How can we test space-time for pseudo-Riemannian-ness?

[TrickyDicky]

is there definition of normality without a metric? I've only seen a definition involving dot product, which uses the metric.

Actually there is, it relies in the fact that it is enough with having an affine connection on a differentiable manifold to define a geodesic thru a point in the manifold and then using a exponential map to the tangent space at that point you get geodesic normal coordinates.

 

[fr.jurain]

However, is there definition of normality without a metric? I've only seen a definition involving dot product, which uses the metric.

Planet Math http://planetmath.org/encyclopedia/RiemannNormalCoordinates.html has one where I don't see any dot product. It is labelled "Riemann normal coordinates" however it relies solely on the existence of geodesics, and coincides with what WikiPedia labels "geodesic normal coordinates" http://en.wikipedia.org/wiki/Normal_coordinates#Geodesic_normal_coordinates.

In both cases, it's clearly stated that
1) changing to them cancels the connection coeffs, which is crucial to be able to identify the connection coeffs with the fictitious forces of old,
2) they're available as soon as the connection is symmetric;
whereas I suspect TrickyDicky intends his "Riemann normal coordinates" to be geodesic normal coordinates in the special case where the connection is Levi-Civita.

 

[fr.jurain]

This doesn't make any sense to me. A (pseudo)-Riemannian manifold (M,g) has one and only one metric. You may define some other tensor field on M, but it is not a "metric tensor m".

A metric tensor on M is just a tensor field required to satisfy certain properties, and certainly you don't need to wait for M to be Riemannian to state what these properties are; cf. the WP definition:
http://en.wikipedia.org/wiki/Metric_tensor
They take care to mark the difference with *the* metric tensor in GR,
http://en.wikipedia.org/wiki/Metric_tensor_(general_relativity)
which can't invoked here since the situation we're discussing is inherently outside the GR sandbox.

Certainly it doesn't make physical sense to call two different tensors "metrics" on the same spacetime manifold.

Oh it most definitely does; no less than, say, to have both a gravitational potential and an EM (=electromagnetic) 4-potential defined on the same space-time manifold. Here goes.

We can make physical measurements by 2 fundamentally different mechanisms (among others):
1) interferometry; make light or a microwave emitted by electrons interfere with itself, and use the resulting pattern as a ruler or a clock;
2) weighing a test mass; tune the Lorentz force to balance fictitious forces acting on the mass.
The 1st mechanism relies on EM only, and gives us experimental access to the fundamental metric.
The 2nd mechanism relies on EM, and also on something extraneous; from times immemorial to this day this 2nd mechanism has been our only experimental access to the "fictitious forces", those we model as a linear connection now that space-time is a manifold. As far as I can fathom, which is pretty little, this mechanism is no longer used when we establish a unit of time, which thus has become only EM-based (CAUTION! not sure about that bit).

The 2nd mechanism is almost the very definition of the "fictitious forces" after digestion of the EP: they're but what puts test masses in motion when we've made sure the Lorentz forces acting on them are 0.

At this point, we have 2 mathematical objects modelling the physical situation: the fundamental metric tensor and the connection. The only constraint imposed on the latter is that normal coordinates be available, so that an EP can be recovered from the theory under construction. A side, maybe snide, observation here: in view of PAllen's post (with emphasis added by me):

I know it is trivial to define geodesics with a connection (in fact I prefer the definition for GR since the variation doesn't really guarantee any extremal properties, so why bother). However, is there definition of normality without a metric? I've only seen a definition involving dot product, which uses the metric.

and TrickyDicky's:

That is my point, to measure angles (for instance right angles) you need a metric. and to set up normal coordinates you need to be able to define othogonality. Edit:here I refer specifically to Riemannian normal coordinates.

it looks like Pr Einstein or his zealots did a nice reflex-conditioning job to let the physics folks turn 1st thing to schmor... sorry, to Levi-Civita connections, when time has come to account for the EP. Why bother indeed, if Levi-Civita's is the only casino in town.

