# I need to calculate δR: R is Ricci scalar

1. ### sourena

14
I need to calculate δR: R is Ricci scalar

2. ### nicksauce

1,275
This should be shown in most GR textbooks...

3. ### sourena

14
This is not a homework.

894
5. ### sourena

14

Do you have any problem with these equations:

Last edited: Aug 29, 2010
7. ### sourena

14
No, I don't have problem with these equations, but I have problem to calculate this equation from them:
δR=Rab δgab+gab δgab -∇a ∇b δgab

First step:
The term $$\nabla_\sigma \left( g^{\mu\nu}\delta\Gamma^{\sigma}_{\mu\nu}-g^{\mu\sigma}\delta\Gamma^{\rho}_{\rho_\mu}\right)$$
is of the form $$\nabla_\sigma A^\sigma$$.
But $$\nabla_\sigma A^\sigma=|\det g]^{-\frac12}\partial_\sigma (A^\sigma |\det g|^{\frac12})$$
for any vector field $$A^\sigma$$. This is handy.

P.S. For some reason my display is not displaying correctly one character in your first term on the right. So, I do not know what is exactly the formula you would like to have.

Last edited: Sep 7, 2010
9. ### JustinLevy

894
Alright, I tried to work it out, but it looks like I got a sign error somewhere. I wrote everything out in more detail than was probably necessary, so where this occurs can stand out.

Okay, so we at least have a starting point we can agree on:
$$\delta R = R_{ab} \delta g^{ab} + g^{ab} \delta R_{ab}$$
$$\delta R_{ab} &= [\nabla_c \delta \Gamma^c_{ab} - \nabla_b \delta \Gamma^c_{c a}]$$

As in wikipedia noting that $$\delta \Gamma^\lambda_{\mu\nu}\,$$ is actually the difference of two connections, it should transform as a tensor. Therefore, it can be written as
$$\delta \Gamma^\lambda_{\mu\nu}=\frac{1}{2}g^{\lambda d}\left(\nabla_\mu\delta g_{d\nu}+\nabla_\nu\delta g_{d\mu}-\nabla_d\delta g_{\mu\nu} \right)$$

and substituting in the equation, after playing with it in excruciating detail one finds:
\begin{align*} \delta R_{ab} &= [\nabla_c \delta \Gamma^c_{ab} - \nabla_b \delta \Gamma^c_{c a}] \\ &= [\nabla_c \frac{1}{2}g^{c d}\left(\nabla_a\delta g_{db}+\nabla_b\delta g_{da}-\nabla_d\delta g_{ab} \right) -\nabla_b \frac{1}{2}g^{c d}\left(\nabla_c\delta g_{da}+\nabla_a\delta g_{dc}-\nabla_d\delta g_{ca} \right)] \\ &= \frac{1}{2}g^{c d}[\left(\nabla_c\nabla_a\delta g_{db}+\nabla_c\nabla_b\delta g_{da}-\nabla_c\nabla_d\delta g_{ab} \right) -\left(\nabla_b\nabla_c\delta g_{da}+\nabla_b\nabla_a\delta g_{dc}-\nabla_b\nabla_d\delta g_{ca} \right)] \end{align*}
swap the derivative order on the fourth term
http://en.wikipedia.org/wiki/Covariant_derivative#Examples
$$\nabla_b\nabla_c\delta g_{da} = \nabla_c\nabla_b\delta g_{da} + R^{e}{}_{dbc} \delta g_{ea} + R^{e}{}_{abc} \delta g_{de}$$
\begin{align*} g^{ab}\delta R_{ab} &= g^{ab} \frac{1}{2}g^{c d}[\left(\nabla_c\nabla_a\delta g_{db}-\nabla_c\nabla_d\delta g_{ab} \right) -\left(R^{e}{}_{dbc} \delta g_{ea} + R^{e}{}_{abc} \delta g_{de} + \nabla_b\nabla_a\delta g_{dc}-\nabla_b\nabla_d\delta g_{ca} \right)] \\ &= \frac{1}{2}g^{c d}[\left(\nabla_c\nabla^b \delta g_{db}-\nabla_c\nabla_d g^{ab} \delta g_{ab} \right) -\left(R^{e}{}_d{}^a{}_c \delta g_{ea} + R^{eb}{}_{bc} \delta g_{de} + \nabla_b\nabla^b \delta g_{dc}-\nabla^a \nabla_d\delta g_{ca} \right)] \\ &= \frac{1}{2}[\left(\nabla^d\nabla^b \delta g_{db}-\nabla^d\nabla_d g^{ab} \delta g_{ab} \right) -\left(R^{eca}{}_c \delta g_{ea} + R^{eb}{}_b{}^d \delta g_{de} + \nabla_b\nabla^b g^{c d} \delta g_{cd} -\nabla^a \nabla^c\delta g_{ca} \right)] \\ &= \frac{1}{2}[\left(\nabla^d\nabla^b \delta g_{db}-\nabla^d\nabla_d g^{ab} \delta g_{ab} \right) -\left(R^{ea}\delta g_{ea} - R^{de} \delta g_{de} + \nabla_b\nabla^b g^{c d} \delta g_{cd} -\nabla^a \nabla^c\delta g_{ca} \right)] \\ &= \nabla^a\nabla^b \delta g_{ab} - g^{ab} \nabla^c\nabla_c \delta g_{ab} \end{align*}

