When time is not orthogonal to phi, that means there is a preferred direction in space? How?

[Pencilvester]

TL;DR

I’m confused about something Schutz says about spherically symmetric spacetimes.

In section 10.1 of "A First Course in GR" by Schutz, in the paragraph directly above equation 10.5, Schutz says "we must have that a line $r$ = const., $\theta$ = const., $\phi$ = const. is also orthogonal to the two-spheres. Otherwise there would be a preferred direction in space. This means that ... $g_{t \phi} = 0$" (I added the bold.) I don't understand this statement. For example, if we consider the unit 2-sphere in time (I'm ignoring $r$), the line element is $ds^2 = -dt^2 + d\theta^2 + \sin^2 \theta ~ d\phi^2$. Now, if I just set the $\phi$ coordinate spinning so that we have a new coordinate $\phi'$ such that $\phi' = \phi + t$, then we have essentially the same metric, but with $g_{t \phi'} = - \sin^2 \theta \neq 0$. I don't see how setting $\phi$ spinning makes it so that there is a preferred direction in space.

 

[Dale]

TL;DR Summary: I’m confused about something Schutz says about spherically symmetric spacetimes.

I don't see how setting ϕ spinning makes it so that there is a preferred direction in space.

There is a direction parallel to the spin axis and directions that are not.

 

[PeterDonis]

if I just set the $\phi$ coordinate spinning so that we have a new coordinate $\phi'$ such that $\phi' = \phi + t$, then we have essentially the same metric

You're thinking of it backwards. Schutz's point is that if the spacetime is spherically symmetric, it must be possible to choose coordinates so that the $t$, $r$ subspace is orthogonal to the $\theta$, $\phi$ subspace. That doesn't mean you can't also find coordinates in which $g_{t \theta}$ or $g_{t \phi}$ are nonzero. It just means you don't have to.

That said, I actually don't find Schutz's presentation here very enlightening. A better argument can be found in, for example, Misner, Thorne, & Wheeler, and basically goes like this: if the spacetime is spherically symmetric, so that it can be foliated everywhere by geometric 2-spheres, then the subspace formed by factoring out the 2-spheres, so to speak, must be everywhere orthogonal to the 2-spheres, because if it weren't, we could use the non-orthogonality to define a vector field on any 2-sphere that was everywhere non-vanishing--and the hairy ball theorem says it is impossible to define a vector field on a 2-sphere that is everywhere non-vanishing. So we must be able to find coordinates such that the $t$, $r$ subspace has no "cross terms" in the metric with the $\theta$, $\phi$ subspace.

 

[Pencilvester]

There is a direction parallel to the spin axis and directions that are not.

This makes sense, thanks. I think my problem was I was connoting some sort of spherical asymmetry with the phrase “preferred direction in space”, which contradicts the whole premise of the section.

 

[Pencilvester]

I actually don't find Schutz's presentation here very enlightening. A better argument can be found in, for example, Misner, Thorne, & Wheeler

This is enlightening, thanks!

 

[PeterDonis]

I think my problem was I was connoting some sort of spherical asymmetry with the phrase “preferred direction in space”

That is what Schutz means, yes: if the spacetime is spherically symmetric, there cannot be a "preferred direction in space". So anything that would imply a preferred direction in space is inconsistent with spherical symmetry. I'm just not sure that "preferred direction in space" is the best way to describe what the consequences of non-orthogonality between the $t$, $r$ subspace and the $\theta$, $\phi$ subspace would be (although as @Dale pointed out, if the non-orthogonality is due to spin, then the spin axis is a preferred direction in space).

 

[cianfa72]

Schutz's point is that if the spacetime is spherically symmetric, it must be possible to choose coordinates so that the $t$, $r$ subspace is orthogonal to the $\theta$, $\phi$ subspace.

That means it does exist a coordinate chart where, at any point/event, any linear combination of $\{ \partial_t, \partial_r \}$ is orthogonal to any linear combination of $\{ \partial_{ \theta}, \partial_{\phi} \}$ (i.e. all cross terms $g_{t, \theta}, g_{t, \phi}, g_{r, \theta}, g_{r, \phi}$ vanish).

if the spacetime is spherically symmetric, so that it can be foliated everywhere by geometric 2-spheres, then the subspace formed by factoring out the 2-spheres, so to speak, must be everywhere orthogonal to the 2-spheres, because if it weren't, we could use the non-orthogonality to define a vector field on any 2-sphere that was everywhere non-vanishing--and the hairy ball theorem says it is impossible to define a vector field on a 2-sphere that is everywhere non-vanishing.

From a mathematical point of view, if spacetime was spherically symmetric, whether the above conditions were not met, why there would be a non-vanishing smooth vector field defined on each 2-sphere foliating it ?

 

[DrGreg]

From a mathematical point of view, if spacetime was spherically symmetric, whether the above conditions were not met, why there would be a non-vanishing smooth vector field defined on each 2-sphere foliating it ?

You could take the orthogonal projection of $\partial_t$ onto the 2-sphere.

 

[jbergman]

You could take the orthogonal projection of $\partial_t$ onto the 2-sphere.

I don't think that quite follows from what was stated. You could only do that if $\partial_t$ was everywhere non-orthogonal or in other words nowhere orthogonal. That is a different statement from the original claim that if $\partial_t$ was not orthogonal at some points then you could construct a nowhere vanishing vector field on the 2-sphere.

 

[PeterDonis]

From a mathematical point of view, if spacetime was spherically symmetric, whether the above conditions were not met

That would be a contradiction.

 

[PeterDonis]

I don't think that quite follows from what was stated. You could only do that if $\partial_t$ was everywhere non-orthogonal or in other words nowhere orthogonal. That is a different statement from the original claim that if $\partial_t$ was not orthogonal at some points then you could construct a nowhere vanishing vector field on the 2-sphere.

The two claims are equivalent. The 2-spheres are orbits of Killing vector fields, or, to put it another way, the orthogonality or lack thereof of any vector field to any given 2-sphere must be rotationally invariant. So if $\partial_t$ is non-orthogonal to a 2-sphere at one point, it must be non-orthogonal at every point.

The derivation in MTW that I was referring to is in Box 23.3. It contains a more detailed statement of the argument than I have given here.

 

[cianfa72]

Spherically symmetric spacetime is not the same as maximally symmetric spacetime. In the latter case (e.g. Minkowski spacetime) there are $4\cdot 5/2 = 10$ continuous symmetries (10 KVFs) while in the former there are only 3 spacelike KVFs (plus one timelike KVF whether it is also stationary).

 

[PeterDonis]

Spherically symmetric spacetime is not the same as maximally symmetric spacetime.

Did anyone say it was?

 

[cianfa72]

Did anyone say it was?

No of course.

 

[PeterDonis]

No of course.

Then what's your point? This thread isn't about a maximally symmetric spacetime, it's about a spherically symmetric spacetime.