I Clock synchronization for ring-riding observers on rotating disk

[cianfa72]

TL;DR Summary

About clock synchronization for Langevin ring-riding observer

Hello,

reading the wiki entry for Langevin observers on rotating disk - Born_coordinates I'm struggling with the following quoted sentence: But as we see from Fig. 1, ideal clocks carried by these ring-riding observers cannot be synchronized.

I do not grasp why, starting from the figure, those ring-riding clocks cannot be synchronized.

Thanks.

 

[ergospherical]

You can synchronise the co-rotating clocks locally but not globally. To see this, let $t$ be the time coordinate of an inertial observer $O$ at rest in the centre (whose worldline is the dotted green line in the figure). Let $\{ O_{\lambda} \}$ be a closed, one-parameter ($\lambda \in [0,1]$) family of co-rotating observers such that $O_0 = O_1$.

Let $p_{\lambda}$ be an event on the worldline of $O_{\lambda}$, and let $p_{\lambda + d\lambda}$ be an event on the worldline of $O_{\lambda + d\lambda}$ which is simultaneous with $p_{\lambda}$ as measured by $O_{\lambda}$. Then $\mathbf{u}_{\lambda} \cdot (p_{\lambda + d\lambda} - p_{\lambda}) = 0$, where $\mathbf{u}_{\lambda}$ is the 4-velocity of $O_{\lambda}$. Can you use this to write down the time interval $dt$ between $p_{\lambda}$ and $p_{\lambda + d\lambda}$ as measured by the inertial observer $O$? Then, what is the line integral of $dt$ around the simultaneity curve containing all the $\{ p_{\lambda} \}$, and can it be zero?

 

[cianfa72]

Let $p_{\lambda}$ be an event on the worldline of $O_{\lambda}$, and let $p_{\lambda + d\lambda}$ be an event on the worldline of $O_{\lambda + d\lambda}$ which is simultaneous with $p_{\lambda}$ as measured by $O_{\lambda}$. Then $\mathbf{u}_{\lambda} \cdot (p_{\lambda + d\lambda} - p_{\lambda}) = 0$, where $\mathbf{u}_{\lambda}$ is the 4-velocity of $O_{\lambda}$.

What do you mean with $p_{\lambda}$ ? Is it just the "position vector" of an event on $O_{\lambda}$ worldline ?

 

[Ibix]

Summary:: About clock synchronization for Langevin ring-riding observer

I do not grasp why, starting from the figure, those ring-riding clocks cannot be synchronized.

Make a cut parallel to the axis of the cylinder and flatten the cylinder on to the table. The blue line should now be recognisable as the $x'$ axis of an observer whose $t'$ axis is the red line. The problem is that if you roll your cylinder back up again, the blue line crosses the red one several times. So there are several timelike separated events on the red line that are "at the same time" by this method, which doesn't make sense.

 

[cianfa72]

Yeah, just some arbitrary event on that worldline.

ok thus $\mathbf{u}_{\lambda} \cdot (p_{\lambda + d\lambda} - p_{\lambda}) = 0$ basically means the spacelike vector $(p_{\lambda + d\lambda} - p_{\lambda})$ is orthogonal to the 4-velocity of $O_{\lambda}$.

From a physical point of view, how do locally synchronize neighboring clocks ?

 

[DrGreg]

From a physical point of view, how do locally synchronize neighboring clocks ?

Standard Einstein synchronisation. Reflect a light signal from A to B and back to A again. Then set B's clock so that the time at the reflection event at B is halfway between the emission and reception events at A.

This works, to a good enough approximation, if A and B are close to each other. (Mathematically, you should consider the limit as the distance between A and B tends to zero.)

 

[cianfa72]

Standard Einstein synchronisation. Reflect a light signal from A to B and back to A again. Then set B's clock so that the time at the reflection event at B is halfway between the emission and reception events at A.

This works, to a good enough approximation, if A and B are close to each other. (Mathematically, you should consider the limit as the distance between A and B tends to zero.)

ok that's fine. AFAIU this synchronization procedure actually defines the integral curve of the $p_3$ spacelike vector of the frame field for Langevin ring-riding observers (the blue curve in the wiki link) as having "the same time".

