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Is there a frame of reference where events A and B are simultaneous?

  1. Apr 21, 2012 #1
    Hi there!
    I'm working on a couple of problems regarding simultaneity.

    For one I'm given events X_a = (ct,x,y,z) = (2,1,-3,2) and X_b = (6,3,-1,5)
    I want to find out if there's a frame of reference where the events are simultaneous.
    I find the change vector X_b - X_a = (Δct,Δx,Δy,Δz) = (4,2,2,3)

    I quickly find out that it's impossible to have the events simultaneous with a pure boost in the x, y and z direction. So I used the Lorentz-transformation matrix for a boost in any direction I found on Wikipedia.

    I get the following equation γ(4-2β_x - 2β_y - 3β_z) = 0 or 4c^2 = 2v_x + 2v_y + 3v_z

    Here we obviously have 0 < β_i < 1 and |v_i| < c

    So I can't exactly make one v_i a pivot and the rest free variables. Any ideas?
     
  2. jcsd
  3. Apr 21, 2012 #2
    Use the spacetime interval S2 = (Δct2,-Δx2,-Δy2,-Δz2).

    If S2 is positive then the interval is timelike and there is a reference frame where the events happen at the same place but no reference frame where they happen simultaneously.

    If S2 is negative then the interval is spacelike and there is a reference frame where the events happen simultaneously but no reference frame where they happen at the same place.

    In your case the spacetime interval is S2 = 16-17 = -1 so there exists an inertial reference frame where the events occur simultaneously.
     
  4. Apr 22, 2012 #3
    We have established that there exists an inertial reference frame where the events are simultaneous, but it turns out there are many solutions. The spacetime interval is invariant under a Lorentz transformation, so in a transform from S to S' it follows that Δct2-Δx2-Δy2-Δz2 = Δct'2,-Δx'2,-Δy'2,-Δz'2.

    In our case ΔS2 = -1 and in a reference frame where the events are simultaneous, Δct'2 = 0, so the change vector must satisfy -1 = 0 -Δx'2-Δy'2-Δz'2 or equivalently 1 = Δx'2+Δy'2+ Δz'2. There are many solutions where Δi' is real. Some obvious possible solutions are (Δx',Δy',Δz') = (1,0,0) or (0,1,0) or (0,0,1) or (-1,0,0) or (√(1/3),√(1/3),√(1/3)) or (-√(1/2),-√(1/2),0) or (-√(1/2),√(1/2),0) etc.
     
    Last edited: Apr 22, 2012
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