Lie derivative of hypersurface basis vectors along geodesic congruence

[Pencilvester]

TL;DR Summary

I’m having trouble with a statement in Poisson’s “A Relativist’s Toolkit” in regards to the Lie derivative along a geodesic congruence.

Hello PF, here’s the setup: we have a geodesic congruence (not necessarily hypersurface orthogonal), and two sets of coordinates. One set, $x^\alpha$, is just any arbitrary set of coordinates. The other set, $(\tau,y^a)$, is defined such that $\tau$ labels each hypersurface (and presumably also represents the proper time of a specific geodesic, $\gamma$, to which all the hypersurfaces are orthogonal) and $y^a$ is assigned to label the geodesics themselves, with the basis vectors $\partial_{y^a}$ lying tangent to the hypersurfaces. Poisson then defines $$u^\alpha = \left({\frac{\partial x^\alpha}{\partial \tau}}\right)_{y^a}$$and$$e^\alpha_a = \left({\frac{\partial x^\alpha}{\partial y^a}}\right)_{\tau}$$and then says that $\mathcal{L}_u e^\alpha_a =0$.

This is where I was having trouble, but I think I just realized what I was missing as I was typing, and I just want to make sure. I was thrown by the fact that he equates the basis vector $\partial_\tau$ with a vector $\mathbf u$—a letter that is typically reserved for 4-velocity. But in general, $\partial_\tau$ at a point P will not equal the 4-velocity of the geodesic passing through P, yes? Does anyone know why he chose the letter u?

 

[Pencilvester]

But now I’m having trouble with the next thing he says, which is “we also have $u_\alpha e^\alpha_a =0$ holding on $\gamma$ (and only $\gamma$)”. How is that not true in general?

 

[strangerep]

I’m having trouble with a statement in Poisson’s “A Relativist’s Toolkit” in regards to the Lie derivative along a geodesic congruence.

As I've told others previously, you should give the relevant page and equation number(s) when asking such questions.
(Here, you're just lucky that I happen to be familiar with Poisson's book. For those playing along at home it's p44.)

I was thrown by the fact that [Poisson] equates the basis vector $\partial_\tau$ with a vector $\mathbf u$

No, that's not what he does. He performs $\partial x^\alpha/\partial \tau$ while holding $y^\alpha$ constant, i.e., staying on one particular curve. His "$u$" is indeed the 4-velocity.

But now I’m having trouble with the next thing he says, which is “we also have $u_\alpha e^\alpha_a =0$ holding on $\gamma$ (and only $\gamma$)”. How is that not true in general?

Because $u^\alpha$ (for a fixed $y^\alpha$) is defined along 1 particular curve $\gamma$. If you move off that curve, you've changed the $y^\alpha$ coords.

 

[Pencilvester]

As I've told others previously, you should give the relevant page and equation number(s) when asking such questions.

I’m sorry, I have not seen many of your posts, so I was unaware of this rule. I will be sure to do so in the future.

No, that's not what he does. He performs ∂xα/∂τ∂xα/∂τ\partial x^\alpha/\partial \tau while holding yαyαy^\alpha constant, i.e., staying on one particular curve. His "uuu" is indeed the 4-velocity.

From page 43:

Select a particular geodesic $\gamma$ from the congruence, and on this geodesic, pick a point P at which $\tau = \tau_P$. Construct, in a small neighborhood around P, a small set $\delta \Sigma (\tau_P)$ of points P’ such that ... at each point P’, $\tau$ is also equal to $\tau_P$... $\gamma$ intersects $\delta \Sigma (\tau_P)$ orthogonally. (There is no requirement that other geodesics do, as the congruence may not be hypersurface orthogonal.) We shall call $\delta \Sigma (\tau_P)$ the congruence’s cross section around the geodesic $\gamma$, at proper time $\tau = \tau_P$.

To me, it sounds like the only hypersurface at which the proper times of all the geodesics is guaranteed to be equal is $\delta \Sigma (\tau_P)$. Doesn’t the fact that the congruence may not be hypersurface orthogonal imply that $\tau$, while still a good parameterization of $\gamma$, will not be equal to the proper time on other geodesics at other hypersurfaces (i.e. the proper time on geodesic $\lambda$ at $\delta \Sigma (\tau_Q)$ doesn’t necessarily equal $\tau_Q$)?

 

[PeterDonis]

I was unaware of this rule.

It's not a rule that @strangerep made up personally. It's part of the general PF rules about giving references. It doesn't make much sense to give a reference to an entire book when you are asking about a particular equation on a particular page. You should always make your references as specific as possible.

 

[Pencilvester]

It's not a rule that @strangerep made up personally. It's part of the general PF rules about giving references. It doesn't make much sense to give a reference to an entire book when you are asking about a particular equation on a particular page. You should always make your references as specific as possible.

Got it. I guess I thought I wasn’t leaving anything important out of the context that I gave, but it’s silly of me to think that I’d be a good judge of that seeing as I’m the one who’s having the trouble understanding what was said in the book.

 

[strangerep]

To me, it sounds like the only hypersurface at which the proper times of all the geodesics is guaranteed to be equal is $\delta \Sigma (\tau_P)$. Doesn’t the fact that the congruence may not be hypersurface orthogonal imply that $\tau$, while still a good parameterization of $\gamma$, will not be equal to the proper time on other geodesics at other hypersurfaces (i.e. the proper time on geodesic $\lambda$ at $\delta \Sigma (\tau_Q)$ doesn’t necessarily equal $\tau_Q$)?

