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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?

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  • $\begingroup$ 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 $\endgroup$
    – Yukterez
    Commented Mar 25, 2023 at 0:53
  • $\begingroup$ @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. $\endgroup$
    – David Shaw
    Commented Mar 25, 2023 at 13:35

1 Answer 1

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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.

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  • $\begingroup$ But the $X$ and $V$ here is not independent, how to understand the $(\frac{\partial}{\partial X})_V$ here? $\endgroup$
    – David Shaw
    Commented Mar 27, 2023 at 19:58
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    $\begingroup$ @Xiao $X$ and $V$ form a perfectly good pair of independent coordinates. $\endgroup$
    – TimRias
    Commented Mar 27, 2023 at 22:38
  • $\begingroup$ It seems they are not orthogonal, and ss $X=\frac{1}{2}(V+W)$, it seems we just simply have $(\frac{\partial}{\partial X})_V=(\frac{\partial}{\partial W})_V$. $\endgroup$
    – David Shaw
    Commented Mar 28, 2023 at 13:54
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    $\begingroup$ @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). $\endgroup$
    – TimRias
    Commented Mar 28, 2023 at 14:14

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