Simultaneity in Special Relativity - 2

by

Ardeshir Mehta

Ottawa, Canada
Thursday, October 25, 2001



Einstein's Theory of Special Relativity claims that if two events are simultaneous in an inertial frame of reference, then they cannot be simultaneous in another inertial frame of reference which is moving uniformly and rectilinearly at a velocity v relative to the first inertial frame of reference.

But if the Lorentz transformation equations are correct, this claim results in a clear contradiction, as follows:

  1. Let there be an inertial frame of reference -- which we shall designate as I -- in which there are two clocks C1 and C2, separated from one another spatially, and synchronised: so that whenever the clock C1 indicates a moment in time t1, the other indicates a moment in time t2 such that t1 = t2 = t.

  2.  
  3. Let there be another inertial frame of reference -- which we shall designate as I' -- also moving rectilinearly and uniformly at a velocity v relative to I, in which there are two more spatially separated clocks C'1 and C'2, which are also synchronised: that is, whenever the clock C'1 indicates a moment t' coinciding with the moment t indicated by the clock C1, the clock C'2 also indicates the same moment t' indicated by the clock C'1.

  4.  
  5. Now when the clock C1 indicates any particular moment t1, the moment t1 must be related to the moment t'1 indicated by the clock C'1 by the Lorentz transformation equation

  6.  
      t'1 = (t-vx/c2)/(1-v2/c2)0.5


    ... where x is the distance, as measured by an observer in I, between clock C1 and clock C'1, and the moment t = t1 is that indicated by the clock C1 ... and c is of course the speed of light.
     

  7. So when C1 and C2 both indicate a particular moment t = t1 = t2, C'1 indicates a particular moment t'1.

  8.  
  9. And when the clock C2 indicates the same moment t2 = t = t1 as is indicated by Clock C1, the moment t2 must be related to the moment t'2 indicated by the clock C'2 by the Lorentz transformation equation

  10.  
      t'2 = (t-vy/c2)/(1-v2/c2)0.5


    ... where y is the distance, as measured by an observer in I, between clock C2 and clock C'2, and the moment t is, again, that indicated by the clock C2 ... and c is again the speed of light.
     

  11. So when the clocks C1 and C2 both simultaneously indicate a particular moment t = t1 = t2, the clock C'2 indicates a particular moment t'2.

  12.  
  13. Now unless x = y above -- which is highly unlikely, though of course not impossible -- it is clear that t'1 will not be equal to t'2 above: both of them being indicated, of course, when both the clocks C1 and C2 indicate the single moment in time t = t1 = t2.

  14.  
  15. But this contradicts Point No. 2. above, according to which whenever C'1 indicates any particular moment t', C'2 must also indicate the same moment t', since both C'1 and C'2 are synchronised.


Or expressed in table form: 
 
 

According to Point No. 2 above, when:


C1 indicates
C2 indicates
C'1 indicates
C'2 indicates
Such that
t1 = t2 = t
t2 = t1 = t
t'1
t'2
t'1 = t'2

 

According to Points Nos. 3, 5 and 7 above, when:


C1 indicates
C2 indicates
C'1 indicates
C'2 indicates
Such that
t1 = t2 = t
t2 = t1 = t
t'1
t'2
t'1 =/= t'2 *

 

It is of course abundantly clear that the above two tables contradict one another in their fifth columns.
 
 
 
 

Any comments? e-mail me.
 
 



 
 

* At least, not necessarily.