Re: How do you know that acceleration is absolute? <eom>
- From: "Dirk Van de moortel" <dirkvandemoortel@xxxxxxxxxxxxxxxxxxxxxxxxxxx>
- Date: Sat, 30 Apr 2005 08:48:03 GMT
"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. We still can do that for
the velocity of course: we can still pick some value of u
that makes v' zero.
hth.
Dirk Vdm
.
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