Transformation of basis vector in Kruskal coordinate

Question

For Schwarzschild solution:

$\mathrm{d} s^2=\left(1-\frac{2m}{r}\right)dt^2 - \left(1-\frac{2m}{r}\right)^{-1}\mathrm{d} r^2-r^2(\mathrm{d}\theta^2+sin^2\theta \mathrm{d}\phi^2),$

Introduce tortoise coordinate:

$ v=t+r_*, \quad w=t-r_*,$ where $r_*=r+2mln(r-2m)$,

the metric becomes

$\mathrm{d} s^{2}=\left (1-\frac{2m}{r}\right )\mathrm{d} v \mathrm{d} w-r^{2}\left(\mathrm{d} \theta^{2}+\sin ^{2} \theta \mathrm{d} \phi^{2}\right).$

Next, define two new coordinate

$ V= e^{v/4m},\quad W=e^{w/4m},$

and then choosing

$T=\frac{1}{2}(V+W), \quad X=\frac{1}{2}(V-W),$

the following metric is obtained

$\mathrm{d} s^{2}= \frac{16 m^{2}}{r} e^{ (-r/2 m)} \mathrm{d} T^{2}-\frac{16 m^{2}}{r} e^{ (-r/2 m)} \mathrm{d} X^{2} -r^{2}\left(\mathrm{d} \theta^{2}+\sin ^{2} \theta \mathrm{d} \phi^{2}\right).$

If I have a vector $\partial/\partial r$, and I want to get it component in $\partial/\partial V$ and $\partial/\partial X$, what is the corresponding coefficient, can I calculate $\partial V /\partial r$ and $\partial X/\partial r$?

It seems there are two ways to decompose $\frac{\partial}{\partial r}$ onto $\frac{\partial}{\partial V}$ and $\frac{\partial}{\partial X}$

(1) directly decompose $\frac{\partial}{\partial r}$ onto $\frac{\partial}{\partial V}$ and $\frac{\partial}{\partial X}$:

$\frac{\partial}{\partial r}=\frac{\partial V}{\partial r}\frac{\partial}{\partial V}+\frac{\partial X}{\partial r}\frac{\partial}{\partial X}$.

(2) decompose $\frac{\partial}{\partial r}$ onto $\frac{\partial}{\partial V}$ and $\frac{\partial}{\partial W}$, and combine then to form a $\frac{\partial}{\partial X}$:

$\frac{\partial}{\partial r}=\frac{\partial V}{\partial r}\frac{\partial}{\partial V}+\frac{\partial W}{\partial r}\frac{\partial}{\partial W}=\frac{\partial V}{\partial r}\frac{\partial}{\partial V}+\frac{\partial W}{\partial r}(\frac{\partial}{\partial V}-2\frac{\partial}{\partial X})$.

It could be viewed that the coefficient of $\frac{\partial}{\partial V}$ doesn't match, so one of them must be wrong, which is right?

Answer 1

When specifying partial derivatives it is important to specify what coordinate is being keeping fixed in addition to what coordinate is being varied. Without this information the partial derivative can mean any number of things. When using a fixed coordinate system, it is usually taken as implicit that the quantities held fixed are the other coordinates, without further notation. However, when mixing coordinates from different systems it becomes vital to specify explicitly what the other quantity(ies) being kept fixed is/are.

One typical notation for this is to add the quantities kept fixed as a subscript on parentheses, e.g. $$ \left(\frac{\partial}{\partial r}\right)_{t},$$ means the partial derivative giving the change in the $r$ direction while keeping $t$ fixed. (For sake of conciseness we can forget about the angular coordinates.) Similarly, $$ \left(\frac{\partial}{\partial r}\right)_{v},$$ means the partial derivative giving the change in the $r$ direction while keeping $v$ fixed. These are not the same directions, instead they are related by $$ \left(\frac{\partial}{\partial r}\right)_{t}= \left(\frac{\partial}{\partial r}\right)_{v} +\left(\frac{\partial v}{\partial r}\right)_{t}\left(\frac{\partial}{\partial v}\right)_{r}.$$

Applying this to the manipulation you are trying (1) becomes

$$ \left(\frac{\partial}{\partial r}\right)_{t} = \left(\frac{\partial V}{\partial r}\right)_{t}\left(\frac{\partial}{\partial V}\right)_{X} + \left(\frac{\partial X}{\partial r}\right)_{t}\left(\frac{\partial}{\partial X}\right)_{V} ,$$ while (2) becomes \begin{align} \left(\frac{\partial}{\partial r}\right)_{t} &= \left(\frac{\partial V}{\partial r}\right)_{t}\left(\frac{\partial}{\partial V}\right)_{W} + \left(\frac{\partial W}{\partial r}\right)_{t}\left(\frac{\partial}{\partial W}\right)_{V} \\ &= \left(\frac{\partial V}{\partial r}\right)_{t} \left[ \left(\frac{\partial}{\partial V}\right)_{X} + \left(\frac{\partial X}{\partial V}\right)_{W}\left(\frac{\partial}{\partial X}\right)_{V} \right] + \left(\frac{\partial W}{\partial r}\right)_{t} \left(\frac{\partial X}{\partial W}\right)_{V} \left(\frac{\partial}{\partial X}\right)_{V} \\ &= \left(\frac{\partial V}{\partial r}\right)_{t} \left(\frac{\partial}{\partial V}\right)_{X} + \left[ \left(\frac{\partial V}{\partial r}\right)_{t} \left(\frac{\partial X}{\partial V}\right)_{W} + \left(\frac{\partial W}{\partial r}\right)_{t} \left(\frac{\partial X}{\partial W}\right)_{V} \right]\left(\frac{\partial}{\partial X}\right)_{V} \\ &= \left(\frac{\partial V}{\partial r}\right)_{t} \left(\frac{\partial}{\partial V}\right)_{X} + \left(\frac{\partial X}{\partial r}\right)_{t} \left(\frac{\partial}{\partial X}\right)_{V}, \end{align} where in the second line we used that $\left(\frac{\partial V}{\partial W}\right)_{V}=0$, because we are keeping $V$ fixed. We thus see that in fact (1) and (2) are equal.