In fact $p_3$ spacelike vector field is orthogonal to each Langevin observer worldline (timelike) tangent vector and thus it represents locally "the space at the same time". However, for the reasons pointed out by @Ibix, we would be in trouble getting different values of time for timelike separated event belonging on each Langevin observer worldline.

In other words the ring-rider Langevin timelike congruence is not static: does not exist a family of spacelike hypersurfaces orthogonal to each event on that congruence foliating the flat (Minkowski) spacetime.

 

[cianfa72]

In other words the ring-rider Langevin timelike congruence is not static: does not exist a family of spacelike hypersurfaces orthogonal to each event on that congruence foliating the flat (Minkowski) spacetime.

Thinking again about it, I found the following argument on Landau book "The Classic Theory of Field" section 99. It seems it should always be possible to choose a reference system (chart) that allows clocks to be synchronized at different points in space.

In this specific case (ring-riding Langevin observer on rotating ring) does the condition on Landau book basically amount to select the Minkowski chart since the flat spacetime ?

 

[Ibix]

I don't have Landau so I don't know if this is quite what you are looking for. But yes, you can always find a coordinate chart that matches your local inertial frame at your location. Minkowski, Rindler, whatever you like. What you can't generally do is insist that the coordinate chart maps nicely on to the local inertial frames of some arbitrary collection of observers. It can be done for inertial or Rindler observers, but not Langevin observers.

 

[cianfa72]

But yes, you can always find a coordinate chart that matches your local inertial frame at your location. Minkowski, Rindler, whatever you like. What you can't generally do is insist that the coordinate chart maps nicely on to the local inertial frames of some arbitrary collection of observers. It can be done for inertial or Rindler observers, but not Langevin observers.

here with local inertial frame at your location do you mean the frame basis having for the timelike vector the tangent vector to "my" worldline (at each event) and as spacelike vectors in the basis a set of independent spacelike vectors belonging to the orthogonal complement of the timelike vector ?

 

[Ibix]

Yes.

 

[cianfa72]

ok, so the point of having a family of spacelike hypersurfaces everywhere orthogonal to a timelike congruence (the set of timelike worldlines of a given collection of observers) is related to the existence of a family of hypersurfaces that "match up" with the local notion of "space at the same time" as defined from each observer in the given congruence at each location, I guess...

Furthermore from this thread if the timelike congruence is not static then even if it is hypersurface orthogonal -- namely it does exist a family of spacelike hypersurfaces orthogonal to the congruence foliating the spacetime -- the geometry on each spacelike hypersurface will not be the same.

Does it make sense ?

 

[PeterDonis]

In this specific case (ring-riding Langevin observer on rotating ring) does the condition on Landau book basically amount to select the Minkowski chart since the flat spacetime ?

Yes. In other words, you "synchronize" the ring-riding observer clocks using the clock at the center of the ring, which is at rest in the underlying Minkowski chart.

 

[PeterDonis]

if the timelike congruence is not static then even if it is hypersurface orthogonal -- namely it does exist a family of spacelike hypersurfaces orthogonal to the congruence foliating the spacetime -- the geometry on each spacelike hypersurface will not be the same.

Yes. Or, to put it another way, if the spacetime is stationary--which is what is required to have a set of spacelike hypersurfaces on which the geometry is always the same--then in order for there to be a timelike congruence which is hypersurface orthogonal, the spacetime must be not just stationary, but static--i.e., the congruence can't be "rotating" like the Langevin congruence is. (The more technical way of saying this is that the congruence must have zero vorticity. In fact, the zero vorticity condition is completely general: any timelike congruence that is hypersurface orthogonal must have zero vorticity, by the Frobenius theorem, even if the spacetime is not stationary so the geometry on each hypersurface is different.)

 

[cianfa72]

Yes. Or, to put it another way, if the spacetime is stationary--which is what is required to have a set of spacelike hypersurfaces on which the geometry is always the same--then in order for there to be a timelike congruence which is hypersurface orthogonal, the spacetime must be not just stationary, but static--i.e., the congruence can't be "rotating" like the Langevin congruence is.

