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I am trying to understand geometric calculus and apply it to physics. In this sense, I was reading Alan Macdonald's book "Vector and Geometric Calculus", and stumbled upon the Fundamental Theorem of Geometric Calculus, which states that:

$ \int_M d^m \mathbf{x} \partial F = \oint_{\partial M} d^{m-1} \mathbf{x} F $

where $M$ is an oriented and bounded $m$-dimensional manifold with boundary $\partial M$, F is a continuous multivector field and $ d^m \mathbf{x} $ is the infinitesimal pseudoscalar of the tangent space to $M$ at $\mathbf{x}$.

Considering that I don't have a full course in differential geometry, I can't really prove that this could be valid in special relativity, which is where I want to apply the theorem.

Given this, I'd like to know if this theorem holds in special relativity, and how each term in the integral is expressed in the Minkowski metric.

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The fundamental theorem itself is independent of metric. Illustrating by example, given a two parameter surface $$\mathbf{x} = \mathbf{x}(a,b) \quad (=\gamma_\mu x^\mu(a, b)),$$ for which the area element is $$\begin{aligned}d^2 \mathbf{x}&= \left( { \frac{\partial {\mathbf{x}}}{\partial {a}} \wedge \frac{\partial {\mathbf{x}}}{\partial {b}} } \right) da db \\ &= (\mathbf{x}_a \wedge \mathbf{x}_b) da db \\ &= d\mathbf{x}_a \wedge d\mathbf{x}_b,\end{aligned}$$ and for which the vector derivative is (no summation convention here) $$\partial = \mathbf{x}^a \partial_a + \mathbf{x}^b \partial_b,$$ the integral expands as $$\begin{aligned}\int d^2 \mathbf{x} \partial F&=\int da db (\mathbf{x}_a \wedge \mathbf{x}_b) \left( { \mathbf{x}^a \partial_a + \mathbf{x}^b \partial_b } \right) F \\ &=\int da db \left( { -\mathbf{x}_b \partial_a F + \mathbf{x}_a \partial_b F } \right) \\ &=\int da db \left( { -\frac{\partial {}}{\partial {a}} \left( { \mathbf{x}_b F} \right) + \frac{\partial {}}{\partial {b}}\left( { \mathbf{x}_a F} \right) } \right)-\int da db \left( { -\frac{\partial {}}{\partial {a}} \left( { \mathbf{x}_b } \right) + \frac{\partial {}}{\partial {b}}\left( { \mathbf{x}_a } \right) } \right) F\\ &=\int -d\mathbf{x}_b {\left.{{F}}\right\vert}_{{\Delta a}} +d\mathbf{x}_a {\left.{{F}}\right\vert}_{{\Delta b}},\end{aligned}$$ where the second integral was killed by equality of mixed partials ($ -\frac{\partial {}}{\partial {a}} \frac{\partial {\mathbf{x} }}{\partial {b}} + \frac{\partial {}}{\partial {b}} \frac{\partial {\mathbf{x}}}{\partial {a }} = 0 $.)

The curvilinear coordinates $ \mathbf{x}_a, \mathbf{x}_b $ and the reciprocal frame vectors $ \mathbf{x}^a, \mathbf{x}^b $, do the heavy lifting in this expansion, and don't require any explicit mention of the metric.

The metric can be brought into the mix here explicitly if desired, since the (Minkowski four-) gradient relates the curvilinear coordinates associated with the parameterization: $$\begin{aligned}\mathbf{x}^a &= \nabla a = \gamma^\mu \partial_\mu a \\ \mathbf{x}^b &= \nabla b = \gamma^\mu \partial_\mu b.\end{aligned}$$ (summation convention here.)

To prove the fundamental theorem for $ m > 2 $, which determines the form of $ d^{m-1} \mathbf{x} $, a simular procedure is required. The results of interest for Minkowski space are (assuming parameters $ a, b, c, d $) $$\begin{aligned}\int d^1 \mathbf{x} \partial F &= {\left.{{F}}\right\vert}_{{\Delta a}} \\ \int d^2 \mathbf{x} \partial F &=-\int d\mathbf{x}_b {\left.{{F}}\right\vert}_{{\Delta a}}+\int d\mathbf{x}_a {\left.{{F}}\right\vert}_{{\Delta b}} \\ \int d^3 \mathbf{x} \partial F &=\int d\mathbf{x}_a \wedge d\mathbf{x}_b {\left.{{F}}\right\vert}_{{\Delta c}}+\int d\mathbf{x}_b \wedge d\mathbf{x}_c {\left.{{F}}\right\vert}_{{\Delta a}}+\int d\mathbf{x}_c \wedge d\mathbf{x}_a {\left.{{F}}\right\vert}_{{\Delta b}} \\ \int d^4 \mathbf{x} \partial F &=\int d\mathbf{x}_a \wedge d\mathbf{x}_b \wedge d\mathbf{x}_c {\left.{{F}}\right\vert}_{{\Delta d}}-\int d\mathbf{x}_d \wedge d\mathbf{x}_a \wedge d\mathbf{x}_b {\left.{{F}}\right\vert}_{{\Delta c}}+\int d\mathbf{x}_c \wedge d\mathbf{x}_d \wedge d\mathbf{x}_a {\left.{{F}}\right\vert}_{{\Delta b}}-\int d\mathbf{x}_b \wedge d\mathbf{x}_c \wedge d\mathbf{x}_d {\left.{{F}}\right\vert}_{{\Delta a}}\end{aligned}$$

