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I've fallen accross the following curious property (in p.10 of these lectures): in order to be able to apply Stokes theorem in Lorentzian manifolds, we must take normals to the boundary of the volume we integrate on that are :

  • inward pointing (with respect to the interior of the volume I guess, as usually), if the boundary is timelike (ie tangent vectors are so)
  • outward pointing, if the boundary is spacelike.

Why is it so? What would be wrong if we had a sphere on two-dimensional Minkowski spacetime and define a normal as if the metric was euclidean? (I've checked Carroll's book "Spacetime and Geometry", where I also found the above statement, but no explanation, and the example he gives (p. 456) still puzzles me, as I do not understand why in the last line of eq.E.19, the Q appearing do not have the same sign as the one of their definition on the preceding page.)

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  • $\begingroup$ OK. But still, I'm puzzled by the definition of inward and outward vector. Your previous definition talked just about existence of a certain curve, but if this curve can be defined across your surface and you go along your curve by restricting your parametrization either before encountering the surface or after, an inward vector can then be outward (I made a sketch of what I mean here. Or I missed something. $\endgroup$ – faero Mar 1 '16 at 23:17
  • $\begingroup$ In your second picture the curve is not a map into $M$. It maps "outside" of the manifold. $\endgroup$ – Ryan Unger Mar 1 '16 at 23:26
  • $\begingroup$ Right, this definition is now clear. $\endgroup$ – faero Mar 1 '16 at 23:30
  • $\begingroup$ Which book of Carrolls is this? $\endgroup$ – Mozibur Ullah Mar 2 '16 at 0:03
  • $\begingroup$ I updated my answer with the correct proof. $\endgroup$ – Ryan Unger Mar 2 '16 at 16:47
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This is version two of my proof. The OP discovered a sign error in my first attempt that revealed my argument to be circular. The correct proof is below.

Not surprisingly, this has to do with the signature of the spacetime metric not being positive definite. Furthermore, this issue is very subtle. It should be noted that this result is quoted partially in Hawking-Ellis, The large-scale structure of spacetime (1973), correctly in Wald, General Relativity (1984), incorrectly in Carroll, Spacetime and Geometry (2003) and hinted at in Straumann, General Relativity (2013). None of these contain a proof. The linked lecture notes have the correct result. Carroll has the "outward" and "inward" directions switched. The book by Gourgoulhon, Special Relativity in General Frames, contains a partial proof in $4$ dimensions, which we adopt here for $n$ dimensions.

We first state the preliminaries. Proofs can be found in the excellent book by Lee, Introduction to Smooth Manifolds (2013). We have provided the theorem numbers for convenience. (At certain points, results in Riemannian geometry must be generalized with some care.)

Let $M$ be a smooth $n$-manifold with boundary $\partial M$. Let $\mathrm{d}$ be the exterior derivative on $M$, $i_X$ the interior derivative wrt. the vector field $X$ and $L_X$ the Lie derivative wrt. $X$. Let $g$ be a pseudo-Riemannian metric on $M$ with Levi-Civita connection $\nabla$ and $(x^\mu)$ be oriented coordinates on $M$. Let $\{E_\mu\}_{\mu=1}^n$ denote an orthonormal frame. The canonical orientation is a top-degree differential form $\mu$ for which $\mu(E_1,\dotsc,E_n)=1$.

We next define "outward" and "inward". A vector $v\in T_pM$ ($v$ should not lie entirely in $T_p\partial M$) is said to point outward if there exists a curve $\gamma:(-\epsilon,0]\to M$ such that $\gamma(0)=p$ and $\dot\gamma(0)=v$. It is said to point inward if there exists a curve $\gamma:[0,\epsilon)\to M$ such that $\gamma(0)=p$ and $\dot\gamma(0)=v$. (Here $\epsilon$ is some positive number.)

