# Integration by parts in curved space time

1. Nov 19, 2015

### naima

In this thread, ramparts asked how integration by parts could be used in general relativity.
suppose you have
$\int_M (\nabla^a \nabla_a f) g .Vol$
Can it be written like
$\int_M (\nabla^a \nabla_a g) f .Vol$ plus a boundary integration term (by integrating twice by parts)?

I think thay it is possible.
look at Benini's paper page 18. I would like to understand the math behind that.

Last edited: Nov 19, 2015
2. Nov 19, 2015

### JorisL

I like to think of it in another way which I learned from fzero a while back.

First you can use that $\nabla^a\nabla_a = \Box$ with $\Box$ the d'Alembertian.
Next write
$\int_M\left[\Box (fg) \right]\text{Vol} = \int_M\left[\Box (f)g + \Box (g)f \right]\text{Vol}$.

Rewriting this as
$\int_M\left[\Box (f)g \right]\text{Vol} = \int_M\left[\Box (fg) \right]\text{Vol} - \int_M\left[ \Box (g)f \right]\text{Vol}$

This is a great way to work with this stuff. Especially when you have differential p-forms in there.

3. Nov 19, 2015

### naima

Are you sure that you consider the metric $h_{\mu \nu}$ in $\nabla^a \nabla_a$?

4. Nov 19, 2015

### JorisL

I don't understand what you mean.
Are the latin indices a subset of the greek ones?
I need a little more context to give a detailed response.

One remark I can make is $\nabla_af = \partial_a f$ for a scalar function f.

5. Nov 19, 2015

### naima

let us take the simpler case of 1+1 space time with this metric
$\begin{matrix} a & b \\ b & c \end{matrix}$
We have to calculate
$\int_M f(t,x) \begin{matrix} (\partial_t & \partial_x)\end{matrix} \begin{matrix} a & b \\ b & c \end{matrix} \begin{matrix} (\partial_t \\ (\partial_x \end{matrix} g(t,x) \sqrt{-ac+b^2}dt.dx$
and in a curved space a, b, and c can depend on t and x.
So it is not obvious to say "i integrate twice by parts"

6. Nov 19, 2015

### JorisL

In fact you forget that the covariant derivatives are chosen in such a way that $\nabla_a g^{bc} = 0$.
That should resolve that part of your confusion.

7. Nov 19, 2015

### haushofer

The covariant derivative obeys the Leibnitz rule and is metric compatible. So yes, performing integration by parts you get your result.

8. Nov 19, 2015

### naima

Can you write the integral just with the $\partial^2_{tt}$ term?(it will be the same for the other terms)
thanks

9. Nov 19, 2015

### naima

I read something that could help:
$g^{\mu \nu} \nabla_\mu \nabla_\nu = (1/\sqrt{-g}) \partial_\mu (\sqrt{-g} \partial^\mu)$
it is here
so the square root disappears in Vol.My notation in #5 for the product of covariant derivative and contravariant derivative was incorrect.

Last edited: Nov 19, 2015
10. Nov 19, 2015

### samalkhaiat

The covariant divergence of a vector field $V^{a} (x)$ is given by the identity

$$\sqrt{-g} \ \nabla_{a}V^{a} = \partial_{a} \left(\sqrt{-g} V^{a} \right) .$$

Let $\Omega$ be some n-dimensional region with boundary $\partial \Omega$. Let $\Phi(x)$ and $\Psi(x)$ be smooth scalar fields. Set

$$V^{a}(x) = g^{ab}\ \nabla_{b}\Phi (x) = g^{ab}\ \partial_{b}\Phi (x) .$$

Put (2) in (1)

$$\sqrt{-g} \ g^{ab} \ \nabla_{a}\nabla_{b}\Phi (x) = \partial_{a} \left( \sqrt{-g} \ g^{ab} \ \partial_{b}\Phi (x) \right) .$$

Multiply (3) with $\Psi(x)$ and integrate over the region $\Omega$

$$\int_{\Omega} d^{n} x \ \sqrt{-g} \left( \nabla^{a}\nabla_{a}\Phi (x) \right) \ \Psi(x) = \int_{\Omega} d^{n} x \ \partial_{a} \left( \sqrt{-g} \ g^{ab} \ \partial_{b}\Phi (x) \right) \ \Psi(x) .$$

Integrate the RHS by part

$$\begin{equation*} \int_{\Omega} d^{n} x \ \sqrt{-g} \left( \nabla^{a}\nabla_{a}\Phi \right) \ \Psi = \int_{\Omega} d^{n} x \ \partial_{a} \left( \Psi \ \sqrt{-g} \ g^{ab} \ \partial_{b}\Phi \right) - \int_{\Omega} d^{n} x \ \partial_{b}\Phi \left( \sqrt{-g} \ g^{ab} \ \partial_{a}\Psi \right) . \end{equation*}$$

