# Clarification on integrating over 3+1 spacetime

Of the following 4 equations (containing some arbitrary function $f^{\nu}$): \begin{equation} \int d^4x \partial_\nu f^{\nu} = f^{\nu} \\ \int d^4x \nabla_\nu f^{\nu} = f^{\nu} \\ \int d^4x \sqrt{-g} \partial_\nu f^{\nu} = f^{\nu} \\ \int d^4x \sqrt{-g} \nabla_\nu f^{\nu} = f^{\nu} \end{equation} are lines #1 and #4 correct (and lines #2 and #3 are incorrect)?

I think this is true because line 1 is an integral over a simple minkowski spacetime (the volume element here is $d^4x$ and the derivative cancels the integral).

In line 4 where we have a more interesting spacetime (described by metric $g_{\mu\nu}$ instead of Minkowski metric $\eta_{\mu\nu}$) the operator $\nabla_{\nu}$ replaces the derivative $\partial_{\nu}$ and the volume element $d^4x$ is replaced by a more general volume element $d^4x\sqrt{-g}$.

I would like to know: 1) If I am correct that lines 1 and 4 are true and 2) is my explanation sufficient, or is there a more elegant way to understand why this is.

• Besides, all of these look false. – Ryan Unger Jun 16 '17 at 21:32
• None of these equations are correct. – Prahar Jun 16 '17 at 21:33
• @Bob - No. They are just completely wrong. – Prahar Jun 16 '17 at 21:34
• @Bob - There is nothing to elaborate here. I have no idea what type of equation you are even trying to write down. What do you mean by $\int d^4 x$ here? Is it a definite integral over a certain volume? If so, you should be writing $\int_V d^4 x$. Do you instead mean it as an indefinite integral of some sort? – Prahar Jun 16 '17 at 21:38
• An integral is a number, and your RHS's are functions. How could they be equal? – Ryan Unger Jun 16 '17 at 21:54

For the Minkowski space, $\displaystyle \int_{spacetime} d^4x \partial_\nu f^\nu =0$ assuming the $f^\nu$ vanishes at infinity sufficiently rapidly. Because it would be the surface integration of $f^\nu$ over the surface which is boundary to the whole of spacetime (by Gauss's theorem).

Similarly, in a general spacetime, $\displaystyle \int_{spacetime} d^4x \sqrt{-g}\nabla_\nu f^\nu =0$.

If you take the Minkowskian space then $\sqrt{-g} = 1$, $\nabla_\nu=\partial_\nu$ and thus, for Minkowski space, you can also write (if you really want to I mean) $\displaystyle \int_{spacetime} d^4x \nabla_\nu f^\nu =0 = \displaystyle \int_{spacetime} d^4x \sqrt{-g} \partial_\nu f^\nu$.

• Thanks for the response. If I write $f^\nu=g^{\mu\nu} A_\mu \delta\phi$ (where $A_\mu$ is combination of terms that contains only one downstairs index), than would it be correct to say $\int d^4x \partial_\nu f^\nu = 0$ since at the boundary we expect any variation ($\delta\phi$) to be zero? – Bob Jun 16 '17 at 22:10
• (Assuming that $\phi$ is a scalar) Yes if you demand $\delta \phi$ to be zero (and $A^\nu$ to be finite) on the boundary and the boundary is finite. If the boundary is infinite then you should demand $f^\nu$ to drop sufficiently rapidly at large distances. You can't simply say that vanishing $\delta \phi$ at the boundary can save you because in that case the boundary area would be infinite. So, you need to demand rapid enough fall of $f^\nu$ so that you can be sure of vanishing surface integration term. – Dvij Mankad Jun 16 '17 at 22:24

These equations are nonsensical. Assuming these are definite integrals, the integrals evaluate to a single number, not to a function. I think you want the LHS volume integrals to be over some finite volume $V$ and the RHS to be a surface integral over the boundary of that volume.

Based on your comments the identity you are looking for is the following $$\int_{\cal V} d^4 x \sqrt{-g} \nabla_\mu f^\mu = \int_{\cal V} d^4 x \partial_\mu \big( \sqrt{-g} f^\mu \big) = \int_{\partial \cal V} d^3 y \sqrt{|h|} \, n_\mu f^\mu$$ Here, $\cal V$ is the closed 4-volume over which you are performing the integral and $\partial \cal V$ is the boundary of this volume.

$n^\mu$ is the outward pointing unit normal vector (if the $\partial \cal V$ is time-like, then $n^2 = +1$ and if $\partial \cal V$ is spacelike then $n^2 = -1$. If $\partial\cal V$ is null, then it's a little bit trickier).

$y^i$ are arbitrarily chosen coordinates on $\partial \cal V$.

$h$ is the determined of the induced metric on $\partial \cal V$ (Again things are a little trickier if $\partial \cal V$ is a null hypersurface).

• Basically, it sounds like I just need to apply Gauss' theorem to the left hand side of my 4 equations? (At this point I agree we should ignore the nonsense I wrote on the RHS.) – Bob Jun 16 '17 at 22:45
• yes. The identity I wrote is Stokes' theorem and this is what you need to use. – Prahar Jun 16 '17 at 22:46
• thanks. I appreciate you taking the time to help me understand better – Bob Jun 16 '17 at 22:48
• You need to take $n$ pointing inward if $\partial\mathcal V$ is timelike. – Ryan Unger Jun 17 '17 at 0:08