# Shortest space-like path

• I
• Bosko

#### Bosko

Gold Member
I consider the geometry around a massive homogeneous static spherical object (for example, a neutron star)
This system is static, so I am not interested in the space-time distance between two events.
I am interested in the spatial distance (without the influence of time) and the shortest path between two points?

Questions:

1. If a light source was turned on at point A, did the ray of light that first reaches point B has traveled along the shortest path? ( Fermat's principle )

2. Is the (null geodesic) path of the ray of light the (locally) shortest path between a two points?
The null geodesic is defined as ##{g_{\mu\nu}} \frac {dx^{\mu}} {ds} \frac {dx^{\nu}} {ds}=0 ##.
Can I remove time ##\mu =0, \nu =0## in my static (pure space-like) geometry in order to get non-zero distance?
$$d(A,B)=L=\int_P \sqrt{g_{\mu\nu} dx^{\mu} dx^{\nu}} \,ds$$

3. Is the length of that path the spatial distance between those two points?

Last edited:
Bosko said:
This system is static, so I am not interested in the space-time distance between two events.
I am interested in the spatial distance (without the influence of time) and the shortest path between two points?
You cannot ignore the time dimension here, nor avoid the spacetime interval.

A point in space is a timelike worldline in spacetime, and the distance between the two points is the spacelike spacetime interval between two events, one on each worldline. But which two? They have to be simultaneous if the interval is to be interpreted as a distance, and simultaneity depends on our arbitrary choice of time coordinate. The length of any spacelike geodesic between between the two worldlines could reasonably be considered the distance between the two points.
(It may be easier to see this as a generalization of length contraction in special relativity: the length of an object is the distance between where its endpoints are at the same time).
1. If a light source was turned on at point A, did the ray of light that first reaches point B has traveled along the shortest path? ( Fermat's principle )
2. Is the (null geodesic) path of the ray of light the (locally) shortest path between a two points?
We’d need a more precise definition of “shortest path” to answer that question. Fermat had it easier because he was assuming absolute time and absolute simultaneity.
The null geodesic is defined as ##{g_{\mu\nu}} \frac {dx^{\mu}} {ds} \frac {dx^{\nu}} {ds}=0 ##.
Can I remove time ##\mu =0, \nu =0## in my static (pure space-like) geometry in order to get non-zero distance?
$$d(A,B)=L=\int_P \sqrt{g_{\mu\nu} dx^{\mu} dx^{\nu}} \,ds$$
Sure, as long as you’re doing the special case in which ##dx^0## is zero and you’ve chosen your coordinates such that ##x^0## is the only timelike coordinate and the metric has no off-diagonal terms involving it. But this rather defeats the purpose of using tensors in the first place)
3. Is the length of that path the spatial distance between those two points?
In general, no.

cianfa72, Bosko and Ibix
Nugatory said:
A point in space is a timelike worldline in spacetime, and the distance between the two points is the spacelike spacetime interval between two events, one on each worldline. But which two? They have to be simultaneous if the interval is to be interpreted as a distance, and simultaneity depends on our arbitrary choice of time coordinate. The length of any spacelike geodesic between between the two worldlines could reasonably be considered the distance between the two points.
The only requirements for the two events picked along the two timelike worldlines is to be spacelike separated. In SR the requirement above is clear; in GR I'm not sure if it is actually well-defined since there could be multiple geodesics of different type joining the two events.

cianfa72 said:
The only requirements for the two events picked along the two timelike worldlines is to be spacelike separated.
It is not. It depends on which two events you pick. You will get different results depending on the two events you pick. This is basically what length contraction is about.

cianfa72
Orodruin said:
It is not. It depends on which two events you pick. You will get different results depending on the two events you pick. This is basically what length contraction is about.
Yes, definitely. Maybe I was unclear but that was my point too. The other point I was trying to make is that I'm not sure if the notion of "spacelike separated events" in GR is actually well-defined or is not.

cianfa72 said:
Yes, definitely. Maybe I was unclear but that was my point too. The other point I was trying to make is that I'm not sure if the notion of "spacelike separated events" in GR is actually well-defined or is not.
Two events have spacelike separation if they are not within eachother’s light cones.

