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Lorentz vs space time interval

  1. Sep 18, 2010 #1
    A question about the relationship between the formulas found using the Lorentz transform and the invariance of the space time interval.

    Two events A and B occur at the same time and different space locations in system S, where A and B are at rest and at distance x.
    The system S' moves with velocity v with respect to S.

    I want to find the time interval t' that S' measures between A and B.

    1. By using the Lorentz transform one finds immediately:

    [tex]t' = -\gamma\frac{v}{c^{2}}x[/tex]

    2. By using the invariance of the space time interval:

    [tex]{(ct')^{2} - x'^{2} = - x^{2}}[/tex]

    and using the contacted [tex]x' = \frac{x}{\gamma}[/tex]

    one finds (neglecting the negative square...)

    [tex]t' = -\frac{v}{c^{2}}x[/tex]

    So here the gamma is missing.... Must I understand that:

    With the Lorentz transform I find the value of the time interval 'assigned' at S' as measured by S (with the time dilation due to the relative velocity between S and S')


    With the invariance of the space time interval I find this time interval as measured by S' (analogously to the use of the contracted x' in the expression of the interval)?

    Have I right understood?
    Thank you
    Last edited: Sep 18, 2010
  2. jcsd
  3. Sep 19, 2010 #2
    I don't quite understand the wording of your question at the end, but think I can see where the anomaly in your calculation comes from. (NOTE: I'll use units where c = 1 to make the equations simpler; c can be replaced afterwards, using dimensional analysis.)

    If we use the Lorentz transformation, setting t = 0,

    [tex]t'=\gamma(t-vx)=-\gamma v x[/tex]

    just as you found. So far so good... And, again setting t = 0,

    [tex]x'=\gamma(x-vt)=\gamma x[/tex]



    [tex](t')^2=v^2(x')^2=\gamma^2 v^2 x^2[/tex]

    [tex](t')=\pm \gamma x v[/tex]

    You got inconsistent results because you put [itex]x[/itex] and [itex](x')^2[/itex] the wrong way round in the equation after you wrote "and using the contacted".
  4. Sep 19, 2010 #3
    Thank you; have you please the patience to help me again?

    My aim is to find t' using the invariance of the space time interval.
    This interval calculated in S is x2.
    This interval in S' should be x'2 - t'2.
    It should be x'2 - t'2 = x2.

    S' moves wrt S at velocity v, so, as I qualitatively understand, it should be x'<x.

    You obtained the right result by putting [tex] x' = \gamma x > x [/tex].

    I think I have here a HUGE misunderstanding :eek:!

    Can you or somebody else please help me?
  5. Sep 19, 2010 #4


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    This is a common misconception. I had the same problem when I first learned it. You are thinking that x' should be less than x, because of length contraction. This is incorrect. In fact, for the case you stated x' >x, as Rasalhague showed. But then what about length contraction? How does that work? To understand this, you need to understand that events that are simultaneous in one frame are not necessarily simultaneous in the other frame. So, consider a rod of length L in the unprimed frame. The two ends of the bar have the following unprimed coordinates ( all coordinates are (t,x)):

    [tex]E1 = (0,0) E2 = (0,L) [/tex]

    So in the unprimed frame, the length of the bar is the difference of the two x-coordinates, which is L. Now, transform these into primed coordinates using the Lorentz transformation, and you find:

    [tex]E1' = (0,0) E2' = ({\beta} {\gamma} L, {\gamma} L)[/tex]

    Now, the length of the bar in the primed frame is not just the difference in the x coordinates, because the two events are now at different times . To find the length of the bar in the primed frame, I need to find coordinates of the two ends at the same value of t'. You can do this by asking what time in the unprimed frame gives a value of t' = 0 in the primed frame. These new events will be:

    [tex]E1= (0,0) E3 = (-\beta L,L) E1' = (0,0) E3' = (0,\gamma (1-\beta^2)L) = (0,\frac{L}{\gamma})[/tex]

    Now we can find the length of the bar in the primed frame by just taking the difference of the x-coordinates between E1' and E3', which are now at the same time in the primed frame. As expected, now L'<L, indicating length contraction. Does this help?
  6. Sep 19, 2010 #5
    Thank you phyzguy, it helps a lot!

    Now I've understood my mistake and I can go on, and find other mistakes... :smile:

    For example, from

    {(ct')^{2} - x'^{2} = - x^{2}}

    can one really find t' without using the Lorentz transformations? I thought yes, but how to?

    Ciao e grazie
    Last edited: Sep 19, 2010
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