I Global simultaneity surfaces - how to adjust proper time?

[PeterDonis]

Coming back to your post #39, you mean since from $\omega = f dt$ we get $d\omega = df \wedge dt$ then the latter is in the form $d\omega = \alpha \wedge \omega/f$ in which $\alpha = df$ is exact, nevertheless it is not in the form $d\omega = \alpha \wedge \omega$.

No. If $\omega = f dt$, then $d \omega = df \wedge dt$, which is in the form $d \omega = \alpha \wedge \omega$ with $\alpha = df / f = d g$ with $g = \ln f$, so $\alpha$ is exact.

 

[cianfa72]

No. If $\omega = f dt$, then $d \omega = df \wedge dt$, which is in the form $d \omega = \alpha \wedge \omega$ with $\alpha = df / f = d g$ with $g = \ln f$, so $\alpha$ is exact.

Ah ok. $d\omega = df \wedge dt = f/f df \wedge dt = df /f \wedge fdt = \alpha \wedge \omega$

 

[romsofia]

$\omega \wedge d\omega = 0$, seems kind of trivially true, no? If $\omega$ is a one form, then it is in the form of $\omega = h dx$ where $h,dx$ are just place holders. By product rule, i get that $d\omega = dh \wedge dx + h d^2x$ since d^2 = 0, the 2nd term vanishes. thus, we're left with $\omega \wedge d \omega = h dx \wedge dh \wedge dx \rightarrow -h dx \wedge dx \wedge dh = 0$

Another interesting thing is that $\omega \wedge d\omega = d\omega \wedge \omega$ because, in general, $\alpha \wedge \beta = (-1)^{pq} \beta \wedge \alpha$ for p,q form $\alpha, \beta$ Now, this is where $\omega \wedge \omega = 0$ comes from, if $\omega$ is a one form. You'd get $\omega \wedge \omega = -\omega \wedge \omega$ which is only true if it's zero.

Conclusion I make: The only times $\omega \wedge d\omega \neq 0$ FOR ONE FORMS $\omega$ is if you have a one form where $\omega \wedge \omega \neq 0$ but I'd have to see an example of that (if it exists).

 

[PeterDonis]

$\omega \wedge d\omega = 0$, seems kind of trivially true, no?

No.

If $\omega$ is a one form, then it is in the form of $\omega = h dx$

No. Not all 1-forms can be expressed in this way.

Another interesting thing is that $\omega \wedge d\omega = d\omega \wedge \omega$ because, in general, $\alpha \wedge \beta = (-1)^{pq} \beta \wedge \alpha$ for p,q form $\alpha, \beta$

No. You left out a minus sign. Since the rank of $d \omega$ is one greater than the rank of $\omega$, we have that one of $p, q$ is even and one is odd, so their product is odd and $-1^{pq} = -1$, not $+1$.

Conclusion I make: The only times $\omega \wedge d\omega \neq 0$ FOR ONE FORMS $\omega$ is if you have a one form where $\omega \wedge \omega \neq 0$

It is impossible to have $\omega \wedge \omega \neq 0$ for any form; the wedge product of any form with itself vanishes by the definition of wedge product. However, this does not mean $\omega \wedge d \omega \neq 0$ is impossible, because all the other claims you have made, on which your conclusion here is based, are also wrong.

 

[romsofia]

No. You left out a minus sign. Since the rank of $d \omega$ is one greater than the rank of $\omega$, we have that one of $p, q$ is even and one is odd, so their product is odd and $-1^{pq} = -1$, not $+1$.

Even times odd is always even, so it would be what I said it was. $\omega$ is one form, $d\omega$ is a two form, thus $\omega \wedge d\omega = (-1)^{1*2} d\omega \wedge \omega = d\omega \wedge \omega$, but that's besides the point, I didn't think this problem through.

I was sloppy with my one form, as I thought it was trivial, but you're right, it isn't trivial. I looked into my notes, and you can find a discussion on this in full detail in "Lectures on differential geometry" by Shlomo Steinberg which my notes have a mixing between calling $\omega \wedge d\omega = 0$ an integrability condition, then diving into "Darboux's theorem" starting at page 130 in this book (or i guess, just go to wiki: https://en.wikipedia.org/wiki/Darboux's_theorem , https://en.wikipedia.org/wiki/Integrability_conditions_for_differential_systems)

So yes, my conclusion was wrong, sorry OP, I should have thought it through more!

An example that may be helpful for OP since you're looking into integrable systems:

Suppose you wanted find the necessary and sufficient conditions for a scalar function $e^\phi$ to be a integrating for some one form $\omega$. This means you're looking for $d(e^\phi \omega) = 0$ using our integrability condition $\omega \wedge d\omega = 0$, and expanding $d(e^\phi \omega) = d\omega + d\phi \wedge \omega = 0$ If we put $\omega$ in "normal form" which looks like $\omega = x^1dx^2+x^3dx^4+...+dx^n$. We can conclude that there can't be a $dx^n$ term because then $\omega \wedge d\omega$ would have random terms like $dx^n \wedge dx^1 \wedge dx^2$. We also conclude that it can't have more than a single monomial because we'd have random terms like $x^1 dx^2 \wedge dx^3 \wedge dx^4$

So, we can say that $\omega = x^1 dx^2$ and find the integrating factor to be $\frac{1}{x^1}$

(I believe this example comes from "Applied differential geometry" by Burke)

 

[romsofia]

Some interesting lecture notes I found on these theorems, chapter 4 is where you should start, but lecture 21 onward seems the most fruitful if looking for a global application:

https://stanford.edu/~sfh/257A.pdf

I'll take a look at them deeper in the future, but maybe this can help.

 

[PeterDonis]

Even times odd is always even

Oops, yes, you're right.

 

[cianfa72]

I think the take-home point is the following: Global Frobenius says that the condition $\omega \wedge d\omega = 0$ at every point $p$ of spacetime manifold is equivalent to the existence of maximal integrating immersed submanifolds foliating the overall manifold -- i.e. the entire spacetime manifold can be foliated by the union of those (disjoint) integrating submanifolds (leaves). *Locally* there exist two smooth functions $f$ and $g$ such that $\omega = f dg$. That means locally (i.e. in an open neighborhood of point $p$) the level sets of $g$ are the maximal integrating immersed submanifolds around $p$.

However the point to make clear is that the set of integrating submanifolds may not be given as the level sets of a global smooth function *alone* defined on the overall manifold.

 

[PeterDonis]

@cianfa72 do Sachs & Wu give any examples of frames that meet the various conditions?

 

[cianfa72]

@cianfa72 do Sachs & Wu give any examples of frames that meet the various conditions?

See the relevant pages about synchronization frames. Here $Z$ is a vector field such that $g(Z,Z)=-1$ everywhere (i.e. it defines a unit timelike congruence that fills the entire spacetime) hence $\xi$ is the one-form $\xi = g(Z, \_)$.

'Locally proper time synchronizable' does mean irrotational (i.e. zero vorticity $d\xi \wedge \xi=0$) & geodesic (so the two conditions actually amount to $d\xi=0$) while 'proper time synchronizable' requires a global function $t$ such that $\xi=-dt$. Since the Frobenius theorem, the difference between the above two is really the existence of a global function $t$ from which get the orthogonal spacelike hypersurface foliation.

Sachs & Wu in section 5.3 claim that if the congruence is proper time synchronizable then the observers it represents can synchronize their wristwatches using a radar procedure (basically Einstein synchronization procedure) -- see also the PF thread referenced in the OP.

 

[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.