## local Minkowski space and free falling

Einstein's equivalence principle states that free-falling observers are in local inertial frame, so one can construct a local Minkowski frame everywhere.
So my question is whether the logic can be inversed, does every local Minkowski space represent free-falling? because in vierbein formalism, the vierbein may not be a coordinate frame.
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http://en.wikipedia.org/wiki/Gullstr...A9_coordinates

'Shell obervers' are introduced thus

 Imagine a series of spherical shells held stationary against the force of gravity by rockets outside the event horizon. The shell observer, standing on the spherical shell, can form his own shell frame. When constricted to small enough region, the space-time in the shell frame can be regarded as flat. Special relativity works in the shell frame.
(my bold)

But these observers experience acceleration which holds them at some fixed r. Presumably this means that the flat region is very small. I'm not completely convinced by this and I'd also appreciate some clarification.

Obviously for large r the acceleration will be small, but the article does not make this restriction.
 Actually, I am not sure whether that metric is local Minkowski, for the spatial part is not a sphere, so it may not make a valid counterexample???

Mentor

## local Minkowski space and free falling

To be honest, I'm still not sure what exactly a local inertial/Lorentz/Minkowski frame is. I think it's equivalent to the concept of "normal coordinates", and that Riemannian normal coordinates, Fermi normal coordinates etc., are all special cases of that. If I'm right, there are many different kinds of local inertial frames in curved spacetime.

However, a local inertial frame associated with the world line of a massive particle at a point p must satisfy the requirement that the "0 axis" is the timelike geodesic that's tangent to the world line at p. Also, any normal coordinate system must have a timelike geodesic as its 0 axis, because it's the 00 component of the metric at p in that coordinate system that's -1 instead of +1 like the other non-zero components.

So the answer to the question in #1 is definitely yes. The 0 axis of any local inertial frame (i.e. any kind of normal coordinate system) describes the motion of a massive test particle in free fall.

Mentz, I agree with what you said. If you need your experiments to have a specific maximum deviation from the predictions of SR, you need them to be localized in a smaller region of spacetime for smaller r.

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1. If you can construct a frame ( using the frame-field procedure) with a local Minkowski metric, then SR will work in that frame even if there is acceleration, because SR can cope with that ( all we require for SR is flat spacetime).
2. Not all local Minkowski frames belong to freely falling observers, because the shell observers are obviously not falling.

 Quote by karlzr ...for the spatial part is not a sphere, ...
I'm not sure that is relevant. The observer experiences the local metric which is flat.
The only Schwarzschild observer that I know about that has the $E^3$ spatial slices is the Gullstrand-Painleve free-faller. A different free falling frame, the LeMaitre observer has curved slices but is synchronous like the G-P frame and is at rest wrt the gravitational 'river'. But the G-P local frames are special in that Gallilean relativity applies there, and boosts are changes in velocity relative to the river.

Another interesting free-faller is the Hagihara observer, who is in orbit in the azimuthal plane.

It is tempting to draw another inference from this - although there are many observers that have local Minkowski frames, they can be distinguished from each other in some way. For instance by different experiences of the tidal accelerations. But that is probably irrelevant if we are resticted to a small enough volume.

( I apologise if I've abused the terminology of GR here. In particular, I don't know the difference between 'frame-field' and 'frame')

Mentor
 Quote by Mentz114 2. Not all local Minkowski frames belong to freely falling observers, because the shell observers are obviously not falling.
You need to rethink this. It's like saying that not all inertial frames in SR belong to inertial observers, because a guy in an accelerating rocket has a comoving inertial frame at every point on his world line.

 Quote by Mentz114 I don't know the difference between 'frame-field' and 'frame')
A frame at p is a basis for TpM (the tangent space at p). Alternatively, it's a function from $\mathbb R^4$ into TpM. (The set of all such functions is isomorphic to the set of all bases). A (local) frame field is a function that assigns a frame at p to each point p in some open subset of the manifold.

However, the word "frame" is also used as a synonym for "coordinate system", e.g. in the term "inertial frame" in SR. Note that each coordinate system defines a frame.

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 Quote by Fredrik A frame at p is a basis for TpM (the tangent space at p). Alternatively, it's a function from $\mathbb{R}^4$ into TpM. (The set of all such functions is isomorphic to the set of all bases). A (local) frame field is a function that assigns a frame at p to each point p in some open subset of the manifold.
Thanks Fredrik, I'll study this ( is this from Lee's book ? I've got that and I must look).

