# Integration by parts in spacetime

1. Nov 9, 2015

### naima

In this paper
we have p18 an integral on space time M. The author takes a 3 dimensional space like Cauchy surface $\Sigma$ which separates M in two regions, the future and the past of $\Sigma$. He gets so the sum of two integrals on these regions. He writes then let us integrate each of them by parts. The fact that $\Sigma$ is a boundary for these regions is obvious. the vector n orthogonal to the boundary occurs in the result but i think that this uses the (- +++) metric. What is the formula used to integrate by parts in relativity?

2. Nov 9, 2015

### HallsofIvy

Staff Emeritus
This is not really a matter of "integration by parts in spacetime" or "in relativity", it is just "integration by parts", a mathematical technique. The idea of integration by parts is that $\int u dv= uv- \int v du$. Here, the integrals are of the form $\int f \psi dvol$. What they are doing is taking $u= \psi$ and $dv= f dvol$ while using the physics fact that a conservative force is the derivative of the potential energy function so $v= E_n$ and $du= \nabla_n\psi$.

3. Nov 9, 2015

### naima

You can read that n is the future-pointing unit normal vector field on $\Sigma$
How does n appear in the result (it needs the metric)?
Thanks

Last edited: Nov 9, 2015
4. Nov 10, 2015

### naima

The result is
$\int_\Sigma (\Phi \nabla_n \Psi - \Psi \nabla_n \Phi) d\Sigma$
Here $\nabla_n$ is the Lie derivative along n.
Is there a rule for integation by parts with Lie derivatives?

5. Nov 24, 2015

### Ssnow

Lie derivative obeys to Leibniz rule so I think will be a integration by parts, if boundary doesn't contribute I think it is $\int_{\Sigma}\Phi\nabla_{n}\Psi\,d\,\sigma=-\int_{\Sigma}\Psi\nabla_{n}\Phi\,d\,\Sigma$