What are the implications of a Euclidean interpretation of special relativity?

[RandallB]

Mortimer; Been a while since I peeked in.
Have you decided if the (x₁, x₂, x₃) part the (x₁, x₂, x₃, c'tau') event coordinates for your theory, expects to correlate directly with a x,y,z location in our locally observed reality.

I.E. Have you decided if your theory is background-independent as I suspected. Or do you think you can preserve background-dependence.

 

[Mortimer]

Have you decided if the (x₁, x₂, x₃) part the (x₁, x₂, x₃, c'tau') event coordinates for your theory, expects to correlate directly with a x,y,z location in our locally observed reality.
I.E. Have you decided if your theory is background-independent as I suspected. Or do you think you can preserve background-dependence.

I'm quite convinced now that it is definitely background-independent, including c\tau.

 

[Hurkyl]

I think I'm beginning to figure out why I have a problem with this theory. Don't get me wrong -- I think it's a very neat idea, but the development seems lacking.


In SR, it was very clear what everything "meant". Space-time is a 4-D Minowski space. We can put coordinates \langle ct, x, y, z\rangle onto Minowski space, and that specifies a place and a time. \tau is simply the variable we use to denote a good parameter for a worldline. The expression (d(ct))^2 - (dx)^2 - (dy)^2 - (dz)^2 is the expression for a pseudometric on Minowski space, as represented in orthonormal coordinates. (i.e. an inertial reference frame)


When I read the presentation for Euclidean relativity, I feel like it's doing nothing more than playing around with the equations of SR to put them into a neat form. It talks about a space parametrized by 5 coordinates: \langle ct, x, y, z, c\tau \rangle. The equation (d(ct))^2 = (dx)^2 + (dy)^2 + (dz)^2 + (d(c\tau))^2 is merely an equation of motion -- it is not a metric. In fact, there seems to be no geometry done at all!

Presumably the central equation of motion (d(ct))^2 = (dx)^2 + (dy)^2 + (dz)^2 + (d(c\tau))^2 should be some sort of invariant: for any two "good" choice of coordinates, it should hold in one if and only if it holds in the other. But what does that say about the geometry? Should the expression (d(ct))^2 - (dx)^2 - (dy)^2 - (dz)^2 - (d(c\tau))^2 be considered an invariant? If so, then we've simply moved into a (4+1)-dimensional Minowski space!

Do we break up 5-d space into one-dimensional time and 4-dimensional \langle x,y,z,c\tau \rangle space, ala Newtonian mechanics? If so, then it makes sense to treat the 4-dimensional space as Euclidean. (But then again, if we foliate Minowski space, we can treat each slice as Euclidean as well) Then we have the neat fact that the equations of motion say that all particles travel with the same, fixed speed. But, we've reverted to a theory that has an absolute time parameter, and all of the philosophical problems with that.

Stepping back to special relativity, proper time has only translational symmetry -- a proper duration is an invariant quantity. In other words, \tau doesn't "mix" with space-time coordinates.

In the passage to Euclidean Relativity, we've lost that -- it attempts to treat \tau as just another spatial coordinate. Now that \tau is no longer an invariant, what do clocks measure?


Another crucial feature of special relativity is that photons travel along null geodesics. In other words, worldlines satisfying d(c\tau) = 0 were very special.

When we pass to Euclidean relativity, it seems to me that this crucial feature of special relativity has been entirely lost: it is destroyed by just about any Euclidean transformation of \langle x, y, z, c\tau\rangle-space.

In other words, I think the laws of physics are not invariant under Euclidean transformations! And that's a big problem.

Maybe as I read more, I will be satisfied on these issues, but skimming through the article doesn't make me very optimistic.

 

[Mortimer]

Thanks for your elaborate comments, Hurkyl. They once more set me thinking. It keeps being extremely difficult to switch between the Minkowski and the Euclidean interpretation.

Stepping back to special relativity, proper time has only tanslational symmetry -- a proper duration is an invariant quantity. In other words, \tau doesn't "mix" with space-time coordinates. In the passage to Euclidean Relativity, we've lost that -- it attempts to treat \tau as just another spatial coordinate. Now that \tau is no longer an invariant, what do clocks measure?

This is consistent in all Euclidean 4-vectors. The invariant in Minkowski space-time is no longer the invariant in Euclidean space-time. The original Minkowski time-component becomes the invariant. Also in the Euclidean interpretation, \tau is defined as what a clock reads that moves along with the object. For me it makes more sense that this value is not invariant. After all two observers that move relative to each other will both say that the other clock is moving slow, so how can proper time than be invariant during a transformation between frames?
At one time I came to the conclusion that Minkowski geometry is a pure mathematical thing that merely gives the correct mathematical results when given the correct input. It has no link to physical reality. I've tried to express this in a separate article at http://www.euclideanrelativity.com/4vectors . Here you will find more on the geometrical background of the Euclidean interpretation. It is perhaps easier to initially treat time t as a parameter for tracking worldlines, similar to the way \tau is used in Minkowski geometry. Other endorsers of Euclidean relativity, like e.g. Hans Montanus do that too. The sole reason why I don't is that its designation as a real dimension is used explicitly in a follow-up article that can be found at http://www.euclideanrelativity.com/dim2html .

Another crucial feature of special relativity is that photons travel along null geodesics. In other words, worldlines satisfying d(c\tau) = 0 were very special. When we pass to Euclidean relativity, it seems to me that this crucial feature of special relativity has been entirely lost: it is destroyed by just about any Euclidean transformation of \langle x, y, z, c\tau\rangle-space.

I've realized this too. In section 4 of the article is mentioned:

Equation (18) is in fact based on the universality of light speed and the basis for reasoning is that an object, e.g. a photon, having speed c for an observer in frame x will still have that same speed for an observer in frame x'. This is one of Einstein's original postulates and also in this Euclidean approach it will still be maintained as a valid postulate, which essentially means that the photons velocity vector, as measured from the moving frame, must have rotated along with that frame.

I just accept this as being a mandatory consequence of the Euclidean interpretation. I do not try to explain it in the article. So far, I have not run into anything that would be inconsistent with it.

In other words, I think the laws of physics are not invariant under Euclidean transformations! And that's a big problem.

The point is that, although there are some strange consequences associated with the Euclidean interpretation, I did not find any law that would be violated by them after careful consideration. But being not even graduate level, I cannot rule out to have missed something crucial.

