Does the Trajectory Formula Account for the Speed of Light Delay?

[kleinwolf]

Let suppose we have a frame of reference, in which a particle is decribed by $$\vec{x}(t)$$...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 ?

 

[vanesch]

Let suppose we have a frame of reference, in which a particle is decribed by $$\vec{x}(t)$$...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]

So do you think that the following transformation $$x_o(t_o)=x(t_o-x_o(t_o))$$ 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$$t_o$$ ?

 

[vanesch]

So do you think that the following transformation $$x_o(t_o)=x(t_o-x_o(t_o))$$ 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]

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..)