cianfa72
Orodruin said:
Two events have spacelike separation if they are not within each other’s light cones.
Ok, so they have spacelike separation if does not exist a timelike path (not necessarily a timelike geodesic) joining them.

cianfa72 said:
if does not exist a timelike path (not necessarily a timelike geodesic) joining them.
… or null …
(My emphasis in quote)

cianfa72
Nugatory said:
You cannot ignore the time dimension here, nor avoid the spacetime interval.
Of course. Thanks, but I am looking for the shortest path between two ("parallel vertical") time-like world-lines of two static points A and B.
I am trying to find (on the internet, books ,...) is the shortest path in fact the trajectory of the light ray that first arrives from "light bulb" A to B.
Nugatory said:
A point in space is a timelike worldline in spacetime, and the distance between the two points is the spacelike spacetime interval between two events, one on each worldline. But which two? They have to be simultaneous if the interval is to be interpreted as a distance, and simultaneity depends on our arbitrary choice of time coordinate. The length of any spacelike geodesic between between the two worldlines could reasonably be considered the distance between the two points.
We’d need a more precise definition of “shortest path” to answer that question.
Path of the minimal length .
Like on Earth, distance between two cities. Every path has its length. The path with minimal length is shortest ( circular arc on earth's surface ) path between those two cities.
Nugatory said:
Fermat had it easier because he was assuming absolute time and absolute simultaneity.
Yes, but still, does Fermat's principle can be applied on light as a wave with constant speed? ##c=\frac {proper\,length} {proper\,time}##
First light ray that arrives from light source (point) A to point B has traveled by the shortest path.
Nugatory said:
Sure, as long as you’re doing the special case in which ##dx^0## is zero and you’ve chosen your coordinates such that ##x^0## is the only timelike coordinate and the metric has no off-diagonal terms involving it.
My intention is exactly that :-) ##dx^0=0##
Nugatory said:
But this rather defeats the purpose of using tensors in the first place)
Tensors (matrices / e.g. metric tensor) instead 4x4 becomes 3x3, if I am right.
Because "first row" and "first column" are 0.
Nugatory said:
In general, no.
If you mean that - no every null geodesic (light ray) path is the shortest path - I agree .
Thanks

Last edited:
Bosko said:
Path of the minimal length .
Like on Earth, distance between two cities. Every path has its length. The path with minimal length is shortest ( circular arc on earth's surface ) path between those two cities.
The cities are points on the two-dimensional manifold that is the surface of surface of the earth, so we calculate the distance along a path between them using the metric tensor: ##D^2=\int_P g_{\mu\nu} dx^{\mu} dx^{\nu} \,ds## (and the most convenient coordinates will have ##x^\mu## and ##x^\nu## as latitude and longitude and the ##g_{\mu\nu}## chosen accordingly).

The analogous procedure in four-dimensional spacetime calculates the spacetime interval between two events using the metric of that spacetime and whatever coordinates we find most convenient.
But in your original post you said that that is exactly what you did not mean by “distance”.
So it’s still not clear what you do mean by “distance”.

Nugatory said:
But in your original post you said that that is exactly what you did not mean by “distance”.
So it’s still not clear what you do mean by “distance”.
My interpretation, given that OP was talking about a static spacetime, was that he was talking about the length of the projection of the path on to the static spacelike planes.

I don't think the projections of geodesics on to such planes are geodesics of the planes. Simple reasoning for timelike curves is that there are infinitely many combinations of launch angle and velocity for me to throw you a ball. Those parabolae can't all be extremal length.

Ibix said:
My interpretation, given that OP was talking about a static spacetime, was that he was talking about the length of the projection of the path on to the static spacelike planes.
That's almost (because I'd say "spacelike hypersurfaces" instead of "planes" in a curved spacetime) my interpretation as well, but which static spacelike planes? There's more than one way to slice the spacetime up into spacelike planes, and different slicings will lead to different projected distances.

