I How do non-diagonal indices of a metric allow for local flatness?

[Sciencemaster]

TL;DR Summary

If we have a metric in GR with non-zero non-diagonal indices, how can it be Minkowskian at a point? After all, the Minkowski metric is diagonal, so it doesn't seem like they can be equal at some given coordinate.

I'm having trouble understanding the local flatness of GR. So far, my interpretation was that it meant that the metric tensor at an infinitesimal point in spacetime will be equal to some multiple of the Minkowski metric since that's the metric that preserves the speed of light/spacetime interval. This in turn means that even in a curved spacetime, for a single point, the Lorentz Transformation relates different frames of reference as it is the transformation that keeps C invariant at said point. This all comes from the starting postulate that at every single point, C is an invariant. From here, it seems to me that a metric could only be diagonal, since it needs to be some multiple of the Minkowski metric at each coordinate. I can see how the diagonal indices can be different from one another at one point than somewhere else, since we start with the assumption that the speed of light just needs to be invariant at each infinitesimal point. So it can have different values in different places by changing the time and space components of the metric. What I'm having trouble understanding is, how can we have diagonal indices? Wouldn't that make it non-Minkowskian if you "plug in" a specific coordinate in spacetime and they end up being non-zero?

 

[Ibix]

Rotate and scale coordinates until the metric is diagonal at the event of interest.

 

[Ibix]

Another way to look at it is that it's always possible to pick a timelike vector and three spacelike vectors that are mutually orthonormal at (or more precisely, in the tangent space associated with) a chosen event. You extend those vectors into lines over at least a small region and then you have the beginnings of a coordinate system over a small region around your chosen event, which are guaranteed to be orthonormal coordinates at least at the origin event. So the metric expressed in those coordinates must be the Minkowski metric at least at the origin.

I think this is discussed in more detail in chapter 5 (it might just be the end of chapter 3, but I don't think so) of Carroll's GR lecture notes. I can dig out a more precise reference when I'm not supposed to be doing something else if you need.

 

[PeterDonis]

the Minkowski metric is diagonal

Not the way you mean. The Minkowski metric is diagonal when you choose a particular coordinate chart. But you can also choose a chart on globally flat Minkowski spacetime in which the metric is not diagonal. (For an example, look up Born coordinates.)

In other words, "diagonal" is not an inherent property of a metric. It's only a property of a metric in a particular coordinate chart.

 

[pervect]

TL;DR Summary: If we have a metric in GR with non-zero non-diagonal indices, how can it be Minkowskian at a point? After all, the Minkowski metric is diagonal, so it doesn't seem like they can be equal at some given coordinate.

I'm having trouble understanding the local flatness of GR. So far, my interpretation was that it meant that the metric tensor at an infinitesimal point in spacetime will be equal to some multiple of the Minkowski metric since that's the metric that preserves the speed of light/spacetime interval. This in turn means that even in a curved spacetime, for a single point, the Lorentz Transformation relates different frames of reference as it is the transformation that keeps C invariant at said point. This all comes from the starting postulate that at every single point, C is an invariant. From here, it seems to me that a metric could only be diagonal, since it needs to be some multiple of the Minkowski metric at each coordinate. I can see how the diagonal indices can be different from one another at one point than somewhere else, since we start with the assumption that the speed of light just needs to be invariant at each infinitesimal point. So it can have different values in different places by changing the time and space components of the metric. What I'm having trouble understanding is, how can we have diagonal indices? Wouldn't that make it non-Minkowskian if you "plug in" a specific coordinate in spacetime and they end up being non-zero?

Consider the Euclidean metric of a plane, dx^2 + dy^2 = 0. Formally, this would actually be called a line element, mea culpa. Any linear transformation of this metric, which we can perform algebraically or by other means (such a the tensor tranformation rules) provides an equivalent metric.

In this case, we can generate off-diagonal terms by transformations such as du = dx+dy. This sort of transformation is generally called a skew transformation. The dialgonal term dx dy vanishes because x and y are orthogonal, if we choose two non-orthogonal bases, we can generate off-diagional terms.

