Trying to understand Schutz's AFCIGR, where does the minus sign appear from in the metric tensor?


I understand that this expresses the invariance of the spacetime interval. Schutz says (I think) that the metric is a (0,2) tensor. I assume that means it is the product of two one-forms, so presumably one of these one-forms has a -ve time component. What does that mean? Why don't both one-forms have a -ve time component? Looking at a Minkowski diagram what is a straightforward way to understand/visualise those one-forms? At my level, two things have been multiplied together to give a 4x4 matrix which has -1 in the top left corner. What are those two things and why do they give a -1 time component?

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    $\begingroup$ A comment to the question (v1): Note that in general (and in fact also in the specific case mentioned), a $(0,2)$ tensor is not always a product of co-vectors. Instead, it is a sum of products of co-vectors. (In the specific case the sum contains at least $4$ terms.) In fact, the same is true if the $-1$ had been a $+1$. $\endgroup$ – Qmechanic Jun 23 '11 at 12:15

The metric is a symmetric bilinear form $$ ds^2 = \eta_{ij} dx^i dx^j = -dt^2 + dx^2 + dy^2 + dz^2 $$ Hence the minus sign is not a property of the one-forms $dx^i$ but of the coefficients $\eta_{ij}$

  • $\begingroup$ Thanks for that but I still don't understand where the minus sign comes from. $\endgroup$ – Peter4075 Jun 24 '11 at 6:51
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    $\begingroup$ @user4075: I'm not sure I understand what you mean with "where the minus sign comes from". In SR the space-time is a flat space with a metric of signature (-+++), more or less per definition. There are of course other ways of saying the same thing, eg the isometry group is SO(3,1), but I don't think this answers your question does it? $\endgroup$ – Olof Jun 24 '11 at 8:04
  • $\begingroup$ Olaf: I'm happy to admit I'm out of my depth here (eg I don't know what a "symmetric bilinear form" is, and I've still struggling with one-forms). I assumed (erroneously - thanks Qmechanic) that a (0,2) tensor is always the product of two one-forms, and thus assumed this might explain why there's a minus sign in the metric. Wikipedia (Introduction to special relativity) says, "Minkowski... found that the correct formula was actually quite simple, differing only by a sign from Pythagoras's theorem". So I'm assuming he just sort of discovered it worked. $\endgroup$ – Peter4075 Jun 24 '11 at 14:37

Although one can think of the metric as about infinitesimal distances, it can also be thought of in a dual way, as about differential equations. Thus, with your sign convention for the metric, the wave equation is $$g^{ij}\frac{\partial^2\phi(x)}{\partial x^i\partial x^j}=-\frac{\partial^2\phi(x)}{\partial t\partial t}+\frac{\partial^2\phi(x)}{\partial x\partial x}+\frac{\partial^2\phi(x)}{\partial y\partial y}+\frac{\partial^2\phi(x)}{\partial z\partial z}=0.$$ In this POV, the metric determines the differential equations that are satisfied by the fields that are introduced in the space-time. The Laplace equation has very different properties than does the wave equation.

There are formalisms in which d$t$ is replaced by id$t$, so that the metric can be uniformly positive. However most people keep this as a mathematical trick up one's sleeve, not as a fundamental part of the construction of a theory. There was a period in the development of GR when imaginary time was much more commonly used than it is now.


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