# Relativity and components of a 1-form

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?

• 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. – G. Smith Jul 1 '19 at 19:37
• @G.Smith Typo, I forgot a r in top. – PackSciences Jul 1 '19 at 19:42
• 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. – G. Smith Jul 1 '19 at 19:46

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

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

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

$$\rm dt=\rm dv-\frac{1}{1-2M/r}\rm d r$$

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.