Null basis vectors, metric signatures near Kruskal

[George Keeling]

TL;DR Summary

On the way to Kruskal I meet null coordinates and want to know a bit more. Does the metric signature change in Schwarzschild?

On the way to Kruskal coordinates, Carroll introduces coordinates $\left(v^\prime,u^\prime,\theta,\phi\right)$ with metric equation$$
{ds}^2=-\frac{2{R_s}^3}{r}e^{-r / R_s}\left(dv^\prime du^\prime+du^\prime dv^\prime\right)+r^2{d\Omega}^2
$$
$R_s=2GM$ and we're using a $-+++$ signature and the speed of light $c=1$. The familiar Schwarzschild coordinate $r$ is used in the equation and implicitly defined by $$
u^\prime v^\prime=-\left(\frac{r}{R_s}-1\right)e^{r/R_s}
$$Carroll then writes "Both $v^\prime$ and $u^\prime$ are null coordinates, in the sense that their partial derivatives $\partial/\partial v^\prime$ and $\partial/\partial v^\prime$ are null vectors. There is nothing wrong with this ...".

I suppose that null vectors are ones that lie on the light cone in contrast to timelike and spacelike vectors. Then I have some questions:

1) I can see that in Minkowski space $t$ is timelike because it's axis goes 'up' and $x,y,z$ are spacelike because their axes go 'across'. This sounds like a tautology. Is there a more rigorous way of saying that?

2) How would I calculate $\partial/\partial v^\prime$ and $\partial/\partial v^\prime$ and show they are on a light cone?

3) The diagonal metric components in Minkowski are negative, positive, positive, positive. Does the sign tell you whether the corresponding coordinate is timelike or spacelike? And if the component vanishes, as in the metric above, does that mean the coordinate is null?

4) With the Schwarzschild metric the signs of the diagonal metric $t$ and $r$ components reverse inside the event horizon. I have read that avoiding the singularity is as impossible as avoiding old age. Is it correct to say that inside the event horizon $t$ is spacelike and $r$ timelike? How do you know that decreasing $r$ on the inside is like increasing $t$ on the outside?

5) Does the Schwarzschild metric signature become $-+--$ inside the event horizon? Is the metric signature of the $\left(v^\prime,u^\prime,\theta,\phi\right)$ system $0\ 0++$?

Thanks!

 

[PAllen]

2) Imagine a vector field whose expression is (u',0,0,0) and another whose expression is (0,v',0,0) . These are the vector fields generated by these two coordinates. Their partial derivatives produce simply the vectors (1,0,0,0) and (0,1,0,0). Compute the norm or inner product of these vectors with themselves using the metric. It is then trivial to see that the norm of both is zero, thus they are null coordinates.

1) The distinction between timelike and spacelike coordinates is the sign of norm squared per the metric. If using the convention (-,+,+,+) then a coordinate basis vector with negative norm squared is timelike, positive spacelike. If using the (+,-,-,-) convention, it is obviously the reverse. These conventions are equivalent. You can simply negate a metric expressed in one convention to get a metric expressed in the other. On the other hand, (-,-,+,+) is fundamentally different as is (+,+,+,+) - the last being a true Riemannian 4-manifold.

3) That is true for one of the conventions in use. If the metric is pure diagonal, the sign tells you the character of the corresponding coordinate. However, for a general metric, you must compute the inner product of e.g. (1,0,0,0) with itself to be sure.

4) Though the metric looks the same, these are two wholly different coordinate patches that cannot be connected because they both become singular at the horizon. The claim coordinates switching places is meaningless because they are simply disconnected charts covering different regions. The letters used for coordinates cannot possibly mean anything. Given a metric and signature convention choice, you can compute which coordinate is timelike or spacelike by taking the norm squared of the corresponding coordinate vector. As to choice of which direction of timelike vector is future pointing, this is a choice you can make. For Schwarzschild interior, if you chose -r as the future pointing direction you have a black hole interior. If you choose +r as the future pointing direction, you have a white hole interior.

5) The signature is not coordinate dependent. The signature of a metric is a statement about its character if its value at a point in some coordinate expression is diagonalized. It is an intrinsic feature of the manifold, not dependent on coordinate choice. When one speaks of a Riemannian metric, it is required that the metric expression everywhere diagonalizes to (+,+.+,+). For a Minkowskian manifold, you must have diagonalization to either (-,+,+,+) everywhere OR (+,-,-,-) everywhere. Another way to look at this is to derive an orthonormal set of non null vectors using the metric expression at a point in some coordinates. The signs of the squared norms of these tells you the signature.

 

[Ibix]

1) I can see that in Minkowski space $t$ is timelike because it's axis goes 'up' and $x,y,z$ are spacelike because their axes go 'across'. This sounds like a tautology. Is there a more rigorous way of saying that?

A vector parallel to the $t$ axis is $V^\mu$, where $V^t=1$ and all other components are zero. Calculate $g_{\mu\nu}V^\mu V^\nu$. Repeat for vectors where $V^x$ (etc) is the only non-zero component, and you'll get the opposite sign. Repeat for vectors where $V^t=\pm V^x$ and the other components are zero and you'll get zero. That's the formal general way of determining if a vector is timelike, spacelike, or null.

2) How would I calculate $\partial/\partial v^\prime$ and $\partial/\partial v^\prime$ and show they are on a light cone?

See above.

3) The diagonal metric components in Minkowski are negative, positive, positive, positive. Does the sign tell you whether the corresponding coordinate is timelike or spacelike? And if the component vanishes, as in the metric above, does that mean the coordinate is null?

