Parallel transport vs Lie dragging along a Killing vector field

[cianfa72]

TL;DR Summary

Parallel transport vs Lie dragging when 'transporting/dragging' a curve on a metric manifold along a killing vector field

Hi,

I would like to ask for a clarification about the difference between parallel transport vs Lie dragging in the following scenario.

Take a vector field $V$ defined on spacetime manifold and a curve $C$ on it. The manifold is endowed with the metric connection (I'm aware of it does exist just one unique torsion-free connection compatible with the metric tensor $g_{\mu\nu}$: the connection such that $\nabla g=0$).

The curve $C$ itself defines a vector field thus we can Lie drag or parallel transport it along the flow of vector field $V$ ($V$ congruence curves).

From my understanding for any $V$ we have $\mathcal L_V V =0$ meaning that Lie dragging the vector field $V$ itself from point P to point Q along its congruence curve results in a vector tangent to the congruence curve at Q.

Said that take now the vector field $V$ to be a killing vector field. In that case I believe the Lie dragging along its flow preserves the inner product between the $C$ tangent vector at P and vector $V(P)$ (basically the vector field $V$ evaluated at P) when both 'dragged' at Q.

Consider now the parallel transport along $V$ congruence curve according to the metric connection. For any such curve it preserves the inner product between parallel transported vectors nevertheless the 'parallel transported' tangent vector to the congruence curve at P could be different from tangent vector to the congruence curve at Q (namely when the curve is not a geodesic according $g_{\mu\nu}$).

To fix that we can use the Fermi-Walker transport.

Does it make sense ?

 

[PeterDonis]

The curve $C$ itself defines a vector field

No, it doesn't. It defines tangent vectors along it, but that is not sufficient to define a vector field, since a vector field must be defined on an open neighborhood, and the curve itself is not an open neighborhood.

If you pick a congruence of curves filling an open region of spacetime, of which $C$ is a member, then the tangent vectors of all the curves do define a vector field. I suspect that is what you actually have in mind. But the fact that you need a congruence of curves, not just a single curve, should be made explicit.

 

[troglodyte]

Do we talk about closed curves?Because
A curve is the image of a continuous function from an interval to a topological space(Wikipedia ).Doesn't this mean a curve implies per definition open neighbourhoods?

 

[PeterDonis]

Doesn't this mean a curve implies per definition open neighbourhoods?

The key point is not what is required to have a curve, but what is required to have a vector field. If all you know is the tangent vectors along the curve, you don't have a vector field, because you don't know any vectors off the curve, so you don't know vectors in an open neighborhood.

 

[cianfa72]

The key point is not what is required to have a curve, but what is required to have a vector field. If all you know is the tangent vectors along the curve, you don't have a vector field, because you don't know any vectors off the curve, so you don't know vectors in an open neighborhood.

Ah that's right. Nevertheless I believe the 'Lie dragging' of the $C$ curve along $V$ congruence curves is actually well-defined.

My point is to compare the Lie dragging of the $C$ curve tangent vector from point P to point Q along $V$ congruence curve vs the parallel transport of it on the same congruence curve between same points according the metric connection.

See for instance these your post#2 and post#8

 

[PeterDonis]

I believe the 'Lie dragging' of the $C$ curve along $V$ congruence curves is actually well-defined.

I don't think it is unless the $C$ curve itself is a member of a congruence, so you can define a vector field on which the tangent vectors to $C$ are members. You can't Lie transport a vector. You can only Lie transport a vector field.

See for instance these your post#2 and post#8

As you can see from post #8, I was mistaken in post #2 of that thread in conflating Lie transport with Fermi-Walker transport in the general case. They're not the same.

 

[cianfa72]

I don't think it is unless the $C$ curve itself is a member of a congruence, so you can define a vector field on which the tangent vectors to $C$ are members. You can't Lie transport a vector. You can only Lie transport a vector field.

I could be wrong but starting from this Lecture - 50:00 it seems to me that 'Lie dragging' alone should be well-defined even to drag just a curve $C$ along a congruence of curves (flow of vector field $V$). What instead is not well defined in that case is the 'Lie derivative' for which we need to know the field to be derived in an open neighborhood.

As you can see from post #8, I was mistaken in post #2 of that thread in conflating Lie transport with Fermi-Walker transport in the general case. They're not the same.

