How to Study Fermi-Walker Transport in Minkowski Spacetime

[PeterDonis]

Greg Bernhardt submitted a new blog post

How to Study Fermi-Walker Transport in Minkowski Spacetime

Continue reading the Original Blog Post.

 

[Orodruin]

of the concept of parallel transport, which only works along geodesics, to transport along worldlines that might not be geodesics, i.e., accelerated worldlines

This seems to imply that you cannot make a parallel transport along a non-geodesic, which is not true. I believe what you mean to say is that the timelike tangent of a worldline is not parallel transported along a non-geodesic world line.

 

[PeterDonis]

This seems to imply that you cannot make a parallel transport along a non-geodesic, which is not true. I believe what you mean to say is that the timelike tangent of a worldline is not parallel transported along a non-geodesic world line.

Yes, you're right. I have updated that paragraph accordingly.

 

[Orodruin]

I like the Insight in general. It is always interesting to see how others introduce a subject. Personally, I always like to think about and draw parallels (haha ) to the Riemannian (or even Euclidean) analogue to introduce things like these. The problem with Fermi-Walker transport is that it requires at least three dimensions to make it clear about what ”non-rotating” means. I would typically give the example of a circular path in R^3 (which is essentially the Euclidean version of your Rindler frame) and discuss that the tangent change is proportional to acceleration and in order to keep a vector orthogonal to the path orthogonal to the path you would need to change it accordingly proportional to the velocity. Of course, the different metric signature complicates things a little bit but is no stranger than other signs appearing when taking the step to Lorentzian from Riemannian.

 

[PeterDonis]

The problem with Fermi-Walker transport is that it requires at least three dimensions to make it clear about what ”non-rotating” means.

Not necessarily; the Rindler case in Minkowski spacetime can be reduced to two dimensions and will still show how the spacelike vector has to rotate to stay orthogonal to the timelike vector.

I would typically give the example of a circular path in R^3 (which is essentially the Euclidean version of your Rindler frame) and discuss that the tangent change is proportional to acceleration and in order to keep a vector orthogonal to the path orthogonal to the path you would need to change it accordingly proportional to the velocity.

Wouldn't this still work if you just looked at the plane of the path, i.e., a circular path in $R^2$?

 

[Orodruin]

Not necessarily; the Rindler case in Minkowski spacetime can be reduced to two dimensions and will still show how the spacelike vector has to rotate to stay orthogonal to the timelike vector.

Yes, it is the same in the Euclidean case, but you need the third direction to demonstrate how vectors orthogonal to both velocity and acceleration do not rotate, which was the point of referring to the frame field as somehow being non-rotating. I guess ”minimally rotating” would describe it sufficiently ...

 

[PeterDonis]

you need the third direction to demonstrate how vectors orthogonal to both velocity and acceleration do not rotate

Ah, got it.

 

[Orodruin]

Of course, you could just start with the 2D case to demonstrate the rotation and then state that you keep the rest orthogonal ...
Once that is introduced I would probably compare with your Rindler case and the corresponding hyperbolic rotation as I would probably already have discussed other concepts in a similar manner earlier in the course (such as circles <-> hyperbolae, etc).

 

[vanhees71]

Well, my own understanding of Fermi-Walker transport in Minkowski space, can be found in my SRT FAQ:

https://th.physik.uni-frankfurt.de/~hees/pf-faq/srt.pdf

It's of course clear that "rotation free spatial basis" is meant in a local sense, i.e., for an infinitesimal transport of an arbitrary vector, that is Minkowski perpendicular to the time-like tangent unit vector $u$, it changes only by an rotation free Lorentz transformation, which is the idea of my derivation of the Fermi-Walker transport equation (1.8.6). Of course for a finite transport the spatial vector is necessarily rotating, if the time-like tangent vector changes direction, since you can see it as a composition of many rotation free Lorentz transformations with varying boost directions, and the composition of two rotation free Lorentz boosts in different boost directions is not rotation free anymore. This is the point of the Thomas precession (or mathematically spoken the Wigner rotation).

