View Poll Results: Does the non-contextual description x(t) takes 'c=v_max' into account ?
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Descriptions in physics

by kleinwolf
Tags: descriptions, physics
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kleinwolf
#1
Nov21-05, 11:50 PM
P: 309
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 ?
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vanesch
#2
Nov22-05, 01:01 AM
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P: 6,236
Quote Quote by kleinwolf
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 ?
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.
kleinwolf
#3
Nov22-05, 03:05 AM
P: 309
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] ?

vanesch
#4
Nov23-05, 06:47 AM
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Descriptions in physics

Quote Quote by kleinwolf
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"
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...
kleinwolf
#5
Nov23-05, 12:21 PM
P: 309
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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