Suppose we are given a type $(0,2)$ tensor $T_{\mu\nu}$ in a Minkowski space with $(-,+,+,+)$ signature. Consider a closed 3-dimensional hypersurface $\partial \Omega$ which encloses a volume $\Omega$ in the Minkowski space-time. We have the following identity:$$\int_{\partial\Omega}T_{\mu\nu}n^\nu~dvol_3=0\tag{1}$$where $n^\nu$ is the contravariant normal vector pointing out of the hyperspace $\partial \Omega$. Then by the Divergence Theorem in higher dimensions, we could transform the above integral to $$\int_{\Omega}\text{div}~T_{\mu\nu}~dvol_4 \tag{2}$$

My questions are as follows:

  • Am I doing Gauss' Divergence Theorem (especially with regard to notation) correctly from equation (1) to equation (2)? I figure that as we have a 2-form in front of a contravariant, I might have made a mistake in keeping $T_{\mu\nu}$ as it is. Do we need to swap $T_{\mu\nu}$ to $T_\mu^\nu$ and why?

  • If I am doing (1) to (2) correctly, then what is the index notation expression for $\text{div}~T_{\mu\nu}$? I intend to say it is $\partial_{\nu}T_{\mu\nu}$ but it is in fact $\partial^{\nu}T_{\mu\nu}$. Why is this the case?

  • 1
    $\begingroup$ I don't quite understand the setup to your question. If $\partial \Omega$ fully encloses some region of spacetime, then "$\mathrm d^3x$ integrates over the spatial coordinates ..." doesn't make sense. That would be fine if it were some spacelike slice which is coordinatized by $(x,y,z)$, but a closed boundary isn't a slice. $\endgroup$
    – J. Murray
    Oct 12, 2021 at 17:31
  • $\begingroup$ the statement about $d^3x$ is meant to indicate that time component is not included, the range of integral is written underneath $\int$ which is the boundary of $\Omega$ in spatial$R^3$ $\endgroup$
    – Rescy_
    Oct 12, 2021 at 17:39
  • $\begingroup$ If you are integrating over the boundary, how can you not include the time coordinate? As a lower-dimensional example, that's like looking at the disk in the $(x,y)$-plane and saying that you're going to integrate over its boundary (a circle) without including the $y$-coordinate. $\endgroup$
    – J. Murray
    Oct 12, 2021 at 17:44
  • $\begingroup$ You are right, I will correct my description. instead of $d^3x$ i will change that to $dvol_3$ which is an analogy to surface area element in lower dimensions. The range of integral should then correctly represent what i mean. $\endgroup$
    – Rescy_
    Oct 12, 2021 at 17:54

1 Answer 1


On a $n$-dimensional manifold, you can't integrate anything other than a $n$-form. Therefore if you already have a volume $3$-form on a 3D manifold $\partial\Omega$ you can't also have a non-scalar laying around. Your integration is just poorly defined, that's why you have trouble making sense of what you wrote.

Stokes' Theorem (the more general form of Gauss' Divergence Theorem) states that, if $\mathcal{M}$ is a $n$-dimensional manifold and $\omega\in\Lambda^{n-1}(\mathcal{M} )$, then $$\int_\mathcal{M}d\omega=\int_\mathcal{\partial M}\omega$$ and if $V\in\Lambda^{1}(\mathcal{M} )$ then applying Stokes' Theorem to its hodge dual in the case of flat metric yields the familiar Gauss' Divergence Theorem $$\int_\mathcal{M}\partial_\mu V^\mu d^nx=\int_\mathcal{\partial M}(V\cdot n)d^{n-1}x$$ where $n$ is the normal vector pointing out of $\mathcal{\partial M} $. Notice how in each of these integrals the integrand is always a form of the same degree as the manifold's dimension.

Also, I think what is commonly called "divergence" $\text{div}$ should only be applied to vectors: $$\text{div}(V)=\frac{1}{\sqrt{-g}}\partial_\mu\left(\sqrt{-g}V^\mu\right)$$ but I might be wrong about this last part.


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