Non-Euclidean geometry and the equivalence principle

["Don't panic!"]

As I understand it, a Cartesian coordinate map (a coordinate map for which the line element takes the simple form $ds^{2}=(dx^{1})^{2}+ (dx^{2})^{2}+\cdots +(dx^{n})^{2}$, and for which the coordinate basis $\lbrace\frac{\partial}{\partial x^{\mu}}\rbrace$ is orthonormal) can only be constructed in the neighbourhood of some point on a manifold if that manifold is locally flat, corresponding to the vanishing of the Riemann tensor at the point whose neighbourhood we are considering. Now, a sphere is not locally flat, but in general relativity, the equivalence principle states that within a sufficiently small neighbourhood of a given point special relativity should apply. In special relativity one considers inertial frames, defined as frames in which one can construct an orthonormal canonical basis, such that the metric tensor is given by $\eta_{\mu\nu}=\text{diag}\left(-1,1,1,1\right)$, such that the line element is of the form $ds^{2}=-(dx^{0})^{2}+ (dx^{1})^{2}+(dx^{2})^{2}+(dx^{3})^{2}$.
My question, given the above, is how it is possible to reconcile this when one considers applying these ideas to the surface of the Earth?

Maybe I have misunderstood things, regardless I'm feeling a bit confused. I realize that one can always approximate to Cartesian coordinates within a sufficiently small neighbourhood of a point, but shouldn't this be exact (rather than an approximation) in order for the equivalence principle to work?!

 

[PeroK]

I thought a sphere was locally flat. Certainly the Earth is locally flat and you can quite happily (as we do) use a set of Cartesian $i, j, k$ vectors near each point on the surface. I don't know enough differential geometry to know how this extends in general, though.

 

["Don't panic!"]

I thought a sphere was locally flat. Certainly the Earth is locally flat and you can quite happily (as we do) use a set of Cartesian $i, j, k$ vectors near each point on the surface. I don't know enough differential geometry to know how this extends in general, though.

Ah OK, it's quite possible that I've misread something somewhere. All the same, it would be great to know how this extends in general if somebody knows?!

 

[lavinia]

On any Riemannian or semi-Riemannian manifold deviations from Euclidean measurement are small over small distances. However this does not mean that the manifold is locally flat. For any curvature tensor, measurement in a small enough region approximates Euclidean.For instance, the sphere has constant positive curvature - so is not locally flat anywhere - yet measurement over small distances approximates Euclidean.

In space-time in a small enough region of space and over small enough time intervals, deviations from Minkowski fall below the ability of measuring instruments to detect them and so for practical purposes are Minkowski. However, in a region where curvature is large, the region may be extremely small. But far away from large bodies of mass, the region may be large since there, the curvature is small.

If the curvature tensor were actually flat in a small region of a point then Euclidean measurement would work for instruments of arbitrarily high accuracy.

 

["Don't panic!"]

For any curvature tensor, measurement in a small enough region approximates Euclidean.For instance, the sphere has constant positive curvature - so is not locally flat anywhere - yet measurement over small distances approximates Euclidean.

Does this mean one can, to a good approximation, use Cartesian coordinates and construct local inertial frames?
Also, is it a result of the local flatness theorem, i.e. given a metric with signature $+2$ (using the (-+++) convention) in 4 dimensions, within a sufficiently small neighbourhood of a point $p\in M$ on a (pseudo-)Riemannian manifold $M$, a coordinate system can be found in which the metric is flat to first order [I'm guessing that the "to first order" bit is what leads to the statement that the manifold is approximately Euclidean to within a sufficiently small region of a point?] $g_{\mu\nu}(x)= \eta_{\mu\nu} +O(x^2)$, such that $g_{\mu\nu}(p)= \eta_{\mu\nu}$ and $\partial_{\lambda}g_{\mu\nu}(p)=0$ (but $\partial_{\sigma}\partial_{\lambda}g_{\mu\nu}(p)\neq 0$). Importantly, the connection coefficients vanish, $\Gamma^{\lambda}_{\mu\nu}(p)=0$, such that the geometry is locally flat to first order within a sufficiently small neighbourhood of $p$. One example of such a coordinate system are Riemann normal coordinates. Apologies for stating what is probably the obvious to you, but I just wanted to put it in writing to check whether I'm correct or not?!

 

[lavinia]

Does this mean one can, to a good approximation, use Cartesian coordinates and construct local inertial frames?

Geodesic normal coordinates will be Euclidean up to second order as you said. The Taylor expansion around the origin of the coordinates for the components of the metric is the Kronecker delta plus terms of second order and higher. The second order term involves the curvature tensor so in regions where the curvature is large the approximation deteriorates more rapidly.

