Fundamental question on GR: Basis vector meaning

[DrGreg]

Stop!
Before getting deeper in this discussion, could you tell if my conclusion

Basically a locally inertial observer who follows a geodesics will be locally inertial on all of the points on the geodesic (if he parallel transports his basis vectors),
while an acclerating observer won't, (obvioulsy there can be found a locally inertial frame at every point of the curve, but it won't be the very same observer from which we started from).

is correct??

Er, the way you've phrased this, you seem to be asking "will a locally inertial observer ... be locally inertial?". The answer to that is trivially yes. That can't be what you intended to ask, so could you phrase your question more clearly?

 

[teddd]

Well..that may actually be the question!

That's becaouse i used to consider an observer as locally inertial only at one point on manifold: i thought that you have to recalculate directly all of the basis vector if you wanted to get another locally inertial boserver at another point on the curve.

But there is one particular case, namely the one in which the observer follows a geodesics.
In that case the timelike basis vector is parallel transported along the curve, and so if i pick a locally inertial observer on one point of the geodesic (restating: if i choose the basis set that makes the metric minkowsky to first order, that makes all og the chirstoffel symbols vanish ecc..), and i parallel transport this basis vector set i will actually have a locally inertial frame all along the geodesics.

The question arised when i asked pervect why the curve had to be a geodesics, since the condition of parallel transport can be imposed geometrically (post n.24).
He explained to me that i can parallel transport an orthonormal basis vector set on every curve i want and obtain an orthonormal frame everywhere, but if the curve is not a geodesic, by parallel transporting those vectors, I'm actually changing observer: like he said:

As far as transporting a time basis vector along a curve - you can certainly do that, but if you have an accelerating observer, the basis vector you get by this process isn't the same as the time basis vector of the accelerating observer's clock.

Suppose at t=0 your velocity v=0 in some inertial frame I0, and that at time t=1 your velocity has increased to v=a.

If you parallel transport the time basis vector of your initial velocity at t=0 along the curve, you'll be using the time basis vector of the inertial frame I0. It won't match your new basis vectors at t=1 because your velocity has changed.

When instead the locally inertial observer at a certain point on the manifold follows a geodesics (trough that point) the parallel transporting condition (which here is a more physical and less geometrical condition) is coherent with the observer: the time basis vector all along the curve is the time basis vector of the very same observer's clock (so by this process I'm not changing observer).

So we can say that if you pick a locally inertial observer on a certain point A of a geodesic (thus choosing a basis vector set which makes the christoffel symbols vanish at A), that observer will be locally inertial all along the geodesic itself, thus all of the christoffel vanish at every point on the curve (using in those point the basis vectors parallel transported from point A).
I wanted to be sure that what I've written above is right!

 

[pervect]

An accelerometer measures acceleration relative to a free-falling test mass (actually the acceleration of the casing of the accelerometer relative to a damped mass inside the device explained in simple terms).
You are of course free in your mind to consider this as "absolute".


This heuristic fails when you measure the Earth velocity relative to the CMB (doppler dipole).
Unless you consider vacuum as a material object, like aether fans do.

I don't see what you think fails when you measure the Earth's velocity relative to the CMB. It's still a relative measurement. The frame is specified not by the existence of the CMB, but by its isotropy.

But I suspect we're far apart enough in our thinking that there's not a lot of sense talking about it. But I do feel some small obligation to point out there isn't any incosistency between measuring one's speed relative to the CMB and thinking that velocities are relative.

 

[WannabeNewton]

So we can say that if you pick a locally inertial observer on a certain point A of a geodesic (thus choosing a basis vector set which makes the christoffel symbols vanish at A), that observer will be locally inertial all along the geodesic itself, thus all of the christoffel vanish at every point on the curve (using in those point the basis vectors parallel transported from point A).

You are basically saying that a freely falling object will remain in free fall as long as it is not subject to any other forces which is, as DrGreg pointed out, a trivial point.

 

[atyy]

My guess is tedd is asking whether only a geodesic observer can be locally inertial (in the appropriate sense) along his worldline.

Yes, in the sense of Fermi-normal coordinates, as discussed in the paragraph after Eq 9.16 of http://arxiv.org/abs/1102.0529 .

 

[TrickyDicky]

I don't see what you think fails when you measure the Earth's velocity relative to the CMB. It's still a relative measurement. The frame is specified not by the existence of the CMB, but by its isotropy.

But I suspect we're far apart enough in our thinking that there's not a lot of sense talking about it. But I do feel some small obligation to point out there isn't any incosistency between measuring one's speed relative to the CMB and thinking that velocities are relative.

