Requesting clarification about metric tensor

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  • #26
stevendaryl
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Either with killing vectors or a metric ansatz. Then you solve for zero curvature. If you don’t want assume facts about spherical geometry in formulating the metric ansatz, you can derive these solving a general 2d ansatz for the case of constant positive curvature. In each case, the curvature assumption gives you enough equations to essentially uniquely constrain the metric.
So what's the ansatz?
 
  • #27
PAllen
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So what's the ansatz?
ds2 = A(r) dr2 + B(r) r2 (dθ2 + sin2θ dφ2)

That covers all spherically symmetric 3 manifold geometries. Then you solve for the flat one via the curvature tensor.
 
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  • #28
pervect
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Basically, specifying the metric is equivalent to specifying the coordinate system.

Misner writes in "Precis of General Relativity", https://arxiv.org/abs/gr-qc/9508043

Equation (1) ((an expression for a metric)) defines not only the gravitational field that is assumed, but also the coordinate system in which it is presented. There is
no other source of information about the coordinates apart from the expression for the metric. It is also not possible to define the coordinate system unambiguously in any way that does not require a unique expression for the metric. In most cases where the coordinates are chosen for computational convenience, the expression for the metric is the most efficient way to communicate clearly the choice of coordinates that is being made.
 
  • #29
Orodruin
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Basically, specifying the metric is equivalent to specifying the coordinate system.
I am not sure I agree with this. You can have a coordinate system on a manifold that is not equipped with a metric.
 
  • #30
PAllen
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I am not sure I agree with this. You can have a coordinate system on a manifold that is not equipped with a metric.
Yes, that is part of the standard definition of manifold in terms of atlas of charts!
 
  • #31
pervect
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I am not sure I agree with this. You can have a coordinate system on a manifold that is not equipped with a metric.
Well, for instance, when the IAU specifies the Barycentric Celestial Reference System (BCRS), they do it by giving the line element for the metric tensor. I'd have to agree that if one is dealing with a general manifold without a metric, that that technique wouldn't work. But in GR we do have a pseudo-Riemannian metric, and it's common physics practice to define the coordinates by specifying the metric.

The point Misner is making (at least in my reading of him) is that if you know the metric, you can calculate whatever proper intervals you like. And Misner regards proper intervals as represents the reading of any physical measuring instrument.


Misner said:
One first banishes the idea of an “observer”. This idea aided Einstein
in building special relativity but it is confusing and ambiguous in general
relativity. Instead one divides the theoretical landscape into two categories.

One category is the mathematical/conceptual model of whatever is happening that merits our attention. The other category is measuring instruments
and the data tables they provide.

For GPS the measuring instruments can be taken to be either ideal SI atomic clocks in trajectories determined by known forces, or else electromag-netic signals describing the state of the clock that radiates the signal.
My own paraphrase of what Misner is saying here. Physical readings are defined as being taken according to the SI measurement system. The fundamental SI units are the meter (for distance), the second (for time), and the kg (for mass), but if one know the value of the fundamental constants G and c, there is only one real fundamental unit required, the second. The other SI units can also be derived from the second with the correct information - the value of Boltzman's constant for temperature, the radiation weighting curve for the candela, the elementary charge of the electron for the ampere, and the number of particles in the mole.

Furthermore, we only need (and want) to measure proper time intervals, to avoid introducing human conventions into our definition of what "physical insturements" read. Introducing a synchronization convention takes us into the realm of "conceptual model of what is happning", rather than being what Misner is trying to define (in the imprecise English language) as a 'physical measuring instruement'.
 
  • #32
Nugatory
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Basically, specifying the metric is equivalent to specifying the coordinate system.
I am not sure I agree with this. You can have a coordinate system on a manifold that is not equipped with a metric
Does it suffice to replace "specifying the metric is equivalent to..." with "specifying the metric implies...."? We can have a coordinate system without a metric, and we can view the metric tensor as an abstract geometric object that exists independent of any coordinate system, but putting something to the right of the ##=## sign in ##ds^2=....## does pretty much specify a coordinate system.
 
  • #33
Orodruin
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Does it suffice to replace "specifying the metric is equivalent to..." with "specifying the metric implies...."? We can have a coordinate system without a metric, and we can view the metric tensor as an abstract geometric object that exists independent of any coordinate system, but putting something to the right of the ##=## sign in ##ds^2=....## does pretty much specify a coordinate system.
In general no. If the spacetime has symmetries, then the metric will take the same form in many different coordinate systems. This is quite evident in the case of Minkowski space and our usual treatment of it. In fact, we define Lorentz transformations as exactly those coordinate transformations that preserve the form of the metric.
 
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  • #34
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In general no. If the spacetime has symmetries, then the metric will take the same form in many different coordinate systems. This is quite evident in the case of Minkowski space and our usual treatment of it. In fact, we define Lorentz transformations as exactly those coordinate transformations that preserve the form of the metric.
D'oh. Got it.
 

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