Given $$ds^2=-A(r)dt^2+B(r)dr^2+2C(r)drdt+D(r)r^2(d\theta^2+\sin^2d\phi^2),\tag{23.1}$$ we want to eliminate that cross term $2C(r)drdt$ Upon change of variable such that. This can happen upon $$T(t,r)=t+\psi(r)\tag{23.2}$$ where $$dT^2=dt^2+\psi'^2dr^2+2\psi'drdt.$$and choosing $\psi$ to satisfy $$\frac{d\psi(r)}{dr}=-\frac{C(r)}{A(r)},\tag{23.4}$$ says Blau in his lecture notes (http://www.blau.itp.unibe.ch/GRLecturenotes.html), page 481. Now, let us try to work backwards and see if this helps us eliminate the cross term $2C(r)dtdr$.

We plug $\psi(r)/dr=\psi'=-C/A$ in $dT^2$ above. We get $$dT^2=dt^2+C^2/A^2-2C/Adrdt.$$ Plugging the latter result in $ds^2$, we get $$-Adt^2-C^2/Adr^2+2Cdrdt+Bdr^2+Dr^2d\Omega^2.$$ And thus the cross term is NOT eliminated but rather summed up.

As I am trying to do the details to see if this works, turns out it does not work unless $$\frac{d\psi(r)}{dr}=\frac{C(r)}{A(r)},\hspace{1cm}\text{ dropped the minus sign}.$$ Am I on the right track or am I mistaken here? I am not sure if this is a typo or not.


The book is right, in fact, but we can be sure by checking a simple example. Let's take all your functions ($A, B, C, D$) equal to $1$, and ignore the angular part:

$$ds^2 = -dt^2 + dr^2 +2\,dt\,dr$$

Now the question is whether $\psi = r$ (your version) or $\psi = -r$ (the book's version). Let's say $\psi = \epsilon\, r$, with $\epsilon = \pm 1$. Replacing:

$$ds^2 = -dT^2 -2\epsilon\, dr^2 + 2(\epsilon+1)dT\,dr$$

So we need $\epsilon =-1$, that is, $d \psi / dr = -C/A$.

Edit: You seem to be having trouble with a particular thing. The original coordinates are $(t,r)$. Therefore, finding $dT^2$ in terms of $dt$ and $dr$ is not very useful, since what we want to do is to find $dt^2$ in terms of $dT$ and $dr$! The relation is $dt = dT-\psi' dr$ and so $dt^2 = dT^2 -2\psi' dt\,dr +\psi'^2 dr^2$, which you should be plugging into your original expression for $ds^2$ to find the new metric written in terms of $T$ and $r$.

  • $\begingroup$ But @Javier, if you simply substitute $\psi'=-C/A$ in $dT^2=dt^2+\psi'dr^2+2\psi'drdt$ and then substituting those in the $ds^2$ in my question above, then the cross terms $(2C(r)drdt)$ will add instead of cancelling out. Please if you may consider keeping the coefficients the same instead of setting them equal to 1 so that we read the same page. $\endgroup$ – Beyond-formulas Oct 24 '15 at 23:41
  • $\begingroup$ @Beyond-formulas: how do you plan on substituting that into the metric? The old metric is written in terms of $t$, not $T$. You should solve for $dt$ and plug everything in, you'll get the right sign. $\endgroup$ – Javier Oct 24 '15 at 23:43
  • $\begingroup$ I edited my question with all the calculation-details. $\endgroup$ – Beyond-formulas Oct 25 '15 at 0:06
  • $\begingroup$ Given your edit, there are 2 terms having each $dr^2$ but different coefficients that get summed up; is that correct? $\endgroup$ – Beyond-formulas Oct 25 '15 at 14:36

The book is right. Sticking to the time-radius sector, we are looking at the transformation \begin{align} t & \to T = t + \psi, \\ r & \to r, \end{align} where $\psi$ is a function of only $r$. In the new coordinates, we have \begin{align} \require{cancel} g_{Tr} & = \frac{\partial x^\mu}{\partial T} \frac{\partial x^\nu}{\partial r} g_{\mu\nu} \\ & = \cancelto{1}{\frac{\partial t}{\partial T}} \left(\cancelto{-\psi'}{\frac{\partial t}{\partial r}} g_{tt} + \cancelto{1}{\frac{\partial r}{\partial r}} g_{tr}\right) + \cancelto{0}{\frac{\partial r}{\partial T}} \left(\frac{\partial t}{\partial r} g_{rt} + \frac{\partial r}{\partial r} g_{rr}\right) \\ & = -\psi' g_{tt} + g_{tr}. \end{align} For $g_{Tr}$ to vanish, it is necessary and sufficient to choose $$ \psi' = \frac{g_{tr}}{g_{tt}} = -\frac{C}{A}. $$

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    $\begingroup$ @Javier Haha thanks. This is why I usually avoid sign error questions. $\endgroup$ – user10851 Oct 24 '15 at 23:31
  • $\begingroup$ I actually didn't get the minus sign the first time I did it, and then when I got it right I left a comment and immediately deleted it because I thought I was wrong again! Sign errors are best avoided by Monte-Carlo simulation: do the math many times over and see what sign comes up the most. $\endgroup$ – Javier Oct 24 '15 at 23:34
  • $\begingroup$ @ChrisWhite, please check the edited part of my question and see my point more clearly. $\endgroup$ – Beyond-formulas Oct 25 '15 at 0:07

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