Why is "time orthogonal to space" in inertial reference frames?

[PeterDonis]

I'll try to prove this result at least for the case of only ##t## and ##x## dimensions to start with.

Yes, the generalization to three space dimensions should be straightforward once this is done.

 

[PeroK]

The question is whether "it is orthogonal to space in an IRF" must be true, given the coordinate-free definition of an IRF. Saying that "IRF" means "standard Minkowski coordinates" is not an answer to the question, since it assumes precisely what is being questioned.

Okay. So what's our definition of "orthogonal" here?

 

[robphy]

Physics (or at least our mathematical formulation of it) defines
spacelike-vectors to be orthogonal to timelike-vectors.

No, it doesn't. This statement is much too strong.

Consider the vectors (in the standard Minkowski coordinate chart) (1,0,0,0)(1,0,0,0) and (0.1,1,0,0)(0.1,1,0,0). The first is timelike and the second is spacelike, but they are not orthogonal.

Even if you restrict the statement to coordinate basis vectors, it is still wrong. It is perfectly possible to have coordinates in which spacelike basis vectors are not orthogonal to timelike basis vectors. Indeed, it is perfectly possible, not only to have coordinates with no timelike basis vector at all, as @Nugatory pointed out, but even to have coordinates with no spacelike basis vector at all. All that is required of basis vectors is that they be a set of four vectors that are linearly independent. That is an extremely weak condition compared to orthogonality.

I admit it that this is a little sloppy.

The points I am trying to get across are

• timelike vectors are defined first (since I'll regard observers as more primary)
• given a metric,
  timelike vectors are characterized by [itex] g(\vec t,\vec t)>0 [/itex]
  and unit-timelike vectors are characterized by [itex] g(\hat t,\hat t)=1 [/itex];
  then,
  secondarily,
  given a specific timelike vector [itex] \vec T [/itex]
  a spacelike vector then defined as a vector [itex] \vec v[/itex]
  that is metric-orthogonal to the timelike vector [itex] \vec T [/itex]: [itex] g( \vec T, \vec v)=0 [/itex]
  
  Such vectors are purely spatial according [itex] \hat T [/itex]
  (Physically, "an observer's sense of space is metric-orthogonal to his sense of time".)

I'm not talking about coordinates or basis vectors
or anything else implied by my sloppy articles (the, some, any, all, etc...) or absence of articles.
I'm talking about how
in my timelike-observer-is-primary formulation
"spacelike" only has meaning after a metric is introduced.
(In my subsequent examples of metrics, the choice of metric determines
which events are simultaneous according the given observer
[i.e. which directions are purely spatial according to the given observer].)

Maybe a more accurate sentence is that
"a spacelike vector is defined as a vector that is metric-orthogonal to a timelike-vector [i.e. some timelike vector]".

Further constructions of coordinates and basis vectors, etc,
need a more careful phrasing, as you point out.

 

[PeterDonis]

So what's our definition of "orthogonal" here?

Two vectors are orthogonal if their inner product is zero. That requires a metric (which we are assuming is the Minkowski metric), in order to have an inner product, but it is independent of any particular coordinate choice.

 

[PeterDonis]

timelike vectors are defined first (since I'll regard observers as more primary)

Without a metric you cannot define timelike or spacelike or null at all for vectors. See below.

given a metric,
timelike vectors are characterized by ##g(\vec t,\vec t)>0## and unit-timelike vectors are characterized by ##g(\hat t,\hat t)=1## ;

A note: this is using the timelike signature convention, which is different from previous posts in this thread that have used the spacelike convention (under which we would have ##g(\hat{t}, \hat{t}) = -1## for a timelike unit vector). Either convention is fine, but I think we should agree on sticking to the same one for this discussion.

a spacelike vector then defined as a vector ##\vec v##
that is metric-orthogonal to the timelike vector

No, it isn't. Again, this is much, much too strong. A spacelike vector is any vector with ##g(\vec{v}, \vec{v}) < 0## using your convention (timelike)--or with ##g(\vec{v}, \vec{v}) > 0## if we use the spacelike convention used previously in this thread. And a spacelike unit vector would have ##g(\hat{v}, \hat{v}) = -1## according to the timelike convention, or ##g(\hat{v}, \hat{v}) = 1## according to the spacelike convention. There are an infinite number of vectors meeting those conditions which will not be orthogonal to any chosen timelike vector.

