Re: Christoffel instead of Riemann
From: Doug Sweetser (sweetser_at_alum.mit.edu)
Date: 07/19/04
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Date: Sun, 18 Jul 2004 22:20:42 -0400
Hello:
Dirk's reading of the symbols was on the money. The basic partial
derivative symbolized by a comma (A^mu_,nu) does not transform like a
tensor. It needs the Christoffel symbol of the second kind, which in
LaTeX would be a capital gamma, and I use an L instead.
The Riemann curvature tensor is one complicated object, requiring six
Christoffel symbols. I hope to use just one :-) Even one Christoffel
symbol can be scary.
One of the things the moderator objected to was that he did not see how
the expoential metric could solve the divergence of the covariant
derivative equation. So I will repeat something from an SPR post back
in December where I should exactly how that is done. Only read on if
you like to see the details of a calculation involving a Christoffel
symbol.
doug
Dec. 23, 2003 spr posting...
...Here was the resulting dynamic metric tensor:
| e^-2GM/c^2 R 0 0 0 |
g^uv = | 0 e^2GM/c^2 R 0 0 |
| 0 0 e^2GM/c^2 R 0 |
| 0 0 0 e^2GM/c^2 R |
This metric has been published by others (Watt and Misner,
gr-qc/9910032) where they say it is a "good" metric in the sense that
the first order parameterized post-Newtonian (PPN) coefficients are
the same as for general relativity, so the same weak field tests will
be passed.
Now the Christoffel symbol can be used since both the potential and
metric are defined. Without lifting an electron, d L_w^00/dt must be
equal to zero because nothing in the metric depends on t.
The only thing that needs to be calculated is Del.L_w^i0 A^w. What
are the nuts and bolts of the Christoffel symbol? It turns out to be
three derivatives of the metric one is working with:
L_w^uv A^w = 1/2 g_wb (g^bu,v + g^vb,u - g^uv,b) A^w
That's the general definition, but let's focus on the case at hand:
L_w^i0 A^w = 1/2 g_wb (g^bi,0 + g^0b,i - g^i0,b) A^w
Two of the three terms drop: there is no time derivative of the
metric, so g^bi,0 is zero, and since the metric is diagonal, g^i0,b is
zero:
L_w^i0 A^w = 1/2 g_wb g^0b,i A^w
This simplifies still further. The index b must be zero in g^0b,i,
and that implies that the index w needs to be zero:
L_0^i0 A^0 = 1/2 g_00 g^00,i c^2/G^(1/2)
Presume the standard relationship between metric tensors with lower
versus upper indices:
g_ub g^vb = delta_u^v
Thus g_00 = e^2GM/c^2. Now to take three derivatives of g^00:
d g^00/dx = d e^(-2 GM/c^2 (x^2 + y^2 + z^2)^(1/2)) /dx
= 2 x G M/c^2 R^3 e^-2GM/c^2 R
The exponential term will vanish when multiplied by g_00:
g_00 d g^00/dx = 2 x G M/c^2 R^3
g_00 d g^00/dy = 2 y G M/c^2 R^3
g_00 d g^00/dz = 2 z G M/c^2 R^3
Calculate the divergence:
d(g_00 d g^00/dx)/dx + d(g_00 d g^00/dy)/dy + d(g_00 d g^00/dz)/dz
= d 2 x G M/c^2 R^3/dx + d 2 y G M/c^2 R^3/dy + d 2 z G M/c^2 R^3/dz
= 2 G M/c^2 R^3 - 6 G M x^2/c^2 R^5
2 G M/c^2 R^3 - 6 G M y^2/c^2 R^5
2 G M/c^2 R^3 - 6 G M z^2/c^2 R^5
= 0
If I were a math whiz kid, I should have expected this. I'm not, so
the zero was a surprise. Mass charge is behaving just like electric
charge as far as singular solutions are concerned, even though
Christoffel symbols are being used instead of potentials in this
choice of reference frame.
Examine a metric close to the singular one. In the limit of that
neighbor approaching the exact solution, there should be the mass
charge density, G^(1/2) m/R^3. I'll give a sketch of how this is
done. Work with this g^00:
g^00 = d e^(-2 GM/c^2 (x^2 + y^2 + z^2)^(1/2) + b) /dx
The factor of "b" is what bumps the solution a wee bit off. Plug it
into exactly the same differential equation as the divergence above,
and then take the Taylor series as b goes to zero (I rely on the
program Mathematica to do that correctly). The first term of that
series has a factor of b in it. If one rescales the series by a
dimensionless factor of R/4b, the first term of the series is non-zero
even if b goes to zero. The first term is G^(1/2) m/R^3, the mass
charge density.
Hopefully this demonstrates that the background metric is not fixed as
thought by Prof. Baez. The field equations expressed only in terms of
Christoffel symbols for an electrically neutral system can be solved
for a particular metric which is consistent with experimental tests to
first-order PPN accuracy. The solution is singular, which is true
whether one uses locally covariant coordinates, or the local no change
in metric coordinates as defined in this post. This is quite a
stringent, non-trivial test for general covariance and logical
consistency of the model.
The Maxwell equations are field equations that depend exclusively on
the potential, not the metric. General relativity depends exclusively
on the metric. This model provides a means for gravity to be handled
as a metric or a potential on equal footing.
The model is still a toy because it clearly is in conflict with the
widely accepted notion that gravitational fields gravitate. The toy
is more sophisticated than it might appear at first glance.
doug
quaternions.com
SPR post on deriving the Watt/Misner metric:
http://groups.google.com/groups?hl=en&lr=&ie=UTF-8&oe=UTF-8&threadm=
3efc7c1c%242%40news.sentex.net&rnum=36&prev=/groups%3Fq%3Dgroup:sci.
physics.research%2Bauthor:sweetser%26num%3D30%26hl%3Den%26lr%3D%26ie
%3DUTF-8%26oe%3DUTF-8%26scoring%3Dd%26start%3D30%26sa%3DN
A mathematica notebook for the derivation is here:
http://www.theworld.com/~sweetser/quaternions/notebooks/spr_post.nb
- Next message: Bill Hobba: "Re: Aether is the empty space in which the Universe sits"
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