Undergrad Schwarzschild metric not dependent on time

[kent davidge]

Since it's possible to choose a coordinate chart where the Schwarzschild metric components are dependent on time, why that's not done? Would'nt there be a scenario where such a choice would be useful?

 

[PeroK]

Since it's possible to choose a coordinate chart where the Schwarzschild metric components are dependent on time, why that's not done? Would'nt there be a scenario where such a choice would be useful?

Is that not what's usually done?

 

[kent davidge]

Is that not what's usually done?

Quite the opposite. I myself have never seen time dependent metric components in the Schwarzschild metric.

I suspect this has to do with the physics involved. That the solution is for a static, etc... spacetime, so the most natural choice should be the one with time independent metric components.

 

[PeroK]

Quite the opposite. I myself have never seen components time dependent for the metric.

I suspect this has to do with the physical situation involved. That the solution is for a static, etc... spacetime, so the most natural choice should be the one with time independent metric components.

Why would you want time-dependent components for a static solution? I must admit that's a question I find difficult to comprehend!

Is GR not hard enough without worrying about questions like that?

I assumed you meant opposite!

 

[kent davidge]

Why would you want time-dependent components for a static solution?

I don't know. I was asking if there's a case where one would want it.

Would'nt there be a scenario where such a choice would be useful?

 

[Ibix]

An obvious time-dependent form is one in which the mass is moving. Presumably you could do it - radar coordinates spring to mind as something with practical utility for a passing spaceship. I haven't seen it done. Probably because you'd want a different solution for every possible powered and unpowered trajectory. Easier just to work in regular Schwarzschild coordinates, I should think.

 

[kent davidge]

An obvious time-dependent form is one in which the mass is moving. Presumably you could do it - radar coordinates spring to mind as something with practical utility for a passing spaceship. I haven't seen it done. Probably because you'd want a different solution for every possible powered and unpowered trajectory. Easier just to work in regular Schwarzschild coordinates, I should think.

thank you

 

[PeterDonis]

it's possible to choose a coordinate chart where the Schwarzschild metric components are dependent on time

Why do you think this is the case? Can you give an example?

 

[kent davidge]

Why do you think this is the case?

Because static only means you can find a coordinate system where the metric takes that form taken in the Schwarzschild coordinates. The definition doesn't say that it's the only possibility.

Can you give an example?

Any metric of the form $g_{\alpha \beta}(u^0) du^\alpha du^\beta$ where the "time" coordinate is $u^0$ and $g_{0 0} < 0$ and $\ g_{0, \alpha}, \ g_{i,j} > 0$?

 

[PeterDonis]

The definition doesn't say that it's the only possibility.

That's correct.

Any metric of the form...

I meant a specific example. If you can't find one after searching the literature, that would be relevant to your OP question, would it not?

 

[kent davidge]

I meant a specific example. If you can't find one after searching the literature, that would be relevant to your OP question, would it not?

yes, it would mean that it's not helpful to have such metric.

 

[PAllen]

The metric in Lemaitre coordinates for the Schwarzschild geometry depends on the timelike coordinate. This is a consequence of covering both the horizon interior and exterior in one coordinate patch, and making the outside seem similar to the inside, which is not static. A world line of constant position in these coordinates is the trajectory of a radial free fall from infinity. These coordinates share with Kruskal coordinates the feature that there is one timelike and 3 spacelike coordinates over the whole patch. However, they don’t, and can’t cover the whole Kruskal geometry - they cover one exterior and part of one interior region.

[edit: These coordinates can be quite useful for many questions relating to radial infall because key facts are just read from the coordinates.]

 

[kent davidge]

The metric in Lemaitre coordinates for the Schwarzschild geometry depends on the timelike coordinate. This is a consequence of covering both the horizon interior and exterior in one coordinate patch, and making the outside seem similar to the inside, which is not static. A world line of constant position in these coordinates is the trajectory of a radial free fall from infinity. These coordinates share with Kruskal coordinates the feature that there is one timelike and 3 spacelike coordinates over the whole patch. However, they don’t, and can’t cover the whole Kruskal geometry - they cover one exterior and part of one interior region.

[edit: These coordinates can be quite useful for many questions relating to radial infall because key facts are just read from the coordinates.]

wow , thanks

 

[kent davidge]

These coordinates can be quite useful for many questions relating to radial infall because key facts are just read from the coordinates

@PeroK , now hoppefully you can see why your earlier comment

Is GR not hard enough without worrying about questions like that?

does not make much of a sense... depending on the situation it's actually more easy to have a second "more difficult" approach.

 

[PAllen]

Note, of course, that Kruskal coordinates also have all metric components depend on the timelike coordinate. This is slightly hidden the way they are usually presented using Schwarzschild r. However, r is a function of both the Kruskal timelike and 'radial' coordinates (given as U and V in many sources), so every metric component depends on U and V.

