Global simultaneity surfaces - how to adjust proper time?

[PeterDonis]

See the relevant pages about synchronization frames.

The pages you attached talk about light signals and synchronization using them. They don't give any specific examples of frames that have the various properties. The only examples given of specific spacetimes are of light signals in those spacetimes,not frames.

 

[cianfa72]

The pages you attached talk about light signals and synchronization using them. They don't give any specific examples of frames that have the various properties. The only examples given of specific spacetimes are of light signals in those spacetimes, not frames.

Yes, there are not specific examples of such frames.

About the synchronization procedure employing light signals, I have not a clear understanding where the hypersurface orthogonal condition enters in such synchronization procedure.

 

[PeterDonis]

I have not a clear understanding where the hypersurface orthogonal condition enters in such synchronization procedure.

Exercise 5.2.6 seems to be relevant to this.

 

[cianfa72]

Exercise 5.2.6 seems to be relevant to this.

Ah ok, basically you are saying the curve $\alpha: [0,a) \rightarrow W$ can be understood as one of the geodesics that lies on the spacelike hypersurfaces orthogonal to the given geodesic $\gamma$ in the congruence.

BTW, I believe the condition for 'proper time synchronizable' (i.e. there is a global function $t$ defined on the entire spacetime such that $\xi = - dt$) actually amounts to the existence of a coordinate chart such that the metric tensor $g_{\mu \nu}$ has components $g_{00}=1$ and $g_{0 \alpha}=0$ for $\alpha=1,2,3$.

 

[PeterDonis]

the curve $\alpha: [0,a) \rightarrow W$ can be understood as one of the geodesics that lies on the spacelike hypersurfaces orthogonal to the given geodesic $\gamma$ in the congruence.

That's how I read it, yes.

I believe the condition for 'proper time synchronizable' (i.e. there is a global function $t$ defined on the entire spacetime such that $\xi = - dt$) actually amounts to the existence of a coordinate chart such that the metric tensor $g_{\mu \nu}$ has components $g_{00}=1$ and $g_{0 \alpha}=0$ for $\alpha=1,2,3$.

I believe that's correct, yes. If we generalize to allow $g_{00}$ to be something other than $1$ but still require $g_{0 \alpha} = 0$, then this is the condition for synchronizable.

 

[PeterDonis]

I believe that's correct, yes.

I say "I believe", but actually it's easy to verify.

If we have a global function $t$ such that $\xi = - dt$, then we can easily construct a coordinate chart meeting the conditions: $x^0 = t$ and choose the three $x^i$ (for $i = 1, 2, 3$) such that $\partial / \partial x^i$ is orthogonal to $\partial / \partial t$. Then $\xi = - dt$ ensures that $g_{00} = 1$ (because $t$ must be equal to proper time along integral curves of $\xi$) and the orthogonality condition just mentioned ensures that $g_{0 i} = 0$.

If we have a coordinate chart meeting the conditions, then we choose $\xi = - dx^0$ and we can see that $x^0$ is the global function $t$ that we are looking for.

If we generalize to $\xi = - h dt$ for some function $h$, then we can verify the more general statement I made about a synchronizable congruence; it will turn out that $h = \sqrt{|g_{00}|}$.

 

[cianfa72]

If we have a global function $t$ such that $\xi = - dt$, then we can easily construct a coordinate chart meeting the conditions: $x^0 = t$ and choose the three $x^i$ (for $i = 1, 2, 3$) such that $\partial / \partial x^i$ is orthogonal to $\partial / \partial t$. Then $\xi = - dt$ ensures that $g_{00} = 1$ (because $t$ must be equal to proper time along integral curves of $\xi$) and the orthogonality condition just mentioned ensures that $g_{0 i} = 0$.

ok, so the coordinate chart we get in this way is actually a global coordinate chart, I believe.

 

[PeterDonis]

the coordinate chart we get in this way is actually a global coordinate chart, I believe.

Since this method of obtaining a chart only works for the synchronizable and proper time synchronizable cases, yes, it will be global since those cases require that the conditions are globally valid.

