# The Variation of Christoffel Symbol

1. Jun 23, 2016

### rezkyputra

1. The problem statement, all variables and given/known data
It is shown in Carrol, an Introduction to GR that the variatiom of Christoffel symbols are :

https://scontent-sin1-1.xx.fbcdn.net/v/t34.0-12/13535871_1161725257182772_897443562_n.jpg?oh=df1a6d26aa0b199d4684b5f0785bee20&oe=576ECCCA

But i have no idea how to derive that, any help would be greatly appreciated, thanks

2. Jun 24, 2016

### haushofer

There are two ways: the clever way, and the straightforward way.

The clever way is to notice that the variation of Gamma is a tensor. Because

$$\Gamma^{\rho}_{\mu\nu} = g^{\rho\lambda} \Bigl(\partial_{(\mu} g_{\nu) \lambda} - \frac{1}{2} \partial_{\lambda}g_{\mu\nu} \Bigr) \ \ \ \ (1)$$

you know that

$$\delta \Gamma^{\rho}_{\mu\nu} = g^{\rho\lambda} \Bigl(\partial_{(\mu} \delta g_{\nu) \lambda} - \frac{1}{2} \partial_{\lambda} \delta g_{\mu\nu} \Bigr) + \delta g^{\rho\lambda} \Bigl(\partial_{(\mu} g_{\nu) \lambda} - \frac{1}{2} \partial_{\lambda}g_{\mu\nu} \Bigr)$$

which must be covariant, even though it doesn't look like that because of the partial derivatives! So you should be able to massage this term into an explicitly covariant expression, starting with writing the second term in terms of variations of the metric instead of its inverse. The only possibility (check for yourself! what else?) is an expression which makes the partial derivatives covariant ones:

$$\delta \Gamma^{\rho}_{\mu\nu} = g^{\rho\lambda} \Bigl(\nabla_{(\mu} \delta g_{\nu) \lambda} - \frac{1}{2} \nabla_{\lambda} \delta g_{\mu\nu} \Bigr)$$

The second way is more straightforward and less clever: Do the variation and do the calculation explicitly, using the expression of the Christoffel symbol (1), and plough ahead.

I've written down the calculation in terms of $\delta g_{\mu\nu}$; you should be able to convert that to $\delta g^{\mu\nu}$ by using the inverse relations, i.e. varying

$$g^{\mu\nu}g_{\nu\rho} = \delta^{\mu}_{\rho}$$

Last edited: Jun 24, 2016
3. Jun 24, 2016

### haushofer

Similar reasoning btw can be applied to variations of the Riemann tensor and Ricci tensor.

4. Sep 23, 2016

### demon

Just a related question:
In the variation of the connection we have terms such as:
λδgμν
But, as far as I know, the variation and covariant derivative operators commute. Hence:
λ(δgμν )= δ(∇λgμν)
But the covariant derivatives of the metric are zero, so all these terms should be zero. I suspect there is a flaw in my reasoning, but I cannot put the finger on it. Any comments?

5. Sep 25, 2016

### samalkhaiat

No. The variation operator does not commute the covariant derivative. The variation symbol $\delta$ is a linear derivation operator, i.e., it satisfies the following rules $$\delta (A + B) = \delta A + \delta B ,$$ $$\delta (AB) = (\delta A) B + A (\delta B) ,$$ and $$\delta (\partial A) = \partial (\delta A) .$$ Using these rules, you can easily show that the following relation holds for any tensor field $g_{\mu\nu} (x)$
$$\nabla_{\lambda} \left( \delta g_{\mu\nu} \right) = \delta \left( \nabla_{\lambda}g_{\mu\nu}\right) + g_{\rho \nu} \ \delta \Gamma^{\rho}_{\mu\lambda} + g_{\mu\rho} \ \delta \Gamma^{\rho}_{\lambda\nu} .$$
If $g_{\mu\nu}$ is the pseudo-Riemannian metric tensor, the above equation becomes
$$\nabla_{\lambda} \delta g_{\mu\nu} = g_{\rho \nu} \ \delta \Gamma^{\rho}_{\mu\lambda} + g_{\mu\rho} \ \delta \Gamma^{\rho}_{\lambda\nu} . \ \ \ \ \ \ (1)$$
Now, to solve this equation for $\delta \Gamma$, we use the (extremely useful) trick of Riemann: permute the indices and write three copies of eq(1), then add two of the equations and subtract the third one. This leads to
$$\nabla_{\mu} \delta g_{\nu \lambda} + \nabla_{\nu} \delta g_{\lambda \mu} - \nabla_{\lambda} \delta g_{\mu\nu} = 2 g_{\lambda \rho} \ \delta\Gamma^{\rho}_{\nu\mu} .$$ Contracting both sides with $g^{\sigma \lambda}$, gives you the required answer
$$\delta \Gamma^{\sigma}_{\nu\mu} = \frac{1}{2}g^{\sigma \lambda} \left(\nabla_{\mu} \delta g_{\nu \lambda} + \nabla_{\nu} \delta g_{\lambda \mu} - \nabla_{\lambda} \delta g_{\mu\nu} \right) .$$

Last edited: Sep 25, 2016
6. Sep 26, 2016

### demon

Hi samalkhaiat, thank you very much for your contribution.

