# More Tensors

1. Oct 15, 2009

### latentcorpse

$\nabla_a R_c{}^a + \nabla_b R_c{}^b - \nabla_c R=0$

can be written as $\nabla^a G_{ab}=0$

where $G_{ab}=R_{ab} - \frac{1}{2} R g_{ab}$

now im trying to work back to prove this is true:

$\nabla^a G_{ab}=\nabla^a R_{ab} - \frac{1}{2} \nabla^a (R g_{ab})$

now im stuck, how do i evaluated these derivative operators, do i need to multiply through by some metric such as $g_{ae}$ to lower the a index?

thanks.

2. Oct 15, 2009

### gabbagabbahey

What is the definition of the contravariant derivative? What is the product rule for contravariant/covariant derivatives? What is the definition of $R$?

3. Oct 16, 2009

### latentcorpse

R is the scalar curvature which is the trace of the Ricci curvature i.e. $R=R_a{}^a$

a contravariant derivative is defined by 3.1.14 in Wald where we take $\tilde{\nabla_a}$ to be the contravariant derivative $\nabla^a$ i think (am i ok to just quote eqns out of Wald? i assume from your answer to my other post that you have your own copy?)

one of my main problems here is that as far as i can remember (ive been reading Chapter 3 of Wald) so far ive only come across covariant derivatives. does 3.1.13 not imply that say $\nabla^a$ and $\partial_a$ would act on a type (k,l) tensor to give the same result? how can this be?
or in fact they dont because we use Christoffel symbols for the partial derivative case yes?

anyway how does using 3.1.14 help here because we'll get an answer in terms of $C^c{}_{ab}$ which we dont know anything about?

Last edited: Oct 16, 2009
4. Oct 16, 2009

No, the $$\tilde{\nabla_a}[/itex] in 3.1.14 can be any type of deriavtive. In the special case that [tex]\tilde{\nabla_a}=\partial_a[/itex], 3.1.14 defines the usual covariant derivative $\nabla_a$. A contravariant derivative is simply defined by $\nabla^a=g^{ab}\nabla_b$. Well, I don't have my own copy, but I do have easy access to a nice digital Library, so yes; quoting equations from Wald works fine for me. 5. Oct 16, 2009 ### latentcorpse oh. ok. i knew that $\tilde{\nabla_a}$ could be any type of derivative but i thought 3.1.14 defined how we got $\nabla_a$ knowing $\tilde{\nabla_a}$ and $\tilde{\nabla_a}$ didn't necessarily have to be $\partial_a$. in fact when $\tilde{\nabla_a}=\partial_a$ doesn't 3.1.15 define how to get the covariant derivative? ok so i have the defn of R as the trace of Ricci tensor, $\tilde{\nabla_a}=g^{ab} \nabla_b$ and the product rule for derivatives is the Leibnitz rule described on p31. so, $\nabla^a R_{ab}=g^{ac} \nabla_c R_{ab}=\nabla_c R^c{}_b$ and, $\frac{1}{2} \nabla^a (R g_{ab})= \frac{1}{2} ( \nabla^a R) g_{ab} + \frac{1}{2} R \nabla^a g_{ab}=\frac{1}{2} (g^{ac} \nabla_c R) g_{ab} + \frac{1}{2} R g^{ac} \nabla_c g_{ab}$ how do i proceed? i guess expanding $R=R_d{}^d$ might help but i dont see how the d index will be involved in anything? 6. Oct 16, 2009 ### gabbagabbahey 3.1.15 is just an application of 3.1.14 for [tex]\tilde{\nabla_a}=\partial_a$$ on a contravariant tensor field of rank one. The general definition of the covariant derivative (Remember, the covariant derivative is the derivative operator $\nabla_a$ naturally associated with the metric $\nabla_cg_{ab}=0$) is 3.1.14 with $$\tilde{\nabla_a}=\partial_a[/itex] and $C^a{}_{bc}=\Gamma^a{}_{bc}$. You'll want to use the symmetry property of the Ricci tensor before applying the last step. What does $\nabla_c g_{ab}$ equal? What does $g^{ab}g_{ac}$ equal? 7. Oct 16, 2009 ### latentcorpse (i) well the symmetry property is just that $R_{ab}=R_{ba}$ but why do i need to use that? can't i just use the metric to raise the a index on the Ricci tensor? or does the covariant derivatice interfere? (ii) should i expand $\nabla_c g_{ab}$ using 3.1.14? i couldn't think what to do with this term. (iii) i know that i get a delta when teh two metrics are next to each other and i was going to do that but again i was afraid the derivatives/tensors in between tehm might cause problems...is it ok to just move the metric about freely in any expression? if so then that last term would be $\frac{1}{2}R \nabla_c \delta^c{}_b=\frac{1}{2} R \nabla_b$ where i assume $\delta^c{}_b=\delta^b{}_c$ by symmetry of Kronecker delta. but that doesn't look right at all. sorry. just getting used to working with the maths behind GR is taking a while... 8. Oct 16, 2009 ### gabbagabbahey There's nothing wrong with what you did in your previous post, its just more useful to you if you use the symmetry condition to get $\nabla^a R_{ab}=g^{ac} \nabla_c R_{ab}=g^{ac} \nabla_c R_{ba}=\nabla_c R_b{}^{c}$ so that the result is the same form as the first two terms in your premise, $\nabla_a R_c{}^a + \nabla_b R_c{}^b - \nabla_c R=0$ You could, but you'll probably find it easier if you just use equation 3.1.22 instead As long as there are no derivatives or other operators acting on it, you are free to move any tensor around when in component form. [tex]T^{a_1,a_2,\ldots a_i}{}_{b_1,b_2,\ldots b_j}U^{c_1,c_2,\ldots c_k}{}_{d_1,d_2,\ldots d_l}=U^{c_1,c_2,\ldots c_k}{}_{d_1,d_2,\ldots d_l}T^{a_1,a_2,\ldots a_i}{}_{b_1,b_2,\ldots b_j}$$

