Whose Clock Slows Down in Relativity?

[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.