Congruence of curves in spacetime

[cianfa72]

TL;DR Summary

About the parametrization of congruent curves in spacetime

Hi, I asked a similar question in Differential Geometry subforum.

The question is related to the symmetries of spacetime. Namely take a spacetime with a timelike Killing vector field (KVF). By definition it is stationary.

Now consider a curve $\alpha$ and translate it along the KVF by a given amount (i.e. apply to it one of the one-parameter isometries $\phi_t$ associated to the timelike KVF). The curve $\beta$ one gets this way is by definition congruent to $\alpha$.

My question is: do $\alpha$ and $\beta$ have the same shape/parametrization in a chart adapted to the timelike KVF (i.e. in a chart where the integral curves of timelike KVF are at rest) ?

In other words: $\beta$ is represented by a coordinate time translation of $\alpha$ in that adapted chart.

Thanks.

 

[PeterDonis]

do $\alpha$ and $\beta$ have the same shape/parametrization

Isn't this what "congruent" means?

in a chart adapted to the timelike KVF (i.e. in a chart where the integral curves of timelike KVF are at rest) ?

"Congruence" is independent of any choice of coordinate chart.

 

[PeterDonis]

$\beta$ is represented by a coordinate time translation of $\alpha$ in that adapted chart.

This is a different question from your others, since it is specific to the particular chart you have chosen, whereas, as I commented in my previous post just now, "congruence" is an invariant.

That said, it should be easy to derive the fact quoted above from the properties of congruence as it has been specified in the OP.

 

[cianfa72]

This is a different question from your others, since it is specific to the particular chart you have chosen, whereas, as I commented in my previous post just now, "congruence" is an invariant.

Of course, the notation of "congruence" as defined in OP is by definition invariant (I took it from https://jhavaldar.github.io/assets/diffgeobook.pdf - definition 3.5).

That said, it should be easy to derive the fact quoted above from the properties of congruence as it has been specified in the OP.

Using the OP definition of "congruence" - namely $\alpha$ and $\beta$ are congruent if there is a spacetime's isometry $\phi_t$ such that $\beta = \phi_t(\alpha)$, the following statement

$\beta$ is represented by a time translation of $\alpha$ in that adapted chart.

comes from the fact that the isometry $\phi_t$ is represented by a time translation of an amount $t$ in the adapted chart, right?

 

[PeterDonis]

he notation of "congruence" as defined in OP is by definition invariant

Yes.

comes from the fact that the isometry $\phi_t$ is represented by a time translation of an amount $t$ in the adapted chart, right?

Yes.

 

[cianfa72]

I believe a similar claim can be done for a curve $\alpha$ Lie dragged along the integral curves of a nonvanishing vector field $X$. Take the vector field $V$ such that $\alpha$ is a member of its integral curves and $[V,X]=0$. Then in the coordinate chart adapted to $(V,X)$ - it exists since they form a coordinate basis - the Lie dragged version of $\alpha$ (say $\beta$) along $X$ is the translation of $\alpha$ by a given amount $t$.

 

[PeterDonis]

I believe a similar claim can be done for a curve $\alpha$ Lie dragged along the integral curves of a nonvanishing vector field $X$. Take the vector field $V$ such that $\alpha$ is a member of its integral curves and $[V,X]=0$. Then in the coordinate chart adapted to $(V,X)$ - it exists since they form a coordinate basis - the Lie dragged version of $\alpha$ (say $\beta$) along $X$ is the translation of $\alpha$ by a given amount $t$.

The operation you describe is of course well-defined, but in the general case I don't think the curve $\beta$ will be congruent to $\alpha$ in the sense you define in your OP. I think that will only be the case if the congruence of integral curves of the vector field $X$ [Edit: corrected] has zero expansion and shear.

 

[cianfa72]

The operation you describe is of course well-defined, but in the general case I don't think the curve $\beta$ will be congruent to $\alpha$ in the sense you define in your OP.

Ok yes, since the "Lie dragging along $X$" operation might not actually be an isometry.

I think that will only be the case if the congruence of integral curves of the vector field $V$ has zero expansion and shear.

Did you mean the integral curves of $X$ ? Since $X$ is the vector field the curve $\alpha$ is Lie dragged along.

 

[PeterDonis]

Did you mean the integral curves of $X$ ? Since $X$ is the vector field the curve $\alpha$ is Lie dragged along.

Ah, yes, I mixed them up. Will fix the post.

