Integral curves of (timelike) smooth vector field

[cianfa72]

TL;DR Summary

About the integral curves of a non-zero smooth vector field; can they cross?

Hi,
suppose you have a non-zero smooth vector field $X$ defined on a manifold (i.e. it does not vanish at any point on it).

Can its integral curves cross at any point ? Thanks.

Edit: I was thinking about the sphere where any smooth vector field must have at least one pole (i.e. at least a point where it vanishes). In this case the integral curves actually cross each other.

 

[Orodruin]

No.

No integral curves of any smooth vector field cross each other.

In order to cross, the curves would need to have distinct tangents at the crossing point, but that would mean at least one of the curves would not be an integral curve.

 

[cianfa72]

No integral curves of any smooth vector field cross each other.

In the case of the sphere, if we take the meridian curves they are not the integral curves of a smooth vector field that vanishes at North and Sud poles ?

In order to cross, the curves would need to have distinct tangents at the crossing point, but that would mean at least one of the curves would not be an integral curve.

ok, therefore any smooth vector field defines integral curves that foliate the entire manifold (i.e. each smooth vector field defines a congruence of integral curves) ?

 

[Orodruin]

In the case of the sphere, if we take the meridian curves they are not the integral curves of a smooth vector field that vanishes at North and Sud poles ?

I thought a bit about that. How are you planning to get the meridian lines out of the poles? Either all of them end/begin at the pole or the field is not smooth (because to get away from the pole you would need tangents going both into and out of the pole).

If you are fine with having a full circle where the field is zero you can probably technically get around that, but the flow lines will not be uniquely determined by the first order differential equations.

 

[PeterDonis]

In the case of the sphere, if we take the meridian curves they are not the integral curves of a smooth vector field that vanishes at North and Sud poles ?

That's correct, they are not. The vector field you describe does not "vanish" at the North and South poles; it is undefined at the North and South poles. Which means it is not defined on the entire manifold $S^2$. It is only defined on the manifold $S^2$ minus the poles. On the latter manifold, the integral curves in question never cross.

 

[Orodruin]

That's correct, they are not. The vector field you describe does not "vanish" at the North and South poles; it is undefined at the North and South poles. Which means it is not defined on the entire manifold $S^2$. It is only defined on the manifold $S^2$ minus the poles. On the latter manifold, the integral curves in question never cross.

You can quite easily define a vector field that is zero at the poles and where the meridians arrive in finite curve parameter (although they will not have constant length tangents). Compare to the vector field $-\vec x$ with $\vec x$ being the position vector in $\mathbb R^2$. (An example on the sphere would be $\sin(\theta)\partial_\theta$.)

 

[PeterDonis]

An example on the sphere would be $\sin(\theta)\partial_\theta$.

Isn't $\partial_\theta$ undefined at the poles?

 

[Orodruin]

Isn't $\partial_\theta$ undefined at the poles?

Yes, but that is just a coordinate issue. Define the field to be zero at the poles and you have a smooth field.

 

[PeterDonis]

Define the field to be zero at the poles and you have a smooth field.

Probably this is just an issue of ordinary language being imprecise. As I understand it, you are saying that the vector field can be defined to be smooth (in the sense that its norm is a smooth function on the manifold and the nonzero vectors have directions that vary smoothly), but it still will not have well-defined integral curves at the poles. Is that correct?

 

[Orodruin]

Probably this is just an issue of ordinary language being imprecise. As I understand it, you are saying that the vector field can be defined to be smooth (in the sense that its norm is a smooth function on the manifold and the nonzero vectors have directions that vary smoothly), but it still will not have well-defined integral curves at the poles. Is that correct?

Right. Since the field is zero, you would need the second derivative to know where it goes and the flow equation does not provide that. The particular field I gave however has no flow curves going into the south pole as it points towards it everywhere. The only curves satisfying the flow equation end in that pole.

