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

[Shirish]

I'm reading about the geometry of spacetime in special relativity (ref. Core Principles of Special and General Relativity by Luscombe). Here's the relevant section:

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Minkowski space is a four-dimensional vector space (with points in one-to-one correspondence with those of $\mathbb{R}^4$) spanned by one timelike basis vector, $\vec e_t$, and three spacelike basis vectors, $\vec e_x, \vec e_y, \vec e_z$. While any four linearly independent vectors can constitute a basis, in IRFs we require time to be orthogonal to space.
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I'm not clear what is meant by this. By "time", does the author mean the $[1,0,0,0]$ basis vector and by "space", does he mean the $[0, 1, 0, 0], [0, 0, 1, 0], [0, 0, 0, 1]$ basis vectors? If that's the case, this is no different from the Euclidean $\mathbb{R}^4$ case - I could've just said that "we require $x$ to be orthogonal to the other dimensions", where by "$x$" I mean the corresponding standard basis vector. So I'm guessing that's not what the author meant.

If by "time" and "space" the author isn't referring to corresponding standard basis vectors, then I'm not sure what's stopping me from taking an arbitrary basis of $\mathbb{R}^4$ as a basis of Minkowski Space as well. Why do we require "time" to be orthogonal to "space"? (whatever the quoted words mean in this context) What is even meant by the statement?

Is there a proper mathematical justification for this?

 

[Orodruin]

By "time", does the author mean the [1,0,0,0][1,0,0,0][1,0,0,0] basis vector and by "space", does he mean the [0,1,0,0],[0,0,1,0],[0,0,0,1][0,1,0,0],[0,0,1,0],[0,0,0,1][0, 1, 0, 0], [0, 0, 1, 0], [0, 0, 0, 1] basis vectors?

Yes.

If that's the case, this is no different from the Euclidean R4R4\mathbb{R}^4 case

No. The metric in Minkowski space is pseudo-Riemannian while the metric in Euclidean space is Riemannian.

If by "time" and "space" the author isn't referring to corresponding standard basis vectors, then I'm not sure what's stopping me from taking an arbitrary basis of R4R4\mathbb{R}^4 as a basis of Minkowski Space as well. Why do we require "time" to be orthogonal to "space"? (whatever the quoted words mean in this context) What is even meant by the statement?

Nothing is stopping you from choosing an arbitrary basis. It will just not be a coordinate system that is what we refer to as Minkowski coordinates. Just as you could pick any coordinate system in Euclidean space (such as spherical coordinates), they just will not be Cartesian coordinates.

 

[PeroK]

While any four linearly independent vectors can constitute a basis, in IRFs we require time to be orthogonal to space.

Note that, strictly speaking, there is no concept of "orthogonal" in a vector space. To define orthogonal you need an inner product (or a metric). The orthogonality of time and space (if we can put it like that) is really a statement about the metric you choose to impose on your vector space.

 

[Ibix]

Why do we require "time" to be orthogonal to "space"?

You can indeed use any set of basis vectors you like, and in general relativity we frequently do, but we only call ones with four orthogonal basis vectors "frames". So orthogonal basis vectors, one timelike and three spacelike, is a definition of a frame.

There's a natural reason for defining your time axis orthogonal to your spatial axes. It corresponds to using the worldlines of clocks at rest with respect to you as a timelike direction and defining "space" by Einstein-synchronising those clocks and using the set of events with equal times as "space at a given time". This gives you a homogeneous and isotropic notion of physics, which is nice and simple. Or as simple as it gets, anyway...

 

[Shirish]

Note that, strictly speaking, there is no concept of "orthogonal" in a vector space. To define orthogonal you need an inner product (or a metric). The orthogonality of time and space (if we can put it like that) is really a statement about the metric you choose to impose on your vector space.

Totally agree with you that orthogonality requires us to define inner product, which in turn requires us to specify a metric. But I don't see how "the orthogonality of time and space" (not sure if we should put it like that) is a statement about the Minkowski metric. All that the Minkowski metric specifies is that $-c\Delta t^2+\Delta x^2+\Delta y^2+\Delta z^2$ is a constant. Without additional assumptions or without adopting a certain convention, I don't see where orthogonality springs up from.

