Understanding the Riemann Tensor and its Properties in Differential Geometry

[latentcorpse]

i need to show that $R_{abc}{}^{e} g_{ed} + R_{abd}{}^{e} g_{ce}=(\nabla_a \nabla_b - \nabla_b \nabla_a) g_{cd} = 0$

ok well i know that $R_{abc}{}^{d} \omega_d=(\nabla_a \nabla_b - \nabla_b \nabla_a) \omega_c$

so i reckon that $R_{abc}{}^{e} g_{ed} = (\nabla_a \nabla_b - \nabla_b \nabla_a) g_{cd}$
and
$R_{abd}{}^e g_{ce}=(\nabla_a \nabla_b - \nabla_b \nabla_a) g_{dc}$

but $g_{cd}=g_{dc}$ so when i add those terms, surely is should get
$R_{abc}{}^{e} g_{ed} + R_{abd}{}^{e} g_{ce}= 2(\nabla_a \nabla_b - \nabla_b \nabla_a) g_{cd}$
so why is there no factor of 2 in the book's answer?

and then, how do i get the whole thing to go to 0?

 

[gabbagabbahey]

ok well i know that $R_{abc}{}^{d} \omega_d=(\nabla_a \nabla_b - \nabla_b \nabla_a) \omega_c$

Are you sure about this identity? When I calculate it out I get

$$2\nabla_{[a}\nabla_{b]}\omega_c=(\nabla_a \nabla_b - \nabla_b \nabla_a) \omega_c=-R^{d}{}_{cab}\omega_d$$

so i reckon that $R_{abc}{}^{e} g_{ed} = (\nabla_a \nabla_b - \nabla_b \nabla_a) g_{cd}$
and
$R_{abd}{}^e g_{ce}=(\nabla_a \nabla_b - \nabla_b \nabla_a) g_{dc}$

I don't think it works like that; $g_{cd}$ is a symmetric tensor, not a vector.

 

[latentcorpse]

it says in the book "General Relativity" by Wald that

$\nabla_a \nabla_b \omega_c - \nabla_b \nabla_a \omega_c=R_{abc}{}^{d} \omega_d$?

 

[gabbagabbahey]

it says in the book "General Relativity" by Wald that

$\nabla_a \nabla_b \omega_c - \nabla_b \nabla_a \omega_c=R_{abc}{}^{d} \omega_d$?

Hmmm... yes, it seems Wald uses that as the definition of the curvature tensor. Most texts I've seen define the Riemann tensor by the equation:

$$R^a{}_{bcd}=\partial_c\Gamma^a_{bd}-\partial_d\Gamma^a_{bc}+\Gamma^e_{bd}\Gamma^a_{ec}-\Gamma^e_{bc}\Gamma^a_{ed}$$

From this definition, and a straight forward computation of the covariant derivatives I get:

$$(\nabla_a \nabla_b - \nabla_b \nabla_a) \omega_c=-R^{d}{}_{cab}\omega_d$$

So it appears, that for the two definitions to be equivalent (assuming I haven't messed up the calculation), we require $R_{abc}{}^{d}=-R^{d}{}_{cab}$...I guess it might be easily provable using the metric to raise and lower the appropriate indices, together with equation 3.2.15 from Wald...

Anyways, it seems like a straight forward application of equation 3.2.12 is all that is required for your problem ($g_{cd}$ is a second rank covariant tensor, not a covariant vector)...

 

[latentcorpse]

the sums on the RHS of 3.2.12 are confusing me, surely i just want that RHS to be $R_{abc}{}^e g_{ed} + R_{abd}{}^e g_{ce}$.

could i use 3.2.13 to re write $R_{abc}{}^e g_{ed}$ as $-R_{bac}{}^e g_{ed}$? then it would be in the form of the RHS of 3.2.13 and both the sums would just run form 1 to 1.