The biggest problem in getting to appreciate the Euclidean interpretation is to get rid of the Minkowski way of thinking that has been taught in all relativity courses for nearly 100 years now. In particular when you have grown accustomed to it over the years, it's hard to avoid it in the Euclidean interpretation. For me it's often the other way around. It's hard to avoid the Euclidean way of thinking in Minkowski discussions and I am aware that this is in fact a "handicap".

 

[Hurkyl]

In other words, I think the laws of physics are not invariant under Euclidean transformations! And that's a big problem.

I wanted to elaborate more upon this one. I will use the case of a photon again.


Suppose we had a worldline (along with its proper time) defined by the equations:

x = ct
y = 0
z = 0
\tau = 0

One orientation-preserving Euclidean motion is the one that cyclically permutes the y-, x-, and c\tau-axes. So, in Euclidean relativity, we have the equally valid coordinatization:

x' = 0
y' = 0
z' = 0
\tau' = t

which is, of course, a perfectly good worldline according to special relativity, just not that of a photon.

Also in the Euclidean interpretation, \tau is defined as what a clock reads that moves along with the object.

which essentially means that the photons velocity vector, as measured from the moving frame, must have rotated along with that frame.

These weird problems suggest that they are coordinate-dependent ideas, and not true geometric entities.

But, it struck me how you can maintain Euclidean symmetries without any of these weird problems.

Instead of insisting that \tau be what clocks measure, which breaks Euclidean symmetry, you could introduce a preferred direction into 4-d space.

In other words, you postulate the existence of a 4-dimensional vector q.

Then, instead of defining \tau to be what clocks measure, you could say that clocks measure \mathbf{v} \cdot d\mathbf{q}. (Where v is the velocity 4-vector). In other words, clocks measure displacement along the q direction.

Similarly, photons would be moving in a direction perpendicular to q.

This restores Euclidean symmetry, because q would rotate along with the coordinates.

Going back to my original example:

x = ct
y = 0
z = 0
\tau = 0
In this <x,y,z,c\tau> coordinate chart, we have

q=<0,0,0,1>

Now, if I applied the Euclidean rotation, I result in

x' = 0
y' = 0
z' = 0
\tau' = t

with q=<0,1,0,0>

And we see that the velocity vector <0,0,0,c> is perpendicular to the preferred direction <0,1,0,0>.


Maybe this is what your article said, but I feel better having written it my way.



By the way, allow me to suggest that telling people they have their understanding of SR wrong is not the right way to get people interested! When I see statements like:

Figure 3 shows the background of this multiplication and it is clear that the components of the Minkowski 4-vector can have no physical meaning.

I get quite put-off. When I read this paper, I'm strongly inclined to think of reasons why the author has no clue about special relativity than attempt to appreciate the new approach.

 

[Mortimer]

I'm afraid I do not fully get the yeast of your idea with the 4-dimensional vector q. It seems like it is 1 dimension short for it to work .

What exactly it is that a clock reads in Euclidean space-time has always been nagging a bit in the background and, frankly, I have ignored this somewhat. You remarks have made it nagging harder and make me wonder if it is not actually t that plays the role of "Euclidean proper time". It would preserve the symmetry that you are looking for and would be consistent with the overall approach but I feel that this might somewhere be inconsistent with experimental observations of time dilation effects.
I'm not completely ready with this yet. I'll have to let it settle.

By the way, allow me to suggest that telling people they have their understanding of SR wrong is not the right way to get people interested!

I appreciate your warning. It has of course never been my intention to say that the Minkowski approach is wrong. The Euclidean approach might however ease people's understanding of relativistic phenomena. Whatever comes out of the Euclidean approach should be also reachable via the Minkowski approach, but probably less intuitively. In the meantime I have adjusted here and there some text in the 4-vector article and hope it is less offensive now.

 

[CarlB]

Dear Hurkyl,

As a supporter of Euclidean relativity, I feel I should join in the fray.

When I read the presentation for Euclidean relativity, I feel like it's doing nothing more than playing around with the equations of SR to put them into a neat form. It talks about a space parametrized by 5 coordinates: \langle ct, x, y, z, c\tau \rangle. The equation (d(ct))^2 = (dx)^2 + (dy)^2 + (dz)^2 + (d(c\tau))^2 is merely an equation of motion -- it is not a metric. In fact, there seems to be no geometry done at all!

The geometry is missing only because all particles, massive or not, are being constrained to travel at speed c. If you want to add geometry to this, a cool way is to assume a Clifford algebra for the 5 dimensions, but with all particles traveling at equal speeds, the only geometry you need is that of straight lines.

The Euclidean rotations are ugly in that they correspond to boosts or rotations depending on orientation. But this is only a superficial problem. That is, physics is divided into kinematics, the art of divining where things go after they are set into motion, and dynamics, the art of divining why things are set in motion. As long as you look at Euclidean relativity as a kinematical theory it is perfect. But kinematics need not involve energy. All kinematics needs is straight lines. It is in dynamics where Euclidean relativity needs work, but that is a subject for quantum mechanics, not classical mechanics, I think.

Probably the best argument for Euclidean relativity, I think, is that it gives a physical reason why it is that two distinct observers can agree on a characteristic of an object. That is, they can agree on how fast the object is aging. The problem that the two observers need not agree on how old the object is can be alleviated by assuming that the proper time dimension is cyclic and small. The aging of an object is proportional to how many times its world line makes a circuit around the hidden dimension. This has other consequences, so I think I am one of the few Euclidean relativity people who goes down this route.

But, we've reverted to a theory that has an absolute time parameter, and all of the philosophical problems with that.

I think that the choice of reference frame amounts to one of the gauges that quantum mechanics is afflicted with. That is, the details of a calculation depend on the choice of reference frame, but the final results of the calculation do not. That in itself suggests that standard quantum mechanics has an "ontological" problem with relativity.

Probably the best author on this subject is the famous physicist David Bohm, who, in his book, the "The Undivided Universe" provides a justification for assuming a preferred reference frame.

Stepping back to special relativity, proper time has only translational symmetry -- a proper duration is an invariant quantity. In other words, \tau doesn't "mix" with space-time coordinates. In the passage to Euclidean Relativity, we've lost that -- it attempts to treat \tau as just another spatial coordinate. Now that \tau is no longer an invariant, what do clocks measure?