"spacelike" only has meaning after a metric is introduced.

So do "timelike" and "null". Without a metric vectors don't even have a squared norm at all, so "timelike", "spacelike", and "null" can't even be defined.

(In my subsequent examples of metrics, the choice of metric determines
which events are simultaneous according the given observer
[i.e. which directions are purely spatial according to the given observer].)

Here you use the term "purely spatial", which I assume means "orthogonal to the tangent vector to the observer's worldline". Which is fine, but it's a much more restrictive condition than "spacelike" and should not be confused with the latter.

 

[PeroK]

Two vectors are orthogonal if their inner product is zero. That requires a metric (which we are assuming is the Minkowski metric), in order to have an inner product, but it is independent of any particular coordinate choice.

I found that here:

https://mathworld.wolfram.com/MinkowskiMetric.html

But that appears to assume Minkowski coordinates. Is there a coordinate-free way to define it?

 

[PeterDonis]

The points I am trying to get across are

Perhaps I should outline how I think the key points for this discussion are formulated in SR.

We start with the physical observation that time behaves differently from space: for example, we measure time with clocks but we measure space with rulers; we can move in arbitrary directions in space but can only move forward in time; and so on.

If we take a coordinate-free, geometric approach to constructing a mathematical model to reflect the above physical observation, then we recognize at once that spacetime cannot have a Euclidean metric, since such a metric would only have one type of vector and there would be no way to reflect the physical distinction between time and space described above. Instead, we see that spacetime must have a Minkowskian metric (where here the term "Minkowskian" refers strictly to the coordinate-free geometric object, not any particular choice of coordinates), or more precisely a pseudo-metric since it is not positive definite.

Such a metric will allow vectors to have three types of squared norm: timelike (negative squared norm using the spacelike signature convention I will adopt since it is what was used for most of the previous posts in this thread), spacelike (positive squared norm), and null (zero squared norm). Timelike vectors represent tangent vectors to the worldlines of observers, and null vectors represent tangent vectors to the worldlines of light rays.

Then, given any timelike vector, there will be a subspace of all spacelike vectors consisting of those which are orthogonal to that timelike vector. Physically, it makes sense to consider this subspace of spacelike vectors as describing "space" at an instant of "time" for an observer who has the given timelike vector as the tangent to their worldline at an instant of time by their clock. It is easy to show that choosing three mutually orthogonal spacelike vectors in this subspace to form a set of four mutually orthogonal vectors (and normalizing as required to make all four of them unit vectors), and constructing coordinates over the entire spacetime by using the particular event where we picked the vectors as an origin, and parallel transporting all four of them to all other events in spacetime, and considering the set of four vectors so obtained at each event as the coordinate basis vectors at that event, is sufficient to make the construction satisfy the requirements for an IRF.

The question at issue in this thread is whether that particular construction is also necessary to satisfy the requirements for an IRF.

 

[PeterDonis]

Is there a coordinate-free way to define it?

The definition I gave is coordinate-free. You don't need to make any particular choice of coordinates to define a metric and an inner product. If you want it written out in tensor form, the inner product of two vectors ##\vec{u}## and ##\vec{v}## is ##g(\vec{u}, \vec{v}) = g_{ab} u^a v^b##. That will be valid in any coordinate chart.

 

[Shirish]

The definition I gave is coordinate-free. You don't need to make any particular choice of coordinates to define a metric and an inner product. If you want it written out in tensor form, the inner product of two vectors ##\vec{u}## and ##\vec{v}## is ##g(\vec{u}, \vec{v}) = g_{ab} u^a v^b##. That will be valid in any coordinate chart.

I think I'm starting to get the picture.

So then in the case of Minkowski space, the metric is already predefined as ##\text{diag}(-1,1,1,1)##. All we have to do is use the conditions for constancy of free particle velocity, and we can try to come up with that set of coordinates which satisfy this requirement. It will turn out that any such set of coordinates is such that they will have timelike and spacelike basis vectors (set of coordinates in which there are only spacelike or in which there are only timelike basis vectors won't do), and the timelike and spacelike basis vectors are orthogonal. Naturally such a set of coordinates will define an IRF in Minkowski space.