 

[cianfa72]

My question is somewhat related to an argument due to Feynman in his lectures 42-7 "The curvature of spacetime". He tries to make a rectangle in spacetime. The chart in Fig 42-18 should be for instance a map of Schwarzschild spacetime in Schwarzschild coordinates (note that the timelike coordinate $t$ is laid down horizontally). He suggests to follow two (timelike) worldlines at constant height A and B respectively for the same amount of proper time (i.e. as measured by co-located clocks at A and B). He claims that since the rectangle (as drawn in that chart) doesn't close then the spacetime must be curved.

I'm not sure whether his argument actually makes sense.

 

[PeterDonis]

I'm not sure whether his argument actually makes sense.

It does. Failure of a rectangle (or more generally a quadrilateral) to close is indeed an indication of a curved manifold. See, for example, Misner, Thorne & Wheeler, section 11.4.

 

[cianfa72]

It does. Failure of a rectangle (or more generally a quadrilateral) to close is indeed an indication of a curved manifold. See, for example, Misner, Thorne & Wheeler, section 11.4.

I've seen it, section 9.6 Box 9.2 and "The road to Reality" book section 14.5-6 from R. Penrose.

The point seems to me that segments making the quadrilateral must be geodesics while curves of constant Schwarzschild coordinate $t$ or $r$ (drop or make constant $\theta,\phi$) are not.

 

[PeterDonis]

segments making the quadrilateral must be geodesics

Yes, that's how the standard analysis is done.

curves of constant Schwarzschild coordinate $t$ or $r$ (drop or make constant $\theta,\phi$) are not.

Correct for curves of constant $r$, but not for curves of constant $t$; those are geodesics, so the radial spacelike "sides" of Feynman's "rectangle" in spacetime are geodesics already, only the timelike sides as he defined them are not.

However, the "rectangle" in spacetime in Feynman's example can just as easily be formed using timelike geodesics that start and end at the same $r$ coordinates as the timelike sides Feynman used; if you use geodesics of the same length, the resulting quadrilateral will still not close.

 

[cianfa72]

However, the "rectangle" in spacetime in Feynman's example can just as easily be formed using timelike geodesics that start and end at the same $r$ coordinates as the timelike sides Feynman used; if you use geodesics of the same length, the resulting quadrilateral will still not close.

A "rectangle" (or better a parallelogram) on a manifold (spacetime) must have opposite geodesic sides parallel according the affine connection defined on it. Are each pair of opposite (geodesic) sides in your modified/adapted argument parallel?

 

[PeterDonis]

A "rectangle" (or better a parallelogram) on a manifold (spacetime) must have opposite geodesic sides parallel

In a curved spacetime it is in general impossible to satisfy this requirement. The whole point is that geodesics that are parallel at one point do not stay parallel. That is why most textbooks, like MTW, use the term "quadrilateral". Feynman probably would have too if he had thought more carefully about it.

Are each pair of opposite (geodesic) sides in your modified/adapted argument parallel?

Only along the radial line where each of them achieves its maximum altitude. As above, it is impossible for them to both remain parallel and be geodesics.

 

[cianfa72]

In a curved spacetime it is in general impossible to satisfy this requirement. The whole point is that geodesics that are parallel at one point do not stay parallel.

You mean nearby geodesics that are parallel at one point do not stay parallel (otherwise they would coincide everywhere whether they passed and were parallel at the same point).

That is why most textbooks, like MTW, use the term "quadrilateral". Feynman probably would have too if he had thought more carefully about it.

I saw the usage of that term in MTW Box 9.2, 10.2 and 11.4. It uses "quadrilateral" when considering vector fields $\mathbf u, \mathbf v$ starting from a point P and evaluating the closure. I believe that the argument on the failure of the closure which implies manifold curvature makes sense only if one considers parallelograms (not just quadrilateral): start from a point P and try to make a parallelogram of geodesic segments using the notion of parallelism given by the affine connection (same affine parameter amount along each geodesic side segment).

Only along the radial line where each of them achieves its maximum altitude. As above, it is impossible for them to both remain parallel and be geodesics.

Ok, when both timelike geodesics reach their maximum altitude (along the radial line) at different values of $r$ coordinate (height from ground) their tangents are parallel according to the definition given by metric compatible's affine connection (Levi-civita).