 

[gabbagabbahey]

Well, let's see if I can clarify equation 3.2.12 for you a little;

$$\begin{aligned}(\nabla_a \nabla_b - \nabla_b \nabla_a) T^{c_1\ldots c_k}{}_{d_1\ldots d_l}=&-\sum_{i=1}^{k}R_{abe}{}^{c_i}T^{c_1\ldots e\ldots c_k}{}_{d_1\ldots d_l}\\&+\sum_{j=1}^{l}R_{abd_j}{}^{e}T^{c_1\ldots c_k}{}_{d_1\ldots e\ldots d_l}\end{aligned}$$

The first sum looks like

$$\sum_{i=1}^{k}R_{abe}{}^{c_i}T^{c_1\ldots e\ldots c_k}{}_{d_1\ldots d_l}=R_{abe}{}^{c_1}T^{ec_2c_3\ldots c_k}{}_{d_1\ldots d_l}+R_{abe}{}^{c_2}T^{c_1ec_3c_4\ldots c_k}{}_{d_1\ldots d_l}+R_{abe}{}^{c_3}T^{c_1c_2ec_4c_5\ldots c_k}{}_{d_1\ldots d_l}+\ldots +R_{abe}{}^{c_k}T^{c_1c_2c_3\ldots e}{}_{d_1\ldots d_l}$$

And the second sum looks like

$$\sum_{j=1}^{l}R_{abd_j}{}^{e}T^{c_1\ldots c_k}{}_{d_1\ldots e\ldots d_l}=R_{abd_1}{}^{e}T^{c_1\ldots c_k}{}_{ed_2d_3\ldots d_l}+R_{abd_2}{}^{e}T^{c_1\ldots c_k}{}_{d_1ed_3d_4\ldots d_l}+R_{abd_3}{}^{e}T^{c_1\ldots c_k}{}_{d_1d_2 ed_4d_5\ldots d_l}+\ldots R_{abd_l}{}^{e}T^{c_1\ldots c_k}{}_{d_1d_2d_3\ldots e}$$

Does that clarify things for you?

If so, just apply that to the tensor $T^{c_1\ldots e\ldots c_k}{}_{d_1\ldots d_l}=g_{cd}$, which has zero contravariant indices ($k=0$, making the first sum zero) and two covariant indices ( $l=2$ and $d_1=c$ and $d_2=d$)...

 

[latentcorpse]

ok. i understand the expansion of the sum. i guess i didn't realize that e was adopting the position of $c_i$ and $d_j$ respectively.

why does the first sum vanish? i get
$\sum_{i=1}^{k}R_{abe}{}^{c_i}g_{cd}$
is it because there are no contravariant indices and so the e index can't be placed anywhere on $g_{cd}$ and so there's nothing to sum over? if so, how do i explain this in maths?

if i did expand the sum however i'd get something like
$R_{abe}{}^{c_1}g_{cd}+R_{abe}{}^{c_2}g_{cd}+...+R_{abe}{}^{c_k}g_{cd}$
now i think I am ok to say that $c \neq c_i \forall i \in [1,k] \cap \mathbb{Z}$. this would mean there would be no contraction in each of the terms. if there's no contraction, does the action of the metric just cause things to vanish? i.e. $X^a g_{bc}=0$ because there's no contraction?
are either of the above correct? if not could you please explain why this sum disappears?

would the "j sum" come out to be:
$R_{abd_1}{}^eg_{ed}+R_{abd_2}{}^eg_{ce}$?
then contraction with the metric gives
$=R_{abd_1d}+R_{abd_2c}$

 

[gabbagabbahey]

ok. i understand the expansion of the sum. i guess i didn't realize that e was adopting the position of $c_i$ and $d_j$ respectively.

why does the first sum vanish? i get
$\sum_{i=1}^{k}R_{abe}{}^{c_i}g_{cd}$
is it because there are no contravariant indices and so the e index can't be placed anywhere on $g_{cd}$ and so there's nothing to sum over? if so, how do i explain this in maths?

$k$ is the number of contravariant indices...if $k=0$, then surely any sum of the form $$\sum_{i=1}^k \text{some stuff}$$ is automatically zero; since there are no terms in the sum.

would the "j sum" come out to be:
$R_{abd_1}{}^eg_{ed}+R_{abd_2}{}^eg_{ce}$?
then contraction with the metric gives
$=R_{abd_1d}+R_{abd_2c}$

Why do you have a $d_1$ and $d_2$ in that expression?

