# Transformation of basis vector in Kruskal coordinate

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?

• In this link your T,X is called u,w so don't confuse that with your v,w: Kruskal Szekeres transformation to r,t Commented Mar 25, 2023 at 0:53
• @Yukterez I think my question is slightly different. I showed two ways to decompose $\frac{\partial}{\partial r}$ onto $\frac{\partial}{\partial V}$ and $\frac{\partial}{\partial X}$, and they lead to two different formula, I just confused which way is right. Commented Mar 25, 2023 at 13:35

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 $$\left(\frac{\partial}{\partial r}\right)_{t}= \left(\frac{\partial}{\partial r}\right)_{v} +\left(\frac{\partial v}{\partial r}\right)_{t}\left(\frac{\partial}{\partial v}\right)_{r}.$$
$$\left(\frac{\partial}{\partial r}\right)_{t} = \left(\frac{\partial V}{\partial r}\right)_{t}\left(\frac{\partial}{\partial V}\right)_{X} + \left(\frac{\partial X}{\partial r}\right)_{t}\left(\frac{\partial}{\partial X}\right)_{V} ,$$ 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.
• But the $X$ and $V$ here is not independent, how to understand the $(\frac{\partial}{\partial X})_V$ here? Commented Mar 27, 2023 at 19:58
• @Xiao $X$ and $V$ form a perfectly good pair of independent coordinates. Commented Mar 27, 2023 at 22:38
• It seems they are not orthogonal, and ss $X=\frac{1}{2}(V+W)$, it seems we just simply have $(\frac{\partial}{\partial X})_V=(\frac{\partial}{\partial W})_V$. Commented Mar 28, 2023 at 13:54
• @Xiao Coordinates do not need to be orthogonal ($v$ and $w$ are not orthogonal). And yes $\left(\frac{\partial}{\partial X}\right)_{V}$ is proportional to $\left(\frac{\partial}{\partial W}\right)_{V}$. (With the factor -1/2). Commented Mar 28, 2023 at 14:14