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Tensor bases

  1. Dec 10, 2012 #1
    I found this discussion online:


    The author tell me to verify that eq. (18) follows from (13) and (17).

    I'm not getting how that works on the basis of what he's given me so far. Take, for example the first expression.

    [itex]\textbf{g}=g_{\mu \nu}\tilde{\textbf{e}}^\mu \otimes \tilde{\textbf{e}}^\nu[/itex]


    [itex]g_{\mu \nu} \equiv \textbf{g}(\vec{\textbf{e}}_\mu , \vec{\textbf{e}}_\nu) = \vec{\textbf{e}}_\mu \cdot \vec{\textbf{e}}_\nu[/itex]

    From the definitions already given, [itex]\textbf{g}[/itex] is a tensor that maps two vectors into a scalar. [itex]\tilde{\textbf{e}}^\mu \otimes \tilde{\textbf{e}}^\nu[/itex] is a collection of tensors taking two vectors as operands such that [itex]\tilde{\textbf{e}}^\mu \otimes \tilde{\textbf{e}}^\nu(\vec{A},\vec{B})=A^\mu B^\nu[/itex]. I can use (12) in the article to contract those values with [itex]\vec{\textbf{e}}_\mu \cdot \vec{\textbf{e}}_\nu[/itex], wave my hands vigorously and claim linearity will allow me to treat that as [itex]\textbf{g}(\vec{A}, \vec{B})[/itex], which demonstrates the assertion.

    But I don't see how [itex]\left\langle \tilde{\textbf{e}}^\mu , \vec{\textbf{e}}_\nu \right\rangle ={\delta^{\mu}}_\nu [/itex] does anything for me with the available definitions.

    Am I missing something here?

    BTW, is there a tutorial on how to format mathematical expression on the forum?
  2. jcsd
  3. Dec 10, 2012 #2


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    You need to use formula (6) to evaluate this <animal>

    [tex] \tilde{e}^{\mu}\otimes\tilde{e}^{\nu}\left(\vec{e}_{\alpha},\vec{e}_{\beta}\right) [/tex]

    twhich takes you to the first formula in (18).

    There's a tutorial about the LaTex code in the <Administrative> section of the forums. https://www.physicsforums.com/showthread.php?t=617567
    Last edited: Dec 10, 2012
  4. Dec 10, 2012 #3

    George Jones

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    Write [itex]\textbf{g}[/itex] as a general tensor expanded in terms of the basis [itex]\left\{\tilde{\textbf{e}}^\mu \otimes \tilde{\textbf{e}}^\nu\right\}[/itex],

    [tex]\textbf{g} = a_{\mu \nu} \tilde{\textbf{e}}^\mu \otimes \tilde{\textbf{e}}^{\nu}.[/tex]
    What does [itex]g_{\alpha \beta} = \textbf{g} \left( \textbf{e}_\alpha, \textbf{e}_\beta \right) [/itex] then give you?

    [edit]Didn't see dextercioby's similar answer.[/edit]
  5. Dec 10, 2012 #4
    [tex] \tilde{e}^{\mu}\otimes\tilde{e}^{\nu}\left(\vec{e}_{\alpha},\vec{e}_{\beta}\right) [/tex] appears to give me [tex]\left\langle \tilde{e}^{\mu}, \vec{e}_{\alpha}\right\rangle \left\langle \tilde{e}^{\nu}, \vec{e}_{\beta}\right\rangle = {\delta^{\mu}}_{\alpha}{\delta^{\nu}}_{\beta}[/tex]


    [itex]g_{\mu \nu}{\delta^{\mu}}_{\alpha}{\delta^{\nu}}_{\beta}=g_{\alpha \beta}[/itex]

    How does that demonstrate that [itex]g_{\mu \nu}\tilde{\textbf{e}}^\mu \otimes \tilde{\textbf{e}}^\nu(\vec{A},\vec{B})=\textbf{g}(\vec{A},\vec{B})[/itex]? I'll have to think about this a spell. If is express [itex]\vec{A}[/itex] and [itex]\vec{B}[/itex] in terms of the basis, I will get the traditional contraction of two vectors with the metric tensor.
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