Anyway: we're under strictly 0 mathematical obligation to let the physical connection, the one we detect by weighing, be the same as the Levi-Civita connection derived from the fundamental metric; although there are some empirical signs that they're very close to one an other. It's only natural to ask: "what are the mathematical consequences if they are equal?". Well, that was snide, too; it sure is more natural today than it was in the 1900's. The answer I understand to be "you're technically forced to adopt GR".

The next most-natural question is: "could the connection be Levi-Civita, yet not be the one derived from the fundamental metric?". In theories that answer "yes", you have 2 metrics: one is the fundamental metric, a thing of EM-only descent; the other is the gravitational potential, a thing of EM+inertial descent.
These theories have the advantage that they salvage & recycle the bulk of GR's theoretical results, at the same time allowing test results some slack, where GR only conceives null-results.

 

[Dale]

Oh it most definitely does; no less than, say, to have both a gravitational potential and an EM (=electromagnetic) 4-potential defined on the same space-time manifold. Here goes.

We can make physical measurements by 2 fundamentally different mechanisms (among others):
1) interferometry; make light or a microwave emitted by electrons interfere with itself, and use the resulting pattern as a ruler or a clock;
2) weighing a test mass; tune the Lorentz force to balance fictitious forces acting on the mass.

This would imply two different laws, not two different metrics. I.e. g_{\mu\nu} J^{\mu} J^{\nu} and g_{\mu\nu} K^{\mu} K^{\nu}

Two different metrics would mean that the same physical measurement by the same mechanism would yield two distinct results, which doesn't make physical sense.

 

[fr.jurain]

This would imply two different laws, not two different metrics. I.e. g_{\mu\nu} J^{\mu} J^{\nu} and g_{\mu\nu} K^{\mu} K^{\nu}

Sorry, this time it's me not making of sense of what you wrote.

Two different metrics would mean that the same physical measurement by the same mechanism would yield two distinct results, which doesn't make physical sense.

2 different metrics mean different physical measurements, by 2 different mechanisms, at the same point yield 2 different results whereas GR *posits* they must be equal.
1st physical measurement at point e: establish the metric at e the Pavillon de Breteuil's way;
2nd physical measurement at point e: define small loops around e, not hesitating to let 1st physical measurement at e help you specify them; general covariance sees to it it makes sense. Measure Gamma's along these loops, the balancing way. Discover the Gamma's can be integrated along the loops as Christoffel prescribed, publish paper "Einstein was right! 1st direct measurement of grav potential". Read paper 1 week later "Einstein was wrong! Discrepancies between grav potential and fundamental metrics".

 

[Dale]

Sorry, this time it's me not making of sense of what you wrote.

2 different metrics mean different physical measurements, by 2 different mechanisms, at the same point yield 2 different results whereas GR *posits* they must be equal.

No, two different metrics would be g_{\mu\nu}J^{\mu}J^{\nu} and h_{\mu\nu}J^{\mu}J^{\nu}.

If you have some other mechanism by a different physical law and also by a different metric then you have g_{\mu\nu}J^{\mu}J^{\nu} \ne h_{\mu\nu}K^{\mu}K^{\nu}, which you can always attribute entirely to differences between J and K with g=h. In order to definitively attribute the difference to g and h you must use the same physical law and find that g_{\mu\nu}J^{\mu}J^{\nu} \ne h_{\mu\nu}J^{\mu}J^{\nu} which doesn't make physical sense.

Also, this does not make mathematical sense. If you have a (pseudo)-Riemannian manifold (M,g) then the term "metric" refers to g (hence "the" metric since there is only one). If you have some other tensor h which is not equal to g then h is not the metric. If you have some manifold (M,g,h,...) where g, h, etc. are all "metrics" on M then (M,g,h,...) is by definition not a (pseudo)-Riemannian manifold.

 

[fr.jurain]

No, two different metrics would be g_{\mu\nu}J^{\mu}J^{\nu} and h_{\mu\nu}J^{\mu}J^{\nu}.

Yeah right, that's what I mean: same manifold, same point, same coordinates so same tangent vectors; for I assume this is what your J^{\mu}'s are; and YET: different lengths obtain.
A bit like in "Heck, my altimeter says we're at 4792m on the summit where IGN's gravimeters said 4810m". See? One point, one and same physical path to get there, 2 different altitudes obtained by different kinds of measurements.