To be more specific I would slightly modify the sentence (bold is mine):
if the spacetime is stationary--which is what is required to have a set of spacelike hypersurfaces on which the geometry is always the same--then in order for there to be a KVF timelike congruence which is hypersurface orthogonal, the spacetime must be not just stationary, but static.

In fact, as said there in post#64, in FRW spacetime for example there is no timelike KVF, but there is still the family of worldlines of "comoving" observers that, even if the FRW spacetime is not even stationary, is hypersurface orthogonal however (the family of spacelike 3-surfaces of constant FRW coordinate time is orthogonal to the congruence of "comoving" worldlines).

 

[PeterDonis]

To be more specific I would slightly modify the sentence (bold is mine):

Yes, the modification is fine. I was implicitly thinking of the KVF timelike congruence (i.e., the congruence formed by the integral curves of the timelike KVF) anyway.

 

[cianfa72]

The more technical way of saying this is that the congruence must have zero vorticity. In fact, the zero vorticity condition is completely general: any timelike congruence that is hypersurface orthogonal must have zero vorticity, by the Frobenius theorem

Some time ago I read the book "The Road to Reality" - R. Penrose. I believe what you pointed out is basically what he said in section 12.3

A timelike congruence (implicitly) defines a covector on each point along the congruence (i.e. a covector field defined on the entire spacetime). Geometrically that covector field defines locally on the tangent space "attached" on each point an element of spacelike hyperplane orthogonal to the congruence worldline passing there (i.e. orthogonal to the congruence worldline's timelike tangent vector in that point).

Hence the question is: does it exist a scalar field $\phi$ such that its gradient (or external derivative) covector field $d\phi$ is the same as the covector (implicitly) defined there by the timelike congruence ? The answer relies on the Frobenius condition/theorem about the timelike congruence.

 

[PeterDonis]

I believe what you pointed out is basically what he said in section 12.3

I don't think so. See below.

A timelike congruence (implicitly) defines a covector on each point along the congruence (i.e. a covector field defined on the entire spacetime).

No, it doesn't, it defines a vector at each point (the tangent vector of the worldline in the congruence that passes through that point).

Geometrically that covector field defines locally on the tangent space "attached" on each point an element of spacelike hyperplane orthogonal to the congruence worldline passing there (i.e. orthogonal to the congruence worldline's timelike tangent vector in that point).

If you know the metric, then the tangent vector at the given point defines an orthogonal covector at that point (since the metric is what defines orthogonality). If the congruence is timelike, then the orthogonal covector will define a spacelike hyperplane in the tangent space at the given point. But the tangent vector is what comes first, not the covector.

Hence the question is: does it exist a scalar field $\phi$ such that its gradient (or external derivative) covector field $d\phi$ is the same as the covector (implicitly) defined there by the timelike congruence ? The answer relies on the Frobenius condition/theorem about the timelike congruence.

I don't see what a scalar field has to do with anything we've discussed in this thread. Certainly no such field is required to analyze the rotating disk problem.

 

[cianfa72]

If you know the metric, then the tangent vector at the given point defines an orthogonal covector at that point (since the metric is what defines orthogonality). If the congruence is timelike, then the orthogonal covector will define a spacelike hyperplane in the tangent space at the given point. But the tangent vector is what comes first, not the covector.

ok, but actually we know the metric since our model of spacetime is a Lorentzian manifold. Therefore the vector field from the timelike congruence actually give us a spacelike hyperplane (the orthogonal complement) in the tangent space at each point.

From my understanding (sorry I'm not an expert) basically we're looking for the conditions for the existence of a family of 3-d spacelike submanifolds $N$ of the Lorentzian manifold $M$ such that its tangent space at each point $p$ is the same as the 3-d vector subspace defined by the above spacelike hyperlanes (covector field). That's actually the hypersuface orthogonal condition for the timelike congruence.

Frobenius theorem establishes the (iff) conditions for the covector field under which such submanifolds $N$ do exist. Then from the theorem you can find locally on each neighborhood a scalar function $t$ (i.e. a coordinate function) such that the submanifold $N$ is defined by $t=0$. Globally I believe that does mean there is a global function (a scalar field) defined on $M$ such that the family of 3-d spacelike submanifolds $N$ are actually level set of ($t=const$).

