Re: How do you know that acceleration is absolute? <eom>


"Jack Martinelli" <jack@xxxxxxxxxxxxxx> wrote in message news:7CZce.2510$HL2.42@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
>
> "Dirk Van de moortel" <dirkvandemoortel@xxxxxxxxxxxxxxxxxxxxxxxxxxx> wrote
> in message news:7DHce.77345$oT7.5041223@xxxxxxxxxxxxxxxxxxxxxxxx
> >
> > "Jack Martinelli" <jack@xxxxxxxxxxxxxx> wrote in message
> > news:Yqyce.1925$HL2.245@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
> >>
> >
> > Let's work in one dimension only.
> >
> > Suppose a particle is described in one inertial frame S as
> > a function x(t) and in another frame S' as a function x'(t).
> > Suppose frame S' is moving with a constant relative
> > velocity u w.r.t. frame S.
> > In Galilean Relativity you have the relation between the
> > distance coordinates given by
> > x' = x - u t
> >
> > The velocity v of the particle in the two frames is then given by
> > in S: v = dx/dt
> > in S': v' = dx'/dt = d(x- u t)/dt = dx/dt - u = v - u
> > So you see that velocity of the particle depends on the frame:
> > v' = v - u
> > i.o.w. velocity is "relative".
> >
> > The accerelation however:
> > in S: a = dv/dt
> > in S': a' = dv'/dt = d(v- u)/dt = dv/dt = a
> > So you see that acceleration of the particle does not depend
> > on the frame:
> > a' = a
> > i.o.w. acceleration is "absolute".
> >
> > In special relativity it is a bit more complicated since S
> > describes the particle as x(t) and S' describes it as x'(t').
> > Each frame uses its own 'time'.
> > Supposing a relative velocity u between the frames again,
> > the coordinate transformation is now given by the equations
> > x' = g (x - u t)
> > t' = g (t - u x/c^2)
> > where
> > g = 1 /sqrt(1-u^2/c^2)
> >
> > A similar but bit more complicated exercise as before gives
> > the result for velocity of the particle:
> > v' = (v - u)/(1 - u v/c^2)
> > and for the acceleration:
> > a' = a / ( g (1- u v/c^2) )^3
> >
> > The difference with Galilean relativity is of course that it is
> > not true that
> > a' = a
> > but you can still say that the accelaration is absolute in the
> > sense that if it is non-zero in one frame, it is also nonzero
> > in every other frame, i.o.w. you cannot transform it away
> > by choosing another inertial frame.
>
> Why another inertial frame? Why not another accelerating frame?

Inertial frames are easier to make the calculations and
show a point. That's why I used two of them, and a
particle in one dimensional motion along the line of
relative motion of the frames.
But sure, take the most extreme case where you would
sit on the accelerated particle itself. You will find that its
coordinates are pretty constant, so as seen in your own
frame, it would have no acceleration. But then *you*
would fee the proper acceleration yourself (like we
say, as measured in the momentarily comoving inertial
frame at each point of your worldline), and again be
forced to conclude that the particle is 'absolutely'
accelerated together with yourself. All this is of course
in the context of Galilean and special relativity.
In the context of general relativity gravity is modeled as
spacetime curvature, and "acceleration due to gravity"
is taken as the 'natural way'. In such a natural free-falling
frame (that used to be considered as being accelerated,
but that is now considered as an inertial frame, provided
it is sufficiently small in space and time), you could say
that a particle falling with you has zero acceleration. But
then again you and the particle are following a geodesic
in a curved spacetime that is shaped by the mass
distribution, which is, since you obviously can't transform
a planet away, nicely absolute again :-)

>
> > We still can do that for
> > the velocity of course: we can still pick some value of u
> > that makes v' zero.
>
> Thanks Dirk, that was the explanation I was looking for.


Glad to hear that :-)
Cheers,

Dirk Vdm

>
> Regards,
>
> Jack Martinelli
>



.

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