For any given path, tau is still an invariant.

Another crucial feature of special relativity is that photons travel along null geodesics. In other words, worldlines satisfying d(c\tau) = 0 were very special.

When we pass to Euclidean relativity, it seems to me that this crucial feature of special relativity has been entirely lost: it is destroyed by just about any Euclidean transformation of \langle x, y, z, c\tau\rangle-space.

Yes, these kinds of transformations are boosts and you can't boost a photon. But this is just a dynamical issue, not a kinematical one.

In other words, I think the laws of physics are not invariant under Euclidean transformations! And that's a big problem.

Well, with Euclidean relativity, you end up with different laws of physics and different symmetries. There are still symmetries. And like I mentioned above, I don't think Euclidean relativity works very well for classical mechanics, for pretty much the same reasons you stated. As for quantum mechanics, well if you assume that the tau dimension is cyclic, that cyclicity will make it different to a theory based on waves and that is enough to give you a different kinematics.

Carl

 

[robphy]

Yes, these kinds of transformations are boosts and you can't boost a photon. But this is just a dynamical issue, not a kinematical one.

Sorry for jumping in here... but what do you mean by "you can't boost a photon"?

 

[Hurkyl]

Hrm; I don't quite think you got the point of my objection. (or maybe you did and your response went over my head. ) So let me try it again! This time I have an analogy.


I could say "I want to do Newtonian mechanics on a Euclidean 4-D space-time." So things are written in (t, x, y, z) coordinates, and I have a metric dt² + dx² + dy² + dz², and all is good.

But when I start writing down the physics, we find that they are all coordinate-dependent things.

And then someone tries to do a rotation in the (t, x)-plane, and I have to jump in and tell them "you can't do that! Time always has to be the first coordinate!"

Well, it's clear I'm not doing Newtonian mechanics in a Euclidean way at all! While I've declared that I'm working in Euclidean 4-space, the laws of physics are not geometric at all. They depend on the choice of coordinates instead of entirely upon Euclidean geometric concepts.


This is the heart of my current problem with Euclidean relativity: physics done in (x, y, z, \tau) space doesn't seem geometric at all. In particular, if:

For any given path, tau is still an invariant.

then things certainly are not geometric, because no coordinate displacement can be invariant under Euclidean motions! (which preserve all of the geometry)

It feels just like my Newtonian geometer -- we say space is Euclidean, but we jump in and forbid anyone from trying to rotate in the (x, \tau) plane!


Now, I would feel much better if \tau wasn't a coordinate at all. We say that we are really working in (t, w, x, y, z) coordinates, where w is just another spatial coordinate. Then, maybe \tau could be a (global) vector which points in the proper time direction.

by assuming that the proper time dimension is cyclic and small.

or, maybe the \tau-direction is determined from the geometry as the direction that best points "around" the loop.

But either way, treating \tau as "just another Euclidean coordinate" seems to be the wrong, just like my Newtonian geometer.

(I actually like the curled dimension too, but for a different reason: since only 4 dimensions seem to matter for telling when two particles bump into each other! And the fact that only d\tau matters -- not the actual value yourself)

The aging of an object is proportional to how many times its world line makes a circuit around the hidden dimension.

This one bothers me a little, though. It either means that \tau is only defined for closed loops, or that its calculation is dependent upon splitting space into 3 unfurled + 1 looped dimensions.

Yes, these kinds of transformations are boosts and you can't boost a photon. But this is just a dynamical issue, not a kinematical one.

I will restate my point specifically for this comment:

If we declare (x, y, z, \tau)-space to be Euclidean, then we can make this transformation. And when the geometry says "we can make this rotation" but the physics says "no you can't", then I say that the geometry isn't appropriate for the physics!

 

[CarlB]

Sorry for jumping in here... but what do you mean by "you can't boost a photon"?

I mean that you can't change the velocity of a photon. If you transform to a boosted coordinate system, you still have a photon traveling at speed c.

Boosting, as an operation on massive particles, is both kinematical and dynamical. It is kinematical in that it changes the velocity of a particle. It is dynamical in that it changes the particle's energy. With massless particles, by contrast, boosting changes only the dynamics.

This is all in the context of my claim that Euclidean relativity, as applied to (other than quantum) mechanics, does great with kinematics but doesn't do dynamics very well. Dynamics is simply not a symmetry under boosts, kinematics is.

Carl

 

[CarlB]

It feels just like my Newtonian geometer -- we say space is Euclidean, but we jump in and forbid anyone from trying to rotate in the (x, \tau) plane!

Ah, now I see. There is some disagreement in the field, but a few of us think that Euclidean relativity implies a preferred reference frame. It's not talked about much, because the results of physics calculations are known, by extensive experiment, to be approximately Lorentz symmetric. But philosophically, Lorentz's theory of spacetime included a preferred reference frame.

What this boils down to is that sure, you can make boost type rotations, but when you do, the result will be a different assumption for the preferred reference frame. Like I said above, this amounts to a choice of gauge. The preferred reference frame is assumed to be there, but it is not (yet) possible for us to detect it.

But either way, treating \tau as "just another Euclidean coordinate" seems to be the wrong, just like my Newtonian geometer.

When one assumes a preferred reference frame, this not only locks down the tau coordinate, it also locks down all the others. You can't rotate x into y any more than you can rotate x into tau. The choice of x, y, and z is a gauge choice. The results of your physics do not depend on how you choose these axes, but your calculations along the way definitely do.

What I'm saying here is that 3-d rotations are a property of physics, they are not a property of space-time itself. If they were, centrifugal force wouldn't work.

This one bothers me a little, though. It either means that \tau is only defined for closed loops, or that its calculation is dependent upon splitting space into 3 unfurled + 1 looped dimensions.

Yes, this should bother you. However, as far as actual calculations, the effect is negligible if the circumference of the closed loop is sufficiently small. To measure short time periods requires high energies, so if the time light takes to travel around that hidden dimension is on the order of the Plank time, it's (WAY) outside our experimental range.

Another objection to a cyclic hidden dimension is that when you apply waves to it, you end up with quantized wave numbers around the hidden dimension. The saving assumption is that only the lowest energy waves are of interest and these are the ones that have just one wave length around. These waves then travel at just a hair under the speed of light, and one assumes they correspond to the massless handed chiral particles. In other words, applying waves to a hidden dimension implies that you have to break the electron up into massless states.

