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I Requesting clarification about metric tensor

  1. Dec 17, 2017 #21

    Orodruin

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    It is not a derivation, it is a definition. As @stevendaryl said, there are many possible metrics on any manifold. However, there is one particular metric that will make your manifold equivalent to a Euclidean space and you can define it according to ##g_{rr} = 1##, ##g_{\theta\theta} = r^2##, ##g_{\varphi\varphi} = r^2\sin^2(\theta)##, and the remaining off-diagonal components equal to zero. It is then easy to show that this is a flat manifold and deduce all other properties (after the appropriate extension to ##r = 0##) that define a Euclidean space.
     
  2. Dec 17, 2017 #22
    OK, thanks, I think I get it. I also realized that I used "derived" when I meant "defined".

    A loose analogy would be the scalar, temperature, being defined (or described) as, say, "cold", but it derives a concrete value when you specify the units (or coordinates). In Fahrenheit it would be 68, and 20 in centigrade, and 292 in kelvin, but the "coldness" is independent of the units being used.

    Having said that, isn't the inertia tensor of a distribution of masses ##I_{ab}=m_i(\delta_{ab}r_i^2 - r_a r_b)##, coordinate-free? Is there one such definition for metric tensor, which is a little more concrete than the abstract definition that "it describes the geometry of the space"? In this thread, you had said the following and I suspect that the definition of metric tensor would be an elaboration of this:

    "A metric tensor is just a (0,2) tensor on the manifold that is (a) symmetric, (b) positive definite, (c) if ##g(X,X)=0## then ##X=0##. The metric defines distances and angles on the manifold, which includes defining concepts such as inner products between tangent vectors. By definition the components are given by ##g_{ij} = g(\partial_i,\partial_j)##."
     
  3. Dec 19, 2017 #23

    PAllen

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    So, you could assume spherical symmetry and no curvature. That would specify standard spherical coordinates up to a scale factor. You don’t ever involve Cartesian coordinates.
     
  4. Dec 19, 2017 #24

    stevendaryl

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    How, mathematically, do you state spherical symmetry?
     
  5. Dec 19, 2017 #25

    PAllen

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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.
     
  6. Dec 19, 2017 #26

    stevendaryl

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    So what's the ansatz?
     
  7. Dec 19, 2017 #27

    PAllen

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    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.
     
    Last edited: Dec 20, 2017
  8. Dec 20, 2017 #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

     
  9. Dec 20, 2017 #29

    Orodruin

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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.
     
  10. Dec 20, 2017 #30

    PAllen

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    Yes, that is part of the standard definition of manifold in terms of atlas of charts!
     
  11. Dec 20, 2017 #31

    pervect

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    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.


    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'.
     
  12. Dec 20, 2017 #32

    Nugatory

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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.
     
  13. Dec 20, 2017 #33

    Orodruin

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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.
     
  14. Dec 20, 2017 #34

    Nugatory

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    D'oh. Got it.
     
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