Lie dragging vs Fermi-Walker transport along a given vector field

[cianfa72]

TL;DR Summary

Lie dragging vs Fermi-Walker transport along a given vector field that is a KVF. Are the two the same?

We had a thread long time ago concerning the Lie dragging of a vector field $X$ along a given vector field $V$ compared to the Fermi-Walker transport of $X$ along a curve $C$ through a point $P$ that is the integral curve of the vector field $V$ passing through that point.

We said that the latter is basically the same as the Lie dragging along $V$ of $X$ when $V$ is a Killing vector field.

Can you explain the above claim in more detail? Thank you.

Edit: the affine connection used from Fermi-Walker transport is of course the Levi-Civita connection associated to the metric tensor.

 

[PeterDonis]

Can you explain the above claim in more detail?

Write down the equations for the two types of transport and compare them for the case of a KVF.

 

[cianfa72]

Write down the equations for the two types of transport and compare them for the case of a KVF.

Lie dragging of a vector field $X$ along a vector field $V$ means employ a family of diffeomorphisms $\phi_t$ generated from $V$ and take the pushforward $\phi_t^*X$.

If $V$ is a KVF by definition $(\mathcal L_V g)_{ab} =0$ in abstract index notation. Then for the Levi-civita connection $\nabla_a$, we have (note that it commutes with raising and lowering indices): $$(\mathcal L_V g)_{ab} = g_{cb}(\nabla_a V^c) + g_{ac}(\nabla_b V^c) = \nabla_a V_b + \nabla_b V_a = 0$$ To take things simple suppose now the integral curves of $V$ are geodesics too. The equation of parallel transport according $\nabla_a$ of $X$ along the integral curve of $V$ is$$V^a\nabla_aX^b = 0$$ How can we get the result ?

 

[PeterDonis]

Lie dragging of a vector field $X$ along a vector field $V$ means employ a family of diffeomorphisms $\phi_t$ generated from $V$ and take the pushforward $\phi_t^*X$.

Ok.

If $V$ is a KVF by definition $(\mathcal L_V g)_{ab} =0$ in abstract index notation. Then for the Levi-civita connection $\nabla_a$, we have (note that it commutes with raising and lowering indices): $$(\mathcal L_V g)_{ab} = g_{cb}(\nabla_a V^c) + g_{ac}(\nabla_b V^c) = \nabla_a V_b + \nabla_b V_a = 0$$

Ok.

To take things simple suppose now the integral curves of $V$ are geodesics too.

If $V$ is a KVF then, except for very rare cases, its integral curves will not be geodesics. So you've just eliminated virtually all of the cases of interest. Why would you do that?

Moreover, there is no need to assume that the integral curves of $V$ are geodesics anyway. There is a perfectly good equation for Fermi-Walker transport of $X$ along integral curves of $V$. Write it down and compare it with the equation for Lie transport, using the fact that $V$ is a KVF.

 

[vanhees71]

Fermi-Walker transport of a vector field $F$ along a (time-like) curve $\gamma$ with four-velocity $U$ is defined by the differential equation
$$\mathrm{D}_U^{(\text{FW})} F=\mathrm{D}_s F - (F,\mathrm{D}_s U) U+(F,U) \mathrm{D}_s U=0.$$
It describes the transport of the vector field $F$ along the curve in such a way that it's locally non-rotating. $\mathrm{D}_s$ is the usual covariant derivative of the vector field along the curve $\gamma$ and $(A,B)=g_{\mu \nu} A^{\mu} B^{\nu}$ is the pseudo-scalar product.

Since geodesics are defined by $\mathrm{D}_s U=0$, Fermi-Walker transport is the same as parallel transport if and only if $\gamma$ is a (time-like) geodesic.

 

[cianfa72]

Fermi-Walker transport is the same as parallel transport if and only if is a (time-like) geodesic.

Yes, of course.

So we can rewrite the above as:$$\mathrm{D}_V^{(\text{FW})} X=V^a\nabla_a X^b - (X^cV^eg_{cd}\nabla_eV^d)V^b + (X^aV^cg_{ac}) V^d\nabla_d V^b$$
Suppose now the Lie bracket $[X,V]$ vanishes (i.e. $X$ is Lie dragged along the vector field $V$):
$$[X,V] = V^a\nabla_aX^b - X^a\nabla_aV^b = 0$$
With the conditions in post #3 (i.e. $V$ is a KVF) the above $\mathrm{D}_V^{(\text{FW})}X$ should vanish as well. Why ?

[Post edited by a Mentor at OP's request]

 

[cianfa72]

I'm stuck at previous post, can you kindly help me? Thanks.

 

[PeterDonis]

I'm stuck at previous post, can you kindly help me?