 You need to rethink this. It's like saying that not all inertial frames in SR belong to inertial observers, because a guy in an accelerating rocket has a comoving inertial frame at every point on his world line.
I don't understand. It's possible to be an accelerating observer in SR. SR includes non-inertial frames. The important thing is that spacetime is flat for all onservers in Minkowski spacetime.

I just noticed this. If a free-falling frame field is set up so that

$$g_{\mu\nu}=\Lambda_{\mu}^a\Lambda_{\nu}^b \eta_{ab}$$

then, if $L$ is a Lorentz transformation

$$L_{\mu}^\alpha L_{\nu}^\beta g_{\alpha\beta}=g_{\mu\nu}=(L\cdot \Lambda)_{\mu}^a(L\cdot \Lambda)_{\nu}^b \eta_{ab}$$

which I interpret as 'if a valid frame-field is boosted, the result is still a valid frame-field'.

Does this also mean that all free-falling frames are connected by a LT ?
 Recognitions: Science Advisor On an accelerated worldline, you can set up Fermi normal coordinates, which (-1,1,1,1) at the origin. However, unless the wordline is non-accelerating, the first derivatives of the metric will not vanish at the origin, and so the coordinates will not be "locally inertial". Even on a non-accelerating wordline, the second derivatives of the metric in "locally inertial coordinates" will not vanish at the origin, unless spacetime is truly flat. Compare eq 23.86 and 24.15 of http://www.pma.caltech.edu/Courses/p...2006/text.html Also section 3.2 of http://www.emis.de/journals/LRG/Articles/lrr-2004-6/

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 Quote by atyy On an accelerated worldline, you can set up Fermi normal coordinates, which (-1,1,1,1) at the origin. However, unless the wordline is non-accelerating, the first derivatives of the metric will not vanish at the origin, and so the coordinates will not be "locally inertial". Even on a non-accelerating wordline, the second derivatives of the metric in "locally inertial coordinates" will not vanish at the origin, unless spacetime is truly flat. Try section 3.2 of http://www.emis.de/journals/LRG/Articles/lrr-2004-6/
How does this square with the assertion that in the shell observer's local frame SR works ?

The worldline of this observer will be parallel to the source's worldline, so presumably it is not an accelerating worldline ?

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 Quote by Fredrik To be honest, I'm still not sure what exactly a local inertial/Lorentz/Minkowski frame is. I think it's equivalent to the concept of "normal coordinates", and that Riemannian normal coordinates, Fermi normal coordinates etc., are all special cases of that. If I'm right, there are many different kinds of local inertial frames in curved spacetime.
I believe the terminology is that Fermi normal coordinates are the "proper reference frame" of an arbitary observer, while Riemann normal coordinates are the "local inertial frame" of a free falling observer. So the local inertial frame is the proper reference frame if the observer is free falling.

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 Quote by Mentz114 How does this square with the assertion that in the shell observer's local frame SR works ? The worldline of this observer will be parallel to the source's worldline, so presumably it is not an accelerating worldline ?
I'm not sure, but off the top of my head, isn't the shell observer (is this the guy hovering outside the event horizon) analogous to Rindler coordinates?

I Googled and got to Wikipedia's shell coordinates http://en.wikipedia.org/wiki/Gullstr...A9_coordinates which says this guy is firing rockets, so he is accelerated.

Edit: I saw your earlier post. OK, so I working off Wikipedia's equations, the shell metric contains rs and r. If one takes drs/dr and evaluates it at r=rs, does one get drs/dr=0 (I haven't tried, but I would guess no).

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 Quote by atyy I'm not sure, but off the top of my head, isn't the shell observer (is this the guy hovering outside the event horizon) analogous to Rindler coordinates?
If this is the case then it justifies 'SR works'.

Referring to my speculations above about boosted frames, mathematically one can turn a shell observer into a FF observer by boosting in the -r direction. So the shell observers must belong to the class of inertial observers, since that is the very definition of the class.

 I saw your earlier post. OK, so I working off Wikipedia's equations, the shell metric contains rs and r. If one takes drs/dr and evaluates it at r=rs, does one get drs/dr=0 (I haven't tried, but I would guess no).
I think I might have forgotten that. I'll have to check if it's relevant to my assertions.

I conjectured earlier

 Does this also mean that all free-falling frames are connected by a LT ?
But it's not true. There are lots of counter examples.

[I've edited a formula in post#7 adding a dot to indicate the matrix product]

 Tags freefall, local inertial frame, vierbein

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