Re: Theory of Relativity
- From: "Ken S. Tucker" <dynamics@xxxxxxxxxxxx>
- Date: 30 Apr 2005 21:59:28 -0700
Tom Roberts wrote:
> Ken S. Tucker wrote:
> > The fundamental assumption of relativity is
> > *absolute _spatial_ motion does not exist*,
>
> Well, that's a new approach AFAIK....
Actually not, it's the juxtaposition of saying
spatial motion is relative.
> It has a fundamental problem: how does one define "motion" so this
even
> makes sense?
Yes, that's a problem.
> For this approach to be fruitful, I'm pretty sure you'll have to
define
> "motion" as _timelike_ -- then your assertion "Absolute spatial
motion
> does not exist" is trivial.
Yes, it needs to be defined to vanish, but the
equation to do that is NOT trivial.
> > Beginning with the well known
> > ds^2 = g_uv dx^u dx^v , u,v,w={0,1,2,3}.
>
> OK. But please remember what you seem to have forgotten:
Why do you claim somethings forgotten? What is it that
you do not understand?
....
> > We can use association provided the covariant
> > derivative,
> > g_uv;w = 0.
>
> That merely states that your choice of covariant derivative is
> compatible with the metric. But no matter, you don't use the
covariant
> derivative anywhere.
No sorry, if one uses g_uv or g^uv in association
then one must stipulate g_uv;w=0, that's often
overlooked. It's a common error for beginners but
now you know.
> > Then by association,
> > ds^2 = dx_u dx^u.
>
> I'm not sure what you mean by "association".
By association use,
dx_u = g_uv dx^v
....
> > Expanding to time and space gives,
> > ds2 = dx_0 dx0 + dx_i dx^i , i,j={1,2,3}.
....
> > Expanding to time and space gives,
> > ds^2 = dx_0 dx^0 + dx_i dx^i , i,j={1,2,3}.
> > The absolute spatial motion I'll define by
> > dx_i dx^i = Absolute spatial motion.
> Defining it so does not make it so:
Agreed, good of you to notice. This is where
physical intuition is required, but you'll
need to follow the math I'll keep simple.
In 3D
dr^2 = dx_i dx^i , {i,j=1,2,3}.
That "dr" is an invariant displacement, aka
an invariant displacement, aka an absolute
displacement. When we go to 4D we can divide
that dr by dt to define "absolute motion",
hence,
dr/dt = invariant motion in 4D.
Relativity is quite clear on this point,
invariant motion cannot be measured, therefore
it vanishes, hence,
dr^2= dx_i dx^i =0
in 4D.
....
> And your expression is not invariant over coordinate
> transforms (if it were, you could claim it is "absolute" in
> a restricted sense).
That's strictly correct mathematically, however not
any or all substitutions for the metric are true
in spacetime. Most begin with the approximation
gii=-1, g00=1
for no other reason except to get from
ds^2 = g_uv dx^u dx^v
to
ds^2 = dt^2 - dx^2
mindlessly, mainly because it's easy to teach.
But that's not how it really should done.
The metrics are defined by a field equations,
to constrain those that a physically viable, so
U_i=0
is the first constraint.
> > Absolute spatial motion cannot exist, IOW's
> > it vanishes, hence,
> > dx_i dx^i =0.
> Imagine using the {dx^i} and the {dx_i} to represent the differential
> between two corners of my desk, measured simultaneously in the
inertial
> rest frame of the desk. Then dx_i dx^i is simply the distance between
> them -- manifestly not zero.
Tom, you aren't going to learn 4D spacetime and
relativity by looking at the corner of your desk,
it's not that simple.
It's easy to draw a 4D diagram on paper just using
x and t with oblique axes.
Regards
Ken
.
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- From: Tom Roberts
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