# Coordinate time vector field: Schwarzschild vs Gullstrand-Painleve

• I
cianfa72
TL;DR Summary
About the coordinate time vector field of Schwarzschild geometry in Schwarzschild vs Gullstrand-Painleve coordinate chart
Hi,

I was reading this insight schwarzschild-geometry-part-1 about the transformation employed to rescale the Schwarzschild coordinate time ##t## to reflect the proper time ##T## of radially infalling objects (Gullstrand-Painleve coordinate time ##T##).

As far as I understand it, the vector field ##{\partial} / {\partial t}## is actually the same as ## {\partial} / {\partial T}##.

Does it sound right ? Thank you.

2022 Award
Yes. But you'll find that the spacelike coordinate bases aren't orthogonal to it, unlike in Schwarzschild coordinates (##\partial_r## isn't, anyway).

cianfa72
Yes. But you'll find that the spacelike coordinate bases aren't orthogonal to it, unlike in Schwarzschild coordinates (##\partial_r## isn't, anyway).
ok, starting from the transformation ##T= t - 2M \left [ -2 \sqrt {r/2M} + ln \left( \frac {\sqrt {r/2M} +1} {\sqrt {r/2M} -1} \right) \right ]## how do we get ## {\partial} / {\partial T} = {\partial} / {\partial t}## ?

2022 Award
Generally, you write down the Jacobean matrix, the matrix whose ##i,j##th element is ##\partial x^i/\partial x'^j## and apply it to a vector ##U'^j## in order to transform it from the ##x'^j## coordinate system to the ##x^i## system. In this case ##x^i## are the upper case coordinates and ##x'^j## are the lower case ones, the elements of the Jacobean are ##\partial T/\partial t## (etc), and you would apply it to the vector (1,0,0,0) (assuming ##t## is your first coordinate) and you'll get (1,0,0,0) out.

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ergospherical
Is there somewhere a definition of these coordinates?

ergospherical
I found them mentioned here: https://jila.colorado.edu/~ajsh/talks/heraeus_ajsh_19/gullstrandpainleve.html
but I think what I wrote beforehand is fine. For the given relationship ##f(t,T,r) = 0## then ##\dfrac{\partial }{\partial t} = \dfrac{\partial T}{\partial t} \dfrac{\partial}{\partial T} + \dfrac{\partial r}{\partial t} \dfrac{\partial}{\partial r} = \dfrac{\partial T}{\partial t} \dfrac{\partial}{\partial t}##, given ##r## and ##t## are independent.

cianfa72
Generally, you write down the Jacobean matrix, the matrix whose ##i,j##th element is ##\partial x^i/\partial x'^j## and apply it to a vector ##U'^j## in order to transform it from the ##x'^j## coordinate system to the ##x^i## system. In this case ##x^i## are the upper case coordinates and ##x'^j## are the lower case ones
I believe the Jacobian matrix of elements ##\partial x^i/\partial x'^j## should actually transform vectors ##U^j## to ##U'^j##. Anyway the elements of the first column should be ##(1,0,0,0)^T## hence the vector ## {\partial_T}## transforms in ##{\partial_t}## (i.e. ##{\partial_t} = {\partial_T} ##).

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• Ibix
2022 Award
Indeed - corrected above.

Edit: And put it back again - see #14

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ergospherical
I thought you had it right the first time, ##U^i = \dfrac{\partial x^i}{\partial {x'}^j} {U'}^j##.

• Ibix
2022 Award
I thought you had it right the first time, ##U^i = \dfrac{\partial x^i}{\partial {x'}^j} {U'}^j##.
I'm going to bed. I'll look at it in the morning...

ergospherical
I should probably do the same, but I noticed that season 2 of the Witcher has come out - so sleep can wait for a bit. • Ibix
Mentor
As far as I understand it, the vector field ##{\partial} / {\partial t}## is actually the same as ## {\partial} / {\partial T}##.
Yes. The only coordinate basis vector field that changes from Schwarzschild to Painleve coordinates is ##\partial / \partial r##.

