Whose Clock Slows Down in Relativity?

[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:

$$g= \left( \begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array} \right)$$

and a non-orthogonal metric:
$$h= \left( \begin{array}{cccc} 1 & 1 & 0 & 0 \\ 1 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array} \right)$$

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:

$$g= \left( \begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array} \right)$$

and a non-orthogonal metric:
$$h= \left( \begin{array}{cccc} 1 & 1 & 0 & 0 \\ 1 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array} \right)$$

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?