On considering this, I realized that the original claim I made cannot be true.

Let's compute a simple example: the Rindler congruence in 1+1 Minkowski spacetime. The 4-velocity field, in Minkowski coordinates (I use them because all the connection coefficients are zero so covariant derivatives are just partial derivatives, which makes the math easier), is:

$$
U = \frac{X}{\sqrt{X^2 - T^2}} \partial_T + \frac{T}{\sqrt{X^2 - T^2}} \partial_X
$$

Note that, when expressed in terms of coordinate basis vectors as partial derivatives, as here, this formula also gives us the directional derivative along $U$, i.e., $\partial_U = U^a \partial_a$, because, as above, covariant derivatives in this spacetime and this chart reduce to partial derivatives.

The proper acceleration of this 4-velocity field is:

$$
A = \frac{T}{X^2 - T^2} \partial_T + \frac{X}{X^2 - T^2} \partial_X
$$

The Fermi-Walker derivative along $U$ is

$$
D^{(FW)}_U V = \partial_U V - ( V \cdot A ) U + ( V \cdot U ) A
$$

The Lie derivative along $U$ is

$$
L_U V = \partial_U V - \partial_V U
$$

Note that I've switched some notation around to avoid confusion when using $X$ and $T$ for the Minkowski coordinates.

We can subtract the Lie derivative equation from the Fermi-Walker derivative equation to get:

$$
D_U^{(FW)} V - L_U V = \partial_V U - ( V \cdot A ) U + ( V \cdot U ) A
$$

Note that imposing the condition that either $D_U^{(FW)} V$ or $L_U V$ vanish independently doesn't really help; we still end up with basically the same condition.

Even without trying to compute the RHS of the above in detail, one can see that it can't possibly vanish for all vectors $V$. So even for this simple example, Fermi-Walker transport cannot be identical to Lie transport for all vectors $V$. And the vector field $U$ above is a Killing vector field, so we have a counterexample to my original claim.

 

[cianfa72]

Ah ok, here $X$ and $T$ are the Minkowski coordinates in flat spacetime. And $V$ is actually a generic vector field in an open neighborhood.

 

[PeterDonis]

here $X$ and $T$ are the Minkowski coordinates in flat spacetime. And $V$ is actually a generic vector field in an open neighborhood.

Yes.

 

[cianfa72]

The same conclusion applies in general also for Fermi-Walker transport vs Lie transport of a generic vector field $V$ along a curve/congruence that is a timelike KVF geodesic.

 

[PeterDonis]

The same conclusion applies in general also for Fermi-Walker transport vs Lie transport of a generic vector field $V$ along a curve/congruence that is a timelike KVF geodesic.

I would think so, yes. For that case the proper acceleration $A$ vanishes so the equation I wrote for the difference in the two derivatives reduces to

$$
D^{(FW)}_U V - L_U V = \nabla_V U
$$

which would not be expected to vanish for an arbitrary vector field $V$.

 

[cianfa72]

So coming back to this old thread the correct way to build a congruent curve w.r.t. a given curve $C$ along the flow of a vector field $V$ is actually the Lie dragging.

If we have a vector field $V$ there is an implied family of diffeomorphisms $\phi_t$. The inner product of tangent vectors Lie dragged by means of pushforward $\phi_t^{*}$ is left invariant by definition. If then $V$ was a KVF the Lie derivative along it would vanish.

 

[PeterDonis]

So coming back to this old thread the correct way to build a congruent curve w.r.t. a given curve $C$ along the flow of a vector field $V$ is actually the Lie dragging.

I don't think that is true; at the very least it depends on what you consider to be a "congruent" curve.

If then $V$ was a KVF the Lie derivative along it would vanish.

I don't think this is true either.

I suggest looking at the example I used in post #8 and computing the above claims for that example.

 

[cianfa72]

I don't think this is true either.

Sorry, I meant the Lie derivative of the inner product of two vector fields along a KVF. If $\phi_t^{*}$ is an isometry then I would expect It vanishes.

 

[PeterDonis]

I meant the Lie derivative of the inner product of two vector fields along a KVF.

The inner product of two vector fields is a scalar function, and for scalars the Lie derivative reduces to the directional derivative along the integral curves. So it is expected that the Lie derivative and the covariant derivative (which is what generates the isometry) will coincide for this special case.

 

[cianfa72]

The inner product of two vector fields is a scalar function, and for scalars the Lie derivative reduces to the directional derivative along the integral curves.

Yes that's true. Nevertheless that (Lie) derivative along a KVF might or might not be actually null.

 

[PeterDonis]

Yes that's true. Nevertheless that (Lie) derivative along a KVF might or might not be actually null.

