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In all the sources I’ve been able to find, the Minkowski metric appears ad hoc, or is defined analogously to the euclidean metric. I’d love to see an argument why this metric (time coordinates positive, space coordinates negative) must follow from the constancy of the speed of light. It is clear that the Minkowski metric is preserved under the hyperbolic transformation of space-time, but likely others are as well. Why this particular metric and not something else.

Consider the determinant function of an n by n matrix. It has a god awful mathematical form involving the sum of n ! terms. Yet all you need to get the (unique) formula are a few postulates — the determinant of the identity matrix is 1, the determinant is a linear function of its rows (or its columns), interchanging any two rows of the determinant reverses the sign of the determinant, etc. etc. This basically determines the (unique) formula of the determinant. I’d really like to see the Minkowski metric come out of something like that.

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marked as duplicate by John Rennie, DavePhD, Kyle Kanos, Brandon Enright, Valter Moretti May 11 '14 at 20:15

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  • $\begingroup$ We don't need the constancy of the speed of light. The existence of an invariant speed follows from the principle of relativity: arxiv.org/abs/physics/0302045 We deduce the most general space-time transformation laws consistent with the principle of relativity. Thus, our result contains the results of both Galilean and Einsteinian relativity. ... We also argue why Galilean and Einsteinian versions are the only possible embodiments of the principle of relativity. $\endgroup$ – Alfred Centauri May 10 '14 at 1:30
  • $\begingroup$ possible duplicate of Special Relativity Second Postulate $\endgroup$ – John Rennie May 10 '14 at 7:06
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" I’d love to see an argument why this metric (time coordinates positive, space coordinates negative) must follow from the constancy of the speed of light. ... Why this particular metric and not something else.".

Well, there is nothing mysterious, or even unusual about the signs of the metric. If you take a look Einstein's "Relativity: The Special and General Theory", you will find in the Appendix I (just before equation (10)) that in order to derive the equation, Einstein simply started off with the Pythagorean Theorem in 3D, which he put like this:

$$r = \sqrt{x^2 + y^2 + z^2} = ct$$

So what he proposed (as well as Minkowski himself) is to show the vector of light traveling in a three dimensional space.

He then squared both sides and moved $c^2t^2$ to the left changing the sign. This gave him the signature [$-, +, +, +$] Obviously, you are equally free to move the $x^2 + y^2 + z^2$ to the right, in which case you will get the signature reversed [$+, -, -, -$]. Apparently, that's what Minkowski did.

The same equation was then build for the primed (O') frame of reference, and both frames were compared.

Now, if you want to compare two different frames of reference, two different spaces, without actually measuring different values, you need to find something they have in common. This common "thing" as assumed by both Minkowski and Einstein was the speed of light, which allowed them to use the equations $c=x/t$ and $c=x'/t'$.

But then, the constancy of light is not necessarily the cause for the particular signs for the metric Minkowski used. As you can see above, the signs would change the same way if you assume that $c'\neq c$. What you do need, however, to develop the metric the way Minkowski and Einstein did, is a way to present the vector in space. They did it through the concept of speed of light (which obviously entails the concept of time).

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  • $\begingroup$ Thanks a lot, I've never had the temerity to read Einstein directly, but his writing is clear as a bell. Also I didn't know about Bartleby either $\endgroup$ – user46221 May 10 '14 at 23:34
  • $\begingroup$ Sure. But then, the book I referred you to only glosses the derivation of the metric. You should try and follow the whole derivation to see more interesting things. Especially the part referring to space. $\endgroup$ – bright magus May 11 '14 at 10:15
  • $\begingroup$ Another common thing, is that all objects move across the 4 dimensional environment known as space-time at the same magnitude of motion as which the speed of light has across across space. $\endgroup$ – Sean Jan 17 '16 at 1:38
  • $\begingroup$ @Sean, "which the speed of light has across across space" the concept of speed requires time. One does not exist without the other. $\endgroup$ – bright magus Jan 18 '16 at 13:20

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