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The signature of Minkowski spacetime is 2, as is explained here: metric signature explanation. The signature is related to the form the fundamental equations take, but I'm not totally clear on the link here. Furthermore I believe there is a link between second order PDEs and the metric of spacetime. I was wondering if anyone could clarify the links here, or could provide more details of how it goes. I have been reading from the following essay: http://www.fqxi.org/data/essay-contest-files/Callender_FQX.pdf

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I read the answer claiming the signature is 2 but I have never heard it referred to as 2. I have always seen the (1,3) notation (or -,+,+,+ or +,-,-,-). It doesn't make sense to add the diagonal elements and say the signature is 2. A flat 2 dimensional plane space would also have signature 2 by that method which certainly is not the same as (1,3). –  FrankH Mar 19 '13 at 19:15
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Obviously it's necessary to give the dimension and the signature, rather than just the signature. –  Rhys Mar 20 '13 at 9:37
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2 Answers 2

Generically, the signature of the metric becomes important when writing down field equations because certain differential operators one may want to write down depend on the choice of metric. For example, take the Laplace-Beltrami operator on a manifold with metric $g_{ij}$ $$ \Delta f = \frac{1}{\sqrt{|g|}}\partial_i\Big(\sqrt{|g|}g^{ij}\partial_j f\Big) $$ Using cartesian coordinates in Euclidean space $\mathbb R^4$ where $g_{ij} = \delta_{ij}$ one obtains $$ \Delta_{\mathbb R^4}f = (\partial_t^2+\partial_x^2 + \partial_y^2 + \partial_z^2 )f $$ while in cartesian coordinates on Minkowski space $\mathbb R^{3,1}$ one obtains $$ \Delta_{\mathbb R^{3,1}}f = (-\partial_t^2+\partial_x^2+\partial_y^2+\partial_z^2)f $$ Notice, in particular, that the sign in front of the $t$-derivative term changed from a $+$ sign to a $-$ sign because of the change in the signature of the metric.

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Here is one type of answer:

The 'fundamental equation' for a massless field $\phi$ is $$ g^{\mu\nu}\partial_\mu \partial_\nu \phi = 0~, $$ where $g^{\mu\nu}$ is the (inverse) metric. In Euclidean space, this is just Laplace's equation $$ \vec\nabla^2\phi = 0~. $$ To uniquely specify a solution to this equation in some region $V$ of space, you must give the value of the field $\phi$ at all points on the boundary. (These are the simplest boundary conditions, anyway).

In Lorentzian space*time*, on the other hand, the equation is $$ -\partial_t^2\phi + \vec\nabla^2\phi = 0~. $$ This is very different. Now a unique solution is obtained in some spacetime region $V$ by specifying the value of $\phi$ (and its time derivative) at all points on some Cauchy surface for $V$ -- a surface with the property that any (say) past-directed light ray beginning in $V$ intersects it at least once.

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Interesting - I never thought about it that way. So basically you're using the (single-number) signature as a probe of the elliptical (signature = dimension) vs. hyperbolic (signature < dimension) nature of this equation? –  Chris White Mar 20 '13 at 5:06
    
Yes; I think this is important. Although I'm now slightly worried about my answer. What I've written is certainly the way that the boundary conditions for the two equations arise most naturally in physics, but it's not the full story. My knowledge of the theory of partial differential equations is patchy at best... –  Rhys Mar 20 '13 at 9:46
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