Linearising Christoffel symbols

[chartery]

Carroll linearising by perturbation $g_{\mu\nu}=\eta_{\mu\nu}+h_{\mu\nu}$ has: (Notes 6.4, Book 7.4)

$\Gamma^{\rho}_{\mu\nu}=\frac{1}{2}g^{\rho\lambda}\left( {\partial_{ \mu}}g_{\nu\lambda}+{\partial_{ \nu}}g_{\lambda\mu}-{\partial_{ \lambda}}g_{\mu\nu}\right)=\frac{1}{2}\eta^{\rho\lambda}\left( {\partial_{ \mu}}h_{\nu\lambda}+{\partial_{ \nu}}h_{\lambda\mu}-{\partial_{ \lambda}}h_{\mu\nu}\right)$

This must mean that ${\partial_{ \mu}}h_{\nu\lambda}$ is taken to be of same order as $h^{\rho\lambda}$
I can't find a justification anywhere, so I guess everyone thinks it self-evident.
Is it certain that a weak gravitational field cannot vary quickly or 'strongly' ?

 

[Orodruin]

It is essentially the matrix equivalent of
$$
\frac{1}{1+x} \simeq 1 - x
$$
for small $x$

 

[chartery]

It is essentially the matrix equivalent of
$$
\frac{1}{1+x} \simeq 1 - x
$$
for small $x$

Sorry, you've lost me. Were you referring to $g^{\mu\nu}=\eta^{\mu\nu}-h^{\mu\nu}$?

My problem was how to know that the partial derivative (i.e. variation) of a small item was necessarily also small.
if $\frac{1}{1+x}$ is how I should think of $\partial_{\mu}$ here, I'm afraid I need extra guidance.

 

[vanhees71]

It becomes more clear if you write
$$g_{\mu \nu}=\eta_{\mu \nu} + \epsilon h_{\mu \nu} \qquad (1)$$
with $\epsilon=const.$ and taking all quantities of interest only up to linear order in $\epsilon$.

For the inverse metric you have
$$g^{\mu \nu}= \eta^{\mu \nu} -\epsilon h^{\mu \nu}+\mathcal{O}(\epsilon^2)$$
with
$$h^{\mu \nu}=\eta^{\mu \rho} \eta^{\nu \sigma} h_{\rho \sigma}, \qquad (2)$$
because then
$$(\eta^{\mu \nu} - \epsilon h^{\mu \nu}+\mathcal{O}(\epsilon^2))(\eta_{\nu \sigma} + \epsilon h_{\nu \sigma}= \delta_{\sigma}^{\mu} - \epsilon {h^{\mu}}_{\sigma}) + \epsilon {h^{\mu}}_{\sigma}+\mathcal{O}(\epsilon^2) = \delta_{\sigma}^{\mu} + \mathcal{O}(\epsilon^2).$$
For the Christoffels you plug (1) and (2) in the definition equation and immediately see that they are of order $\epsilon$ and given at this order by the equation in the OP.

 

[Orodruin]

I may have read your OP a bit quick. There are a couple of things to consider here.

The foremost one is that the derivatives of eta vanish and the derivatives you have left are multiplied by essentially eta+O(h). Regardless of the derivatives of h are in terms of size, multiplying them by h is going to give you something of higher order than multiplying by eta.

Apart from that, further assumptions on slow variations etc are common.

 

[chartery]

I may have read your OP a bit quick. There are a couple of things to consider here.

The foremost one is that the derivatives of eta vanish and the derivatives you have left are multiplied by essentially eta+O(h). Regardless of the derivatives of h are in terms of size, multiplying them by h is going to give you something of higher order than multiplying by eta.

Apart from that, further assumptions on slow variations etc are common.

Sorry @vanhees71 I can't get the multiple quote insert to work!Yes, my problem was being sure that $h^{\rho\lambda}{\partial_{ \mu}}h_{\nu\lambda}$ terms were order $h^2$

It makes sense that $\epsilon h^{\rho\lambda} {\partial_{ \mu}}\epsilon h_{\nu\lambda}$ would be order $\epsilon^2$

Many thanks

 

[Orodruin]

Sorry @vanhees71 I can't get the multiple quote insert to work!Yes, my problem was being sure that $h^{\rho\lambda}{\partial_{ \mu}}h_{\nu\lambda}$ terms were order $h^2$

It makes sense that $\epsilon h^{\rho\lambda} {\partial_{ \mu}}\epsilon h_{\nu\lambda}$ would be order $\epsilon^2$

Many thanks

I mean, your worry is partially justified. There is nothing a priori stopping $\partial h$ to be order $1/\epsilon$ in the above. However, it will always be the case that - regardless of the order of the derivative - the h-term in front will be one order higher than the leading one.

 

[chartery]

I mean, your worry is partially justified. There is nothing a priori stopping $\partial h$ to be order $1/\epsilon$ in the above. However, it will always be the case that - regardless of the order of the derivative - the h-term in front will be one order higher than the leading one.

Sorry for gap. I can see Vanhees understands, though it seems to me if $\partial h$ is order $1/\epsilon$ then $\epsilon h^{\rho\lambda} {\partial_{ \mu}}\epsilon h_{\nu\lambda}$ is only order $\epsilon$ but needs to be order $\epsilon^2$ to be ignored in OP equation?

 

[Orodruin]

Sorry for gap. I can see Vanhees understands, though it seems to me if $\partial h$ is order $1/\epsilon$ then $\epsilon h^{\rho\lambda} {\partial_{ \mu}}\epsilon h_{\nu\lambda}$ is only order $\epsilon$ but needs to be order $\epsilon^2$ to be ignored in OP equation?

If that is the case then the leading term in OP's equations is $\mathcal O(1)$, not $\mathcal O(\epsilon)$. Therefore, to leading non-trivial order, terms $\mathcal O(\epsilon)$ should be ignored.

 

[chartery]

If that is the case then the leading term in OP's equations is $\mathcal O(1)$, not $\mathcal O(\epsilon)$. Therefore, to leading non-trivial order, terms $\mathcal O(\epsilon)$ should be ignored.

Durr... Got fixated on second term of $\eta^{\rho\lambda} {\partial_{ \mu}}\epsilon h_{\nu\lambda} - \epsilon h^{\rho\lambda} {\partial_{ \mu}}\epsilon h_{\nu\lambda}$ (just in case someone of similar density looking up).
Many thanks.