 

[PeterDonis]

ctually we know the metric since our model of spacetime is a Lorentzian manifold.

That tells us the metric of the tangent space at any point, yes: it's just the Minkowski metric of flat Minkowski spacetime.

Therefore the vector field from the timelike congruence actually give us a spacelike hyperplane (the orthogonal complement) in the tangent space at each point.

In the tangent space, yes.

the 3-d vector subspace defined by the above spacelike hyperlanes (covector field).

No. "Spacelike hyperplane" and "covector field" are not the same. You appear to be under the mistaken impression that a "covector field" can have more than one dimension. It can't. A single covector field cannot describe a 3-dimensional manifold, such as a spacelike hypersurface orthogonal to a given vector field.

Frobenius theorem establishes the (iff) conditions for the covector field under which such submanifolds $N$ do exist.

No. The Frobenius theorem gives a condition on the vector field. It says nothing whatever about covector fields. (The theorem can also be formulated using differential forms, but that's also not the same thing as covector fields.)

Then from the theorem you can find locally on each neighborhood a scalar function $t$ (i.e. a coordinate function) such that the submanifold $N$ is defined by $t=0$. Globally I believe that does mean there is a global function (a scalar field) defined on $M$ such that the family of 3-d spacelike submanifolds $N$ are actually level set of ($t=const$).

You can find such a scalar "time coordinate", and a corresponding foliation, for congruences which are not hypersurface orthogonal. For example, the standard Minkowski time coordinate $t$ for the inertial frame in which the center of the rotating disk is at rest is such a scalar, and the surfaces of constant $t$ are such a foliation, for the congruence of Langevin observers. Of course this foliation is not orthogonal to the congruence, but it still meets the conditions you are giving here. In other words, the conditions for such a global time coordinate and accompanying foliation to exist are much more general than "hypersurface orthogonal" is.

 

[ergospherical]

No. The Frobenius theorem gives a condition on the vector field. It says nothing whatever about covector fields. (The theorem can also be formulated using differential forms, but that's also not the same thing as covector fields.)

A differential one-form and a co-vector field are the same thing, and what Frobenius says is that for a co-vector field $\omega$ satisfying $\omega \wedge d\omega = 0$ at every point then $\omega = f dg$ for some functions $f,g$. In other words the co-vector field $\omega$ is orthogonal to level hypersurfaces of $g$.

 

[PeterDonis]

A differential one-form and a co-vector field are the same thing

But a differential 3-form, which is what you need to describe a 3-dimensional spacelike hypersurface, is not the same as a covector field.

 

[PeterDonis]

What do you mean by “describe”?

In general, a single covector field does not pick out a set of spacelike hypersurfaces. A covector field $d g$ that is the exterior derivative of a scalar does (the level surfaces of $g$), but not all covector fields are the exterior derivatives of scalars.

A 3-form, however, does pick out a set of spacelike hypersurfaces (because at any point it is the volume form of some spacelike hypersurface), whether it is the exterior derivative of something or not.

 

[cianfa72]

But a differential 3-form, which is what you need to describe a 3-dimensional spacelike hypersurface, is not the same as a covector field.

I believe the point @ergospherical is making is actually the same as explained by Penrose in section 12.3 Fig 12.7: a co-vector field defines locally a (n-1)-dimensional hyperplane element. Therefore in the case of 4d spacetime a co-vector field actually defines a 3-d hyperplane (basically a 3-d vector subspace of the 4-d tangent space at each point).

 

[martinbn]

But a differential 3-form, which is what you need to describe a 3-dimensional spacelike hypersurface, is not the same as a covector field.

The set of vectors on which a one form vanishes is a 3d distribution. Frobenius, in the differential form version, gives the condition on the one form to garantee that the distribution has integral hypersurfases.

 

[PeterDonis]

I believe the point @ergospherical is making is actually the same as explained by Penrose in section 12.3 Fig 12.7: a co-vector field defines locally a (n-1)-dimensional hyperplane element. Therefore in the case of 4d spacetime a co-vector field actually defines a 3-d hyperplane (basically a 3-d vector subspace of the 4-d tangent space at each point).