For a worldline of hyperbolic motion of a particle it's easy to see. The worldline starting for $\tau=0$ at $x^1=0$ is given by
$$x(\tau)=\begin{pmatrix} \sinh(\alpha \tau)/alpha \\ [\cosh(\alpha \tau)-1]/\beta. \end{pmatrix}$$
The tangent vector along the curve is simply the four-velocity,
$$u(\tau)=\dot{x}(\tau)=\begin{pmatrix} \cosh(\alpha \tau) \\ \sinh(\alpha \tau) \end{pmatrix}.$$
The dot stands for derivatives wrt. proper time, and I use the (+---) convention with $c=1$ for this posting (otherwise I get confused with the signs, although it's no problem to switch to the east-coast convention of course). The acceleration is
$$a(\tau)=\ddot{x}(\tau)=\begin{pmatrix} \alpha \sinh(\alpha \tau) \\ \alpha \cosh(\alpha \tau) \end{pmatrix}.$$

Now let's start with the basis $e_0(0)=(1,0)^T$, $e_1(0)=(0,1)^T$ at $\tau=0$. All we need to get the 2-bein along the hyperbolic world line is the Fermi-Walker transport of $e_1(0)$ along this world line. Let's define $e_1(\tau)=(a(\tau),b(\tau)^T$. Then the FW-transport equation,
$$\dot{e}_1=a (u \cdot e_1)-u (a \cdot e_1),$$
reads
$$\begin{pmatrix} \dot{a} \\ \dot{b} \end{pmatrix} = \begin{pmatrix} \alpha b \\ \alpha a \end{pmatrix}.$$
The solution of course leads to
$$e_1(\tau)=\frac{1}{\alpha} a(\tau).$$
Indeed $e_1$ is not along $e_1(0)=(0,1)^T$ anymore, although locally not rotating in the sense of the definition of Fermi-Walker transport.

Of course, in this case the spatial coordinate system (4-bein) along the hyperbolic world line is not rotating at all since all infinitesimal boosts are in one direction, here chosen as the $x^1$-direction, and thus here we have no Wigner rotations.

 

[cianfa72]

A clarification about covariant derivative and Fermi-Walker transport: AFAIK parallel transport as defined by an affine connection (covariant derivative operator) makes sense only for a vector field $X$ along a curve.

Does Fermi-Walker transport make sense even only for vectors (i.e not just for vector fields) to be transported along a curve ?

 

[Orodruin]

A clarification about covariant derivative and Fermi-Walker transport: AFAIK parallel transport as defined by an affine connection (covariant derivative operator) makes sense only for a vector field $X$ along a curve.

Does Fermi-Walker transport make sense even only for vectors (i.e not just for vector fields) to be transported along a curve ?

Parallel transport is well defined for any vector (or tensor) along a curve. The same with FW transport. If you want to apply the covariant derivative to a full field you can extend the field smoothly to a neighborhood around the curve, but the transport will not depend on the extension.

 

[cianfa72]

Parallel transport is well defined for any vector (or tensor) along a curve. The same with FW transport.

Ah ok, so the rule that a geodesic parallel transports its tangent vector i.e. $\nabla_{X} X = 0$ actually involves only the set of tangent vectors $X$ defined along the curve (geodesic) and not off of it.

 

[cianfa72]

Btw, if we take a vector field $X$, evaluate it on a point P on a generic curve that is orbit of a Killing vector field $V$ and FW transport it along that curve up to point Q, it should result the same as the Lie transport of $X$ along the (KVF) $V$.

Does it make sense ?

 

[martinbn]

Btw, if we take a vector field $X$, evaluate it on a point P on a generic curve that is orbit of a Killing vector field $V$ and FW transport it along that curve up to point Q, it should result the same as the Lie transport of $X$ along the (KVF) $V$.

Does it make sense ?

Why?

 

[cianfa72]

Why?