I would say precisely when there is differences in thinking (wich BTW I don't think are so far apart) is when forum debates are worth IMHO.

Nothing fails when measuring Earth's velocity wrt CMB, and there is no inconsistency in thinking that objects velocities are relative IMO.
What I pointed out is that you are considering the CMB an object and maybe you didn't realize it. Do you consider the CMB frame a material object?

 

[Ben Niehoff]

The CMB is a bunch of photons, so of course it's a material object.

2.7 K photons are arriving at our location from all directions. When we look at their power spectrum, we find a predominant dipole anisotropy that could be transformed away by a change in velocity. Therefore we conclude that the CMB measurements define a preferred frame: the one in which the dipole anisotropy vanishes. Hence we see that we have some nonzero velocity relative to this frame.

 

[TrickyDicky]

The CMB is a bunch of photons, so of course it's a material object.

a substance that pervade all the universe's space vacuum would you say?

2.7 K photons are arriving at our location from all directions. When we look at their power spectrum, we find a predominant dipole anisotropy that could be transformed away by a change in velocity. Therefore we conclude that the CMB measurements define a preferred frame: the one in which the dipole anisotropy vanishes. Hence we see that we have some nonzero velocity relative to this frame.

I see, preferred frame you say? How is it preferred?, looks like we haven't much choice with this one frame. It seems to me we are taking something that can't have motion like the CMB frame or vacuum as an absolute motion reference for the Earth's velocity, that appears more like an absolute frame than a preferred frame. Maybe you don't make distinctions between preferred and absolute frame of reference?

 

[WannabeNewton]

It is a preferred frame in that it is the frame in which the isotropy and homogeneity of the observed universe is made apparent. Note that we can choose frames in which space - like hypersurfaces do NOT have the properties of isotropy and homogeneity, and these frames would of course not be the CMB frame but we choose the CMB frame because it directly agrees with what we experimentally observe for our universe.

 

[TrickyDicky]

we choose the CMB frame because it directly agrees with what we experimentally observe for our universe.

Right, obviously we can chose whatever frame we want, I'm pointing out precisely that we choose this specific one for empirical reasons, and in this sense it looks like a natural choice rather than a "preferred" choice, even if any frame can be called preferred, if you choose it.
Can you see the difference?

 

[Ben Niehoff]

a substance that pervade all the universe's space vacuum would you say?

You say that like it's a bad thing...

I see, preferred frame you say? How is it preferred?, looks like we haven't much choice with this one frame. It seems to me we are taking something that can't have motion like the CMB frame or vacuum as an absolute motion reference for the Earth's velocity, that appears more like an absolute frame than a preferred frame. Maybe you don't make distinctions between preferred and absolute frame of reference?

The way the word "preferred" is intended in this context is simply that there is a physical situation that picks out a particular frame as being special. I don't mean "preferred" in the sense that I like this frame better than others, for example.

The laws of physics are Lorentz-invariant. But this does not mean that any specific physical configuration must be Lorentz-invariant. For example, if I put a massive body out in empty space, I've broken Lorentz invariance! Because now there is a preferred frame where the massive body is at rest. The physical situation has only SO(3) rotational symmetry, not full SO(3,1) Lorentz symmetry. If I put two massive bodies in space with zero angular momentum, the symmetry is further broken to SO(2). And if I give them orbital angular momentum, then there is no continuous symmetry left at all.*

* Correction: If they are in a circular orbit, then there is a continuous (affine) U(1) symmetry that is a combination of time translation and rotation in the plane of the orbit. This symmetry takes the double helix of their worldlines onto itself.

 

[TrickyDicky]

You say that like it's a bad thing...

Haha, not at all

The way the word "preferred" is intended in this context is simply that there is a physical situation that picks out a particular frame as being special. I don't mean "preferred" in the sense that I like this frame better than others, for example.

The laws of physics are Lorentz-invariant. But this does not mean that any specific physical configuration must be Lorentz-invariant. For example, if I put a massive body out in empty space, I've broken Lorentz invariance! Because now there is a preferred frame where the massive body is at rest. The physical situation has only SO(3) rotational symmetry, not full SO(3,1) Lorentz symmetry. If I put two massive bodies in space with zero angular momentum, the symmetry is further broken to SO(2). And if I give them orbital angular momentum, then there is no continuous symmetry left at all.*

* Correction: If they are in a circular orbit, then there is a continuous (affine) U(1) symmetry that is a combination of time translation and rotation in the plane of the orbit. This symmetry takes the double helix of their worldlines onto itself.

This is all fine, but these are artificial situations, all begin with "if I put such and such...".
What I'm saying is that the frame we encounter in the CMB is a "natural" frame, we don't have to put anythig, we just measure it.