 

[Shirish]

You can indeed use any set of basis vectors you like, and in general relativity we frequently do, but we only call ones with four orthogonal basis vectors "frames". So orthogonal basis vectors, one timelike and three spacelike, is a definition of a frame.

There's a natural reason for defining your time axis orthogonal to your spatial axes. It corresponds to using the worldlines of clocks at rest with respect to you as a timelike direction and defining "space" by Einstein-synchronising those clocks and using the set of events with equal times as "space at a given time". This gives you a homogeneous and isotropic notion of physics, which is nice and simple. Or as simple as it gets, anyway...

Do you mean that it's a matter of convention that we select time and space basis vectors (in Minkowski space) in such a way that the time basis vector is orthogonal to space basis vectors?

 

[PeroK]

Totally agree with you that orthogonality requires us to define inner product, which in turn requires us to specify a metric. But I don't see how "the orthogonality of time and space" (not sure if we should put it like that) is a statement about the Minkowski metric. All that the Minkowski metric specifies is that $-c\Delta t^2+\Delta x^2+\Delta y^2+\Delta z^2$ is a constant. Without additional assumptions or without adopting a certain convention, I don't see where orthogonality springs up from.

If we take our spacetime vectors as $(ct, x, y, z)$, then the inner product that defines the metric is:
$$(ct_1, x_1, y_1, z_1) \cdot (ct_2, x_2, y_2, z_2) = -c^2t_1t_2 + x_1x_2 + y_1y_2 + z_1z_2$$
That in turn implies that $(ct, 0, 0, 0)$ is orthogonal to $(0, x, 0, 0)$ etc.

 

[Shirish]

If we take our spacetime vectors as $(ct, x, y, z)$, then the inner product that defines the metric is:
$$(ct_1, x_1, y_1, z_1) \cdot (ct_2, x_2, y_2, z_2) = -c^2t_1t_2 + x_1x_2 + y_1y_2 + z_1z_2$$
That in turn implies that $(ct, 0, 0, 0)$ is orthogonal to $(0, x, 0, 0)$ etc.

But would this not hold for any arbitrary diagonal metric? The reason I'm confused is that this orthogonality between time and space dimensions seems to have been presented as a feature distinguishing Minkowski metric from Euclidean. The above conclusion of $(ct, 0, 0, 0)$ being orthogonal to $(0, x, 0, 0)$ should apply in the case of Euclidean metric as well.

Maybe I'm overthinking and this "time orthogonal to space" phrase is ill-posed.

 

[PeroK]

But would this not hold for any arbitrary diagonal metric? The reason I'm confused is that this orthogonality between time and space dimensions seems to have been presented as a feature distinguishing Minkowski metric from Euclidean. The above conclusion of $(ct, 0, 0, 0)$ being orthogonal to $(0, x, 0, 0)$ should apply in the case of Euclidean metric as well.

It would, but $c^2dt^2 + dx^2 + dy^2 + dz^2$ doesn't work out as a model of the spacetime we live in.

The difference between Minkowski space and Euclidean is the minus sign on the $c^2dt^2$.

 

[PeterDonis]

Do you mean that it's a matter of convention that we select time and space basis vectors (in Minkowski space) in such a way that the time basis vector is orthogonal to space basis vectors?

Yes, it's a convention. Note that what you quoted in the OP says "in IRFs". What it's actually defining there is not what Minkowski spacetime is, but what an IRF is.

would this not hold for any arbitrary diagonal metric?

The metric being diagonal is dependent on your choice of coordinates. It's easy to find coordinates in which the metric of Minkowski spacetime is not diagonal. But those coordinates would not describe an IRF.

this orthogonality between time and space dimensions seems to have been presented as a feature distinguishing Minkowski metric from Euclidean

No, it's being presented as a feature distinguishing an IRF from other types of coordinate charts.