Some Results. $\partial M$ is a smooth embedded codimension $1$ submanifold and has a uniquely defined (unit) normal vector. (Thm. 5.11, Prop. 15.33.) The notion of "outward" and "inward" is an equivalence relation on the set of normal vectors. A normal vector $N$ is inward iff $-N$ is outward. (Prop. 5.41, Prop. 15.33.) $\mu$ is a canonical orientation iff $\mu=\sqrt{\lvert\det g_{\mu\nu}\rvert}\,\mathrm{d}x^1\wedge\cdots\wedge\mathrm{d}x^n$. (Prop. 15.31.) If $\omega$ is an orientation, $\iota:S\hookrightarrow M$ is the boundary of a subset of $M$ and $Y$ is a vector that points out of $S$, then $\iota^*(i_Y\omega)$ is a consistently oriented orientation on $S$. (Prop. 15.24) The divergence $\operatorname{div}X:=\nabla_\mu X^\mu$ satisfies $\operatorname{div}X\,\mu=L_X\mu$. (Page 425.) Cartan's formula: $L_X=i_X\circ\mathrm{d}+\mathrm{d}\circ i_X$. (Thm. 14.35.)

Now let $(\mathcal{M},g)$ be the spacetime manifold and metric. Now $\partial \mathcal{M}$ is the boundary and $\iota:\partial \mathcal{M}\hookrightarrow \mathcal{M}$ is the inclusion. See Sect. 2.7 of Hawking-Ellis for a discussion of

Hypersurfaces. If $\iota^*g=:h$ is Riemannian, $\partial\mathcal{M}$ is said to be spacelike and $N$ is timelike. If $h$ is Lorentzian, $\partial\mathcal{M}$ is said to be timelike and $N$ is spacelike.

In the following, let $\tilde\mu$ be the canonical orientation on $(\partial\mathcal{M},h)$ and let $N$ be, for now, the outward pointing unit normal. (Since $g$ is Lorentzian, this means that $\langle N,N\rangle=\pm1$.)

By the Cartan formula, for any vector field $X$, $$\operatorname{div}X\,\mu=L_X\mu=\{\mathrm{d},i_X\}\mu=\mathrm{d}(i_X\mu),$$ and by Stokes' theorem $$\int_\mathcal{M}\operatorname{div}X\,\mu=\int_{\partial \mathcal{M}}\iota^*(i_X\mu).$$

It turns out that $\iota^*(i_X\mu)=i_X\mu\restriction_{\partial\mathcal{M}}$ is not so easy to determine.

It is clear if $\partial \mathcal{M}$ is not null, then we have the smooth orthogonal splitting $$T_p\mathcal{M}=T_p\partial \mathcal{M}\oplus \operatorname{span}N,$$ for all $p\in\partial\mathcal{M}$. So let $$ X=X^\top+X^\bot, X^\top\in T_p\partial\mathcal{M},X^\bot\in\operatorname{span}N.$$ Then $X^\bot=\alpha N$ and $X=X^\top+\alpha N$. Taking the inner product with $N$ gives $$\langle N,X\rangle=\alpha\langle N,N\rangle=\sigma\alpha\implies X=X^\top+\sigma\underbrace{\langle N,X\rangle}_\beta N,$$ where $\sigma=+1$ if $N$ is spacelike and $\sigma=-1$ is $N$ is timelike. Thus $$i_X\mu=i_{X^\top}\mu+\sigma\beta i_N\mu.$$

Note now that technically everything we are doing is being pulled back onto $\partial\mathcal{M}$, which is equivalent to the restriction to $\partial\mathcal{M}$. Let $\{E_\mu\}_{\mu=2}^n$ be an orthonormal frame on $\partial \mathcal{M}$. Then $X^\top$ is a linear combination of these vectors, $X^\top=\sum_{\mu=2}^n c_\mu E_\mu$ and $$i_{X^\top}\mu(E_2,\dotsc,E_n)=\sum_{\mu=2}^nc_\mu\mu(E_\mu,E_2,\dotsc,E_n)=0,$$ due to the total antisymmetry of $\mu$. (If any two slots of a totally antisymmetric tensor are filled by the same vector, it vanishes.) By linearity we can extend this to all vectors in $T_p\partial\mathcal{M}$. Thus $i_{X^\top}\mu\restriction_{\partial \mathcal{M}}\equiv 0$ and we have $$i_X\mu=\sigma\beta i_N\mu\quad(\text{on $\partial\mathcal{M}$}).$$

