Coord. Time Vector Field: Schwarzschild vs Gullstrand-Painleve

[cianfa72]

TL;DR

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.

 

[Ibix]

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 ${\partial} / {\partial T} = {\partial} / {\partial t}$ ?

 

[Ibix]

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.

 

[ergospherical]

Is there somewhere a definition of these coordinates?

 

[Ibix]

They're discussed in Peter's Insight linked in the OP, or there's https://en.wikipedia.org/wiki/Gullstrand–Painlevé_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. ${\partial_t} = {\partial_T}$).

 

[Ibix]

Indeed - corrected above.

Edit: And put it back again - see #14

 

[ergospherical]

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

 

[Ibix]

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.

 

[PeterDonis]

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

 

[vanhees71]

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]

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).