# Exploring Null Basis Vectors, Metric Signatures Near Kruskal

• A
• George Keeling
In summary: How would I calculate ##\partial/\partial v^\prime## and ##\partial/\partial v^\prime## and show they are on a light cone?The norm of a vector is the product of its lengths. You can calculate the norm of a vector by taking the product of each component with the metric.
George Keeling
Gold Member
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!

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.

Last edited:
George Keeling and PeterDonis
George Keeling said:
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.
George Keeling said:
2) How would I calculate ##\partial/\partial v^\prime## and ##\partial/\partial v^\prime## and show they are on a light cone?
See above.
George Keeling said:
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.
George Keeling said:
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
George Keeling said:
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.

George Keeling
Ibix said:
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.

vanhees71, George Keeling, PAllen and 1 other person
Thanks for the insights! The one about the ##t,r## coordinates being different inside and outside the event horizon is very good.
PAllen said:
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.
Ibix said:
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.

PAllen said:
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!

George Keeling said:
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.

George Keeling
George Keeling said:
Everything after that is fine. But why does a coordinate 'generate' a vector field? Ibix neatly skipped that part!
You can skip this step.

PAllen said:
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)##.

Ibix, George Keeling, PAllen and 2 others

## 1. What is the purpose of exploring null basis vectors?

The purpose of exploring null basis vectors is to understand the structure of a vector space and its basis. Null basis vectors are vectors that span the null space of a matrix, which contains all the vectors that are mapped to zero by the matrix. By exploring null basis vectors, we can gain insight into the linear transformations and properties of a matrix.

## 2. How does the concept of metric signatures relate to Kruskal's theorem?

Metric signatures are used to classify the types of spacetime in general relativity, while Kruskal's theorem is a mathematical theorem that describes the structure of black holes. The concept of metric signatures is used in the proof of Kruskal's theorem to show that the spacetime inside a black hole is different from the spacetime outside the black hole.

## 3. Can you explain the significance of metric signatures in physics?

Metric signatures play a crucial role in physics, particularly in general relativity. They help us understand the geometry of spacetime and the curvature of the universe. Metric signatures also allow us to classify different types of spacetime, such as flat, curved, or even spacetimes with multiple time dimensions.

## 4. How do null basis vectors and metric signatures intersect in mathematics?

Null basis vectors and metric signatures both involve the concept of linear transformations and vector spaces. In mathematics, null basis vectors are used to find the null space of a matrix, while metric signatures are used to classify the types of vector spaces. They intersect in the sense that they both provide insight into the structure and properties of vector spaces and their transformations.

## 5. What are some practical applications of understanding null basis vectors and metric signatures?

Understanding null basis vectors and metric signatures has many practical applications, particularly in the fields of physics and engineering. For example, in physics, they are used to study the curvature of spacetime and the properties of black holes. In engineering, they can be applied to problems involving linear transformations, such as image and signal processing. Additionally, understanding these concepts can also help in the development of new mathematical models and theories.

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