I Does Light Travel the Shortest Path in Curved Space-Time Around a Neutron Star?

[Bosko]

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?

 

[Nugatory]

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]

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.

 

[Orodruin]

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]

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.

 

[Orodruin]

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]

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.

 

[Orodruin]

if does not exist a timelike path (not necessarily a timelike geodesic) joining them.

… or null …
(My emphasis in quote)

 

[Bosko]

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.

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.

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.

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 :-) $dx^0=0$

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.

In general, no.

If you mean that - no every null geodesic (light ray) path is the shortest path - I agree .
Thanks

 

[Nugatory]

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”.

 

[Ibix]

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.

 

[Nugatory]

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.

 

[pervect]

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.

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.

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]

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.

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.

 

[Ibix]

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?

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...

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...

 

[Orodruin]

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.

 

[cianfa72]

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).

 

[Bosko]

...
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 $L(P_{min})$

 

[Bosko]

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?

 

[Orodruin]

Sorry, what means OP?

Original Post or Original Poster (ie, you in this case).

 

[Bosko]

....
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.

 

[Ibix]

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.

 

[PeterDonis]

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.

 

[pervect]

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.

 

[cianfa72]

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).

 

[Orodruin]

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.

 

[Orodruin]

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.

 

[pervect]

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.

 

[cianfa72]

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).

 

[Ibix]

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.

 

[cianfa72]

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).

 

[Orodruin]

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 …

 

[Ibix]

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.

 

[PeterDonis]

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.

 

[Ibix]

Not how I read it, but fine, edited above. I understand the caveat.

 

[PeterDonis]

edited above

Btw, I'm not sure @Nugatory disagrees either. I think he just made a terminology slip. See below.

which static spacelike planes? There's more than one way to slice the spacetime up into spacelike planes

Yes, but only one of those slicings is orthogonal to the timelike KVF. And since "static" means the timelike KVF is hypersurface orthogonal, the obvious meaning of "static spacelike planes" is the slicing that is orthogonal to the timelike KVF.

 

[Bosko]

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$$\frac{d^2r}{d\lambda^2}=\frac{GM}{r(r-2GM)}\left(\frac{dr}{d\lambda}\right)^2+\frac{L^2(r-2GM)}{r^4}$$
...

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

What I am looking for ...
Is there any physical experiment proving that light speed slows down in gravitation field? (not for local observer . For him is constant c.)

Yes.

I'm not sure what the original poster is struggling with. It might be illuminating for him to take a simple case. ...

I am a mathematician , not physicist . I am trying to understand properties of geometry around heavy spherical object e.g. a neutron star or a black hole.
It looks to me there has to be two photon spheres .
I asked question about spacial distance because that (sub)space is full Reimannian metric space . (only distance from a point to itself is zero)
Space time in GR is pseudo-Reimannian (All events-points on a light-cone are on zero distance)

Thanks

 

[pervect]

The question of the uniqueness of the timelike Killing vector field in the Schwarzschild metric is something I've glossed over a bit, because I haven't really worked out anything rigorous. But I believe that you can construct a vector field that is timelike in a certain region by adding together multiples of the kvf for $\partial / \partial t$ and some multiple of $\partial / \partial \phi$ in the usual Schwarzschild metric with coordinates $(t, r, \theta, \phi)$ to generate another kvf. This new field won't be timelike everywhere, though, it will only be timelike in a certain region. This isn't exhaustive - there are two other space-like Killing vector fields according to wiki, that basically correspond to the freedom of choice in chosing the equatorial plane. Probably something needs to be added about smoothness or differentiability of the KVF, but I'm not sure how to word that.

 

[Ibix]

Is there any physical experiment proving that light speed slows down in gravitation field?

That isn't a well defined question because "space" is a matter of choice, even in static spacetimes. So speed is also a matter of choice except when it is measured locally, in which case it's a measure of the inner product of the four velocities of the objects.

That said, Shapiro delay of radar pulses bounced off Venus is usually interpreted as demonstrating that the coordinate speed of light varies.

 

[PeterDonis]

Is there any physical experiment proving that light speed slows down in gravitation field?

There can't be because "light speed slows down in a gravitational field" is not a coordinate-free, invariant statement. So you're going to have to be more explicit about exactly what experimental result you are interested in.

I am a mathematician , not physicist . I am trying to understand properties of geometry around heavy spherical object e.g. a neutron star or a black hole.

"Light speed" is not a property of geometry. It's a property of particular choices of coordinates.

The light cones are a property of the spacetime geometry. But it doesn't seem like you're interested in the spacetime geometry because it's not Riemannian. Which means you're leaving out a lot of information that is highly relevant to the physics involved.

It looks to me there has to be two photon spheres .

Why? This is false; there is only one photon sphere in Schwarzschild spacetime.

I asked question about spacial distance because that (sub)space is full Reimannian metric space . (only distance from a point to itself is zero)

Yes, but, as has already been pointed out, the geodesics of this space are not the same as the curves you get when you project spacetime geodesics into this space.

 

[PeterDonis]

I believe that you can construct a vector field that is timelike in a certain region by adding together multiples of the kvf for $\partial / \partial t$ and some multiple of $\partial / \partial \phi$ in the usual Schwarzschild metric with coordinates $(t, r, \theta, \phi)$ to generate another kvf.

