Re: Metric Tensor of Flat Space-Time
From: Alex Green (dralexgreen_at_yahoo.co.uk)
Date: 08/29/04
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Date: 29 Aug 2004 14:06:57 -0700
glhansen@steel.ucs.indiana.edu (Gregory L. Hansen) wrote in message news:<cgsl4n$rs2$1@hood.uits.indiana.edu>...
> In article <42c8441.0408290248.62fa70dc@posting.google.com>,
> Alex Green <dralexgreen@yahoo.co.uk> wrote:
> >glhansen@steel.ucs.indiana.edu (Gregory L. Hansen) wrote in message
> >news:<cgnj1q$3hp$1@hood.uits.indiana.edu>...
>
> >> >Riemannian geometry). I used my text books because there are few good
> >> >analyses of this on the net but:
> >> >http://www.astro.ku.dk/~cramer/RelViz/text/geom_web/node2.html
> >> >captures the general idea, it is well worth a visit). Each of the
> >> >coefficients is the sum of a set of further coefficients:
> >> >
> >> >g00= (dT/dt dT/dt + dX/dt dX/dt + dY/dt dY/dt + dZ/dt dZ/dt)
> >> >g11= (dT/dx dT/dx + dX/dx dX/dx + dY/dx dY/dx + dZ/dx dZ/dx)
> >> >g22= (dT/dy dT/dy + dX/dy dX/dy + dY/dy dY/dy + dZ/dy dZ/dy)
> >> >g33= (dT/dz dT/dz + dX/dz dX/dz + dY/dz dY/dz + dZ/dz dZ/dz)
> >>
> >> It didn't quite say this.
> >
> >True, this was my mistake, thank you.
> ...
> >
> >Summing up what you've said above. It seems that the analysis of space
> >time curvature involves 3 simple steps:
> >
> >1. Select your assumption or 'metric'. In this case Pythagoras or
> >Minkowski.
>
> The metric is the Pythagorean theorem generalized to an arbitrary manifold
> and coordinate system.
>
> >2. Express intervals on each coordinate as differentials.
> >3. Insert these into the original metric.
> >
> >In the case of flat, orthogonal space-time it seems the following
> >results occur.
> >
> >If you start with the basic postulate of relativity as the metric
> >(ds^2 = dx^2 + dy^2 +dz^2 - dt^2) then the metric tensor has (diag
> >-1,1,1,1) and g00 is -1*dT/dt dT/dt so time is real.
> >
> >If you start with Pythagoras: ds^2 = dx^2 + dy^2 +dz^2 + dt^2 then, in
> >the case of flat space-time the 3 steps give g00 = dT/dt dT/dt. So to
> >get back to the fundamental proposition of Relativity we must either
> >assume
> >t=it'
> >OR
> >dT/dt = (sqrt -1)
>
> No. dT/dt = 1/sqrt(1-v^2/c^2).
This is looking at the problem of the nature of the coefficients of
the metric tensor from a different standpoint from Gauss' analysis and
brings us back to the reality of something moving along the space-time
plane. ds^2 is a path from the origin to a point on the surface and as
you point out there will be a Lorentz transformation so long as dT and
dt differ. However, in flat space-time if the Minkowski metric is
used then g00 = dT/dt DT/dt and dT=dt so dT/dt equals 1 when the
space-time surface becomes the same as the tangent plane. When the
plane and surface are merged they are both moving at the same velocity
and become a single coordinate system.
See:
http://www.users.globalnet.co.uk/~lka/tensors.htm
for a diagram for Gauss' analysis.
If the Pythagorean metric is used and imaginary time is assumed then
again dT=dt when the surface coordinates merge with those of the
tangent plane.
If the Pythagorean metric is used and imaginary time is assumed on the
plane but real time is assumed on the surface then there are two
different coordinate systems. What would the velocity of the tangent
plane in the observer's coordinate system be relative to the surface?
What happens with T=t? Do you say
> dt/dt=-1? That would be impossible.
In the third analysis dT/dt*dT/dt = -1 so dT/dt= (sqrt -1)
Even if t were imaginary, the
> sqrt(-1) would cancel out of the differential.