Hmm...
maybe there isn't an error. Does
$\delta g_{ab} = - \delta g^{ab}$ ?
I'm too tired to check right now.

Not exactly.

Use:

$$0=\delta ( \delta^a_b) =\delta (g^{ac}g_{cb})= ....$$

Now use the Leibniz rule, calculate what you need.

11. ### JustinLevy

894
Thanks.

Alright, so
$$0=\delta ( \delta^a_b) =\delta (g^{ac}g_{cb})= g_{cb} \delta g^{ac} + g^{ac} \delta g_{cb}$$
$$g_{cb} \delta g^{ac} = - g^{ac} \delta g_{cb}$$

Thus manipulating the previous posts result gives
\begin{align*} g^{ab}\delta R_{ab} &= \nabla^a\nabla^b \delta g_{ab} - g^{ab} \nabla^c\nabla_c \delta g_{ab} \\ &= \nabla^a\nabla_c g^{bc} \delta g_{ab} - \nabla^c\nabla_c g^{ab} \delta g_{ab} \\ &= - \nabla^a\nabla_c g_{ab} \delta g^{bc} + \nabla^c\nabla_c g_{ab} \delta g^{ab} \\ \end{align*}
which is what sourena wanted to find:
$$g^{ab}\delta R_{ab} = g_{ab} \nabla^c\nabla_c \delta g^{ab} - \nabla_a\nabla_b \delta g^{ab}$$

...

I assume your post #8 hint leads to the result faster, but I'm not sure how to apply that. Once we've converted from covariant to ordinary coordinate derivative, how do we recognize the result as terms with two covariant derivatives?

The trick with ordinary derivatives is useful when you calculate under the integral. There, in variational calculus, when you assume that variations vanish at the boundary (usually at infinity), you need a true divergence of a vector field, and not some "covariant one". You discard such terms converting volume integral into surface integrals using ordinary rules of the calculus.

13. ### sourena

14
Dear JustinLevy
Sorry for being late to answer your posts. I value your hard work to obtain this expression a great deal. Thank you very much for your time and care.

Last edited: Sep 29, 2010
14. ### sourena

14
I wanted to calculate this:
$$g^{ab}\delta R_{ab} = g_{ab} \nabla^c\nabla_c \delta g^{ab} - \nabla_a\nabla_b \delta g^{ab}$$
or
$$\delta R = R_{ab} \delta g^{ab} + g_{ab} \nabla^c\nabla_c \delta g^{ab} - \nabla_a\nabla_b \delta g^{ab}$$