Yes, as long as you form the linear combination with constant coefficients (i.e., the coefficients are the same at every event). Any linear combination of kvfs with constant coefficients is also a kvf.

This new field won't be timelike everywhere, though, it will only be timelike in a certain region.

More precisely, it won't be timelike in the same region (everywhere outside the horizon) that $\partial / \partial t$ is. It will be timelike in a smaller region (basically from just outside the horizon to just inside a finite value of $r$--exactly which value will depend on what constant coefficients you choose to form the linear combination). Heuristically, the outer boundary of this region is the point at which an object following the integral curve of the KVF at that point would have to move at the speed of light (i.e., the integral curve becomes null).

Also, this new kvf won't be hypersurface orthogonal. Hypersurface orthogonality is the unique property that picks out $\partial / \partial t$ from all the other possible timelike KVFs in this spacetime.

there are two other space-like Killing vector fields according to wiki, that basically correspond to the freedom of choice in chosing the equatorial plane.

Yes, there is actually a three-parameter group of KVFs that you can form by the process above, since the spacelike KVFs arising from spherical symmetry form a three-parameter group.

Probably something needs to be added about smoothness or differentiability of the KVF, but I'm not sure how to word that.

This is taken care of by the linear combination process; if the kvfs you are combining meet all those conditions, so will any linear combination of them with constant coefficients.

 

[Bosko]

...
Why? This is false; there is only one photon sphere in Schwarzschild spacetime.
...

Yes you are right for the Schwarzschild spacetime outside the object.
Sorry, I didn't explain well this my opinion about the two photon spheres.

Imagine a neutron star with a photon sphere.
We have space-time outside and inside (internal Schwarzschild solution). I meant the inner and outer photon spheres. Of course that inner photon sphere is only a space-time region within e.g. neutron stars probably without photons orbiting it.

 

[Bosko]

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.

Thanks

 

[PeterDonis]

Imagine a neutron star with a photon sphere.
We have space-time outside and inside (internal Schwarzschild solution). I meant the inner and outer photon spheres. Of course that inner photon sphere is only a space-time region within e.g. neutron stars probably without photons orbiting it.

I see, the inner photon sphere is inside the neutron star, and inside that inner photon sphere, an idealized test object that could follow a geodesic path without being disturbed by the neutron star's matter could follow a circular geodesic orbit (with the limiting case being the geodesic at $r = 0$ that just stays at rest there).

 

[Bosko]

I see, the inner photon sphere is inside the neutron star, and inside that inner photon sphere, an idealized test object that could follow a geodesic path without being disturbed by the neutron star's matter could follow a circular geodesic orbit (with the limiting case being the geodesic at $r = 0$ that just stays at rest there).

Hahaha ... thanks anyway

I even drew a rough cross-section of Flamm's paraboloid with the interior Schwarzschild solution added.

The inner photon sphere must definitely be there even though there probably aren't any photons. Certainly, if someone made a circular tunnel and photons would move through it in circles around the center.

That region of space-time between the inner and outer photon sphere seems to me to have very unusual and interesting properties.

 

[PeterDonis]

That region of space-time between the inner and outer photon sphere seems to me to have very unusual and interesting properties.

Yes, although I don't think any known neutron stars are compact enough for that region to be present. As far as I know all neutron stars have $r > 3M$ at their surfaces, so there is no photon sphere anywhere.

 

[Bosko]

Yes, although I don't think any known neutron stars are compact enough for that region to be present. As far as I know all neutron stars have $r > 3M$ at their surfaces, so there is no photon sphere anywhere.

it's a shame :-) this whole discussion goes to waste.

Is there any chance that a neutron star is compact enough to form a photon sphere around it, or at least for a moment as it collapses into a black hole?

I was interested in the shortest paths between two points everywhere in the surrounding area, but especially in that area.

 

[PeterDonis]

Is there any chance that a neutron star is compact enough to form a photon sphere around it

The Buchdahl Theorem limit on how compact a static object can be is $r > (9/4) M$ at the object's surface, which is inside the photon sphere. So in principle a compact object like a neutron star could have a photon sphere around it, at least if we just look at GR. The question would be whether there is any state of matter that has an equation of state that allows a stable equilibrium in that size range.

or at least for a moment as it collapses into a black hole?

A collapsing object is not static, so while the exterior geometry (i.e., the vacuum region around the object) will have a photon sphere, the interior geometry won't be the same as in the static case, and I don't think there will be a meaningful photon sphere inside the object.

 

[PeterDonis]

I was interested in the shortest paths between two points everywhere in the surrounding area, but especially in that area.

The same general statement we gave for the exterior region would apply in this area as well: the geodesics of the space (i.e., of a spacelike slice of constant time) will not, in general, be the same as the projections of spacetime geodesics into the space.

In the interior of the object, note that the spatial geometry is that of a 3-sphere, not a Flamm paraboloid.

 

[Bosko]

...

In the interior of the object, note that the spatial geometry is that of a 3-sphere, not a Flamm paraboloid.

The transition from the outside (Flamm's paraboloid) to the inside (it also looks like a paraboloid) should be smooth.

There are inflection points on the surface of the object