On Jan 25, 1:16 pm, "Koobee Wublee" wrote:
On Jan 25, 9:36 am, "Igor" wrote:

Here's a good link that shows how the Einstein tensor is derivable from
the Bianchi identities:

http://www.mth.uct.ac.za/omei/gr/chap6/node14.html

You got to be kidding me with that convoluted derivation of the field
equations.

Not only are they kidding you. They're laughing at you. All the time.

On Thu, 25 Jan 2007 12:36:18 -0500, Igor wrote:

Here's a good link that shows how the Einstein tensor is derivable from
the Bianchi identities:

http://www.mth.uct.ac.za/omei/gr/chap6/node14.html


Cool, thanks.

Unfortunately though, that doesn't actually answer my question, which is
how to prove that G is identical with the G that's defined as a
contraction on the double dual of the curvature tensor.

Either way, thanks though.

Psy-Kosh

On Thu, 25 Jan 2007 13:48:17 -0500, JanPB wrote:


Ugh. It looks like a bit of index gymnastics. The problem is not to
derive Einstein's field equation but to verify the usual relationship
between G^a_b and R^a_b given the definitions of both as contractions
of certain tensors (*Riemann* in the first case and the plain Riemann
in the second, where "*" is the Hodge star).


Yeah, the field equation doesn't get dealt with until later anyways. (chap
17, I think)

I think I'll pass... :-)


awww. Thanks for looking, anyways.

(any hints/ideas how I might procede though?)

Psy-Kosh

On Jan 25, 10:39 am, "Igor" wrote:
On Jan 25, 1:16 pm, "Koobee Wublee" wrote:

On Jan 25, 9:36 am, "Igor" wrote:

Here's a good link that shows how the Einstein tensor is derivable from
the Bianchi identities:

http://www.mth.uct.ac.za/omei/gr/chap6/node14.html
You got to be kidding me with that convoluted derivation of the field
equations.Not only are they kidding you. They're laughing at you. All the time.

I know I am.

Psy-Kosh wrote:
What the subject says. Working through Gravitation(MTW), waited until I
was in mid chap 14 before going back and having another look at this bit,
but am still stumped here. (ie, page 326, problem 13.12b)

Tried some index jiggerpokery but am still stumped. Any help would be
appreciated, thanks.


Use the fact that the Riemann tensor is totally
antisymmetric on all of its indicies to find the
possible contractions. For example,

R_abcd = -R_bacd

Contract the first two indicies,

g^ab R_abcd =? -g^ab R_bacd

Since you contracted out the first two indices, you have

R^cd = -R_cd

Which is only true if that tensor is zero. Since for any
pair of adjacent indicies, you can permute the indicies
to get the form above, contractions on any pair of adjacent
indicies gives 0. Now you just need to consider contractions
on the first & third, second & fourth, etc., and make the
same argument to show that those contractions all give the
same tensor up to a sign. The choice of sign is convention and
defined by contracting the first & third indicies.

Koobee Wublee wrote:
On Jan 24, 11:11 am, "Igor" wrote:

The Einstein tensor has zero covariant divergence. Ricci doesn't.

What does that mean?

Well, Mr. One and Only, self-proclaimed Disciple of Riemann,
it means, D^u G_uv = 0.


IIRC, the Einstein tensor is derivable from a contraction on the
Bianchi identities of the Riemann Curvature. [...]

Bullsh*t! The Einstein tensor can only be derived from the
Einstein-Hilbert Lagrangian.

Stop punishing yourself. The only way you could understand
this is if you could live longer than a giant redwood tree and
were smarter than a giant redwood tree.

On Fri, 26 Jan 2007 02:52:47 -0500, bergeron wrote:
Use the fact that the Riemann tensor is totally
antisymmetric on all of its indicies to find the
possible contractions. For example,

R_abcd = -R_bacd

Contract the first two indicies,

g^ab R_abcd =? -g^ab R_bacd

Since you contracted out the first two indices, you have

R^cd = -R_cd

Which is only true if that tensor is zero. Since for any
pair of adjacent indicies, you can permute the indicies
to get the form above, contractions on any pair of adjacent
indicies gives 0. Now you just need to consider contractions
on the first & third, second & fourth, etc., and make the
same argument to show that those contractions all give the
same tensor up to a sign. The choice of sign is convention and
defined by contracting the first & third indicies.


Thank you.

But how do I go about deriving the relationship between that and the
tensor produced by a contraction of the double dual of the curvature
tensor. ie, how do I show that

R^b_d = G^b_d+1/2*R*delta^b_d

Based on the definition of G as a contraction of the double dual of the
curvature tensor?

Psy-Kosh

Psy-Kosh wrote:

[...]
But how do I go about deriving the relationship between that and the
tensor produced by a contraction of the double dual of the curvature
tensor. ie, how do I show that

R^b_d = G^b_d+1/2*R*delta^b_d

Based on the definition of G as a contraction of the double dual of the
curvature tensor?

The double dual is defined in terms of the curvature tensor and the
Levi-Civita tensor. When you contract two indices, you are contracting
indices on a pair of Levi-Civita tensors. Write this out, and then
use the identities of exercise 3.13 of MTW.

Steve Carlip

On Fri, 26 Jan 2007 12:52:16 -0500,
wrote:
R^b_d = G^b_d+1/2*R*delta^b_d

Based on the definition of G as a contraction of the double dual of the
curvature tensor?

The double dual is defined in terms of the curvature tensor and the
Levi-Civita tensor. When you contract two indices, you are contracting
indices on a pair of Levi-Civita tensors. Write this out, and then
use the identities of exercise 3.13 of MTW.


I'll see if that's enough for me to work it out.

Thank you

Psy-Kosh

On Thu, 25 Jan 2007 09:36:18 -0800, Igor wrote:



Here's a good link that shows how the Einstein tensor is derivable from
the Bianchi identities:

http://www.mth.uct.ac.za/omei/gr/chap6/node14.html

I took a look at that page. It sounded very familiar.

Some of the text on that page is lifted word for word from Schutz's 1985
book, "A first course in general relativity", page 174. The derivation
given on the web page is the same as Schutz's, very slightly altered.

Example: Text from the web page by Peter Dunsby, 1996 (cut and paste from
the browser window):

"The tensor G^uv is constructed only from the Riemann
tensor and the metric, and it is automatically divergence free as an
identity. It is called the Einstein tensor, since its importance for
gravity was first understood by Einstein. We will see in the next chapter
that Einstein's field equations for General Relativity are..."

Text from the book by Schutz, 1985 (typed in from the book):

"The tensor G^ab is constructed only from the Riemann tensor and the
metric, and is automatically divergence free as an identity. It is called
the Einstein tensor, since its importance for gravity was first
understood by Einstein. (In fact we shall see that the Einstein field
equations for GR are..."

I looked at a couple other pages. Dunsby seems to have been "heavily
inspired", shall we say, by Schutz; his derivations follow along line by
line with Schutz and his sentences typically differ slightly from what
Schutz wrote.

Does Dunsby acknowledge his debt to Schutz anywhere on that site?


--
Nospam becomes physicsinsights to fix the email
I can be also contacted through http://www.physicsinsights.org