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Metric tensor rank (1,1)

  1. Aug 30, 2015 #1
    I read in many books the metric tensor is rank (0,2), its inverse is (2,0) and has some property such as
    ##g^{\mu\nu}g_{\nu\sigma}=\delta^\mu_\sigma## etc. My question is: what does ##g^\mu_\nu## mean?! This tensor really confuses me! At first, I simply thought that ##g^{\mu\nu}\delta_{\mu\sigma}=g^\nu_\sigma##, but I realized it is not true. Is ##g^\mu_\nu## a metric, again? I mean can I write something like ##g^\mu_\nu x^\nu=x^\mu##?
     
    Last edited by a moderator: Aug 30, 2015
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  3. Aug 30, 2015 #2

    Mentz114

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    Please fix your Latex by replacing the '$' with a pair of '#'. Like this

    ##g^{\mu\nu}g_{\nu\sigma}=\delta^\mu_\sigma##
     
  4. Aug 30, 2015 #3

    Orodruin

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    I have used my magic Mentor powers to fix the LaTeX in your original post.

    What do you mean by ##\delta_{\mu\nu}##? The Kronecker delta tensor is a (1,1) tensor and a priori has one covariant and one contravariant index. Generally, you would not talk about ##g^\mu_\nu##, by definition ##g^{\mu\sigma}g_{\sigma\nu} = \delta^\mu_\nu##.
     
  5. Aug 30, 2015 #4
    Thanks for LaTeX. I had some problems with that.. I know that the Kronecker delta tensor is ##\delta^\mu_\nu##, but we can consider ##\delta_{\mu\nu}## as the identity matrix, can't we? Anyway, my question is what does ##g^\mu_\nu## mean? How one can define it on a Riemannian manifold? or just we can't define it?!
     
    Last edited: Aug 30, 2015
  6. Aug 30, 2015 #5

    DrGreg

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    For any type-2 tensor [itex]\textbf{A}[/itex][tex]
    A^\mu{}_\nu = g^{\mu\sigma}A_{\sigma\nu}
    [/tex]Now put [itex]\textbf{A} = \textbf{g} [/itex] and what do you get?
     
  7. Aug 30, 2015 #6
    Yes, the answer is ##\delta^\mu{}_\nu##. So, one can simply write ##g_{\mu\nu}\delta^{\nu\sigma}=g_{\mu\sigma}##, right? By the way, can one define a new metric (and not new coordinate system) such that ##g'_{\mu\nu}=A_\mu{}^\sigma g_{\sigma\nu}## where ##{\bf A}## is an arbitrary position independent tensor which and not necessarily ##\delta_{\mu}{}^\nu##?!
     
  8. Aug 30, 2015 #7

    Orodruin

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    No, definitely not. Even if it has the correct components in one coordinate system, it will not in another one. If you want any kind of meaning for ##g^\mu_\nu## you must define it to be equal to ##\delta^\mu_\nu##.
     
  9. Aug 30, 2015 #8

    Orodruin

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    Big no no, the indices do not even match in that expression.
     
  10. Aug 30, 2015 #9

    Nugatory

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    No. You may get a new tensor that way, but there's no particular reason to expect it to be the metric for some manifold.
     
  11. Aug 30, 2015 #10
    O.K, I understand ##\delta_{\mu\nu}## is not a tensor, but can't we define ##\delta_{\mu\nu}##? For example, can't we write ##\partial_\mu \partial_\nu x^2=2\delta_{\mu\nu}##? I know it is not a definition..but I think that we use ##\delta_{\mu\nu}##..
     
  12. Aug 30, 2015 #11

    Orodruin

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    What do you intend ##x^2## to be here? ##\delta_{\mu\nu}## is simply an object you will not see apart from for Cartesian tensors.
     
  13. Aug 30, 2015 #12

    stevendaryl

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    You can certainly pick a particular coordinate system, and define a tensor [itex]A_{\mu \nu}[/itex] in that coordinate system to be 1 if [itex]\mu = \nu[/itex] and 0 otherwise. However, it won't have that property in other coordinate systems. So, it would be misleading to call it [itex]\delta_{\mu \nu}[/itex].

    In contrast, [itex]g^\mu_\nu[/itex] has the property that it is 1 if [itex]\mu = \nu[/itex] and 0 otherwise, and this property holds in every coordinate system.
     