15. ### haushofer

999
Do you know how to derive

$$\nabla_{\mu}\delta\Gamma^{\lambda}_{\nu\rho} = \frac{1}{2}g^{\lambda\alpha}[\nabla_{\mu}\nabla_{\nu}\delta g_{\rho\alpha} + \nabla_{\mu}\nabla_{\rho}\delta g_{\nu\alpha} - \nabla_{\mu}\nabla_{\alpha}\delta g_{\nu\rho}]$$
?
This can be shown directly by varying the connection, but you can also guess the form of it; varying the connection gives a tensor, and the partial derivatives in it can only become covariant ones (convince yourself if you're not!). You can plug this in the Palatini equation, which gives

$$\delta R_{\mu\nu\rho}^{\ \ \ \ \lambda} = \frac{1}{2}g^{\lambda\alpha}[\nabla_{\mu}\nabla_{\nu}\delta g_{\rho\alpha} + \nabla_{\mu\rho}\delta g_{\nu\alpha}-\nabla_{\mu}\nabla_{\alpha}\delta g_{\nu\rho}] - \frac{1}{2}g^{\lambda\alpha}[\nabla_{\nu}\nabla_{\mu}\delta g_{\rho\alpha}+\nabla_{\nu}\nabla_{\rho}\delta g_{\mu\alpha} - \nabla_{\nu}\nabla_{\alpha}\delta g_{\mu\rho}]$$

Does this help?

16. ### haushofer

999
Do you know how to derive

$$\nabla_{\mu}\delta\Gamma^{\lambda}_{\nu\rho} = \frac{1}{2}g^{\lambda\alpha}[\nabla_{\mu}\na_{\nu}\delta g_{\rho\alpha} + \nabla_{\mu}\na_{\rho}\delta g_{\nu\alpha} - \nabla_{\mu}\na_{\alpha}\delta g_{\nu\rho}]$$

This can be shown directly, but you can also guess the form of it; varying the connection gives a tensor, and the partial derivatives in it can only become covariant ones (convince yourself if you're not!). You can plug this in the Palatini equation, which gives

$$\delta R_{\mu\nu\rho}^{\ \ \ \ \lambda} &=& \frac{1}{2}g^{\lambda\alpha}[\nabla_{\mu}\nabla_{\nu}\delta g_{\rho\alpha} + \nabla_{\mu}\nabla_{\rho}\delta g_{\nu\alpha}-\nabla_{\mu}\nabla_{\alpha}\delta g_{\nu\rho}] - \frac{1}{2}g^{\lambda\alpha}[\nabla_{\nu}\nabla_{\mu}\delta g_{\rho\alpha}+\nabla_{\nu}\nabla_{\rho}\delta g_{\mu\alpha} - \nabla_{\nu}\nabla_{\alpha}\delta g_{\mu\rho}]$$

17. ### yuiop

The equation and mathematics posted in this thread are very impressive, but I always wonder how useful they are (or what the point of them is) if these sort of calculations seemingly can not be used to answer "simple" questions like the ones posed here https://www.physicsforums.com/showthread.php?t=431712 and here https://www.physicsforums.com/showthread.php?p=2901038#post2901038 ?

This is one reason I have never made the effort to really try and learn the apparatus of tensors, because of the seeming limited applicability of these formalisms. Is it that tensors are "overkill" to solve the questions posed in those links (which have not yet been solved), like using a sledgehammer to crack a walnut, or is it just that the people that like to play with tensors never look at or try to solve the "simple" questions?

My real question is this. If a person makes an effort to learn tensors would they able to answer those "simple" questions and is it worth the effort to learn these formidable looking formalisms, if all you want to do is solve the sort of simple questions posed in those threads? (which up to now are seemingly unsolvable).

There is another simple question: "What is the purpose of life?" For an engineer his purpose can be, for instance, to learn about elasticity and to apply his knowledge. And when you start learning elasticity stuff - you will soon find that you can't go too far without tensors. So it all depends on your purpose.

19. ### yuiop

So does that mean tensors cannot be used for the sort of questions posed in those threads or that they would be overkill?
It might be worth noting that the first thread quoted, specifically mentions Ricci and Weyl curvature.

I imagine that for an engineer for whom tensors are their bread and butter, then they probably think and even dream in terms of tensors and I assume they could answer those sort of questions in a trice and yet they do not. Why is that? Are the questions too simple (and there are presumably quicker less daunting methods of solving them) or are they outside the normal domain of the application of tensors?

Last edited: Sep 29, 2010