Mine was just an intuitive argument: along the Killing vector field $V$ the metric $g$ does not change by definition ($\mathcal L_V g = 0$). The inner product (defined by $g$ on each tangent space) between the vector fields $X$ and $V$ both evaluated at point P should be the same as the inner product of the Lie transport of $X$ at point Q and the vector field $V$ evaluated at the same point Q (Lie transport of $X$ along the integral orbit of $V$ from P to point Q).

I think the above should be actually the same as the FW transport along the integral orbit of $V$ through P connecting it with point Q.

 

[Orodruin]

To prove (or disprove) the statement, you should consider whether or not the FW derivative along V being zero implies that the Lie derivative is zero (or not).

 

[cianfa72]

To prove (or disprove) the statement, you should consider whether or not the FW derivative along V being zero implies that the Lie derivative is zero (or not).

We know that FW transport does not change the inner product between vectors being transported and that the FW transport of $V$ along itself (i.e. along one of the integral orbits of $V$) is always null.

Suppose the FW derivative of vector field $X$ along V ((i.e. along the integral orbit that connect point P to point Q) is null at Q. That basically means the FW transported vector at Q equals the vector field $X$ evaluated at Q.

Since $\mathcal L_V V = 0$ the Lie transport of vector field $V$ along $V$ equals the vector field $V$ evaluated at Q hence the FW transport of $V$ along $V$ is the same as the Lie transport of $V$ along $V$. Now if $V$ is a KVF for the underlying manifold then the Lie transport and the FW transport of a vector field $X$ along $V$ give the same result.

Hence a zero FW derivative along the KVF $V$ implies Lie derivatives along it is zero too.

 

[Orodruin]

Now if V is a KVF for the underlying manifold then the Lie transport and the FW transport of a vector field X along V give the same result.

Why? You cannot assume what you wanted to show.

 

[martinbn]

@cianfa72 If you replace Fermi-Walker with parallel transport, in what you wrote, what changes? The argument you are presenting seems to work for both, so it cannot be right.

 

[cianfa72]

@cianfa72 If you replace Fermi-Walker with parallel transport, in what you wrote, what changes? The argument you are presenting seems to work for both, so it cannot be right.

In case of parallel transport along a generic integral orbit of $V$, $\nabla_V V \neq 0$ hence the vector you get by the parallel transport of $V$ along itself from point P to point Q is not the same as its Lie transport ($\mathcal L_V V = 0$).

In case of geodesic congruence as integral orbits of $V$, FW transport and parallel transport are the same.

 

[cianfa72]

Why? You cannot assume what you wanted to show.

As said before mine was an intuitive reasoning: my idea is that in case of a KVF vector field $V$ the diffeomorphism defined by its flow from point P to Q actually results in an orthonormal transformation between tangent vector spaces defined at both point (let me say it is just a 'rotation').

 

[Orodruin]

As said before mine was an intuitive reasoning: my idea is that in case of a KVF vector field $V$ the diffeomorphism defined by its flow from point P to Q actually results in an orthonormal transformation between tangent vector spaces defined at both point (let me say it is just a 'rotation').

The tangent spaces are different so it cannot be a rotation per se. It is also not sufficient to say that the FW and Lie transports of $V$ give the same result and that the inner products between $X$ and $V$ after FW/Lie transports are the same (it is only sufficient in a two-dimensional manifold) - fixing the transport of $V$ and the inner product of the transported $X$ and $V$ only constraints the component of $X$ in the direction of $V$ (and the conservation of the norm of $X$ further constrains $X$ only up to a rotation in $N-1$ dimensions).

 

[cianfa72]

It is also not sufficient to say that the FW and Lie transports of $V$ give the same result

You really mean the FW/Lie transport of $V$ along $V$ give the same result, I think.

and that the inner products between $X$ and $V$ after FW/Lie transports are the same (it is only sufficient in a two-dimensional manifold) - fixing the transport of $V$ and the inner product of the transported $X$ and $V$ only constraints the component of $X$ in the direction of $V$ (and the conservation of the norm of $X$ further constrains $X$ only up to a rotation in $N-1$ dimensions).

