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Tensor equation in Dirac's 1975 book

  1. Sep 17, 2013 #1
    Dirac has equation 3.4 as:

    x[itex]^{\lambda}_{,\mu}[/itex]x[itex]^{\mu}_{,\nu}[/itex]=g[itex]^{\lambda}_{\nu}[/itex]

    Shouldn't that have a 4 on the right side?

    x[itex]^{\lambda}_{,\mu}[/itex]x[itex]^{\mu}_{,\nu}[/itex]=(4?)g[itex]^{\lambda}_{\nu}[/itex]
     
  2. jcsd
  3. Sep 17, 2013 #2
    Nope. let me open the expression for you


    [itex]\Sigma _ \mu \frac{\partial x^ \lambda}{\partial x ^ \mu} \frac{\partial x^ \mu}{\partial x ^ \nu}=\delta ^ \lambda _ \nu = g ^ \lambda _ \nu [/itex]
     
  4. Sep 17, 2013 #3
    Let me open it some more

    [itex]
    \Sigma _ \mu \frac{\partial x^ \lambda}{\partial x ^ \mu} \frac{\partial x^ \mu}{\partial x ^ \nu}=\Sigma _ \mu \delta ^ \lambda _ \mu \delta ^ \mu _ \nu =\delta ^ \lambda _ \nu = g ^ \lambda _ \nu [/itex]
     
  5. Sep 17, 2013 #4
    ? Wouldn't the first term, for example, be:

    [itex]\frac{\partial x^{0}}{\partial x^{0'}}[/itex][itex]\frac{\partial x^{0'}}{\partial x^{0}}[/itex]+[itex]\frac{\partial x^{0}}{\partial x^{1'}}[/itex][itex]\frac{\partial x^{1'}}{\partial x^{0}}[/itex]+[itex]\frac{\partial x^{0}}{\partial x^{2'}}[/itex][itex]\frac{\partial x^{2'}}{\partial x^{0}}[/itex]+[itex]\frac{\partial x^{0}}{\partial x^{3'}}[/itex][itex]\frac{\partial x^{3'}}{\partial x^{0}}[/itex]=4?

    And all the off-diagonals be 0 of course.
     
  6. Sep 17, 2013 #5

    Bill_K

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    Science Advisor

    Consider the special case of the identity transformation, where all the new coordinates are the same as the old ones. That is, x0' = x0, x1' = x1, x2' = x2, and x3' = x3. Plug this into your equation, and I think you'll see that the result is 1, not 4.
     
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