 

[latentcorpse]

that's what is in the eqn 3.2.12. should i just be d? I am confused as to why it would be though?

 

[gabbagabbahey]

You changed the $d_1$ and $d_2$ to $c$ and $d$ respectively for the metric on the RHS (since $c$ is the first covariant index and $d$ is the second for the $g_{cd}$ on the LHS), so why wouldn't you do the same for the indices of the Riemann tensor?

 

[latentcorpse]

ok. so the j sum is:

$\nabla_a g_{cd}=-R_{abc}{}^e g_{ed} - R_{abd}{}^e g_{ce}=-R_{abcd} - R_{abdc}$

what would be the next step in the proof from here?

 

[gabbagabbahey]

ok. so the j sum is:

$\nabla_a g_{cd}=-R_{abc}{}^e g_{ed} - R_{abd}{}^e g_{ce}=-R_{abcd} - R_{abdc}$

what would be the next step in the proof from here?

Errm... Don't you mean $(\nabla_a \nabla_b - \nabla_b \nabla_a) g_{cd}=R_{abc}{}^e g_{ed} + R_{abd}{}^e g_{ce}=R_{abcd} +R_{abdc}$?

If so, just use the fact that $\nabla_a g_{cd}=0$ (equation 3.1.22).

 

[latentcorpse]

so the final proof that it all equals 0 just follows from
$\nabla_a \nabla_b g_{cd} - \nabla_b \nabla_a g_{cd}=\nabla_a (0) - \nabla_b (0) = 0$

 

[gabbagabbahey]

Yup!

 

[latentcorpse]

kl. thanks.

i thought i'd post another question here as its on the same stuff but that way i wouldn't have to start a new thread.

if you look at p45 of Wald, I've got to eqn 3.3.13 and then it says we use the formula for the Christoffel symbol to show that 3.3.13 is just the geodesic eqn.
clearly Wald things this is just so blatantly obvious to the reader that there was no point in doing the calculation...i hate it when textbooks do this.
i can't get anywhere with this.

any advice?

 

[gabbagabbahey]

if you look at p45 of Wald, I've got to eqn 3.3.13 and then it says we use the formula for the Christoffel symbol to show that 3.3.13 is just the geodesic eqn.
...
any advice?

Hmmm... You might want to multiply 3.3.13 by $g^{\sigma\beta}$ and sum over $\beta$...

clearly Wald things this is just so blatantly obvious to the reader that there was no point in doing the calculation...i hate it when textbooks do this

Textbooks aren't the only place you find this type of thing. All throughout the scientific literature, it is a common occurrence for authors to only provide important details of how they did a calculation (just enough for the reader to reproduce the calculation themselves if they are so inclined) and not include all the nitty gritty details. It's really only in introductory level texts that detailed calculations are frequently given, in higher level material it is assumed that the reader/student will reproduce any important calculations themselves; the purpose of which is two-fold:

(1) It saves space and writing for the author

(2)It keeps the reader/student actively involved in the studying process and allows them to slowly master the material as they go along simply by reproducing calculations that are omitted.

So, I think its unfair to say that Wald simply thins the result is blatantly obvious. He merely assumes that at this point in the readers studies, they have enough sense to take the time and work out the details for themselves.

I'm disappointed with the number of professors who never preach this practice of working out the details of a calculation to their students.

 

[latentcorpse]

ok. that gives

$0=- \sum_{\alpha,\beta} g_{\alpha \beta} g^{\alpha \beta} \frac{d^2 x^{\alpha}}{d t^2} - \sum_{\alpha, \beta, \lambda} g^{\alpha \beta} \frac{ \partial g_{\alpha \beta}}{\partial x^{\lambda}} \frac{d x^{\lambda}}{dt} \frac{d x^{\alpha}}{dt} + \frac{1}{2} \sum_{\alpha, \beta, \lambda} \frac{\partial g_{\alpha, \lambda}}{\partial x^{\beta}} \frac{d x^{\alpha}}{dt} \frac{d x^{\lambda}}{dt} g^{\alpha \beta}$
now if I am summing over alpha and beta $g_{\alpha \beta} g^{\alpha \beta}=n$ and so the first term will be
$-\sum_{\alpha} n \frac{d^2 x^{\alpha}}{d t^2}$
im not sure how to simplify the other two terms though. how does the metric act on derivatives?

