Whose Clock Slows Down in Relativity?

[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.