Also, this does not make mathematical sense. If you have a (pseudo)-Riemannian manifold (M,g) then the term "metric" refers to g (hence "the" metric since there is only one). If you have some other tensor h which is not equal to g then h is not the metric. If you have some manifold (M,g,h,...) where g, h, etc. are all "metrics" on M then (M,g,h,...) is by definition not a (pseudo)-Riemannian manifold.

Are you kidding me? I'm afraid not, although arguably I'd deserve it. So:
If you have some manifold M on which g, h, etc. are defined and they're all metric tensors (w/out double quotes) by any account sensible folks might think of, then of course (M, g), (M, h),... are (pseudo-)Riemannian manifolds. What else.
Are we clear now? Can we step back to the problem?

 

[Dale]

Yeah right, that's what I mean: same manifold, same point, same coordinates so same tangent vectors; for I assume this is what your J^{\mu}'s are; and YET: different lengths obtain.

This is exactly the situation which does not make sense physically. You cannot possibly get two different results for one physical measurement.

A bit like in "Heck, my altimeter says we're at 4792m on the summit where IGN's gravimeters said 4810m". See? One point, one and same physical path to get there, 2 different altitudes obtained by different kinds of measurements.

The altimeter and the gravimeter are based on different physical principles so the two tensors would represent different physical quantities obtained by different physical laws. In my notation above E.g. the altimiter reading would be based on J and the gravimeter reading would be based on K. Any differences would be wholly attributable to the differences between J and K and not due to the metric.

If you have some manifold M on which g, h, etc. are defined and they're all metric tensors (w/out double quotes) by any account sensible folks might think of, then of course (M, g), (M, h),... are (pseudo-)Riemannian manifolds.

Yes, but if g is not equal to h then (M,g) and (M,h) are entirely different (pseudo-)Riemannian manifolds. A (pseudo-)Riemannian manifold is not just a manifold by itself, but a manifold equipped with a metric. If you equip it with different metrics then you have different mathematical objects. So g is the metric for (M,g) and h is the metric for (M,h), but g is not a metric for (M,h) nor is h a metric for (M,g).

 

[atyy]

Bimetric theories are reviewed in http://arxiv.org/abs/grqc/0502097 .

Are they related to metric-affine theories?

 

[fr.jurain]

This is exactly the situation which does not make sense physically. You cannot possibly get two different results for one physical measurement.

Gee... How do you understand this then?

2 different metrics mean different physical measurements, by 2 different mechanisms, at the same point yield 2 different results whereas GR *posits* they must be equal.

Please quote the post where I claimed otherwise.

The altimeter and the gravimeter are based on different physical principles so the two tensors would represent different physical quantities obtained by different physical laws. In my notation above E.g. the altimiter reading would be based on J and the gravimeter reading would be based on K. Any differences would be wholly attributable to the differences between J and K and not due to the metric.

Continuing with the metaphor, " the altimiter reading would be based on J and the gravimeter reading would be based on K" makes sense if, when discussing altimeter readings we are required to use isobares, and when discussing gravimeter readings we are required to use isopotentials. Yet it ignores the fact that we can make precise, physically meaningful, statements, without using either. Natives from Chamonix will instantly locate "la Voie Royale, the regular climb from Le Nid d'Aigle to the summit", or trace it on a particular map (of the Massif du Mont-Blanc in the current instance) for the rest of us.
The path is the same up to scale, whether it is traced on a meteorological map or on a map of the grav. anomaly; same J's, different "lengths".

So g is the metric for (M,g) and h is the metric for (M,h), but g is not a metric for (M,h) nor is h a metric for (M,g).

Glad to read it. Let's go to business then.

Bimetric theories are reviewed in http://arxiv.org/abs/grqc/0502097 .

Are they related to metric-affine theories?

As far as I understand (which is not much actually), § 2.1.1 describes the precise situation on which DaleSpam is choking: g is the fundamental metric, tau is the difference between g and the gravitational potential, Einsteinian GR is a theory where tau must be exactly zero.
And I understand "metric-affine theory" as an umbrella word for the above situation and others, where there is a fundamental metric on the 1 hand, and an affine connection on the other hand; the "bi-metric" situation being the case above, where the affine connection in question derives from the metric g+tau. Again don't get we wrong: I don't know these articles by heart, so their authors might start in horror at what I write.