To work out the above example in more detail re-write the metric as (dx+dy)^2 - 2 dx dy = du^2 - 2 dx (du -dx) and gather the terms together to generate a metric in terms of du and dx.

Your question is more about the Lorentzian metric dx^2 - dt^2, or a higher dimensional version therof. Two dimensions is simplest, though. If you have a local metric of the form dx^2 - dt^2, a linear transformation such as du = dx + $\alpha$ dt will generate off-diagional terms. It is the convention that time be orthogonal to space that makes the off-diagional terms usually vanish. This is closely related to innumberable and usually confused threads about "the one way speed of light", but I'd rather not turn this thread into one of those, we have more than enough of them already :).

 

[Sciencemaster]

Another way to look at it is that it's always possible to pick a timelike vector and three spacelike vectors that are mutually orthonormal at (or more precisely, in the tangent space associated with) a chosen event. You extend those vectors into lines over at least a small region and then you have the beginnings of a coordinate system over a small region around your chosen event, which are guaranteed to be orthonormal coordinates at least at the origin event. So the metric expressed in those coordinates must be the Minkowski metric at least at the origin.

I think this is discussed in more detail in chapter 5 (it might just be the end of chapter 3, but I don't think so) of Carroll's GR lecture notes. I can dig out a more precise reference when I'm not supposed to be doing something else if you need.

Thank you, but I think I found them myself. When you get a chance (seriously, no rush), is this a copy of the right one: https://arxiv.org/pdf/gr-qc/9712019
Chapter 2 is the only one that discusses local flatness by name (thanks, ctrl+F!), and it shows that proof that expands the metric transformation to 2nd order. Chapter 5 does go over how the Euclidean metric on #R^3# "induces" #dS^2=d\theta^2+sin^2(\theta)d\phi^2# on a unit sphere. In essence, my understanding is that the Euclidean metric is the same as the spherical coordinate version under a coordinate substitution. Similarly, other metrics are the same even if you coordinate transform them, and you can usually (if not always) transform into a coordinate system where the metric is diagonal, similarly to what Pervect said. That being said, I'm still just a bit unsure about non-diagonal indices using specifically (t,x,y,z) coordinates. There are metrics with off-diagonal terms using this coordinate system, even though the Minkowski metric is diagonal in this system specifically. I get that you can find coordinates where some metric is diagonal (For example, #du = dx+dy#, #dv = dy# may give a metric that's diagonal in terms of (u,v) but not in terms of (x,y)). However, the invariant speed of light is explicitly in terms of (x,t), being a velocity, so I'm unsure about how some different coordinate system being Minkowskian "fits" with this.

 

[renormalize]

However, the invariant speed of light is explicitly in terms of (x,t), being a velocity, so I'm unsure about how some different coordinate system being Minkowskian "fits" with this.

Are you familiar with "light-cone coordinates" in Minkowski spacetime?
(See https://en.wikipedia.org/wiki/Light-cone_coordinates.)

 

[Dale]

the invariant speed of light is explicitly in terms of (x,t), being a velocity, so I'm unsure about how some different coordinate system being Minkowskian "fits" with this.

It doesn’t. The speed of light is only invariant in the standard inertial coordinates. In non-inertial coordinates the speed of light may be something else even in flat spacetime.

 

[Nugatory]

However, the invariant speed of light is explicitly in terms of (x,t), being a velocity, so I'm unsure about how some different coordinate system being Minkowskian "fits" with this.

Minkowski spacetime and Minkowski coordinates are different things. The spacetime is what it is no matter what coordinates we use to label events within it.

Stating the invariant speed of light in terms of $x$ and $t$ is inherently coordinate-dependent. The coordinate-independent statement is that light moves on paths of zero interval: $\int g_{\mu\nu}dx^\mu dx^\nu = 0$ (and note that we make no distinction between diagonal and off-diagonal elements - they’re all there).