Possibly. Note, however, that the Frobenius theorem as @ergospherical stated it (which is the "differential form" formulation, and is different from the "vector field" formulation in which I stated it) says that $\omega = f d g$; that means it is not (quite) the gradient of a scalar (as the end of your post #18 implied). It is one scalar ($f$) times the gradient of another ($g$).

 

[PeterDonis]

The set of vectors on which a one form vanishes is a 3d distribution.

Hm, yes.

 

[PeterDonis]

the Frobenius theorem as @ergospherical stated it (which is the "differential form" formulation, and is different from the "vector field" formulation in which I stated it)

A good exercise, btw, is to satisfy yourself that the two formulations are equivalent.

 

[cianfa72]

Note, however, that the Frobenius theorem as @ergospherical stated says that $\omega = f d g$; that means it is not (quite) the gradient of a scalar (as the end of your post #18 implied). It is one scalar ($f$) times the gradient of another ($g$).

Maybe I'm wrong: why we can't just define a new scalar function let's say $h$ such that $\omega = f d g = d h$ ?

 

[martinbn]

Maybe I'm wrong: why we can't just define a new scalar function let's say $h$ such that $\omega = f d g = d h$ ?

$dh$ is closed(even exact), so you have $ddh=0$. On the other hand $xdy$ is not $d(xdy)=dx\wedge dy \not = 0$.

 

[cianfa72]

$dh$ is closed(even exact), so you have $ddh=0$. On the other hand $xdy$ is not $d(xdy)=dx\wedge dy \not = 0$.

So the existence of a scalar function $h$ such that $dh=f(x)dx$ is always true just in 1-dimension, I believe.

 

[PeterDonis]

the existence of a scalar function $h$ such that $dh=f(x)dx$

I'm not sure what you mean by this. The notation $dg$ in $\omega = f dg$ indicates the exterior derivative of the scalar function $g$, not the "differential" as it is used in ordinary calculus. And the $f$ in $\omega = f dg$ is not a function of $g$; both $f$ and $g$ are scalar functions on spacetime. So $dh = f(x) dx$ doesn't really make sense in the context of this discussion.

 

[cianfa72]

I'm not sure what you mean by this. The notation $dg$ in $\omega = f dg$ indicates the exterior derivative of the scalar function $g$, not the "differential" as it is used in ordinary calculus. And the $f$ in $\omega = f dg$ is not a function of $g$; both $f$ and $g$ are scalar functions on spacetime. So $dh = f(x) dx$ doesn't really make sense in the context of this discussion.

Surely, my point was as follows: take the set $\mathbf R$ as 1-dimensional manifold with identity map as coordinate function $x$. $f$ is a scalar function on $\mathbf R$ and $dx$ the exterior derivative of the coordinate function $x$. In this case I believe $fdx$ is basically the same as the differential $f(x)dx$ in the ordinary calculus therefore taking $h$ as the integral of $f$ we get $dh=fdx$.

 

[cianfa72]

A differential one-form and a co-vector field are the same thing, and what Frobenius says is that for a co-vector field $\omega$ satisfying $\omega \wedge d\omega = 0$ at every point then $\omega = f dg$ for some functions $f,g$. In other words the co-vector field $\omega$ is orthogonal to level hypersurfaces of $g$.

orthogonal in the sense of vanishing of the co-vector field $\omega=fdg$ on the vector fields belonging to the 3d distribution (vectors in the tangent space at each point on the level hypersurfaces of $g$) ?

 

[PeterDonis]

take the set $\mathbf R$ as 1-dimensional manifold with identity map as coordinate function $x$. $f$ is a scalar function on $\mathbf R$ and $dx$ the exterior derivative of the coordinate function $x$.

$x$ (or $g$ in the formula $\omega = f d g$) is just a scalar; there is no requirement that it be a "coordinate function".

In this case I believe $fdx$ is basically the same as the differential $f(x)dx$ in the ordinary calculus

You believe incorrectly, as I have already said. In ordinary calculus, $dx$ just means an infinitesimal interval of the variable $x$, which is taken to zero in the limit. It is a completely different concept from the exterior derivative of a function.