Let $\{E_\mu\}_{\mu=2}^n$ be as above, then $\{N,\{E_\mu\}_{\mu=2}^n\}$ is an orthonormal frame on $\mathcal{M}$. (One might have to adjust the sign of an $E_\mu$ so that $\{N,\{E_\mu\}_{\mu=2}^n\}$ is consistently oriented.) Then $$i_N\mu(E_2,\dotsc,E_n)=\mu(N,E_2\dotsc,E_n)=1,$$ which implies that $i_N\mu$ is the canonical orientation $\tilde\mu$ by Prop. 15.24 and Prop. 15.31 of Lee. Note once again that we take $N$ to be pointing outwards here. Putting everything together, we obtain $$\int_{\mathcal{M}}\operatorname{div}X\,\mu=\sigma\int_{\partial\mathcal{M}}\langle N,X\rangle\,\tilde\mu.$$

We now have two cases.

Case 1. $N$ is spacelike. Then $\sigma=+1$ and we are done.

Case 2. $N$ is timelike. Now $\sigma=-1$. But then $$-\int_{\partial\mathcal{M}}\langle N,X\rangle\,\tilde\mu=\int_{\partial\mathcal{M}}\langle -N,X\rangle\,\tilde\mu=\int_{\partial\mathcal{M}}\langle N',X\rangle\,\tilde\mu,$$ where $N'$ is the inward pointing normal vector, by Prop. 5.41 in Lee.

Note that changing the sign of $N$ does not alter $\tilde\mu$, since for it to be consistently oriented we took the normal to be pointing outwards. You could write $\tilde\mu=i_{-N'}\mu$ if you wanted to.

Prop 15.31 in Lee gives $$\mu=\sqrt{\lvert g\rvert }\,\mathrm{d}x^1\wedge\cdots\wedge\mathrm{d}x^n,\quad\tilde\mu=\sqrt{\lvert h\rvert}\,\mathrm{d}y^1\wedge\cdots\wedge\mathrm{d}y^{n-1},$$ where $(y^1,\dotsc,y^{n-1})$ are consistently oriented coordinates on $\partial\mathcal{M}$.

We have thus shown: $$\int_\mathcal{M}\nabla_\mu X^\mu\sqrt{\lvert g\rvert }\,\mathrm{d}x^1\wedge\cdots\wedge\mathrm{d}x^n=\int_{\partial\mathcal{M}}N_\mu X^\mu\sqrt{\lvert h\rvert}\,\mathrm{d}y^1\wedge\cdots\wedge\mathrm{d}y^{n-1},$$ where $N$ points inward if $\partial\mathcal{M}$ is spacelike and outward if $\partial\mathcal{M}$ is timelike. Note that more generally, one can take the integrals over any $n$-dimensional $\mathcal{U}\subset\mathcal{M}$ with boundary $\partial\mathcal{U}$. This is because $(\mathcal{U},g\restriction_\mathcal{U})$ is a spacetime in its own right, and the proof given above works for $\mathcal{M}$ replaced by $\mathcal{U}$ and $g$ by $g\restriction_\mathcal{U}$.

As for the example on page 456 of Carroll: The region $R$ is bounded by spacelike hypersurfaces, so we must take the normal vector pointing inwards. Then the normal on $\Sigma_2$ is future directed and the normal on $\Sigma_2$ is past directed. But he is defining his integrals $\int_{\Sigma_1}$ and $\int_{\Sigma_2}$ with future directed normal vectors. So the integral $\int_{\Sigma_2}\star J$ picks up a minus sign from the past-directed normal.

