About the solution of the parallel transport equation

[Jianbing_Shao]

TL;DR Summary

parallel transport equation

If a vector moves along a particular curve $l$ from point $x_0$ to point $x$ on a manifold whose connection is $\Gamma^i_{jk}(x)$,
then the vector field we get obviously satisfy the pareallel transport equations:
$$\partial_kv^i(x)+\Gamma^i_{jk}(x)v^j(x)=0$$
Because $[\Gamma^i_{jk}(x) , \Gamma^i_{jk}(x')]\neq 0$, so the solution should be written in the form of a path-ordered product:
$$v(x)=\mathcal{P}\left\{\exp{\left(\int^l-\Gamma^i_{jk}(x)dx^k\right)}\right\} v(x_0)$$
The definition of path-ordered product is just like the time-ordered product in quantum mechanics.
Then we usually use Dyson expansion to calculate the path-ordered procuct:
$$\mathcal{P}\left\{\exp{\left(\int^l-\Gamma^i_{jk}(x)dx^k\right)}\right\} =I+\sum^{\infty}_{k=1}(-1)^k\int_{x_0}^x\int_{x_0}^{x_k}\cdots\int_{x_0}^{x_2}\Gamma(x_k)\cdots \Gamma(x_1)dx_k\cdots dx_1$$
To an arbitary connection field $\Gamma^i_{jk}(x)$, The higher-order terms in this series don't always converge to 0. So I want to know if we have any other way to find an exact solution to the parallel transport equation.

 

[jambaugh]

I'm confused by what you're trying to do. Specifically you are referring to a vector field when talking about parallel transport of a vector. Since the propagation of a vector at one point to another is path dependent there is no unique solution to a field.

I don't recognize your parallel transport equations as such but would rather expect something like:
$$d v^i +\gamma^i_{jk} v^k dx^j = 0$$
or in parametric form $\dot{v}^i +\gamma^i_{jk}v^k \dot{x}^j = 0$ where the dots are parameter derivatives. The equations you wrote would manifest first by considering a vector field with components $v^i(x)$ which is invariant under parallel transport under a specific displacement field $u^i(x)$ so that
$$\left[ \partial_j v^i(x) + \gamma^i_{jk}v^k(x)\right]u^j = 0$$
Then the global satisfaction of the equations you wrote would imply a field which is invariant under all flows for any arbitrary field $u^i(x)$. I would imagine only the zero vector field could satisfy such except under very restricted geometries.

To me, obviously the emergence of Berry phase for parallel transport along a closed path invalidates what you seem to be attempting since it seems to me you are implicitly assuming all Berry phases are zero.

Parallel transport is not the same as a propagator. Propagators propagate the boundary conditions for a constraint on a field (scalar or vector/tensor/spinor/gauge). Hence you can propagate the time slice of a field subject to a dynamics constraint to another time slice. It is in this context that one might invoke a Dyson expansion of the Green's function integral. Is this what you're trying to do?

I'm also a bit confused by your commutator of the connection components. Those components are real numbers and commute. I think you mean the commutator for the matrices formed by those components, namely $[\Gamma_k]$ the matrix with row j and column i component $\Gamma^i_{jk}$ then you can take the non-trivial commutator:
$$\left[ [\Gamma_k],[\Gamma_{k'}] \right] \ne 0$$
I assume this is what you meant, in which case you should work in this single matrix algebra or re-express your "commutator" as a component contraction relation, e.g. "It is not the case that...
$\Gamma^i_{jk}\Gamma^j_{\ell k'} - \Gamma^i_{jk'}\Gamma^j_{\ell k} \ne 0$
for all $j,\ell$.

I think you might find it useful to work in this matrix algebra when you try to sensibly express exponentiation of the differential transformations constituting parallel transport of the vectors along a given path.

 

[Jianbing_Shao]

Thanks for your reply!

I'm confused by what you're trying to do. Specifically you are referring to a vector field when talking about parallel transport of a vector. Since the propagation of a vector at one point to another is path dependent there is no unique solution to a field.

Of course parallel transport of a vector on a manifold with non-zero curcature is path dependent, so I only want to find a solution if given a particular path $l$, so if $x_0, x \in l$. we can get $v(x)$ from a given $v(x_0)$.

