Centrifugal and Coriolis forces covariant formulation

[WannabeNewton]

Are you sure? Shouldn't the factor in front be $\gamma$, not $\gamma^2$?

According to http://en.wikipedia.org/wiki/Born_coordinates#Langevin_observers_in_the_cylindrical_chart it looks to be $\gamma^2 \omega_0$

Also, the sign should be negative because $\vec{\omega}\times (\vec{\omega}\times \vec{r})$ should point radially inward; $\vec{\omega} \times \vec{r}$ points in the $\vec{\phi}$ direction (i.e., tangentially in the direction of rotation), and $\vec{\omega} \times \vec{\phi}$ points in the minus $\vec{r}$ direction.

Sorry yes wherever you see $\vec{\omega}\times (\vec{\omega}\times \vec{r})$ it should be $\vec{\omega}\times (\vec{r} \times \vec{\omega})$, that was a careless mistake on my part.

 

[PeterDonis]

According to http://en.wikipedia.org/wiki/Born_coordinates#Langevin_observers_in_the_cylindrical_chart it looks to be $\gamma^2 \omega_0$

Ah, yes, that's right, the "local" vorticity is $\gamma^2 \omega_0$; we get $\gamma \omega_0$ only if we apply a factor of $\gamma^{-1}$ to get the vorticity "relative to infinity".

I wonder if the paper you linked to is implicitly defining $\vec{\omega}$ to be "relative to infinity" instead of "local". I'm going to have to take more time to read through it and check various statements as I go.

Sorry yes wherever you see $\vec{\omega}\times (\vec{\omega}\times \vec{r})$ it should be $\vec{\omega}\times (\vec{r} \times \vec{\omega})$

Yes, I see that that's how the paper defines it, so the signs are OK.

 

[WannabeNewton]

So Peter I'm still quite confused, the same questions are lingering, particularly (4). With regards to (4), I just can't get an intuition for why the centrifugal force would vanish. So say we choose some congruence of inertial observers following orbits of the flat space-time time-like Killing field $\xi^{\mu}$. Then we know the acceleration $a^{\mu} = 0$ and the vorticity $\xi_{[\gamma}\nabla_{\mu}\xi_{\nu]} = 0$. We equip each inertial observer with a Lorentz frame $\{e_{\alpha}\}$ such that the spatial axes of each Lorentz frame rotate relative to local gyroscopes with an arbitrary but constant angular velocity $\Omega^{\mu}$. This defines our extended reference frame.

Now this certainly contrasts with what the paper calls the rigidly rotating frame, which consists of the congruence of observers following orbits of $\eta^{\mu} = \xi^{\mu} + \omega \psi^{\mu}$ each carrying a Lorentz frame $\{p_{\alpha}\}$ as in: http://en.wikipedia.org/wiki/Born_coordinates#Langevin_observers_in_the_cylindrical_chart and as we know $\mathcal{L}_{\eta}p_{\alpha} = 0$ so the $p_{i}$ are rigidly anchored to the circulating observers of $\eta^{\mu}$ (hence the characterization "rigidly rotating frame"); as we know the local frame rotation in this case is directly given by $\eta_{[\gamma}\nabla_{\mu}\eta_{\nu]}$. The acceleration $a^{\mu}$ of the circulating observers gives rise to a centrifugal force, also according to the paper, of the form $\vec{F} = \vec{\omega}\times(\vec{r}\times \vec{\omega})$. Now this takes advantage of the fact that there exists a preferred origin for this circulating congruence because we employ a radial position vector $\vec{r}$; the preferred origin is (up to a $z$ translation) the point lying on the axis at which $\psi^{\mu} = 0$. The example of the rigidly rotating frame corresponds directly (up to time dilation factors) to the standard example of a uniformly rotating frame in Newtonian mechanics wherein the centrifugal force takes the form $\vec{F} = \vec{\Omega}\times(\vec{r}\times \vec{\Omega})$ where again we make use of a preferred origin in order to define $\vec{r}$ in conformity with our usual intuitions about the centrifugal force.

However, coming back to the example of the congruence of inertial observers each with an identically rotating Lorentz frame, there is no preferred origin we can pick. There is a preferred axis but since we just have a bunch of observers hovering in free space, each carrying an identically rotating tetrad, there is no possible origin to choose over another. This is why I'm having trouble understand intuitively why the centrifugal force vanishes in this example. Clearly the extended reference frame so defined fails to be rigidly rotating since the spatial axes of neighboring observers in this congruence are not locked onto one another so there doesn't seem to be any connection to the usual Newtonian notion of centrifugal force which is defined for rigidly rotating frames. Furthermore there is, as already noted, no preferred origin in this example hence no meaningful notion of a radial position vector $\vec{r}$. Say I choose an arbitrary observer $\mathcal{O}$ from this congruence as a reference observer and define the origin $O$ of the extended reference frame as the location of $\mathcal{O}$. The main problem is I can't actually visualize an arbitrary inertial motion through this extended reference frame because any inertial motion through this extended frame that I do visualize, in order to see physically if there is a centrifugal force, instead ends up being motion through the rigidly rotating frame centered on $O$, congruent to my Newtonian intuition of centrifugal force and more generally any kind of motion relative to a coordinate system (or extended reference frame) as specifically applying only to rigid ones.

So how does one visualize and interpret the vanishing of the centrifugal force in the above example? The paper provides some illustrations in the Appendix (wherein these examples are discussed) but the illustrations don't help me in the least bit.

 

[PeterDonis]

With regards to (4), I just can't get an intuition for why the centrifugal force would vanish.

Consider the simplest case, a freely falling particle that is at rest relative to the observer whose axes are rotating with angular velocity $\Omega$. In the rotating frame described by these axes, the freely falling particle is orbiting with angular velocity $- \Omega$ at a constant radius. But in the nomenclature of the paper, "centrifugal force" is a force that changes the particle's radius; it accelerates it (in the coordinate sense) "outward" (increasing radius) with respect to the observer. So there can't be any centrifugal force for this case.

(The case of a freely falling particle moving at a constant speed relative to the observer with rotating axes is just an obvious extension of the above.)

 

[WannabeNewton]

Consider the simplest case, a freely falling particle that is at rest relative to the observer whose axes are rotating with angular velocity $\Omega$. In the rotating frame described by these axes, the freely falling particle is orbiting with angular velocity $- \Omega$ at a constant radius. But in the nomenclature of the paper, "centrifugal force" is a force that changes the particle's radius; it accelerates it (in the coordinate sense) "outward" (increasing radius) with respect to the observer. So there can't be any centrifugal force for this case.

Wait why does the centrifugal force, as defined in the paper, have to change the particle's radius? 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.

I'm having trouble intuitively differentiating the rigidly rotating congruence and the congruence of inertial observers with rotating tetrads.
Thanks.

 

[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.