By the way, another feature of a hidden dimension dates back to the observation of de Broglie that matter waves have phase velocities that exceed the speed of light. If you take into account the hidden dimension, the phase velocity of those waves drops back down to c. It is only when you ignore the hidden dimension that you find that their phase velocity is greater than c.

The analogy on the beach is the fact that when waves are approaching the shore from a direction close to perpendicular to the beach, the breakers move up and down the coast at a speed far in excess of the wave velocity in the ocean itself. If all you were aware of was the beach, you would think that breakers moved faster than the speed of sound in water.

Here's a translation of de Broglie's original announcement, see the 3rd paragraph for the note about phase velocities in matter waves:
http://www.davis-inc.com/physics/broglie/broglie.shtml

Carl

 

[member 11137]

Ah, now I see. There is some disagreement in the field, but a few of us think that Euclidean relativity implies a preferred reference frame. It's not talked about much, because the results of physics calculations are known, by extensive experiment, to be approximately Lorentz symmetric. But philosophically, Lorentz's theory of spacetime included a preferred reference frame.

What this boils down to is that sure, you can make boost type rotations, but when you do, the result will be a different assumption for the preferred reference frame. Like I said above, this amounts to a choice of gauge. The preferred reference frame is assumed to be there, but it is not (yet) possible for us to detect it.

When one assumes a preferred reference frame, this not only locks down the tau coordinate, it also locks down all the others. You can't rotate x into y any more than you can rotate x into tau. The choice of x, y, and z is a gauge choice. The results of your physics do not depend on how you choose these axes, but your calculations along the way definitely do.

What I'm saying here is that 3-d rotations are a property of physics, they are not a property of space-time itself. If they were, centrifugal force wouldn't work.

Carl

It is an amazing thing to observe the permanent research of simplicity in the human history. The essay of Mortimer is a supplementary one. Things would be so much easy to write, understand and explain if there were ... easy; but they are not; unfortunately.

Another item that comes repetitivly is the question of a preferred frame. I also do have my opinion on the subject. The unique possibility that I can imagine for a preferred family of frames to exist would be the following: the human brain can discover a set of frames that would be greater than the set of frames effectively realized in the nature; in the physics. Then and only then, we could write: physics takes place in a preferred family of frames. That's for the logical side of the question.

Now for this preferred family to exist, this would also mean that our brain can imagine series of transformations realized in the nature and resulting exclusively in the possibility to jump from one preferred frame to another preferd frame. To prove that we would have discover a greatest set than the set realized in the nature, we must discover transformations that allow to go from one preferd frame to another anyone that is not realized in the nature.

To sum up, mathematics must generate more possible frames than physics do realize. With other words, there must exist holes in the mathematic structure corresponding to nothing in the reality.

Just some ideas to, I hope it, help in a better comprehension of this question.

 

[kmarinas86]

Yes, If the other planet would be stationary with respect to Earth, its proper time would run synchronous with that of Earth (apart from possible gravitational time dilation differences). Actually there is no difference between this planet and the asteroid in the example. The asteroid is also stationary with respect to Earth.
I'm very careful with this because the missile's individual elementary particles will obviously not turn each into their corresponding anti-particles. That would potentially violate conservation laws (of e.g. electrical charge). In the article, I'm merely suggesting a possible relation of the negative timespeeds that can occur with the existence of anti-particles that are also sometimes suggested to kind of run backwards in time (in e.g. Feynman diagrams). In the original text this remark was just a footnote but the editor of the journal preferred all footnotes to be placed in the main text.

I have attached two pictures that might clarify why it is necessary to take the negative root.
The first picture is the same as Fig. 3 in the article but I have added the vector X, which is the time-velocity vector of the third object (later on the missile). In this picture, X runs along the positive x_4 axis.
In the second picture, the actual situation as described with the spaceship and the missile is reflected. The frame x&#039;&#039; has now rotated beyond \pi/2 relative to frame x and vector X now runs along the negative x_4 axis. Vector W is the spatial velocity of the missile (as observed from Earth) with magnitude v_m. The magnitude of X is \chi_m=\sqrt{c^2-v_m^2} but when calculating \tau_m from this value one must obviously take the negative value in order to correctly account for the direction of the time-velocity vector X along the negative x_4 axis.

You might want to check this out:

http://www.livescience.com/technology/060518_light_backward.html

 

[Mortimer]

Weird...
Anyway, it's not related to the addition formula that comes out of my article as far as I can see.

 

[RandallB]

You might want to check this out:

http://www.livescience.com/technology/060518_light_backward.html

Weird that such an inconsistent interpretation should be put forward by a University.
Even Boyd said "no information is truly moving faster than light," Yet Boyd in his animated analogy is showing a detectable max point of a light pulse skipping over a real distance in what appears to be real time. In effect two parts of one pulse are existing simultaneously at what is a space-like separation.

This boils down to a simple SR simultaneity problem where events separated by to much space-time can be made to appear " simultaneous" by selecting an impossible reference frame. Namely one that is A) moving in the other direction and B) moving BACKWARDS in time. I don't even think C) FTL that he claims is even needed. Fast as light going backwards in time should do.
More like the assumption of small scale backwards in time movement of positrons equivalent to electrons problems in Feynman diagrams than an observation of something real.

I think it much more likely that some rather broad assumptions presuming a real backwards time frame to insert transformed measurements from our reference frame to create the display shown.

The kind of loose interpretation of things I hope your trying to avoid Mortimer.

 

[bda]

Hi, I registered specifically to post in this thread; let me use Hurkyl's post as point of departure

In SR, it was very clear what everything "meant". Space-time is a 4-D Minowski space. We can put coordinates \langle ct, x, y, z\rangle onto Minowski space, and that specifies a place and a time. \tau is simply the variable we use to denote a good parameter for a worldline. The expression (d(ct))^2 - (dx)^2 - (dy)^2 - (dz)^2 is the expression for a pseudometric on Minowski space, as represented in orthonormal coordinates. (i.e. an inertial reference frame)

I propose we go back to basics and define evrything, both in Minkowski and Euclidean spaces. I don't feel at ease assuming synchronized clocks and measuring rods because I think the concepts are ill defined. How can one insure that a moving clock is synchronized with a stationary clock, even if they happen to be at the same place at a given instant? Synchronizing distant stationary clocks poses similar problems and measuring rods assume synchronized clocks at both ends.