 

[Dale]

Gee... How do you understand this then?

2 different metrics mean different physical measurements, by 2 different mechanisms, at the same point yield 2 different results whereas GR *posits* they must be equal.

I understand it as an incorrect assertion for the reasons given above. GR does not posit that two different physical measurements by two different mechanisms at the same event must be equal, and finding that they are different does not imply there are two metrics.

 

[Dale]

As far as I understand (which is not much actually), § 2.1.1 describes the precise situation on which DaleSpam is choking: g is the fundamental metric, tau is the difference between g and the gravitational potential, Einsteinian GR is a theory where tau must be exactly zero.

Yes, equation 1 is a good one to look at, perhaps it will help me understand your meaning.

In equation 1, tau is a Lorentz-violating term, so my understanding from your point 3 in your OP and our initial exchange is that you think that even with tau=0 (local Lorentz symmetry) there is still some possibility that spacetime may not be pseudo-Riemannian. I.e. that pseudo-Riemannian-ness means something other than SR holds locally.

I do not agree with that. I agree with the cited paper that any other background field is a Lorentz violating term. I think that "SR holds locally" means that spacetime is pseudo-Riemannian and vice versa and that tests for Lorentz invariance are tests for pseudo-Riemannian-ness.

 

[fr.jurain]

I understand it as an incorrect assertion for the reasons given above. GR does not posit that two different physical measurements by two different mechanisms at the same event must be equal,

Ah, that's the main contention then.
Measurement #1: interferometry. Yields components of the fundamental metric. Allows SR approximation in an infinitesimal neighborhood of event p, i. e. in the tangent space at p.
Measurement #2: gravimetry. Yields coeffs of the actual connection, the one Nature chose to implement at p.
I contend Einsteinian GR posits the connection obtained by #2 is equal to the Levi-Civita connection defined by the metric obtained by #1; e. g. equality is not a necessary consequence of an EP.
It is generally recognized that both connections are exceedingly difficult to tell experimentally from the connection of a strictly Minkowskian space-time+Newtonian gravitation, and there's nothing particularly wonderful if it's one order of approximation more difficult to test whether there is really equality.

and finding that they are different does not imply there are two metrics.

We agree on that; that the actual connection derives from *some* metric is also posited a priori, and need not be confirmed by future experiments. Quite naturally it's even more difficult to test than GR's thesis (i. e. than the thesis that it derives from the fundamental metric), since it's a relaxation of it.
The meaning of my post, and arguably the letter too, was that if we posit it, then we salvage the bulk of Einsteinian GR's results while allowing some slack in the test results.

 

[fr.jurain]

In equation 1, tau is a Lorentz-violating term, so my understanding from your point 3 in your OP and our initial exchange is that you think that even with tau=0 (local Lorentz symmetry) there is still some possibility that spacetime may not be pseudo-Riemannian. I.e. that pseudo-Riemannian-ness means something other than SR holds locally.

I do not agree with that. I agree with the cited paper that any other background field is a Lorentz violating term. I think that "SR holds locally" means that spacetime is pseudo-Riemannian and vice versa and that tests for Lorentz invariance are tests for pseudo-Riemannian-ness.

Sorry, for some reason my browser wasn't showing me this part when I posted last. More on this later.

 

[Dale]

Ah, that's the main contention then.
Measurement #1: interferometry. ...
Measurement #2: gravimetry. ...

Maybe you can be more explicit. Please show the full tensorial expressions for precisely the measurements that you are considering. From what you have written here I still don't see anything that I haven't already rebutted.

 

[atyy]

Theories written without the Levi-Civita connection are not necessarily bimetric. http://arxiv.org/abs/1007.3937 shows that at least in some cases, you can rewrite it as the Einstein field equation with an additional matter field.

 

[fr.jurain]

Maybe you can be more explicit. Please show the full tensorial expressions for precisely the measurements that you are considering. From what you have written here I still don't see anything that I haven't already rebutted.