If we choose, as we often do, to use Minkowski coordinates on Minkowski spacetime, we will find that the coordinate velocity of a light signal moving along such a path is $c$, and this will also be the case for any coordinates related to our original choice by a Lorentz transformation.

But if we choose coordinates that are related by some other transformation (for example, the origin is rotating) the coordinate velocity of our light signal may not be $c$. But it’s the same spacetime, we’re just labeling the events in it differently, and the line integral along the path of a light signal still comes out zero.

 

[Sciencemaster]

Fair enough. In that case, though, what exactly *does* locally flat/Minkowskian mean? It seems to me like the Minkowski metric can be anything (that follows the requirements of a metric like being symmetric and such) at a single point (i.e. rather than indices being dependent on coordinates like #R#, we plug in values at a point) if we choose the right coordinate system. As such, any metric would be locally equal to the Minkowski one with the proper coordinates defined. This description seems trivial, and pretty wrong, though. I'm not sure what constraint local flatness places on a metric/what it means to say that its equivalent to the Minkowski metric at a point, though. Also, I was under the impression that the Minkowski metric "enforces" the speed of light. After all, it transforms under a hyperbolic transformation. If C is only invariant in some choices of coordinates, does that mean that a metric is Minkowskian in that there is always some coordinate system where C is invariant, but not all systems (i.e. non-inertial ones where C isn't invariant) are necessarily locally equivalent to a form of the Minkowski metric on their own?

 

[Dale]

Fair enough. In that case, though, what exactly *does* locally flat/Minkowskian mean?

It means that at any event in spacetime there exists a set of coordinates where the Christoffel symbols all vanish as do their first derivatives.

Basically, you can always find a set of coordinates that at a given event the metric is Minkowski, and at nearby points the deviations from Minkowski are 2nd order or higher in those coordinates.

 

[Nugatory]

what exactly *does* locally flat/Minkowskian mean?

In coordinate-independent language, it means that the Riemann curvature tensor is zero at that point (and "close enough" to zero nearby).

 

[PeterDonis]

In coordinate-independent language, it means that the Riemann curvature tensor is zero at that point (and "close enough" to zero nearby).

No, that's not what it means. @Dale's explanation in post #11 is correct, and what he says is true when the Riemann curvature tensor is nonzero. The key point is that, if you choose the right coordinates at a given point, the Riemann curvature tensor being nonzero only shows up in the second derivatives of the metric coefficients at the chosen point: the metric coefficients themselves are Minkowski, and their first derivatives at that point are all zero. (The coordinates in question are, of course, Riemann normal coordinates, and their properties are explained in many GR textbooks.)

 

[Ibix]

No, that's not what it means. @Dale's explanation in post #11 is correct, and what he says is true when the Riemann curvature tensor is nonzero. The key point is that, if you choose the right coordinates at a given point, the Riemann curvature tensor being nonzero only shows up in the second derivatives of the metric coefficients at the chosen point: the metric coefficients themselves are Minkowski, and their first derivatives at that point are all zero. (The coordinates in question are, of course, Riemann normal coordinates, and their properties are explained in many GR textbooks.)

Which, incidentally, means that "locally flat" is a bit of a misnomer.

The analogue in spherical geometry is that you can lay your house out using Euclidean geometry, but not a city, and you can't use Euclidean geometry on a Mercator map. It's not really locally flat, it's that you can always pick an orthonormal coordinate system at a point, and the fact that the first derivatives of the metric are zero at that point and the coordinate functions are smooth means that the error grows slowly. So Euclidean (or Minkowski) geometry only has small errors if used in a small region around a point (or event).

 

[Sciencemaster]

Alright. In that case, if local flatness just means that a metric's Christoffel symbols are 0 to first order (on a small enough scale), wouldn't that mean they'd be just as much Minkowskian as any other totally flat metric? For example, #\delta_{\mu\nu}# has 0 for all of its derivatives, but is not Minkowskian (in the sense that none of its indices have the opposing sign). Is there anything special about saying that a metric is "locally *Minkowskian*" in the context of flatness?
Again, my mind goes to the invariance of C, but now I'm not sure if that has anything to do with flatness. The connection in my mind being that the Minkowski metric is so intertwined into SR, a theory fundamentally about flat spacetime and invariance in it. Of course, it could always 'built in' to the theory somewhere else, like the S=0 path length thing for any metric with absolutely no connection to the Minkowski metric whatsoever.