  14. Aug 30, 2015 #13
    As far as I understood, the best way to consider ##\delta_{\mu\nu}## is just simply the flat metric.
     
  15. Aug 30, 2015 #14

    Orodruin

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    This is not correct either, there are several metrics which are flat but not equal to ##\delta_{\mu\nu}##, consider all of the metrics in different curvilinear coordinate systems in a Euclidean space. It is also not always possible to chose a flat metric on any given manifold.
     
  16. Aug 30, 2015 #15

    robphy

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    ##\delta^a{}_b## is probably best thought of as the "index substitution operator" [independent of a metric]
    ( https://www.google.com/search?q="index+substitution+operator" ).

    Given a metric, we can raise and lower indices...
    ##\delta_a{}_b\equiv\delta^c{}_a g_{cb}=g_{ab} ##
    ...maybe thought of as an operator on a tensor like ##A^b## that does index-lowering, then substitution
    ...but this reveals itself to be the metric after index-substitution...

    Note: ##\delta_a{}_b## is just a convenient abbreviation for ##\delta^c{}_a g_{cb}##, although it should probably be abandoned in favor of ##g_{ab}##.
     
  17. Sep 1, 2015 #16

    vanhees71

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    Tip to the OP: Take some book on linear algebra and read about bilinear forms and "Sylvester's Law" of inertia and the signature of fundamental forms (pseudo-metrics or metrics of pseudo-Euclidean and Euclidean vector spaces):

    https://en.wikipedia.org/wiki/Sylvester's_law_of_inertia

    Then note that in the Ricci calculus it is crucial whether you have upper and lower indices. The place of the index indicates how it transforms under changes of the basis of the vector space and its co-basis.

    For differential geometry in pseudo-Riemann spaces and General Relativity you consider tangent and co-tangent spaces. For holonomous bases the transformation laws for tensor components are easy to remember. If you change from coordinates ##q^{\mu}## to ##\tilde{q}^{\mu}## contravariant tensors transform like the increments ##\mathrm{d} q^{\mu}##:
    $$\mathrm{d} \tilde{q}^{\mu} = \mathrm{d} q^{\nu} \frac{\partial \tilde{q}^{\mu}}{\partial q_{\nu}},$$
    i.e., contravariant vector components transform like
    $$\tilde{V}^{\mu} = \frac{\partial \tilde{q}^{\mu}}{\partial q_{\nu}} V^{\nu} ={T^{\mu}}_{\nu} V^{\nu}.$$
    As you see ##{T^{\mu}}_{\nu}## forms the Jacobian matrix of the coordinate transformation (local diffeommorphism).

    Covariant vector components (i.e., components of a one-form) transform as the partial derivatives of a scalar field, i.e., because of
    $$\tilde{\partial}_{\mu} \Phi:=\frac{\partial \Phi}{\partial \tilde{q}^{\mu}} = \frac{\partial \Phi}{\partial q^{\nu}} \frac{\partial q^{\nu}}{\partial \tilde{q}^{\mu}},$$
    i.e., covariant vector components transform like
    $$\tilde{V}_{\mu} = V_{\nu} \frac{\partial q^{\nu}}{\partial \tilde{q}^{\mu}}=V_{\nu} {U^{\nu}}_{\mu}.$$
    Now ##{U^{\nu}}_{\mu}## is the inverse Jacobian, because
    $${T^{\mu}}_{\nu} {U^{\nu}}_{\rho} = \frac{\partial \tilde{q}^{\mu}}{\partial q^{\nu}} \frac{\partial q^{\nu}}{\partial \tilde{q}^{\rho}} = \frac{\partial \tilde{q}^{\mu}}{\partial \tilde{q}^{\rho}} = {\delta^{\mu}}_{\rho}.$$
    The contravariant and covariant vector components transform contragrediently to each other, as it must be, because you want the contraction between a 1-form and a vector, i.e., ##V_{\mu} W^{\mu}## to be a scalar, i.e.,
    $$V_{\mu} W^{\mu} = \tilde{V}^{\mu} \tilde{W}_{\mu}.$$
    The transformation laws for higher-rank tensor follow now immideately by the definition that the transform like Kronecker products of vectors and 1-forms.
     
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