Bosko said:
I consider the geometry around a massive homogeneous static spherical object (for example, a neutron star)
This system is static, so I am not interested in the space-time distance between two events.
I am interested in the spatial distance (without the influence of time) and the shortest path between two points?

Well, the space-time geometry inherently includes time, so to exclude time what you need to do is define a projection operator that projects the 4-d space-time manifold into a 3-d spatial manifold in order to answer your question.

In general, there is no unique projection operator, but in the case of the static geometry, the time-like symmetry (the time-like killing vector field) specifies a projection operator via the requirement that it preserves the symmetry.

It may seem like a fine point, but sloppy thinking about what you are doing can cause confusion.

1. If a light source was turned on at point A, did the ray of light that first reaches point B has traveled along the shortest path? ( Fermat's principle )

I don't have a detailed calculation, but I believe the answer to this is "no".

To sketch the required detailed calculations, specify a static space-time Schwarzschild geometry. There are several choices, I will chose to use isotrorpic Schwarzschild coordinates (t,x,y,z). (If this isn't clear, I could provide more detail, but hopefully it's not necessary). Then the projection operator maps a point (t,x,y,z) in the spacetime manifold to the point (x,y,z) in the spatial manifold, which is a quotient manifold. If you prefer to use other coordinates, feel free. The isotropic coordinates I used are conceptually simple (IMO), but troublesome to work through the mathematics.

You'll find that the path of the light beam is "bent" in these coordinates, not a straight line. This is referred to commonly as the "deflection of light by gravity".

2. Is the (null geodesic) path of the ray of light the (locally) shortest path between a two points?

Locally, yes. Ignoring my point about sloppy thinking, I'll just say that while light is bent by gravity, locally it is so insignificant it doesn't matter.

You might consider a different example. Consider Rindler coordinates on an accelerating elevator, again using (t,x,y,z) coordinates for the space-time manifold and (x,y,z) as coordinates for the spatial submanifold. You will again see that light follows a curved path, but that radar methods give the correct notion of distance for nearby points on the elevator.

The null geodesic is defined as ##{g_{\mu\nu}} \frac {dx^{\mu}} {ds} \frac {dx^{\nu}} {ds}=0 ##.
Can I remove time ##\mu =0, \nu =0## in my static (pure space-like) geometry in order to get non-zero distance?
$$d(A,B)=L=\int_P \sqrt{g_{\mu\nu} dx^{\mu} dx^{\nu}} \,ds$$

3. Is the length of that path the spatial distance between those two points?

It's not exactly the problem you are interested in, but some of the questions you ask are covered in the context of a rotating platform in https://arxiv.org/abs/gr-qc/0309020, "The Relative Space: Space Measurements on a Rotating Platform". I would say that the paper in question solves a problem that is actually harder than the one you are interested in - but I'm not aware of any other formal reference that touches on your question offhand.

arxiv said:
We can introduce the local spatial geometry of the disk, which defines the
proper spatial line element, on the basis of the local optical geometry. To this
end we can use the radar method[18], [24].

Thus, they do something similar to what you describe - they use light to determine the local spatial geometry.

Another quote.

arxiv said:
That is, the relative space is the ”quotient space” of the world tube of the
disk, with respect to the equivalence relation RE, among points and space plat-
forms belonging to the lines of the congruence Γ.

Instead of using the world tubes of the rotating disk, as in the paper, we use the world tubes of the "static observers" in your example. The important point is that in order to answer the question, we need to define some projection operator to reduce the 4-d space-time manifold to a 3-d manifold. A congruence of worldlines is one way to define the projection operator - every worldline in the 4 dimensonal congruence is mapped to a single point in the "quotient space", which is the 3d projection of the 4d manifold. Thus we map some curve (t,x,y,z) representing a worldline (defined by the set of points with t varying and x,y,z fixed) to a single point in the "quotient space", namely (x,y,z).