 

[Shirish]

Yes, it's a convention. Note that what you quoted in the OP says "in IRFs". What it's actually defining there is not what Minkowski spacetime is, but what an IRF is.

Thank you! But then why would "space" be orthogonal to "time" in IRFs? Per my understanding, the defining feature of IRFs is that free particles have straight worldlines, which isn't related to the metric.

I'm doubtful on how the structure of spacetime geometry (more specifically, the Minkowski metric) has anything to do with the fact that it's an IRF, OR how "space" not being orthogonal to "time" could indicate a non-inertial frame.

Is there a mathematical chain of reasoning that connects the two?

 

[Nugatory]

Do you mean that it's a matter of convention that we select time and space basis vectors (in Minkowski space) in such a way that the time basis vector is orthogonal to space basis vectors?

Sort of. The choice of basis vectors is conventional; we can choose any four linearly independent vectors to form our basis and whether they are orthogonal or not clearly depends on our choice. So yes, it is conventional.

However, you have constrained the choice (probably without noticing) by assuming that we have a time basis vector and three space vectors. What you are calling "the time basis vector" is the tangent to the worldline of some inertially moving clock, and having made that choice it would be perverse (at least in flat spacetime) not to choose space vectors orthogonal to it and to each other. This is the point that @Ibix was making in post #4 above.

But you don't have to a time basis vector at all... Consider the two-dimensional Minkowski space that we use for space-time diagrams; we naturally use $x$ and $t$ to define our basis and call one of the them space direction and the other the time direction. But we could define coordinates $a=x+t$ and $b=x-t$ and choose basis vectors parallel to the $a$ and $b$ axes; and now we have neither a time vector nor a space vector in our basis.

 

[PeroK]

Thank you! But then why would "space" be orthogonal to "time" in IRFs? Per my understanding, the defining feature of IRFs is that free particles have straight worldlines, which isn't related to the metric.

Let's start there. You could also stop there and decide to proceed without ever trying to define any geometric relationship between space and time. As I understand it, Einstein was initially reluctant to accept Minkowski's idea that SR could be modeled as a 4D spacetime. Your first option is to proceed with SR without looking for any 4D geometrical basis for the theory.

Suppose, however, you want to try to encapsulate the laws of SR geometrically by considering a 4D spacetime. You must consider what you mean by the length of a spacetime interval. Using an IRF is the simplest, so you consider those first. If a particle moves from $(t_1, x_1)$ to $(t_2, x_2)$, then how do you define the spacetime distance between those points? The thing you want is that this spacetime distance is the same in all IRF's. That's the defining characteristic of a geometrical theory of spacetime.

By whatever process, you discover that $-c^2(t_2 - t_1)^2 + (x_2 - x_1)^2$ is invariant. You may have already derived the Lorentz transformation and prove it using that. Or, you may do some experiments and measure spacetime coordinates between events in different frames.

In any case, $-c^2(t_2 - t_1)^2 + (x_2 - x_1)^2$ turns out to be an invariant quantity. And that, as far as I can see, is essentially the whole story. Whatever further mathematical terminology you use to describe what you've discovered is immaterial.

 

[kent davidge]

Per my understanding, the defining feature of IRFs is that free particles have straight worldlines, which isn't related to the metric.

It isn't related to the metric, but it is related to the coordinates. In Minkowski coordinates, a free particle has constant speed and also light travels with constant speed in any direction. But this is exactly what is found in an inertial frame, by definition. Therefore Minkowski coordinates describe inertial frames.

And if you are wondering, it can be shown that Minkowski coordinates are the only coordinates in which that is possible.

 

[PeterDonis]

You may have already derived the Lorentz transformation

You can only derive the Lorentz transformation in the form you give it (that it leaves $dt^2 - dx^2$ invariant) if you have already assumed you are using standard inertial coordinates. So you can't use this to argue that standard inertial coordinates are the only possible ones for an IRF.

 

[PeroK]

So you can't use this to argue that standard inertial coordinates are the only possible ones for an IRF.

Who's suggesting that?

 

[PeterDonis]

Who's suggesting that?

