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the following line element defines the Kahn-Penrose metric with coordinates $(u,v,x,y)$ and constraints $u \geq 0$, $v < 0$

$$ds^2=-2dudv+(1-u)^2dx^2+(1+u)^2dy^2$$

If we restrict ourselves to the following convention, in which the Minkowski metric $\eta$ is mostly plus $(-,+++)$ and $ds^2>0$ for space like intervals.

The reasoning goes as follows: in order to determine the nature of a coordinate, we produce some small but otherwise arbitrary variation of that coordinate while keeping the rest constant (which can be done if we assign algebraically the valor $0$ to the $x^μ$ not being varied. Then the sign of $ds^2$ will tell the space-time-null character of the coordinate in question.

My question is if the following statements are correct:

  1. $x$ and $y$ are space like coordinates.
  2. $v$ is a space like coordinate
  3. $u$ is a time like coordinate
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  • $\begingroup$ Can you explain why you believe your statements? $\endgroup$ – Prahar Jun 15 '13 at 0:07
  • $\begingroup$ The argument goes as follows: in order to determine the nature of a coordinate, we produce some small but otherwise arbitrary variation of that coordinate while keeping the rest constant (which can be done if we assign algebraically the valor $0$ to the $dx^{\mu}$ not being varied. Then the sign of $ds^2$ will tell the space-time-null character of the coordinate in question. $\endgroup$ – Jorge Lavín Jun 15 '13 at 0:11
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    $\begingroup$ @Nivalth you should probably add that to the question. $\endgroup$ – David Z Jun 15 '13 at 4:56
  • $\begingroup$ @Trimok I think $u$ and $v$ are the null coordinates constructed from Minkowski space $t$ and $z$ rather than $t$ and $x$? $\endgroup$ – twistor59 Jun 15 '13 at 7:33
  • $\begingroup$ Obviously $t= \frac{u + v}{2}$, and $z= \frac{u - v}{2}$ are respectively time-like and space-like coordinates (with the constraint z>0), so how can you state that u or v has a precise behaviour ? (@twistor59 :thanks for the correction..) $\endgroup$ – Trimok Jun 15 '13 at 8:35
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The rule is : With a diagonal metrics, the character of the coordinate $x^i$, is given by the sign of $g_{ii}$

Setting $t= \frac{u + v}{2}$ and $z= \frac{u - v}{2}$, your metrics becomes (with the constraint $z >0$) :

$$ds^2= -dt^2 + dz^2+(1-t-z)^2dx^2+(1+t+z)^2dy^2$$

So, the character of $t$ as a time-like coordinate and $z$ as a space-like coordinate (with the constraint $z >0$) appears clearly. And $x$ and $y$ are space-like coordinates too.

The variables $u = t + z$ and $v = t - z$ are called light-cone coordinates, you could say that they are light-like coordinates. They are very often used in general relativity and string theory. They are also called null coordinates, because for instance, if $dx=dy=0$, then you have $ds^2 = - 2dudv$, so each particle with $U=$Constant or $V=$Constant corresponds to $ds^2=0$, that is a light-like interval (a light ray).

But you cannot say that one of the coordinates $u, v$ is a time-like coordinate, and the other a space-like coordinate.

In Schwarzchild metric, it is not correct to say that $r$ and $t$ could change their sign. The Schwarzchild metric describes a gravitational field only for $r>r_S$. It is a limited description (which does not cover the entire manifold), and you have to use Kruskal-Szekeres coordinates to have a glbal view of the black hole. Mathematically, "These coordinates have the advantage that they cover the entire spacetime manifold of the maximally extended Schwarzschild solution and are well-behaved everywhere outside the physical singularity."

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  • $\begingroup$ Thank you Trimok, very clear and insightful. I did my calculations wrong because I didn't take into account that term $2dudv$ is zero for both the $u$ and $v$ variations and hence $ds^2=0$ so they are light-like. $\endgroup$ – Jorge Lavín Jun 17 '13 at 13:09

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