ok, starting from the transformation ##T= t - 2M \left [ -2 \sqrt {r/2M} + ln \left( \frac {\sqrt {r/2M} +1} {\sqrt {r/2M} -1} \right) \right ]## how do we get ## {\partial} / {\partial T} = {\partial} / {\partial t}## ?
A quicker way (I think) than the brute force way others have described is to note the following: the difference ##T - t## is a function of ##r## only, and ##r## is the same in both charts (i.e., any point in spacetime has the same value of ##r## in both charts), and the integral curves of ##\partial / \partial t## are curves of constant ##r##. Therefore, the difference ##T - t## is constant along any integral curve of ##\partial / \partial t##. I believe this is sufficient to show that the integral curves of ##\partial / \partial T## must be identical to the integral curves of ##\partial / \partial t##.

cianfa72
I thought you had it right the first time, ##U^i = \dfrac{\partial x^i}{\partial {x'}^j} {U'}^j##.
Oops, I think @Ibix was right. I confused the use of Jacobian matrix ##\left ( \dfrac{\partial x^i}{\partial {x'}^j} \right )##.

Namely ##\left ( \dfrac{\partial x^i}{\partial {x'}^j} \right )## transforms the coordinate basis vectors ##\left \{{\partial} / {\partial x^i} = {\partial_i} \right \}## associated to ##x^i## coordinate system into ##\left \{ {\partial} / {\partial x'^i} = {\partial_i'}\right \}## associated to ##x'^i## coordinate system. From the point of view of vector components in those coordinate basis the transformation rule is the other way around (as @Ibix said in post #4). Nevertheless the end result is fine.

Sorry for the confusion Last edited:
• Ibix
Gold Member
2022 Award
It's easy to remember, if you always work with invariant objects, e.g., the vectors themselves. For the here considered "holonomous coordinates" the mnemonics is
$$\vec{A}=A^i \partial_i = A^{\prime j} \partial_j' =A^{\prime j} \frac{\partial x^i}{\partial x^{\prime j}} \partial_i,$$
from which you read off for the components
$$A^i=A^{\prime j} \frac{\partial x^i}{\partial x^{\prime j}}$$
or the other way
$$A^i \partial_i = A^i \frac{\partial x^{\prime j}}{\partial x^i} \partial_j'=A^{\prime j} \partial_j',$$
from which
$$A^{\prime j} = A^i \frac{\partial x^{\prime j}}{\partial x^i}.$$
This is of course consistent with the previous formula, because
$$\frac{\partial x^{\prime j}}{\partial x^i} \frac{\partial x^i}{\partial x^{\prime k}} = \frac{\partial x^{\prime j}}{\partial x^{\prime k}}=\delta_k^j.$$
The vector components thus transform contravariantly.

For forms you have the corresponding dual basis ##\mathrm{d}^i##. Any one-form ##L## thus has components wrt. the

$$\tilde{L}=L_i \mathrm{d}^i =L_j' \mathrm{d}^{\prime j}=L_j' \frac{\partial x^{\prime j}}{\partial x^i} \mathrm{d} x^i=L_i \frac{\partial x^i}{\partial x^{\prime j}} \mathrm{d} x^{\prime j},$$
from which
$$L_i = L_j' \frac{\partial x^{\prime j}}{\partial x^i} \; \Leftrightarrow \; L_j' = L_i \frac{\partial x^i}{\partial x^{\prime j}}.$$
So the components of a linear form ##\tilde{L}## transform covariantly.

• Ibix and cianfa72
2022 Award
Sorry for the confusion No, I should have got pen and paper and checked instead of trying to do it in my head. We seem to be all sorted now (and I've un-un-corrected my post above).

cianfa72
For forms you have the corresponding dual basis ##\mathrm{d}^i##
At this level no metric is involved -- i.e. we are really talking of a smooth manifold with a tangent & cotangent (dual) vector spaces defined on each point of it.

In other words at this level there is not a natural isomorphism between vector & covector (dual) space at each point. To realize it we need to pick a basis in the vector space (e.g. the coordinate/holonomic basis associated to a given coordinate chart and the corresponding basis in the dual space).

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• vanhees71