Insights How to Study Fermi-Walker Transport in Minkowski Spacetime

[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

 

[PeterDonis]

My point was that to understand the FWT concept, it is simpler to carry out the analysis in flat spacetime

Which, as has already been pointed out, is exactly what the article under discussion does. If other references try to start the analysis in curved spacetime, that is irrelevant for this discussion.

You have now been banned from further posting in this thread.

 

[cianfa72]

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).

Just to be complete: there is a third equation namely
$$ \dot X^t = X^t \partial_t V^t = X^t$$ for which $X^t=0$ identically is a solution.

Therefore the solution vector field $X$ 'evaluated' at the point $(0,0,0)$ matches the vector $X^x=1,X^t,X^y=0$ we started with. Hence the vector field $X$ (i.e. its components $X^t,X^x,X^y$) is function of $s$ coordinate only.

Is the above correct ? Thanks.

 

[Orodruin]

Just to be complete: there is a third equation namely
$$ \dot X^t = X^t \partial_t V^t = X^t$$ for which $X^t=0$ identically is a solution.

The component $V^t$ is equal to one. Therefore its derivative is zero.

 

[cianfa72]

The component $V^t$ is equal to one. Therefore its derivative is zero.

Yes sure, we get $\dot X^t = 0$ therefore $X=const=0$ identically. As said in #37 the solution vector field $X$ is a function of coordinate $t$ only (i.e. $s$ along the curve), right ?

 

[Orodruin]

Yes sure, we get $\dot X^t = 0$ therefore $X=const=0$ identically. As said in #37 the solution vector field $X$ is a function of coordinate $t$ only (i.e. $s$ along the curve), right ?

Technically $X$ is not a full field, but only defined along the integral curve, which makes it a function of the curve parameter such that $s\mapsto X(s) \in T_{\gamma(s)}M$, where $\gamma(s)$ is the point the parameter maps to along the curve.

 

[cianfa72]

Technically $X$ is not a full field, but only defined along the integral curve, which makes it a function of the curve parameter such that $s\mapsto X(s) \in T_{\gamma(s)}M$, where $\gamma(s)$ is the point the parameter maps to along the curve.

So basically the solution $X$ is not defined in a open neighborhood of each point along the integral curve. What has been done is to solve the three differential equations along the integral curve $t=0$.

 

[Orodruin]

So basically the solution $X$ is not defined in a open neighborhood of each point along the integral curve. What has been done is to solve the three differential equations along the integral curve $t=0$.

The integral curve is $x = y = 0$, $t$ is the only varying coordinate.

 

[cianfa72]

The integral curve is $x = y = 0$, $t$ is the only varying coordinate.

Oops, yes (i.e. let me say $t$ becomes the curve parameter $s$). Thank you.