Does this sound correct?

 

[PeterDonis]

in the case of Minkowski space, the metric is already predefined as ##\text{diag}(-1,1,1,1)##.

No. The metric is a coordinate-independent geometric object. Its representation in one particular choice of coordinates is ##\text{diag}(-1,1,1,1)##. But its representation will be different in other coordinates.

All we have to do is use the conditions for constancy of free particle velocity, and we can try to come up with that set of coordinates which satisfy this requirement.

If you say that the metric is represented by ##\text{diag}(-1,1,1,1)##, you have already chosen coordinates that satisfy the requirements for an IRF.

The question is whether that is the only possible choice of coordinates that satisfies those requirements. You cannot prove anything about that by assuming the metric is represented by ##\text{diag}(-1,1,1,1)##. You need to consider other possible representations in other possible coordinate charts and see if it is possible for any of them to satisfy the requirements for an IRF.

Does this sound correct?

No. See above.

 

[PeroK]

The definition I gave is coordinate-free. You don't need to make any particular choice of coordinates to define a metric and an inner product. If you want it written out in tensor form, the inner product of two vectors ##\vec{u}## and ##\vec{v}## is ##g(\vec{u}, \vec{v}) = g_{ab} u^a v^b##. That will be valid in any coordinate chart.

I can see that for a general metric. But, then, this specific Minkowski metric must be defined by some property. Is is that the (Ricci) curvature is zero everywhere?

There must be something that tells you how to distinguish this metric from all others.

 

[PeterDonis]

The metric is a coordinate-independent geometric object. Its representation in one particular choice of coordinates is ##\text{diag}(-1,1,1,1)##. But its representation will be different in other coordinates.

You need to consider other possible representations in other possible coordinate charts and see if it is possible for any of them to satisfy the requirements for an IRF.

Here is an exercise that may help to clarify the above statements of mine.

Consider Rindler coordinates [1]. This is a different coordinate chart on Minkowski spacetime, in which the metric of Minkowski spacetime has a different representation that is not ##\text{diag}(-1,1,1,1)##.

However, this coordinate chart is non-inertial. Try showing this by showing that this chart does not meet either of the two requirements for an IRF: free particles do not have constant coordinate velocity, and light does not always have speed ##c##.

[1] https://en.wikipedia.org/wiki/Rindler_coordinates

 

[PeterDonis]

this specific Minkowski metric must be defined by some property. Is is that the (Ricci) curvature is zero everywhere?

Not just the Ricci tensor; the Riemann tensor is zero everywhere. (The Ricci tensor is zero in any vacuum spacetime, including all of the curved ones like Schwarzschild or Kerr.)

 

[PeroK]

Not just the Ricci tensor; the Riemann tensor is zero everywhere. (The Ricci tensor is zero in any vacuum spacetime, including all of the curved ones like Schwarzschild or Kerr.)

Okay, thanks. So that's the defining characteristic here. I don't see how you can progress without that - or, alternatively, specify the metric in some known coordinate system.

 

[PeterDonis]

I don't see how you can progress without that

It is certainly necessary since you would need to check that any other coordinate chart you considered satisfied the condition (compute the Riemann tensor for the metric as expressed in the chart and verify that it vanishes).

 

[Shirish]

Here is an exercise that may help to clarify the above statements of mine.

Consider Rindler coordinates [1]. This is a different coordinate chart on Minkowski spacetime, in which the metric of Minkowski spacetime has a different representation that is not ##\text{diag}(-1,1,1,1)##.

However, this coordinate chart is non-inertial. Try showing this by showing that this chart does not meet either of the two requirements for an IRF: free particles do not have constant coordinate velocity, and light does not always have speed ##c##.

[1] https://en.wikipedia.org/wiki/Rindler_coordinates

One final thing: what mathematical background would one need to prove the result we want to (regarding the orthogonality requirement of an IRF)? Would linear algebra (up to basic knowledge of covectors / vectors and transformation properties of tensors) suffice? Or would it require more advanced stuff like differential geometry and require me to know fancy concepts like Ricci curvature, etc.? I'll accordingly brush up on the needed mathematical prerequisites.

