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

[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

 

[PeterDonis]

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

Yes, it's smooth. I'm not sure where you are getting "looks like a paraboloid" from. My understanding is that the spatial geometry of the interior solution is a 3-sphere.

There are inflection points on the surface of the object

If you mean an inflection point of the "slope" of the spatial geometry in an embedding diagram, yes.

 

[A.T.]

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

The interior geometry is rather spherical:
https://en.wikipedia.org/wiki/Interior_Schwarzschild_metric#Visualization

... should be smooth.

To some degree of continuity. The curvature switches from negative to positive.

 

[vanhees71]

But that's not a "black-hole solution" but the original Schwarzschild solution for a static incompressible fluid of constant density.

 

[PeterDonis]

But that's not a "black-hole solution" but the original Schwarzschild solution for a static incompressible fluid of constant density.

The OP said they were interested in both.

 

[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.

For which observer does the speed of light slows down entering a stronger gravitational field?

Let's imagine...

Let two rays of light emitted at point B arrive at (detector) point A at the same time.
Does $L_A=L_B$ for both observers?

For example, $t_A=20 ns, L_A=6 m$ and $t_B=10 ns, L_B=3 m$.
If $L_{AB}$ is removed from both paths, we get $2L_A=2L_B$ and $L_A=L_B$.
(Speed of light = 30 cm, or foot, per nano second)

The speed of light appears to be constant for any observer regardless of the gravitational field.
I'm interested in whether it slows down as approaches the photon sphere (or the event horizon).

 

[PeterDonis]

Let two rays of light emitted at point B arrive at (detector) point A at the same time.

How is that even possible? Two different rays would be emitted from point B at different times, and that means they would arrive at point A at different times.

 

[PeterDonis]

Does $L_A=L_B$ for both observers?

Yes. The time taken for light to travel is what is different for the two observers. The simplest way to test that is for each observer to send a round-trip light signal that reflects off a mirror at the other observer's location, and then compare the round-trip travel times by the two observers' clocks.

 

[Bosko]

How is that even possible? Two different rays would be emitted from point B at different times, and that means they would arrive at point A at different times.

Both ray are emitted from the same source and at same time from point B ( see image)

 

[PeterDonis]

Both ray are emitted from the same source and at same time from point B ( see image)

Then what's the point of having two rays? The two rays aren't telling you anything that one ray wouldn't tell you.

 

[Bosko]

Then what's the point of having two rays? The two rays aren't telling you anything that one ray wouldn't tell you.

By adjusting the LA, you can set them to arrive at the same time.
Then $L_A=L_B$ for both observers.