>
The dT/dt = (sqrt -1) gives rise to g00 = -1 because g00 = dT/dt dT/dt
> You still write Lorentz transforms like x'=gamma(x-vt), right? t is as
> real as x is.
>
> t is real, T is real, the ratio is real, the differential is real.
Yes, if the Minkowski metric is used as the assumption behind the
analysis (as it should be). I am not sure that I have made it clear
that I am exploring the metric tensor mathematically rather than
physically, suggesting that there are 3 mathematical alternatives (at
least) for the value of g00. The correct one is g00 = -1 on the basis
of an assumption of the Minlowski metric (ds^2 = -dt^2 + dx^2...). But
2 other formulations are possible mathematically, if we assume
imaginary time on the plane and the surface we have g00 = 1 and the
Pythagorean metric is (ds^2 = +idt^2 + dx^2 + dy^2 + dz^2). If we
assume imaginary time on the plane and real time on the surface the
space-time interval is:
(ds^2 = +(-1)dt^2 + dx^2 + dy^2 + dz^2) and g00 is -1.
> It
> doesn't have to do with the definition of the position vector, it has to
> do with the manifold, and how the vector is related to its dual. The
> -1 is in the distance relationship, not in the coordinate.
>
> It's not so different from plane polar coordinates with the metric
>
> ds^2 = dr^2 + r^2 d(theta)^2
>
> But you're not going to say d(THETA)/d(theta)=r. It probably equals 1,
> because THETA is probably theta plus an offset. It just so happens that
> in order to convert an angular displacement to an arc length, you need to
> multiply your d(theta) by the distance from the center of rotation.
>
> So then if you have a differential displacement (dr,d(theta)) and want the
> magnitude, you can't simply dot the vector into itself,
> sqrt((dr)^2+(d(theta))^2). theta doesn't even have the same units as r.
> To find the magnitude you must first find the dual to the vector, which is
>
> (dr, r^2 d(theta) )
>
> Then you can multiply the vector by its dual to find the differential
> length, and integrate along a path.
>
> In the same way, in special relativity, dt isn't a length in Minkowski
> space, but the metric tells you how to convert it into a length. It's not
> the coordinate that has any kind of imaginary quality attached to it, it's
> the distance relationship on that manifold that does. So you could say
> distances along the t axis are imaginary. I mean, nobody ever writes it
> that way. They always write ds^2 = -c^2 dt^2 + dx^2 and leave the
> interval squared.
Yes, this is correct when the Minkowski metric is assumed. In fact I
would be unhappy about using ict at all if we are talking about real
time. If people do this I suspect they are wandering into an implicit
assumption of a Pythagorean metric rather than a Minkowski metric.
>
> And then, if you're using Cartesian coordinates, you could write a
> position vector as (x,ict) and ignore everything about the metric, about
> the distinction between a vector and its dual. But then when you do a
> Lorentz transform you take the ic out again and use just plain ol' t. Or
> you would have to put appropriate ic terms into the Lorentz transforms to
> cancel it out. What you did with (x,ict), even if you didn't know it,
> was symmetrize the vector and its dual and convert t into a distance as
> you wrote out the vector, instead of using the metric later to find
> convert to distance.
>
> >
> >True. However, I am faintly disturbed by the possibility that mixed
> >real and imaginary coordinates seem to give the same result when a
> >Pythagorean metric is assumed as when real time coordinates and a
> >Minkowskian metric are assumed. In fact the two sets of assumptions
> >seem to lead to identical expressions for the space-time interval and
> >identical metric tensors even though they seem to involve quite
> >different assumptions.
>
> There's always more than one way to do things. But ict doesn't
> generalize. And if you're going to use non-Cartesian coordinates, you
> have to pay attention to the metric anyway.
Yes, ict doesn't but I have never come across the other formulation
with mixed real and imaginary coordinates. It's a mathematical
curiosity that gives the value for g00 as -1 as in the standard case
and also converts the Pythagorean metric to the correct formula for
space-time.
Best Wishes
Alex Green
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