 

[gabbagabbahey]

No, you don't want to multiply by $$g^{\alpha\beta}$$. Since $\alpha$ is already being summed over, it is essentially a dummy variable. You can't take $$g^{\alpha\beta}$$ inside the sums like that.

Instead, multiply by $$g^{\sigma\beta}$$ or $$g^{\rho\beta}$$ or something like that...and then sum over $\beta$...

 

[latentcorpse]

sorry. i should have spooted i had three $\alpha$ indices present. looking at this i'd get

$- \sum_{\beta} g^{\sigma \beta} \sum_{\alpha} g_{\alpha \beta} \frac{d^2 x^{\alpha}}{dt^2}-\sum_{\beta} g^{\sigma \beta} \sum_{\alpha, \lambda} \frac{\partial g_{\alpha \beta}}{\partial x^{\lambda}} \frac{d x^{\lambda}}{dt} \frac{d x^{\alpha}}{dt} + \frac{1}{2} \sum_{\beta} g^{\sigma \beta} \sum_{\alpha, \lambda} \frac{\partial g_{\alpha \lambda}}{\partial x^{\beta}} \frac{d x^{\alpha}}{dt} \frac{d x^{\lambda}}{dt}$

how can i sum over $\beta$ unless I am able to move the $g^{\sigma \beta}$ metric inside the other summations - that way i can put the metrics together and do something constructive...

 

[gabbagabbahey]

Why wouldn't you be able to move [tex]g^{\sigma\beta}[/itex] inside the other sums? None of them are summing over either $\sigma$ or $\beta$.

 

[latentcorpse]

gd point!

$-\sum_{\alpha, \beta} \frac{d^2 \sigma}{dt^2}- \sum_{\alpha, \beta \lambda} g^{\sigma \beta} \frac{\partial g_{\alpha \beta}}{\partial x^{\lambda}} \frac{d x^{\lambda}}{dt} \frac{d x^{\alpha}}{dt} + \frac{1}{2} \sum_{\alpha, \beta \lambda} g^{\sigma \beta} \frac{\partial g_{\alpha \lambda}}{\partial x^{\beta}} \frac{d x^{\alpha}}{dt} \frac{d x^{\lambda}}{dt}$

where i made a delta function on the first term. how does the metric operate on the 2nd and 3rd terms when the other metric is "caught up" in apartial derivative?

 

[gabbagabbahey]

I think the fact that $\Sigma$'s are being used to explicitly show summations (instead of using the Einstein summation convention) is confusing you:

$$\sum_{\alpha,\beta} g^{\sigma \beta} g_{\alpha \beta} \frac{d^2 x^{\alpha}}{dt^2}=\sum_{\alpha}\delta^\sigma}{}_{\alpha}\frac{d^2 x^{\alpha}}{dt^2}=\frac{d^2 x^{\sigma}}{dt^2}$$

You'll want to compare the other two terms to equation 3.1.30 and do some jiggling around...

 

[latentcorpse]

hmm. yeah but even without that sum in front, there's no $\frac{d^2 x^{\sigma}}{dt^2}$ term in 3.1.30 is there?

also for hte other terms, i assume i leave the metric terms alone as they appear to be in the same form as 3.1.30. so it's a question of getting rid of the "dx/dt" terms.
am i allowed to cancel the $dx^{\lambda}$ terms and then id be left with $\frac{d^2 x^{\sigma}}{dt^2}$ which i could remove from every term as a common factor...or am i heading in hte wrong direction again?

 

[gabbagabbahey]

hmm. yeah but even without that sum in front, there's no $\frac{d^2 x^{\sigma}}{dt^2}$ term in 3.1.30 is there?

No, but that term is present in the geodesic equation 3.3.5

also for hte other terms, i assume i leave the metric terms alone as they appear to be in the same form as 3.1.30. so it's a question of getting rid of the "dx/dt" terms.