 

[Ibix]

Is there anything special about saying that a metric is "locally *Minkowskian*" in the context of flatness?

It means that the errors from treating a region of spacetime as Minkowski spacetime become small as you make the region smaller. That's basically why SR works at all on Earth - you need quite careful experimentation to detect the deviation from flat.

I think the point about "locally Minkowski" is (a) the manifold has all the necessary stuff like a metric, connection, etc, and (b) it is locally Minkowski, not locally Euclidean (or any other metric signature), so retains a sense of time plus three spatial dimensions.

 

[PeterDonis]

if local flatness just means that a metric's Christoffel symbols are 0 to first order in a particular coordinate chart

You keep ignoring the qualifier I put in bold in the quote above. It's crucial. It means you're getting fixated on the properties of a particular coordinate chart, instead of invariant properties of the spacetime geometry. That's why you keep getting confused.

Is there anything special about saying that a metric is "locally *Minkowskian*" in the context of flatness?

"Locally Minkowskian" has to do with the signature of the metric as well as local flatness. "Minkowskian" is a different signature from "Euclidean".

my mind goes to the invariance of C, but now I'm not sure if that has anything to do with flatness.

It doesn't; the light cone structure of the spacetime (which is the correct general way of thinking about "the invariance of C") is a different property from the curvature tensor.

 

[cianfa72]

It doesn’t. The speed of light is only invariant in the standard inertial coordinates. In non-inertial coordinates the speed of light may be something else even in flat spacetime.

I think you actually mean the one-way speed of light (OWSOL) is only invariant in the standard inertial coordinates (assuming of course flat spacetime).

 

[PeterDonis]

I think you actually mean the one-way speed of light (OWSOL) is only invariant in the standard inertial coordinates (assuming of course flat spacetime).

No, he doesn't. For example, if you transform to Rindler coordinates, the round trip speed of light is not always $c$ in those coordinates.

 

[Dale]

I think you actually mean the one-way speed of light (OWSOL) is only invariant in the standard inertial coordinates (assuming of course flat spacetime).

As @PeterDonis mentioned also the coordinate two way speed of light can vary in non-inertial charts. Although that does also depend on how you define things. But with pretty much any choice of definitions you can come up with some coordinate system where the speed will vary.

 

[cianfa72]

As @PeterDonis mentioned also the coordinate two way speed of light can vary in non-inertial charts.

Ah ok, I think what I put in bold is the key point. What is actually invariant are the light cones and therefore the light's velocity as tangent vector to light cones' boundary in spacetime.

But with pretty much any choice of definitions you can come up with some coordinate system where the speed will vary.

Yes, as special case Anderson coordinates (underlying flat spacetime) can be classified as inertial (timelike geodesic paths occur with zero coordinate acceleration), yet the OWSOL isn't isotropic even though the coordinate two-way speed of light is the constant $c$.

 

[Dale]

What is actually invariant are the light cones and therefore the light's velocity as tangent vector to light cones' boundary in spacetime.

Yes.

Also, any four-velocity is a 4D tangent vector to a worldline. So that vector is the same vector in different frames, although its components will be different. That means as a geometric object they can be considered to be the same thing in different frames. That is usually said to be "covariant" rather than "invariant" though as a matter of terminology.

 

[cianfa72]

In Anderson coordinates the expression of the flat spacetime metric $ds^2$ includes mixed term in $\kappa dtdx$. For $\kappa=0$ (i.e. standard inertial coordinates) they vanish, hence a distinctive property/feature of the standard inertial coordinates is that the timelike coordinate is orthogonal to spacelike coordinates in spacetime (better, orthogonal in the tangent space at any event/point in spacetime).