If you've diagonalised your metric then you are using one timelike and three spacelike coordinates and you can read off the timelike coordinate from the signs, yes. If you've diagonalised your metric and you've got zeros on the diagonal then something's gone wrong - your metric is singular which mean something's broken somewhere.

4) With the Schwarzschild metric the signs of the diagonal metric $t$ and $r$ components reverse inside the event horizon. I have read that avoiding the singularity is as impossible as avoiding old age. Is it correct to say that inside the event horizon $t$ is spacelike and $r$ timelike? How do you know that decreasing $r$ on the inside is like increasing $t$ on the outside?

People usually use different coordinate labels for the coordinates inside the event horizon to avoid confusion. The patch inside the horizon is discontinuous from the patch outside, so you can't identify a particular coordinate in interior Schwarzschild coordinates with one outside, but you are correct that the coordinates with metric coefficients with the same functional form have different timelike and spacelike natures inside and outside the hole

5) Does the Schwarzschild metric signature become $-+--$ inside the event horizon? Is the metric signature of the $\left(v^\prime,u^\prime,\theta,\phi\right)$ system $0\ 0++$?

Order of coordinates isn't significant, except that by convention you put the timelike one (if any) either first or last. So typically you'd say that the signature was +--- both inside and out. -+-- would just be a slightly non-standard way of writing it, that's all.

 

[Nugatory]

People usually use different coordinate labels for the coordinates inside the event horizon to avoid confusion.

But not enough people do... which accounts for the prevalence of the "switching places" myth. In https://arxiv.org/abs/0804.3619 the author says "Unfortunately, this fact is overlooked sometimes..." which I consider to be something of an understatement.

 

[George Keeling]

Thanks for the insights! The one about the $t,r$ coordinates being different inside and outside the event horizon is very good.

If the metric is pure diagonal, the sign tells you the character of the corresponding coordinate. However, for a general metric, you must compute the inner product of e.g. (1,0,0,0) with itself to be sure.

If you've diagonalised your metric then you are using one timelike and three spacelike coordinates and you can read off the timelike coordinate from the signs, yes.

So I'm trying to imagine a non-diagonal metric where the norm of a coordinate is not like the diagonal component. and I have this:
The vector of the ith coordinate is $V=\left(\delta_{io},\delta_{i1},\delta_{i2},\delta_{i3}\right)$. It's norm is $g_{\mu\nu}V^\mu V^\nu=g_{ii}$. So the norm of $\ V$ is always the same sign (or 0) as the diagonal component of the metric, off diagonal components don't seem to come into it.

 

[George Keeling]

2) Imagine a vector field whose expression is (u',0,0,0) and another whose expression is (0,v',0,0) . These are the vector fields generated by these two coordinates

Everything after that is fine. But why does a coordinate 'generate' a vector field? Ibix neatly skipped that part!

 

[PAllen]

Thanks for the insights! The one about the $t,r$ coordinates being different inside and outside the event horizon is very good.


So I'm trying to imagine a non-diagonal metric where the norm of a coordinate is not like the diagonal component. and I have this:
The vector of the ith coordinate is $V=\left(\delta_{io},\delta_{i1},\delta_{i2},\delta_{i3}\right)$. It's norm is $g_{\mu\nu}V^\mu V^\nu=g_{ii}$. So the norm of $\ V$ is always the same sign (or 0) as the diagonal component of the metric, off diagonal components don't seem to come into it.

Yes, for a coordinate basis vector, whether it is lightlike or not can be read from corresponding diagonal metric components. I was thinking of a general vector. However, for a basis vector that is not lightlike, the sign of diagonal metric element does not tell you whether it is spacelike or timelike without first determining the metric signature by diagonalization. For example, the diagonal of the metric could be (0,0,1,1) at some point (with nonzero off diagonal components as well) and without diagonalization, you don’t know whether the last two coordinates are timelike or spacelike.

 

[PAllen]

Everything after that is fine. But why does a coordinate 'generate' a vector field? Ibix neatly skipped that part!

You can skip this step.

 

[PeterDonis]

Imagine a vector field whose expression is (u',0,0,0) and another whose expression is (0,v',0,0) . These are the vector fields generated by these two coordinates. Their partial derivatives produce simply the vectors (1,0,0,0) and (0,1,0,0).

I'm not sure this is correct as you state it. In a curved spacetime, I don't think coordinate 4-tuples are vectors, so $(u', 0, 0, 0)$ and $(0, v', 0, 0)$ are not vector fields. The vectors $(1, 0, 0, 0)$ and $(0, 1, 0, 0)$ are thus not partial derivatives of vector fields.

I would state it this way: any curve can be expressed in parameterized form as $x^\mu (s)$, i.e., as four functions of the curve parameter, one for each coordinate. The tangent vector $V$ to this curve can then be written:

$$
V = \frac{d}{ds} = \frac{du'}{ds} \frac{\partial}{\partial u'} + \frac{dv'}{ds} \frac{\partial}{\partial v'} + \frac{d \theta}{ds} \frac{\partial}{\partial \theta} + \frac{d \phi}{ds} \frac{\partial}{\partial \phi}
$$

where I have made use of the isomorphism between vectors and directional derivatives to express both the vector $V$ itself and each of the basis vectors as derivatives. The components of the vector are therefore

$$
\left( \frac{du'}{ds}, \frac{dv'}{ds},\frac{d \theta}{ds}, \frac{d \phi}{ds} \right)
$$

The vector $\partial / \partial u'$ itself is therefore simply the tangent vector to a curve along which the only coordinate that is changing is $u'$, and along which we have adopted the coordinate $u'$ itself as the curve parameter; thus this vector has components $(1, 0, 0, 0)$. And similarly for $\partial / \partial v'$ with the coordinate $v'$ to obtain the components $(0, 1, 0, 0)$.