I fail to follow you on the derivation of:

λ(δgμν)=δ(∇λgμν)+gρν δΓρμλ+gμρ δΓρλν

I'd be very grateful if you could show the intermediate steps.

I made a mistake asserting the the covariant derivative and the variation commute. This is true for the partial derivative and the variation though (see e.g. Hobson p529), as you also pointed it out in your post. But I still have a problem with ∇λ(δgμν), and hopefully you can help me with it. My reasoning is: the variation consists of taking an alternative path for the integration, where instead of gμν we have g'μν:

gμν → g'μν=gμν+δgμν ⇒ δgμν = g'μν - gμν

If we take covariant derivatives of δgμν:

λ(δgμν) = ∇λ(g'μν - gμν) = ∇λ(g'μν) - ∇λ(gμν)

But the covariant derivative of the metric is zero, therefore ∇λ(δgμν) must be zero, mustn't it?

Last edited: Sep 26, 2016
7. Sep 26, 2016

### samalkhaiat

From the definition $$\delta g_{\mu\nu} (x) \equiv \bar{g}_{\mu\nu}(x) - g_{\mu\nu}(x) \ ,$$ the following properties follow immediately:
$$\delta \left(f(x) + g(x) \right) = \left(\bar{f} + \bar{g}\right)(x) - \left( f + g\right)(x) = \delta f(x) + \delta g(x) \ , \ \ \ (1)$$
$$\delta \left(f(x) g(x)\right) = \bar{f}(x) \bar{g}(x) - f(x)g(x) \approx f (x) \ \delta g(x) + g(x) \ \delta f(x) \ , \ \ \ (2)$$
$$\delta \left( \partial g(x) \right) = \partial \bar{g}(x) - \partial g(x) = \partial \left( \bar{g} - g \right)(x) = \partial \left(\delta g(x)\right) \ , \ \ \ (3)$$ and (more importantly) the object $\delta g_{\mu\nu}(x)$ is a tensor (being the difference between two tensors at the same $x$). So, like any other tensor, you can calculate the covariant derivative of $\delta g_{\mu\nu}$ from
$$\nabla_{\lambda} \delta g_{\mu\nu} = \partial_{\lambda} \delta g_{\mu\nu} - \Gamma^{\rho}_{\mu\lambda} \ \delta g_{\rho \nu} - \Gamma^{\rho}_{\lambda\nu} \ \delta g_{\mu\rho} \ ,$$ Now, on the RHS, apply property (3) to the first term, property (2) to the remaining $(\Gamma \delta g)$-terms, and finally use (1) to collect the terms with total variations:
$$\nabla_{\lambda} \delta g_{\mu\nu} = \delta \left( \partial_{\lambda}g_{\mu\nu} - \Gamma^{\rho}_{\mu\lambda}g_{\rho\nu} - \Gamma^{\rho}_{\lambda\nu}g_{\mu\rho} \right) + g_{\rho\nu} \ \delta\Gamma^{\rho}_{\mu\lambda} + g_{\mu\rho} \ \delta\Gamma^{\rho}_{\lambda\nu} \ .$$ Now, the first term on the RHS is just $\delta \left( \nabla_{\lambda} g_{\mu\nu}\right)$, and so you arrive at
$$\nabla_{\lambda} \left(\delta g_{\mu\nu}\right) = \delta \left( \nabla_{\lambda}g_{\mu\nu} \right) + g_{\rho\nu} \ \delta\Gamma^{\rho}_{\mu\lambda} + g_{\mu\rho} \ \delta\Gamma^{\rho}_{\lambda\nu} \ . \ \ \ \ \ (4)$$
Of course, nothing in the above derivation say that $g_{\mu\nu}$ is the Riemann metric tensor. What we have shown is the fact that equation (4) holds for arbitrary rank-2 tensor.

No. The metricity condition holds only for the unperturbed metric $g_{\mu\nu}$, i.e., $\nabla g = 0$ does not mean that $\nabla \bar{g} = 0$. In fact, small deformation of the Riemannian geometry means that $$\nabla g = 0 \ \Rightarrow \ \nabla (\bar{g} - \delta g) = 0 \ \Rightarrow \ \nabla \bar{g} = \nabla \delta g \ .$$

8. Sep 27, 2016

### demon

Samalkhaiat,

Thank you very much for both explanations!

Best regards

Last edited: Sep 27, 2016