I rather prefer Bondi's approach, which you can find in several textbooks (see for instance Ray d'Inverno) but also in my own paper http://www.arxiv.org/abs/physics/0201002 . The idea is that you have just one clock and you can send a radar pulse which is reflected by any distant object; your clock allows you to time the send and receive instant. The argument is then extended by letting the radar pulse bunce back and forth.

If we call t_0 to the send instant and t_2 to the receive instant, the time and position coordinates of the distant object are defined by

t_1 = (t_0 + t_2)/2
x_1 = (t_2 - t_0)/2

I've assumed c = 1.

By letting the pulse bounce back and forth we get a succession of even t instants which allows the calculation of the odd t and x, for the distant object. By plotting the (x,t) pairs with equal indices we draw the worldline of the distant object. This is the way Bondi and d'Inverno introduce special relativity and I find it much more manageable than the standard approach. Now for Euclidean relativity.

Define the new coordinate

\tau_1 = \sqrt{t_0 \ast t_2}

We can now plot the (x, \tau) pairs, to get a different worldline. The two worldlines are related by

d t^2 = dx^2 + d \tau^2,

as I demonstrate in the cited reference.

The equation (d(ct))^2 = (dx)^2 + (dy)^2 + (dz)^2 + (d(c\tau))^2 is merely an equation of motion -- it is not a metric. In fact, there seems to be no geometry done at all!

Why is it not a metric? It is telling us that space is Euclidean and so distances are measured according to Pythagoras theorem.

Presumably the central equation of motion (d(ct))^2 = (dx)^2 + (dy)^2 + (dz)^2 + (d(c\tau))^2 should be some sort of invariant: for any two "good" choice of coordinates, it should hold in one if and only if it holds in the other. But what does that say about the geometry? Should the expression (d(ct))^2 - (dx)^2 - (dy)^2 - (dz)^2 - (d(c\tau))^2 be considered an invariant? If so, then we've simply moved into a (4+1)-dimensional Minowski space!

Here I diverge from Rob. There are two invariants: dt and d \tau, which means that frame axes are orthogonal only for a given preferred frame. I know people will now ask what is the preferred frame attached to but I will have to defer the answer to that to a later post; there's a lot to be said before we can come to that.

Another crucial feature of special relativity is that photons travel along null geodesics. In other words, worldlines satisfying d(c\tau) = 0 were very special.

It is just as important in Euclidean relativity; it means photons travel in 3D space, preserving the fourth coordinate.

By the way, this brings me to one of my strong points of divergence with Rob. Rob says that photons travel on null geodesics of X_4[\itex] but null geodesics need mixed signature spaces; as far as I can understand Rob&#039;s spaces have all plus signatures and so they cannot have null geodesics. Maybe Rob will like to clarify this point.<br /> <br /> <br /> Jose

 

[Hurkyl]

I don't feel at ease assuming synchronized clocks and measuring rods because I think the concepts are ill defined.

They can be well-defined, though. For example, Einstein's convention says that another clock is stationary and synchronized WRT your one-clock iff:

(stationary) All of the x's are equal.
(synchronized) The time on the other clock at the time when the radar pulse reflects off of it is always equal to the corresponding t.

And it's a happy fact of SR that, for an inertial one-clock and for each possible value of x, it is possible to have an identical clock at that location synchronized with your one-clock. (in this sense)

Quote:
The equation (d(ct))^2 = (dx)^2 + (dy)^2 + (dz)^2 + (d(c\tau))^2 is merely an equation of motion -- it is not a metric. In fact, there seems to be no geometry done at all!

Why is it not a metric? It is telling us that space is Euclidean and so distances are measured according to Pythagoras theorem.

If it is telling us that \langle c\tau, ct, x, y, z \rangle-space is Euclidean, then it would be assigning the length (\Delta c\tau)^2 + (\Delta ct)^2 + (\Delta x)^2 + (\Delta y)^2 + (\Delta z)^2 to the displacement 5-vector \langle \Delta c\tau, \Delta ct, \Delta x, \Delta y, \Delta z \rangle.

But that's not what this equation does. It is an equation of motion -- it says that, along any worldline, the displacement along the ct-axis is equal to the displacement in the (Euclidean) 4-D \langle c\tau, x, y, z \rangle slices.

(at least, that's what it does if I assume you are using the Euclidean metric on the \langle c\tau, x, y, z \rangle slices)

There are two invariants: dt and d\tau, which means that frame axes are orthogonal only for a given preferred frame.

Which I see to be a big problem, because this is contrary to the whole name "Euclidean Special Relativity".

First off, it means that you are not working in a grand Euclidean space-time. Instead, you are doing something more akin to Newtonian mechanics, with Euclidean 3-D space, and two (independent) time coordinates.

I say "independent" because there is no geometric relationship between them. There is only equations of motion -- equations that express a relationship between how a worldline moves in 3-D space to how it moves in the two time dimensions.

Secondly, it seems like you're not doing any sort of relativity at all!

 

[bda]

They can be well-defined, though. For example, Einstein's convention says that another clock is stationary and synchronized WRT your one-clock iff:

(stationary) All of the x's are equal.
(synchronized) The time on the other clock at the time when the radar pulse reflects off of it is always equal to the corresponding t.

I don't reject Einstein's convention but I find Bondi's more natural and easier to work with; if I have only one measuring instrument I can avoid all synchronizations. That Bondi's approach works for special relativity is a fact generally accepted.

If it is telling us that \langle c\tau, ct, x, y, z \rangle-space is Euclidean, then it would be assigning the length (\Delta c\tau)^2 + (\Delta ct)^2 + (\Delta x)^2 + (\Delta y)^2 + (\Delta z)^2 to the displacement 5-vector \langle \Delta c\tau, \Delta ct, \Delta x, \Delta y, \Delta z \rangle.

It is telling us that \langle x, y, z, \tau \rangle is 4D Euclidean space and dt is the interval of this space. Let me go one step back.

My approach has differences to Rob's; I say: let us accept without discussion that \langle t, x, y, z, \tau \rangle is 5D with signature (-++++), that is, the interval in this space is given by the quadratic form:

(ds)^2 = -(dt)^2 +(dx)^2 + (dy)^2 + (dz)^2 + (d \tau)^2

This will be a first principle, to which we add a second one: all motion is along null geodesics of 5D space (ds)^2 = 0.