I thought that was the argument of your post #13.

 

[PeroK]

I thought that was the argument of your post #13, that you need standard inertial coordinates to make $dt^2 - dx^2$ invariant under Lorentz transformations, therefore standard inertial coordinates are the only possible ones for an IRF.

My argument was simply that by considering those coordinates we get an invariant spacetime interval. Once you have that there should be no argument about "time being orthogonal to space" in an IRF. Although, it remains to be seen what happens under a general Lorentz Transformation. I guess that would still have to be shown.

PS Or, starting with the OP and the assumption that time is orthogonal to space, it remains to be seen whether that definition of an IRF matches any prior definition: like a free particle moving in a straight line.

 

[Shirish]

It isn't related to the metric, but it is related to the coordinates. In Minkowski coordinates, a free particle has constant speed and also light travels with constant speed in any direction. But this is exactly what is found in an inertial frame, by definition. Therefore Minkowski coordinates describe inertial frames.

And if you are wondering, it can be shown that Minkowski coordinates are the only coordinates in which that is possible.

Since I don't mind looking stupid, I'll confess I'm still confused and I'll tell you my thought process. Let's just assume the $x$ and $t$ coordinates for now.

In Minkowski coordinates, a free particle has constant speed and also light travels with constant speed in any direction.

This is what's not obvious to me. I thought this should be "In IRFs" instead of "In Minkowski coordinates". I thought that a coordinate system is just a mathematical construction we use to describe whatever happens in an IRF. So even if I used some random coordinate system to describe IRFs - sure, the free particle trajectory could potentially look very different/weird and the math would be much more complicated, but just selecting a different coordinate system to describe the IRF wouldn't change the physical fact that free particle and light travel at constant speed in an IRF.

The physical fact shouldn't get altered based on what mathematical construction we use to define the IRF, right?

And if you are wondering, it can be shown that Minkowski coordinates are the only coordinates in which that is possible.

Could you give some guidance on how it can be shown? Could you prove it/give an example/provide link to some reference? I'd be really thankful.

 

[Ibix]

just selecting a different coordinate system to describe the IRF wouldn't change the physical fact that free particle and light travel at constant speed in an IRF.

Can you define "constant speed" without reference to a coordinate system?

The point about an inertial coordinate system is that $\partial \vec x/\partial t=\mathrm{const}$ (a statement about rate of change of three coordinates with respect to the fourth along some worldline) corresponds to inertial motion (a physical, measurable state). You are not obligated to use them, but a lot of maths is simpler if you do.

 

[robphy]

Do you mean that it's a matter of convention that we select time and space basis vectors (in Minkowski space) in such a way that the time basis vector is orthogonal to space basis vectors?

(Apologies that I did not read all of the replies since this post I am quoting from.)

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

Here's how Minkowski describes this...

We decompose any vector, such as that from O to x, y, z, t into four
components x, y, z, t. If the directions of two vectors are, respectively, that
of a radius vector OR from O to one of the surfaces \mp F = 1, and that of
a tangent RS at the point R on the same surface, the vectors are called
normal to each other. Accordingly,
c^2tt_1 − xx_1 − yy_1 − zz_1 = 0
is the condition for the vectors with components x, y, z, t and x_1, y_1, z_1, t_1 to
be normal to each other.

In other words,
locate the intersection of
an observer's 4-velocity with the unit-hyperbola (the Minkowksi circle) centered at the tail of the observer's 4-velocity.
The tangent line to that hyperbola is Minkowski-perpendicular to that observer's 4-velocity.
That observer's x-axis is drawn through the tail of her 4-velocity, parallel to that tangent line.

The "intuition" to have is that

the tangent to the "circle" in that geometry
is orthogonal to the radius vector.​

(If you feel better, you can add pseudo- and other such adjectives.)
Physically,

"Space" is orthogonal to "Time".​

This idea, motivated by Euclidean geometry, works in the Galilean spacetime geometry and Minkowski spacetime geometry.

