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Descriptions in physics

Does the non-contextual description x(t) takes 'c=v_max' into account ?

  1. Yes

    0 vote(s)
  2. No

    2 vote(s)
  1. Nov 21, 2005 #1
    Let suppose we have a frame of reference, in which a particle is decribed by [tex] \vec{x}(t) [/tex]...I don't give, on purpose, the context, if it is a classical framework, or a relativistic one...(Like somebody new, that does not even know what those word mean)...The question is : does this trajectory, which is, after some discussion with the non-knower, as seen (or oberseved or measured ??..not measured, because this would mean a radar like method...at least I suppose) by an observer at the origin, hence at x=0. So the question is : does this take into account the delay imposed by the speed of light limit, after another discussion with the same person who has to give an answer to the question ?
    Do you think it takes the delay into account ?
    Last edited: Nov 21, 2005
  2. jcsd
  3. Nov 22, 2005 #2


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    No, normally not. The standard usage of x(t) is to take it to be the *instantaneous* value of x at t. In non-relativistic physics, that's unambiguous, in relativistic physics, the t is the time parameter of the "obvious" foliation of spacetime (usually observer-related when observers are moving). But it does normally NOT take into account the delay for x to be observed. You can do that, of course, but that means t is not parametrising a spacelike foliation anymore, but is parametrising light cones.

    As in non-relativistic physics, the lightcone is flat, it coincides with the spacelike foliation, so both notions are identical. But in relativistic physics, clearly the cone is not flat.
  4. Nov 22, 2005 #3
    So do you think that the following transformation [tex] x_o(t_o)=x(t_o-x_o(t_o))[/tex] could eventually be a transformation of a "time foliation" (in which I don't really understand that global time valid on the whole space)...to a sort of "observed trajectory" (the observer reamains at O)...or is there another formula, because i get trapped in kind of circular reasoning...: the third time argument should be t ot[tex] t_o[/tex] ?
  5. Nov 23, 2005 #4


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    Well, in string theory, one often works with the transformation

    x+ = x + t
    x- = x - t
    y = y
    z = z

    I'm forgetting right now the name of this coordinate system... it has a specific name...
  6. Nov 23, 2005 #5
    Thanks, I suppose it is linked to the old-fashioned : retarded or advanced terms...which correspond two the two x+, x- (with c=1 of course) (depending on your consider observer->observed, or observed->observer transformation i suppose : if you observe x at t, it was at t-x (it is not there at 't'), whereas if it is at x in t, it will be observed at t+x...but do you know for the formula above, because it is not only an event coordinate transformation, it involves the whole trajectory (hence a dependence between x and t, or other way expressed : x+ and x-)...? (it seems to be an implicit notion..)
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