 

[cianfa72]

Then $\xi = - dt$ ensures that $g_{00} = 1$ (because $t$ must be equal to proper time along integral curves of $\xi$)

Just to do the complete calculation. We start with a unit timelike vector field $Q$ i.e. $g(Q,Q)=-1$. Then $\xi$ is defined by $\xi= g(Q, \_)$. Our goal is integrate the one-form $\xi=-dt$ over the integral orbits of $Q$.

Assume $\gamma: I \rightarrow M$ is an orbit parametrization based on proper time $u$. Now the pullback of $\xi$ by $\gamma$ is exactly $-du$, so the integral of $\xi$ along any integral orbit $\gamma$ is $-\Delta u=\int_{\gamma} \xi =- \Delta t$. On the other hand along any integral orbit we get $du = \sqrt g_{00} dt$ hence $g_{00}=1$.

For a 'synchronizable' congruence we get $- \Delta u = \int_{\gamma} \xi =- \int_{t_1}^{t_2} hdt$ hence $h=\sqrt g_{00}$.

Is the above correct ? Thank you.

 

[PeterDonis]

Is the above correct ?

Yes.

 

[cianfa72]

ok, the latter part of post #69 is in fact consistent to assume the function $h$ to be $h>0$ everywhere (as done from Sachs and Wu in their definition of synchronizable congruence).

 

[cianfa72]

BTW I was thinking that their definition of locally proper time synchronizable (i.e. $d\xi=0$) is actually the same as the existence *locally* of a function $f$ such that $\xi= df$ -- in force of Poincaré lemma.

 

[PeterDonis]

BTW I was thinking that their definition of locally proper time synchronizable (i.e. $d\xi=0$) is actually the same as the existence *locally* of a function f such that $\xi= df$ -- in force of Poincaré lemma.

Yes, for 1-forms this local equivalence holds.

 

[cianfa72]

I was thinking about the following trivial example involving a sphere as discussed here Synchronous reference frame.

Consider the set of latitude circles. They can be described as the level sets of a global function $g$ defined on the sphere in a such way that the difference between its values on different level sets is the length of any segment of great circle between them (geodesic segment) and orthogonal to them.

By the logic discussed above we can build a global coordinate chart for the sphere such that $g_{00}=1$ and $g_{01}=0$ (it turns out in an absurd since we know it does not exist at all).

I believe the problem is that such globally defined function $g$ actually does not exist. If there was then $\omega=dg$ would define a smooth unit vector field $Q$ such that $\omega(Q)=-1$ not vanishing at any point. We know it cannot exist -- see Hair ball theorem.

 

[PeterDonis]

Consider the set of latitude circles. They can be described as the level sets of a global function $g$ defined on the sphere

No, they can't. The poles are not "level sets"; they're single points. Single points won't work as "level sets" since each point in a "level set" has to have an open neighborhood that is also in the "level set".

I believe the problem is that such globally defined function $g$ actually does not exist.

The one you defined does not work, as above. I think the sphere in general does not work for this kind of global construction because it's a compact manifold.

If there was then $\omega=dg$ would define a smooth unit vector field $Q$ such that $\omega(Q)=-1$ not vanishing at any point. We know it cannot exist -- see Hair ball theorem.

Yes, this is another reason why the sphere would not work for this kind of global construction. One can still do it on an open subset of the sphere, as @PAllen said in the thread you linked to.

 

[cianfa72]

No, they can't. The poles are not "level sets"; they're single points. Single points won't work as "level sets" since each point in a "level set" has to have an open neighborhood that is also in the "level set".

Ah ok, so this is not actually a counterexample. The claim that if exists a smooth global function $g$ such that $\omega = dg$ (i.e. a unit vector field $Q$ defined from $\omega(Q)=-1$) then there is a *global* coordinate chart such that $g_{00}=1$ and $g_{0\alpha}=0$ holds true.

 

[PeterDonis]

this is not actually a counterexample.

Correct.

 

[PeterDonis]

Formally, $d \alpha = \beta \wedge \omega$ is another possibility to make the first term vanish, yes. However, I believe it is ruled out because continuing to apply $dd = 0$ leads to an infinite regress.

