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We know that $$ds^2 = g_{\mu\nu}dx^{\mu}dx^{\nu},$$ Can you say how to calculate $ds$?

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closed as not a real question by Manishearth, Qmechanic, Emilio Pisanty, David Z Dec 5 '12 at 3:33

It's difficult to tell what is being asked here. This question is ambiguous, vague, incomplete, overly broad, or rhetorical and cannot be reasonably answered in its current form. For help clarifying this question so that it can be reopened, visit the help center.If this question can be reworded to fit the rules in the help center, please edit the question.

I don't understand what you are asking here. – Bernhard Dec 3 '12 at 7:42
@FrankH we can show $ds^2$ in term $g_{\mu\nu}dx^{\mu}dx^{\nu},$ what about $ds$? isn't it possible to show it in same term? – Neo Dec 3 '12 at 8:35
@Neo Please edit your question, as it should be clear without reading comments. Make sure that it is absolutely clear what you are asking and what you do and don't understand. – Bernhard Dec 3 '12 at 9:43
Indeed, if you know $ds^2$, you may calculate $ds$ as the square root of $ds^2$ except that $ds^2$ may be both positive and negative (and zero) so $ds$ may be both real or pure imaginary and it's more sensible and natural not to take the square root. – Luboš Motl Dec 3 '12 at 12:43
Dear @Neo: In this context, it is always implicitly understood that $ds^2$ means $(ds)^2$, not $d(s^2)$. So all there is left to do is to take the square root, as Lubos Motl writes. – Qmechanic Dec 3 '12 at 21:44

if you are trying to measure an infinitesimal proper distance (ds) for a particle, it is convenient to specify the 4 dimensional coordinates of the particle as functions of an arbitrary parameter that I will call $\xi$. Thus the particle's position for each value of $\xi$ will be: $x^{\mu}(\xi)$. Then for an infinitesimal change of the parameter $\xi$ the changes of the coordinates will be:

$dx^{\mu} = \frac{dx^\mu(\xi)}{d\xi} d\xi$

So the infinitesimal proper distance is:

$ds = \sqrt{(ds)^2} = \sqrt{g_{\mu\nu}(x)\frac{dx^\mu(\xi)}{d\xi}d\xi \frac{dx^\nu(\xi)}{d\xi}d\xi}$

and therefore:

$ds = \sqrt{g_{\mu\nu}(x)\frac{dx^\mu(\xi)}{d\xi} \frac{dx^\nu(\xi)}{d\xi}} \ \ d\xi$

This could be integrated from, say $\xi_0$ to $\xi_1$ to get the proper distance from the point $x^{\mu}(\xi_0)$ to $x^{\mu}(\xi_1)$ like this:

$s = \int ds = \int_{\xi_0}^{\xi_1}\sqrt{g_{\mu\nu}(x)\frac{dx^\mu(\xi)}{d\xi} \frac{dx^\nu(\xi)}{d\xi}} \ \ d\xi $

Note that all these equations apply to flat Minkowski spacetime in any arbitrary coordinate system (including, for example, Cartesian or polar coordinates). It also applies to arbitrarily curved spacetime with any kind of coordinate system.

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This is the correct answer. – twistor59 Dec 4 '12 at 7:39
@FrankH I am totally confused by your answer – Neo Dec 4 '12 at 7:46
@neo, what line or statement is confusing. You do get that the equation you wrote is for $(ds)^2$ not $d(s^2)$, don't you? – FrankH Dec 4 '12 at 13:31
@neo, is it the parameterisation by $\xi$ that is confusing? Try thinking of the $\xi$ as being like the "time" (t) in classical Newtonian mechanics. In Newtonian mechanics the position (x,y,z) are three functions of time (t). This only works if there is absolute time (t) as in Newtonian mechanics. In relativity t is another part of the coordinate system and in the most general case (such as space like events or tachyons) different observers will disagree on the order of events so it is convenient to use an arbitrary parameter to specify (x,y,z,t) as four functions of that parameter. – FrankH Dec 4 '12 at 13:45
First of all in my answer, I never wrote any expression involving $t^2, x^2, y^2$ or $z^2$. You are misunderstanding where the $^2$ applies. The equation is $(ds)^2 = g_{\mu\nu} (dx^{\mu}) (dx^{\nu})$ and I hope you understand the repeated indices notation means this is a sum over $\mu$ and $\nu$ varying over 0, 1, 2 and 3. – FrankH Dec 4 '12 at 14:19

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