 

[ergospherical]

You believe incorrectly, as I have already said. In ordinary calculus, $dx$ just means an infinitesimal interval of the variable $x$, which is taken to zero in the limit. It is a completely different concept from the exterior derivative of a function.

I don't agree with that. If $f$ is a $C^{\infty}$ function (0-form) then its exterior derivative $df$ really is just the differential of $f$.

 

[cianfa72]

$x$ (or $g$ in the formula $\omega = f d g$) is just a scalar; there is no requirement that it be a "coordinate function".

Anyway in 1-dimensional case the differential one-form $fdg$ should be closed:
$d(fdg)=df\wedge dg=\left ( \frac {df} {dx}\right ) dx \wedge \left ( \frac {dg} {dx} \right ) dx = 0$, since $dx \wedge dx =0$.

From what you said this does not mean there exist a scalar function $h$ such that $dh=fdg$, however.

 

[PeterDonis]

I don't agree with that. If $f$ is a $C^{\infty}$ function (0-form) then its exterior derivative $df$ really is just the differential of $f$.

No, it isn't. The gradient, which is the 1-form exterior derivative of a scalar function, is not the same as the differential, which is an infinitesimal interval of $f$ that gets taken to zero in a limit. A 1-form is not the same as an infinitesimal interval.

 

[PeterDonis]

$d(fdg)=df\wedge dg=\left ( \frac {df} {dx}\right ) dx \wedge \left ( \frac {dg} {dx} \right ) dx = 0$, since $dx \wedge dx =0$.

Wrong. The first part of what you wrote, $d (f dg) = df \wedge dg$, is correct (since the other term $f \wedge ddg$ vanishes since $dd$ always vanishes). However, the rest is wrong. It is an example of the error I have already described, thinking that a coordinate differential $dx$ in an expression like $df / dx$ is the same thing as the gradient of a scalar function. It isn't.

You say you are not an expert. That is evidently true based on your posts. I strongly suggest spending some time with a textbook on the subject before posting further.

 

[PeterDonis]

Wrong.

Btw, it also seems like you are imagining a 1-dimensional case; but the ring-riding observer scenario cannot be analyzed in just one dimension. It requires at least three (two space dimensions and one time dimension). So even if you discover special cases of some formulas for 1 dimension, they will not generalize to the case that is the topic of this thread.

 

[PeterDonis]

it also seems like you are imagining a 1-dimensional case; but the ring-riding observer scenario cannot be analyzed in just one dimension.

Not only that, but any expression like $df \wedge dg$, or for that matter $d (f dg)$ itself, is only meaningful in at least 2 dimensions (since those expressions are 2-forms and 2-forms require at least 2 dimensions).

 

[ergospherical]

No, it isn't. The gradient, which is the 1-form exterior derivative of a scalar function, is not the same as the differential, which is an infinitesimal interval of $f$ that gets taken to zero in a limit. A 1-form is not the same as an infinitesimal interval.

Why do you say this? The modern approach is indeed to treat differentials as 1-forms [not as numbers greater than zero but less than any standard real number, à la "nonstandard analysis"].

Let $f$ be a $C^{\infty}$ function on $M$, then there is a 1-form $df$ defined by the relation $df(X) = Xf$. You can check it is a 1-form by showing it is linear, $df(X+Y) = (X+Y)f = Xf + Yf = df(X) + df(Y)$ and also $df(gX) = (gX)(f) = gX(f) = g df(X)$. Then $df$ is called the differential of $f$.

For example, let $f : \mathbb{R}^n \longrightarrow \mathbb{R}$, and consider a vector field $X = X^i \partial_i$. Then $df(X) = Xf = X^i \partial_i f$. Also, $(\partial_i f dx^i) X = \partial_i f X x^i = X^i \partial_i f$. Therefore$$df = \partial_i f dx^i$$which is nothing but the chain rule of calculus.

 

[PeterDonis]

The modern approach is indeed to treat differentials as 1-forms [not as numbers greater than zero but less than any standard real number, à la "nonstandard analysis"].

Infinitesimals in nonstandard analysis are still not the same as 1-forms. If you think they are, please give a reference.