 

[PeroK]

@Shirish I won't interrupt your thread any more. I was hoping my questions would help you, which I why I asked them. I would have hoped you could learn SR without getting into these deep conceptual waters. From a pedagogical point of view I think this treatment is asking a lot. Rindler coordinates, Riemann tensor, coordinate-free descriptions of spacetime. These are things you need to wrestle with eventually. But, up front I think it's a lot to ask of the student.

 

[robphy]

Without a metric you cannot define timelike or spacelike or null at all for vectors. See below.

Yes, again I'm a little sloppy.

I was going to write a more complete reply...
but saw that it takes too long to nail down the ideas, and will go off-topic for this thread.

I'll stand by my intuitive characterization
that
"spacelike" only has meaning after a metric is introduced,
in the sense that
given an observer's worldline on a position-vs-time graph,
the direction that is "purely spatial" to it
is determined by a metric [say, determined by experiment]
as the tangent line to a conic that describes
the set of point-events equidistant from the origin.
Then
the set of proper metric-preserving transformations will
lead to characterizing a subset of vectors to be "spacelike" from such purely-spatial vectors.

For more details of this viewpoint (still under development),
https://www.aapt.org/doorway/Posters/SalgadoPoster/Salgado-GRposter.pdf#page=6

 

[Shirish]

@Shirish I won't interrupt your thread any more. I was hoping my questions would help you, which I why I asked them. I would have hoped you could learn SR without getting into these deep conceptual waters. From a pedagogical point of view I think this treatment is asking a lot. Rindler coordinates, Riemann tensor, coordiante-free descriptions of spacetime. These are things you need to wrestle with eventually. But, up front I think it's a lot to ask of the student.

Oh no you weren't interrupting, no worries It's a chance for all of us (especially me haha) to clarify concepts. I want to go into deep conceptual waters sooner or later for sure. I might be getting ahead of myself at times and that might slow me down, but I guess that's a downside of self-teaching.

 

[PeterDonis]

what mathematical background would one need to prove the result we want to (regarding the orthogonality requirement of an IRF)?

Ordinary algebra.

Work in 1+1 spacetime for simplicity, as you suggested.

Since we already have one coordinate chart that satisfies the requirements, we can re-formulate our question as follows: what transformations could we make from this chart to some other chart, that would preserve the two properties required for an IRF (free particle worldlines are straight lines, and light always has speed ##c##)? What constraints can we show must hold for such transformations?

If you find that the required constraints narrow down all of the possible transformations to just the Lorentz transformations in the standard form for the Minkowski chart, you're done: you've proven that only one particular kind of coordinate chart, the standard ##(t, x)## Minkowski chart in which the metric is ##\text{diag}(-1, 1)##, will work, since the standard Lorentz transformations starting from such a chart can only take you to the same kind of chart.

The constraints that can be made on the transformations from the two IRF requirements can be expressed using just ordinary algebra.

 

[PeterDonis]

I'll stand by my intuitive characterization
that
"spacelike" only has meaning after a metric is introduced,

This was not in dispute. I agree with it.

given an observer's worldline on a position-vs-time graph,
the direction that is "purely spatial" to it
is determined by a metric [say, determined by experiment]
as the tangent line to a conic that describes
the set of point-events equidistant from the origin.
Then
the set of proper metric-preserving transformations will
lead to characterizing a subset of vectors to be "spacelike" from such purely-spatial vectors.

This, however, appears to be based on the reference you linked to, which is, as you note, still under development, and is off topic for this thread. (The reference in general is beyond a "B" level discussion anyway; it's at least "I" and possibly "A".)

 

[Shirish]

Since we already have one coordinate chart that satisfies the requirements, we can re-formulate our question as follows: what transformations could we make from this chart to some other chart, that would preserve the two properties required for an IRF (free particle worldlines are straight lines, and light always has speed ##c##)? What constraints can we show must hold for such transformations?

If you find that the required constraints narrow down all of the possible transformations to just the Lorentz transformations in the standard form for the Minkowski chart, you're done

So this is something I managed to do assuming that the transformation from chart ##(t,x)## to ##(t',x')## is linear, constancy of light speed and isotropy/homogeneity of space. We know that in chart ##(t,x)##, by construction, the time and space basis vectors are orthogonal (equipped with ##\text{diag}(-1,1,1,1)## metric). Again because of the way we constructed it, free particles have straight worldlines and light has 45 degree worldlines.