for any tensor fields $T$ and $U$, even if $TU\neq UT$, and even if there are repeated indices/implied summations, since each individual component of a tensor field is a scalar field, and scalar fields commute.

How did $R$ get in front of the derivative operator?

Last edited: Oct 16, 2009
9. Oct 17, 2009

### latentcorpse

ok. i get the symmetry condition bit and the use of 3.1.22 definitely helps.

why does that TU=UT when in component form even if TU doesn't equal UT when not in component form?

i moved that R in front of the derivative operator in post 5 when i used the product rule for derivative operators (see point 2 on p31 of Wald)

10. Oct 17, 2009

### gabbagabbahey

Because each individual component of a tensor field is a scalar field, and scalar fields commute. You are multiplying some component of some tensor field (which will be a scalar field) with some component of another tensor field (which will also be a scalar field)....That's basic tensor algebra.

Okay, I see now that you were talking about the last term; $\frac{1}{2}Rg^{ac}\nabla_cg_{ab}\neq\frac{1}{2}R\nabla_c (g^{ac}g_{ab})$ since the derivative acts only on $g_{ab}$ in the first expression. However, using $\nabla_cg_{ab}=0$, this term can be easily simplified!

11. Oct 17, 2009

### latentcorpse

$\frac{1}{2} \nabla^a (R g_{ab})= \frac{1}{2} ( \nabla^a R) g_{ab} + \frac{1}{2} R \nabla^a g_{ab}=\frac{1}{2} (g^{ac} \nabla_c R) g_{ab} + \frac{1}{2} R g^{ac} \nabla_c g_{ab}$
the second term vanishes giving
$\frac{1}{2} (g^{ac} \nabla_c R) g_{ab}$
hmmm.....i assume the covariant derivative acts by product rule to give me 2 terms but i dont see how?
if i make $R=R_d{}^d$ then it still doesn't help

also im still not quite following the TU=UT thing. surely if you can move it about in component form so that TU=UT then, when not in component form, TU=UT also?

12. Oct 17, 2009

### gabbagabbahey

Good...

No, the derivative operator acts only on $R$ in this expression, so $\frac{1}{2} (g^{ac} \nabla_c R) g_{ab}=\frac{1}{2} g_{ab}g^{ac} \nabla_c R=\frac{1}{2}\delta^c_b\nabla_cR=\frac{1}{2}\nabla_b R$

Let's look at something a little more familiar to you.... consider the vector cross product between two Euclidean 3D vector fields $\textbf{u}$ and $\textbf{v}$....In general, $\textbf{u}\times\textbf{v}\neq\textbf{v}\times\textbf{u}$ (The vector cross product is one type of tensor product, so in abstract notation this is equivalent to saying $uv \neq vu$ ).... Try a few examples (Like say $\textbf{u}=(2x+3y)\mathbf{\hat{e}_1}-6\mathbf{\hat{e}_2}$ and $\textbf{v}=(x-4y)\mathbf{\hat{e}_1}+(6x+2z)\mathbf{\hat{e}_2}-\mathbf{\hat{e}_3}$ where $\{\mathbf{\hat{e}_1},\mathbf{\hat{e}_2},\mathbf{\hat{e}_3}\}$ are the usual Cartesian unit vectors) and convince that yourself despite this, $u_iv_j=v_ju_i$ for all $i,j\in\{1,2,3\}$...

Last edited: Oct 17, 2009
13. Oct 17, 2009

### latentcorpse

ok so i have

$\frac{1}{2} \nabla_b R=\frac{1}{2} \nabla_b R_c{}^c=...$

i assume im meant to use 3.1.14 here.

doesn't that delta raise the b index also?

14. Oct 17, 2009

### gabbagabbahey

While this is true; it's unessecary....your equation is now

$\nabla^a G_{ab}=\nabla_c R_b{}^{c}-\frac{1}{2} \nabla_b R$

Multiply the whole thing by two, and use the fact that $2\nabla_c R_b{}^{c}=\nabla_c R_b{}^{c}+\nabla_c R_b{}^{c}=\nabla_a R_b{}^{a}+\nabla_c R_b{}^{c}$ since $c$ is a dummy index...now compare to your premise...

15. Oct 17, 2009

### latentcorpse

ok so

$\nabla^a G_{ab} = \nabla_a R_b{}^a + \nabla_c R_b{}^c - \nabla_b R$

not quite the same. is it to do with relabelling the indices now?
if i relabel $b \stackrel{\leftrightarrow}{} c$

then $\nabla_a R_c{}^a + \nabla_b R_c{}^b - \nabla_c R$

is that ok?

16. Oct 17, 2009

Yup!