The Lie derivative of the inner product of vectors will be, because the Lie derivative reduces to the covariant derivative along the curve for scalars, and for a KVF, the covariant derivative of the inner product of vectors does vanish, since the KVF generates an isometry and an isometry preserves inner products.

 

[cianfa72]

The Lie derivative of the inner product of vectors will be, because the Lie derivative reduces to the covariant derivative along the curve for scalars, and for a KVF, the covariant derivative of the inner product of vectors does vanish

I tried to check this by explicit calculation as follows
$$L_V(X^ag_{ab}Y^b) = g_{ab}Y^b L_VX^a + X^a(g_{ab}L_VY^b + Y^bL_Vg_{ab})$$
The last term vanishes by definition of $V$ being a KVF, so we get $$L_V(X^ag_{ab}Y^b) = g_{ab}Y^b L_VX^a + X^ag_{ab}L_VY^b = g_{ab}Y^b [V,X^a] + X^ag_{ab} [V,Y^b]$$
Now, how we get the expected result (namely that the above derivative vanishes) ?

 

[PeterDonis]

I tried to check this by explicit calculation

I've looked at this and thought about what being a Killing vector field actually implies, and I think the claim I made was too strong. Obviously the metric is unchanged along an integral curve of a KVF, but that does not necessarily imply that all inner products of vectors are unchanged, because the vectors themselves might change along the curve. Even a KVF does not leave all vectors unchanged along its integral curves.

Your formulas do show that if the vectors $X$ and $Y$ are Lie transported along an integral curve of a KVF (i.e., $L_V X$ and $L_V Y$ both vanish), their inner product will be unchanged. But that is a much weaker claim than the one I made before.

 

[cianfa72]

Obviously the metric is unchanged along an integral curve of a KVF, but that does not necessarily imply that all inner products of vectors are unchanged, because the vectors themselves might change along the curve. Even a KVF does not leave all vectors unchanged along its integral curves.

I'm not sure that for a given family of diffeomorphisms $\phi_t$, Lie dragging of a pair of vector fields $X$ and $Y$ along it preserves their inner product (even in case $\phi_t$ is the flow of a KVF).

 

[PeterDonis]

I'm not sure that for a given family of diffeomorphisms $\phi_t$, Lie dragging of a pair of vector fields $X$ and $Y$ along it preserves their inner product (even in case $\phi_t$ is the flow of a KVF).

Aren't you just repeating what you quoted from my post #20?

 

[cianfa72]

Aren't you just repeating what you quoted from my post #20?

I was actually talking about the Lie dragging of a pair of vector fields $X$ and $Y$ starting from a point along the flow of a given vector field $V$, not the Lie derivative of their inner product. Lie dragging is well-defined (think of it as the pushforward of $X$ and $Y$ by means of the family of diffeomorphisms $\phi_t^*$ associated to the vector field $V$). The Lie dragged vector fields along $V$ do not preserve their inner product even if $V$ is a KVF.

 

[PeterDonis]

I was actually talking about the Lie dragging of a pair of vector fields $X$ and $Y$ starting from a point along the flow of a given vector field $V$, not the Lie derivative of their inner product.

Go read the last sentence of what you quoted from my post #20. It talks about precisely this.

 

[cianfa72]

Sorry, I was reading the lecture notes from S. Carrol about Lie derivative. He claims that the inner product between a geodesic's tangent vector and a KVF is conserved along the geodesic (eqn 5.43 p. 140).

He starts from the following ($V$ is a KVF, $U$ is the geodesic's tangent vector and $\nabla$ is any symmetric covariant derivative): $$U^{\nu}\nabla_{\nu}(V_{\mu}U^{\mu}) = U^{\nu}U^{\mu}\nabla_{\nu}V_{\mu} + V_{\mu}U^{\nu}\nabla_{\nu}U^{\mu}$$
Then he says that, because the Killing equation 5.42, the first term on RHS vanishes. The Killing equation is: $$\nabla_{(\mu}V_{\nu)}=0$$
Why the first term on RHS vanishes ? Thanks.

 

[PeterDonis]

Why the first term on RHS vanishes ?

Killing's equation says that the tensor $\nabla V$ is antisymmetric. But the tensor (dyad) $U U$ is symmetric. What happens if you contract a symmetric tensor with an antisymmetric tensor?

 

[cianfa72]

What happens if you contract a symmetric tensor with an antisymmetric tensor?

$$U^{\nu}U^{\mu}\nabla_{\nu}V_{\mu}= - U^{\mu}U^{\nu}\nabla_{\mu}V_{\nu}$$ Since RHS and LHS are opposite they vanish. Thank you.