I'm attempting to understand Einstein's theory and I'm sure my question has an answer there. I'm not arguing for or against Einstein's theory, I'm just trying to understand what it says. Since my goal is to understand Einstein's theory, that is the answer I am looking for. I don't know what the answer is but people with a better understanding should know. Another way to ask the question is: If the gravitational field is spatially uniform (a field equivalent to acceleration of a reference frame), do the Einstein equations imply that the zero-zero component of the metric tensor is 1 (or -1 depending on the convention)?

The short answer is that the individual components of the metric tensor don't have any physical significance, because their values can be changed by a general coordinate transformation. So the answer to the question is, No. The components of the metric tensor are for the gravitational field what the scalar and vector potential are for the electromagnetic field, the gauge transformations are, just, different.

Thank you Stam Nicolis . I understand what you said. Even within a reference frame with a given motion, there are still choices of what coordinates (e.g., spherical coordinates versus rectangular in a Euclidean 3D space) can be used. What I would like to say is "the coordinates in the accelerating reference frame would be Lorentz rectangular if there was no acceleration." I realize that this intuitive statement needs some work to have an unambiguous meaning so I need to do some more thinking on this. But thanks for getting this thinking started.

Maybe it is enough to say that the coordinate that is chosen to be the time coordinate in the accelerating system is what is measured by the clock that is moving with the accelerating system. Would that be enough to make the zero-zero component of the metric tensor equal to 1 (or -1 depending on the convention)?

No; the correct statement is that, at any spacetime point, it's possible to choose coordinates such that the metric tensor, as a whole, takes the form it would have in flat spacetime, namely diag(1,-1,-1,-1) (or diag(-1,1,1,1), depending on the convention)). But, unless spacetime is flat, it's not possible to find such coordinates globally.

There's nothing special about g00. The special property that a metric tensor has and that's relevant for this discussion is that a spacetime is either geodesically complete, or it isn't. If it's geodesically complete, this means that it doesn't have singularities. All physical objects propagate in spacetime along time-like curves (if they're massive) or light-like curves (if they're massless). If the spacetime has a singularity, the proper time will be finite; if it doesn't have a singularity, it will be infinite.

There's a conjecture that all spacetime singularities are hidden behind horizons, so, even in spacetimes with singularities, there do exist observers that don't have singularities in their future-these are a certain class of accelerating observers, whereas free falling observers and other classes of accelerating observers do have the singularity inevitably in their future (if it's a future singularity, as occurs for black holes; the cosmological singularity is a past singularity).

Accelerating observers also have horizons, incidentally-and that's an invariant statement.

At horizons-in spacetime, always-what happens is that the sign of g00 changes. However horizons are ``coordinate artifacts'', they depend on the observer=coordinate system, they aren't invariants. Singularities are invariants.

Let me ask the question this way. How do we justify the claim that proper time of a given point on the world line is equal to the coordinate time of a clock that travels along the world line? The justification is simple when there is no acceleration so the Lorentz metric applies. I don't know how to justify this claim when there is acceleration and some mysterious (to me, i.e., I don't know what it is) metric applies.

The significance of the metric is that, when relating an increment of proper time, denoted here as dT, to an increment of coordinate time dt between nearby points on the frames own world line, the equation is

(dT)2 = g00 (dt)2

This is the reason that I need to show that g00 = 1 in order to conclude that proper time equals coordinate time for the accelerating clock. This is clearly true without acceleration so the metric is the Lorentz metric. What about with acceleration?

By showing that the identification is a gauge choice, that fixes the symmetry known as reparametrization invariance of the worldline. Cf., for instance, here: https://web.physics.ucsb.edu/~phys230A/w2016/Lecture_notes_files/pointpart.pdf

Acceleration simply means in this context that the metric isn't flat. No problem, the action remains invariant under reparametrization of the worldline whether the metric is flat or not.

“…I question this because of the equivalence between acceleration and gravity, ……”

Acceleratton and gravity aren’t equivalent, that is fundamentally impossible. Acceleration is a kinematical effect, while Gravity, as that is rigorously scientifically shown in the Shevchenko-Tokarevsky’s 2007 Planck scale informational physical model, is fudamenatlly nothing else than some fundamental Nature force, more see

https://www.researchgate.net/publication/365437307_The_informational_model_-_Gravity_and_Electric_Forces

- which acts in the Matter’s fundamentally absolute, fundamentally flat, fundamentally continuous, and fundamentally “Cartesian”, (at least) [4+4+1]4D spacetime with metrics (at least) (cτ,X,Y,Z, g,w,e,s,ct).

That gravity and acceleration aren’t equivalent is observed experimentally, see papper

“Muons, gravity and time: in https://arxiv.org/abs/1508.02339, and from the equivalence principle postulate any number of really physically senseless concequence follow, see example in https://www.researchgate.net/publication/322798185_The_informational_model_twin_paradox, about Tolman solution of twin paradox that is based on this principle.

Recent SS post in https://www.researchgate.net/post/NO43Doubts_about_General_Relativity_8-How_is_Energy-Momentum_of_Gravitational_Field_Expressed_How_is_It_Transferred_How_is_It_Exchanged/2 - is relevant to this thread question

Cheers

Sergey Shevchenko : I wish you would keep your delusions off of my threads. I have asked you before to please stay off of my threads. You always add a lot of clutter.

Thank you for your help Stam Nicolis . I still need a little more help. I understand that there is flexibility in choosing what we want to call a time coordinate. My problem is that it seems to me that this flexibility is removed when we decide to use a traveling cesium clock (for example) to define local time coordinates. We have now made a definite choice. My understanding is that, with this specific choice for the time coordinate, coordinate time of a point on the world line is equal to proper time. That is easy to justify when there is no acceleration and the Lorentz metric applies. More generally, an increment of proper time, denoted here as dT, is related to an increment of coordinate time dt between nearby points on the clocks own world line (so dxi = 0 when i = 1,2,3) by

(dT)2 = g00 (dt)2

How do we show that dT = dt for the definite choice in which time is measured by a cesium clock when there is acceleration?

Stam Nicolis : Maybe I just now got it with help from the article you gave a link for. Not only is the time coordinate flexible (but I claim that this flexibility is removed when we use a clock) but the parameterization that defines proper time is also flexible. We can change parameterization, equivalent to a change in the metric tensor. So when we say that the proper time is the time displayed on the cesium clock (or any clock that keeps the same time, i.e., that is a respectable clock), it is because we selected a parametrization that makes that happen. Am I correct?

Yes, your last statement is correct. The important point however is that such a choice of parametrization is always available and, in fact, nothing can depend on making a different choice. It's just particularly convenient for many practical issues.

Stam Nicolis : As is usual with the questions I submit to RG, the best answer came from you. Thanks again. Too bad you don't get paid for educating other people on RG.

L.D. Edmonds,

I do get paid for teaching at my university and answering these questions helps me in that :) Beyond, as Sherlock Holmes remarked, ``wok is its own reward''... Thank you for your questions, which help me check how well I understand things and get better at it!

"I question this because of the equivalence between acceleration and gravity"

The key point is that the lapse of time in an accelerated system is linked to the frequency shift of a photon that moves from the accelerated frame to the local frame, where we consider the situation.

It is not obvious that the zero/zero metric tensor component just reflects the key point, because we must consider a path to check for the frequency shift.