Repeat integration by part on the second integral on the RHS

$$\begin{equation*} \begin{split} \int_{\Omega} d^{n} x \ \sqrt{-g} \left( \nabla^{a}\nabla_{a}\Phi (x) \right) \ \Psi (x) =& \int_{\Omega} d^{n} x \ \partial_{a} \left( \sqrt{-g} \ g^{ab} \left( \Psi (x) \ \partial_{b}\Phi (x) - \Phi (x) \ \partial_{b} \Psi (x) \right) \right) \\ & + \int_{\Omega} d^{n} x \ \Phi (x) \ \partial_{a} \left( \sqrt{-g} \ g^{ab} \ \partial_{b} \Psi (x) \right) . \end{split} \end{equation*}$$

Using Stokes' theorem, the first integral on the RHS can be converted to integral over the boundary. In the second integral, use (3)

$$\begin{equation*} \begin{split} \int_{\Omega} d^{n} x \ \sqrt{-g} \left( \nabla^{a}\nabla_{a}\Phi (x) \right) \ \Psi (x) =& \int_{\partial \Omega} d \Sigma^{a} \ \sqrt{-g} \ \left( \Psi \ \partial_{a}\Phi - \Phi \ \partial_{a} \Psi \right) \\ & + \int_{\Omega} d^{n} x \ \sqrt{-g} \ \Phi (x) \ \left( \nabla^{a}\nabla_{a} \Psi (x) \right) . \end{split} \end{equation*}$$

Finally, if the scalar fields vanish on the boundary, you get

$$\int_{\Omega} d^{n} x \ \sqrt{-g} \left( \nabla^{a}\nabla_{a}\Phi (x) \right) \ \Psi (x) = \int_{\Omega} d^{n} x \ \sqrt{-g} \ \Phi (x) \ \left( \nabla^{a}\nabla_{a} \Psi (x) \right) .$$

Last edited: Nov 26, 2015
11. Nov 20, 2015

### naima

Thank you very much for this answer.
In fact in the paper after two integrations by parts the only remaining things are the integrals on the boundary. (the remaining integral on $\Omega$ contains something which vanishes because it obeys the generalized Klein Gordon equation.
Thanks again

12. Nov 20, 2015

### samalkhaiat

You welcome. I did not look at any paper, I just answered your original question.

13. Nov 23, 2015

### naima

I have no problem before that.

If we start from a manifold M having some boundary $\partial M$ it maydescribed by a set of overlapping maps. An open set V of the manifold which does not intersect $\partial M$ can be mapped to $R^4$. In this map we can integrate by parts but we have no natural $\Sigma^a$ boundaries. Have we to consider boundaries at infinity and what are $d\Sigma$ at infinity?

