# Local inertial frame

Hello,

in general relativity we introduce local inertial frames to be such frames where the laws of special relativity holds. Let ξα the coordinates in the local inertial frame, so we get ds²=ηαβαβ. If we switch the frame of reference to coordinates xμ : ξαα(x0,x1,x2,x3) and with gμν(x)=ηαβ ∂ξα/∂xμ ∂ξβ/∂xν we get:

ds²=gμν(x) dxμdxν

I don't understand why it isn't possible to find a transformation to get ds²=ηαβαβ on the whole or almost the whole mannifold? Because gμν(x) is still the same on the whole mannifold?

Thanks
Neutrino

bcrowell
Staff Emeritus
Gold Member
ds²=gμν(x) dxμdxν
Just a side issue -- you should have ##x^\mu## and ##x^\nu## here.

I don't understand why it isn't possible to find a transformation to get ds²=ηαβαβ on the whole or almost the whole mannifold?
In some cases it is possible to find such a transformation. Those would be cases in which the spacetime is flat (although possibly with an unusual topology). But not all spacetimes are flat, so this is not possible in all cases. As an analogy, you can't define x,y coordinates on the surface of the earth such that ##ds^2=dx^2+dy^2##, because geometry on the earth's surface isn't Euclidean.

Because gμν(x) is still the same on the whole mannifold?
I'm not following you here. The same as what?

I'm not following you here. The same as what?

I mean the same mathematical function of x on the whole manifold.

haushofer
You can write the metric in terms of Vielbeins. These Vielbeins are the local coordinate transformations which, at every point on the manifold, tell you how to transform to a flat metric. As such you will measure no gravitational field, but will accelerate ('free fall') at every point according to the equivalence principle.

So the Vielbein rewrites, locally, curvature into acceleration. Only in certain cases this also can be done globally. Compare this with a velocity. If a n object has a constant velocity v, you can choose an observer with constant velocity such that the object is standing still wrt this observer. This would be a global choice. But if the object changes its velocity at every point in spacetime (v = v(t,x) ) and you want an observer such that the object is standing still in his frame, this observer also needs a velocity depending on (t,x); a constant (global) velocity won't work.

Hope this helps :)

This would be a global choice. But if the object changes its velocity at every point in spacetime (v = v(t,x) ) and you want an observer such that the object is standing still in his frame, this observer also needs a velocity depending on (t,x); a constant (global) velocity won't work.

But could this transformation definied on a arbitrary chart of the manifold or only very locally around a point?

haushofer
But could this transformation definied on a arbitrary chart of the manifold or only very locally around a point?
Well, with coordinate transformations you have to be careful: many times they are not defined for the full manifold (if not, you can try to do an analytic extension). But you can calculate e.g. the Vielbein (up to a Lorentz transformation, of course) of the Schwarzschild solution in the usual radial coordinates as an example :) This example is not valid for the full manifold (it breaks down after the event horizon), but it is certainly not only defined 'very locally around a point', just as the Schwarzschild solution in radial coordinates itself is not.

stevendaryl
Staff Emeritus
I don't understand why it isn't possible to find a transformation to get ds²=ηαβαβ on the whole or almost the whole mannifold? Because gμν(x) is still the same on the whole mannifold?

I always find it helpful to look at lower-dimensional analogues of curved spacetime. The simplest curved space is the two-dimensional surface of a sphere. Suppose on the Earth that you start at the North Pole, and walk south 10,000 kilometers (to the equator), turn left and walk east 10,000 kilometers, and then turn left again and walk north 10,000 kilometers. You'll end up back where you started, at the North Pole. If you were walking a flat 2-D surface, then you would have traveled 3 sides of a square of length 10,000 kilometers, and you'd still be 10,000 kilometers from where you started. This difference between the surface of the Earth and a flat surface is an intrinsic difference between the two spaces, independent of whatever coordinate system you use. But whatever coordinate system you use has to include this curvature in the way that distances are calculated. So on the surface of the Earth, you can't possibly find coordinates $x,y$ where distances are measured by $ds^2 = dx^2 + dy^2$.

Nugatory
So on the surface of the Earth, you can't possibly find coordinates x,yx,y where distances are measured by ds2=dx2+dy2
But for the local inertial frames there must exist transformations to a flat spacetime? And the question is now: Is this possible (to find such a transformation) only in each point or on the whole chart of a manifold?

DrGreg
Gold Member
But for the local inertial frames there must exist transformations to a flat spacetime? And the question is now: Is this possible (to find such a transformation) only in each point or on the whole chart of a manifold?
Each local inertial frame is only an approximation to the local spacetime. Given an event E you can always find a coordinate system in which the components of the metric are exactly equal to the Minkowski components ##\eta_{\alpha \beta}## at the event E, but only approximately equal in a small region around the event.

Analgously, for any point on the Earth's surface you can draw an almost-exactly-to-scale-map on a flat piece of paper for a smallish region around the point (e.g. a 10-mile square), but you can't draw an exactly-to-scale-map of the whole Earth's surface on a flat piece of paper.

fresh_42
Mentor
But for the local inertial frames there must exist transformations to a flat spacetime? And the question is now: Is this possible (to find such a transformation) only in each point or on the whole chart of a manifold?
The manifold is locally euclidean which means every point has an open neighborhood where it behaves like a flat space. It might be the case that the local transformations / local coordinates / Vielbeins are the same (in terms of formulas) at each point but that doesn't make the manifold itself globally flat. The earth in stevendaryl's post is a good example. Your road maps work pretty well on short distances and each page of it looks kind of similar (as a map) but that doesn't flatten the earth. By the way: those open neighborhoods together are called an atlas, too.