Okay... Only there's a bit of the way you'll have to do by yourself. Which is to realize:
1) all of this is done on one and same manifold M, in a neighborhood of one 4-point e, so we pick one set of coordinates around e and write everything in the basis of tangent vectors.
2) not everything that is measured is a tensor, so the best we can do is to show their components in the chosen basis.

Measurement #1: interferometry; yields g_{ab}, the components of the fundamental metric;
Measurement #2: gravimetry; yields \Gamma^a_{bc} the components of the connection in the chosen basis, possibly under the postulate that the connection is symmetric.
These are smooth fields, meaning that we imagine we could get a continuous record of these quantities with arbitrary precision all along any finite set of paths from e to some other point e', and numerically compute derivatives, to any order and with arbitrary precision as well.

What properties the result could verify is anyone's guess at this time; so let's speculate
1) we compute the Ricci tensor: R_{ab} =<br /> \partial_{d}{\Gamma^d_{ba}} - \partial_{b}\Gamma^d_{da}<br /> + \Gamma^d_{de} \Gamma^e_{ba}<br /> - \Gamma^d_{be}\Gamma^e_{da} <br /> and discover that it is symmetric,
2) we compute its partial derivatives and discover that they verify R_{db}\Gamma^d_{ca} + R_{da}\Gamma^d_{cb} - \partial_{c}{R_{ab}} = R_{ab}\phi_{c}<br /> for a certain set of components \phi_{c}, which are those of a tensor by the way since the LHS is a covariant derivative;
3) taking derivatives again, we discover that \phi is curl-free: \partial_{b}{\phi_{a}} = \partial_{a}{\phi_{b}} <br />, so that \phi is the 4-gradient of some scalar field for which we choose the expression e^{-\rho},
4) we numerically integrate \phi to obtain \rho;
then, with but moderate amazement, we'll find that <br /> \Gamma^a_{bc}=\frac{1}{2}h^{ad} \left(<br /> \partial_{b}h_{dc} + \partial_{c}h_{db} - \partial_{d}h_{bc}\right)\, <br /> where h is defined as h_{ab}=R_{ab} / \rho and of course h^{ab}h_{bc}=\delta^a_c
The whole point being that none of the hypotheses in the above sorites mentions g. In other words, h, the gravitational potential, can obtain from the \Gamma^a_{bc}'s alone. It is a peculiarity of Einsteinian GR that it be equal to g.

 

[Dale]

Measurements are always scalars, please show from first principles what pair of measurements you think could disagree to demonstrate two metrics.

Also, can you reply to my earlier post? At a minimum, do you agree that all tests for pseudo-Riemannian-ness are tests for local Lorentz invariance, or are you thinking that it is possible to have local Lorentz invariance and still not be pseudo-Riemannian?

 

[fr.jurain]

Measurements are always scalars.

This does not even look like genuine stupidity. I give up.

 

[fr.jurain]

so that \phi is the 4-gradient of some scalar field for which we choose the expression e^{-\rho},

Sorry, it's -\ln{\rho}, not e^{-\rho}.

 

[Dale]

This does not even look like genuine stupidity. I give up.

There is no need to be rude, particularly since I am right. A measuring device produces a number, and that number is always the same regardless of what coordinate system you are using, therefore the number is a scalar.

 

[fr.jurain]

A measuring device produces a number, and that number is always the same regardless of what coordinate system you are using, therefore the number is a scalar.

Hem... How do you measure the position of a particle (in a Euclidean setting)?

 

[Dale]

Hem... How do you measure the position of a particle (in a Euclidean setting)?

Usually with three individual measurements, unless the particle is otherwise constrained.

In any case, the numbers returned by the measuring apparatus, whether one or multiple, are the same in all coordinate systems. So they transform as scalars.

 

[fr.jurain]

Usually with three individual measurements, unless the particle is otherwise constrained.

In any case, the numbers returned by the measuring apparatus, whether one or multiple, are the same in all coordinate systems. So they transform as scalars.

Yes, vectors and other multi-dimensional quantities can be measured. You do it by specifying a reference frame, and then, you get simultaneous readings -scalars assuredly, present and the same for all to see, whatever their position- which are the components of what you want measured, a 3-dim vector in the above case, in that frame, and only in that frame.

