Line element in Kerr coordinates at singularity

[chartery]

From paper 'A brief introduction to Kerr spacetime' ( https://arxiv.org/pdf/0706.0622 )

setting m->0 in the line element in Kerr coordinates gives, equation 7 :
$$ \text{d}s_0^{2} = -\left( \text{d}u + a\sin^{2}\theta \text{d}\phi \right)^{2}+2\left( \text{d}u + a\sin^{2}\theta \text{d}\phi \right)\left( \text{d}r + a\sin^{2}\theta \text{d}\phi \right)+\left( r^{2} +a^2 \cos^{2}\theta\right)\left( \text{d}\theta^{2} + \sin^{2}\theta \text{d}\phi \right) $$
At the singularity $\left( r=0\text{, } \theta=\frac{\pi}{2}\right)$ the author gets: $~~~~\text{d}s_0^{2}|_{singularity}= -\text{d}u^{2}+a^{2}\text{d}\phi^{2} ~~~~$, equation 19.

But when I do the substitution I get additional cross terms $~~~\text{d}u\text{d}r~~$ and $~~a\text{d}r\text{d}\phi~~~$.
These must obviously be zero, but I cannot see why. Must $\text{d}r$ vanish if $r$ is zero/constant ?

 

[Ibix]

In a hurry, so without looking: is that meant to be the line element along the singularity? The singularity has constant $r$ all the way around.

Also, I think Carroll considers the $m\rightarrow 0$ case near the end of chapter 7 of his lecture notes - you could take a look for asecond opinion.

 

[chartery]

In a hurry, so without looking: is that meant to be the line element along the singularity? The singularity has constant $r$ all the way around.

My understanding was that the line element is equivalent to the metric, the (curvature) singularity of the metric is at $\left( r=0\text{, } \theta=\frac{\pi}{2}\right)$, and is circular. The problem is with the $\text{d}r$ or cross terms in these coordinates.

Also, I think Carroll considers the $m\rightarrow 0$ case near the end of chapter 7 of his lecture notes - you could take a look for asecond opinion.

He is using Boyer-Lindquist coordinates, and I am not knowledgeable enough for the swapping to be helpful.

 

[PeterDonis]

Must $\text{d}r$ vanish if $r$ is zero/constant ?

Obviously yes, since $r$ being constant means it doesn't change, and $dr$ must be zero if $r$ doesn't change.

 

[chartery]

Obviously yes, since $r$ being constant means it doesn't change, and $dr$ must be zero if $r$ doesn't change.

Thanks. I was just uneasy about the simplicity of that deduction in circumstances where the metric was becoming singular and applying to a basis $\text{d}r$ (as opposed to the differential $dr$).

 

[JimWhoKnew]

At the singularity $\left( r=0\text{, } \theta=\frac{\pi}{2}\right)$ the author gets: $~~~~\text{d}s_0^{2}|_{singularity}= -\text{d}u^{2}+a^{2}\text{d}\phi^{2} ~~~~$, equation 19.

But when I do the substitution I get additional cross terms $~~~\text{d}u\text{d}r~~$ and $~~a\text{d}r\text{d}\phi~~~$.
These must obviously be zero, but I cannot see why. Must $\text{d}r$ vanish if $r$ is zero/constant ?

The vertical line in Equation 19 means a restriction of $~ds^2_0~$ to the 2D surface $~r=0~,~\theta=\pi/2~$ . From this line element you can deduce that in Cartesian coordinates it is the cylinder surface given by $~z=0~,~x^2+y^2=a^2~$ ($~ds^2_0~$ is a line element in Minkowski spacetime which has no singularities, of course).

 

[PeterDonis]

a basis $\text{d}r$ (as opposed to the differential $dr$).

I don't understand what you mean by this.

 

[chartery]

The vertical line in Equation 19 means a restriction of $~ds^2_0~$ to the 2D surface $~r=0~,~\theta=\pi/2~$ . From this line element you can deduce that in Cartesian coordinates it is the cylinder surface given by $~z=0~,~x^2+y^2=a^2~$ ($~ds^2_0~$ is a line element in Minkowski spacetime which has no singularities, of course).

Also thanks. Reducing the dimension of a metric (by fixing an index) as a route to geometric insight got a bit lost for me trying to follow the abstruse coordinate manipulations!

