I try to calculate the line element in Kruskal coordinates, these coordinates use the Schwarzschild coordinates but replace $t$ and $r$ by two new variables. $$ T = \sqrt{\frac{r}{2GM} - 1} \ e^{r/4GM} \sinh \left( \frac{t}{4GM} \right) \\ X = \sqrt{\frac{r}{2GM} - 1} \ e^{r/4GM} \cosh \left( \frac{t}{4GM} \right) $$ Wikipedia shows the result of the line element. $$ ds^2 = \frac{32 G^3M^3}{r} e^{-r/2GM} (-dT^2 + dX^2) + r^2d\Omega^2 $$

I tried to calculate the metric tensor using $ds^2 = g_{ij} \ dx^i dx^j$. As $T$ and $X$ show no dependence in $\theta$ and $\phi$, the $d\Omega$ seems to make sense, but the calculation of the first component of $g$ was not working.

$$ g_{tt} = J^TJ = \frac{\partial T}{\partial t} \frac{\partial T}{\partial t} + \frac{\partial X}{\partial t} \frac{\partial X}{\partial t}\\ = \frac{1}{32} \left( \frac{r}{GM} - 2 \right) \frac{ e^{\frac{1}{2} \frac{r}{GM}}}{G^2M^2} \left( \cosh^2 \left( \frac{t}{4GM} \right) + \sinh^2 \left( \frac{t}{4GM} \right) \right) $$

Is this the right way to compute the line elements?

What would be better way to calculate the line elements (maybe starting with the Schwarzschild-coordinates)?


I don't think you can drive the line element with the jacobian $J$

The Kruskal-Szekeres line element

Beginning with the Schwarzschild line element: \begin{align*} &\boxed{ds^2 =\left(1-\frac{r_s}{r}\right)\,dt^2-\left(1-\frac{r_s}{r}\right)^{-1}\,dr^2-r^2\,d\Omega^2}\\\\ r_s &:=\frac{2\,G\,M}{c^2} \,,\quad \text{for 2 dimension space}\\ ds^2 & =\left(1-\frac{r_s}{r}\right)\,dt^2-\left(1-\frac{r_s}{r}\right)^{-1}\,dr^2 \end{align*} Step I)
\begin{align*} &\text{for} \quad ds^2=0\\ 0&=\left(1-\frac{r_s}{r}\right)\,dt^2-\left(1-\frac{r_s}{r}\right)^{-1}\,dr^2\,,\Rightarrow\\ \left(\frac{dt}{dr}\right)^2&=\left(1-\frac{r_s}{r}\right)^{-2}\,,\Rightarrow \quad t(r)=\pm\underbrace{\left[r+r_s\ln\left(\frac{r}{r_s}-1\right)\right]}_{r^*}\\ &\Rightarrow\\ \frac{dr^*}{dr}&=\left(1-\frac{r_s}{r}\right)^{-1}\,,\quad \frac{dr}{dr^*}=\left(1-\frac{r_s}{r}\right)\,,&(1) \end{align*} Step II) \begin{align*} &\text{New coordinates}\\ u & =t+r^* \\ v & =t-r^*\\ &\Rightarrow\\ t&=\frac{1}{2}(u+v)\,,\quad dt=\frac{1}{2}(du+dv)\\ r^*&=\frac{1}{2}(u-v)\,,\quad dr^*=\frac{1}{2}(du-dv)\\ dr&=\left(1-\frac{r_s}{r}\right)\,dr^*=\frac{1}{2}\,\left(1-\frac{r_s}{r}\right) (du-dv) \quad\quad(\text{With equation (1)})\\ \Rightarrow \end{align*} \begin{align*} ds^2 &=\left(1-\frac{r_s}{r}\right)\,du\,dv \end{align*} Step III) \begin{align*} r^* & =\left[r+r_s\ln\left(\frac{r}{r_s}-1\right)\right]= \frac{1}{2}(u-v)\,\Rightarrow\\ \left(\frac{r}{r_s}-1\right)&=\exp\left(-\frac{r}{r_s}\right) \,\exp\left(\frac{1}{2\,r_s}(u-v)\right)\\ \left(1-\frac{r_s}{r}\right)&=\frac{r_s}{r}\left(\frac{r}{r_s}-1\right)\\ \,\Rightarrow\\\\ ds^2&=\frac{r_s}{r}\,\exp\left(-\frac{r}{r_s}\right) \,\exp\left(\frac{1}{2\,r_s}(u-v)\right)\,du\,dv \end{align*} Step IV) \begin{align*} &\text{New coordinates}\\ U= & -\exp\left(\frac{u}{2\,r_s}\right) \,,\quad \frac{dU}{du}=-\frac{1}{2\,r_s}\,\exp\left(\frac{u}{2\,r_s}\right)\\ V= & \exp\left(-\frac{v}{2\,r_s}\right) \,,\quad \frac{dV}{dv}=-\frac{1}{2\,r_s}\,\exp\left(-\frac{v}{2\,r_s}\right)\\ \,\Rightarrow\\\\ ds^2&=\frac{4\,r_s^3}{r}\exp\left(-\frac{r}{r_s}\right) \,dU\,dV \end{align*} Step V) \begin{align*} &\text{New coordinates}\\ U & =T-X\,,\quad dU=dT-dX \\ V & =T+X\,,\quad dV=dT+dX\\ \,\Rightarrow\\\\ &\boxed{ds^2=\frac{4\,r_s^3}{r}\exp\left(-\frac{r}{r_s}\right) \left(dT^2-dX^2\right)} \end{align*}