 

[PeterDonis]

You mean nearby geodesics that are parallel at one point do not stay parallel

They don't have to be nearby, they just have to be different geodesics. "At one point" was probably a bad choice of words; the geodesics do not pass through the same point in spacetime. Unfortunately there is no short phrase in English that describes the situation being discussed. But the following is an example: consider the worldlines of two test objects in free fall in the spacetime around a non-rotating planet. At some instant of coordinate time in the rest frame of the planet, the two objects are both at rest, relative to each other and the planet, along the same radial line at different altitudes. Then the worldlines of the two objects are parallel at that instant of coordinate time. But they weren't parallel before that instant of coordinate time, and they won't be parallel after that instant of coordinate time. That's true regardless of their separation in altitude.

By the way, at the end of your post you recognize what I've just said, so I don't understand why you seem to be ignoring it in what I quoted just above.

(otherwise they would coincide everywhere whether they passed and were parallel at the same point).

See above.

I believe that the argument on the failure of the closure which implies manifold curvature makes sense only if one considers parallelograms

You are wrong. As I've already said, if the manifold is curved it is impossible to draw a true parallelogram with geodesics. So only considering parallelograms can't possibly be right.

when both timelike geodesics reach their maximum altitude (along the radial line) at different values of $r$ coordinate (height from ground) their tangents are parallel according to the definition given by metric compatible's affine connection (Levi-civita).

Yes. See above.

 

[cianfa72]

But the following is an example: consider the worldlines of two test objects in free fall in the spacetime around a non-rotating planet. At some instant of coordinate time in the rest frame of the planet, the two objects are both at rest, relative to each other and the planet, along the same radial line at different altitudes. Then the worldlines of the two objects are parallel at that instant of coordinate time.

Ok, you are assuming Schwarzschild spacetime in Schwarzschild coordinates (the planet is at rest in such coordinates and the two objects have zero coordinate velocity at some instant of coordinate time).

As I've already said, if the manifold is curved it is impossible to draw a true parallelogram with geodesics. So only considering parallelograms can't possibly be right.

Sorry for sloppiness (indeed it is impossible to draw it in a curved manifold). As in MTW the relevant quadrilateral starting at point P is built by parallel transporting vectors $\mathbf u$ and $\mathbf v$ at P along the geodesics $\gamma$ and $\delta$ up to points Q and R respectively by the same amount of affine parameter $\Delta \lambda$ (i.e. $\gamma$ and $\delta$ are the geodesics built from P having as tangents vectors $\mathbf v$ and $\mathbf u$ respectively). From Q and R build $\delta'$ and $\gamma'$ respectively and follow them by the same amount of their affine parameter $\Delta \lambda$. This should be the quadrilateral that doesn't close in a curved manifold.

 

[PeterDonis]

you are assuming Schwarzschild spacetime in Schwarzschild coordinates

Yes, since that's what was used in the Feynman argument you referenced.

This should be the quadrilateral that doesn't close in a curved manifold.

Yes. And calling it a "parallelogram" is obviously wrong.

 

[cianfa72]

I was thinking about the following: in flat spacetime if one takes two vector fields $\mathbf u$ and $\mathbf v$ whose integral curves are geodesics (i.e. straight lines in Minkowski geometry) then their commutator vanishes, hence as in MTW Box 10.2 the "quadrilateral closure" $$\nabla_{\mathbf u} {\mathbf v} - \nabla_{\mathbf v} {\mathbf u} = [\mathbf u, \mathbf v] = 0$$ in other words the "special quadrilateral" as defined in #18 actually closes (it is a "special" quadrilateral according to the "rules" used to build it).

 

[PeterDonis]

I was thinking about the following

In other words, in flat spacetime the curvature is zero. Yes, indeed. Isn't this obvious?

 

[cianfa72]

In other words, in flat spacetime the curvature is zero. Yes, indeed. Isn't this obvious?

Yes, note furthermore that in that case $$\nabla_{\mathbf u} {\mathbf v} = \nabla_{\mathbf v} {\mathbf u} = 0$$

 

[PeterDonis]

Yes, note furthermore that in that case $$\nabla_{\mathbf u} {\mathbf v} = \nabla_{\mathbf v} {\mathbf u} = 0$$

Yes, again, this is obvious in flat spacetime because all vector fields that are tangent to geodesics commute. The fact that global inertial frames can be constructed centered on any event and with any timelike geodesic passing through that event as the time axis is a consequence of this.

 

[cianfa72]

Yes, again, this is obvious in flat spacetime because all vector fields that are tangent to geodesics commute.

Yes, indeed. To me a point that should be emphasized in textbooks is that the "quadrilateral" used to show the curvature of manifold/spacetime is actually a "special" quadrilateral built according to the "rules/constraints" described earlier in this thread.