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Say we are in 3 dimensions and use $(-++)$. If we have the metric $$ds^2=-dt^2+dr^2+r^2df^2(t),$$ then what is the third coordinate if the first two were $t$ and $r$? $$X^iX_i=-t^2+r^2+?$$

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First of all, the first equation for $ds^2$ is only valid if $f$ is nothing else than the azimuthal angle $\phi$.

Second, if you are evaluating $X_i X^i$, the squared distance from the origin without any infinitesimals, then it is exactly equal to $-t^2+r^2$ and nothing else. The polar coordinate $r$ is chosen as $\sqrt{x^2+y^2}$ so its square already includes two of the three terms one encounters in Cartesian coordinates.

There is nothing wrong if the formula for a squared distance from a particular point (the origin) just has two terms.

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  • $\begingroup$ Thanks and I'm very sorry about this stupid question. So is it that $\partial_a X\cdot\partial_bX=-\partial_at\partial_bt+\partial_ar\partial_br+r^2\partial_af \partial_b f$? Why is the last term so? (Or is it that you're saying this term shouldn't exist?) $\endgroup$ – user46348 Aug 12 '14 at 11:06
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    $\begingroup$ No, your question and guess are perfectly right! But $\partial_a X\cdot \partial_b X$ - which is similar to $dX\cdot dX=ds^2$ - is something else and behaves differently than $X\cdot X$ without the derivatives. The two formulae (first and second in your original question) are just very different. $\endgroup$ – Luboš Motl Aug 12 '14 at 17:51
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    $\begingroup$ Forget your 3D spacetime. Take just the two ++ coordinates. Then $X\cdot X$ is just the inner product of the vector $X$ with itself which is simply $r^2$, the squared distance from the origin. The distance from the origin is called the radial $r$ coordinate. But $dX\cdot dX=ds^2$ is something else. It's the squared length of an infinitesimal vector $d\vec X$ which is located everywhere. So it's not the squared length of a line interval starting in the origin $(0,0)$, which was the case of $X\cdot X$. Now the tiny line interval $dX$ is at any point $X$ and it has nonzero length even ... $\endgroup$ – Luboš Motl Aug 12 '14 at 17:54
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    $\begingroup$ even if $d|\vec X|=0$ i.e. if the tiny infinitesimal line interval $d\vec X$ goes tangentially along a circle with a fixed $r$, i.e. if it goes in the angular direction. Because the length of the arc given by angle $d\phi$ is simply $r\cdot d\phi$, the formula for $ds^2$ must contain this squared, i.e. $r^2\cdot d\phi^2$. At the end, the opposite-signature time coordinate is added to both expressions in the obvious way. $\endgroup$ – Luboš Motl Aug 12 '14 at 17:55

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