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Lorentz Transformation

  1. Jun 9, 2015 #1
    Dear PF Forum,
    First, I'd like to thanks this forum for helping this much and so far.
    I have a question about Lorentz Transformation. Lots of questions actually :smile:

    Instead of using t and x, I'd like to use ta and xa, and instead of using t' and x' I'd like to use tb and xb
    So here is the equation.
    ##t_b = \gamma(t_a - \frac{v_ax}{c^2})##
    ##x_b = \gamma(x - v_at_a)##
    ##y_b = y_a##, won't be used
    ##z_b = z_a##, won't be used

    ##\gamma \text{ is } \frac{1}{\sqrt{1-\frac{v_a^2}{c^2}}}##
    Before I go any further, can I just use 2 dimensions?
    1 time and 1 spatial (x), without y and z?
    And after this thread, I'd like to go back to my previous threads to understand them
    Twin Paradox asymmetry
    Twin Paradox symmetry
    Motion in space
    Lorentz and Doppler
    Universe Frame of Reference.
    But before those, I'd like to understand Lorentz first.
    Last edited: Jun 9, 2015
  2. jcsd
  3. Jun 9, 2015 #2

    Vanadium 50

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    You know, by starting six threads on slight variations of the same subject, you maximize the chances for confusion. It's also a good idea to get the basics understood first before firing off half a dozen slightly different complications.
  4. Jun 9, 2015 #3
    Actually, no.
    Those threads actually almost a month ago.
    I asked one question, then I realize that there's more that I have to learn.
    I started another, still I realize that it's less basic. I want to get to the most basic knowledge.
    I start this one because of this in someone thread that I read

    Last edited by a moderator: May 7, 2017
  5. Jun 9, 2015 #4


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    This might help:

  6. Jun 9, 2015 #5


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  7. Jun 9, 2015 #6
    Excelent, excelent, thanks.
  8. Jun 10, 2015 #7
    Thanks A.T, but could you tell me.
    I want to study Lorentz formula
    Boost only in x direction.
    Can I use just 2 dimensions, before I ask further?
    I want to understand it.
    Can I just use
    ##t_b = \gamma(t_a - \frac{vx}{c^2})##
    this only?
    ##x_b = \gamma(x_a - vt_a)##
    Last edited: Jun 10, 2015
  9. Jun 10, 2015 #8
    Better use (see post https://www.physicsforums.com/threads/length-contraction.817911/page-4#post-5137038):

    ##x' = \gamma(x - vt)##
    ##t' = \gamma(t - vx/c^2)##

    BTW, the other dimensions are really easy:

    y' = y
    z' = z

    From that follows, as explained in the other thread, for ta=tb:
    ##x'_b - x'_a = \Delta x' = \gamma \Delta x##
    That means that according to how clocks are synchronized in S, lengths in S' appear length contracted.

    Note that for relatively small speeds (say <0.001c or < 300 km/s) γ ≈ 1 so that:
    x' ≈ x - vt
    t'≈ t - vx/c2
    That's just to highlight relativity of simultaneity in the Lorentz transformations :oldeyes:
  10. Jun 10, 2015 #9
    Perhaps 1 spatial dimension (plus 1 time dimension) is enough for boost in x direction?
    So that the calculation is somewhat simpler.
  11. Jun 10, 2015 #10
  12. Jun 11, 2015 #11


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    As long as you're happy to consider everything moving in one line, x and t is fine.

    This is why trains are popular for SR examples.
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