14. Nov 25, 2015

### samalkhaiat

What do you mean by “can do integration by part”? You can always do that. If you are asking about the application of Stokes’ theorem, then one can say few words about it. Basically, it is all about the topological properties of $\partial\Omega$. In order to avoid complications, which can arise from the “corners” of $\partial\Omega$, one can take $\partial\Omega$ to be a “tube” consisting of two spacelike hypersurfaces $\Sigma_{t_{1}}$ and $\Sigma_{t_{2}}$ having the same topology, and a connecting timelike element $\mathcal{T}$
$$\partial\Omega = \Sigma_{t_{1}} \cup \Sigma_{t_{2}} \cup \mathcal{T} .$$
It is usually assumed that spacetime is foliated into spacelike hypersurfaces $(\Sigma_{t})_{t \in \mathbb{R}}$ defined by $x^{0}= t = \mbox{constant}$, where $t : M^{(1,n-1)} \to \mathbb{R}$ is a well defined time function throughout space time $$M^{(1,n-1)}= \bigcup_{t \in \mathbb{R}}\Sigma_{t}.$$
($\Sigma_{t_{1}}$ and $\Sigma_{t_{2}}$ are the initial and final Cauchy surfaces). This induces another foliation on the boundary element $\mathcal{T}= \mathbb{R} \times \mathcal{S}^{(n-2)}$, where $\mathcal{S}^{(n-2)}$ are $(n-2)$-dimensional spacelike surfaces given by $$\mathcal{S}^{(n-2)} = \partial \Sigma_{t} = \Sigma_{t} \cap \mathcal{T}.$$ (The boundary $\partial \Sigma_{t}$ of each $\Sigma_{t}$ (leaf) lies in the boundary element $\mathcal{T}$; the intersections of $\Sigma_{t}$ leaves with $\mathcal{T}$ induce a foliation of $\mathcal{T}$)
The surface element differential $d\Sigma_{\mu}$ on $\partial \Omega$ is given by
$$d\Sigma_{\mu} = d\Sigma \ n_{\mu} = d^{n-1}x \ \sqrt{|g^{(n-1}|} \ n_{\mu} ,$$ where $d\Sigma$ is “volume” form on $\partial\Omega$, that is to say that the $\epsilon$-tensor on $\partial\Omega$ is given by $$\epsilon_{\mu_{1}\mu_{2}\cdots \mu_{n-1}} = n^{\nu}\epsilon_{\nu \mu_{1}\mu_{2}\cdots \mu_{n-1}},$$
$n^{\mu}$ is the outward pointing normal to $\partial\Omega$, and $g^{(n-1)}$ is the determinant of the induced metric on $\partial\Omega$. Now, we have all the ingredients to do Stokes’ theorem
$$\int_{\Omega \subset M} \epsilon \ (\nabla_{\mu}V^{\mu}) = \int_{\partial \Omega} (\epsilon \cdot V) ,$$ or, explicitly
\begin{align*} \int_{\Omega} d^{n}x \ \sqrt{|g|} \ \nabla_{\mu}V^{\mu} &= \int_{\partial\Omega} d^{n-1}x \ \sqrt{|g^{(n-1)}|} \ n_{\mu}V^{\mu} \\ &= \left( \int_{\Sigma_{t_{1}}} + \int_{\Sigma_{t_{2}}} + \int_{\mathcal{T}} \right) d^{n-1}x \ \sqrt{|g^{(n-1)}|} \ n_{\mu}V^{\mu} . \end{align*}
Okay, now take
\begin{align*} n_{\mu}(\Sigma_{t_{1}}) &= (-1, 0, 0, \cdots , 0) \\ n_{\mu}(\Sigma_{t_{2}}) &= (1, 0, 0, \cdots , 0) , \ \mbox{and} \\ n_{\mu}(\mathcal{T}) &= (0, 1, 1, \cdots , 1) , \end{align*}
and use $\mathcal{T} = \mathbb{R} \times \mathcal{S}^{(n-2)}$ to obtain
$$\begin{equation*} \begin{split} \int_{\Omega} d^{n}x \ \sqrt{|g|} \ \nabla_{\mu}V^{\mu}(x) =& \int_{\Sigma_{t_{1}}}^{\Sigma_{t_{2}}} d^{(n-1}\vec{x} \ \sqrt{|g^{(\Sigma_{t})}|} \ V^{0}(x^{0} , \vec{x}) \\ & + \int_{t_{1}}^{t_{2}} dx^{0} \ \int d^{(n-2)}\vec{x} \ \sqrt{|g^{(\mathcal{S})}|} \ n_{j}V^{j}( x^{0}, \vec{x}) . \end{split} \end{equation*}$$
If the vector $V^{\mu}$ is built out of physical fields, we are most likely to having
$$|\vec{x}|^{n-2} V^{j} \to 0 , \ \ \mbox{as} \ \ |\vec{x}| \to \infty .$$ Therefore
$$\int_{\Omega} d^{n}x \ \sqrt{|g|} \ \nabla_{\mu}V^{\mu}(x) = \int_{\Sigma_{t_{1}}}^{\Sigma_{t_{2}}} d \vec{x} \ \sqrt{|g^{(\Sigma_{t})}|} \ V^{0}(t , \vec{x}) .$$ Furthermore, if it is covariantly conserved $\nabla_{\mu}V^{\mu}=0$, we obtain a time-independent charge
\begin{align*} Q(t) &= \int_{\Sigma_{t_{1}}} d \vec{x} \ \sqrt{|g_{(\Sigma_{t})}|} \ V^{0}(t,\vec{x}) \\ &= \int_{\Sigma_{t_{2}}} d \vec{x} \ \sqrt{|g_{(\Sigma_{t})}|} \ V^{0}(t,\vec{x}) \\ &= \int_{\Sigma_{t}} d \vec{x} \ \sqrt{|g_{(\Sigma_{t})}|} \ V^{0}(t,\vec{x}) . \end{align*}
All of this should be familiar from Noether theorem and/or from the word go in general relativity. The starting statement in GR is that of the invariance of the H-E action under the group of general coordinate transformations, i.e. under a diffeomorphism generated by the Lie derivative along a vector field $\xi^{\mu}$ that is tangent to the boundaries:
$$\mathscr{L}_{\xi} \left( \int_{\Omega} d^{4}x \sqrt{|g|} R \right) = \int_{\Omega} \mathscr{L}_{\xi} (d^{4}x \sqrt{|g|}) R + \int_{\Omega} d^{4}x \ \sqrt{|g|} \mathscr{L}_{\xi}(R) ,$$
or
\begin{align*} \delta S &= \int_{\Omega} d^{4}x \sqrt{|g|} \left( R \nabla_{\mu}\xi^{\mu} + \xi^{\mu} \nabla_{\mu}R \right) \\ &= \int_{\Omega} d^{4}x \ \nabla_{\mu}(\xi^{\mu} \sqrt{|g|}R) . \end{align*}
Using, Stokes' theorem, we get
$$\begin{equation*} \begin{split} \delta S =& \int_{\Sigma_{1}}^{\Sigma_{2}} d^{3}x \sqrt{|g^{(3)}(\Sigma)|} \ ( n_{\mu} \xi^{\mu} ) R \\ & + \int_{\mathcal{T}} d^{3}x \sqrt{|g^{(3)}(\mathcal{T})|} \ ( n_{\mu} \xi^{\mu}) R , \end{split} \end{equation*}$$
And this vanishes because of $(\xi^{\mu}n_{\mu})|_{\partial \Omega} = 0$.

Last edited: Nov 26, 2015