(All three are among the nine Cayley-Klein geometries in two dimensions...
the others include hyperbolic and elliptic geometry and,
for relativity, their Lorentzian analogues (called the de Sitter spacetimes) and
their Galilean limits (called the Newton-Hooke spacetimes).)

You can play around with this idea in my visualization (time runs upwards)
https://www.desmos.com/calculator/r4eij6f9vw

- The unit-hyperbola (the "Minkowski circle") is in blue.
This figure is unchanged by a Lorentz boost.
- The red dotted line is the [timelike] observer-worldline.
- The red tangent line is the prototype for "simultaneous events according to
the red-observer".
- The red-observer's [spacelike] x-axis is drawn parallel to that tangent line.

Here is the E= +1, Minkowski case:

Play around with the E-slider to change signature to see the Galilean and Euclidean analogues.
Here E is approaching 0, the Galilean case (displaying the opening of the light-cone):
(Note how the tangent lines associated with different radii will coincide when E=0.
That's "absolute simultaneity".
Further note, that:
Generally, the tangents do not coincide.
Our "common-sense" Galilean "absolute simultaneity" case is the exception... not the [general] rule.
)

Here is E= -1, the Euclidean case:

 

[PeterDonis]

My argument was simply that by considering those coordinates we get an invariant spacetime interval.

You can have an invariant spacetime interval in any coordinates.

 

[Shirish]

The metric being diagonal is dependent on your choice of coordinates. It's easy to find coordinates in which the metric of Minkowski spacetime is not diagonal.

This much I completely understand.

But those coordinates would not describe an IRF.

I didn't get this. Any set of coordinates that we use can potentially be used to describe an IRF, can't it? The behavior of worldlines or the math might get complicated, but can't potentially any set of coordinates be used to describe an IRF?

No, it's (orthogonality between time and space dimensions) being presented as a feature distinguishing an IRF from other types of coordinate charts.

But IRF itself isn't a coordinate chart? By IRF in this case, do you mean the standard inertial coordinates (i.e. the typical ones we use in spacetime diagrams), or have I completely misunderstood this point?

 

[PeterDonis]

starting with the OP and the assumption that time is orthogonal to space

But that's the wrong place to start from to address the OP's question. The right place to start from is to not assume that time is orthogonal to space, since that's precisely the question at issue.

 

[PeterDonis]

Any set of coordinates that we use can potentially be used to describe an IRF, can't it?

No. For a counterexample, consider Rindler coordinates on Minkowski spacetime. Observers at rest in these coordinates are not inertial.

IRF itself isn't a coordinate chart?

It depends on what definition of "IRF" you are using. (But it's worth noting that if you define "IRF" to mean a particular kind of coordinate chart, then your previous claim that any set of coordinates can be used to describe an IRF would make no sense.)

So let's take a step back: what is your definition of "IRF"?

 

[PeterDonis]

By IRF in this case, do you mean the standard inertial coordinates

The question isn't what I mean, it's what you mean. You're the one asking questions about "IRF". What "IRF" are you asking about?

 

[PeterDonis]

By IRF in this case, do you mean the standard inertial coordinates (i.e. the typical ones we use in spacetime diagrams)

To answer this question as you ask it, normally the term "inertial reference frame" in SR is used as a synonym for "standard Minkowski coordinate chart in which the metric takes the form $ds^2 = - dt^2 + dx^2 + dy^2 + dz^2$ (in units where $c = 1$)". But that doesn't mean that's the only possible meaning for the term "IRF".

We could also ask, since you referenced a specific book: how does that book define "IRF"?

 

[Shirish]

No. For a counterexample, consider Rindler coordinates on Minkowski spacetime. Observers at rest in these coordinates are not inertial.
It depends on what definition of "IRF" you are using. (But it's worth noting that if you define "IRF" to mean a particular kind of coordinate chart, then your previous claim that any set of coordinates can be used to describe an IRF would make no sense.)

So let's take a step back: what is your definition of "IRF"?

From what I know, an IRF is a set of coordinates in which one can perform measurements of the motion of other bodies. The property of an IRF is that free particles have a constant velocity and the speed of light is $c$. That's about as precise as I can get.

