# Deriving EFE from the Bianchi Identities

1. Aug 23, 2011

### jfy4

Hi,

I ran this by my friend awhile ago and I'm not sure how to feel about it still... Consider the following:

We know
$$\nabla_\alpha G^{\alpha\beta}=0\implies \partial_\alpha G^{\alpha\beta}+\Gamma^{\beta}_{\gamma\delta}G^{ \gamma \delta}+\Gamma^{\delta}_{\delta\gamma}G^{ \gamma \beta}=0$$
Then is it possible to take as the definition of the divergence of the SEM as
$$\partial_\alpha T^{\alpha\beta}=-(\Gamma^{\beta}_{\gamma\delta}G^{\gamma\delta}+ \Gamma^{\delta}_{\delta\gamma}G^{\gamma\beta})?$$
This seems so to me since if I were clairvoyant I would know that $\mathbf{G}=\mathbf{T}$. Anyways, given that, I have
$$\partial_{\alpha}G^{\alpha\beta}-\partial_{\alpha}T^{\alpha\beta}=\partial_{\alpha}(G^{\alpha\beta}-T^{\alpha\beta})=0$$
Now, is there a way to know if the divergence of something being zero implies that the something is zero in order to justify the final step $\mathbf{G}=\mathbf{T}$?

Thanks,

2. Aug 23, 2011

### WannabeNewton

Pardon me for asking, but I do not understand this step. If you have $$\partial _{\alpha }G^{\alpha \beta } = - (\Gamma ^{\alpha }_{\alpha \delta }G^{\delta \beta} + \Gamma ^{\beta }_{\alpha \delta }G^{\alpha \delta })$$ how do you justify the replacement $\partial _{\alpha }G^{\alpha \beta } \to \partial _{\alpha }T^{\alpha \beta }$?

3. Aug 23, 2011

### jfy4

Hey!

I actually took the above as the definition of the divergence of the SEM, there wasn't really a rigorous step taken there... I imagined that given EFE ($G_{\alpha\beta}=T_{\alpha\beta}$) and taking the divergence of both sides
$$\partial_\alpha G^{\alpha\beta}=\partial_{\alpha}T^{\alpha\beta}$$
and then substituting into the Bianchi identities I could get away with it...

Do you think that is legit?

Thanks,

4. Aug 23, 2011

### Ben Niehoff

In curved space,

$$\partial_\alpha G^{\alpha \beta}$$

is not the divergence! The divergence is given by

$$\nabla_\alpha G^{\alpha \beta}$$

5. Aug 23, 2011

### WannabeNewton

Oh, ok. I'm trying really hard but I can't think of anything that would imply $G^{\alpha \beta } = T^{\alpha \beta }$ from $\partial _{\alpha }(G^{\alpha \beta } - T^{\alpha \beta }) = 0$. As far as I know it just implies that $$\int_{\Sigma }(G^{\alpha \beta } - T^{\alpha \beta })dS_{\alpha } = const.$$ which doesn't seem like much help at all.

6. Aug 23, 2011

### jfy4

Yes I know, I was using it poorly, sorry, I thought the meaning would be clear enough though. I just don't know what else to call in English the quantity $\partial_{\alpha}A^{\alpha\beta}$, if not the divergence. Thanks.

7. Aug 24, 2011

### Ben Niehoff

I guess I don't see why you are interested in the quantity $\partial_\alpha G^{\alpha \beta}$, since it is not a tensor and therefore does not represent any geometrical object. It does not behave well under coordinate transformations.

I also don't see how it is relevant to "deriving" Einstein's equations. You can't "derive" Einstein's equations anyway; you have to postulate them. $G_{\alpha \beta} = \frac12 T_{\alpha \beta}$ is not the only conceivable way matter can be coupled to curvature. In fact, Einstein's gravitational theory was neither the first nor the only geometrical theory of gravity; rather, it is Einstein's theory that has passed the tests of several experiments.

8. Aug 24, 2011

### jfy4

I'm not particularly interested in $\partial_\alpha G^{\alpha\beta}$, it just happens to be involved in the work above.

Your continual close attention to my vocabulary usage continues to serve you well. The above isn't a derivation in a strict mathematical sense, there is some postulating going on about the relationship of mathematical quantities to physical quantities, to be sure, it took place in the line "take as the definition of $\partial_\alpha T^{\alpha\beta}$ as etc..." So you are right again that this isn't a derivation in a strict sense.

As for the relavence of $\partial_\alpha G^{\alpha\beta}$, it simply plays a role such that I was hoping at the end of the derivation to be able to have two divergences equal zero, and then hopefully be able to declare the thing in parenthesis as equal to zero from that, that's it.

I understand sometimes I am not the clearest, but you seem to be misunderstanding my english and not the math. In the OP the "derivation" was informal and WannabeNewton seemed to understand what I was going for, except for my definition of $\partial_\alpha T^{\alpha\beta}$, which after I noted he got directly what I was going for. Is my math so unclear that we have to pay attention to all the informal usage of some of my english?