The dx/dt terms are still present in the geodesic equation.

$$- \sum_{\alpha, \beta, \lambda} g^{\alpha \beta} \frac{ \partial g_{\alpha \beta}}{\partial x^{\lambda}} \frac{d x^{\lambda}}{dt} \frac{d x^{\alpha}}{dt} + \frac{1}{2} \sum_{\alpha \beta, \lambda} \frac{\partial g_{\alpha, \lambda}}{\partial x^{\beta}} \frac{d x^{\alpha}}{dt} \frac{d x^{\lambda}}{dt} g^{\alpha \beta}=-\sum_{\alpha,\beta,\lambda}\left(g^{\alpha \beta} \frac{ \partial g_{\alpha \beta}}{\partial x^{\lambda}} -\frac{1}{2}\frac{\partial g_{\alpha \lambda}}{\partial x^{\beta}}\right)\frac{d x^{\alpha}}{dt}\frac{d x^{\lambda}}{dt}$$

You want to compare [tex]\sum_{\beta}g^{\alpha \beta} \frac{ \partial g_{\alpha \beta}}{\partial x^{\lambda}} -\frac{1}{2}\frac{\partial g_{\alpha \lambda}}{\partial x^{\beta}}[/itex] to equation 3.1.30

 

[latentcorpse]

i think i follow so i have

$\frac{d^2 x^{\sigma}}{dt^2} - \sum_{\alpha,\beta,\lambda}\left(g^{\alpha \beta} \frac{ \partial g_{\alpha \beta}}{\partial x^{\lambda}} -\frac{1}{2} g^{\alpha \beta} \frac{\partial g_{\alpha \lambda}}{\partial x^{\beta}}\right)\frac{d x^{\alpha}}{dt}\frac{d x^{\lambda}}{dt}$

now i want to use the fact that
$g^{\alpha \beta} \frac{\partial g_{\alpha \beta}}{\partial x^{\lambda}} = \frac{1}{2}g^{\alpha \beta} \frac{\partial g_{\alpha \beta}}{\partial x^{\lambda}}+\frac{1}{2}g^{\alpha \beta} \frac{\partial g_{\alpha \beta}}{\partial x^{\lambda}}$
and then relabel the indices on the second term somehow - can u give me a reason for this? I'm not sure why i can do this? also what do i rearrange them to?i know i can get rid of the alpha sum

anyway that let's me write it as the Christoffel symbol and should lead to the answer eventually...

 

[gabbagabbahey]

i think i follow so i have

$\frac{d^2 x^{\sigma}}{dt^2} - \sum_{\alpha,\beta,\lambda}\left(g^{\alpha \beta} \frac{ \partial g_{\alpha \beta}}{\partial x^{\lambda}} -\frac{1}{2} g^{\alpha \beta} \frac{\partial g_{\alpha \lambda}}{\partial x^{\beta}}\right)\frac{d x^{\alpha}}{dt}\frac{d x^{\lambda}}{dt}$

Close, you should have

$-\frac{d^2 x^{\sigma}}{dt^2} - \sum_{\alpha,\beta,\lambda}\left(g^{\sigma \beta} \frac{ \partial g_{\alpha \beta}}{\partial x^{\lambda}} -\frac{1}{2} g^{\sigma \beta} \frac{\partial g_{\alpha \lambda}}{\partial x^{\beta}}\right)\frac{d x^{\alpha}}{dt}\frac{d x^{\lambda}}{dt}=0$When you compare that to your geodesic equation, you should see that if you can show that

$$\Gamma^\sigma{}_{\alpha\lambda}=\sum_{\beta}\left(g^{\sigma \beta} \frac{ \partial g_{\alpha \beta}}{\partial x^{\lambda}} -\frac{1}{2} g^{\sigma \beta} \frac{\partial g_{\alpha \lambda}}{\partial x^{\beta}}\right)$$

you'll be set.

 

[latentcorpse]

yes, so i need to somehow group the first two terms of 3.1.30 together so that 1/2+1/2=1 but they have different indices - how is this possible? i thought i had to relabel dummy indices but then the indices on the metric will change also and i won't be able to take it out as a common factor in the formula for the christoffel symbol.

