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Lorentz Transformation

  1. Aug 8, 2007 #1
    So I'm looking at some Lorentz transformation equations and it says
    x'=[tex]\gamma[/tex](x-vt)
    t'=[tex]\gamma[/tex](t-vx/c[tex]^{2}[/tex])
    y'=y
    z'=z

    I'm assuming the values for y', y, z' and z only hold true when the inertial frames of S and S' are moving at a relative velocity in the x-direction. With this being said, what would the transformations be if the inertial frames were in an xy or xyz direction? Thanks in advance.

    -Pat
     
  2. jcsd
  3. Aug 8, 2007 #2
    lorentz transformation plane motion

    As far as I know the situation is known as plane motion. Your question is answered in

    Relativistic motion in a plane
    Byron L. Coulter
    Am. J. Phys. 48, 633 (1980) Full Text: [ PDF (486 kB) GZipped PS Order ]
    I think it can be simplified.
     
  4. Aug 9, 2007 #3
    If [itex] (\mathbf{r},t) [/itex] are space-time coordinates of an event in the reference frame O, and the reference frame O' moves with velocity [itex] \mathbf{v} = c \vec{\theta} \theta^{-1} \tanh \theta [/itex] with respect to O, then space-time coordinates [itex] (\mathbf{r}',t') [/itex] of the same event in O' can be obtained by formulas

    [tex] \mathbf{r}' = \mathbf{r} + \frac{\vec{\theta}}{\theta}(\mathbf{r} \cdot \frac{\vec{\theta}}{\theta}) (\cosh \theta - 1) - \frac{\vec{\theta}}{\theta} ct \sinh \theta [/tex]

    [tex] t' = t \cosh \theta - (\mathbf{r} \cdot \frac{\vec{\theta}}{\theta}) \frac{\sinh \theta}{c} [/tex]

    These formulas are derived by the same procedure as momentum-energy Lorentz transformations (see eq. (4.2) - (4.3) in http://www.arxiv.org/physics/0504062)

    Eugene.
     
  5. Aug 9, 2007 #4

    Meir Achuz

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    [tex]t'=\gamma(t-{\vec r}\cdot{\vec v}/c^2)[/tex]
     
    Last edited: Aug 9, 2007
  6. Aug 9, 2007 #5

    Meir Achuz

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    The previous reply is three equations, I couldn't get Latex to do line by line.
    r_\parallel and r_\perp are parallel and perp to v.
     
  7. Aug 9, 2007 #6

    CompuChip

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    And if you know matrix multiplication,
    [tex]B = \begin{pmatrix} \gamma & -\gamma \beta_1 & -\gamma \beta_2 & -\gamma \beta_3 \\
    -\gamma \beta_1 & 1 + \frac{(\gamma - 1)\beta_1^2}{\beta^2} & \frac{(\gamma - 1)\beta_1\beta_2}{\beta^2} & \frac{(\gamma - 1)\beta_1\beta_3}{\beta^2} \\
    -\gamma \beta_2 & \frac{(\gamma - 1)\beta_1\beta_2}{\beta^2} & 1+\frac{(\gamma - 1)\beta_2^2}{\beta^2} & \frac{(\gamma - 1)\beta_2\beta_3}{\beta^2} \\
    -\gamma \beta_3 & \frac{(\gamma - 1)\beta_1\beta_3}{\beta^2} & \frac{(\gamma - 1)\beta_2\beta_3}{\beta^2} & 1 + \frac{(\gamma - 1)\beta_3^2}{\beta^2}
    \end{pmatrix}[/tex]
    where [itex]\beta = (\beta_1, \beta_2, \beta_3)[/itex] is a unit vector in the direction of the relative velocity, and
    [itex]x' = B x[/itex] for [itex]x = (c t, x, y, z)[/itex] and similar for the transformed system [itex]x'[/itex].

    Source: Jackson, Classical Electrodynamics, chapter 11.7
     
  8. Aug 9, 2007 #7
    I had a feeling there was no simple answer, thanks guys.
     
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