according to him) is 81.2, she (according to her) is 50.3, and L = 40 ly.

(And for the following 2 years of +1g, you get similar behavior at the midpoint, with beta = 0, L = 39.4 ly, and her age (according to both siblings) being 51.5 when he is 29.)

Mike Fontenot

 

[GrayGhost]

yuiop,

While it is true that the longer pathlength thru euclidean 4-space (ie paper-length hypothenuse) is actually a shorter pathlength thru Minkowskian 4-space (ie < ict), I personally would not recommend presenting it that way in a beginners relativity classroom. We refer to "worldline length" very often in discussions of relativity theory, and in the case of the spacetime interval, it's meaning is global as opposed to POV-restricted. Folks would not (always) qualify "Euclidean vs Minkowskian" every time they spoke of "the length of the interval". So although what you say is correct, IMO this would likely lead to student confusion. Maybe not in a perfect world, but in reality, there'd likely be needless routine debates of "longer vs shorter".

Per the Minkowski metric, no qualification is necessary, because s is numerically nothing more than the moving observer's proper time readout t' ... where t' < t ... which is true for all. In fact, when we talk about worldline lengths in general, we always refer to the comparison of durations experienced by both clocks. The length ict' under a euclidean metric does not represent time readings of moving clocks. So my main concern would be ensuring folks do not think that the length of ict' in a euclidean space is the length of the spacetime interval s as defined by Minkowski, because folks will refer to the ict' length "as the spacetime interval". Yes?

Gray Ghost

 

[GrayGhost]

Mike Fontenot,

OK. I was interested in knowing the numbers before the brother is 26. Basically, he accelerates off from earth, goes inertial, then (reverses thrusters at) -1g to gradually commence the direction reverseal. At some point thereafter, he (for the first time) arrives back in the Earth frame at the turnabout point ...

what was the separation & both ages at the moment he first attained a state of rest with his sister?

GrayGhost

 

[Mike_Fontenot]

I was interested in knowing the numbers before the brother is 26.
[...]
what was the separation & both ages at the moment he first attained a state of rest with his sister?

Starting at a separation L of 37.6 ly, when he (the traveler) is 21, she (according to him) is 4.2, she (according to her) is 40.6, and velocity beta = +0.968, then:

After 1 year at -1g, you get beta = +0.774, he is 22, she (according to him) is 12.3, she(according to her) is 43.2, and L = 40 ly.

After 2 years at -1g, you get beta = 0, he is 23, she (according to both siblings) is 44.4, and L = 40.5 ly.

After 3 years at -1g, you get beta = -0.774, he is 24, she (according to him) is 76.5, she(according to her) is 45.6, and L = 40 ly.

After 1 year at +1g, you get beta = 0, he is 25, she (according to both siblings) is 46.8, and L = 39.4 ly.

After 2 years at +1g, you get beta = +0.774, he is 26, she (according to him) is 17, she(according to her) is 47.9, and L = 40 ly.

Mike Fontenot

 

[GrayGhost]

Mike Fontenot,

OK then. Thanx for the extra data. I just wanted to make sure that the sister was reasonably younger than age 17 at the initial turnabout point, because you showed her at age 17 later on during a reversal on the return leg. That said, on the surface, I don't see anything fishy with your age declarations at each event. I'll assume that you integrated the proper time correctly for the accelerating observer. The inertial parts are fine, and generally speaking, your accelerating intervals look in order.

Have you ever tried plotting the data on graph from spreadsheet?

GrayGhost

 

[GrayGhost]

...the particle that travels the greatest distance through Euclidean 4 space ( [ (ct²+x²+y²+z² ]^1/2 )/c has the shortest path length (proper time)
in Minkowski 4 space ( [ (ct²-x²-y²-z² ]^1/2 )/c. Does that seem reasonable with my original 3 conditions? I welcome any corrections to my terminology or conclusions here .

yuiop,

I been looking over your posts again on this matter, and something has occurred to me.

Since we do not begin with an all-real euclidean system, I don't believe you can say that (ct')²=(ct)²+(vt)². All you can say is that (ict')²=(ict)²+(vt)². They are not the same thing. Applying Pathagorus' theorem to the complex system, "requires" that distances follow the Minkowski metric, because imaginary axes are rotated (90 deg) causing the hypothenuse to become a non-hypothenuse. IOWs, the system we start with is not a real euclidean system, it's instead a complex euclidean system with imaginaries. This complex euclidean system IS the Minkowskian system from the get-go.

So ...

(ict')²+(vt)² = -(ct)²+(vt)²​

where timelike is negative since v<c
where spacelike is positive since v>c​

It cannot be said that ...