 

[PeterDonis]

Why would you want time-dependent components for a static solution?

Schwarzschild spacetime is only static outside the event horizon.

 

[kent davidge]

Schwarzschild spacetime is only static outside the event horizon.

but by a change of coordinate chart we can make it static again, correct? (but I guess in this new chart it becomes non-static outside the event horizon)

 

[PeterDonis]

but by a change of coordinate chart we can make it static again, correct?

Whether or not a spacetime, or a region of spacetime, is static is an invariant geometric property of the spacetime, independent of your choice of coordinates. The correct definition of "static" is that there is a timelike Killing vector field. For any spacetime with a Killing vector field, you can find a coordinate chart in which the metric coefficients are independent of one of the coordinates (that coordinate is $t$ in standard Schwarzschild coordinates on Schwarzschild spacetime). But that Killing vector field might not be timelike; in Schwarzschild coordinates, $t$ is only timelike outside the horizon.

 

[Ibix]

But that Killing vector field might not be timelike; in Schwarzschild coordinates, $t$ is only timelike outside the horizon.

Just to expand on this - there are three four Killing vector fields in Schwarzschild spacetime. Two Three correspond to rotations. The third fourth is timelike outside the horizon, and is what many coordinate systems use as "time" there, but is spacelike inside.

Corrections per the next post.

 

[PeterDonis]

there are three Killing vector fields in Schwarzschild spacetime

No, there are four. Three correspond to spatial rotations (heuristically, you can think of them as rotations about three mutually perpendicular spatial axes). The fourth is the one you describe.

 

[Ibix]

No, there are four.

Indeed - can't count on a Friday night.

 

[stevendaryl]

No, there are four. Three correspond to spatial rotations (heuristically, you can think of them as rotations about three mutually perpendicular spatial axes). The fourth is the one you describe.

Okay, I'm a little puzzled now. If you think about Schwarzschild spacetime, you're in some region, what directions can you travel and leave the metric tensor components unchanged? Well, you can vary $t$, $\theta$ or $\phi$ in the usual Schwarzschild coordinates. Those correspond to three independent Killing vector fields. So how do those fields relate to your rotations?

Well, if you're sitting at the point $x=R, y=0, z=0$ then traveling in the $\phi$ direction is equivalent to rotating about the $z$ axis. Traveling in the $\theta$ direction is equivalent to rotating about the $y$ axis. But at that particular point, there is no direction corresponding to rotating about the $x$ axis. It's a no-op.

I would say that there are at every point, only three independent Killing vectors. The time direction is always one. But then the two other ones correspond to directions you can travel while leaving $r$ and $t$ unchanged. Since the surface of points with constant $r$ and $t$ is a sphere, it's a 2-D space, and so there are only two independent Killing vectors.

That's what I would think.

 

[PAllen]

@stevendaryl ,
I have been confused about rotational killing vectors in the past. However I think the following makes it crystal clear that ther are three, not two:

http://www.physics.usu.edu/Wheeler/GenRel2013/Notes/GRKilling.pdf

 

[vanhees71]

Well, you have to distinguish between the manifold and the coordinates used to parametrize (parts of) it.

E.g., usual 3D Euclidean affine space is isotropic. This is evident, using Cartesian coordinates. The metric is given by
$\mathrm{d}s^2 = \mathrm{d}x^2+ \mathrm{d} y^2 + \mathrm{d} z^2,$
which is invariant under the full rotation group $\mathrm{SO}(3)$.
To see this you can use the Lagrangian for geodesics (in the "square form" to make life easy)
$$L=\frac{1}{2} \dot{\vec{x}}^2.$$
Applying Noether's theorem to it you indeed get the "angular momenta"
$$\vec{J}=\vec{x} \times \dot{\vec{x}}$$
as the generators of rotation. In the languauge of differential geometry they are "Killing vectors", defining the symmetry transformation via the corresponding Lie derivatives.

Now if you use spherical coordinates, the Lagrangian reads
$$L=\frac{1}{2} (\dot{r}^2 + r^2 \dot{\vartheta}^2 + r^2 \sin^2 \vartheta \dot{\varphi}^2).$$
Now it seems as if there's only one rotational symmetry left, because only $\varphi$ is cyclic. Of course, the symmetries of Eulidean space haven't changed at all, but the introduction of the polar axis breaks rotational symmetry in the sense that the symmetry isn't manifest anymore in the coordinates used to describe the manifold.

