Frame fields and congruences of worldlines

[cianfa72]

[Moderator's note: Spin-off from previous thread due to more advanced discussion than is appropriate for the other "B" level thread.]

Note that it can be proven that, under fairly general assumptions, senses #1 and #2 of "rotation" above are equivalent if we make appropriate choices about how to match them up.

Sorry, what do you mean by the last part of the above claim ? Thanks.

 

[PeterDonis]

what do you mean by the last part of the above claim ?

If we have a set of worldlines, we can pick one of them (usually the worldline of the center of mass) and use spacelike separation vectors between it and nearby worldlines to find a set of orthonormal spacelike vectors at each point along the chosen worldline, such that each vector always points at the same nearby worldline. So we can get #2 from #1. And if the set of worldlines has zero vorticity (i.e., "non-rotating" in sense #1), the orthonormal spacelike vectors obtained by the procedure just described will be Fermi-Walker transported along the chosen worldline (i.e., "non-rotating" in the sense of #2). But if the set of worldlines has nonzero vorticity, we will not be able to find a set of Fermi-Walker transported orthonormal spacelike vectors by the above process.

Conversely, if we have a worldline within the object's "world tube" (again this will usually be the worldline of the object's center of mass) and a set of three orthonormal spacelike vectors at each point along it, we can construct nearby worldlines by going out the same increment of affine parameter along each of the spacelike geodesics defined by the spacelike vectors. We can then rotate the spacelike vectors at each point by the same angle and repeat the process, and we can do the same for other values of affine parameter. If we do this for all possible angles of rotation of the spacelike vectors, we will end up with a set of worldlines that fills the "world tube" of the object, and the "rotation" in the sense of #2 of the spacelike vectors (i.e., whether or not they are Fermi-Walker transported) will match up with the "rotation" in the sense of #1 of the set of worldlines (i.e., whether or not the vorticity is zero).

 

[cianfa72]

If we have a set of worldlines, we can pick one of them (usually the worldline of the center of mass) and use spacelike separation vectors between it and nearby worldlines to find a set of orthonormal spacelike vectors at each point along the chosen worldline, such that each vector always points at the same nearby worldline. So we can get #2 from #1.

Ok, you mean take the orthogonal complement to the chosen worldline's 4-velocity at each point P along it. Then any nearby worldline in the set will be "connected" by a particular spacelike geodesic from P (this is the result of exponential mapping from P along a particular spacelike vector on the tangent space at P). This process results in a set of spacelike vectors at each point along the chosen worldline (basically a frame field along it).

And if the set of worldlines has zero vorticity (i.e., "non-rotating" in sense #1), the orthonormal spacelike vectors obtained by the procedure just described will be Fermi-Walker transported along the chosen worldline (i.e., "non-rotating" in the sense of #2). But if the set of worldlines has nonzero vorticity, we will not be able to find a set of Fermi-Walker transported orthonormal spacelike vectors by the above process.

Ok, the frame field along the chosen worldline will be Fermi-Walker along it only if the set/congruence of worldlines has zero vorticity. In any case the frame field one gets this way match up with "rotation" in sense #1.

Conversely, if we have a worldline within the object's "world tube" (again this will usually be the worldline of the object's center of mass) and a set of three orthonormal spacelike vectors at each point along it, we can construct nearby worldlines by going out the same increment of affine parameter along each of the spacelike geodesics defined by the spacelike vectors. We can then rotate the spacelike vectors at each point by the same angle and repeat the process, and we can do the same for other values of affine parameter. If we do this for all possible angles of rotation of the spacelike vectors, we will end up with a set of worldlines that fills the "world tube" of the object, and the "rotation" in the sense of #2 of the spacelike vectors (i.e., whether or not they are Fermi-Walker transported) will match up with the "rotation" in the sense of #1 of the set of worldlines (i.e., whether or not the vorticity is zero).

Ok, yes.

 

[PeterDonis]

@cianfa72 yes, your statements in post #3 are correct.

 

[cianfa72]

In post #3 I used the term frame field. Actually it is an uncountable set of spacelike vectors at each point P along the chosen worldline.

 

[PeterDonis]

In post #25 I used the term frame field. Actually it is an uncountable set of spacelike vectors at each point P along the chosen worldline.

If you add the tangent vector field for the set of worldlines, you have a frame field.

 

[cianfa72]

If you add the tangent vector field for the set of worldlines, you have a frame field.

Sorry, a frame field is a set of 4 vector fields defined in this case along the chosen worldline orthonormal each other at each point. As far as I can understand, the procedure described in post #2 results in a uncountable set of spacelike vectors at each point along the chosen worldline. Can we call it a frame field?

 

[PeterDonis]

a uncountable set of spacelike vectors at each point

Sorry, I didn't see this part before. It's wrong, and I don't understand where you're getting it from.

At each individual point on the worldline you have three orthonormal spacelike vectors, and since they are, as you pointed out, in the orthogonal complement of the worldline at that point, they are orthogonal to the worldline's tangent vector, so adding the tangent vector to the set gives an orthonormal tetrad at that point. So we have an orthonormal tetrad at each point on the worldline, i.e., a frame field.

 

[PeterDonis]

the procedure described in post #24

Says:

a set of orthonormal spacelike vectors at each point along the chosen worldline

Such a set would consist of three vectors, not an uncountable number. So again I don't know where you are getting "an uncountable number" from.

 

[cianfa72]

Such a set would consist of three vectors, not an uncountable number. So again I don't know where you are getting "an uncountable number" from.

Ah ok, you mean consider only three nearby worldlines to the chosen worldline in order to pick/define a set of 3 spacelike vectors at each point along it (i.e. a frame field adding the chosen worldline's 4-velocity at each point).