With Matrices and Vectors

The Kruskal-Szekeres line element

Beginning with : \begin{align*} ds^2 & =a\,du\,dv\\ &\Rightarrow\\ g&=\frac{1}{2}\begin{bmatrix} 0 & a \\ a & 0 \\ \end{bmatrix}\\\\ q'&=\begin{bmatrix} du \\ dv \\ \end{bmatrix}\,,\quad q=\begin{bmatrix} u \\ v \\ \end{bmatrix} \,,\quad a=\left(1-\frac{r_s}{r}\right) \end{align*} Step I) \begin{align*} R&= \begin{bmatrix} \frac{1}{2}(u+v) \\ \frac{1}{2}(u-v) \\ \end{bmatrix} \,\Rightarrow\quad J_1=\frac{dR}{dq}= \begin{bmatrix} \frac{1}{2} & \frac{1}{2} \\ \frac{1}{2} & -\frac{1}{2} \\ \end{bmatrix}\\\\ ds^2=&a\,q'^T\,J_1^T\,\eta\,J_1\,q'=a\,du\,dv \end{align*} where $\eta= \begin{bmatrix} 1 & 0 \\ 0 & -1 \\ \end{bmatrix}\\\\$

Step II) \begin{align*} a&\mapsto {\it r_s}\,{{\rm e}^{-{\frac {r}{{\it r_s}}}}}{{\rm e}^{1/2\,{\frac {u-v }{{\it r_s}}}}}{r}^{-1} \\\\ ds^2&=a\,du\,dv={{\it du}}^{2}{\it r_s}\,{{\rm e}^{-1/2\,{\frac {2\,r-u+v}{{\it r_s}}}}} {r}^{-1}-{{\it dv}}^{2}{{\rm e}^{1/2\,{\frac {2\,r-u+v}{{\it r_s}}}}}r{ {\it r_s}}^{-1} \end{align*} Step III) \begin{align*} R & = \begin{bmatrix} -\exp\left(\frac{u}{2\,r_s}\right) \\ \exp\left(-\frac{v}{2\,r_s}\right) \\ \end{bmatrix}\,,\Rightarrow\quad J_2=\frac{dR}{dq}=\begin{bmatrix} -\frac{2\,r_s}{\exp\left(\frac{u}{2\,r_s}\right)} & 0 \\ & -\frac{2\,r_s}{\exp\left(-\frac{v}{2\,r_s}\right)} \\ \end{bmatrix}\\\\ ds^2=&q'^T\,J_2^T\,J_1^T\,g\,J_1\,J_2\,q'= \frac{4\,r_s^3\,\exp\left(-\frac{r}{r_s}\right)}{r}\,du\,dv \end{align*} Step IV \begin{align*} R & = \begin{bmatrix} u-v \\ u+v \\ \end{bmatrix}\,,\Rightarrow\quad J_3=\frac{dR}{dq}=\begin{bmatrix} 1 & -1 \\ 1 & 1 \\ \end{bmatrix}\\\\ ds^2=&q'^T\,J_3^T\,J_2^T\,J_1^T\,g\,J_1\,J_2\,J_3\,q' = \frac{4\,r_s^3\,\exp\left(-\frac{r}{r_s}\right)}{r}\left( du^2-dv^2 \right) \end{align*}

  • $\begingroup$ In the part "With matrixes and vectors" , how do you get the $\frac{1}{2}a$ term in the Jacobian of Step 1. It is not present in either $R$ or $q'$ $\endgroup$ – Alexander Cska Aug 6 '19 at 15:54
  • $\begingroup$ this is wrong, i will correct it. Thank you $\endgroup$ – Eli Aug 6 '19 at 15:58
  • $\begingroup$ @Alexander I hope it is now o.k ? $\endgroup$ – Eli Aug 6 '19 at 16:30
  • $\begingroup$ i think that there is a minus sign missing in the second row. This is due to the derivatives of the linear function $\frac{1}{2}\left ( u-v \right )$ $\endgroup$ – Alexander Cska Aug 7 '19 at 7:12
  • $\begingroup$ Yes sorry for that $\endgroup$ – Eli Aug 7 '19 at 7:13

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