The two principles together define 4D space without a metric (null displacement does that). We are allowed to pull to the left hand side of the equation any of the terms on the rhs; if we pull (d \tau)^2 we get Minkowski 4-space and if we pull (dt)^2 we get Euclidean 4-space.

This operation is entirely analogous to what we are all used to in special relativity when we say that light is restricted to the light cone. We are then defining 3-space without a metric and we give it one by writing

(dt)^2 = (dx)^2 + (dy)^2 + (dz)^2

which is the basis of Fermat's principle. What I propose is that we do optics with one extra dimension.

Secondly, it seems like you're not doing any sort of relativity at all!

Let us go one step at a time. I tried to show above that special and Euclidean relativity are both daughters of the 5D null displacement principle; geodesics of one space can be one to one maped to the other space, although points of one space cannot be maped to the other.


Jose

 

[Hurkyl]

It is telling us that \langle x, y, z, \tau \rangle is 4D Euclidean space and dt is the interval of this space. Let me go one step back.

If dt is merely the interval, then it can't be a physical coordinate too! That's fine if you don't want t to be a physical coordinate (though that does raise some problems)... but everything I've read so far suggests that you really do want t as a physical coordinate.

My approach has differences to Rob's; I say: let us accept without discussion that \langle t, x, y, z, \tau \rangle is 5D with signature (-++++), that is, the interval in this space is given by the quadratic form:

(ds)^2 = -(dt)^2 +(dx)^2 + (dy)^2 + (dz)^2 + (d \tau)^2

This will be a first principle, to which we add a second one: all motion is along null geodesics of 5D space (ds)^2 = 0.

This makes me wonder what you plan to gain; ESR is advertised on the basis that it works in Euclidean space... but here all you've done is trade in 3+1 Minowski space for a 4+1 Minowski space!

Or did you not mean to suggest that this approach is an ESR approach?

The two principles together define 4D space without a metric

No, they don't. They define a 5D space with a metric. And if we take any 4D plane through this space, it is automatically equipped with a metric! From Minowski 4+1 space, we can easily obtain Euclidean 4-space, or Minowski 3+1 space, by simply projecting away one of the coordinates.

We are allowed to pull to the left hand side of the equation any of the terms on the rhs; if we pull (d \tau)^2 we get Minkowski 4-space and if we pull (dt)^2 we get Euclidean 4-space.

But this doesn't produce any 4-spaces at all. You are simply rewriting your 5D equation of motion in a form that formally resembles the metrics on Minowski 3+1- and Euclidean 4-space respectively.

 

[bda]

This makes me wonder what you plan to gain; ESR is advertised on the basis that it works in Euclidean space... but here all you've done is trade in 3+1 Minowski space for a 4+1 Minowski space!

Or did you not mean to suggest that this approach is an ESR approach?

I plan to place special and Euclidean relativity side by side, being able to translate between the two. One always gains perspective by looking at a problem from diferent angles. When the 5D null displacement principle is replaced by the more fundamental concept of 5D monogenic functions we get into QM but I cannot jump into that straight away

No, they don't. They define a 5D space with a metric. And if we take any 4D plane through this space, it is automatically equipped with a metric! From Minowski 4+1 space, we can easily obtain Euclidean 4-space, or Minowski 3+1 space, by simply projecting away one of the coordinates.



But this doesn't produce any 4-spaces at all. You are simply rewriting your 5D equation of motion in a form that formally resembles the metrics on Minowski 3+1- and Euclidean 4-space respectively.

By placing myself on the null cone I can no longer use s as an affine parameter and I am allowed to choose either t or \tau for parameters along null geodesics (I could have made other choices, of course). I agree with you that this does not produce 4-spaces; what produces 4-space is the null cone. The two alternative ways of working with null geodesics are formally equivalent to working in flat Minkowski or Euclidean 4-spaces. I believe if one wants to be formally correct there is a lot more to be said about these operations but that will not alter the substance that we have a means of mapping 4D Minkowski geodesics to Eucliedean 4D ones; so far I am not claiming anything else.

If the readers of this thread will put up with me I will develop the theory in forthcoming posts; I need some feedback on whether or not people want me to do that because I don't want to impose myself on anybody


Jose

 

[Mortimer]

By the way, this brings me to one of my strong points of divergence with Rob. Rob says that photons travel on null geodesics of X_4 but null geodesics need mixed signature spaces; as far as I can understand Rob's spaces have all plus signatures and so they cannot have null geodesics. Maybe Rob will like to clarify this point.

Hello Jose,
Excuse me for reacting so late but I only now returned from work.

As I see it, the "null geodesic" is defined as the path along which the tangent vector has norm 0 in a geometry with metric (-+++), hence the word "null". The norm 0 results from the ds^2=0 in the geodesic.

When this is translated to Euclidean relativity with metric (++++), ds^2 is not zero (it then equals (cdt)^2 for the photon which >0) but the displacement in the \tau dimension is zero. So strictly spoken one could not speak of a "null geodesic" any more; it is a timelike geodesic according to the original definition from Minkowski space-time. I have maintained the use of the familiar term in Euclidean relativity because it is generally associated with the path of massless particles and referring to that as "timelike" in Euclidean relativity would likely cause a lot of confusion.

It would probably be best however to introduce a whole new term in Euclidean relativity for the path of massless particles and I am inclined to suggest something like "3D geodesics" versus "4D geodesics" of mass carrying particles.


Rob

 

[Hurkyl]

As I've suggested, many of my problems are just with the advertising/naming/etc -- I've made those objections and I'll try to stop harping on them.


One thing I wanted to point out is that there is no natural way to map worldlines in 3+1-space to worldlines in 4-space. What you need to work with are "pointed" worldlines; that is, you have to choose a point on the worldline acts as the origin. (e.g. for a worldline in 3+1-space, you have to choose what point corresponds to \tau = 0 before you can map it to 4-space)

By placing myself on the null cone
...
I believe if one wants to be formally correct there is a lot more to be said about these operations but that will not alter the substance that we have a means of mapping 4D Minkowski geodesics to Eucliedean 4D ones; so far I am not claiming anything else.

I have a problem with the journey, but not the destination. I have no problem with mapping pointed geodesics back and forth between 3+1-space and 4-space (and up to geodesics in 4+1-space)... I just have a problem with the way you go about doing it. I don't know if it will matter, so I won't say any more about it right now.