I don't recognize your parallel transport equations as such but would rather expect something like:
$$dv^i+\gamma^i_{jk}v^kdx^j=0$$

I think my differential equations are equivalent to your equations above. Ssuppose the vector moves a Infinitesimal displacement $dx_1$, Then $dv^i=\partial_1v^i dx_1$, then we can get
$$(\partial_1v^i+\gamma^i_{j1}v^i)dx^1=0$$
Repeat this step, then we can get:
$$(\partial_kv^i+\gamma^i_{jk}v^j)dx^k=0$$
Because to an arbitrary infinitesmal displacement $\Sigma dx^i$ these equations are true, so we can get:
$$\partial_kv^i+\gamma^i_{jk}v^j=0$$

So my question is if given a particular path $l$ on the manifold, and two points $x_0,x \in l$, then how to get $v(x)$ from a given $v(x_0)$ using parallel transport equation. To me, the answer is a path-orderd production(ordered exponentials), but such ordered exponentials is too complex to be calculated, so we hardly can get an exact result, So I am confused if it is the only way to solve the problem, and if we can get an exact solution of the problem.
Or perhaps in general relativity this question is not very important.

 

[jambaugh]

The two equations are not equivalent for the same reason that a non-trivial function can have a zero directional derivative in a particular direction without having a zero gradient in general.

 

[jambaugh]

To answer your more general question, the problem resolves as this, you parameterize the curve, you express the parallel transport as a ordinary differential equation on your vector. And you solve it by the means available to solve ODE's.

As a first order linear ODE analogous to say Schrodinger's equation you attack it just like those... as you described with a Dyson series expansion. But you are not propagating a field in space-time, you are propagating a single vector forward along a parametric path. You are solving an ODE not a PDE. Possibly I misunderstood your notation, if you are using $x_k$ to indicate your parameter, or now your series of parameters as you subdivide the path in the Dyson series expansion.

 

[Jianbing_Shao]

To answer your more general question, the problem resolves as this, you parameterize the curve, you express the parallel transport as a ordinary differential equation on your vector. And you solve it by the means available to solve ODE's.

But to a parallel transport equation
$$dv^i(x)+\gamma^i_{jk}(x)v^j(x)dx^k=0$$
We always have such relations:
$$dv^i(x)= \frac{\partial v^i(x)}{\partial x^j}dx^j=\frac{\partial v^i(x)}{\partial x^j}\frac{dx^j}{d λ}d λ$$
and
$$\gamma^i_{jk}(x)v^j(x)dx^k=\gamma^i_{jk}(x)v^j(x)\frac{dx^k}{d λ}d λ$$
If we integrate the equation along the curve $l$, then the result
$$\int_l dv^i=-\int_l \gamma^i_{jk}(x)v^j(x)dx^k$$
does not depend on a choose of a particular parameter. In fact if $[\gamma(x) , \gamma(x)]= 0$, the genaral solution of the differential equations if very simple, it is $v(x_0)e^{-\int_l \gamma^i_{jk}(x)v^j(x)dx^k}$.
But if $[\gamma(x) , \gamma(x')]\neq 0$, then the order of Infinite number of infinitesimal operators $e^{- \gamma^i_{jk}(x)v^j(x)dx^k}$ is very important, so we must introduce the comcept of orderer exponentials. so the solution should be written in the form of ordered exponentials. just as we did in quantum mechanics.
So in a word, because of the non-commutativity of $\gamma(x)$ at different points, when dealing with problem of the parallel transport of a vector along a curve, we also should use ordered exponentials.

 

[jambaugh]

Yes, I agree. This can be viewed in another format by defining the Hausdorff product (non-commutative sum):
$$A\boxplus B \equiv C: e^A e^B = e^C$$
See the Baker-Campbell-Hausdorff formula for the expansion in terms of commutators and then define what I suppose you would call a "Baker-Campbell-Hausdorff-Riemann sum" and its limit as the ordered integral:
$$U =\int_{t_0}^{t_1}\kern -18pt\to \Gamma(t)dt \equiv \lim_{N\to\infty} \mathop{\boxplus}_{k=1}^N \Gamma(t_k^*)\Delta t^*$$
(Note in the "Hausdorf sum" one must index from right to left, or reverse the convention I used here. But, anyway, it is an ordered "sum" since the "addition" is not commutative.)

Upon expanding in terms of pair-wise commutators one would recover the Dyson series expansion. (Note: I'm drinking bourbon right now and I've never seen this put together this way per se, but I have seen the pieces and I think is a nicely compact way to express the general method.)

Here $\Gamma(t)$ is your Lie algebra element, i.e. putting the $v^i$ components into a column matrix the components of the $\Gamma(t)$ matrices would be $\Gamma dt=[\gamma^i_{jk}dx^k] = [\gamma^i_{jk}\dot{x}^k(t)]dt$. Then $[v^j(x(t_1))] = U[v^j(x(t_0)]]$ where we are parallel transporting the vector $v$ along the path with parametric derivative $\dot{x}$. This should fit it to your parallel transport case. (my t = your lambda).

