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Lorenz transformations for two parallel inertial systems

  1. Aug 17, 2011 #1
    Thanks for reading!

    1. The problem statement, all variables and given/known data
    I have been given a proof for the lorentz transformations (which I only partly understand) that relied on the two relativity postulates (equivalence of inertial systems and the speed of light being constant) for the case of two standard inertial systems: the system S' moved with velocity v along the x axis, and the systems are parallel.
    Now I've been asked to generalize the transformation for the case the the systems are parallel, but the velocity isn't necessarily in the x axis. (a general v vector).
    I need to show, that in this case, this equation holds:

    1. [itex]\vec{r'}[/itex] = [itex]\vec{r}[/itex] + ([itex]\frac{\vec{v}\cdot\vec{r}}{v^{2}}[/itex]([itex]\beta[/itex] - 1) - [itex]\beta[/itex]t)[itex]\vec{v}[/itex]

    2. t' = [itex]\beta[/itex](t - ([itex]\frac{\vec{v}\cdot\vec{r}}{v^{2}}[/itex]))

    2. Relevant equations

    The development of the transformation for the simpler case (where v = (v,0,0) ) arrives to the point where it shows that the size x2 + y2 + z2 - c2t2 is invariant. It doesn't to that point use the assumption that the systems are parallel.
    It is then sugested that from the fact that the space is isotropic y = y' and z = z'.
    This is however no longer the case. So that's where, I guess, I need to work with new tools.

    3. The attempt at a solution

    By using [itex]\vec{r}[/itex]2 = x2 + y2 + z2 I've tried to make an analogical development, treating |[itex]\vec{r}[/itex]| as x was. I just don't seem to derive the equation, and I don't know how to use the fact that the systems are parallel.

    I'm pretty rusty with algebra, I must say, haven't touched it for almost 2 years.

    I'd appreciate any help!

    Tomer.
     
  2. jcsd
  3. Aug 17, 2011 #2

    vela

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    Rather than trying to derive the general transformation from scratch, I would try expressing the transformation as a rotation to align the velocity with the +x axis, followed by a boost in the +x direction, followed by a rotation back to the original axes.
     
  4. Aug 18, 2011 #3
    Thanks for the answer.

    When you say "expressing the transformation", do you mean the expression I've written above? Or the transformation for the standard systems?

    I am really rusty :-)
     
  5. Aug 18, 2011 #4

    vela

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    The one you're trying to derive now. The idea is that you know how to calculate r' and t' when the velocity v lines up with the x-axis, so you use a spatial rotation to bring the original system into that case, do the boost, and then rotate back to the original axes because you want the answer in terms of those axes. That said, I haven't actually worked the problem out, so there might be a simpler way to solve it. I'm just saying this is what I'd try first.
     
  6. Aug 18, 2011 #5

    vela

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    Actually, now that I've looked at the problem a bit more, it's probably easier to resolve [itex]\vec{r}[/itex] into a component parallel to the velocity and one perpendicular to the velocity:[tex]\vec{r} = \vec{r}_\parallel + \vec{r}_\perp[/tex]The boost will mix t and [itex]r_\parallel[/itex] but will leave [itex]\vec{r}_\perp[/itex] unchanged, so you'll have
    \begin{align*}
    \vec{r_\perp} &\to \vec{r}_\perp \\
    \vec{r_\parallel} &\to \vec{r'}_\parallel \\
    t &\to t'
    \end{align*}
    and [itex]\vec{r}' = \vec{r}_\perp + \vec{r}'_\parallel[/itex].

    By the way, are you sure those equations you're trying to derive are correct? You should have factors of [itex]\gamma[/itex] in there somewhere.
     
    Last edited: Aug 18, 2011
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