Here you are saying that the inner product of vector fields $X$ and $V$ both evaluated at point P is the same as the inner product of their FW/Lie transports along $V$ at point Q.

fixing the transport of $V$ and the inner product of the transported $X$ and $V$ only constraints the component of $X$ in the direction of $V$ (and the conservation of the norm of $X$ further constrains $X$ only up to a rotation in $N-1$ dimensions).

Ah ok (so my argument makes sense only for 2D manifold).

 

[Orodruin]

So, based on post #22, can you write down a counter example to your assertion?

 

[cianfa72]

So, based on post #22, can you write down a counter example to your assertion?

For example on a 3D Riemann manifold we can image a metric field $g$ with a 'geodesic' symmetry (i.e. a KVF field $V$ having geodesics as integral orbits) such that the Lie transport of a vector field $X$ along $V$ from point P to point Q differs from the FW/parallel transport of $X$ along the (geodesic) integral orbit of $V$ from P to Q by just a rotation in 2 dimension.

However I've some difficulty to write down a counter example from an analytic point of view.

 

[Orodruin]

Here’s one:

Consider three-dimensional Minkowski space and the symmetry transformation:
$$
t \to t + s, \quad x \to x \cos(\omega s) + y \sin(\omega s), \quad y \to y \cos(\omega s) - x \sin(\omega s).
$$
The corresponding tangent field $V$ is given by
$$
V^t = 1, \quad V^x = - \omega y, \quad V^y = \omega x.
$$
Now consider the transports along the time axis $x = y = 0$ from $t = 0$ to $t = 1$ along which $V^t = 1$ is the only non-zero component of $V$. The relevant integral curve is therefore $(s,0,0)$.

Take the vector $X^x = 1$, $X^t = X^y = 0$ at the origin. The Fermi-Walker transport of this vector along the curve $(s,0,0)$ is simply $X^\mu = (0,1,0)$ because the time-axis is a geodesic and Fermi-Walker transport reduces to parallel transport in regular Minkowski coordinates, which means constant components.

However, the Lie derivative of $X$ if you want to Lie transport it is given by $(\mathcal L_V X)^a = [V,X]^a = \dot X^a - X^b \partial_b V^a = 0$, leading to the two non-trivial equations
$$
\dot X^x = X^x \partial_x V^x + X^y \partial_y V^x = - \omega X^y, \qquad
\dot X^y = \omega X^x
$$
with the solution $X^x = \cos(\omega s)$ and $X^y = \sin(\omega s)$ and therefore $X^x = \cos(\omega)$ and $X^y = \sin(\omega)$ for $s = 1$, which is generally different from (0,1,0).

 

[cianfa72]

Just to clarify myself some points as follows:

First one if we calculate the derivative of $x \cos(\omega s) + y \sin(\omega s)$ w.r.t. $s$ ed evaluate it for $s=0$ we get
$V^x = \left. \omega [-x \sin(\omega s) + y \cos(\omega s)] \right |_{s=0} = \omega y$

and similarly from $y \cos(\omega s) - x \sin(\omega s)$ we get
$V^y = \left. \omega [-y \sin(\omega s) - x \cos(\omega s)] \right |_{s=0} = - \omega x$

Then I tried to prove myself that the above map is really a symmetry for the 3D Minkowski spacetime. I calculated the Jacobian matrix $J$ of the (active) transformation as:

$$ J = \begin{bmatrix}
1 & 0 & 0 \\
0 & cos(\omega s) & sin(\omega s) \\
0 & -\sin(\omega s) & cos(\omega s) \\
\end{bmatrix}$$
This is actually an orthogonal matrix $J^{-1} = J^T$ hence the pullback $\phi_{*}$ of the diagonal metric tensor ($\eta_{\mu \nu}$ in standard Minkowski coordinates) does not change.

About the Lie derivative $(\mathcal L_V X)^a = [V,X]^a = \dot X^a - X^b \partial_b V^a$ my understanding is that the first RHS term $\dot X^a$ represents the derivatives of vector field $X$ components w.r.t. the $t$ Minkowski coordinate (basically it is the expansion of $\nabla_V X$ vector field along $V^t=1$).

