Centrifugal and Coriolis forces covariant formulation

[PeterDonis]

The expression for the centrifugal force, as defined in the paper, in the case of the rigidly rotating frame yields something non-vanishing that doesn't change the radius of a freely falling particle at rest with respect to the background inertial observers (hence in circular orbit in the rigidly rotating frame), just like in the Newtonian case.

Are you sure? The rigidly rotating congruence has nonzero proper acceleration, unlike the inertial congruence with rotating tetrads. The motion of a freely falling particle in the rigidly rotating frame is produced by the proper acceleration, not the centrifugal force, at least as I'm intuitively thinking about them. Doesn't the gravitoelectric force G include a proper acceleration term as well as a centrifugal force term?

 

[WannabeNewton]

The paper defines the gravitoelectric force as $\vec{G} = -\vec{a}$ where $\vec{a}$ is the proper acceleration of the rigidly rotating congruence. It then says (see p.24: http://arxiv.org/pdf/1207.0465v2.pdf) that $\vec{G} = \vec{\omega}\times(\vec{r}\times \vec{\omega})$ where $\vec{\omega}$ is the vorticity of the rigidly rotating congruence (this expression still doesn't make sense to me for the reasons discussed prior but that's just an aside). The paper then calls this the centrifugal force (same page) and as you can see it does not necessarily change the radius (we can have an inertial observer who in the rigidly rotating frame undergoes uniform circular motion maintained by the counter-acting centrifugal and Coriolis forces). The expression for the equation of motion of a freely falling particle, relative to the rigidly rotating frame, is given on p.25, c.f. eq (60).

Thanks.

 

[PeterDonis]

as you can see it does not necessarily change the radius (we can have an inertial observer who in the rigidly rotating frame undergoes uniform circular motion maintained by the counter-acting centrifugal and Coriolis forces).

But that means the centrifugal force alone *would* change the radius; it's only the fact that it's counteracted by another force that keeps the radius from changing. At least, that's the way I would interpret what I think the paper is trying to say; but it's possible that I'm misinterpreting it, or that the paper itself is not entirely consistent in its interpretation.

 

[WannabeNewton]

But that means the centrifugal force alone *would* change the radius; it's only the fact that it's counteracted by another force that keeps the radius from changing.

I see what you're saying now but I still cannot get an intuition for why there is no centrifugal force in the aforementioned example. Perhaps what I describe below will shed light on why I'm confused as to why the centrifugal force vanishes. The problem is, I can't for the life of me visualize inertial motion in the extended reference frame defined by the congruence of inertial observers and their uniformly rotating spatial axes. Say I choose any inertial observer from this congruence. If I try to visualize the motion of a freely falling particle relative to this extended reference frame then what I end up visualizing is motion through the rigidly rotating frame centered on the chosen observer instead. My problem again is that the frame does not rotate rigidly (one cannot have a rigidly rotating extended frame of inertial observers of course but that's besides the point) and this makes it hard for me to visualize free fall kinematics relative to this extended frame.

For example, take a particle that is at rest relative to the inertial observers in the above congruence. According to the paper's definitions, there is no gravitoelectric (centrifugal) field in this frame and there is a gravitomagnetic (Coriolis) field arising from the Fermi-Walker rotation $\vec{\Omega}$ of the spatial axes of each inertial observer in the congruence. This in turn gives rise to a Coriolis force acting on inertial particles given by $\vec{F}_{c} = \gamma \vec{U}\times \vec{\Omega}$. For the particle at rest relative to the congruence, $\vec{U} = 0$ and so there is neither a Centrifugal nor a Coriolis force acting on this particle in this extended reference frame-it simply hovers in place in this frame. For future clarity I will denote this congruence by $\xi^{\mu}$ as before.

