Computing stuff to order O(m)
jimbobjames said:
And expanding like you suggested we get:
-c^2(1-2GM/rc^2) \, dt^2 + (1+2GM/rc^2) \, dr^2
which seems to me to be flat Minkowski space plus a delta.
To find out, compute the curvature, up to first order in the parameter M. And for gosh sake do set G=c=1; I can see you're getting confused in part by worrying about them and making a minor error.
Let's try a Riemannian analogue. Consider the coframe field
<br />
\sigma^1 = \sqrt{1+m \, f} \, dx, \; \;<br />
\sigma^2 = \frac{1}{\sqrt{1+m\, g}} \, dy<br />
where f,g are functions of x,y. The metric is
\sigma^1 \otimes \sigma^1 + \sigma^2 \otimes \sigma^2 <br />
= (1 + m \, f) \, dx^2 + \frac{dy^2}{1 + m \, g}<br />
We compute the curvature a la Cartan, keeping only terms which are first order in the parameter m. So
<br />
\sigma^1 = \left( 1+ m \, f/2 \right) \, \dx, \; \;<br />
\sigma^2 = \left (1- m\, g/2 \right) \, dy<br />
Taking the exterior derivative of the cobasis one-forms, we find
<br />
d \sigma^1 = \frac{m \, f_y \, dy \wedge dx}{2} <br />
= \frac{-m \, f_y}{2} \, dx \wedge dy<br />
= \frac{-m \, f_y}{2} \, dx \wedge \sigma^2<br />
(remember, we are only keeping terms first order in m), and likewise
<br />
d \sigma^2 = \frac{-m g_x}{2} \, dy \wedge \sigma^1<br />
where subscripts denote
partial differentiation. Comparing with
<br />
d\sigma^1 = -{\omega^1}_2 \wedge \sigma^2, \; \;<br />
d\sigma^2 = -{\omega^2}_1 \wedge \sigma^1<br />
we find
{\sigma^1}_2 = m \, \frac{f_y \, dx + g_x \, dy}{2}
Taking the exterior derivative of this, we find the curvature two-form
<br />
{\Omega^1}_2 = m \, \frac{g_{xx}-f_{yy}}{2} \, dx \wedge dy<br />
= m \, \frac{g_{xx}-f_{yy}}{2} \, \sigma^1 \wedge \sigma^2<br />
(remember, we are only keeping terms linear in m!), from which we read off the Gaussian curvature
K = m \, \frac{g_{xx}-f_{yy}}{2}
which is valid to first order in the parameter m.
jimbobjames said:
Chris, I've been thinking about what you wrote but I'm still not clear on it. Are you saying that just because (in the weak field approximation) g00 provides the parallel to Newtonian Gravity
Actually, I
didn't say that!
jimbobjames said:
I thought that what we know and love as "gravity" here on earth, was almost entirely related to the time curvature in the Schwarzschild metric only.
"time curvature"?
jimbobjames said:
The curvature related to the dr^2 coefficient is relatively insignificant. Am I wrong with this?
Yes, that's why I mentioned the static wf metric (using a Cartesian type chart, incidently, not a polar spherical chart).
I second the recommendation of any expositions by
John Baez, incidently, but I am not sure the particular book you bought is the best book for learning gtr first time around. You might try something like D'Inverno or Schutz first. See http://math.ucr.edu/home/baez/RelWWW/HTML/reading.html for more recommendations.
The few video lectures I could find so far (from Kip Thorne and Sean Carroll) were excellent, but unfortunately only skim the surface.