 

[PeterDonis]

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) $(1, 0, 0, 0)$ and $(0.1, 1, 0, 0)$. 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.

 

[PeterDonis]

The property of an IRF is that free particles have a constant velocity and the speed of light is $c$.

Yes, that's a good definition, and note that, in itself, it doesn't assume any particular choice of coordinates; instead, it places requirements on whatever choice of coordinates you are going to make to describe an IRF, since "velocity" and "speed" here must mean coordinate velocity and speed.

So the question you are asking can be rephrased as follows: given these requirements for an IRF and coordinates that describe it, can we show that the timelike coordinate basis vector in any coordinate chart that meets the requirement must be orthogonal to all of the spacelike coordinate basis vectors?

 

[Ibix]

"velocity" and "speed" here must mean coordinate velocity and speed.

This is covered under speed being coordinate speed, but possibly worth emphasising that "speed of light" here means the one-way speed of light.

 

[Shirish]

Yes, that's a good definition, and note that, in itself, it doesn't assume any particular choice of coordinates; instead, it places requirements on whatever choice of coordinates you are going to make to describe an IRF, since "velocity" and "speed" here must mean coordinate velocity and speed.

So the question you are asking can be rephrased as follows: given these requirements for an IRF and coordinates that describe it, can we show that the timelike coordinate basis vector in any coordinate chart that meets the requirement must be orthogonal to all of the spacelike coordinate basis vectors?

Thanks! I'll try to prove this result at least for the case of only $t$ and $x$ dimensions to start with. One more clarification - you mentioned a requirement that the timelike basis vector should satisfy. The only requirement that I can make out is that the speed of free particles is constant, which I guess translates to a requirement on both spacelike and timelike basis vectors?

"that the timelike coordinate basis vector in any coordinate chart that meets the requirement must be orthogonal to all of the spacelike coordinate basis vectors?"
Did you mean:
"that the timelike and spacelike coordinate basis vectors in any coordinate chart that meet the requirement must be orthogonal to each other?"

Thanks so much for your patience so far!

 

[PeroK]

But that's the wrong place to start from to address the OP's question. The right place to start from is to not assume that time is orthogonal to space, since that's precisely the question at issue.

I thought the question is now "why isn't time always orthogonal to space"? Given that it is orthogonal to space in an IRF.

 

[PeterDonis]

I thought the question is now "why isn't time always orthogonal to space"?

I don't think so. See below.

Given that it is orthogonal to space in an IRF.

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.

See my post #30.

 

[PeterDonis]

you mentioned a requirement that the timelike basis vector should satisfy.

Well, it's also a requirement on the three spacelike basis vectors, since orthogonality is a property of pairs of vectors, not single vectors.

The only requirement that I can make out is that the speed of free particles is constant

That and that the speed of light is $c$.

which I guess translates to a requirement on both spacelike and timelike basis vectors?

Coordinate speeds will involve both the timelike basis vector and the spacelike basis vectors, yes.

"that the timelike coordinate basis vector in any coordinate chart that meets the requirement must be orthogonal to all of the spacelike coordinate basis vectors?"
Did you mean:
"that the timelike and spacelike coordinate basis vectors in any coordinate chart that meet the requirement must be orthogonal to each other?"

These look to me like two different ways of saying the same thing. Unless you intend the second way to impose the orthogonality requirement on the spacelike basis vectors between themselves--i.e., that they must all be orthogonal to each other, not just to the timelike basis vector. Whether you want to impose that additional requirement is up to you; the requirement that seems to me to be relevant to the title question of this thread is orthogonality between the timelike basis vector and the spacelike basis vectors.

 

[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 g(\vec t,\vec t)>0
  and unit-timelike vectors are characterized by g(\hat t,\hat t)=1;
  then,
  secondarily,
  given a specific timelike vector \vec T
  a spacelike vector then defined as a vector \vec v
  that is metric-orthogonal to the timelike vector \vec T: g( \vec T, \vec v)=0
  
  Such vectors are purely spatial according \hat T
  (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).