(Gap in proof: To prove that the chart ##(t',x')## also has the ##\text{diag}(-1,1,1,1)## metric). Once we know that even the new chart ##(t',x')## has ##\text{diag}(-1,1,1,1)## metric (say ##M##), then we know that under a linear transformation, ##v_1^TMv_2## is invariant. Also, the timelike basis vector in the old chart will be mapped to a timelike basis vector in the new one, and similarly for spacelike basis vector (again assuming ##\text{diag}(-1,1,1,1)## metric for the new chart). So
$$0 = \mathbf{e}_t^T M \mathbf{e}_x = \mathbf{e}_t^T L^T M L \mathbf{e}_x = \mathbf{e'}_t^T M \mathbf{e'}_x$$
since ##L^TML=M##, which proves orthogonality of timelike and spacelike basis vectors in the new chart as well.

So I guess this shows that the only transformation mapping IRFs to IRFs, i.e. the Lorentz transformation, preserves the orthogonality between timelike and spacelike vectors. The only gap is to show that the chart ##(t',x')## also has the ##\text{diag}(-1,1,1,1)## metric.

 

[PeterDonis]

assuming that the transformation from chart ##(t,x)## to ##(t',x')## is linear, constancy of light speed and isotropy/homogeneity of space

The first (linear) shouldn't be an assumption; you should be able to prove that it's required based on the requirements of an IRF.

The second is fine since it's a requirement of an IRF.

The third shouldn't be an assumption either; you should also be able to prove that it's required based on the requirements of an IRF.

Without proofs of the first and third assumptions based on the requirements of an IRF, what you are doing is not proving anything.

 

[PeterDonis]

the timelike basis vector in the old chart will be mapped to a timelike basis vector in the new one

No, it won't. This is certainly not true for a standard Lorentz transformation from one standard Minkowski chart to another. (If you had left out the word "basis" the second time you used it, your statement would be correct; but it's false with the second "basis" included.)

similarly for spacelike basis vector

Same remark as above: basis vectors in the old chart will not be basis vectors in the new chart. Their signature (timelike or spacelike) and norm will be preserved, but not the fact of them being basis vectors.

 

[Shirish]

The first (linear) shouldn't be an assumption; you should be able to prove that it's required based on the requirements of an IRF.

The second is fine since it's a requirement of an IRF.

The third shouldn't be an assumption either; you should also be able to prove that it's required based on the requirements of an IRF.

Without proofs of the first and third assumptions based on the requirements of an IRF, what you are doing is not proving anything.

Linearity of the transformation follows from the fact that in IRFs free particles have constant speed. The chart ##(t,x)## is rectilinear, so the free particle worldline is straight in it. Assuming that the new chart ##(t',x')## representing another IRF is also rectilinear, that particle's worldline will still be straight due to constant speed. Map from straight to straight worldlines implies linearity. (I'm not sure what we can do if we don't assume that the chart ##(t',x')## is rectilinear.

Third point: There's no preferred location in space nor a preferred direction, no matter where we are or which direction we're facing, we shouldn't be able to distinguish between IRFs, since all IRFs are assumed to be physically equivalent via the principle of relativity. So hopefully this is an acceptable physical justification of the homogeneity and isotropy assumptions.

(If you had left out the word "basis" the second time you used it, your statement would be correct; but it's false with the second "basis" included.)

Same remark as above: basis vectors in the old chart will not be basis vectors in the new chart. Their signature (timelike or spacelike) and norm will be preserved, but not the fact of them being basis vectors.

Yes, my mistake. I should've said that spacelike (timelike) basis vectors are mapped to spacelike (timelike) basis vectors in the new chart. But this puts another hole in my argument - now I can only say that the images of the old chart's basis vectors are orthogonal. This doesn't enable me to say anything about the new chart's basis vectors - whether both have the same or different signatures, or the inner product of those.

 

[pervect]

Apologies in advance if I'm addressing points already raised, I've only skimmed the thread.

Experimentally, time being orthogonal to space is the same as saying that we use Einstein clock synchronization convention, as a way to get "fair" or "isotropic" clock synchronization.

When we make time orthogonal to space, physics work the way we usually expect. Light moves the same in both directions, and so do electron beams with the same energy. It doesn't matter to the physics which way the light or the electron beam is going, it's speed is the same.

