I have a question regarding Misner, Charles W.; Thorne, Kip S.; Wheeler, John Archibald (1973), Gravitation ISBN 978-0-7167-0344-0. It is a book about Einstein's theory of gravitation.

At page 313, the exercise 13.2. "Practice with Metric" presents a four-dimensional manifold in spherical coordinates + $v$ that has a line element $$ds^2 = - (1-2 M/r) dv^2 + 2 dv dr+r^2 (d\theta^2 + sin^2 \theta d\phi^2).$$

The question (b) is:

Define a scalar field $t$ by $$t \equiv v - r - 2M \ln((r/2M)-1)$$What are the covariant and contravariant compoenents of the 1-form $dt$ (equal to u tilde)? What is the squared length $u^2$of the corresponding vector? Show that $u$ is timelike in the region $R > 2M$.

My attempt:

First differentiate to get the 1-form $dt$:

$$dt = dv - dr - dr/2M \cdot \frac{1}{(r/2M)-1} = dv - dr (1+\frac{1}{r-2M})$$

However, the correction tells that $u_r = -1/(1-2M/r)$ which is not equivalent to what I wrote. Where is my mistake?

I understand that the squared length of $u$ comes from the non-zero term v covariant and contravariant: $1\cdot -1/(1-2M/r)$ and the r, $\phi$ and $\theta$ terms have zero components in the contravariant terms.

Now to prove that it is timelike in a certain region, I need to do the dot product of dt with the spatial components and find zero? For the angles, it seems rather trivial, but for $r$, I am not sure how to show that $dt \cdot dr = 0$. Could someone help me please?

  • $\begingroup$ Your expression for $dt$ is dimensionally inconsistent. It doesn’t make sense to write $1+\frac{1}{r-2M}$ since $r$ and $M$ are lengths. $\endgroup$ – G. Smith Jul 1 '19 at 19:37
  • $\begingroup$ @G.Smith Typo, I forgot a r in top. $\endgroup$ – PackSciences Jul 1 '19 at 19:42
  • $\begingroup$ Where would that come from? What belongs on top is the $2M$ that Leah’s answer pointed out. My point is that the dimensional inconsistency should have been a clue that you didn’t differentiate correctly. $\endgroup$ – G. Smith Jul 1 '19 at 19:46

You forgot the $2M$ multiplied by the $\ln(r/2M-1)$.

$\rm {d} $$t$$=\rm {d} $$v$$-\rm {d} $$r$$-\frac{2M}{r/2M-1}\cdot\frac{1}{2M}\rm d$$r$

$\rm d$$t$$=\rm d$$v-$$\big(1+\frac{1}{r/2M-1}\big)\rm d $$r$

$\rm d$$t$$=\rm d$$v-$$\frac{1}{1-2M/r}\rm d $$r$

You now have $u_v = 1,u_r=-1/(1-2M/r), u_\theta=u_\varphi=0$

$u^v=g^{v \mu}u_\mu=1\cdot u_r=-1/(1-2M/r)$ and $u^r=g^{r \mu}u_\mu=1\cdot u_v+(1-2M/r)\cdot u_r=1-1=0$

$u^{\mu} u_{\mu}= -1/(1-2M/r)$ is negative in the region $r>2M$ and hence timelike.

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