An observer, say in a free fall frame, would set up coordinates with his measuring instruments and if his frame is small enough, it will appear to him as flat up to the accuracy of his instruments.

 

["Don't panic!"]

Geodesic normal coordinates will be Euclidean up to second order as you said. The Taylor expansion around the origin of the coordinates for the components of the metric is the Kronecker delta plus terms of second order and higher. The second order term involves the curvature tensor so in regions where the curvature is large the approximation deteriorates more rapidly.

So is this what is meant by the informal statement that "the local geometry around a point is approximately Euclidean" (meaning that we can use special relativity)? So, for example, if we consider a sphere, then at a point we can construct a geodesic normal coordinate system such that sufficiently near that point the geometry is Euclidean (to second order) and as such constitutes a local inertial frame, with metric $\eta_{\mu\nu}$. Is it correct to say that the existence of a geodesic normal coordinate system around a point enables one to construct an orthonormal basis, such that the metric is $\eta_{\mu\nu}$? Would the coordinate system be considered a local Cartesian coordinate system (i.e. are Cartesian coordinate systems defined by the form of the metric, $ds^{2}=dx^{2}+dy^{2}+dz^{2}$)?

 

[lavinia]

So is this what is meant by the informal statement that "the local geometry around a point is approximately Euclidean" (meaning that we can use special relativity)? So, for example, if we consider a sphere, then at a point we can construct a geodesic normal coordinate system such that sufficiently near that point the geometry is Euclidean (to second order) and as such constitutes a local inertial frame, with metric $\eta_{\mu\nu}$. Is it correct to say that the existence of a geodesic normal coordinate system around a point enables one to construct an orthonormal basis, such that the metric is $\eta_{\mu\nu}$? Would the coordinate system be considered a local Cartesian coordinate system (i.e. are Cartesian coordinate systems defined by the form of the metric, $ds^{2}=dx^{2}+dy^{2}+dz^{2}$)?

If the curvature is non-zero a Cartesian coordinate system can not be exact. But it will be close in a small enough neighborhood. An attempt to construct a Cartesian coordinate system would nearly work.

If one takes a two dimensional plane in the tangent space at a point,$p$, and exponentiates it, one gets a local two dimensional surface. On this surface, one can compare the circumference of polar circles to the Euclidean formula, $2πr$, where $r$ is the length of the radial geodesic from $p$ to the circle. The formula is, if I remember correctly,

$C = 2πr - π/3Kr^3$ + terms of higher order in $r$. Here $K$ is the sectional curvature of the tangential plane (Gauss curvature of the surface) at $p$. The formula shows that the Euclidean circumference is close to the actual for small $r$.

 

["Don't panic!"]

If the curvature is non-zero a Cartesian coordinate system can not be exact. But it will be close in a small enough neighborhood. An attempt to construct a Cartesian coordinate system would nearly work.

Does this follow because within a small enough neighbourhood of a given point one can construct geodesic coordinates for which the geometry is flat to second order, $g_{\mu\nu}=\eta_{\mu\nu}+O(x^{\alpha}x_{\alpha})$, and so this coordinate system is approximately Cartesian (up to second order)? Sorry to keep re-iterating on this point, it's just to clear up in my head that this is the reason?!

 

[lavinia]

Let's do this in geodesic polar coordinates. It suffices to take a surface since one can always exponentiate a plane in the tangent space at a point and restrict to the resulting local surface.

In polar coordinates at a point, $p$, on a surface the metric takes the form, $ds^2 = dx^2 + G(x,y)dy^2$ Here the coordinate, $x$ ,is the distance along a radial geodesic starting at $p$ and $y$ is the angular coordinate of the circle normal to radial geodesics of constant length.

In terms of the metric tensor, $g_{11} = 1$, $g_{12} = 0$, $g_{22} = G(x,y)$.

The function, $G(x,y)$ has two properties at $p$: $G = 0$ and $(G^{1/2})_{x} = 1$.

For instance, in Euclidiean polar coordinates, $G(x,y) = x^2$ although $r$ (for "radius") is usually used to denote the radial coordinate.
So for Euclidean polar coordinates , the metric is $g_{11} = 1$, $g_{12} = 0$, and $g_{22} = x^2$ and one has $ds^2 = dx^2 + x^2dy^2$

For each fixed value of $y$ one can develop the Taylor Series for $G^{1/2}$.