On the other hand, consider an extended rigidly rotating frame centered on an inertial observer; when I say extended I mean as usual that the central inertial observer carries a set of spatial axes which rotate with angular velocity $\vec{\Omega}$ and we have a congruence of observers circulating this inertial observer with angular velocity $\vec{\Omega}$ each carrying a set of spatial axes rigidly locked onto neighboring spatial axes in the congruence through Lie transport. This setup then allows us to extend the uniformly rotating spatial axes of the inertial observer into a rigidly rotating grid extended globally up to the light barrier. Intuitively, this is what I picture when I imagine a frame corotating with the spatial axes of the inertial observer. Now, a particle at rest relative to this inertial observer will execute uniform circular motion through the extended rigidly rotating frame. I will denote this congruence by $\eta^{\mu}$ also as before.

Now if we take any inertial observer in $\xi^{\mu}$, the observer's spatial axes rotate in exactly the same was as those of the central inertial observer in $\eta^{\mu}$. However if we consider an inertial particle infinitesimally separated from the observer at rest with respect to the respective congruence, then in the case of $\xi^{\mu}$ it just hovers in place with respect to the associated extended reference frame whereas in the case of $\eta^{\mu}$ it moves in a circle through the associated extended reference frame; in the latter case I can easily picture the particle as moving in a circle through the extended rigid grid of points which rotate relative to the local compass of inertia at the rate $\omega^{\mu} = \epsilon^{\mu\nu\alpha\beta}\eta_{\nu}\nabla_{\alpha}\eta_{\beta}$ but I cannot properly visualize the hovering of the particle in the former case because the extended frame in that case isn't rigidly rotating.

Now I can understand that since there is no preferred origin in $\xi^{\mu}$, the particle won't know what to circle about and so just sits there whereas in the case of $\eta^{\mu}$ we've clearly defined a preferred origin by singling out the central inertial observer. But as noted the chosen inertial observer from $\xi^{\mu}$ has a set of spatial axes that rotate in the exact same way as the central inertial observer in $\eta^{\mu}$ and we've considered an inertial particle infinitesimally separated from the observer so is there a better way to understand intuitively and visualize why the particle just hovers in place in $\xi^{\mu}$ as opposed to circling around, without appealing to the lack of a preferring origin?

Thanks for the help!

 

[PeterDonis]

I can't for the life of me visualize inertial motion in the extended reference frame defined by the congruence of inertial observers and their uniformly rotating spatial axes.

The problem may be that you are assuming that there *is* such a thing as "the extended reference frame" in question. I don't think there is, because each individual inertial observer in the congruence has rotating spatial axes, so the spatial axes of different inertial observers don't stay aligned. In other words, the uniformly rotating spatial axes are *not* the same as Lie transported spatial axes that are fixed with respect to neighboring members of the $\xi^{\mu}$ congruence. (Note that the two *are* the same for the $\eta^{\mu}$ congruence, the usual "rigidly rotating" congruence.) So you can't use the uniformly rotating spatial axes to define an extended reference frame in which members of the $\xi^{\mu}$ congruence are at rest; the spatial axes of such an extended frame would have to stay aligned with respect to Lie transport, and the uniformly rotating spatial axes of each individual member of the congruence don't do that.

 

[WannabeNewton]

My intuitions agree with everything you've said, Peter, but the paper constantly refers to the entire congruence (in particular the congruence of inertial observers with rotating axes) as a reference frame so I didn't know how else to interpret it.

See, for example, Appendix A of p.57: http://arxiv.org/pdf/1207.0465v2.pdf

 

[PeterDonis]

the paper constantly refers to the entire congruence (in particular the congruence of inertial observers with rotating axes) as a reference frame

They use the term "frame", yes, but I'm not sure they mean by it what we normally think of as a "reference frame". The normal usage of the latter term assumes a correspondence between a frame field on a region of spacetime and a coordinate chart on the same region, but that's what I don't think exists for the "frame" defined by a congruence of inertial observers with uniformly rotating spatial axes. As far as I can tell, such a correspondence is not necessary for the conclusions of the paper to be true; but the paper's usage of the term "frame" does appear to me to be misleading.