Analgously, for any point on the Earth's surface you can draw an almost-exactly-to-scale-map on a flat piece of paper for a smallish region around the point (e.g. a 10-mile square), but you can't draw an exactly-to-scale-map of the whole Earth's surface on a flat piece of paper.

So if iÍ take two charts for the earth (I identify the earth with S²): It won't be possible to find such a transformation on a whole chart?

DrGreg
Gold Member
So if iÍ take two charts for the earth (I identify the earth with S²): It won't be possible to find such a transformation on a whole chart?
Your question seems little ambiguous to me. But what I am saying is, on flat paper you can draw an almost-accurately-to-scale map of London, say, but if you use the map's transformation and try to extend it to cover a much larger area, or even the whole of the Earth's surface, the further from London you are, the less accurate the map will be. Either the map scale or the map angles, or both, will vary across different parts of the map. Which is another way of saying the map's metric won't be exactly Euclidean (but it will be approximately Euclidean in the London area).

To characterize an atlas of the S² I only need two maps (one includes the south pole another the nord pole). And so the question is: If i choose an arbitrary atlas is it possible to define such a transformation (to minkowski spacetime) to every chart oft the atlas? Or must I choose a point of the manifold and a small open subset around this point for the transformation (so independent of the choice of atlas)?

Ibix
2020 Award
Am I right to say that the vielbeins in the earth are basically an infintesimal set square dropped on the ground at my feet? Whereas the coordinate basis is a set square dropped on my Mercator projection (or whatever chart I am using)?

DrGreg
Gold Member
Am I right to say that the vielbeins in the earth are basically an infintesimal set square dropped on the ground at my feet? Whereas the coordinate basis is a set square dropped on my Mercator projection (or whatever chart I am using)?
Yes.

DrGreg
Gold Member
To characterize an atlas of the S² I only need two maps (one includes the south pole another the nord pole). And so the question is: If i choose an arbitrary atlas is it possible to define such a transformation (to minkowski spacetime) to every chart oft the atlas? Or must I choose a point of the manifold and a small open subset around this point for the transformation (so independent of the choice of atlas)?
You can only do it point-by-point (your second option).

To give an example: the most obvious coordinates to use for (almost all of) a unit sphere are spherical coordinates (latitude and longitude) in which$$ds^2 = d\theta^2 + \cos^2 \theta \, d\phi^2$$
In fact, on the Equator ##\theta = 0## this simplifies to$$ds^2 = d\theta^2 + d\phi^2$$so this coordinate system is the locally Euclidean coordinate system on the Equator, which is approximately Euclidean near the Equator.

You can't find a coordinate system ##(e, n)## for which
$$de^2 + dn^2 = d\theta^2 + \cos^2 \theta \, d\phi^2$$everywhere.

pervect
Staff Emeritus
Hello,

in general relativity we introduce local inertial frames to be such frames where the laws of special relativity holds. Let ξα the coordinates in the local inertial frame, so we get ds²=ηαβαβ. If we switch the frame of reference to coordinates xμ : ξαα(x0,x1,x2,x3) and with gμν(x)=ηαβ ∂ξα/∂xμ ∂ξβ/∂xν we get:

ds²=gμν(x) dxμdxν

I don't understand why it isn't possible to find a transformation to get ds²=ηαβαβ on the whole or almost the whole mannifold? Because gμν(x) is still the same on the whole mannifold?

Thanks
Neutrino

The reason is rather similar to the reason why you can't use a flat paper map to give an accurate to-scale representation of distances on the globe. That (hopefully simpler and more familiar) problem involves only the spatial metric rather than the space-time metric, but the principles are the same.

Suppose you have a space-time metric ##g_{\mu\nu} = \eta_{\mu\nu}##. Then, what is the Riemann curvature tensor of this metric? You can do a formal calculation (see for instance the Wiki formula https://en.wikipedia.org/w/index.php?title=Riemann_curvature_tensor&oldid=667714359)
$$R^\rho{}_{\sigma\mu\nu} =\partial_\mu\Gamma^\rho{}_{\nu\sigma} - \partial_\nu\Gamma^\rho{}_{\mu\sigma} + \Gamma^\rho{}_{\mu\lambda}\Gamma^\lambda{}_{\nu\sigma} - \Gamma^\rho{}_{\nu\lambda}\Gamma^\lambda{}_{\mu\sigma}$$

To apply this, you need to find the Christoffel symbols for your metric, but since the Christoffel symbols ##\Gamma_{\rho\mu\nu}## are a sum of partial derivatives of the metric, and all of the metric coefficients are constants, ##\Gamma_{\rho\mu\nu} = 0##. When you compute ##\Gamma^{\sigma}{}_{\mu\nu} = g^{\rho\sigma} \, \Gamma_{\rho\mu\nu} ## they're still zero.

So, you wind up with a zero Riemann curvature tensor, when you substitute in the zero values for the Christoffel symbols into the Wiki expression for the Riemann in terms of the Christoffel symbols. This should be expected, it says that ##\eta_{\mu\nu}## is "flat", you can parallel transport a vector around a loop and have it come back unchanged, just like you can do on a flat plane, which has the spatial metric ##ds^2 = dx^2 + dy^2## (or the space-time metric ##-dt^2 + dx^2 + dy^2##).

Now, the question becomes - what coordinate transformation turns an all-zero Riemann tensor into a tensor with non-zero components? The answer to that is simple too - it doesn't happen. The tensor transformation laws say that if all the components of a tensor are zero in one basis, they're zero in any basis. So no such a transformation exists.

Therefore, no matter what transformation you try, you can't transform zero-curvature metric(one with a vanishing Riemann tensor) into a metric with non-zero curvature, or vice-versa.