Orodruin, cianfa72 and Nugatory
Nugatory said:
but which static spacelike planes?
Presumably those which are orthogonal to the timelike Killing field. It is ambiguous only if there are several timelike Killing fields, such as in Minkowski space.
Ibix said:
I don't think the projections of geodesics on to such planes are geodesics of the planes. Simple reasoning for timelike curves is that there are infinitely many combinations of launch angle and velocity for me to throw you a ball. Those parabolae can't all be extremal length.
The loophole of this argument is of course that the OP is asking about null curves, not timelike ones. However, it should be pretty straightforward to check by looking a particular example such as the Schwarzschild metric.

Nugatory said:
That's almost (because I'd say "spacelike hypersurfaces" instead of "planes" in a curved spacetime) my interpretation as well, but which static spacelike planes?
Orodruin said:
Presumably those which are orthogonal to the timelike Killing field.
That was my thought. You can construct a lot of hypersurfaces (point taken on that one) that are invariant under time translation, but isn't the very definition of a static spacetime that there exist a set of hypersurfaces that are everywhere orthogonal to the timelike KVF? Would you call the other surfaces "static surfaces"?

Apologies if this starts another "are non-orthogonal frames on flat spacetime inertial frames" debate...
Orodruin said:
The loophole of this argument is of course that the OP is asking about null curves, not timelike ones.
Indeed. It's easy enough to get the geodesic equations for the 3-surface, but I don't think the affine parameter is the same as the projection of the affine parameter in the 4d case, so I think there's more work than I have time to do on my morning commute...

Ibix said:
but isn't the very definition of a static spacetime that there exist a set of hypersurfaces that are everywhere orthogonal to the timelike KVF? Would you call the other surfaces "static surfaces"?
Yes.

The caveat is of course if there are several timelike Killing fields such as in Minkowski space. This is what leads to the relativity of simultaneity in the first place.

Nugatory said:
There's more than one way to slice the spacetime up into spacelike planes, and different slicings will lead to different projected distances.
Why you talk of spacelike planes ? Such slices (e.g. leaves of a foliation) are generally spacelike hypersurfaces not just planes (where with planes I mean Euclidean spaces).

Nugatory said:
...
But in your original post you said that that is exactly what you did not mean by “distance”.
So it’s still not clear what you do mean by “distance”.
For ##dx^0=0## and all paths P between static points A and B minimum of
$$L(P)=\int_P \sqrt{g_{\mu\nu} dx^{\mu} dx^{\nu}} \,ds$$
for shortest path ##P_{min}## distance is ##L(P_{min})##

Last edited:
Ibix said:
My interpretation, given that OP was talking about a static spacetime, was that he was talking about the length of the projection of the path on to the static spacelike planes.

I don't think the projections of geodesics on to such planes are geodesics of the planes. Simple reasoning for timelike curves is that there are infinitely many combinations of launch angle and velocity for me to throw you a ball. Those parabolae can't all be extremal length.
Sorry, what means OP?

Bosko said:
Sorry, what means OP?
Original Post or Original Poster (ie, you in this case).

Bosko
pervect said:
....
The important point is that in order to answer the question, we need to define some projection operator to reduce the 4-d space-time manifold to a 3-d manifold. A congruence of worldlines is one way to define the projection operator - every worldline in the 4 dimensonal congruence is mapped to a single point in the "quotient space", which is the 3d projection of the 4d manifold. Thus we map some curve (t,x,y,z) representing a worldline (defined by the set of points with t varying and x,y,z fixed) to a single point in the "quotient space", namely (x,y,z).
System is static and all worldlines are "vertical" . I am looking for ##dx^0(time)=0## paths between A and B.

