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Tensor calculation / Lorentz transformation

  1. Oct 1, 2016 #1
    Hi,
    I have trouble understanding why the following relations hold true. Given the Minkowski metric [itex] \eta_{\alpha\beta}=diag(1,-1,-1,-1) [/itex] and the line segment [itex] ds^2 = dx^2+dy^2+dz^2[/itex], then how can i see that this line segment is equal to [itex] ds^2 = \eta_{\alpha\beta}dx^\alpha dx^\beta [/itex]. Further, we want the line segment to be unchanged under this metric. And i don't understand why the following equivalences hold true: [itex] ds^2 = ds'^2 [/itex] if and only if [itex] c^2d\tau^2 = c^2d\tau'^2[/itex]
    and [itex] \Lambda^{\alpha}{}_{\gamma} \Lambda^{\beta}{}_{\delta} \eta_{\alpha}{\beta} = \eta_{\gamma}{\delta} \iff \Lambda^T \eta \Lambda = \eta

    [/itex]
    Thank you.
     
    Last edited: Oct 1, 2016
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  3. Oct 1, 2016 #2

    Ray Vickson

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    Why would you think that ##dx^2+dy^2+dz^2 = c^2 dt^2 - dx^2-dy^2-dz^2##?
     
  4. Oct 1, 2016 #3
    if the first relation is true (which I of course believe but do not understand) then [itex] ds^2 = dx^2 + dy^2 + dz^2 = \eta_{\alpha\beta}dx^\alpha dx^\beta = c^2dt^2 -dr^2[/itex], since [itex] dx^{\alpha} = (cdt,dx,dy,dz) [/itex] is the 4-vector and we treat [itex] dx^{\alpha}dx^{\beta} [/itex] like the scalar product to get [itex] (c^2 dt^2,dx^2,dy^2,dz^2) [/itex].
    After multiplying this with the metric eta we would get your equation, is that correct?
     
  5. Oct 1, 2016 #4

    Nugatory

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    That's not the right definition of the interval; it's supposed to be (using your sign convention for the metric) ##ds^2 = dt^2-dx^2-dy^2-dz^2##.

    The quantity [itex] dx^2+dy^2+dz^2[/itex] is something different. It's the square of the spatial separation in a particular coordinate system between two events that happen to have the same time coordinate in that coordinate system. It is not unchanged under Lorentz transformations, and it has no physical significance except when the coordinate system is such that ##dt=0## so that we can interpret it as the square of a spatial distance.
     
    Last edited: Oct 1, 2016
  6. Oct 2, 2016 #5

    vanhees71

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  7. Oct 2, 2016 #6
    Ah, i see. We just define it this way. Thank you so much. But in equation (1.2.8), in the last equality, how do I formally see that we can swap [itex] \bar{e_\mu} [/itex] with [itex] \Lambda^{\nu}{}_{\sigma} [/itex]? I mean is it not a matrix multiplication?
     
  8. Oct 2, 2016 #7

    vanhees71

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    The ##{\Lambda^{\nu}}_{\sigma}## is a number and by definition thus commutes with vectors. This of course holds also under the sum, which is implied through the Einstein summation equation.
     
  9. Oct 2, 2016 #8
    Thank you all for taking your time helping me. I think i understand this part now. Thanks.
     
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