 

[CarlB]

One thing I wanted to point out is that there is no natural way to map worldlines in 3+1-space to worldlines in 4-space. What you need to work with are "pointed" worldlines; that is, you have to choose a point on the worldline acts as the origin. (e.g. for a worldline in 3+1-space, you have to choose what point corresponds to \tau = 0 before you can map it to 4-space).

That is also one of my problems with that way of doing it. That is, if you are going to talk about a Euclidean relativity, it should map to the coordinates I plug into my dad's GPS device.

My solution for this is to make the tau dimension be very small, so that errors in converting coordinates from one version to the other may be ignored. Sometimes it seems that Euclidean relativity is the sort of heresy that reminds one of Tolstoy. Everyone happy with relativity is the same, while everyone unhappy with it is different.

Carl

 

[bda]

I believe we are on the right track now, so let us take a few more steps.

One thing I wanted to point out is that there is no natural way to map worldlines in 3+1-space to worldlines in 4-space. What you need to work with are "pointed" worldlines; that is, you have to choose a point on the worldline acts as the origin. (e.g. for a worldline in 3+1-space, you have to choose what point corresponds to \tau = 0 before you can map it to 4-space)

Quite right. I would like to rephrase that to make sure we understand exactly where we are: Although geodesics can be mapped from 3+1- to 4-space and back, the same thing cannot be done with points, that is, if three geodesics cross at one point in 3+1-space they will normally not have a common crossing point in 4-space; this has important consequences.

What is known as an event in special relativity, a set of particular values for \langle t, x, y, z \rangle, cannot usually be univocally translated into a set of particular values for \langle x, y, z, \tau \rangle. This is, I think, why Carl proposes that the \tau be curled up in a tight helix, but he will have to explain that himself; I am sticking to flat spaces. Furthermore, it must also be noted that only the spacelike part of 3+1-space is mapped to the full 4-space. We could, if needed, make a separate mapping from the timelike part to another full 4-space.

I said above that points in 3+1-space don't usually map to points in 4-space but now remember how I modified Bondi's approach to define both t and \tau:

t = (t_0 + t_2)/2

\tau = \sqrt{t_0 \ast t_2}

As we move the distant object closer to the observer t_0 \rightarrow t_2 and so \tau \rightarrow t. So, for the observer, there is no problem in translating from one space to the other. This means, for instance, that I can map a collision event from one space to the other, because it happens at one single point in spacetime and I can place myself, as observer, at that point. Please see ArXiv: physics/0201002 for a collision discussion.

Now let us think a little about how we translate distant or moving objects. We must restrict the discussion to objects moving on geodesics (straight lines); I will say a little bit about dynamics below. The observer's worldline is the t axis in 3+1-space and the \tau axis in 4-space; as we have seen the two measurements coincide for the observer. If an object's worldline crosses the observer's at any point, this point can be mapped from one space to the other; since the worldlines are also mapped, we just need to translate time intervals measured in 3+1-space into distances measured on the worldline in 4-space.

The problem is a bit more tricky if the object's worldline never crossed the observer's. The two definitions above should always solve the problem but there is one philosophical argument I would like to advance: If the Universe is expanding from a big bang it can be argued that all worldlines must have crossed at some time in the past, so justifying the synchronism of all clocks.

Dynamics is a problem and I would say it does not map from one space to the other. I've given some attention to the different dynamics in the two spaces and you can read about in the paper I mentioned above. However I don't think we should care too much about those differences because special relativity, as we know, is not the final answer to dynamics. When we come to discussing equivalence between general relativity and a generalization of ESR, I will deal with dynamics problems.

Jose

 

[bda]

Sometimes it seems that Euclidean relativity is the sort of heresy that reminds one of Tolstoy. Everyone happy with relativity is the same, while everyone unhappy with it is different.

I can see your point but I don't think it apllies here. I wrote about an alternative way of looking at problems but I did not say relativity was wrong. My position about this is that no physical theory is ever final and so no physical theory is ever absolutely right, which does not mean it is wrong. We must accept that every theory has an application domain; sometimes we may think that the application domain is wider than it really is and I believe that to be the case with a large number of physicists in relation to GR.

Jose

 

[Hurkyl]

\tau = \sqrt{t_0 \ast t_2}

I just noticed a problem with this -- it has absolutely nothing to do with the proper time along a worldline! There are several ways to see it (such as looking at two worldlines that cross twice), but I'll address it this way:

The two worldlines are related by

d t^2 = dx^2 + d \tau^2

Let's compute it:

d t_1 = d\left( \frac{t_0 + t_2}{2} \right) = \frac{dt_0 + dt_2}{2}

d x_1 = d\left( \frac{t_2 - t_0}{2} \right) = \frac{dt_2 - dt_0}{2}

d \tau = d \sqrt{t_0 t_2} = \frac{1}{2\sqrt{t_0 t_2}} (t_0 dt_2 + t_2 dt_0)

So (dt_1)^2 - (dx_1)^2 looks nothing like (d \tau)^2! (this is not a rigorous disproof, but I was merely going for a demonstration)


What your \tau computes is the proper time along the straight-line path from the origin of your coordinate system to the reflection event. (Complete with the problems when the event is space-like separated from the origin. I.E. when t_0 &lt; 0 and t_2 &gt; 0)

 

[bda]

This message was edited to correct a serious mistake; the editing is clealy marked with bold.

Dear Hurkyl,

I'm very happy because finally someone is really paying attention to what I write

(...)

I just noticed a problem with this -- it has absolutely nothing to do with the proper time along a worldline! There are several ways to see it (such as looking at two worldlines that cross twice), but I'll address it this way:

Put in those terms I can't but agree with you. Actually \tau is a line integral in 3+1-space and so its value depends on the specific worldline where it is evaluated. Conversely, the same happens with t in 4-space. But remember I am applying to specific objects moving on geodesics (...);
for simplicity let us consider a geodesic going through the origin, for which it will be x = v t ; using the definitions for x and t we can then write

t_2 - t_0 = v (t_2 + t_0)

now square that

(t_2)^2 + (t_0)^2 -2 t_2 t_0 = v^2 (t_2)^2 + (t_0)^2 + 2 t_2 t_0

arrange the terms

(t_2 t_0)^2 = (1 - v^2) \frac{(t_2)^2 + (t_0)^2}{4}

now replace back with the definitions

\tau^2 = (1 - v^2) x^2

This shows that \tau is indeed proper time. Now, if the worldlines don't cross at the origin we can always make a vertical translation of the x axis and it will still be (d \tau)^2 = (1- v^2)(d x)^2. In two dimensions
the worldlines will always cross but in 4D they may not cross. As I argued in a previous post, all worldlines will have crossed at some time in the past if the Universe is expanding from a big bang.

