Couple questions about time re-scaling function in Birkhoff’s theorem

[Pencilvester]

Towards the end of proving Birkhoff's theorem, you have a line element of the form $$ds^2 = -b(t) \cdot (1 - a/r) dt^2 + (1 - a/r)^{-1} dr^2 + r^2 d\Omega^2$$where $a$ is some constant and $b$ is a (positive) function of $t$. We are free to define a new coordinate $t'$ such that $dt' = \sqrt{b(t)} dt$ to eliminate $b$ from the metric. My question is, since $b(t)$ just represents your freedom to re-scale the time coordinate however you want and could be anything (positive), in order to get rid of it (or at least make it less visible), would we be equally justified in simply choosing that $b(t) = 1$?

And while we’re on the subject, would we also be equally justified in requiring $b$ to be positive for the reason that we want our manifold to be Lorentzian, not Reimannian?

 

[PeterDonis]

since $b(t)$ just represents your freedom to re-scale the time coordinate however you want and could be anything (positive), in order to get rid of it (or at least make it less visible), would we be equally justified in simply choosing that $b(t) = 1$?

That's what you are doing when you say $dt' = \sqrt{b(t)} dt$. You are specifying a coordinate chart in which $b(t) = 1$.

would we also be equally justified in requiring $b$ to be positive for the reason that we want our manifold to be Lorentzian, not Reimannian?

That's what you had to do to get the metric to the form you wrote at the start of your post.

See this Insights article:

https://www.physicsforums.com/threads/a-short-proof-of-birkhoffs-theorem-comments.828225/

I used $f(t)$ in that article for what you are calling $b(t)$.

 

[Pencilvester]

That's what you are doing when you say $dt' = \sqrt{b(t)} dt$. You are specifying a coordinate chart in which $b(t) = 1$.

I think I see what you're saying. I was originally thinking that by choosing, for example, $b(t) = t^2 + 1$, then performing the coordinate transformation where $dt' = \sqrt{b} ~ dt$, this would mean $b(t) \neq 1$ but we're still able to make it go away with the change of coordinates. But you're saying that choosing a particular $b$ can be done via coordinate transformation, i.e. in performing this particular coordinate transformation, we have a new $b$ such that $b = (t^2 + 1)/(t^2 + 1) = 1$, yes?

That's what you had to do to get the metric to the form you wrote at the start of your post.

See this Insights article:

https://www.physicsforums.com/threads/a-short-proof-of-birkhoffs-theorem-comments.828225/

I used $f(t)$ in that article for what you are calling $b(t)$.

I have read your insights article on the subject, and found it very helpful and influential in my own comprehension of the proof. In the process of replying, I think I’ve worked out everything I was confused about. The direct reason you give for $b$ being positive is that your coordinate transformation will take the square root of $b$. I can see now that if the transformation takes complex values, it will change the signature of the metric to ++++. Given my recent thread, you may remember this was a new idea to me in the context of coordinate transformations, so I was a little thrown, but I'm getting the grasp of it now. Another thing that threw me was this: At the beginning of your insights article, you mention how the unknown metric components that MTW use in their proof are positive from the outset, but in your proof you don't make this requirement. But to get from $$ds^2 ~ = ~ -j ~ (dt')^2 ~ + ~ 2k ~ dt' ~ dr ~ + ~ l ~ dr^2 ~ + ~ r^2 ~ d\Omega^2$$to$$ds^2 ~ = ~ -f ~ dt^2 ~ + ~ h ~ dr^2 ~ + ~ r^2 ~ d\Omega^2$$ you need this transformation: $\sqrt{f} ~ dt = \sqrt{j} ~ dt' - k/\sqrt{j} ~ dr$ (and to define $h$ to be $h \equiv l + k^2 / j$). Because of the way I had it written, I was thinking if $b$ had to be positive because of a square root, then surely $f$ and $j$ must be as well, but if I move $f$ to the RHS, it becomes obvious that the only requirement would be that the signs of $j$ and $f$ match. So with these last few threads, I think I am all squared away now! Thanks to everyone who helped me!

 

[PeterDonis]

I was originally thinking that by choosing, for example, $b(t) = t^2 + 1$

You don't "choose" what $b(t)$ is originally. It is just whatever it happens to be when you put the line element in that particular form.

then performing the coordinate transformation where $dt' = \sqrt{b} ~ dt$, this would mean $b(t) \neq 1$ but we're still able to make it go away with the change of coordinates.

