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Would Einstein object?

  1. Dec 2, 2008 #1
    Consider the relative position of the inertial reference frames I and I’ as detected from I when the standard synchronized clocks of that frame read t. The reference frames I and I’ are in the arrangement which leads to the Lorentz transformations of the space-time coordinates of the same event. The distance between of the origins O and O’ is at that very moment V(t-0). Let M(x) and M’(x’) be two points located on the permanently overlapped OX(O’X’) axes, located ate the same point in space, when detected from I and I’ respectively. The length (x’-0) is a proper length in I. Measured from I it is the Lorentz contracted length (x’-0)/g where g is the Lorentz factor. Adding only lengths measured by observers from I the result is
    (x’-0)/g=(x-0)-V(t-0). (1)
    We obtain the “inverse” of (1) by changing the sign of V and interchanging the corresponding primed physical quantities with unprimed ones i.e.
    (x-0)/g=(x’-0)+V(t-0) . (2)
    Solving the simultaneous equations (1) and (2) for t and t’ respectively we obtain
    (t-0)/g=(t’-0)+V(x’-0)/cc (3)
    (t’-0)/g=(t-0)-V(x’-0)/cc (4)
    Confronted with the “four line” derivation of the Lorentz transformations presented above would Einstein object?
     
  2. jcsd
  3. Dec 2, 2008 #2

    DrGreg

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    The derivation is correct (except equation (2) which should have t' instead of t, and equation (4) which should have x instead of x', which I'm sure were just typing errors when you posted this).

    In an argument like this I think it's important to state clearly at the beginning what you are going to assume. In this case you assume

    1. The length contraction formula (which you would have had to derive elsewhere, before this)

    2. Einstein's 1st postulate is used when you argue that equation (2) follows from equation (1)

    3. Einstein's 2nd postulate is also used as you use the same value of c for both observers.

    When you say "Let M(x) and M’(x’) be two points located on the permanently overlapped OX(O’X’) axes, located at the same point in space, when detected from I and I’ respectively." what you really mean is "M(x) is a point located on the OX axis detected from I, M'(x') is a point located on the O'X' axis detected from I', and both points coincide at time t in I and time t' in I'".
     
  4. Dec 2, 2008 #3
    Thank you.
    Should I say "coincide (in space??? at time t in I and time t' in I'."
     
  5. Dec 4, 2008 #4

    DrGreg

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    When I refer to "a point on the OX axis" I think that clearly means a point in space, so when I say "when two points coincide" it can't mean anything else but "coincide in space".

    If you want to talk about a "point" in spacetime, we usually call that an "event" rather than a "point". A point in space corresponds to a worldline in spacetime, so the event being referred to is where the worldlines of M(x) and M'(x') cross.

    If you want to think of M as being an event rather than a point, it would be better to use a notation such as M(t,x) and M(t',x').
     
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