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Negative clock reading

  1. Jun 29, 2008 #1
    are clocks, displaying a negatve time, compatible with special relativity?
  2. jcsd
  3. Jun 30, 2008 #2


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    If you mean... "can a clock-reading be a negative number?",
    I would say "yes, it is compatible with special relativity".
  4. Jun 30, 2008 #3
    Bernhard, its good to see you posting again. How have you been?
    Negative numbers often represent quantities that refer to prior quantities. I.e. a negative value of height merely represents a distance from a surface which is below ground level. A negative time would merely represent a time before the event where you set your clock to read zero. So if you set you clock to read t = 0 at 1:00am then 12:30am would have a negative value of the time. This is important to understand given the nature of non-simultaneity since if the clocks are synchronized in S then the clocks in S' which are moving relative to S will not be synchronized. The Lorentz transformation of events will then give negative values of time. One must know how to interpret these values.

    Best wishes

  5. Jul 3, 2008 #4
    Thanks Pete
    Consider please the clocks C(0) and C(x) and a source of light S(0) at rest in the I frame.
    When C(0) reads t(e) S(0) emits a synchronizing light signal in the positive direction of the OX axis. It arrives at the location of clock C(x) when it reads t(E). Equation
    t(E)=t(e)+x/c (1)
    Does it hold for negative and positive values of t(e)? With what consequences?
  6. Jul 4, 2008 #5


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    Pete doesn't seem to be around anymore. I don't know why.
  7. Jul 4, 2008 #6


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    A negative clock reading simply indicates an event that occurred before another event that occurs when the clock reads zero.

    The actual value of a clock means nothing by itself, it only makes sense when you compare it with another clock value; it is the difference between two clock values that is significant.

    In the example you give, it doesn't matter if the times are negative. However, you may need to think what it means for x to be negative. The words that you used imply that is impossible, because you say the light goes in the positive OX direction.

    From some of the previous threads you have posted about similar equations, it might make sense to insist that the equation is still true for negative x, but then the description of what this means would need to change: you would have to consider a signal towards S(0) for negative x, but away from S(0) for positive x. On the other hand you could modify the equation to be

    [tex] t_E = t_e + \frac {|x|}{c} [/tex]​

    but that would have consequences on what you derive from that equation.

    Or you could simply insist that x is always positive and then no problem arises.
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