# Centrifugal and Coriolis forces covariant formulation

1. Feb 17, 2014

### WannabeNewton

Hi guys. I was reading a paper in which a calculation was done to show that in Schwarzschild space-time, if we consider a time-like circular orbit 4-velocity $u^{\mu} = \gamma(\xi^{\mu} + \omega \eta^{\mu})$ where $\xi^{\mu}$ is the time-like Killing field, $\eta^{\mu}$ is the axial Killing field, and $\omega$ the angular velocity of the circular orbit, then the vorticity $u_{[\gamma}\nabla_{\nu}u_{\nu]} = 0$ for all possible values of $\omega$ if the orbit lies on the photon radius $r = 3M$. Then the paper directed me to the following short paper, http://www.dmf.unisalento.it/~giordano/allow_listing/AJP000936.pdf, which explains intuitively how for a circular orbit with the above $u^{\mu}$ at the photon radius, the "centrifugal force" on a spacecraft in the orbit vanishes for all values of $\omega$.

The reason I put centrifugal force in scare quotes is the paper never actually defines what it means by "the centrifugal force on the spacecraft" in a circular orbit of any allowed angular velocity $\omega$ at $r = 3M$! The centrifugal force, both in GR and in Newtonian mechanics, only arises in rotating frames wherein the rotation is relative to local gyroscopes. Therefore what rotating frame is the paper referring to when it talks about "the centrifugal force on the spacecraft"? If we go to the rest frame of a spacecraft in circular orbit at the photon radius then from the first paper we know that $u_{[\gamma}\nabla_{\nu}u_{\nu]} = 0$ for all possible values of $\omega$ and since $u^{\mu}$ follows an orbit of the Killing field $\xi^{\mu} + \omega \eta^{\mu}$ (taking $\omega$ to be constant throughout space-time) this implies that the rest frame of the spacecraft is non-rotating for all values of $\omega$. Hence there would be no centrifugal forces in the rest frame of the spacecraft for all allowed angular velocities at the photon orbit. Is this what the paper means when it claims that "the centrifugal force on the spacecraft" vanishes for all angular velocities at the photon orbit?

I would like to think so but then I came upon the following paper, http://arxiv.org/pdf/gr-qc/9808036v1.pdf, which managed to confuse me beyond repair. The relevant part of the paper, as far as this thread is concerned, is section 3. The paper attempts to give covariant formulations of the familiar inertial forces from Newtonian mechanics, particularly the centrifugal and Coriolis forces. Immediately however I don't get the basis of the covariant formulation because it's done relative to an irrotational congruence. If the congruence is irrotational then why would there be centrifugal or Coriolis forces arising relative to it? As is well known, these inertial forces only arise in rotating reference frames.

For example, let's apply the formula (40) for the centrifugal force relative to the irrotational time-like Killing field of a static space-time, in particular flat space-time. More precisely, let the static space-time be flat space-time and consider a rigidly rotating disk relative to a global inertial frame with origin at the center of the disk. Then the irrotational time-like Killing field $\xi^{\mu}$ corresponds to orbits of inertial observers at rest in this global inertial frame. The 4-velocity field of observers at rest on the disk is given as usual by $u^{\mu} = \gamma (\xi^{\mu} + \omega \eta^{\mu})$. Then $e^{\psi} = \gamma$ and $e^{\alpha} = r$ hence applying (40) we get $Z = -\frac{1}{2}\gamma^2 \omega^2 r \partial_r$ but this is not the centrifugal force on the observers at rest on the disk. Rather it's the centripetal force acting on said observers. Obviously it would make no sense for there to be a centrifugal force because we've done our calculation relative to an irrotational congruence and in particular in a non-rotating global inertial frame in flat space-time.