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  • $\begingroup$ Thank you for your detailed answer! I only have a few questions though: - Why is the contraction of $\mu$ with $\tilde{X}$ zero? - How does this definition of in and outward agree with the intuitive notion of those? For example, take a radial curve in $R^3$ meeting normally $S^2$, starting from 0 and a normal outward vector $n$ to the sphere. Then you can either restrict this curve to $]-\epsilon, 0]$ or to $[0,\epsilon[$ such that its tangent is $n$. $\endgroup$ – faero Mar 1 '16 at 20:56
  • $\begingroup$ I have found a few typos I think: $\mu$ should start from 1 in the definition of orthonormal frame and $v \in T_pM - T_p\partial M$ in the definition of in/outward vectors. Given your definition of $\alpha$, shouldn't there be a sign in your sixth equation? $\endgroup$ – faero Mar 1 '16 at 20:56
  • $\begingroup$ @faero On the intuition of outwards/inwards: Suppose we have some curve going along the inside of $M$ that eventually hits $\partial M$. We call this curve $\gamma(t)$. We define $t$ so that at $t=0$ $\gamma$ hits $\partial M$. So clearly $\gamma$ is "moving towards" $\partial M$ for "small" times less than $0$. So its tangent vector is pointing "towards" $\partial M$ as $t\to 0$. At $t=0$ we say the tangent, now "on" $\partial M$ is pointing "out". Inwards works the same way. $\endgroup$ – Ryan Unger Mar 1 '16 at 21:19
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in order to apply the stokes theorem to Lorentz manifold we must take normals at the boundary ...

The general stokes theorem for differential forms is valid for any orientable manifold with a boundary:

$$\int_D d\omega = \int_{\partial D} \omega$$

A metric is not req'd; so is valid for a Lorentz manifold - a manifold with a metric of signature (n,1).

Equation E19 in Carrolls Spacetime and Geometry directly follows from this:

Imagine a four-dimensional spacetime region R, defined between two spatial hyper-surfaces E1, E2; ....; the part of the boundary connecting the two hyper-surfaces is assumed to be off to infinity where all the fields vanish and can be ignored.

The conservation law E16 gives:

$$\int_R d(*J)= 0$$

Then directly by Stokes above

$$\int_{\partial R} *J$$

Now the hyper-surface R is made of two components E1 & E2 of opposite orientations, so:

$$\int_{\partial E1} *J - \int_{\partial E2} *J$$

Which gives us equation E19:

$$=Q1-Q2$$

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  • $\begingroup$ That general stokes theorem is phrased in terms of differential forms, not normal vectors. The question is about how the sign of the normal vector in a "vector calc" form of it has to be chosen on a Lorentzian manifold (and here you see why the metric might matter - to say a vector is "normal" to something involves using a metric). $\endgroup$ – ACuriousMind Mar 1 '16 at 23:12
  • $\begingroup$ @curiousmind: sure - that's what I wrote ie differential forms - the linked notes is written in the language of vector analysis; if the language is translated into vectors, and we're talking about an embedded manifold then normal vectors will matter. $\endgroup$ – Mozibur Ullah Mar 1 '16 at 23:17
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    $\begingroup$ This does not answer the question. OP is specifically asking about normal vectors, this post does not mention them at all. $\endgroup$ – Ryan Unger Mar 1 '16 at 23:21
  • $\begingroup$ @ocelo7: the question opens with 'in order to apply the stokes theorem to Lorentz manifolds'; it's that bit that I'm answering; to make it clear, I'll add that. $\endgroup$ – Mozibur Ullah Mar 1 '16 at 23:23
  • $\begingroup$ This still does not answer the question. $\endgroup$ – Ryan Unger Mar 1 '16 at 23:39

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