How to Study Fermi-Walker Transport in Minkowski Spacetime

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PeterDonis
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Greg Bernhardt submitted a new blog post

How to Study Fermi-Walker Transport in Minkowski Spacetime
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  • #2
Orodruin
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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.
 
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PeterDonis
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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.
 
  • #4
Orodruin
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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 :rolleyes:) 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.
 
  • #5
PeterDonis
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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##?
 
  • #6
Orodruin
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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 ...
 
  • #7
PeterDonis
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you need the third direction to demonstrate how vectors orthogonal to both velocity and acceleration do not rotate

Ah, got it.
 
  • #8
Orodruin
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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).
 
  • #9
vanhees71
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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.
 
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