9. Aug 24, 2011

### Ben Niehoff

It's not just your vocabulary I'm criticizing. The mathematical process in your first post makes no sense, mainly because you started writing equations that are not true in all coordinate systems. The vocabulary you were using (such as "divergence") signaled to me that you didn't understand this subtlety.

It may be true that

$$\partial_\alpha G^{\alpha \beta} = \partial_\alpha T^{\alpha \beta}$$
in some coordinate systems. But there is no reason it should be true in general! So it would hardly be a sensible thing to write down in any attempt to "derive" anything.

I was curious what you meant by "deriving" in the first place, because your mathematical steps make no sense at all, and I don't see what you are trying to accomplish here. I was hoping you could explain in words, but you seem to think there is no value in explaining things in words.

10. Aug 24, 2011

### jfy4

I'm grateful for your criticism, but I can't know your criticism without you saying it. By addressing my signals of a lack of understanding, I was left with the impression I spoke wrong, but did the math correctly... I would have rathered you simply point to the step in the math where it went sour.

Once again thank you, now if you will I still don't entirely understand.

Starting over, given $G_{\alpha\beta}=T_{\alpha\beta}$ in all frames, consider $\partial_\alpha (G^{\alpha\beta})$, why can I not replace $G^{\alpha\beta}$ with $T^{\alpha\beta}$ because of their equality?

Secondly, going back to my OP, I actually didn't use that formula in the derivation, I used it on the side to help predict the outcome of my "derivation". With that in mind, that couldn't have contributed to the fault in my derivation since it wasn't even in it. However, I did claim that $\partial_\alpha T^{\alpha\beta}=-(\Gamma^{\beta}_{\gamma \delta}G^{\gamma \delta}+\Gamma^{\delta}_{\delta \gamma}G^{\gamma \delta})$. Does this claim make no sense?

Thanks,

11. Aug 24, 2011

### WannabeNewton

I think what Ben is saying (forgive me Ben if I have misinterpreted you =p) is that $$\triangledown _{\alpha }G^{\alpha \beta } = \partial _{\alpha }G^{\alpha \beta } + \Gamma ^{\alpha }_{\alpha \delta }G^{\delta \beta} + \Gamma ^{\beta }_{\alpha \delta }G^{\alpha \delta } = 0$$ is a valid tensor equation so it can be applied to any coordinate system but $\partial _{\alpha }G^{\alpha \beta } = \partial _{\alpha }T^{\alpha \beta }$ is NOT a tensor equation (the quantities on either side do not transform as tensors) so it does not need to be true for all coordinate systems. Therefore, the replacement $\partial _{\alpha }G^{\alpha \beta } \to \partial _{\alpha }T^{\alpha \beta }$ in $$\partial _{\alpha }G^{\alpha \beta } = - (\Gamma ^{\alpha }_{\alpha \delta }G^{\delta \beta} + \Gamma ^{\beta }_{\alpha \delta }G^{\alpha \delta })$$ is not valid because this statement is true for all coordinate systems but the statement involving the replacement need not be.

12. Aug 24, 2011

### jfy4

Thanks,

I can accept that. I thought I had the whole tensor thing down... it seems weird to me though that $g^{\alpha\beta}\partial_\alpha G_{\beta\gamma}$ equals something, but $g^{\alpha\beta}\partial_\alpha T_{\beta\gamma}$ equals something else, even though $T_{\alpha\beta}$ is $G_{\alpha\beta}$ (in the appropriate units}.

I guess I have some relearning to do...

13. Aug 24, 2011

### WannabeNewton

Yeah it does come off as weird but remember that the directional derivatives at some point on the manifold make up the basis for the tangent space at that point and this basis need not be orthonormal for all coordinate systems. If the bases change relative to each other during a coordinate change then your equation involving the directional derivatives need not be true anymore.

Last edited: Aug 24, 2011
14. Aug 24, 2011

### jfy4

If I may ask one more question,

Starting with
$$G_{\alpha\beta}=T_{\alpha\beta}$$
then
$$G_{\alpha\beta}-T_{\alpha\beta}=0$$
taking your statement above, does this mean that
$$\partial^{\alpha}(G_{\alpha\beta}-T_{\alpha\beta})=\partial^{\alpha}(0)=0$$ doesn't hold? This seems odd to me since it is the quantity in parenthesis which is zero, not the "divergence."

Last edited: Aug 24, 2011
15. Aug 24, 2011

### PAllen

Mostly I was ignoring this thread, but I will say:

If you are able to posit that:

partial (G) = partial (T)

in all coordinate systms (despite using partial rather than covariant derivative, and assuming somehow you know G and T are tensors), then it must be true that G=T. However, this is useless for a derivation, since it is really assuming the conclusion.

16. Aug 24, 2011

### WannabeNewton

Oh, if the value in the parenthesis is known to be a scalar then yeah you can just replace the partial derivative with a covariant derivative and say its true for all coordinate systems. But if you didn't know they were equal then that equation doesn't have to hold under a change of basis.

17. Aug 24, 2011

### robphy

Since the divergence is a type of derivative,
"if two quantities have equal derivatives
(so that the difference of those derivatives is zero,
or, by linearity of the derivative,
that the derivative of the difference of the two quantities is zero),
are the two quantities equal?"