You're thinking of it wrong. "Making $b(t) = 1$" means changing what $b(t)$ is so that it equals 1. A better way to say it is that you are finding a coordinate transformation that takes $b(t)$ in the old coordinates to $b' (t')$ in the new coordinates, such that $b' (t') = 1$. But an even better way to think of it is that you are removing the function $b(t)$ from that component of the metric, i.e., you are removing any variation of the metric components with $t$, by finding an appropriate coordinate transformation.

you're saying that choosing a particular $b$ can be done via coordinate transformation, i.e. in performing this particular coordinate transformation, we have a new $b$ such that $b = (t^2 + 1)/(t^2 + 1) = 1$, yes?

I don't think that's the best way to think of it, since it seems to be confusing you. See above.

to get from $$ds^2 ~ = ~ -j ~ (dt')^2 ~ + ~ 2k ~ dt' ~ dr ~ + ~ l ~ dr^2 ~ + ~ r^2 ~ d\Omega^2$$to$$ds^2 ~ = ~ -f ~ dt^2 ~ + ~ h ~ dr^2 ~ + ~ r^2 ~ d\Omega^2$$

I don't discuss that part at all in the Insights article. If you want to see how the metric gets to the form where it is diagonal, i.e., no $t$, $r$ cross term, that's a different discussion, which is not limited to Birkhoff's Theorem: the metric for any spherically symmetric spacetime can be written in the form I give at the start of the Insights article. MTW discuss this in Box 23.3.

 

[Pencilvester]

You're thinking of it wrong. "Making $b(t) = 1$" means changing what $b(t)$ is so that it equals 1. A better way to say it is that you are finding a coordinate transformation that takes $b(t)$ in the old coordinates to $b' (t')$ in the new coordinates, such that $b' (t') = 1$.

Ok, ok, I think I get what you’re saying, so let me attempt to paraphrase again, and you can tell me if I’m getting there. The line element that includes $b$ as an unspecified function is expressed the way it is because you’re using a specific coordinate system, namely a specific coordinate $t$ (in keeping with my original post), a coordinate which encapsulates the ambiguity of the freedom to rescale it while maintaining spherical symmetry. If we want $b$ to take some specific form, then we would perform a simple coordinate transformation to achieve this. Am I getting it?

I don't discuss that part at all in the Insights article.

I know, I wrote that just in case anyone in the future has the same confusion I did. Part of my confusion came from using the square root as the reason for requiring $b$ to be positive in the context of a proof which allows $g_{tt}$ to be positive or negative (which is in your article). I just wanted to state the confusion and how I cleared it up for myself (as well as how I created that confusion for myself in the first place).
As an aside, I really do need to examine MTW’s box 23.3, but I think I need a little break to work on some hyperbolic geometry and crochet problems.

 

[PeterDonis]

The line element that includes $b$ as an unspecified function is expressed the way it is because you’re using a specific coordinate system, namely a specific coordinate $t$ (in keeping with my original post), a coordinate which encapsulates the ambiguity of the freedom to rescale it while maintaining spherical symmetry.

It seems strange to call this a "specific" coordinate $t$, since the whole point is that it's not specific; there are an infinite number of choices for $t$ that are all consistent with that form of the line element--each one corresponding to a different function $b(t)$. So the reason the function $b(t)$ is there is that we have not gotten specific about the $t$ coordinate; we haven't yet used our freedom to rescale it.

If we want $b$ to take some specific form, then we would perform a simple coordinate transformation to achieve this.

I suppose this would be true for any specific form of $b(t)$, yes, but I don't see what the point would be.

 

[Pencilvester]

It seems strange to call this a "specific" coordinate $t$, since the whole point is that it's not specific; there are an infinite number of choices for $t$ that are all consistent with that form of the line element--each one corresponding to a different function $b(t)$.

Oooookay, $\theta$ and $\phi$ are standard angular coordinates on a 2-sphere; they are well defined. We explicitly defined $r$ to be the circumference of any given 2-sphere divided by $2\pi$; it is also well defined. We made a few demands of orthogonality on the coordinates, but otherwise we said nothing specific about $t$. I get it now. The line element in that form reveals your freedom to make a choice of any well defined* $t$ (or at least any that doesn’t mess with its orthogonality to the other coordinates), but there is only one choice that removes the metric’s dependence on it, namely the one where $dt’ = \sqrt{b}~dt$. Thank you for bearing with me!

*[Edit] well-defined might not be the best phrase because there’s not really anything wrong with how we defined $t$ up to that point. Maybe specifically defined would be better?

 

[PeterDonis]

well-defined might not be the best phrase because there’s not really anything wrong with how we defined up to that point. Maybe specifically defined would be better?

Yes.