In Schwarzschild space-time the irrotational time-like Killing field $\xi^{\mu}$ corresponds to the integral curves of the observers at rest in the gravitational field. The rest frames of these observers are non-rotating. Therefore there should be no centrifugal forces arising in these rest frames. If we applied (40) relative to $\xi^{\mu}$ in Schwarzschild space-time for a 4-velocity $u^{\mu} = \gamma(\xi^{\mu} + \omega \eta^{\mu})$ correspond to a circular orbit then presumably $Z$ would again resemble the centripetal force on this orbit. So what gives? Why is this paper calling this the centrifugal force? And on that note why would the above example in flat space-time yield a centrifugal force when it clearly gives the centripetal force instead?

What's really confusing to me is in section 5.1 of the paper (bottom of p.12), it is proven that $Z$ vanishes for any and all circular orbits at the photon radius in Schwarzschild space-time but as discussed directly above since $Z$ is defined relative to the non-rotating observers at rest in the gravitational field, it shouldn't correspond to a centrifugal force since the centrifugal force only shows up in rotating frames. If like in the flat space-time case $Z$ corresponds rather to a centripetal force then I'm having trouble seeing intuitively why $Z$ would vanish for circular orbits at the photon radius since these orbits most certainly have a centripetal acceleration!

2. Feb 17, 2014

### Staff: Mentor

The term "centrifugal force" is often used sloppily. I posted a formula on my PF blog for the proper acceleration of an arbitrary congruence of the kind you're talking about, with some discussion of three regimes with different behavior (the photon orbit at $r = 3M$ being one of the three--the others are, as you might guess, $r > 3M$ and $r < 3M$), and how it relates to a "centrifugal force" interpretation. See here:

https://www.physicsforums.com/blog.php?b=4327 [Broken]

I'll look through the papers you posted to see how their interpretation relates to the one in my blog post.

Last edited by a moderator: May 6, 2017
3. Feb 17, 2014

### Staff: Mentor

I think so, yes (and I adopted basically the same interpretation in my blog post).

Is it? The general congruence they are working with has a nonzero angular velocity, which in static (i.e., Schwarzschild) spacetime corresponds to nonzero vorticity (except at the photon orbit). The only special case where the congruence would be irrotational at any radius would be the case with zero angular velocity, but in this case their $Z$ goes to zero so indeed there is no centrifugal force relative to this irrotational congruence. See further comments below.

Yes, it is--more precisely, it is in the natural terminology of the observers on the disk. It wouldn't be *called* "centrifugal force" by inertial observers, but it would by the observers at rest on the rotating disk. So this is just a matter of terminology; the paper is using the natural terminology of the actual observers it's talking about in each particular case, which means the natural terminology of observers following the congruence it's talking about in each particular case.

Perhaps that isn't the same congruence that things are expressed "with reference to", but that seems like a non-issue to me. Physically, observers at rest on the disk follow a congruence with nonzero vorticity, hence it seems perfectly reasonable to say that they measure a nonzero centrifugal force. In any case, that's how the paper appears to me to be using terms, so there's no disagreement about the physics, only (possibly) about the terms being used to describe it.

4. Feb 17, 2014

### WannabeNewton

Thanks! I'll take a look through it.

On p.6 of the paper (very start of section 3) they define the decomposition into centrifugal, gravitational, and Coriolis forces relative to a globally hypersurface orthogonal unit time-like vector field $n^{\mu}$ which is equivalent to $n^{\mu}$ being irrotational. See also expression (30) and the paragraph preceding it.

Perhaps you can clarify something for me then because it's not clear as of yet, at least to me. The paper defines relative to an irrotational congruence the decomposition into the Coriolis and centrifugal forces associated with a circular orbit. So again taking the example of Schwarzschild space-time this would be the decomposition into Coriolis and centrifugal forces associated with a circular orbit relative to the congruence of static observers. Does this mean the Coriolis and centrifugal forces are observed in the rest frames of these static observers, which makes no sense to me since these frames are non-rotating, or are they as observed in the rest frames of the observers in the circular orbits? It's not at all clear to me from the paper's definitions although only the latter really makes sense as per the usual definitions of centrifugal and Coriolis forces. It would also harmonize with the fact that the observers in circular orbit at the photon radius have non-rotating rest frames for all angular velocities, due to vanishing vorticity at all angular velocities, and hence would observer no centrifugal or Coriolis forces in their rest frames.