 

[cianfa72]

The particular field I gave however has no flow curves going into the south pole as it points towards it everywhere. The only curves satisfying the flow equation end in that pole.

The field $\sin(\theta)\partial_\theta$ refers to the physics convention I believe.

Do you mean it has no flow curves going through the south pole ?

 

[Orodruin]

The field $\sin(\theta)\partial_\theta$ refers to the physics convention I believe.

We are in the physics section after all …

Do you mean it has no flow curves going through the south pole ?

Indeed. They arrive in finite time, but cannot get away from there because the field points to the pole everywhere around.

 

[cianfa72]

So, if we restrict to just non-zero smooth vector fields defined on a manifold (if they actually do exist), can we get around and get the net result that the flow curves do not cross each other and foliate the entire manifold ?

 

[vanhees71]

I thought a bit about that. How are you planning to get the meridian lines out of the poles? Either all of them end/begin at the pole or the field is not smooth (because to get away from the pole you would need tangents going both into and out of the pole).

If you are fine with having a full circle where the field is zero you can probably technically get around that, but the flow lines will not be uniquely determined by the first order differential equations.

See

https://en.wikipedia.org/wiki/Hairy_ball_theorem

 

[PeterDonis]

See

https://en.wikipedia.org/wiki/Hairy_ball_theorem

The hairy ball theorem says you cannot have an everywhere nonvanishing vector field on a 2-sphere. @Orodruin is talking about a vector field that does vanish at the poles, so it is not everywhere nonvanishing and the hairy ball theorem does not prohibit it.

 

[Orodruin]

The hairy ball theorem says you cannot have an everywhere nonvanishing vector field on a 2-sphere. @Orodruin is talking about a vector field that does vanish at the poles, so it is not everywhere nonvanishing and the hairy ball theorem does not prohibit it.

Indeed, as I stated, you can construct a field where all meridian lines (except those lying on the great circle where the field is zero) satisfy the flow equations if they have a variable flow speed. An example would be $\sin^2(\theta) \cos(\phi)\partial_\theta$. This in no way contradicts the hairy ball theorem.

The flow line for $\phi = 0$ would be given by $\dot\theta = \sin^2(\theta) = 1-\cos(2\theta)/2$. It goes between the poles in finite time and the corresponding flow line for $\phi = \pi$ goes in the other direction between the poles in the same time. Concatenating these lines gives a curve that follows the meridian great circle and goes around in finite time. The meridian great circle that does not do this is given by $\phi = \pm \pi/2$ as the field is zero on that curve.

 

[PeterDonis]

Concatenating these lines gives a curve that follows the meridian great circle and goes around in finite time.

Wouldn't the flow stop at the poles, though? How can it go around the whole circle in finite time if it stops at the poles?

 

[Orodruin]

Wouldn't the flow stop at the poles, though? How can it go around the whole circle in finite time if it stops at the poles?

It satisfies the flow equations with zero derivative at the poles yet still passes the poles for the same reason $y = x^3$ passes $y = 0$ with zero derivative.
Edit: Even the second derivative is zero, but the third one isn’t and allows passage in finite parameter change.

Things with zero derivative are nasty.

 

[PeterDonis]

for the same reason $y = x^3$ passes $y = 0$ with zero derivative.

In this case, $y$ passes $y = 0$ because $x$ is assumed to continue to increase. We can assume that because $y$ is assumed to be a function of $x$, i.e., the behavior of $y$ cannot affect the behavior of $x$; it's the other way around.

But in the case of the vector field $\sin^2 \theta \ \partial_\theta$ along the $\phi = 0$ curve, the length of the vector--the analogue of $y$--does affect the behavior of the curve parameter, i.e., which point on the curve you are at--the analogue of $x$. If the vector has zero length, that means you are no longer moving to a different value of the curve parameter, i.e., to a different point along the curve. So you can't just assume that you are moving to a different point, since that contradicts the vector having zero length.