If we don't make time orthogonal to space, both the speed of light and the speed of electron beams of a specified energy depends on the direction. This is most obvious for high energy electron beams, as they approach the speed of light as one increases their energy per electron. Thus if the behavior of light is not isotropic, neither is the behavior of other things. See for instance and/or the peer reviewed paper associated with the video. The general topic is known as "isotropy". Sometimes people appear to think isotropy applies only to light, but it applies to other physical phenomenon as well.

If the idea that the speed of identical objects (identical because they have a known energy, or momentum) moving in different directitons is uncomfortable, then one should stick with isotropic coordinates and/or inertial frames, which implies that time must be orthogonal to space.

 

[PeterDonis]

Linearity of the transformation follows from the fact that in IRFs free particles have constant speed.

Yes.

Assuming that the new chart ##(t',x')## representing another IRF is also rectilinear

I don't think this assumption for the new chart is actually required. If you look at coordinate speeds in non-rectilinear charts, you will see that they can't possibly be constant in all directions. For example, consider ##d\theta / dt## for a free particle in spherical coordinates (and ignore the edge case where the particle happens to be moving along an axis that makes ##\theta## constant). So constant coordinate speed in all directions already implies a rectilinear chart.

hopefully this is an acceptable physical justification of the homogeneity and isotropy assumptions.

It's a good start, but you spoke in general terms. There is a specific requirement for an IRF that requires isotropy.

I should've said that spacelike (timelike) basis vectors are mapped to spacelike (timelike) basis vectors in the new chart.

Meaning, mapped by the coordinate transformation? Yes, you can think of the transformation that way, but you can also think of it as simply changing the components of vectors, but not the vectors themselves. In other words, the old basis vectors will have more than one nonzero component in the new chart, and the new basis vectors will have more than one nonzero component in the old chart. The question then is how the coordinate transformation can change those components while still preserving the required properties for an IRF.

now I can only say that the images of the old chart's basis vectors are orthogonal.

Orthogonality is a coordinate-independent property, so a pair of vectors that are orthogonal in one chart will be orthogonal in any chart.

This doesn't enable me to say anything about the new chart's basis vectors - whether both have the same or different signatures, or the inner product of those.

You can't assume the new basis vectors are orthogonal, or even that one is timelike and three are spacelike, that's true. You have to prove that, if it is provable.

You do have information about the new basis vectors, though: you know that free particles still have to have straight worldlines, and that light still has to have speed ##c##. And you know that the coordinate transformation has to be linear. So you do have information to work from.

 

[Shirish]

...you know that free particles still have to have straight worldlines, and that light still has to have speed ##c##. And you know that the coordinate transformation has to be linear. So you do have information to work from.

This info is used to derive the Lorentz transformation - in the ##(t,x)\to(t',x')## case, it's ##\mathcal{L}=\begin{bmatrix}\gamma & \beta\gamma\\\beta\gamma & \gamma\end{bmatrix}## (sign of off-diagonal terms depends on relative velocity so let's consider them positive in this case).

Orthogonality is a coordinate-independent property, so a pair of vectors that are orthogonal in one chart will be orthogonal in any chart.

Now this is important info and something I'm embarrassed to admit that I didn't know. I thought if orthogonality is coordinate-independent as you said, then maybe even the inner product is (googling confirmed this). So if we have a couple of vectors ##\mathbf{a},\mathbf{b}## with representations ##[ a ]_{\mathcal{A}}, [ b ]_{\mathcal{A}}## in the initial coordinate chart ##\mathcal{A}## (and since we know that ##\eta=\text{diag}(-1,1)## is the metric representation in ##\mathcal{A}## by construction), their inner product is ##[ a ]^T_{\mathcal{A}}\eta[ b ]_{\mathcal{A}}##.