(ict)²+(vt)² = (ct)²+(vt)²​

That said, IMO it does not seem that you can ever say that the length of ict' on a spacetime diagram is the longest path, even though "it appears as such visually". It would be in an all-real euclidean space, but it's not, it's a complex system with imaginaries.

You disagree?

GrayGhost

 

[GrayGhost]

In the context of the twins paradox we might explain it like this. Calculate the proper time that elapses for each segment of the each twins journey. Add up the elapsed proper times of the segments. The twin that has the least total of proper times, will as a general rule, be the twin that has experienced the least proper time. This is a true and correct statement, but it is not very helpful to anyone. It amounts to this. Question: Which twin experiences the least proper time? Answer: The twin that experiences the least proper time.

Well I'd say it a little differently, and in doing so, I do not believe that problem arises. We cannot measure space as the other moving fellow does. We each take our own measurements. Assuming each twin travels a different pathlength between the 2 events, then the only proof of this is the accrued proper time of each observer. We got to look at clocks, and compare. Given light speed invariant, the rate at which time passes by me per me is the same rate it passes by you per you. Therefore, he who travels the longest path must age the most, because he must accrue the most proper time over the interval.

GrayGhost

 

[Dale]

FYI GrayGhost, the use of ict with a (++++) metric fell out of favor several decades ago. You can certainly find it in many older texts, but it is quite outdated. Now everybody uses ct or even t with a (-+++) metric or a (+---) metric. It doesn't make your comments wrong, just dated.

 

[Dale]

Well I'd say it a little differently, and in doing so, I do not believe that problem arises.

No, I agree with yuiop. What you are saying is true, but it is a tautology.

If 2 observers reside at both events, eg in the twins scenario, then the one who travels the shorter path thru the continuum ages less over the interval, because he accrues less proper time.

Here "path through the continuum" means "spacetime interval", and "ages less" means "experiences less proper time", and "proper time" is "timelike spacetime interval". So your earlier statement is:

"If 2 observers reside at both events, eg in the twins scenario, then the one who travels the shorter spacetime interval experiences less timelike spacetime interval over the interval, because he accrues less timelike spacetime interval".

Which is tautologically true. Similarly, this statement is tautological:

"Therefore, he who travels the longest spacetime interval must experience the most timelike spacetime interval, because he must accrue the most timelike spacetime interval over the interval"

 

[Mike_Fontenot]

Have you ever tried plotting the data on graph from spreadsheet?

No, I never use spreadsheet software. I do rough sketches of a lot of the numerical data, just to see the overall general picture. When more precision is needed, I use my own homemade plotting programs.

In my paper, I have a precise plot of her (the home sibling's) age, versus his (the traveling sibling's) age (according to him), for his age segment from 26 to 30. I was interested in knowing the exact shape of the simultaneity curve during one of the complete "yoyo-type" cycles. The plot confirms what can be shown analytically: that during a segment of constant acceleration, the most rapid relative (positive or negative) ageing by the home sibling (according to the accelerating sibling) occurs as the velocity is passing through zero (in either direction).

If anyone would like to play around with these types of scenarios, I would be happy to email you the executable of my "CADO" program. I have an executable for Microsoft machines, and also for Linux machines. I don't have one for Macs.

If you want the program, send me an email, including a statement that you won't use (or knowingly allow the use of) the program for any commercial purposes. Also specify Microsoft or Linux.

To email me, remove the de-spamming animal name in my address: mlfasfzebra@comcast.net .

Mike Fontenot

 

[yuiop]

I have no disagreement with this as stated. Pedagogical value is a matter of judgment and also differs by individual. My concern is that since only some of Euclidean intuitions carry over, rather than having to define the precise limits on this, just learn to use non-Euclidean metrics directly.

The intuition you want to encourage is 'longer on paper is shorter in time', but you have to add all those qualifiers:

- only between a given pair of point, not for comparing paths between different points
- only if all path segments are less the 45 degrees to the t axis (assuming c=1)
- only if you can not move back in t (while you can move back in x, and in Euclidean geometry there is no 'special coordinate' where moving back is prohibited).

I honestly don't see a net pedagogical benefit by the time you add all the qualifiers.

I would like to take the qualifiers one at a time.

- only if all path segments are less the 45 degrees to the t axis (assuming c=1)

In an introduction to SR and spacetime diagrams, this is as good a time as any, to demonstrate that velocities of real particles or clocks that are greater than the speed of light is incompatible with SR and causes contradictions.

- only if you can not move back in t (while you can move back in x, and in Euclidean geometry there is no 'special coordinate' where moving back is prohibited).

That you can only move forward in coordinate time is a very natural intuition for any lay person that has not not yet been introduced to relativity. If a tutor wrote this problem on the board:

My route to work is 45 kilometers in a straight line. I leave home at 9.00 am. I arrive at the paper shop 30 kilometers down the road at 9.30 am and buy a newspaper. I leave the paper shop at 9.35 am and arrive at work at 8.35 am the same day. What is my average speed?