Last edited:
Bosko said:
System is static and all worldlines are "vertical" . I am looking for ##dx^0(time)=0## paths between A and B.
It's easy enough to write down an equation for geodesics in the spatial plane. If we pick coordinates so that the motion is in the equatorial plane then ##d\theta=0## and the Lagrangian is $$\mathcal{L}=\frac 1{1-2GM/r}\left(\frac{dr}{d\lambda}\right)^2+r^2\left(\frac{d\phi}{d\lambda}\right)^2$$The ##\phi## Euler-Lagrange equation yields ##\frac{d\phi}{d\lambda}=\frac{L}{r^2}##, where ##L## is a constant of the motion, and the ##r## one yields$$\frac{d^2r}{d\lambda^2}=\frac{GM}{r(r-2GM)}\left(\frac{dr}{d\lambda}\right)^2+\frac{L^2(r-2GM)}{r^4}$$
We can do the same for the full 4d case, or just look up the answer (7.43, 7.44, 7.47 and 7.48 in Carroll's lecture notes, with ##\epsilon=0## for the null case). This gives us ##\frac{d\phi}{d\lambda'}=\frac{L'}{r^2}## and$$\left(\frac{dr}{d\lambda'}\right)^2=E^2+\frac{L'^2}{r^3}(r-2GM)$$where ##E## and ##L'## are constants of the motion.

That the ##d\phi## equations are the same form means, I think, that we can choose ##\lambda=\lambda'## and ##L'=kL##, where ##k## is an unknown dimensionless constant. Then we can simply plug the 4d ##r## equation in to the 3d one to eliminate the derivatives. I think that unless that reduces to ##0=0##, the spatial projections of null paths are not extrema of distance in general.

I think that's correct. Let's see what others say.

vanhees71
Ibix said:
the spatial projections of null paths are not extrema of distance in general.
In general they won't be, that's correct. Similar remarks apply to geodesic timelike paths; for example, consider a planet in a circular orbit about the star. Even considering that the geometry of a spacelike slice of constant time is not Euclidean (it's the "Flamm paraboloid" geometry), its geodesics are still not circles concentric on the source.

cianfa72, vanhees71 and Ibix
Bosko said:
System is static and all worldlines are "vertical" . I am looking for ##dx^0(time)=0## paths between A and B.

Write down the space-time metric in the coordinates you prefer. The path of light will be a (null) geodesic of the space-time metric.

Are you familiar with writing and solving the geodesic equations given the metric? It's unclear exactly what help you need.

After solving the space-time problem, the process for finding the path of shortest distance is similar. It is similar because GR uses the Levi-Civita connection, thus geodesics of the spatial metric (not to be confused with the space-time metric) are also the paths of the shortest distance.

Once you have the spatial metric, you can then use the same methods you used previously, to find the path of light through space-time, to find the geodesics (which are also the paths of shortest distance) in "space", once you have the spatial metric.

If you need more than this general information , you'll actually have to show your work. Start by writing down the line element of the space-time metric you are using, and , if possible, how you project the space-time metric to a spatial metric without time.

Bosko
pervect said:
After solving the space-time problem, the process for finding the path of shortest distance is similar. It is similar because GR uses the Levi-Civita connection, thus geodesics of the spatial metric (not to be confused with the space-time metric) are also the paths of the shortest distance.

Once you have the spatial metric, you can then use the same methods you used previously, to find the path of light through space-time, to find the geodesics (which are also the paths of shortest distance) in "space", once you have the spatial metric.
You mean the induced metric from ##g_{\mu \nu}## on the ##t=const## spacelike hypersurfaces. They are Riemannian manifolds of their own (they should be 3D spacelike embedded submanifold of the spacetime manifold).

Take any static spacetime. The metric in such a spacetime may be written in static coordinates ##t## and ##x^i## as
$$ds^2 = f(x)\, dt^2 - d\Sigma^2 = f(x)\, dt^2 - g_{ij} dx^i dx^j$$
where ##d\Sigma^2## is the (Riemannian) metric on the spatial slices defining simultaneous static time.

By the time translation symmetry, it follows that
$$E = g(\partial_t, \dot \gamma) = f(x)\dot t$$
is constant for any spacetime geodesic ##\gamma##. The spatial parts of the spacetime geodesic equations take the form (assuming I did the math right)
$$\bar\nabla_{\dot x} \dot x^j = \frac{E^2}2 g^{jk}\partial_k(1/f)$$
with ##\bar\nabla## representing the affine connection on the spatial slices. For the worldline to correspond to a geodesic when projected on the spatial slices, this should at the very least satisfy the non-affinely parametrised geodesic equations on the spatial slice. This would require the RHS to be proportional to ##\dot x^j##, which is not the case.