 

[stevendaryl]

@stevendaryl ,
I have been confused about rotational killing vectors in the past. However I think the following makes it crystal clear that ther are three, not two:

http://www.physics.usu.edu/Wheeler/GenRel2013/Notes/GRKilling.pdf

Okay, getting back to my intuitive picture: My mistake is cavalierly conflating vectors with vector fields. It is true that at any point in Schwarzschild spacetime, there are only 3 independent vectors that "point" in a direction where the metric components are unchanging. You can take those directions and turn them into vector fields. But there is a 4th vector FIELD that is independent of the other three. It just happens to be that that field vanishes at the point.

My crudely-drawn picture illustrating: If you're at the point indicated by the black dot, there are indeed two independent directions you can travel and remain on the sphere. Those two directions can be expanded into a Killing vector fields. But there is a third independent Killing vector field at that point, indicated by the picture on the right: It points around the point in question. So it corresponds to a field, but not to a vector (since it vanishes at the point).

 

[Ibix]

Okay, getting back to my intuitive picture: My mistake is cavalierly conflating vectors with vector fields.

I confuse myself like this sometimes. I remind myself by thinking of Minkowski spacetime whose spacelike Killing fields are the translations and rotations, making six. There are only three spacelike directions, so three Killing vectors at a point must be dependent. But the fields aren't. There's always a way to linearly combine the fields to make a fourth at a point, but never so that it works everywhere.

 

[vanhees71]

Shouldn't Minkowski space even have 10 killing vectors (or rather Killing fields, to avoid the sloppy use of math language by physicists that again seems to lead to some confusion ;-)) generating the full orthochronous proper Poincare group (space-time translations, Lorentz boosts, and rotations)?

Let's see. For Minkowski space we can choose coordinates (the usual $x^{\mu}=(t,\vec{x})$ ones) such that
$$g_{\mu \nu}=\eta_{\mu \nu}=\text{const}.$$
The Killing equation thus reads
$$\nabla_{\mu} \xi_{\nu} + \nabla_{\nu} \xi_{\mu}=\partial_{\mu} \xi_{\nu} + \partial_{\nu} \xi^{\mu}=0.$$
To solve this, the trick is to take one more (partial) derivative and cyclically premutate the indices
$$\partial_{\rho} \partial_{\mu} \xi_{\nu} + \partial_{\rho} \partial_{\nu} \xi_{\mu} = 0 \; \Rightarrow \; \partial_{\mu} \partial_{\nu} \xi_{\rho} + \partial_{\mu} \partial_{\rho} \xi_{\nu}=0, \quad \partial_{\nu} \partial_{\rho} \xi_{\mu} + \partial_{\nu} \partial_{\mu} \xi_{\rho}=0.$$
Adding the first two of these equations and subtracting the third, finally leads to
$$\partial_{\rho} \partial_{\mu} \xi_{\nu}=0 \; \Rightarrow \; \xi_{\nu} = a_{\nu} + \Omega_{\nu \rho} x^{\rho}.$$
Plugging this again in the killing equation, yields
$$\partial_{\rho} \xi_{\nu} + \partial_{\nu} \xi_{\rho} = \Omega_{\nu \rho} + \Omega_{\rho \nu}=0 \; \Rightarrow \; \Omega_{\rho \nu} = - \Omega_{\nu \rho}.$$
Indeed, as we see the 10 independent killing fields, defined by setting all but one of the 10 independent parameters $a_{\nu}$ and $\Omega_{\mu \nu}$ to zero, indeed generate space-time translations, rotations (parametrized with the 3 independent $\Omega_{\mu \nu}$ with $\mu,\nu \in \{1,2,3 \}$) and boosts (parametrized with the three independent $\Omega_{\mu 0}$, $\mu \in \{1,2,3 \}$).

 

[Ibix]

Shouldn't Minkowski space even have 10 killing vectors (or rather Killing fields, to avoid the sloppy use of math language by physicists that again seems to lead to some confusion ;-)) generating the full orthochronous proper Poincare group (space-time translations, Lorentz boosts, and rotations)?

I believe so. I specified the space-like ones, which I think are six in number. I could have said Euclidean 3-space, I suppose.

 

[vanhees71]

I've just put the proof in my previous posting. For Euclidean 3D affine space, the proof is of course literally the same (as for any pseudo-metric affine space, while the Euclidean spaces are of course the special cases of the true metric affine spaces). In 3D you have 3 independent translations and 3 independent rotations. The symmetry group is the group $\text{ISO}(3)$.

 

[kent davidge]

Two of the Killing fields are easy to know, it follows directly from the independence of the metric on $t$ and $\varphi$ that $(-(2m - r), 0,0,0)$ and $(0,0,0, (r \sin \vartheta)^{-2})$ are the vectors components of the vectors of the two fields at any point $(t,r,\vartheta,\varphi)$.

But I'm afraid the vector components of the remaining vector fields can only be found by solving the system of PDE resulting from the Killing equations. And this is really hard.

Or perhaps we can split the metric into parts and compute only the necessary (unknown) terms for the Killing equations?