Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Space-time interval invariance question

  1. Jan 26, 2009 #1
    Cinsider please the invariance of the space-time interval in an one space dimension approach
    (x-0)2-c2(t-0)2=(x'-0)2-c2(t'-0)2
    My question is: does it hold for arbitrary events (x,t) in I and (x',t') in I?
    Does it hold only in the case when the events are genertated in I and I' by the same light signal (x=ct,t=x/c); (x'=ct',t'=x'/c) or in the case when the events are generated by the same tardyon moving with speed u in I and u' in I' i.e. (x=ut,t=x/u) and (x'=u't', t'=x'/u')?
    Are x and x' the components of a "two" vector or only x=ct, x'=ct' and x=ut, x'=u't', u amd u' being related by the addition law of parallel speeds?
    Thanks for your answer.
     
  2. jcsd
  3. Jan 26, 2009 #2

    clem

    User Avatar
    Science Advisor

     
  4. Jan 26, 2009 #3

    clem

    User Avatar
    Science Advisor

    It holds for any x and t. If x is written as x=ut, then x' will =u't', with u'given by the relativistic velocity addiltion.
     
  5. Jan 26, 2009 #4

    clem

    User Avatar
    Science Advisor

    x and t are two components of a four-vector, as are x' and t'. Writing x=ut implies that a consstant velocity, which is not necessary for t^2-x^2 to be invariant.
     
  6. Jan 26, 2009 #5

    Fredrik

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    The "interval" [itex]-(x^0)^2+\vec x^2[/itex] is invariant because it's the Minkowski space "scalar product" of a four-vector with itself. The "scalar product" (which isn't really a scalar product since the result can be negative) is defined by

    [tex]\langle y,x \rangle=y^T\eta x=-y^0x^0+\vec y\cdot\vec x[/tex]

    This is invariant under Lorentz transformations because all Lorentz transformations satisfy the condition [itex]\Lambda^T\eta\Lambda=\eta[/itex].

    [tex]\langle \Lambda y,\Lambda x\rangle=(\Lambda y)^T\eta (\Lambda x)=y^T\Lambda^T\eta\Lambda x=y^T\eta x=\langle y,x \rangle[/tex]
     
  7. Jan 27, 2009 #6
    [tex]\Delta[/tex]
    Thanks for your answer. Consider please the inertial reference frames I, I' and I" in the standard arrangement. I' moves with velocity V relative to I and I" moves with speed u relative to I and with speed u' relative to I' all speed showing in the positive direction of the overlapped x, x' and x" axes. A rod of proper length L(0) is located along the overlapped axes at rest relative to I". Observers from I measure its Lorentz contracted length
    L=L(0)(1-u2/c2)1/2. (1)
    For observers from I' the length of the same rod is
    L'=L(0)(1-u'2/c2)1/2 (2)
    Eliminating L(0) between (1) and (2) we obtain that the non-proper lengths are related by
    L=L'(1-u2/c2)1/2/(1-u'2)/c21/2) (3)
    Expressing the right side of (3) as a function of u' via the addition law of parallel speeds it becomes
    L=L'(1-V2/c2)1/2)/[1+Vu'/c2] (4)
    resulting that L and L' do not transform via the Lorentz transformation. Under such conditions are L=Dx and L'Dx' the components of a "two vector? Equation (4) suggests that 1/L and 1/L' are. Is there some connection with the concept of wave vector?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?