 

[gabbagabbahey]

Well, its not actually possible, but you can easily show that

$$\sum_{\beta}\left( g^{\sigma \beta} \frac{ \partial g_{\alpha \beta}}{\partial x^{\lambda}} -\frac{1}{2} g^{\sigma \beta} \frac{\partial g_{\alpha \lambda}}{\partial x^{\beta}}\right)=\Gamma^\sigma{}_{\alpha\lambda}+\frac{1}{2} \sum_{\beta}g^{\sigma \beta} \left(\frac{\partial g_{\alpha \beta}}{\partial x^{\lambda}}-\frac{\partial g_{\lambda \beta}}{\partial x^{\alpha}}\right)$$

Substitute that into your equation, and see what happens to the extra term when you sum over $\alpha$ and $\lambda$...

 

[latentcorpse]

ok. i follow the step you just gave me.
subbing in i get

$\sum_{\alpha, \lambda} \Gamma^{\sigma}{}_{\alpha, \lambda} \frac{d x^{\alpha}}{dt} \frac{d x^{\lambda}}{dt} + \frac{1}{2} \sum_{\alpha, \beta, \lambda} g^{\sigma \beta} \left \frac{\partial g_{\alpha \beta}}{\partial x^{\lambda}} - \frac{\partial g_{\lambda \beta}}{\partial x^{\alpha}} \right \frac{d x^{\alpha}}{dt} \frac{d x^{\lambda}}{dt}$

so if i can show the second term vanishes then i have the geodesic eqn.

assuming the partial derivatives don't mess with the way the metrics combine to give Kronecker delta functions then

i can make the second term

$\frac{\partial}{\partial x^{\lambda}} \delta^{\sigma}{}_{\alpha} \left \frac{d x^{\alpha}}{dt} \frac{d x^{\lambda}}{dt} \right - \frac{\partial}{\partial x^{\alpha}} \delta^{\sigma}{}_{\lambda} \left \frac{d x^{\alpha}}{dt} \frac{d x^{\lambda}}{dt} \right = \frac{\partial}{\partial x^{\lambda}} \frac{d x^{\sigma}}{dt} \frac{d x^{\lambda}}{dt} - \frac{\partial}{\partial x^{\alpha}} \frac{d x^{\alpha}}{dt} \frac{d x^{\sigma}}{dt} = \frac{d^2 x^{\sigma}}{dt^2} - \frac{d^2 x^{\sigma}}{dt^2}=0$

which leaves us with

$- \frac{d^2 x^{\sigma}}{dt^2} - \sum_{\alpha, \lambda} \Gamma^{\sigma}{}_{\alpha \lambda} \frac{d x^{\alpha}}{dt} \frac{d x^{\lambda}}{dt}=0 \\ \Rightarrow \frac{d^2 x^{\sigma}}{dt^2} + \sum_{\alpha \lambda} \Gamma^{\sigma}{}_{\alpha \lambda} \frac{d x^{\alpha}}{dt} \frac{d x^{\lambda}}{dt}=0$ QED.

is that ok?

 

[gabbagabbahey]

No,

$$g^{\sigma\beta}\frac{\partial g_{\alpha\beta}}{\partial x^\lambda}\neq\frac{\partial}{\partial x^{\lambda}}\delta^{\sigma}_{\alpha}$$

You can't take $g^{\sigma\beta}$ inside the partial derivative like that.

Instead, all you need to do is realize that when $\alpha$ and $\lambda$ are summed over, they are dummy indices

$$\implies \frac{1}{2} \sum_{\alpha, \beta, \lambda} g^{\sigma \beta} \frac{\partial g_{\alpha \beta}}{\partial x^{\lambda}}\frac{d x^{\alpha}}{dt} \frac{d x^{\lambda}}{dt}=\frac{1}{2} \sum_{\alpha, \beta, \lambda} g^{\sigma \beta}\frac{\partial g_{\lambda \beta}}{\partial x^{\alpha}} \right \frac{d x^{\alpha}}{dt} \frac{d x^{\lambda}}{dt}$$

 

[latentcorpse]

thanks a lot for your help.