 

[Jianbing_Shao]

Here Γ(t)Γ(t)\Gamma(t) is your Lie algebra element, i.e. putting the vivi v^i components into a column matrix the components of the Γ(t)Γ(t)\Gamma(t) matrices would be Γdt=[γijkdxk]=[γijk˙xk(t)]dtΓdt=[γjkidxk]=[γjkix˙k(t)]dt \Gamma dt=[\gamma^i_{jk}dx^k] = [\gamma^i_{jk}\dot{x}^k(t)]dt . Then [vj(x(t1))]=U[vj(x(t0)]][vj(x(t1))]=U[vj(x(t0)]][v^j(x(t_1))] = U[v^j(x(t_0)]] where we are parallel transporting the vector vvv along the path with parametric derivative ˙xx˙\dot{x}. This should fit it to your parallel transport case. (my t = your lambda).

Perhaps this idea is just the same as the Whison loop.
In fact, I find two examples this method can be applied, The first one is the description of the rotation of a rigid body, if $\omega(t)$ is the angular velocity of the rigid body, then we know that the rotation can be described in such a way:
$$\frac{dv(t)}{dt}=\omega(t)\times v(t)=(\omega^i(t)S_i)v(t)$$
Then the rotation can be expressed in a time-ordered product:
$$v(t)=\mathcal{T}\left\{\exp^{\int_{t_0}^t\omega^i(s)S_ids} \right\}v(t_0)$$
Another example is Thomas Precession. In the discussion of Thomas Precession, we know that the instantaneous
rest frames of an accelerating particle are Fermi-Walker coordinates which fulfill the relation:
$$e{(t+\delta t)}=e \exp(-\frac{d\zeta^i(t)}{dt} K_i\delta t) , e_\mu\doteq\frac{\partial}{\partial x^\mu}$$
Obviously the Fermi-Walker coordinates also can be described using a ordered exponentials:
$$e(t)=\mathcal{T}\left\{\exp^{\int_{t_0}^t-\frac{d\zeta^i(s)}{ds} K_ids} \right\}v(t_0)$$
Here I have a question, if we use the ordered exponentials to describe the coordinate transformation between a accelerating system and an inertial coordinate in stead of pure boost . then can we say that the special relativity also can be applied to non-inertial system?

 

[Jianbing_Shao]

Another example is Thomas Precession. In the discussion of Thomas Precession, we know that the instantaneous
rest frames of an accelerating particle are Fermi-Walker coordinates which fulfill the relation:
$$e{(t+\delta t)}=e \exp(-\frac{d\zeta^i(t)}{dt} K_i\delta t) , e_\mu\doteq\frac{\partial}{\partial x^\mu}$$
Obviously the Fermi-Walker coordinates also can be described using a ordered exponentials:
$$e(t)=\mathcal{T}\left\{\exp^{\int_{t_0}^t-\frac{d\zeta^i(s)}{ds} K_ids} \right\}v(t_0)$$

Sorry, I have made some mistakes, The Fermi-Walker coordinates which fulfill the relation:
$$e{(t+\delta t)} ≈ \exp(-\frac{d\zeta^i(t)}{dt} K_i\delta t) e(t) , e_\mu\doteq\frac{\partial}{\partial x^\mu}$$
So it fulfill the differential equation:
$$\frac{de(t) }{dt}=(- \frac{d\zeta^i(t)}{dt} K_i) e(t) $$
It is similar to the eqaution which describes the rotaiton of a rigid body above. So
the Fermi-Walker coordinates also can be described using a ordered exponentials:
$$e(t)=\mathcal{T}\left\{\exp^{\int_{t_0}^t-\frac{d\zeta^i(s)}{ds} K_ids} \right\}e(t_0)$$
So we can find that the coordinate transformation of Fermi-Walker coordinates and inertial coordinat does not depend on the relative velocity $v(t)$, but also depend on the whole changing procesess of $v(t)$ from $v(t_0)$ to $v(t)$, I think this feature perhaps can be called path-dependent.
So I want to know if we using Fermi-Walker coordinates, can we say that the special relativity also can be applied to an arbitrary non-inertial system?

 

[jambaugh]

I haven't yet dug through the mathematical details however one can certainly describe accelerating observers and work with arbitrary non-inertial coordinate systems withing the flat space-time of special relativity. One simply works within the differential geometry of the constant Minkowski metric (when expressed in inertial coordinates) as expressed in whatever coordinate system you choose to utilize. Some additional comments:

• Be wary of describing extensive "rigid bodies" in special relativity. Rigidity is not a covariant property as for example extending a rotating rigid object far enough outward from its axis of rotation would result in FTL motion eventually. The central issue is that simultaneity is frame dependent so different observers will not agree when the components of a "rigid" object arrive at given spatial positions relative to each other.