 

[chartery]

I don't understand what you mean by this.

My inexpert studies had left me with the impression one should be careful about distinction between a basis vector and its associated differential (or partial). I wasn't certain about all the circumstances in which that care was required.

 

[PeterDonis]

istinction between a basis vector and its associated differential (or partial).

$dr$ is not a vector, it's a differential. So there is no such thing as "a basis $dr$" as opposed to "a differential $dr$". There is a coordinate basis vector $\partial / \partial r$, but that's not the same thing as $dr$.

 

[chartery]

$dr$ is not a vector, it's a differential. So there is no such thing as "a basis $dr$" as opposed to "a differential $dr$". There is a coordinate basis vector $\partial / \partial r$, but that's not the same thing as $dr$.

Hmm. I had $\partial_r$ as basis for contravariant vectors and $\text{d}r$ (not $dr$) as basis for covariants. And line element as $ds^2 = g_{\mu\nu}\text{d}x^{\mu}\text{d}x^{\nu}$. Carroll seems to go to a lot of trouble to distinguish gradients from differentials. If not so, then my misunderstandings go possibly too deep for recovery.

 

[PeterDonis]

Carroll

Where does Carroll come into it? The reference you gave in the OP of this thread is a paper by Matt Visser.

distinguish gradients from differentials.

These are two different things, yes. Most texts, even ones that observe the distinction between differentials and 1-forms (covectors), such as Misner, Thorne & Wheeler, write line elements in terms of coordinate differentials, i.e., $ds^2$ is the squared length of a differential line element in spacetime. Writing a "line element" (in quotes because what the formula describes is no longer directly interpretable as a differential line element in spacetime) in terms of tensor products of basis 1-forms requires more care with interpretation.

However, as far as your particular question in this thread is concerned, the distinction doesn't matter. If $r$ is held constant, the differential $dr$ is zero, and the gradient $\text{d} r$ is also zero, because a surface with constant $r$ is a level surface of the 1-form $\text{d} r$. So either way it drops out of the line element.

 

[chartery]

Where does Carroll come into it? The reference you gave in the OP of this thread is a paper by Matt Visser.

My learning source; I thought widely so.

These are two different things, yes. Most texts, even ones that observe the distinction between differentials and 1-forms (covectors), such as Misner, Thorne & Wheeler, write line elements in terms of coordinate differentials, i.e., $ds^2$ is the squared length of a differential line element in spacetime. Writing a "line element" (in quotes because what the formula describes is no longer directly interpretable as a differential line element in spacetime) in terms of tensor products of basis 1-forms requires more care with interpretation.

Yes, I guess that was the implicit question in OP. I presumed that (as in Carroll) the format $\text{d}r$ indicated, or at least implied, that the latter version obtained in Visser's paper. It seemed sensible to understand any difference between physical interpretation, especially at a singularity, and the mathematical manipulations.

However, as far as your particular question in this thread is concerned, the distinction doesn't matter. If $r$ is held constant, the differential $dr$ is zero, and the gradient $\text{d} r$ is also zero, because a surface with constant $r$ is a level surface of the 1-form $\text{d} r$. So either way it drops out of the line element.

Thanks, this was exactly what I was trying to confirm and visualise. Is it possible to say in which types of context the distinction would matter for someone at my level?

 

[PeterDonis]

My learning source; I thought widely so.

This doesn't answer the question I asked.

 

[PeterDonis]

Is it possible to say in which types of context the distinction would matter for someone at my level?

It depends on what you are using the metric for. As I said before, if you write it as a line element, $ds^2$, you are treating it as describing the squared length of a differential line element in spacetime. Strictly speaking, it doesn't even make sense to write a differential line element as $\text{d} s^2$; it's not a 1-form. So if you're using the metric this way, it should be written in terms of differentials, since if the LHS is a differential, the RHS has to be made up of differentials too, otherwise the equation doesn't make sense; you can't add together a bunch of 1-forms to get a differential.

If, OTOH, you insist on writing the metric in terms of 1-forms, then you are viewing it as giving inner products of vectors in the tangent space at a chosen point. In which case it doesn't really make sense to write it as a line element to begin with, since an inner product of vectors is not a line element. What you have is a sum of tensor products of 1-forms, i.e., a (0, 2) tensor, which is a linear map from pairs of vectors to numbers (the numbers are the inner products). Writing this as $\text{d} s^2$ makes no sense either.