The students would immediately object to your traveling from the paper shop to your work in minus one hour. The concept that coordinate time only advances is entirely natural and we have never measured anything that contradicts this and the thermodynamic arrow of time generally supports this.

- only between a given pair of point, not for comparing paths between different points

I am not quite sure what you mean by this last qualifier. If by "comparing paths between different points" you mean comparing elapsed proper times between clocks that are initially co-located and finally co-located, then this is as good as time as any to introduce the students to the concept that you can not compare elapsed proper times of spatially separated clocks in meaningful invariant way. If two clocks are spatially separated, then different observers will disagree on which clock is ticking slower. Until the clocks are re-co-located any comparison of elapsed proper times is just a matter of opinion and is observer dependent.

To GrayGhost. I am not ignoring you. It is just that I believe that Dalespam is doing a better job in responding to your objections than I ever possibly could.

 

[GrayGhost]

FYI GrayGhost, the use of ict with a (++++) metric fell out of favor several decades ago. You can certainly find it in many older texts, but it is quite outdated. Now everybody uses ct or even t with a (-+++) metric or a (+---) metric. It doesn't make your comments wrong, just dated.

Maybe you misunderstood my prior post? Your response suggested that I may not have known that the Minkowski metrics (-+++) and (+---) are both used. I must say though, I was unaware indeed that any (++++) Minkowski metric was ever used, and find that quite suprising ...

Here's what I was saying ...

Yuiop suggested the 4d euclidean length of the path along ict' would be ... ct' = sqrt[ (ct)²+(vt)² ], ie the longest path, which stemmed from yuiop's assumption that (ct')²=(ct)²+(vt)². That's a (++++) euclidean metric. This would be true if the time axes were not imaginary, however they are.

So I merely pointed out that applying Pathagorus' theorem to our particular complex systems cannot produce yuiop's (++++) metric, they can only produce the Minkowskian (-+++ or +---) metrics. Which itself means that one cannot say (ct')²=(ct)²+(vt)² in the very first place, or equivalently ... one cannot say the longest path is ict'.

EDITED: On the other hand, if we ignore the fact that time is represented as imaginary, one can easily say (ct')²=(ct)²+(vt)² resulting in the euclidean metric (++++). But then that is mathemaically improper. Nonetheless, this makes the paper length of ict' the longest, as yuiop pointed out. Are there more benefits to doing this, than not doing this? IMO, I don't think so. I suppose one could always try, see if they can find ways of avoiding any resultant confusion maybe. I have witnessed countless debates over many years as to whether the longest worldline length is the shortest proper time experienced, versus whether the shortest worldline length is the shortest proper time experienced. Few surrender their position. However, I'd also have to submit that most do not qualify as to whether the system is euclidean vs Minkowskian. IOM though the "complex" euclidean system we begin with IS a Minkowskian system from the start.

Am I incorrect on this reasoning?

GrayGhost

 

[GrayGhost]

To GrayGhost. I am not ignoring you. It is just that I believe that Dalespam is doing a better job in responding to your objections than I ever possibly could.

Well, we'll see. No problem though. Thanx for the notice yuiop :)

GrayGhost

 

[yuiop]

Maybe you misunderstood my prior post? Your response suggested that I may not have known that the Minkowski metrics (-+++) and (+---) are both used. I must say though, I was unaware indeed that any (++++) Minkowski metric was ever used, and find that quite suprising ...

Here's what I was saying ...

Yuiop suggested the 4d euclidean length of the path along ict' would be ... ct' = sqrt[ (ct)²+(vt)² ], ie the longest path, which stemmed from yuiop's assumption that (ct')²=(ct)²+(vt)². That's a (++++) euclidean metric. This would be true if the time axes were not imaginary, however they are.

I think what Dalespam was getting at is that the Minkowski (-+++) metric:

s^2 = - (ct)^2 +x^2 + y^2 + z^2

can be written as a Minkowski (++++) metric like this:

s^2 = + (ict)^2 +x^2 + y^2 + z^2

which is no different to the first form that does not include the imaginary term. This seems to be your preferred form.
I am not sure why you are insisting that the time axes should be imaginary. I can write the Minkowski metric as:

c^2(\Delta \tau)^2 = (\Delta t)^2 - (\Delta x)^2 - (\Delta y)^2 - (\Delta z)^2

which is perfectly valid and does not involve any imaginary quantities at all. For any particle moving at the less than the speed of light, (\Delta \tau) is always real for any real value of (\Delta t).

 

[GrayGhost]

yuiop,

Thanx, for the response. I have to run at this instant, so I'll respond a little later. I see what you're saying. I'll try to cut to the chase. thanx.

GrayGhost