 

[ergospherical]

Infinitesimals in nonstandard analysis are still not the same as 1-forms. If you think they are, please give a reference.

I think you parsed what I wrote differently to what I intended. I meant that it’s more common to define a differential as a 1-form. The hyperreals of Robinson are a different approach and what I thought you had in mind.

 

[PeterDonis]

I meant that it’s more common to define a differential as a 1-form.

In expressions like line elements $ds^2$, yes, the coordinate differentials are more properly interpreted as 1-forms, and things like $dx^2$ are more properly interpreted as $dx \otimes dx$, things like $dx dy$ are more properly interpreted as $dx \otimes dy$, etc.

I suppose if you are just dealing with one dimension, since there is only one possible 1-form (modulo a scalar factor), one could establish a correspondence between that 1-form and the coordinate differential $dx$ interpreted in the way one usually does in ordinary calculus (see below) (again modulo a scalar factor). But this will break down as soon as you have more than one dimension.

The hyperreals of Robinson are a different approach and what I thought you had in mind.

Ah, I see. Yes, those provide a rigorous foundation (though not the only possible one--the old epsilon-delta formulation in terms of limits still works) for infinitesimals as they are usually used in ordinary calculus.

 

[cianfa72]

So even if you discover special cases of some formulas for 1 dimension, they will not generalize to the case that is the topic of this thread.

Yes definitely it does not apply to the topic of the thread.

Coming back to the topic: I'm not sure to grasp the difference between the Frobenius condition on the timelike congruence's tangent vector field (zero vorticity) vs the corresponding one on covector field for the congruence hypersurface othogonal property.

Can you help me? Thanks.

 

[pervect]

I don't have the time to write anything much, but I think a discussion of the volume 1-form and the volume 3-form would be helpful here.

MTW's index has a brief mention on pg 133 which I found in the index, but I think there was a much fuller treatment elsewhere which I don't have time to find. The treatment on pg 133 was very terse :(.

Both the volume 1-form and the volume 3-form are useful concepts, and both should be distinguished from the manifolds and vector spaces. Space-time is a 4d manifold, so at every point we can find 4 indepedent vectors that span the space. In an orthonormal basis, these 4 vectors are orthogonal.

Given a 4d spacetime and a timelike congruence as specified by a vector field in this manifold, we can always locally define a 3d manifold with three basis vectors, all of which basis vectors are orthogonal to the vector field defining the congruence. The space spanned by these 3 vectors is the 3d space.

The purpose of the volume 1 form and 3 form is to get a volume element, which allows one to calculate the volume of some region of 3-space.

 

[cianfa72]

What you seek is explained in Appendix B.3 of Wald.

While reading Wald B.3 I got stuck on LHS equation B.3.4 (As far as I can tell Wald uses abstract index notation for tensorial equations)

$\omega_a(Y^b\nabla_bZ^a - Z^b\nabla_bY^a)=-Z^aY^b\nabla_b\omega_a + Y^aZ^b\nabla_b\omega_a$

$\omega_a$ is the co-vector field such that vanishing vector fields belong to the 3d distribution. $Y^a$ and $Z^b$ are generic vector field belonging to the 3d distribution.

Can you please help me? Thanks.

 

[ergospherical]

@cianfa72 note that $\omega_a Y^a = \omega_a Z^a = 0$, so for example$$\nabla_b (\omega_a Z^a) = \omega_a \nabla_b Z^a + Z^a \nabla_b \omega_a = 0$$i.e. $\omega_a \nabla_b Z^a = -Z^a \nabla_b \omega_a$, and the same for $Y^a$.

 

[cianfa72]

@cianfa72 note that $\omega_a Y^a = \omega_a Z^a = 0$, so for example$$\nabla_b (\omega_a Z^a) = \omega_a \nabla_b Z^a + Z^a \nabla_b \omega_a = 0$$i.e. $\omega_a \nabla_b Z^a = -Z^a \nabla_b \omega_a$, and the same for $Y^a$.

ok got it. What about the 'order' of $Z^a$ and $Y^b$ for example in the first term on LHS ? Can we actually 'reverse' their order manteining their abstract index names i.e. $-Y^bZ^a\nabla_b\omega_a$ ?