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I had doubts about the meaning of time being orthogonal to space. I have seen several threads about the topic and my conclusions are as follows (please correct if anything is wrong):

  • Yes, time is orthogonal to space.
  • This is an invariant: if frame A describes its basis vectors x and t as being orthogonal, then all other frames agree that it is so.
  • The way to test it is by carrying out the inner product as per Minkowski metric (or rather pseudometric), which has a negative signature. There is orthogonality if this product is 0.
  • In frame A itself the components of the basis vectors are x = (1,0) and x = (0,1). Actually here the inner product would be 0 no matter if you carry it out with the positive (0 + 0 = 0) or the negative signature (0 - 0 = 0).
  • In turn, a frame A’ moving at v = 0,5 c with regard to A would translate the coordinates of the basis vectors of A into its own coordinates as x' = (0.577, 1.115) and t' = (1.115, 0.577) and, yes, we would confirm that also for A' those vectors are orthogonal because obviously 0.577 * 1.115 - 1.115*0.577 = 0. The same result would not be obtained, though, with an “ordinary” dot product with + sign.

By the way, I am aware of robphy’s usual remark that in spacetime you build up a second axis orthogonal to the first, instead of by drawing the tangent to the point where the first axis intersects the unit circle and parallel-transporting it to the origin, by doing the same thing with the unit hyperbola.

But then my doubt is: if the way to test it and the construction method are different, does this “Minkowski perpendicularity” ultimately mean the same as the usual one?

By the “usual meaning” I mean that the vectors are not only linearly independent but totally independent, i.e. that one has no component at all in the other.

For example, if my problem is that Barcelona is located from Madrid at a distance d = sqrt{x^2 + y^2} and then someone moves the target one unit away in the +X axis, I still have to travel the same number of units in the Y axis.

Extrapolating this to spacetime, I could reason as follows: if my problem is to learn if event A can have an influence over event B, taking into account that the available time between the two of them and the path to be traversed from one to the other are t and x, respectively, even if more time is allowed, the same space will need to be covered or even if the path is enlarged, the available time will not vary... Is this the correct understanding of the concept?

Sierra
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    if frame A describes its basis vectors X and T as (1,0) and (0,1), then all other frames agree that it is so No. Other frames would use different components to describe A’s basis vectors… as you later state. – Ghoster Dec 06 '22 at 00:18
  • @Ghoster Thanks, sloppy language on my side. I edited the question to correct that. – Sierra Dec 06 '22 at 09:20
  • I'm not sure what the specific question is. The definition of orthogonality I use is based on Minkowski's definition in "Space and Time" (see my quoting of that passage in https://physics.stackexchange.com/a/643407/148184 ). – robphy Dec 06 '22 at 16:56
  • @robphy The question simply boils down to (i) whether this hyperbolic orthogonality also means that the dimensions must be "totally independent" and (i) in what sense applies the idea in this context. The question arose for me because I found the negative sign was so weird that I wondered if this would mean that perpendicularity takes here another different meaning. In the end, I have realized that the answer is "yes, total independence is also applicable here" and I have understood in what sense. – Sierra Dec 06 '22 at 23:22
  • I don't know what "totally independent" means. Observer A's 4-velocity dotted (with the Minkowski metric) with her x-axis is zero. However, Observer B's 4-velocity dotted with A's x-axis need not be zero... analogous to what happens with the x- and y-axes and the y'-axis of a rotated set of axes in Euclidean space. Linear-independence is a vector-space issue, independent of metric. The metric does affect the dot-product and some numerical details, but it is still quite analogous with the Euclidean metric. – robphy Dec 07 '22 at 00:57
  • @robphy In the understanding that I now have, it is clear to me that, as you point out, the essence of perpendicularity does not change in the hyperbolic context, just the metric. But I don't know why you introduce in the first part of your comment "velocity", when we are talking about a "displacement". I will post an answer that may clarify my point. – Sierra Dec 07 '22 at 09:29
  • The "4-velocity" is the common name of the unit 4-vector in spacetime that is tangent to the worldline of an observer. In other words, it's the unit vector that points along that observer's time axis. Geometrically speaking, it is dimensionless since it is a unit-vector. It corresponds to "one tick" along the inertial worldline with that 4-velocity. – robphy Dec 07 '22 at 19:50
  • @robphy Ok, thanks, I did not know about that name. I see what you meant, although I have to learn more about 4-velocity. But coming back to the question and the answer, have you come to terms with the expression "total independence"? Do you agree that orthogonality is linear independence but to the maximum extent, i.e not only that one basis vector is not colinear with the other but furthermore that it has no component at all in the other (at least in invariant terms, even if it is defined as 1,0 or 0,1 in its own frame)? – Sierra Dec 07 '22 at 23:30

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I can understand the negative votes to the question, as this is the type of requests about the “deep meaning” or the “intuitive understanding” that some people don’t like…

However, now that I grasped the answer, I think that it may be helpful to share it. The answer is “yes, hyperbola-based or Minkowski perpendicularity shares with circle-based or Euclidean perpendicularity the same essence or generalized meaning”.

The essence of the abstract idea of perpendicularity/orthogonality is that two vectors are “totally independent”, in the sense that they provide entirely different pieces of information (not overlapping in the slightest). This is not indispensable for building a basis. The minimum requirement is only that the basis vectors are at least partially independent or not totally overlapping, which means that each of them adds new information, even if the rest is redundant (we call this “linear independence”, maybe missing in the expression the shade that this independence is only partial). But we do prefer basis vectors with the feature of “total independence” or “orthogonality” because that simplifies the calculations since in that case you don’t have to account for the fact that part of the information is overlapping.

For example, If you must calculate some “spatial distance between spatial points”, it is convenient to choose as references two rods Y and X, where: the Y rod contains spatial points separated only by Y distance and no X distance, whereas the X rod contains others separated only by X distance and no Y distance.

If the problem is instead “a spacetime distance between events”, it is advisable to choose references cT and (simplifying) X, where: cT contains events (spacetime points) separated only by time, since they happen at the same location, and X contains events that are separated only by space, not by any time, since they are all simultaneous.

An interesting insight provided by this logic is that the axes cT and X contain “events”, i.e. spacetime points, even if they are defined in the relevant frame as happening at the same location or the same time, respectively.

Sierra
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