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

  • Thread starter Robbi
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  • #1
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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?
 

Answers and Replies

  • #2
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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.
 
  • #3
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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
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.
 
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