Issues with the variation of Christoffel symbols

[JuanC97]

Hello everyone,
I'm sure a lot of you know that the Christoffel symbols are not tensors by themselves but, their variation is a tensor.
I want to revive a post that was made in 2016 about this: The Variation of Christoffel Symbol and ask again "How is that you can calculate ∇_ρδg_μν if δ{g_μν} is not a tensor?"

You know that δg_μν can be written as (- g_μα g_νβ δg^αβ) since δ{δ_μ^ν}=0 so... at first glance, you could say that δg_μν is not a tensor since you can't lower its indices just by using two metrics BUT the truth is that those indices are not the indices of the variation so you can't just think like that, it is, in fact, a matter of notation as I will explain now:

As far as I've tought, δg_μν stands for "the variation of the components" of the metric ( i.e. δ{g_μν} )
and that's really different from "the components of the variation" ( i.e. (δg)_μν - which, of course, would be tensorial since that's just the difference of two metrics ), take this into account:

δg = δ{g_μνdx^μ⊗dx^ν} = δ{g_μν}dx^μ⊗dx^ν + g_μνδ{dx^μ⊗dx^ν} so (δg)_μν is, in general different from δ{g_μν}.

Now...maybe the problem is just about notation...
The issue that really makes me think twice about it is... What does it really mean that covariant derivative of δg_μν that you found in the formula for the variation of the Christoffel symbols? and, even more, how can you interpret that equation in the proper form?

Could you please rephrase ∇_ρδg_μν in a different notation or in words, or maybe interpreting as multilinear maps?.

Thanks btw.

 

[JuanC97]

UPDATE: I've been reading some thing and made some conclusions, help me to understand this please.
https://anhngq.wordpress.com/2010/0...ariant-derivative-of-tensors-in-a-short-form/.

First of all, by ∇_ρδg_μν we refer to the scalar that results from the evaluation of (∇_∂ρδg)(∂_μ,∂_ν).
Then, you should interpret (∇_∂ρδg)(∂_μ,∂_ν) as the covariant derivative of the metric tensor along ∂_ρ evaluated in the basis given by ∂_μ and ∂_ν .

Using the properties of the covariant derivative you can find that (∇_∂ρδg)(∂_μ,∂_ν) is given by
∂_ρδg(∂_μ,∂_ν) - δg(∇_∂ρ∂_μ,∂_ν) - δg(∂_μ,∇_∂ρ∂_ν)
= ∂_ρδg(∂_μ,∂_ν) - δg( Γ^λ_ρμ∂_λ,∂_ν) - δg(∂_μ, Γ^λ_ρν∂_λ)
= ∂_ρδg(∂_μ,∂_ν) - Γ^λ_ρμδg( ∂_λ,∂_ν) - Γ^λ_ρνδg(∂_μ,∂_λ)
= ∂_ρδg_μν - Γ^λ_ρμδg_λν - Γ^λ_ρνδg_μλ

The conclusion is that ∇_ρδg_μν = ∂_ρδg_μν - Γ^λ_ρμδg_λν - Γ^λ_ρνδg_μλ has to be interpreted as the components of ∇_∂ρδg so... this is a tensor (?)

Hmmm, and... what would happen if I use the same reasoning with δg(∂_μ,∂_ν) = δg_μν ?
As I see it, δg(∂_μ,∂_ν) stands for the components of δg so δg_μν is a tensor (?).

 

[haushofer]

Just my 2 cents: the variation of the metric is the difference between two metric tensors in the same coordinate system and hence a tensor. Take e.g. the Lie derivative of the metric as explicit example.

 

[JuanC97]

Just my 2 cents: the variation of the metric is the difference between two metric tensors in the same coordinate system and hence a tensor. Take e.g. the Lie derivative of the metric as explicit example.

I agree with you when you said that (δg) is a tensor but we have to remember that we also have this two equations from the literature that are always used when computing variations:

(i) (∇_∂ρδg_μν) = ∂_ρδg_μν - Γ^λ_ρμδg_λν - Γ^λ_ρνδg_μλ
states that δg_μν is the component of a tensor and this relation can be deduced from its interpretation as (δg)_μν.
(ii) δg_μν = - g_μα g_νβ δg_αβ
states that δg_μν is not the component of a tensor and this relation can be deduced from its interpretation as δ{g_μν}.

Also, taking into account that
δg = δ{g_μνdx^μ⊗dx^ν} = δ{g_μν}dx^μ⊗dx^ν + g_μνδ{dx^μ⊗dx^ν}
and the second term in the right-hand side is, in general, different from 0, it's clear that (δg)_μν ≠ δ{g_μν} so both interpretations are, in principle, uncompatible. The question is then... Why we usually use both of this formulas?, maybe I am misreading something?.

