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Starting with Euclidean space, suppose I make a number of copies of a coordinate system all coincident at the origin, together with copies of a standard unit length.

For any space interval common to all the coordinate systems, all the observers will count the same minimum number of times their unit length can be laid end to end to one another between this interval. And from this invariant counting process we derive the Euclidean metric in terms of its orthogonal components.

What invariant counting process do we carry out to derive the Minkowski metric in terms of its orthogonal components?

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  • $\begingroup$ I don't understand. How does each observer measuring the same length derive the Euclidean metric? How are the copies of each coordinate system different from one another? Are they merely rotated with respect to each other? $\endgroup$ – Muphrid Jun 26 '14 at 16:54
  • $\begingroup$ @Muphrid yes, they're rotated wrt one another. You derive the metric by laying out the space interval along the x-axis of one coordinate system, and then infinitesimally rotate the other coordinate systems consecutively. Integrating this gets you the metric in terms of the lengths measured along the orthogonal axis in the last coordinate system. $\endgroup$ – Larry Harson Jun 26 '14 at 17:09
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    $\begingroup$ So, you presuppose the rotation transformation, and then it follows that the Euclidean metric is the one that is kept invariant under that rotation transformation. Do you think this process would turn out differently by presupposing the Lorentz transformation instead? $\endgroup$ – Muphrid Jun 26 '14 at 17:10
  • $\begingroup$ @Muphrid Hmm, forget the "infinitesimally rotate" bit in the last comment. It should read "infinitesimally transform, keeping the count invariant". $\endgroup$ – Larry Harson Jun 26 '14 at 19:37
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The invariant process is the speed of light $c$ which should be (measured) the same for every observer in relative motion.

And indeed Einstein's 1905 paper derives from first principles, the Lorentz transformations (i.e the transformations of special relativity) using (what some would derogatorily say) 'high-school mathematics'. Yet the concepts behind these simple relations is the whole point.

If one wants to characterise the geometry of Minkowski space in terms of F.Klein's Erlangen program of groups and invariants, then the geometry of special relativity is exactly this geometry which leaves the Lorentz metric invariant (using a signature of $(+,-,-,-)$):

$$ds^2 = (cdt)^2-dx^2-dy^2-dz^2$$

This is exactly analogous to the Euclidean geometry as the geometry which leaves the quadratic metric invariant (e.g in 4 dimensions, $x,y,z,w$):

$$ds^2 = dx^2 + dy^2 + dz^2 + dw^2$$

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  • $\begingroup$ Yeah, I'm inclined to agree with this view. But this relates the invariance of light-like intervals only, right? $\endgroup$ – Larry Harson Jun 26 '14 at 19:44
  • $\begingroup$ @LarryHarson, no the whole form is invariant, as it is for the euclidean space under (euclidean) coordinate transformations, that is what the proper time is al about, to see it in another way, it is a scalar, as such it is invariant $\endgroup$ – Nikos M. Jun 26 '14 at 19:46
  • $\begingroup$ On the first line, you state the invariant process is the speed of light, which connects light like intervals. $\endgroup$ – Larry Harson Jun 26 '14 at 19:49
  • $\begingroup$ @LarryHarson, correct, what is meant is that the whole metric (as indeed the Lorentz transforms) can be derived by assuming just that $c$ is constant under all acceptable transformations (much like the original Einstein paper), the rest is basd on Erlangen characterization, sorry for any ambiguity $\endgroup$ – Nikos M. Jun 26 '14 at 19:51
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    $\begingroup$ This answer doesn't strike me as in analogy to Larry's question, though: why not consider an object of unit spacetime interval and proceed exactly as in Larry's explanation for Euclidean space? $\endgroup$ – Muphrid Jun 26 '14 at 20:29

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