However, since we don't have access to that privileged origin, we must find a way of synchronizing time measurements on non-crossing worldlines. This is done by third worldline that crosses both. So, if I want to find out the time over another worldline I send out a radar pulse to interact with the distant object and time the send
and receive instants; then use the definitions to get t and \tau.

I will soon have to clarify how I deal with curved worldlines but this is realm of general relativity, not special relativity. Before going into that I want to make sure that the mapping method in flat space is clear.
Jose

 

[Hurkyl]

You don't need GR to handle curved worldlines...


What is the point of #57? I think you're still trying to describe how to go from 3+1-space to 4+1-space... but I think it's much easier than you're making it.

When we're working in 3+1-space, if we pick any pointed worldline, we already know how to assign a proper-time to any point on that worldline. Then, IMHO it's straightforward to lift that worldline into 4+1-space simply by making the new coordinate to be equal to the proper time at that point.

(If I've misunderstood your intent, let me know)

 

[bda]

You don't need GR to handle curved worldlines...

If you allow non-geodesic movement you have then different paths linking any two points in 3+1-space, so you get different evaluations for the proper time difference between them. Each path could then be separately mapped to 4-space but their endpoints would not coincide. I want to make sure I am dealing only with geodesic movement (straight lines).

There are only 4 interactions in nature, but restricting ourselves to the macroscopic world we only need to consider gravity and electromagnetism. The former is described by geodesics if you go to GR and I want to do something similar. EM can also be made geometric by a procedure similar to Kaluza-Klein. So, in short, I want to consider only geodesics when I do dynamics. This is looking ahead to forthcoming posts.

What is the point of #57? I think you're still trying to describe how to go from 3+1-space to 4+1-space... but I think it's much easier than you're making it.

When we're working in 3+1-space, if we pick any pointed worldline, we already know how to assign a proper-time to any point on that worldline. Then, IMHO it's straightforward to lift that worldline into 4+1-space simply by making the new coordinate to be equal to the proper time at that point.

Yes, as long as we stick to geodesics; we will leave curved worldlines to be handled by the metric.

Jose

 

[bda]

Introducing General Euclidean Relativity

I believe we are now ready to start addressing GR and its Euclidean counterpart; this post will introduce the subject.

In a curved space displacements are evaluated by a quadratic form which we write as

<br /> <br /> (d \tau)^2 = m_{\mu \nu} d x^\mu d x^\nu<br /> <br /> [/itex]<br /> <br /> In this expression we use a few conventions. First of all there is the Einstein summation convention which says that repeated indices below and above imply a summation over all possible values. The second convention is that Greek letter indices chosen from \langle \mu, \nu, \lambda \rangle take values in the interval 0 to 3. The expression above thus implies a sum of 16 terms on the rhs. For the non-diagonal elements it is always m_{\mu \nu} = m_{\nu \mu}We are also assuming that m_{00} &amp;gt;0 and m_{mm} &amp;lt;0 (indices <i>m, n, o</i> take values from 1 to 3). This assumption is essential to ensure that tangent space is Minkowski. I&#039;ve used the letter <i>m</i> for the GR metric tensor, rather than the more usual <i>g</i>, because I am very short of letters working in different spaces at the same time.<br /> <br /> If we go one dimension up, as we did in special relativity, we have new indices that go from 0 to 4, chosen from \langle \alpha, \beta, \gamma \rangle and also from 1 to 4, chosen from \langle i, j, k \rangle. In 5D the distance between two points is evaluated by the quadratic form<br /> <br /> &lt;br /&gt; &lt;br /&gt; (ds)^2 = g_{\alpha \beta} dx^\alpha d x^\beta&lt;br /&gt; &lt;br /&gt;<br /> <br /> where now the second member has 25 terms, g_{00}&amp;lt;0 and g_{ii} &amp;gt;0.<br /> <br /> In the most general case all the g_{\alpha \beta} can be different from zero and they can be functions of all the 5 coordinates. I don&#039;t know if we will ever look at this most general case but for now I want to make two restrictions:<br /> <br /> 1 - g_{0i} = g_{i0} = g_{4i} = g_{i4} = 0<br /> <br /> 2 - All the g_{\alpha \beta} are functions only of the 3 coordinates x^m.<br /> <br /> Now suppose we are interested in null geodesics (ds)^2 =0. We can expand the rhs as follows<br /> <br /> &lt;br /&gt; &lt;br /&gt; 0 = g_{00} (dx^0)^2 + g_{mn} dx^m dx^n + g_{44}(dx^4)^2&lt;br /&gt; &lt;br /&gt;<br /> <br /> Now pull (dx^4)^2 to the lhs<br /> <br /> &lt;br /&gt; &lt;br /&gt; (dx^4)^2 = -\frac{1}{g_{44}}(g_{00} (dx^0)^2 + g_{mn} dx^m dx^n)&lt;br /&gt; &lt;br /&gt;<br /> <br /> This has obviously the form of a GR metric and we can assign dx^4 \equiv d \tau and dx^0 \equiv dt. Instead of pulling out (dx^4)^2 we can pull (dx^0)^2 to the lhs<br /> <br /> &lt;br /&gt; &lt;br /&gt; (dx^0)^2 = -\frac{1}{g_{00}}(g_{mn} dx^m dx^n + g_{44} (dx^4)^2)&lt;br /&gt; &lt;br /&gt;<br /> <br /> This has the form of a pseudo-Euclidean metric, that is, a metric with all the diagonal terms positive. Let us write it down<br /> <br /> &lt;br /&gt; &lt;br /&gt; (dt)^2 = e_{ij} dx^i dx^j&lt;br /&gt; &lt;br /&gt;<br /> <br /> where the 4 e_{ii} are positive.<br /> <br /> I think this is enough for now. If it is unclear I will add more explanations before we proceed.<br /> Jose