Sure, If i got it correctly they are the same only in case the vector field $V$ with respect to dragging is done is a Killing vector field of the underlying metric manifold

 

[PeterDonis]

it seems to me that 'Lie dragging' alone should be well-defined even to drag just a curve $C$ along a congruence of curves (flow of vector field $V$). What instead is not well defined in that case is the 'Lie derivative'

I don't see how this would work since "Lie dragging" means "applying the Lie derivative along integral curves of $V$".

If i got it correctly they are the same only in case the vector field $V$ with respect to dragging is done is a Killing vector field of the underlying metric manifold

I believe this is true, yes.

 

[cianfa72]

I don't see how this would work since "Lie dragging" means "applying the Lie derivative along integral curves of $V$".

I was wondering about the possibility to build a quadrilateral in spacetime modeled as a metric manifold with Killing vector field. As discussed here the idea was to start with a curve $C$ at point P (a path in spacetime across event P) and transport it along the orbits (flow) of a Killing vector field from P to point Q to get the $C$ 'congruent' side at Q (suppose that the quadrilateral "base" side was the specific Killing orbit connecting point P with Q)

For generic orbits of Killing vector field (no constraint to be geodesics of the underlying metric manifold) the task is performed by applying Fermi-Walker transport to the $C$ tangent vectors along each orbit.

My understanding was Lie dragging/transport should give same result as well

 

[PeterDonis]

My understanding was Lie dragging/transport should give same result as well.

For a vector field $V$ that is a Killing field, it just so happens that Lie transport of another vector field $W$ along the integral curves of $V$ is equivalent to Fermi-Walker transport of the vectors that are members of $W$ at the appropriate points along the integral curves of $V$. And since Fermi-Walker transport can transport a simple vector along a curve (it doesn't require a vector field), you can Fermi-Walker transport the tangent vector to a curve $C$ at each point of $C$, along the integral curve of a Killing field $V$ that passes through that point, and the result can be viewed, if you're willing to be sloppy, as "Lie transport" of the curve $C$ along the flow of $V$.

However, (a) this only works if $V$ is a Killing vector field (for a non-Killing vector field, Fermi-Walker transport and Lie transport are not the same), and (b) it's still being sloppy because you still aren't transporting a vector field (since you only know the tangent vectors to $C$ on $C$ itself, you don't have a vector field in an open neighborhood) so calling what you are doing "Lie transport" is, strictly speaking, not correct. You have to assume that the result you get from Fermi-Walker transport is the same as what you would get from Lie transport for any vector field on an open neighborhood of $C$ that included the tangent vectors of $C$ for the points on $C$ itself. That might not be true in general.

 

[cianfa72]

For a vector field $V$ that is a Killing field, it just so happens that Lie transport of another vector field $W$ along the integral curves of $V$ is equivalent to Fermi-Walker transport of the vectors that are members of $W$ at the appropriate points along the integral curves of $V$. And since Fermi-Walker transport can transport a simple vector along a curve (it doesn't require a vector field), you can Fermi-Walker transport the tangent vector to a curve $C$ at each point of $C$, along the integral curve of a Killing field $V$ that passes through that point, and the result can be viewed, if you're willing to be sloppy, as "Lie transport" of the curve $C$ along the flow of $V$.

However, (a) this only works if $V$ is a Killing vector field (for a non-Killing vector field, Fermi-Walker transport and Lie transport are not the same), and (b) it's still being sloppy because you still aren't transporting a vector field (since you only know the tangent vectors to $C$ on $C$ itself, you don't have a vector field in an open neighborhood) so calling what you are doing "Lie transport" is, strictly speaking, not correct. You have to assume that the result you get from Fermi-Walker transport is the same as what you would get from Lie transport for any vector field on an open neighborhood of $C$ that included the tangent vectors of $C$ for the points on $C$ itself. That might not be true in general.

ok, got it !

I have another point already discussed here about the notation for Covariant Derivative.
Using Greek letters to denote 'numeric' indices, my understanding is that the expression $\nabla_\alpha X^\mu$ represents actually the components of (1,1) tensor and it is really a shorthand for $(\nabla X)^\mu {}_{\alpha}$.

On the other hand $\nabla_V X$ --where $X$ is a vector field and $V$ is just a vector-- represents instead a vector.

If now assume abstract index notation then the expression $\nabla_a X^b$ (note the indices in Latin) is really a (1,1) tensor in which $\nabla_a$ is just the same as $\nabla$ or in other words we have $\nabla_a X^b \equiv \nabla X$

 

[PeterDonis]

my understanding is that the expression $\nabla_\alpha X^\mu$ represents actually the components of (1,1) tensor and it is really a shorthand for $(\nabla X)^\mu {}_{\alpha}$ .