To make this more concrete, consider the following specific parameterization of a spacetime surface, with a linear timelike path component, and a component that is a boost along the x-direction $$\mathbf{x}(a, b) = \gamma_0 a + \gamma_1 \exp\left( { \gamma_1 \gamma_0 b } \right).$$ Here $ a, b $ are any two of the four possible coordinates $ ct, x, y, z $.

The curvilinear coordinates with respect to parameters $ a, b $ are $$\begin{aligned}\mathbf{x}_a &= \gamma_0 \\ \mathbf{x}_b &= - \gamma_0 \exp\left( { \gamma_1 \gamma_0 b } \right),\end{aligned}$$ so the area element is $$\begin{aligned}d^2 \mathbf{x}&=\left( { \mathbf{x}_a \wedge \mathbf{x}_b } \right) da db \\ &=- {\left\langle{{ \exp\left( { \gamma_1 \gamma_0 b } \right) }}\right\rangle}_{2} da db \\ &=- \gamma_1 \gamma_0 \sinh( b ) da db.\end{aligned}$$ The reciprocal frame vectors are $$\begin{aligned}\mathbf{x}^a &= \mathbf{x}_b \cdot \frac{1}{{ \mathbf{x}_a \wedge \mathbf{x}_b }} = - \frac{\gamma_1 \exp\left( { \gamma_1 \gamma_0 b } \right)}{\sinh(b)} \\ \mathbf{x}^b &= \mathbf{x}_a \cdot \frac{1}{{ \mathbf{x}_b \wedge \mathbf{x}_a }} = - \frac{\gamma_1}{\sinh(b)},\end{aligned}$$ which satisfy $ \mathbf{x}^a \cdot \mathbf{x}_a = \mathbf{x}^b \cdot \mathbf{x}_b = 1 $, and $ \mathbf{x}^a \cdot \mathbf{x}_b = \mathbf{x}^b \cdot \mathbf{x}_a = 0 $, which can be shown explicitly with relative ease using scalar selection operations. The vector derivative is $$\partial = - \frac{\gamma_1}{\sinh(b)} \left( { \frac{\partial {}}{\partial {b}} + \exp\left( { \gamma_1 \gamma_0 b } \right) \frac{\partial {}}{\partial {a}} } \right).$$

This is enough to explicitly express the left and right hand sides of the fundamental theorem identity $$\int d^2 \mathbf{x} \partial F=\gamma_0 \int da db \left( { \frac{\partial {}}{\partial {b}} + \exp\left( { \gamma_1 \gamma_0 b } \right) \frac{\partial {}}{\partial {a}} } \right) F,$$ and $$-\int d\mathbf{x}_b {\left.{{F}}\right\vert}_{{\Delta a}}+\int d\mathbf{x}_a {\left.{{F}}\right\vert}_{{\Delta b}}=\gamma_0 \int db \exp\left( { \gamma_1 \gamma_0 b } \right) {\left.{{F}}\right\vert}_{{\Delta a}}+ \gamma_0 \int da {\left.{{F}}\right\vert}_{{\Delta b}}.$$ As expected by the theorem, these are clearly identical.

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According to the book Geometric Algebra for Physicists, by Doran and Lasenby, "the [fundamental] theorem is still valid in a Lorentzian space" (page 200, in chapter 6). The next section, "Embedded surfaces and vector manifolds" (6.5) might also be of interest to you. Doesn't directly express the integral in terms of the Minkowski metric, though.

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  • $\begingroup$ I'll try to figure if they mean exactly the same thing and work out the math... I am uncertain about my understanding of the theorem, that's why I'd like someone to point out how the terms become in Minkowski metric. $\endgroup$ – Barros Jun 16 '18 at 4:39

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