Is the above correct ? Thank you.

 

[Orodruin]

Looks like I had a sign error, but you get the idea anyway.

my understanding is that the first RHS term X˙a represents the derivatives of vector field X components w.r.t. the t Minkowski coordinate (basically it is the expansion of ∇VX vector field along Vt=1).

It is $V^b \partial_b X^a$ along the curve. The covariant derivative does not appear in the Lie derivative as the Lie derivative is independent from the connection.

 

[cianfa72]

It is $V^b \partial_b X^a$ along the curve. The covariant derivative does not appear in the Lie derivative as the Lie derivative is independent from the connection.

ah ok, so $\dot X^a$ is really the vector field $V^b \partial_b X^a$ evaluated along the points on the integral orbit of $V^t=1 , V^x=V^y=0$.

Btw in #27 I employed a symmetric connection $\nabla$ since for any of it should result $(\mathcal L_V X)^a = [V,X]^a = (\nabla_V X)^a - (\nabla_X V)^a$.

 

[Orodruin]

ah ok, so X˙a is really the vector field Vb∂bXa evaluated along the points on the integral orbit of Vt=1,Vx=Vy=0.

If it was not, say if the commutator depended on derivatives of $X$ in directions other than along the path, then you would not be able to compute the Lie transport without knowing the full field $X$. As it is, you only need to know the initial value of $X$.

Also, it $V^b \partial_b X^a$ is not a vector field. The expression becomes the component of a vector field only when considered along with the second term as well.

Btw in #27 I employed a symmetric connection ∇ since for any of it should result (LVX)a=[V,X]a=(∇VX)a−(∇XV)a.

While this is true, it is not very satisfying as it makes it appear as if the Lie derivative depends on the connection. It does not.

 

[cianfa72]

Sorry, I have some problem to grasp how you get the two non-trivial equations:

$$
\dot X^x = X^x \partial_x V^x + X^y \partial_y V^x = - \omega X^y, \qquad
\dot X^y = \omega X^x
$$
The LHS for each of the above equations should be $(\mathcal L_V X)^a = [V,X]^a = \dot X^a - X^b \partial_b V^a = 0$ evaluated for index $a=x$ and $a=y$ respectively. What about the respective RHS terms, namely $- \omega X^y$ and $\omega X^x$ ?

Thank you.

While this is true, it is not very satisfying as it makes it appear as if the Lie derivative depends on the connection. It does not.

Yes, you are definitely right.

 

[Orodruin]

The LHS for each of the above equations should be (LVX)a=[V,X]a=X˙a−Xb∂bVa=0 evaluated

No. I moved the term $X^b \partial_b V^a$ to the RHS. Hence resulting in
$$
\dot X^a = X^b \partial_b V^a = X^x \partial_x V^a + X^y \partial_y V^a.
$$

 

[cianfa72]

No. I moved the term $X^b \partial_b V^a$ to the RHS. Hence resulting in
$$
\dot X^a = X^b \partial_b V^a = X^x \partial_x V^a + X^y \partial_y V^a.
$$

Yes, sure. The point is where the two terms $- \omega X^y$ and $\omega X^x$ come from ?

 

[Orodruin]

Yes, sure. The point is where the two terms $- \omega X^y$ and $\omega X^x$ come from ?

Evaluating the derivatives of $V$.

 

[vanhees71]

In my opinion the origen of FWT is better understood in flat spacetime. Its generalization to curved space time is then rutinary as explained in this paper https://doi.org/10.1007/s10699-020-09679-9

I think that's a bit overcomplicating the issue. Indeed FWT provides a transport of a tetrad along a time-like worldline which is locally rotation free. That's must simpler derived directly by simply using an infinitesimal Lorentz transformation when moving an "infinitesimal increment" along the worldline:

https://itp.uni-frankfurt.de/~hees/pf-faq/srt.pdf p. 21ff