 

[PeterDonis]

you mean consider only three nearby worldlines to the chosen worldline in order to pick/define a set of 3 spacelike vectors at each point along it

Obviously you can only consider three at a time, because three is the maximum number of spacelike vectors you can have that are mutually orthogonal.

Go read the last part of the second paragraph of post #2 again to see how we deal with the fact that there are an uncountable number of nearby worldlines filling the object's world tube, not just three.

 

[cianfa72]

If we have a set of worldlines, we can pick one of them (usually the worldline of the center of mass) and use spacelike separation vectors between it and nearby worldlines to find a set of orthonormal spacelike vectors at each point along the chosen worldline, such that each vector always points at the same nearby worldline. So we can get #2 from #1. And if the set of worldlines has zero vorticity (i.e., "non-rotating" in sense #1), the orthonormal spacelike vectors obtained by the procedure just described will be Fermi-Walker transported along the chosen worldline (i.e., "non-rotating" in the sense of #2).

This holds true since Fermi-Walker transport is basically parallel transport (according to the metric compatible connection) of tetrad along the chosen worldline plus a boost/rotation to "align" the parallel transported timelike vector in the tetrad to the worldline's 4-velocity at each point. Now if the set of nearby worldlines has zero vorticity (i.e. hypersurface orthogonal) then the set of spacelike vectors in the frame field one gets with the described procedure will result Fermi-Walker transported along the chosen worldline.

 

[PeterDonis]

Fermi-Walker transport is basically parallel transport (according to the metric compatible connection) of tetrad along the chosen worldline plus a boost/rotation to "align" the parallel transported timelike vector in the tetrad to the worldline's 4-velocity at each point.

Yes.

if the set of nearby worldlines has zero vorticity (i.e. hypersurface orthogonal) then the set of spacelike vectors in the frame field one gets with the described procedure will result Fermi-Walker transported along the chosen worldline.

Since you are just repeating what I said in what you quoted, yes, this is true.

 

[cianfa72]

Just a clarification about the following: when one writes down the condition $\nabla_X Y=0$ or $D_X^{(FW)} Y=0$ for either the parallel transport or the Fermi-Walker transport of $Y$ along $X$ respectively, it suffices that $Y$ is defined along the curve that has $X$ as tangent along it. In other words there is no additional requirement that $Y$ has to be defined in an open neighborhood off the curve.

 

[PeterDonis]

Just a clarification about the following: when one writes down the condition $\nabla_X Y=0$ or $D_X^{(FW)} Y=0$ for either the parallel transport or the Fermi-Walker transport of $Y$ along $X$ respectively, it suffices that $Y$ is defined along the curve that has $X$ as tangent along it. In other words there is no additional requirement that $Y$ has to be defined in an open neighborhood off the curve.

That is correct. The Lie derivative, on the other hand, does require both $X$ and $Y$ to be vector fields defined in an open neighborhood around the curve.

 

[cianfa72]

The thing puzzling me is that expanding for instance the parallel transport condition in a coordinate basis $\{ \partial_k\}$ one gets:
$$\nabla_XY = \left (X^jY^i \Gamma^k_{ij} + X^j \frac {\partial Y^k} {\partial {x^j}} \right) \mathbf {\partial}_k$$
where the partial derivatives of $Y$'s components w.r.t. the chart coordinates are involved.

 

[PeterDonis]

expanding for instance the parallel transport condition in a coordinate basis

Will of course give you quantities that will depend on the coordinates you chose. That's why it's important to first write equations in invariant form, so you can see by inspection that they are invariant--and then, if you need to for computation, pick a particular coordinate chart and expand the equations in that chart, because you already know the equation is invariant so if you compute an answer in one chart, it will be the same in any other.

 

[cianfa72]

Will of course give you quantities that will depend on the coordinates you chose.

Yes, definitely. However my point was that, from that expression in any chart, it seems at first sight that, in order to compute the partial derivatives involved, vector field $Y$ must be known in an open neighborhood around each point along the curve with tangent $X$ (i.e. off the curve).

 

[PeterDonis]

it seems at first sight that, in order to compute the partial derivatives involved, vector field $Y$ must be known in an open neighborhood around each point along the curve with tangent $X$ (i.e. off the curve).

Ah, I see. What this is actually telling you is that you cannot use the coordinate basis expansion to compute what you are trying to compute unless you do know $X$ and $Y$ in an open neighborhood. If you don't, you're right, you can't compute the answer this way. But if you do know $X$ and $Y$ in an open neighborhood and you compute the answer this way, you will find that only the components of the partial derivatives of $Y$ along the curve affect the answer, because of the way the partial derivatives of $Y$ are contracted with the components of $X$. So basically this is a very roundabout way of getting an answer that only depends on $Y$ along the curve.

 

[cianfa72]

Ah, I see. What this is actually telling you is that you cannot use the coordinate basis expansion to compute what you are trying to compute unless you do know $X$ and $Y$ in an open neighborhood.

I believe only $Y$ actually has to be known in an open neighborhood since the coordinate basis expansion involves only $Y$'s components partial derivatives.

 

[PeterDonis]

I believe only $Y$ actually has to be known in an open neighborhood since the coordinate basis expansion involves only $Y$'s components partial derivatives.

Yes, you're right, for that expansion $X$ only needs to be known on the curve.

 

[cianfa72]

Basically it goes like the directional derivative of a scalar function, i.e. given $f$ and the curve $\gamma (t)$ parametrized by $t$ the directional derivative at point $P=\gamma (t_0)$ is by definition $$\left. \frac {df(\gamma((t)} {dt}\right |_{t=t_0}$$ $f$ is required to be known only along $\gamma (t)$ even though in its expansion in a coordinate basis $f$'s partial derivatives appear.

 

[PeterDonis]

Basically it goes like the directional derivative of a scalar function

Yes.