Algebra Question: Where Does the 2 Come From?

[binbagsss]

Probably a really stupid question..

$u=t+r+2M ln(\frac{r}{2M}-1)$
From this I get
$\frac{du}{dr}=(1-\frac{2M}{r})^{-1}$

But, 1997 Sean M. Carroll lectures notes get $\frac{du}{dr}=2(1-\frac{2M}{r})^{-1}$ . (equation 7.71).

No idea where this factor of 2 comes from.

Thanks

 

[PeterDonis]

$u=t+r+2M ln(\frac{r}{2M}-1)$

From this I get

$\frac{du}{dr}=(1-\frac{2M}{r})^{-1}$

First, $u$ is not just a function of $r$, so what you are doing here is taking a partial derivative $\partial u / \partial r$ with $t$ held constant. That is, you are looking at how $u$ changes with $r$ along a curve of constant $t$. This is not the same as what Carroll is doing in the equation you refer to (see below).

Second, you might want to check your algebra; the result you are getting for $\partial u / \partial r$ does not look right.

1997 Sean M. Carroll lectures notes get $\frac{du}{dr}=2(1-\frac{2M}{r})^{-1}$ . (equation 7.71).

That equation is a solution for an outgoing null geodesic; it is not the same thing as you were trying to derive in what I quoted above. It is derived by setting $ds^2 = 0$ in equation 7.69 (and also setting $d\Omega^2 = 0$ so the curve is purely radial) and rearranging the resulting equation into an equation for $du / dr$. (Note that there are two ways of doing this, corresponding to whether $dr$ is positive or negative for a positive $du$. If $dr$ is positive, we have an outgoing geodesic; if it is negative, we have an ingoing geodesic. That's why Carroll gives two solutions in equation 7.71.)