Last edited:
Bosko, PeterDonis and Ibix
To add to the above: Consider the case of ##\dot t \simeq 1## in the weak field approximation where ##f = 1 + 2\phi## where ##\phi## is the classical gravitational potential. We would then find:
$$\nabla _{\dot x} \dot x^j \simeq - g^{jk}\partial_k \phi$$
which to leading order is just the classical equation of motion for a test particle in Newtonian gravity.

cianfa72 said:
You mean the induced metric from ##g_{\mu \nu}## on the ##t=const## spacelike hypersurfaces. They are Riemannian manifolds of their own (they should be 3D spacelike embedded submanifold of the spacetime manifold).

Yes.

I'm not sure what the original poster is struggling with. It might be illuminating for him to take a simple case. Express the Euclidean plane in polar coordinates r, theta. Use the line element ##dr^2 + r^2 d\theta^2## , calculate all the Christoffel symbols (manually, but I'd recommend using some of the free symbolic algebra / tensor packages), and write down the geodesic equation. The goal is to demonstrate that the path of shortest distance is a straight line, regardless of whether you use polar coordinates or cartesian coordinates. Some work in transforming the relevant entities (tensors) between polar and Cartesian coordinates might also be needed, depending on the approach.

On the other hand, maybe he's just interested in the curved spatial geometry of a surface of constant t in Schwarzschild coordinates rather than the detailed calculations. There's quite a bit of discussion of this, but I don't have any references handy that just give the results, though I can say that the induced spatial geometry is curved, and not Euclidean.

I initially thought he was mostly interested in the projection process - how a 4d space-time metric induces a 3d metric on a submanifold - but I'm no longer sure about that.

Ibix said:
I don't think the projections of geodesics on to such planes are geodesics of the planes. Simple reasoning for timelike curves is that there are infinitely many combinations of launch angle and velocity for me to throw you a ball. Those parabolae can't all be extremal length.
As far as I can tell you mean the projection of a timelike geodesic onto spacelike hypersurfaces of constant coordinate time ##t## (i.e. the loci of events/points on those spacelike hypersurfaces where the timelike geodesic "pierces" them).

cianfa72 said:
As far as I can tell you mean the projection of a timelike geodesic onto spacelike hypersurfaces of constant coordinate time ##t## (i.e. the loci of events/points on those spacelike hypersurfaces where the timelike geodesic "pierces" them).
Yes, although specifically assuming that the "constant coordinate time" implies the hypersurfaces orthogonal to the KVF (which I thought was implied by calling them "static", but Orodruin and (edit: apparently not) Nugatory disagree). Formally I should probably add that the projection is being done along the timelike KVF. In Schwarzschild spacetime the maths is sketched in #22, and more generally by @Orodruin in #26.

Last edited:
Ibix said:
Formally I should probably add that the projection is being done along the timelike KVF.
Ah ok, so basically the projection is along the timelike KVF onto any of the spacelike hypersurfaces orthogonal to the timelike KVF (they exist and have all the same induced spatial metric since the spacetime is assumed to be static).

Ibix
Ibix said:
which I thought was implied by calling them "static", but Orodruin and Nugatory disagree
I am not sure that is an accurate description of what I said …

Orodruin said:
I am not sure that is an accurate description of what I said …
That's what I read into #16. Perhaps I misunderstood - I concede it depends on what part of my quoted post you were saying "yes" to.

Ibix said:
I concede it depends on what part of my quoted post you were saying "yes" to.
@Orodruin was saying yes to all of what he quoted from you. The caveat he gave was not that "static" doesn't always mean "hypersurface orthogonal". It was that in Minkowski spacetime, there is not just one static timelike KVF, as there is in Schwarzschild spacetime; there are multiple ones! Every inertial frame corresponds to a static timelike KVF. Also every Rindler frame corresponds to a static timelike KVF. There are an infinite number of both types of frames.

vanhees71, cianfa72 and Orodruin
Not how I read it, but fine, edited above. I understand the caveat.