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)?


2 Answers 2


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 $q'$ $\endgroup$ Aug 6, 2019 at 15:54
  • $\begingroup$ this is wrong, i will correct it. Thank you $\endgroup$
    – Eli
    Aug 6, 2019 at 15:58
  • $\begingroup$ @Alexander I hope it is now o.k ? $\endgroup$
    – Eli
    Aug 6, 2019 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 $\frac{1}{2}\left ( u-v \right )$ $\endgroup$ Aug 7, 2019 at 7:12
  • $\begingroup$ Yes sorry for that $\endgroup$
    – Eli
    Aug 7, 2019 at 7:13

I would like to add that, contrary to the other answer, it is possible to find the metric using the Jacobian.

The transformation law for the metric tensor from Schwarzschild coordinates to Kruskal–Szekeres coordinates is as follows:

$$g_{\bar{\alpha}\bar{\beta}}=\frac{\partial x^\alpha}{\partial x^\bar{\alpha}}\frac{\partial x^\beta}{\partial x^\bar{\beta}}g_{\alpha\beta}$$

where barred indices correspond to Kruskal–Szekeres coordinates.

We need to find the Jacobian matrix $\frac{\partial x^\alpha}{\partial x^\bar{\alpha}}$, as you have already done so. The mistake that you made was leaving out the metric terms $g_{\alpha\beta}$. This is a key part of the solution, as we are not transforming from Cartesian coordinates which have the Kronecker delta as its metric.

For example, the calculation for $g_{TT}$ would be as such: \begin{align*} g_{TT}&=\frac{\partial x^\alpha}{\partial T}\frac{\partial x^\beta}{T}g_{\alpha\beta}\\ &=\frac{\partial t}{\partial T}\frac{\partial t}{T}g_{tt}+\frac{\partial r}{\partial T}\frac{\partial r}{T}g_{rr}\\ &=16G^2M^2 e^{-\frac{r}{GM}}\left[-\left(\frac{X}{\frac{r}{2GM}-1}\right)^2 \left(1-\frac{2GM}{r}\right)+\left(-\frac{2GMT}{r}\right)^2 \left(1-\frac{2GM}{r}\right)^{-1}\right]\\ &=16G^2M^2 e^{-\frac{r}{GM}}\left(T^2-X^2\right)\frac{(2GM/r)^2}{1-\frac{2GM}{r}}\\ &=-\frac{32G^3M^3}{r}e^{-\frac{r}{2GM}} \end{align*}

where I used $\frac{\partial t}{\partial T}=\frac{X}{\frac{r}{2GM}-1}$ and $\frac{\partial r}{\partial T}=-\frac{2GMT}{r}$ together with the Schwarzschild metric components in the third line.

I think you can figure out the rest.


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