In short, I would not put a lot of weight on Visser using the $\text{d}$ notation in a line element. To me it's just a notation that is not explained and which doesn't really make sense; it should really be read as differentials, not 1-forms, the way Visser is using it in his paper. If you have a reference by Carroll that actually discusses using the metric for inner products and using the $\text{d}$ notation for that, you need to give that reference; just saying "my learning source" is not enough, since I have no idea what your "learning source" is or what it says.

 

[chartery]

This doesn't answer the question I asked.

Sorry. It didn't occur to me that such a widely recognised (respected?) source would have such a basic concept being contentious, so assumed it was everyone's approach (or at least knowledge), and wouldn't need explanation or specific referencing. The relevant passages are before equation 2.40 and after 2.43 in his book (2.29/2.32 in his notes). So I was reasonably clear about his difference, just not about when or where it would be important.

 

[PeterDonis]

such a widely recognised (respected?) source

What source? How do I know it's "widely recognised" if you won't even tell me what it is? I could guess, but I'm not supposed to guess. You're supposed to tell me. That's what the PF rules that you signed up to when you joined this forum say.

 

[PeterDonis]

The relevant passages are before equation 2.40 and after 2.43 in his book (2.29/2.32 in his notes).

If I guess that you mean Carroll's online lecture notes on GR [1], then observe that Carroll, unlike Visser, never writes $\text{d} s^2$ for the LHS of any line element equation. He does use the $\text{d}$ notation on the RHS, and draws the distinction between "the informal notion of an infinitesimal displacement" and "the rigorous notion of a basis 1-form given by the gradient of a coordinate function". And he then notes that the notation $ds^2$ is "just conventional shorthand for the metric tensor", so a line element equation like (2.29) is not really even an equation, it's a definition.

In other texts such as MTW and Wald, line elements are not written with the $\text{d}$ notation at all, even though both texts discuss the rigorous definition of 1-forms and tensors and state that the metric is a (0, 2) tensor. Wald doesn't really use the exterior derivative $\text{d}$ notion anywhere. MTW does use their version of it, a boldface $\mathbf{d}$, a lot, but not in line elements, because they treat line elements as component equations and they never use the $\mathbf{d}$ notation in component equations, only in abstract tensor equations. (Wald uses abstract index notation for abstract tensor equations, which, as he explains, makes it clearer which slots of the tensors are which, a detail which MTW's notation for abstract tensor equations obscures.)

The upshot of all this is that, as Carroll notes, viewing the metric/line element in terms of differentials is an "informal" notion; the only really rigorous view is the view in terms of tensor products of basis 1-forms. The "informal" view is widely used because it is easier conceptually. But if there's ever any doubt in your mind about which view to use, you should try to use the rigorous view--but then you have to, as I said before, take more care in interpretation.

[1] https://arxiv.org/abs/gr-qc/9712019

 

[chartery]

What source? How do I know it's "widely recognised" if you won't even tell me what it is? I could guess, but I'm not supposed to guess. You're supposed to tell me. That's what the PF rules that you signed up to when you joined this forum say.

In many researches, for instance on Stack Exchange (https://physics.stackexchange.com) and even on Physics Forums (here!), I have never seen Carroll's book (Carroll, S. Spacetime and Geometry) needing a bibliographic reference, and so didn't imagine it would require inspired guesswork. It is difficult for an amateur to avoid getting impressions that some things are 'obvious' to 'everyone'. I can only apologise for the innocent but evidently unacceptable imprecision.

 

[PeterDonis]

I have never seen Carroll's book (Carroll, S. Spacetime and Geometry) needing a bibliographic reference

I have already explained the PF rules to you.

Further, you could have just told me "Carroll's book/lecture notes on GR" in much less time than you have spent posting about why you don't think you have to tell me. That is not respectful of anyone's time, including yours.

didn't imagine it would require inspired guesswork

So you think Carroll's book/lecture notes is the only reference on GR that exists? There couldn't possibly be other textbooks or sources that cover the same topics? There couldn't be any reason why you might need to specify which source you are using, instead of just assuming that if you don't give a source everyone will know it must be Carroll? Particularly when, in the OP to this thread, you gave a different source?

I don't think you have thought through your position very well.

 

[PeterDonis]

The OP question has been answered. Thread closed.