Edit: Even the second derivative is zero, but the third one isn’t and allows passage in finite parameter change.

How would higher derivatives affect this in the case of the vector field? In the case of $y = x^3$, the second derivative still does not affect the behavior of $x$; the behavior of $x$ is just assumed and is not a consequence of anything about the function.

 

[Orodruin]

In this case, $y$ passes $y = 0$ because $x$ is assumed to continue to increase. We can assume that because $y$ is assumed to be a function of $x$, i.e., the behavior of $y$ cannot affect the behavior of $x$; it's the other way around.

But in the case of the vector field $\sin^2 \theta \ \partial_\theta$ along the $\phi = 0$ curve, the length of the vector--the analogue of $y$--does affect the behavior of the curve parameter, i.e., which point on the curve you are at--the analogue of $x$. If the vector has zero length, that means you are no longer moving to a different value of the curve parameter, i.e., to a different point along the curve. So you can't just assume that you are moving to a different point, since that contradicts the vector having zero length.How would higher derivatives affect this in the case of the vector field? In the case of $y = x^3$, the second derivative still does not affect the behavior of $x$; the behavior of $x$ is just assumed and is not a consequence of anything about the function.

The one dimensional case is perfectly analogous to the one of the vector field here when projected onto the curve.

In the one dimensional case you have $y’ = kx^2$ and it exactly the same type of equation you will find along the curve for the curve parameter if you write it down in coordinates that are not singular at the pole.

$x$ is not the analogue of the point you are at, it is the curve parameter. The $y$ direction is what corresponds to the point you are at, with a tangent of zero at $y=0$. $|y’|$ is the length of the tangent.

Edit: I may also be confusing myself with $y$ vs $y’$, but kids have been … resisting … sleep.

 

[PeterDonis]

In the one dimensional case you have $y’ = kx^2$ and it exactly the same type of equation you will find along the curve for the curve parameter if you write it down in coordinates that are not singular at the pole.

Hm. I see what you're saying, but I'll have to think about it some more to see if my intuition just needs to be retrained or if I have a further substantive question.

 

[cianfa72]

It satisfies the flow equations with zero derivative at the poles yet still passes the poles for the same reason $y = x^3$ passes $y = 0$ with zero derivative.

Sorry, if $x$ is the curve parameter in one dimensional case, which is the flow differential equation such that $y = x^3$ is solution of ?

 

[Orodruin]

Hm. I see what you're saying, but I'll have to think about it some more to see if my intuition just needs to be retrained or if I have a further substantive question.

No, you’re right. I think I managed to fool myself with similar confusion. The y’ = -y field clearly has the solution $dy/y = - dx$ so $y = C e^{-x}$, which does not arrive at the poles in finite time.

 

[cianfa72]

So, if we restrict to just non-zero smooth vector fields defined on a manifold (if they actually do exist), can we get around and get the net result that the flow curves do not cross each other and foliate the entire manifold ?

Quoting Wikipedia a non-zero smooth vector field defines a foliation of the entire manifold (when it exists of course)

If the vector field $X$ is nowhere zero then it defines a one-dimensional subbundle of the tangent bundle of $M$ and the integral curves form a regular foliation of $M$.

 

[PeterDonis]

Quoting Wikipedia a non-zero smooth vector field defines a foliation of the entire manifold (when it exists of course)

But it is a foliation by curves, not a foliation by surfaces.

 

[cianfa72]

But it is a foliation by curves, not a foliation by surfaces.

Yes, but it is still a regular foliation by one-dimensional leaves.

 

[jbergman]

In the case of the sphere, if we take the meridian curves they are not the integral curves of a smooth vector field that vanishes at North and Sud poles ?ok, therefore any smooth vector field defines integral curves that foliate the entire manifold (i.e. each smooth vector field defines a congruence of integral curves) ?

The sphere doesn't admit a vector field that is non-zero every where. Integral curves that approach a zero slow down to zero speed so they still never cross.