Yes.

On the other hand $\nabla_V X$ --where $X$ is a vector field and $V$ is just a vector-- represents instead a vector.

Yes. That's easy to see if you write it in components: $V^\alpha \nabla_\alpha X^\mu$. Basically the operator $\nabla_V$ is the contraction of $\nabla$ with $V$, so it is a scalar operator, whereas $\nabla$ itself is a vector (or more precisely covector) operator.

 

[PeterDonis]

If now assume abstract index notation then the expression $\nabla_a X^b$ (note the indices in Latin) is really a (1,1) tensor in which $\nabla_a$ is just the same as $\nabla$ or in other words we have ∇aXb≡∇X

Yes.

 

[cianfa72]

Yes.

However, in any case, we cannot interpret $\nabla_a X^b$ as tensor product between $\nabla_a$ e $X^b$ I believe...

 

[PeterDonis]

we cannot interpret $\nabla_a X^b$ as tensor product between $\nabla_a$ e $X^b$

That's correct, because $\nabla_a$ is not a tensor, it's an operator.

 

[cianfa72]

Another source of confusion (to me) is the following:

Take $\{e_i = {\frac {\partial } {\partial x^i}}\}$ as the set of coordinate base vectors then $\nabla_{e_i} {e_j} \equiv \nabla_i {e_j}$ for each i,j is just a vector (we can write it as linear combination of coordinate base vectors through the Connection symbols).

So when we see $\nabla_i {e_j}$ we are really at an impasse: should interpret it as just a vector for each i,j or as the components of a (0,2) tensor (note Latin letters for indices) ?

 

[PeterDonis]

then $\nabla_{e_i} {e_j} \equiv \nabla_i {e_j}$ for each i,j is just a vector

No, it isn't, it's a confusion.

The notation $\nabla_{e_i}$ means the operator $\nabla_V$ where $V = e_i$. This is a scalar operator, so appying it to any vector produces another vector.

The notation $\nabla_i$ means the operator $\nabla$. This is a vector operator, so appying it to a vector produces a 2nd rank tensor.

 

[cianfa72]

No, it isn't, it's a confusion.

Sometime I have seen it -- for instance covariant derivative notation

The notation $\nabla_{e_i}$ means the operator $\nabla_V$ where $V = e_i$. This is a scalar operator, so appying it to any vector produces another vector.

The notation $\nabla_i$ means the operator $\nabla$. This is a vector operator, so appying it to a vector produces a 2nd rank tensor.

Thus in general as co-vector operator applying $\nabla$ to a (r,s) rank tensor field produces a (r,s+1) rank tensor

Instead as scalar operator applying $\nabla_V$ to a a (r,s) rank tensor field results in a tensor with the same (r,s) rank.

 

[PeterDonis]

Sometime I have seen it -- for instance covariant derivative notation

You are misinterpreting that page if you think it is saying the same thing you were saying in your OP. It's not. It's saying the same thing I am saying.

as co-vector operator applying $\nabla$ to a (r,s) rank tensor produces a (r,s+1) rank tensor

Yes.

as scalar operator applying $\nabla_V$ to a a (r,s) rank tensor results in a tensor with the same (r,s) rank

Yes.

 

[cianfa72]

You are misinterpreting that page if you think it is saying the same thing you were saying in your OP. It's not. It's saying the same thing I am saying.

In the first answer below you can see the abbreviate form $(\nabla_i V )^k$ where $V$ is a vector field and $i$ stand for the i-th coordinate base vector $\frac {\partial } {\partial x^i}$ . To me also it's just confusion

 

[PeterDonis]

In the first answer below you can see the abbreviate form $(\nabla_i V )^k$ where $V$ is a vector field and ii stand for the i-th coordinate base vector $\frac {\partial } {\partial x^i}$ . To me also it's just confusion

I'm not sure what that first answer is actually trying to say, but I think that when it uses the notation

$$
\nabla_{\frac{\partial}{\partial x_i}}
$$

what it actually means is not the scalar operator $\nabla_{e_i}$, i.e., $\nabla_V$ with $V = e_i$, but the vector operator $\nabla$ itself. It is just using an extremely bad notation to express what is supposed to be the $i$ component of $\nabla$. So what it is actually describing is not the vector $\nabla_{e_i} V^k$ but the (1, 1) tensor $\nabla_i V^k$.

In general I don't think that page is a good source for any kind of claim about notation conventions that actually get used in the literature. You need to look at textbooks.