 

[stevendaryl]

It seems to me that there is no real distinction between $(\delta \boldsymbol{g})_{\mu \nu}$ and $\delta g_{\mu \nu}$. As you say,

$\boldsymbol{g} = g_{\mu \nu} dx^\mu \otimes dx^\nu$

So if we vary $\boldsymbol{g}$, you get:

$\delta \boldsymbol{g} = \delta g_{\mu \nu} dx^\mu \otimes dx^\nu + g_{\mu \nu} \delta(dx^\mu \otimes dx^\nu)$

But you're not varying your coordinate system, so $dx^\mu$ isn't varying. So the last term is zero.

 

[stevendaryl]

I think the confusion is that $g^{\alpha \beta}$ should really be written as $(g^{-1})^{\alpha \beta}$. So both $\delta \boldsymbol{g}$ and $\delta (\boldsymbol {g}^{-1})$ are tensors, but they are different tensors. They are related by:

$\delta (\boldsymbol{g}^{-1})^{\mu \nu} = - g^{\mu \alpha} g^{\nu \beta} \delta \boldsymbol {g}_{\alpha \beta}$

In other words, $\delta (\boldsymbol{g}^{-1})_{\mu \nu} \neq \delta \boldsymbol {g}_{\mu \nu}$. They are actually the negatives of each other.

 

[PeterDonis]

In other words, $\delta (\boldsymbol{g}^{-1})_{\mu \nu} \neq \delta \boldsymbol {g}_{\mu \nu}$. They are actually the negatives of each other.

Did you mean to write the indexes on $\delta (\boldsymbol{g}^{-1})_{\mu \nu}$ as lower indexes here? They were upper indexes in your earlier formula.

 

[JuanC97]

Did you mean to write the indexes on $\delta (\boldsymbol{g}^{-1})_{\mu \nu}$ as lower indexes here? They were upper indexes in your earlier formula.

Peter, he is saying that $\, (\delta\boldsymbol{g^{-1}}) \,$ is a tensor, hence:
$(\delta\boldsymbol{g^{-1}})^{\mu \nu} \equiv g^{\mu \alpha} g^{\nu \beta} (\delta\boldsymbol{g^{-1}})_{\alpha \beta}$

but we've found that
$\delta\{ \delta_\mu^\nu \} = \delta\{ g_{\mu\alpha} (\boldsymbol{g^{-1}})^{\alpha\nu} \} = 0 \;\Leftrightarrow\; (\delta\boldsymbol{g^{-1}})^{\mu \nu} = - g^{\mu \alpha} g^{\nu \beta} \delta g_{\alpha \beta} \hspace{0.3cm}$, so $\,\, (\delta\boldsymbol{g^{-1}})_{\alpha \beta} = - \delta g_{\alpha \beta}$

I agree with this point of view but I'm still dealing with the argument of $\delta\{ dx^\mu \} = 0$ and $\delta\{ \partial_\mu \}=0$ since, I think about $\delta$ as a general variation (it could be a variation due to a change in coordinates or anything else) which makes it difficult to me to understand why the basis (co-)vectors would remain unchanged under any kind of variation.

 

[stevendaryl]

Did you mean to write the indexes on $\delta (\boldsymbol{g}^{-1})_{\mu \nu}$ as lower indexes here? They were upper indexes in your earlier formula.

No, I meant for them to be lowered. if $\delta(\boldsymbol g^{-1})$ is a tensor, then we can use $g$ to raise and lower its indices. So

$(\delta(\boldsymbol g^{-1}))_{\mu \nu} \equiv g_{\mu \alpha} g_{\nu \beta} (\delta(\boldsymbol g^{-1}))^{\alpha \beta} = - (\delta \boldsymbol{g})_{\mu \nu}$

 

[stevendaryl]

I agree with this point of view but I'm still dealing with the argument of $\delta\{ dx^\mu \} = 0$ and $\delta\{ \partial_\mu \}=0$ since, I think about $\delta$ as a general variation (it could be a variation due to a change in coordinates or anything else) which makes it difficult to me to understand why the basis (co-)vectors would remaining unchanged under any kind of variation.

You can certainly allow more general variations. But in the context of variational principles in deriving Einstein's field equations, the way that I think of it is this:

• You have a candidate metric, $\boldsymbol{g}$
• You consider a second tensor, $\boldsymbol{\bar{g}}$ that is infinitesimally different than $\boldsymbol{g}$
• You subtract them to get $\delta \boldsymbol{g} \equiv \boldsymbol{\bar{g}} - \boldsymbol{g}$
• You calculate the effect on the action to first-order in $\delta \boldsymbol{g}$.
• Your candidate $\boldsymbol{g}$ is the real metric if the first-order variation vanishes.

 

[haushofer]

In the euler lagrange eqns you only consider functional changes, not coordinate ones. That's why variations and partial derivatives commute. In the Lie derivative you consider the difference between two tensors in the same coordinate system. I can't think of an explicit example of more general ones which I have encountered.