Also, is the Coriolis force as defined in the paper the same as the Newtonian Coriolis force? According to the paper the Coriolis force vanishes identically in static space-times with regards to circular orbits but certainly we can observe Coriolis forces in the rest frames of observers in circular orbits in Schwarzschild space-time if the orbits are not at the photon radius as these frames are rotating, at least if we interpret Coriolis force in the Newtonian sense.

Thanks!

Last edited by a moderator: May 6, 2017
5. Feb 18, 2014

### yuiop

Recall the comment in the Malament paper, where an object following a photon path is effectively travelling in a straight line and thus the vorticity is zero for any object following the photon orbit at any speed. While the object is effectively travelling in a straight line as far as gyroscopes are concerned, the proper centrifugal force does not necessarily vanish, as discussed below.

I have not read the paper yet, but I can say that the "centrifugal force" does not vanish at the photon orbit. Take a look at this equation given by Peter in his blog:

$a = \frac{M}{r^2 \sqrt{1 - 2M/r}} \left[ \frac{1 - \left( \frac{r}{M} - 2 \right) v^2}{(1 - v^2)} \right]$

This is not the centrifugal force, it is the total proper force (per unit test mass) acting on the orbiting object. The total force can be logically broken down into gravitational and centrifugal force components by re-writing the equation (with a sign reversal and inserting test mass m) like this:

$f_{total} = f_{grav}+f_{centrifugal}$

$f_{total} = \left(-\frac{Mm}{r^2} \right)\frac{\gamma}{(1-v^2) } + \left(\frac{mv^2}{r}\right)\frac{\gamma^{-1}}{(1 - v^2)}$

where $\gamma = 1/\sqrt{1 - 2M/r}$. Note that when $v=0$ we are left with only the gravitational force but also note that gravitational force is also a function of v. The signs are such that a negative value means the force is acting inwards and vice versa.

As Peter has demonstrated in his blog, for a geodesic orbit, $f_{total} =0$, but it is easy to see that the centrifugal force has not vanished, but is equal in magnitude to the gravitational force, but acting in the opposite direction. This observation manifests itself as a tidal force detected in a finite sized orbiting object, away from the centre of mass.

When $r=3M$ the total proper force detected by the object is $\frac{m\sqrt{3}}{9M}$ which is independent of v, but not zero. The centrifugal force at $r=3M$ is also not zero but is dependent on v as is the gravitational force acting in the opposite direction, which together result in the velocity independent total proper force of $\frac{m\sqrt{3}}{9M}$. At orbits of $r<3M$ the centrifugal force is not actually negative, it continues to act outwards but since the gravitational force is increased by the gravitational gamma factor and the centrifugal force is decreased by the same factor, the centrifugal force is overwhelmed by the rapidly increasing gravitational force which also increases with increasing tangential velocity.

Just for clarity, the total force is the proper force measured by an accelerometer onboard the orbiting satellite and acts along a line parallel to an instantaneous radial vector pointing towards the centre of the orbit. If the satellite has intrinsic spin relative to the orthonormal basis vectors then the total force will not be acting a constant direction according to an observer at rest with and co spinning with the satellite. Come to think of it, this is one of many ways of defining the intrinsic spin of the orbiting satellite.

The velocity dependence of the force of gravity was discussed in a very old thread in the context of a linearly moving object and also by Matsas in the context of the forces acting on a submarine.

Consider a satellite in a "forced orbit" that is orbiting much faster than the geodesic orbital angular velocity, so it requires rocket thrusters pointing outwards to provide the required inward centripetal force to keep the satellite in circular orbit. This satellite experiences an outward proper centrifugal force. Now if we adjust the intrinsic spin of the satellite in such a way that it is a rest with local gyroscopes, it continues to feel an outward centrifugal force due to its orbital angular velocity, therefore it cannot be true that centrifugal force only arises in frames rotating relative to local gyroscopes. Perhaps there is confusion between centrifugal forces experienced at the extremities of the satellite due to any intrinsic spin and centrifugal force acting on the centre of mass of the satellite due to its orbital angular velocity.