$G(x,y)^{1/2} = G(0,y)^{1/2} + x(G^{1/2})_{x}$ I$_{x=0} + 1/2x^2(G^{1/2})_{xx}$ I$_{x=0} + 1/6x^3(G^{1/2})_{xxx}$ I$_{x=0} + o(x^3)$

Substituting for the first two terms gives

$G(x,y)^{1/2} = x + 1/2x^2(G^{1/2})_{xx}$ I$_{x=0} + 1/6x^3(G^{1/2})_{xxx}$ I$_{x=0} + o(x^3)$

The second term is zero (to be shown below) so the formula is

$G(x,y)^{1/2} = x + 1/6x^3(G^{1/2})_{xxx}$ I$_{x=0} + o(x^3)$

or $G(x,y) = x^2 + 1/3x^4(G^{1/2})_{xxx}$ I$_{x=0} + o(x^4)$ or $G(x,y) = x^2 + o(x^3)$

If the terms in $G$ and its derivatives were zero, one would have the flat Euclidean metric expressed in polar coordinates. Since the deviation is $o(x^3)$ for very small $x$ the polar coordinate grid is close to Euclidean and an observer would not be able to measure the defect. For large enough $x$ the defect would appear in the difference between the circumference of a polar circle and $2πx$.

Curvature Calculations

Since the metric has the form $ds^2 = dx^2 + G(x,y)dy^2$ an orthonormal basis for the tangent space is the vectors, $∂/∂x$ and $G^{-1/2}∂/∂y$.

The dual orthonormal basis is $dx$ and $G^{1/2}dy$. ( In tensor notation this formula is obtained by raising indices.)

The equations for the connection 1-form are

$d(dx) = ω∧G^{1/2}dy$
$d(G^{1/2}dy) = -ω∧dx$

Since $d(dx) = 0$, $ω$ has no components in the $dy$ direction and the second equation gives

$ω = -1/2G^{-1/2}G_xdy$

Taking the exterior derivative of $ω$ one finds that the curvature 2 form is $-(G^{1/2})_{xx}dx∧dy$ so the Gauss curvature is $K = -G^{-1/2}(G^{1/2})_{xx}$ since the area element, $dA$, is $G^{1/2}dx∧dy$

Note that in the flat Euclidean case, $K = 0$ since $G^{1/2} = x$.

In polar coordinates $G$ is zero at the origin of the coordinate system but the limit of the curvature as $x$ goes to zero still exists.

Substituting this formula into the Taylor series above, one gets that the second term is $1/2G^{1/2}K_{p}$ and since $G$ is zero at $p$, the second term is zero.

A similar calculation shows that the third term is $-1/6K_{p}$ so one has

$G^{1/2} = x^2 - 1/6x^3K_{p} + o(x^3)$

The moral is that the defect from flat Euclidean polar coordinates is a function of the curvature at $p$ and goes to zero as $x^3$.

 

[PeroK]

Does this follow because within a small enough neighbourhood of a given point one can construct geodesic coordinates for which the geometry is flat to second order, $g_{\mu\nu}=\eta_{\mu\nu}+O(x^{\alpha}x_{\alpha})$, and so this coordinate system is approximately Cartesian (up to second order)? Sorry to keep re-iterating on this point, it's just to clear up in my head that this is the reason?!

I think you've got it. Here's my thinking. For any given point P:

1) You have a global change of coordinates, which gives the metric certain properties at P:

a) $g_{\alpha \beta}(P) = \eta_{\alpha \beta}$ (Precisely $\eta$, not approximately $\eta$)

b) $\frac{\partial g_{\alpha \beta}}{\partial x^\gamma}(P) = 0$ (Again, precisely 0, not approximately 0)

2) We might as well, while we are at it, choose coordinates such that $P = (0, 0, 0, 0)$ i.e. $x^{\alpha}(P) = 0$

3) This means that (in these coordinates) the metric $g_{\alpha \beta}$ differs from $\eta_{\alpha \beta}$ (globally) by no linear terms in any of the coordinates. I'm not sure how to render this in $O$ notation!

4) If $P \ne 0, \ P = (t_0, x_0, y_0, z_0)$ then we'd be talking about no linear terms in $t- t_0$ etc.

And, as you say, this doesn't make the metric Euclidean or flat near $P$, just flat to first order. But, you can't always make the metric flat to the second order, so this is as good as you can do.

 

[PeroK]

PS To be useful in a physical context, you need some further constraint that shows your coordinate values are small. Being non-linear is one thing, but it makes no difference unless your coordinate values are small. In the free-fall example, your metric would have the curvature of the gravitational field in there somewhere and the deviation from $\eta$ would have the curvature as an inverse factor.

In that case, a linear approximation to $\eta$ might well be good enough. In case of the earth, radius $R$, you may well be able to ignore terms in $\frac{x-x_0}{R}$. What this process shows is that, if need be, you can find coordinates where the deviation from $\eta$ is terms in $(\frac{x-x_0}{R})^2$.

On the other hand, if you studying a football, this approximation might be no use whatever.

Things are either approximately flat at the scale you are working on or not. The coordinates don't change that. So, it seems to me it's really about finding coordinates with certain mathematical properties.