Now (with emphasis added here, not when 1st posted):

Okay... Only there's a bit of the way you'll have to do by yourself. Which is to realize:
1) all of this is done on one and same manifold M, in a neighborhood of one 4-point e, so we pick one set of coordinates around e and write everything in the basis of tangent vectors.
2) not everything that is measured is a tensor, so the best we can do is to show their components in the chosen basis.

Measurement #1: interferometry; yields g_{ab}, the components of the fundamental metric;
Measurement #2: gravimetry; yields \Gamma^a_{bc} the components of the connection in the chosen basis, possibly under the postulate that the connection is symmetric.
These are smooth fields, meaning that we imagine we could get a continuous record of these quantities with arbitrary precision all along any finite set of paths from e to some other point e', and numerically compute derivatives, to any order and with arbitrary precision as well.
[...]
In other words, h, the gravitational potential, can obtain from the \Gamma^a_{bc}'s alone. It is a peculiarity of Einsteinian GR that it be equal to g.

In view of the above, can't you answer your question by yourself?

Measurements are always scalars, please show from first principles what pair of measurements you think could disagree to demonstrate two metrics.

Is there any ambiguity in my post about what is measured and what is integrated from measurement results, what is a metric tensor and what is not, what I claim can disagree whereas Einsteinian GR posits they're equal?

 

[Dale]

Is there any ambiguity in my post about what is measured and what is integrated from measurement results,

Yes, it is completely ambiguous. I have no idea what measuring device you are considering, what physical principle it operates under, and what experiment you are proposing. As far as I can tell you are avoiding answering my question and are simply assuming the consequent.

If you would like to describe how you collect some experimental data and then compute tensors from that, then that is fine, but I am interested in the first step which you have been skipping, the experimental measurements. Those are scalars. You are completely missing the description of how you physically collect those.

 

[fr.jurain]

Is there any ambiguity in my post about what is measured and what is integrated from measurement results,

Yes, it is completely ambiguous.

What a pity. Well: g and \Gamma are obtained by measurement; h is obtained by integration.

I have no idea what measuring device you are considering, what physical principle it operates under, and what experiment you are proposing. As far as I can tell you are avoiding answering my question and are simply assuming the consequent.
If you would like to describe how you collect some experimental data and then compute tensors from that, then that is fine, but I am interested in the first step which you have been skipping, the experimental measurements. Those are scalars. You are completely missing the description of how you physically collect those.

This is not the point; not necessarily an uninteresting or irrelevant one, mark. Let me just remind you that you've had the following info:

We can make physical measurements by 2 fundamentally different mechanisms (among others):
1) interferometry; make light or a microwave emitted by electrons interfere with itself, and use the resulting pattern as a ruler or a clock;
2) weighing a test mass; tune the Lorentz force to balance fictitious forces acting on the mass.

You've also had clear indications that interferometry is how we get g, and weighing how we get \Gamma.

Now, back to the point. My question was:

In view of the above, can't you answer your question by yourself?
Is there any ambiguity in my post about what is measured and what is integrated from measurement results,
what is a metric tensor and what is not, what I claim can disagree whereas Einsteinian GR posits they're equal?

The question in question being:

Measurements are always scalars, please show from first principles what pair of measurements you think could disagree to demonstrate two metrics.

Well? Is it clear to you what I think could disagree? Is it a pair of measurements? Is it clear how I define the two metrics?

 

[Dale]

What a pity. Well: g and \Gamma are obtained by measurement

How, exactly?

You've also had clear indications that interferometry is how we get g, and weighing how we get \Gamma.

You think it is clear, but I don't know what measurements you intend.

Is it clear to you what I think could disagree? Is it a pair of measurements? Is it clear how I define the two metrics?

No, it is not clear to me. That is why I keep asking you so many times to explain your intended measurements.

 

[fr.jurain]

OK, one last try. Remember this?

Gee... How do you understand this then?

2 different metrics mean different physical measurements, by 2 different mechanisms, at the same point yield 2 different results whereas GR *posits* they must be equal.

I understand it as an incorrect assertion for the reasons given above. GR does not posit that two different physical measurements by two different mechanisms at the same event must be equal, and finding that they are different does not imply there are two metrics.

