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I Lorentz transformation validity

  1. Jun 13, 2017 #1
    Is the Lorentz transformation given by the equations
    valid only if the origin of S and S' coincides at t=t'= 0 and the other axis (x,y,z) remains parallel to (x',y',z') respectively?
  2. jcsd
  3. Jun 13, 2017 #2


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    Yes, that is sometimes called the standard configuration.
  4. Jun 13, 2017 #3
    But you can choose whatever coordinate system you want, right? So why complicate things?
  5. Jun 13, 2017 #4
    mi.png r
    Why do we draw parallel to the axes?
    What does ensure that reading off the intercepts this way gives the space and time coordinates?
  6. Jun 14, 2017 #5


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    You could also choose the original frame for your coordinates. What coordinate system you pick is a matter of convenience. It is not always convenient to use the standard configuration.

    Lorentz transformations include rotations and boosts. If you want to include translations in time and space, then you are talking Poincare transformations. If you want a completely arbitrary coordinate system, you are talking general coordinate transformations.

    This is the definition of those coordinates.
  7. Jun 14, 2017 #6


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    This is the same thing we do with an ordinary x-y position graph. Are you asking why an x-t (spacetime) graph works the same way, or are you asking why all graphs work this way?
  8. Jun 14, 2017 #7
  9. Jun 14, 2017 #8

    Mister T

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    A line parallel to the x-axis is a line of constant t.
    A line parallel to the t-axis is a line of constant x.

    If a line parallel to the x-axis is a line of constant t, then that line passes through the t-axis at the value of t where x is zero.
    If a line parallel to the t-axis is a line of constant x, then that line passes through the x-axis at the value of x when t is zero.

    As the others have pointed out, this is the way we define things when we create graphs. It's what they mean, by definition. These points on the axes are called intercepts.
  10. Jun 14, 2017 #9

    Mister T

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    If I understand the discussion correctly, this is not an issue of the choice of coordinate systems. Rather it's the choice of a reference event such that the values of the coordinates equal (0, 0, 0, 0) in all coordinate systems for that event.
  11. Jun 15, 2017 #10
    But that doesn't work in curved spacetime, correct? Because inertial frames are local? Or does it work anyway since all zero's all transform to all zero's?
  12. Jun 15, 2017 #11


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    It works in curved spacetime too, but only in the region of spacetime around the event that is is small enough to be treated as flat - outside of that region the Lorentz transformations don't work at all.

    The word "since" above has things a bit backwards though. Things aren't working out because all zeroes transforms to all zeroes; instead all zeroes transforms to all zeroes because it has to. You've decided to label the same event (0,0,0,0) in both coordinate systems so unless there's a mistake somewhere the transformation between the two coordinate systems has to take (0,0,0,0) in one to (0,0,0,0) in the other because that's what a transformation does. This will be true in flat and curved spacetime, with inertial and non-inertial coordinates.
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