Let ##\eta'## be the metric representation in the new coordinate chart ##\mathcal{B}## (the Minkowski metric doesn't change - its representation changes - something I realized after reading your post and googling related stuff). The representation of ##\mathbf{a}## in ##\mathcal{B}## will be ##[ a ]_{\mathcal{B}}=\mathcal{L}[ a ]_{\mathcal{A}}## (similarly for ##\mathbf{b}##). Then
$$[ a ]^T_{\mathcal{B}}\eta'[ b ]_{\mathcal{B}}=[ a ]^T_{\mathcal{A}}\eta[ b ]_{\mathcal{A}}=[ a ]^T_{\mathcal{A}}\mathcal{L}^T(\mathcal{L}^T)^{-1}\eta\mathcal{L}^{-1}\mathcal{L}[ b ]_{\mathcal{A}}
=[ a ]^T_{\mathcal{B}}(\mathcal{L}^{-1})^T\eta\mathcal{L}^{-1}[ b ]_{\mathcal{B}}
\\\implies \eta'=(\mathcal{L}^{-1})^T\eta\mathcal{L}^{-1}=\eta$$
So this was the missing link. And now since the metric representation is the same even in the new coordinate chart, and since the representation of the new basis vectors in the new chart ##\mathcal{B}## is ##[1,0]## and ##[0,1]##, their inner product is zero.

It's a good start, but you spoke in general terms. There is a specific requirement for an IRF that requires isotropy.

I confess I don't know about that requirement. As far as I've read, isotropy is justified loosely by the fact that the orientation of the relative velocity vector between IRFs can be arbitrary. If there's an even stronger/more specific justification, I'm lost.

 

[PeroK]

Orthogonality is a coordinate-independent property, so a pair of vectors that are orthogonal in one chart will be orthogonal in any chart.

Although that itself does need to be proved at some stage.

 

[PeterDonis]

This info is used to derive the Lorentz transformation

How?

I thought if orthogonality is coordinate-independent as you said, then maybe even the inner product is

It is.

the metric representation is the same even in the new coordinate chart

It is if you use the Lorentz transformation, yes. But you still have to show how you derive the Lorentz transformation given the info you quoted at the top of your post (see my first question above).

I confess I don't know about that requirement.

Sure you do; we've mentioned it repeatedly in this discussion. It's even mentioned in the first quote from my previous post that you give in this post: look at the clause after the comma in the first sentence of that quote.

 

[PeterDonis]

that itself does need to be proved at some stage

It's obvious from the definition of orthogonality and the fact that inner products are coordinate-independent. And before you say that inner products being coordinate-independent has to be proved, that's part of the definition of a metric.

 

[Shirish]

But you still have to show how you derive the Lorentz transformation given the info you quoted at the top of your post (see my first question above).

One thing - assuming that the transformation is Lorentz, does the argument in my post (which shows ##\eta=\eta'## and hence orthogonality of basis in the new chart) seem okay to you?

As for Lorentz, there's a related thread here that I'd created previously in which I showed steps for showing that the transformation is Lorentz: https://www.physicsforums.com/threa...locks-are-synchronized-at-each-origin.989150/

The proof in the top post in that thread is incomplete, but folks at PF helped me complete it using homogeneity and isotropy assumptions (homogeneity to show that without loss of generality, we can assume a linear transformation instead of affine, and isotropy to show that ##\gamma(v)## is an even function of ##v##).

My bad, I probably should've linked that earlier for some context.

 

[PeterDonis]

assuming that the transformation is Lorentz, does the argument in my post (which shows ##\eta=\eta'## and hence orthogonality of basis in the new chart) seem okay to you?

If the transformation is Lorentz, then yes, you're just expressing the well-known fact that the Lorentz transformation preserves the Minkowski form ##\eta## of the metric.

there's a related thread here that I'd created previously in which I showed steps for showing that the transformation is Lorentz

Ah, ok. Yes, if you use the derivation given in that thread, you're basically showing that the transformation has to be Lorentz in order to satisfy the speed of light being ##c## in the new coordinates, which is one of the requirements for an IRF.

 

[Shirish]

If the transformation is Lorentz, then yes, you're just expressing the well-known fact that the Lorentz transformation preserves the Minkowski form ##\eta## of the metric.
Ah, ok. Yes, if you use the derivation given in that thread, you're basically showing that the transformation has to be Lorentz in order to satisfy the speed of light being ##c## in the new coordinates, which is one of the requirements for an IRF.

Yay! So finally my mind is at peace. Now I can happily continue with the rest of the studies (until next time a doubt derails me again )

Thank you so much

 

[PeterDonis]

Thank you so much

You're welcome!