Yes, but,... er,... so what?

I perceived your question to be about differentiation along a given geodesic. In that context, what's happening off the geodesic is irrelevant.

Or perhaps I don't understand what really bugging you.

 

[Pencilvester]

Yes, but,... er,... so what?

Well if $\tau$ is being used as a parameter for hypersurfaces, and it happens to only be the proper time along a single geodesic $\gamma$, then how can $\frac{\partial x^\alpha}{\partial \tau}$ represent the 4-velocity of an arbitrary geodesic? i.e. if, given the way Poisson defines it, $\tau$ changes in the direction of the changing hypersurfaces, and in general, most geodesics will not be orthogonal to those hypersurfaces, how can the 4-velocities which are tangent to those geodesics be simultaneously orthogonal to the hypersurfaces?

 

[strangerep]

Well if $\tau$ is being used as a parameter for hypersurfaces, and it happens to only be the proper time along a single geodesic $\gamma$, then how can $\frac{\partial x^\alpha}{\partial \tau}$ represent the 4-velocity of an arbitrary geodesic? i.e. if, given the way Poisson defines it, $\tau$ changes in the direction of the changing hypersurfaces, and in general, most geodesics will not be orthogonal to those hypersurfaces, how can the 4-velocities which are tangent to those geodesics be simultaneously orthogonal to the hypersurfaces?

OK, let's be a bit more careful with the notation. Let $x^\mu$ be ordinary mutually-independent coordinates on some open set of the spacetime manifold. By "mutually-independent" I mean that $[\partial_{x^\alpha} , \partial_{x^\beta}] f(x) = 0$ for any function $f$ on the manifold.

Now consider a congruence of 1D curves "$\gamma_a$", where $a$ is a 3-tuple labeling each curve. The coordinates of any particular point on any particular curve are denoted as $x^\alpha = \gamma_a^{\;\alpha}(x^\mu)$.

Now change the coordinate system so that one of the new coordinates always increments along the curves of the congruence. I'll call that coordinate $\tau$. If you're still having trouble with this, imagine taking many pieces of string, mark off 1mm intervals along each string , and then drape them all around the room such that every string is near to another, but they never touch each other. The 1mm marks along each string correspond to the $\tau$ coordinate, albeit discretized in this analogy. Just because you've marked out intervals of $\delta\tau = 1$mm along one particular string doesn't preclude the use of the coordinate "$\tau$" along other strings.

Now choose any particular mark on any particular string ("S1", say), and arbitrarily name it as $\tau=0$, and rename other marks along that string accordingly, i.e., $$ \dots, ~ \tau=-2, ~ \tau=-1, ~ \tau=0, ~ \tau=1, ~ \tau=2, ~ \dots $$Then find all the nearest marks on other strings S2, S3, closest to $\tau=0$ on S1. Re-label those marks on each of S2 and S3 as $\tau=0$, and re-label every other mark along S2 and S3 in terms of monotonically-increasing $\tau$ values.

In this way, you've established a consistent coordinate labeling, i.e., $\tau$, along every string. The fact that you started at one particular mark on one particular string doesn't prevent you from extending that labeling consistently to nearby strings.

[Apologies if this explanation sounds like kindergarten level. I suspect this is something that's confusing if you don't understand it, yet becomes trivially obvious when you do.]

 

[PeterDonis]

Now consider a congruence of 1D curves "$\gamma_a$", where $a$ is a 4-tuple labeling each curve.

Shouldn't this be a 3-tuple?

 

[strangerep]

Shouldn't this be a 3-tuple?

Oops,... yes. Fixed now.

(This is what comes from changing one's mind midstream about how to try and explain something.)

 

[Pencilvester]

Then find all the nearest marks on other strings S2, S3, closest to $\tau=0$ on S1. Re-label those marks on each of S2 and S3 as $\tau=0$, and re-label every other mark along S2 and S3 in terms of monotonically-increasing $\tau$ values.

Okay, I think I finally unearthed my misguided assumption. For some crazy reason, I had it in my head that a construction such as the one you just described could only be achieved by starting with a hypersurface orthogonal congruence, and conversely, trying to construct a foliation in this way using a rotating geodesic congruence would be impossible. But this is false, yes?

 

[strangerep]

[...] But this is false, yes?

Since, once again, I'm not sure what's in your head, I'll decline to say yea or nay.

Regardless, I hope you can now move on...

 

[Pencilvester]

a construction such as the one [strangerep] just described [in post #9] could only be achieved by starting with a hypersurface orthogonal congruence, and conversely, trying to construct a foliation in this way using a rotating geodesic congruence would be impossible.

I just want to be absolutely sure—is there anyone who can confirm for me that this idea is crazy-town-banana-pants?

 

[PeterDonis]

a construction such as the one you just described could only be achieved by starting with a hypersurface orthogonal congruence, and conversely, trying to construct a foliation in this way using a rotating geodesic congruence would be impossible

A hypersurface orthogonal congruence is not required for the construction to work; however, I'm not sure it's true that the construction is guaranteed to work for any geodesic congruence. (Note that, in most cases, a rotating congruence will not be geodesic; the only rotating geodesic congruence that I'm aware of is in Godel spacetime.)

 

[Pencilvester]

A hypersurface orthogonal congruence is not required for the construction to work

Phew, glad I’m all straightened out, thanks!

Note that, in most cases, a rotating congruence will not be geodesic; the only rotating geodesic congruence that I'm aware of is in Godel spacetime.

Good point. Noted.