Undergrad Computing Ricci Tensor Coefficients w/ Tetrad Formalism

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SUMMARY

The discussion focuses on computing Ricci tensor coefficients using the tetrad formalism as outlined in "Differentiable Manifolds: A Theoretical Physics Approach" by Castillo. The specific equation discussed is $$d\theta^i = \Gamma^i_{[jk]} \theta^j \wedge \theta^k$$, with particular attention to the coefficient $$\Gamma_{2[12]} = \frac{\sqrt{1-kr^2}}{2r}$$. Participants clarify the reasoning behind the factor of 2 in the coefficient and the summation involved in calculating $$\Gamma^3_{[13]}$$, emphasizing the importance of understanding the antisymmetrization and the properties of exterior derivatives in this context.

PREREQUISITES
  • Understanding of differential geometry concepts, particularly Ricci tensors
  • Familiarity with tetrad formalism and 1-forms
  • Knowledge of exterior derivatives and their properties
  • Experience with Christoffel symbols and their role in non-coordinate bases
NEXT STEPS
  • Study the derivation of Ricci tensor coefficients in various metrics
  • Learn about the properties of exterior derivatives in differential geometry
  • Explore the relationship between Christoffel symbols and connection coefficients in non-coordinate bases
  • Investigate the implications of antisymmetrization in tensor calculus
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Mathematicians, theoretical physicists, and students of general relativity who are working with differential geometry and tensor analysis, particularly in the context of Ricci tensors and tetrad formalism.

snypehype46
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TL;DR
Computing Ricci tensor coefficients using the tetrad formalism
I'm reading "Differentiable manifolds: A Theoretical Physics Approach" by Castillo and on page 170 of the book a calculation of the Ricci tensor coefficients for a metric is illustrated. In the book the starting point for this method is the equation given by:

$$d\theta^i = \Gamma^i_{[jk]} \theta^j \wedge \theta^k$$

where ##\theta^i## are 1-forms.

Then the book proceeds to calculate the coefficients for the following metric:
1621383077968.png


What I don't quite understand is why is the coefficient for ##\Gamma_{2[12]}## is given by ##\frac{\sqrt{1-kr^2}}{2r}##. From the second equation I would have guessed the coefficient is actually ##\frac{\sqrt{1-kr^2}}{r}##, so why do we need to divide by 2? From the second equation I would read $$d\theta^2 = \Gamma_{[12]}^2 \theta^1 \wedge \theta^2$$.

Also I'm not sure how the book actually computed the term for ##\Gamma^3_{[13]}##.
 
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There is summation in the formula

##d\theta^2 = \Gamma^2_{[jk]} \theta^j \wedge \theta^k=\cdots+\Gamma^2_{[12]} \theta^1 \wedge \theta^2+\Gamma^2_{[21]} \theta^2 \wedge \theta^1+\cdots=\cdots+2\Gamma^2_{[12]} \theta^1 \wedge \theta^2+\cdots##
 
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@martinbn what is the meaning of the dots?
 
snypehype46 said:
@martinbn what is the meaning of the dots?
Just the other terms with indeces not 1 and 2.
 
Ok I see, so for the term ##\Gamma^3_{[13]}##, we used the last equation. My point of confusion is why we don't care about the factor involving ##\theta^2 \wedge \theta^3##
 
I'm going to be explicit!

$$d\theta^i = \Gamma^i_{jk} \theta^j \wedge \theta^K$$
$$d\theta^3 = \Gamma^3_{11} \theta^1 \wedge \theta^1 + \Gamma^3_{12} \theta^1 \wedge \theta^2 + \Gamma^3_{13} \theta^1 \wedge \theta^3$$ $$+ \Gamma^3_{21} \theta^2 \wedge \theta^1 + \Gamma^3_{22} \theta^2 \wedge \theta^2 + \Gamma^3_{23} \theta^2 \wedge \theta^3$$ $$ + \Gamma^3_{31} \theta^3 \wedge \theta^1 + \Gamma^3_{32} \theta^3 \wedge \theta^2 + \Gamma^3_{33} \theta^3 \wedge \theta^3$$

Reducing this down (which property of exterior derivatives did i use to reduce to this point?), we get:
$$d\theta^3 =\Gamma^3_{12} \theta^1 \wedge \theta^2 + \Gamma^3_{13} \theta^1 \wedge \theta^3$$ $$+ \Gamma^3_{21} \theta^2 \wedge \theta^1+ \Gamma^3_{23} \theta^2 \wedge \theta^3$$ $$ + \Gamma^3_{31} \theta^3 \wedge \theta^1 + \Gamma^3_{32} \theta^3 \wedge \theta^2$$

Now, we know that ##d\theta^3 = \frac{\sqrt{1-kr^2}}{r} \theta^1 \wedge \theta^3 + \frac{\cot \theta}{r} \theta^2 \wedge \theta^3##

Match the same with the same. Which terms involve ##\theta^1 \wedge \theta^3##? You're not ignoring the ##\theta^2 \wedge \theta^3## terms, it's just that ##\Gamma^3_{13}## doesn't have any ##\theta^2 \wedge \theta^3## terms!

Also, some food for thought, what can you conclude about ##\Gamma^3_{12}, \Gamma^3_{21}## from this equation?
 
A quickie: factors of 2 and antisymmetrization [...] , what's the book's convention? Weight one (anti)symmetrization?
 
I have never worked in a non-coordinate basis before, so I'm having trouble with the basic equation:

##d \Theta^i = \Gamma^i_{|jk|} \Theta^j \wedge \Theta^k##

Is ##\Gamma^i_{jk}## defined for a non-coordinate basis via:

##\nabla_{e_i} e_j = \Gamma^k_{ij} e_k##

If so, then it seems to me that if ##e_\mu## is a coordinate basis, with corresponding connection ##\Gamma^\lambda_{\mu \nu}##, and ##e_i## is some other basis (coordinate or not), then the relationship between the ##\Gamma##s is given by:

##\Gamma^k_{ij} = \Lambda^\nu_i \Lambda^k_\mu \partial_\nu \Lambda^\mu_j + \Lambda^k_\mu \Lambda^\nu_i \Lambda^\lambda_j \Gamma^\mu_{\nu \lambda}##

where ##\Lambda^\mu_i## is the coordinate transformation matrix, and ##\Lambda^i_\mu## is its inverse.

On the other hand, in the covector basis,
##e^k = \Lambda^k_\mu e^\mu = \Lambda^k_\mu d x^\mu##

Taking exterior derivatives of both sides gives:

##d e^k = d(\Lambda^k_\mu d x^\mu) = (\partial_\nu \Lambda^k_\mu) d x^\nu \wedge dx^\mu## (because ##d d x^\mu = 0##)

Re-expressing in terms of the noncovariant basis ##e^i## gives:

##d e^k = \Lambda^\nu_i \Lambda^\mu_j (\partial_\nu \Lambda^k_\mu) e^i \wedge e^j##

So that's not the same thing as ##d e^k = \Gamma^k_{|ij|} e^i \wedge e^j##. Is it?
 
In holonomic bases ("coordinate bases") in a space without torsion (as in standard GR) the connection coefficients are the Christoffel symbols, which are symmetric in the lower components.

That's different for (pseudo)-orthonormal bases of course.
 

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