It's somewhat unfortunate you truncated my explanations right before the moment they had a chance to avoid a misunderstanding. The full quote is:

2 different metrics mean different physical measurements, by 2 different mechanisms, at the same point yield 2 different results whereas GR *posits* they must be equal.
1st physical measurement at point e: establish the metric at e the Pavillon de Breteuil's way;
2nd physical measurement at point e: define small loops around e, not hesitating to let 1st physical measurement at e help you specify them; general covariance sees to it it makes sense. Measure Gamma's along these loops, the balancing way. Discover the Gamma's can be integrated along the loops as Christoffel prescribed, publish paper "Einstein was right! 1st direct measurement of grav potential". Read paper 1 week later "Einstein was wrong! Discrepancies between grav potential and fundamental metrics".

It should be clear by now the 1st physical measurement in question was always meant to yield the g_{ab} at e, and the 2nd was always meant to yield the \Gamma^a_{bc} around e, which of course cannot be compared one for one with the g_{ab}. And so I never intended them to be compared that way, and it takes a good dose of benevolence to accept you could misunderstand me on that count:

GR does not posit that two different physical measurements by two different mechanisms at the same event must be equal.

Dub the statement what you want, in Einsteinian GR \Gamma^a_{bc} = g^{ad}(\partial_{b}g_{dc} + \partial_{c}g_{db} - \partial_{d}g_{bc})/2. Whereas the bulk and the detail of the quote, and of my subsequent posts, is I claim it can be that \Gamma^a_{bc} \ne g^{ad}(\partial_{b}g_{dc} + \partial_{c}g_{db} - \partial_{d}g_{bc})/2, all the while maintaining \Gamma^a_{bc} = h^{ad}(\partial_{b}h_{dc} + \partial_{c}h_{db} - \partial_{d}h_{bc})/2 for some h \ne g. No more, no less than that.
Where in Einsteinian GR g is by definition what gives the tangent space at e its Minkowskian structure, thereby allowing SR to hold locally; and of course it is defined the same in my posts;
and likewise in Einsteinian GR \Gamma is by definition the connection thanks to which Newton's law admits a generally covariant formulation, i. e.
F^a = mA^a = m(\frac{dU^a}{ds} + \Gamma^a_{bc}U^bU^c) with F the resultant of the 4-forces acting on the test particle, m its (constant) mass, U its 4-velocity, and again it is defined the same in my posts. Again: of course, what else could it be.
If it was not clear until stated the way I just did, then sorry; just quote me in full in the future, and in return I'll make an effort to launch math formulas right from the start.

So, in answer to your question:

No, it is not clear to me. That is why I keep asking you so many times to explain your intended measurements.

Well it's really too bad it's not clear; for these are exactly the same measurements that give experimental access to g and \Gamma in Einsteinian GR; so if you know enough of GR to assert what's posited in it and what's not, as in:

GR does not posit that two different physical measurements by two different mechanisms at the same event must be equal

then you must necessarily know what these measurements are. Mustn't you?

 

[Dale]

you must necessarily know what these measurements are. Mustn't you?

Clearly not, which is why I have been asking for several pages. And despite more than ample opportunity, I see no indication that you know either.

Also, you have avoided answering the repeated question about if you agree that all tests for pseudo-Riemannian-ness are tests for local Lorentz invariance, or if you are thinking that it is possible to have local Lorentz invariance and still not be pseudo-Riemannian?

 

[fr.jurain]

Clearly not, which is why I have been asking for several pages. And despite more than ample opportunity, I see no indication that you know either.

Not so fast. Let me remind you that a statement by you is part of the dispute, and despite more than ample indication that we might not have the same meanings in mind when using the same words, you have not clearly committed to one yet.
The disputed part is emphasized like this here (it was not when posted):

I understand it as an incorrect assertion for the reasons given above. GR does not posit that two different physical measurements by two different mechanisms at the same event must be equal, and finding that they are different does not imply there are two metrics.

You chose to assert, not to ask. What did you consider these "two different physical measurements" to be? Were they g and \Gamma, defined as I now define them? Or the same two, defined otherwise? Or "Any two things fr.jurain might claim can obtain from measurement"? Else, what?