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Tensorial Calculation and antisymmetric tensors

  1. Jun 27, 2012 #1
    Hi Friends
    I am reading the following paper
    In the page 4 he says that

    [itex]\tilde{W}_{\mu\nu}=0\Rightarrow V_{\mu}=\partial_{\mu}\lambda[/itex]

    Where [itex]\tilde{W}^{\mu\nu}\equiv\frac{1}{2}\epsilon^{\mu \nu\rho\sigma}W_{\rho\sigma}[/itex] and [itex]W_{\mu\nu}\equiv\partial_{[\mu}V_{\nu]} [/itex] and [itex] \epsilon [/itex] is antisymmetric Levi-Civita tensor.

    The above expression is a general argument and it is not related to the paper. I can not understand how can we drive [itex]V_{\mu}=\partial_{\mu}\lambda[/itex] from [itex]\tilde{W}_{\mu\nu}=0 [/itex]
    Would someone please explain it for me
  2. jcsd
  3. Jun 27, 2012 #2
    Are you sure it doesn't say
    \partial_{\nu} \tilde{W}^{\mu \nu} = 0


    Oh, sorry, I saw that there is a V and a W, and the W is the anti-symmetrized derivative.

    Do you know Stokes' theorem in 4-dimensional space-time?
    Last edited: Jun 27, 2012
  4. Jun 27, 2012 #3
    Yes, I am sure. :( You can check it in the mentioned paper.
  5. Jun 27, 2012 #4
    Unfortunately I do not know. Is it related to Stokes's theorem?
  6. Jun 27, 2012 #5
    First of all, there is a one-to-one correspondence between [itex]\tilde{W}_{\mu \nu}[/itex], and [itex]W_{\mu \nu}[/itex]. You just showed how to find [itex]\tilde{W}[/itex] if you know W. But:
    \epsilon^{\mu \nu \rho \pi} \, \tilde{W}_{\rho \pi} = \frac{1}{2} \epsilon^{\mu \nu \rho \pi} \, \epsilon_{\rho \pi \sigma \tau} \, W^{\sigma \tau} = -\left(\delta^{\mu}_{\sigma} \, \delta^{\nu}_{\tau} - \delta^{\mu}_{\tau} \, \delta^{\nu}_{\sigma} \right) \, W^{\sigma \tau} = -W^{\mu \nu} + W^{\nu \mu} = -2 \, W^{\mu \nu}
    W^{\mu \nu} = -\frac{1}{2} \, \epsilon^{\mu \nu \rho \pi} \, \tilde{W}_{\rho \pi}

    Therefore, if you say [itex]\tilde{W}_{\mu \nu} = 0[/itex], then, so is [itex]W_{\mu \nu} = 0[/itex].
  7. Jun 27, 2012 #6
    Then, you will have:
    \partial_{\mu} V_{\nu} - \partial_{\nu} V_{\mu} = 0

    Integrate this over an arbitrary 2-dimensional surface with an element [itex]df^{\mu \nu} = -df^{\nu \mu}[/itex], and convert it to a line integral over the boundary of the surface. You should get:
    \oint{V_{\mu} \, dx^{\mu}} = 0

    Do you know what this means?
  8. Jun 27, 2012 #7
    Thanks! Now I got it. When [itex]W_{\mu \nu} =0[/itex] means [itex]\partial_\mu V_{\nu} -\partial_{\nu}V_{\mu}=0[/itex]. So for having this expression we must suppose that [itex]V_{\mu} =\partial_\mu\lambda[/itex] where [itex]\lambda[/itex] is a scalar. Because we can change order of derivations [itex]\partial_{\mu}, \partial_{\nu}[/itex]. Is it correct?
  9. Jun 27, 2012 #8
    Would you please explain more about the integral? It seems interesting.
  10. Jun 27, 2012 #9
    No, what you are proving is that [itex]W_{\mu \nu} = 0[/itex] is a necessary condition for [itex]V_{\mu} = \partial_{\mu} \lambda[/itex], which I though is trivial to show (because derivatives commute). But, I was trying to point out that it is also a sufficient condition. Well, locally at least (see Poincare's Lemma).
  11. Jun 27, 2012 #10
    Dear Dickfore, I am really poor on Topology and such kinds of mathematics. Recently I decided to begin studying this topics. Can you please suggest me some good textbooks for self-study. I am thinking about 3rd edition of Frankle's book: "Geometry of Physics".
  12. Jun 27, 2012 #11
    i don't know what to recommend, sorry.
  13. Jun 27, 2012 #12
    Thank you again for your help, my friend!
  14. Jun 28, 2012 #13


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    Frankle is nice, Nakahara is also very good for selfstudy; at least the chapters untill Fibre Bundles. After that it becomes a bit wuzzy.
  15. Jun 28, 2012 #14
    Thanks Haushofer! :)
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