I went back and looked at this again, and I don't think my previous comment, quoted above, was correct.

If $d\alpha = \beta \wedge \omega \neq 0$, then $d d \alpha = 0$ gives $d\beta \wedge \omega + \beta \wedge d \omega = 0$, which in turn gives $\left( d \beta + \beta \wedge \alpha \right) \wedge \omega = 0$.

There are two ways to satisfy that last equality. First, we could have $d \beta = 0$ and $\beta \wedge \alpha = 0$. But $\beta \wedge \alpha = 0$ means $\beta = k \alpha$, and thus $d \beta = 0$ means $dk \wedge \alpha + k d \alpha = dk \wedge \alpha + k \beta \wedge \omega = 0$. Since $\alpha$, $\beta$, and $\omega$ are all linearly independent (by hypothesis, since we have assumed that $d\alpha \neq 0$), the only way to satisfy the last equality is to have $dk \wedge \alpha = 0$ and $d \alpha = 0$. But $d \alpha = 0$ contradicts our hypothesis (just stated in the parentheses above), so this possibility cannot be correct.

The second way to satisfy the last equality in the 2nd paragraph above is to have $d\beta = - \beta \wedge \alpha = \alpha \wedge \beta$. But this is the same equation that is satisfied by $d \omega$ and $\omega$, so we must have $\beta = \omega$ for this case--but then, once again, we get $d\alpha = \beta \wedge \omega = \omega \wedge \omega = 0$, which contradicts our hypothesis. So this possibility cannot be correct either.

In short, there is no way to have $d\alpha = \beta \wedge \omega \neq 0$ without leading to a contradiction.

 

[cianfa72]

In short, there is no way to have $d\alpha = \beta \wedge \omega \neq 0$ without leading to a contradiction.

ok, I believe it is locally the expected result.

By Poincarè lemma locally any closed form is even exact hence *locally* (i.e. in any open neighborhood) $\omega \wedge d\omega=0$ is really equivalent to the existence of functions $f$ and $g$ such that $\omega=fdg$.

 

[PeterDonis]

By Poincarè lemma locally any closed form is even exact hence *locally*

Any closed 1-form is locally exact, yes. This result does not generalize to higher rank forms, though.

In any case, way back in the earlier post I quoted, I was responding to your suggestion that $d \alpha = 0$ was not the only possibility. If there were a way to have $d \omega = \alpha \wedge \omega$ with $d \alpha \neq 0$, then the 1-form $\alpha$ could not be proven to be closed and we could not apply the Poincare lemma. I was simply going back and revisiting and correcting my original claim about why $d \alpha \neq 0$ will not work.

(i.e. in any open neighborhood) $\omega \wedge d\omega=0$ is really equivalent to the existence of functions $f$ and $g$ such that $\omega=fdg$.

This only follows from the Poincare lemma if we can prove that $d \alpha = 0$, where $d \omega = \alpha \wedge \omega$. To do that, we have to rule out the possibility that $d \alpha = \beta \wedge \omega \neq 0$ for some $\beta$. That's what my post #78 is about.

 

[cianfa72]

This only follows from the Poincare lemma if we can prove that $d \alpha = 0$, where $d \omega = \alpha \wedge \omega$. To do that, we have to rule out the possibility that $d \alpha = \beta \wedge \omega \neq 0$ for some $\beta$. That's what my post #78 is about.

Sorry, not sure to understand. You said there is no way to have $d \alpha = \beta \wedge \omega \neq 0$ without leading to a contradiction. Hence it has to be $d \alpha=0$ I believe...

 

[PeterDonis]

You said there is no way to have $d \alpha = \beta \wedge \omega \neq 0$ without leading to a contradiction.

Yes, that was the point of my post #78.

Hence it has to be $d \alpha=0$

Once we have correctly shown that the other possibility leads to a contradiction, yes. As I noted in post #78, I don't think my previous claim about the other possibility was correct; that's why I posted post #78, to correctly show why $d \alpha = \beta \wedge \omega$ does not work.