Metric Tensor of Flat Space-Time

From: Alex Green (dralexgreen_at_yahoo.co.uk)
Date: 08/26/04 

Date: 26 Aug 2004 05:05:53 -0700

The metric of space-time is normally given by the expression:
 
ds^2=gdxdx
 
where s is the space-time interval and the usual indexes should be
assumed. 'g' is the metric tensor which for flat space-time
conventionally has a principle diagonal (-1,1,1,1) or, in field
theory, (1,-1,-1,-1)
 
The most puzzling coefficient in the metric tensor is g00. Anderson
gives a derivation of this at:

http://www.astro.ku.dk/~cramer/RelViz/text/geom_web/node2.html

where g00 is given as (dT/dt dT/dt + dX/dt dX/dt + dY/dt dY/dt + dZ/dt
dZ/dt)

So that the term for time in the metric is:
(dT/dt dT/dt + dX/dt dX/dt ..) dt dt

the overall metric becomes:
ds^2 = dx^2 + dy^2 + dz^2 + (dT/dt)^2 (ct)^2

Conventionally dX/dt, dY/dt,dZ/dt are zero and (dT/dt)^2, is given the
value: -1. In the metric tensor for the flat metric the 'dT'
applies to the observer's change in time and the 'dt' applies to the
change in time on the space-time surface. (See Gauss' analysis of
surfaces where dT would lie on a plane in the observer's coordinate
system that touches the surface and dt is on the surface).

If (dT/dt)^2 is -1 then (dT/dt) may be imaginary. Furthermore dt is
conventionally real so if (dT/dt) is imaginary then the conventional
metric tensor of GR seems to be suggesting that the observer's time
coordinate is imaginary.
 
Is the conventional metric tensor really suggesting that the observer
has an imaginary time axis and the space-time surface has a real time
axis or am I getting this hopelessly wrong? Is this the wrong
interpretation of the metric tensor? If it were the right
interpretation it would possibly have consequences for the
relationship between QM and GR that surely would have been explored in
the literature. Assuming that the interpretation is wrong can you
explain why?

Best Wishes

Alex Green

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