• As for Wilson loops, I am not directly familiar with them but from the quick read on wikipedia that looks to be a simple extension of the concept of residues in complex analysis. One can view complex analysis as a U(1) gauge version of the more general idea. But this is distinct from say Berry phase and the other path dependent effects we are talking about here. The Wilson loop quantities are invariant under perturbations of the paths provided they retain their topological equivalence.

• I am also confused by your reference to Fermi-Walker coordinates and inertial coordinates in the same universe. In the absence of curvature Fermi-Walker coordinates should reduce to inertial coordinates and in the presence of curvature one can only define a set of coordinates as inertial at a given event point or at best along a given world line for a free falling observer.

I think much of my confusion here is my failure to understand your principle objective. Let us pick a concrete example, say relativistic precession akin to Thomas precession. Let me do some math and come back with a second post. [This may take some time]. But to start, let's consider a gyroscope in orbit around the origin in the x-y plane at a radius R using a non-gravitational central force (say a tether) so we can asset full validity of special relativity. Let the gyroscope as seen by the center frame inertial observer initially be at x=R, y=0 with angular momentum in the y direction. I will attempt to first describe the (special) relativistic precession and then construct the connection for a rotating cylindrical coordinate frame and set up the ordered parallel transport integral you are invoking.

 

[vanhees71]

Fermi-Walker coordinates are not necessarily inertial coordinates, because you can define them as a "non-rotating" vierbein along an arbitrary congruence of time-like curves (physically the worldlines of a fluid).

 

[Jianbing_Shao]

• Be wary of describing extensive "rigid bodies" in special relativity. Rigidity is not a covariant property as for example extending a rotating rigid object far enough outward from its axis of rotation would result in FTL motion eventually. The central issue is that simultaneity is frame dependent so different observers will not agree when the components of a "rigid" object arrive at given spatial positions relative to each other.

Here I talk about the "rigid bodies" only in the in the sense of classical mechanics, I want to show that if we take the non-commutative relation into account, then an arbitrary rotation of a "rigid body" should be described using a time-ordered operator.
In the past, if given two operators $A(t), B(t)$ who change with time, we always can get new operator $A(t)B(t)$ or $B(t)A(t)$, if we can introduct a time-ordered product of $A(t), B(t)$ just as we did in QFT：$\mathcal{T}[A(t), B(t)]$. Then we can get a more general product which the direct product $A(t)B(t)$ or $B(t)A(t)$ can be included.
For example, if $A(t)=e^{\int_0^t \omega^1(t)X_1dt}=e^{\theta^1(t)X_1}, B(t)=e^{\int_0^t \omega^2(t)X_2dt}=e^{\theta^2(t)X_2}, X_1,X_2\in \mathcal{so}(3)$. Then $\mathcal{T}[A(t), B(t)]=\mathcal{T}\{e^{\int_0^t (\omega^1(t)X_1+\omega^2(t)X_2)dt}\}$
So I want to find more applications of this comcept in physics.

As for Wilson loops, I am not directly familiar with them but from the quick read on wikipedia that looks to be a simple extension of the concept of residues in complex analysis. One can view complex analysis as a U(1) gauge version of the more general idea. But this is distinct from say Berry phase and the other path dependent effects we are talking about here. The Wilson loop quantities are invariant under perturbations of the paths provided they retain their topological equivalence.

For a Lie algebra valued connection one-form $A(x)=A_\mu^i(x)T_idx^\mu$, if there exsists a curve $C$(closed), the Wilson loop is defined as :$W[C]=\mathcal{P}\{e^{\int_CA_\mu^i(x)T_idx^\mu}\}$. $T_i$ can be the generators of an arbitrary real or complex Lie group.
$U(1)$ is an abelian Lie group. and the Wilson loop $W[C]$ is path dependent partly result from the non commutativity of the generators $T_i's$. So if $T_i\in\mathcal{u}(1)$, It is possible Wilsom loop $W[C]$ is not path dependent.

I am also confused by your reference to Fermi-Walker coordinates and inertial coordinates in the same universe. In the absence of curvature Fermi-Walker coordinates should reduce to inertial coordinates and in the presence of curvature one can only define a set of coordinates as inertial at a given event point or at best along a given world line for a free falling observer.