The centrifugal force that arises due to the intrinsic spin of the object, wherein there is rotation relative to local gyroscopes, is zero at the centre of mass of the object (but non zero elsewhere), while the centrifugal force due to the orbital velocity of the object is non zero at the centre of mass of the object. For an object orbiting at $r=3M$, gyroscopes at rest with the orthonormal basis vectors do not precess, but gyroscopes not at rest with the othornormal basis vectors would precess and if the satellite had intrinsic spin relative to the orthonormal basis vectors would experience centrifugal forces at the extremities of the satellite even at $r=3M$.

What is your understanding of what they mean by irrotational? As I understand the term in the Newtonian fluid dynamics context, irrotational means that a vector at rest wrt the congruence will always point at a fixed distant star and that globally such a congruence would have an angular velocity proportional to $1/r$. Such a congruence does not have to have zero orbital angular velocity, so we can choose an irrotational congruence that is co-orbiting with the object under investigation.

I am beginning to suspect that the paper is using a non standard definition of "irrotational" to mean not rotating relative to orthonormal basis vectors that are at rest with the instantaneous radial vector and instantaneous tangential vector of the the orbit.

From other threads, we know that what does vanish at $r=3M$ is the gyroscopic precession relative to orthonormal basis vectors, where one of the axes always points to the centre of rotation. To understand the centrifugal aspect in the context of the paper you refer to, we have to assume they mean the forces that can be considered a function of velocity. For this interpretation we have to pretend that the total force acting on an orbiting object is the product of $m*a_{grav}*a_{centrifugal}$ rather than the sum $f_{grav}+f_{centrifugal}$ as I described above. By this interpretation pretend the "gravitational acceleration" is:

$a_{grav*} = -\frac{M}{r^2 \sqrt{1 - 2M/r}}$

and independent of v and pretend the "centrifugal acceleration" is:

$a_{centrifugal*} = \frac{(1 - \left( r/M - 2 \right) v^2)}{(1 - v^2)}$

At $r=3M$ the "centrifugal acceleration" has not vanished, but has become independent of v and here $a_{centrifugal*} =1$. For $r<3M$ the "centrifugal acceleration" factor is always positive so the resulting total force is always negative and inwards, while for $r>3M$ the "centrifugal acceleration" factor can take either sign. In this sense centrifugal acceleration reverses at the photon orbit, but I do not feel this is an entirely satisfactory way to define centrifugal acceleration.

Last edited: Feb 18, 2014
6. Feb 18, 2014

### Staff: Mentor

No, we don't; we get that $a$ is independent of $v$, because $\left( r / M - 2 \right) = 1$ so the numerator of the second factor is $1 - v^2$ and cancels the denominator, leaving $a = M / \left( r^2 \sqrt{1 - 2M/ r} \right)$.

Correct. However, I'm not sure about the physical meaning of the split you give into $f_{grav}$ and $f_{centrifugal}$. I think, as I said in my previous post, that this is largely a matter of terminology, how we interpret the term "centrifugal force".

7. Feb 18, 2014

### Staff: Mentor

No, of course not. They are observed by observers following orbits of the 4-velocity field $u^a$, not orbits of the irrotational timelike vector field $n^a$. That's obvious from equation (26) in the paper; the centrifugal and Coriolis forces are only nonzero if $v$ is nonzero, and $v$ is zero for orbits of $n^a$. (But see further comments below on the Coriolis force.)

Again, I think this is a matter of terminology, not physics. You don't like using the term "centrifugal force" unless we write everything in a coordinate chart in which orbits of the actual 4-velocity field $u^a$ are integral curves of the time coordinate. The authors of the paper don't share that dislike, apparently. But everyone agrees on what observers following particular timelike worldlines will observe.

