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Discussion, please

  1. Sep 29, 2006 #1
    The set of events that are simultaneous to a person moving with velocity v is the same set of events that would be occupied by a hypothetical particle moving at
    (c^2)/v
     
  2. jcsd
  3. Sep 29, 2006 #2

    pervect

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    My $.02 - close, but no cigar.

    Particles can't move at velocities faster than light. I would say, instead "on a space-time diagram, the slope of a line representing simultaneous events is c^2/v" to avoid the issue.
     
  4. Sep 29, 2006 #3
    Let I’ be the rest frame of the mentioned person and ( ,t’) and ( ,t’) the space-time coordinates of two simultaneous events. The same events, detected from the I inertial frame relative to which I’ moves with speed V are (x1,t1) and (x2,t2) respectively. We have in accordance with the Lorentz-Einstein transformations
    t'=g(t1-Vx1/cc (1)
    t'=g(t2-V x2/cc (2)
    Combining (1) and (2) we obtain
    x2-x1=cc(t2-t1)/V (3)
    i,e. in I the simultaneous events in K’ take place on the world line described by (3).
     
  5. Sep 30, 2006 #4
    So if such a fictitious particle really existed, it would appear to be everywhere between x1' and x2' simultaneously
     
  6. Sep 30, 2006 #5
    discussion please

    No! A typing error occured due to the fact that the forum has no a formula editor. The two events are simultaneous only at x'1 and x'2 and not elsewhere.
     
  7. Sep 30, 2006 #6
    But consider the formula t'=gamma*(t-vx/cc)

    Set t'=0 and watch the magic happen!
     
  8. Oct 2, 2006 #7
    magic?

    Do you mean that the problem has something in common with events not causaly related?
     
  9. Oct 4, 2006 #8
    Sounds interesting - i'll open a new thread
     
  10. Oct 4, 2006 #9
    Magic -- what Magic??
    And please don't open another thread over this! (You'd do better to delete this entire thread)
    When is the change in t' equal zero? -- when the change in t is zero.
    That is t=0
    What will "x" be? -- the same place or zero!
    So you consider it magic to multiply gamma by zero and get 0!?

    You’re comparing the starting point "zero" to the same starting point "zero" in both time and location.
    ALL reference systems will measure this one event measured as two events as being simultaneous; with the identical separation both in time and distance for all reference frames; zero!
     
  11. Oct 4, 2006 #10
    Hi Randall,

    I must respectfully disagree. If you take the lorentz transfromation for time (t'), and set t'=0, you will come up with the set of events for which t'=0. There are an infinite number of these events. You might have heard this set referred to as the "line of simultaneity".
     
  12. Oct 4, 2006 #11
    cc/v what is that?

    Interesting enough, if you consider, in the uniformly accelerating reference frame, the observers who move with all possible proper accelerations and you try to find out the geometric locus of the points where theirs velocities are the same, you find out that it is a straight line the slope of which is c^2/v.
     
  13. Oct 4, 2006 #12
    SO
    At displaced x' values with t' =0 give non simultaneous times t and x values don't match either, that’s what SR gamma is for --- Standard Ordinary even classical Special Relativity; what magic??
     
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