In the discussion of Thomas precession, if we do some research on a accelerating particle's instantaneous
rest frames, If at time $t$, The particle's instantaneous rest frame $x'^{(t)}$ can be regarded as inertial frame. it means:
$$x'^{(t)}=x\exp(-\zeta^i(t) K_i)$$
and we also know that the coordination between the two neighbouring instantaneous rest frames $x'^{(t+\Delta t)}$ and $x'^{(t)}$ should be a pure boost:
$$x'^{(t+\Delta t)}=x'^{(t)}\exp(-\Delta\zeta^iK_i)=x\exp(-\zeta^i(t) K_i)\exp(-\Delta\zeta^jK_j)$$
Now if the direction of $\zeta^i(t)$ and $\Delta\zeta^i$ is different, then from BCH-formula,
$$\exp(-\zeta^i(t) K_i)\exp(-\Delta\zeta^jK_j) \neq \exp(-\zeta^i(t) K_i-\Delta\zeta^jK_j)$$
It is not a pure boost, it contains rotation, so instantaneous rest frame $x'^{(t+\Delta t)}$ could not be an inertial frame. Also we can draw a conslusion the instantaneous rest frame should be described using ordered product of infinite infinitesimal pure boost.

Now if we know how to describe the instantaneous rest frames of a acclerating observer, then just like we did in classical mechanics, there we can describe a changing in a rotating system, here we can try to find how to describe the changing vector field in the acclerating system, it perhaps can be described using a ordered exponentials, then if the parallel transport of a vector on a manifold also can be expressed with a ordered exponentials, then perhaps we can find something in common between the two things.

 

[Jianbing_Shao]

Fermi-Walker coordinates are not necessarily inertial coordinates, because you can define them as a "non-rotating" vierbein along an arbitrary congruence of time-like curves (physically the worldlines of a fluid).

Thanks for your attention, I think it is because the product of two pure boosts without roataion is not always a pure boost,

 

[Jianbing_Shao]

i really thanks i look for this for a long itime

I also have been thinking about this problem for a long time, and I can't find any article about this problem, but I find a method called product integration in mathematics, It inspired me a lot . and I think it perhaps can be applied to the relativity, If you have interst in this problem. I sincerely hope I can communicate with you .

 

[Jianbing_Shao]

• As for Wilson loops, I am not directly familiar with them but from the quick read on wikipedia that looks to be a simple extension of the concept of residues in complex analysis. One can view complex analysis as a U(1) gauge version of the more general idea. But this is distinct from say Berry phase and the other path dependent effects we are talking about here. The Wilson loop quantities are invariant under perturbations of the paths provided they retain their topological equivalence.

I found a detailed description of Whison loop:

The gauge connection tells you what matrix you multiply by when you move in an infinitesimal direction $\delta x_\alpha$. The SU(2) matrix you rotate by is

$$ M^i_j =I + i A_{\alpha j}^i \delta x^\alpha$$

This is infinitesimally close to the identity, so the A part is in the Lie algebra of SU(2). The "i" is conventional in physics, to make the A matrix hermitian as opposed to anti-hermitian, as is the cleaner convention and the one used in mathematics. This means that A is a linear combination of Pauli matrices. This gives you a concrete representation of the gauge field (suppressing the i,j indices):

$$ A_\alpha = A_{\alpha k}\sigma^k $$

You assumed that the parallel transport is linear in the $\delta x$'s, this is so that the notion is compatible with the notion of space-time as a differential manifold--- if you double the displacement you double the infinitesimal rotation angle. You assume it's infinitesimal by physical continuity.

From this, it is obvious that the parallel transport along a curve is the product of A's along each of the infinitesimal segments that make up the curve:

$$ \prod (I+ A_\alpha^i T_i dx^\alpha) = \mathrm{Pexp}( \int A dx )$$

Where the path-ordered exponential is defined as the limit of the product on the left. This is the nonabelian generalization of the phase acquired by a charge particle in an electromagnetic field along a path.

The gauge field is then a map between curves and SU(N) matrices with the property that if you join paths end-to-end, the matrices multiply. The matrix associated to an infinitesimal closed loop is called the curvature, and it is proportional to the element of area enclosed in the loop. This is identical to general relativity.

In fact, an arbitrary connection $\Gamma^\mu_{\nu\rho}(x)$ can equivalently be described using $A_\mu^i(x)T_i, T_i\in \mathcal{gl}(n.R)$, and we will find that $\mathrm{Pexp}( \int A_\mu^i(x)T_i dx^\mu )\in GL(n.R)$
In general relativity, we demand that $\Gamma^\mu_{\nu\rho}(x)=\Gamma^\mu_{\rho\nu}(x)$, I don't know how this condition confine the choose of $T_i$. and which group $\mathrm{Pexp}( \int A_\mu^i(x)T_i dx^\mu )$ belongs to?