I'm not sure; the paper uses the term "Coriolis-Lense-Thirring force", and says it's identically zero in static spacetime; the Lense-Thirring precession is indeed identically zero in Schwarzschild spacetime, but as you say, the Newtonian Coriolis force certainly isn't. I'll have to think some more about this one.

8. Feb 18, 2014

### WannabeNewton

Right but that's the same reasoning the paper gives as to why the "centrifugal force" also vanishes at the photon orbit for circularly orbiting spaceships there.

I have to disagree there. The rest frame of a spaceship in circular orbit at the photon orbit is non-rotating for any angular velocity so I see no reason why there should be centrifugal forces in this frame. The papers considered are treating these essentially as point particles by the way.

Sure but there are no time-like circular geodesics at the photon orbit so I don't exactly see the relevance.

That's a completely different scenario involving the Lorentz transformation of the gravitational field. There is no such Lorentz transformation involved here, the decomposition is relative to the global rest frame of the static observers i.e. the "absolute Newtonian" rest frame.

What definition of centrifugal force are you using? Perhaps we should clarify that first. The standard definition of the inertial force termed "centrifugal force" is one that arises in rotating frames. Go back to Newtonian mechanics for a second. If you have a particle in uniform circular motion then in the inertial frame centered on the motion there is a centripetal force acting on the particle. If we instead go to the rotating frame centered on the motion in which the particle is at rest, there is a centrifugal force acting on the particle whose magnitude is the same as the centripetal force in the inertial frame but the direction is outwards from the rotation axis. But there is no centrifugal force in the inertial frame-it only arises in rotating frames! That's just bog-standard Newtonian mechanics and it's no different in GR. Mathematically the centrifugal force acting on a particle at a position $\vec{x}$ in a frame is given by $\vec{a}_c = -\vec{\Omega}\times (\vec{\Omega}\times \vec{x})$ where $\vec{\Omega}$ is the rotation of the frame. This Newtonian formula holds equally well in GR.

If the first paper, when it says "centrifugal forces on the spacecraft", means the centrifugal forces arising due to the rotation of the spacecraft rest frame then this clearly vanishes at the photon orbit for reasons already mentioned. The second paper linked in post #1 explicitly defines what it means by gravitational and centrifugal forces and gives formulas for both in the case of static space-times, which includes Schwarzschild space-time. The gravitational force is, as expected, just the gradient of the gravitational potential hence independent of velocity as always. The centrifugal force, in static space-times, is defined by the paper to be proportional to $\nabla_{\mu}(\frac{\eta_{\mu}\eta^{\mu}}{\xi_{\mu}\xi^{\mu}})$ and this certainly does vanish at the photon orbit as shown in proposition 3.2.5. of Malament's text. Said proposition also shows that this is equivalent to non-rotation according to local gyroscopes or a local compass of inertia so this would seem to be a characterization of vanishing centrifugal forces in the rest frame of the spacecraft.

This is entirely equivalent to the usual definition of irrotational in the sense of vanishing vorticity if the time-like tangent vector field involved is a time-like Killing field which for circular orbits in Schwarzschild space-time is certainly the case as it is just a linear combination of the static and axial Killing fields.

Thanks for the reply and help!

EDIT: By the way yuiop before you reply to my comments above take a look at my clarifications in post #10 so that my comments above can be answered by you with a clearer context in mind. I apologize for the lack of lucidity prior.

Last edited: Feb 18, 2014
9. Feb 18, 2014

### yuiop

Thanks for the correction. I am not sure what I was thinking of there. Possibly I was thinking of the total force for a geodesic orbit which is zero. Anyway, I have deleted that glaring error out of my post. The total acceleration when $r=3M$ when all the substitutions are done is $1/(\sqrt{27}M)$ as mentioned later in the post.

I cannot think of any other way to sensibly make a split. Certainly it seems absurd to call $a_total$ the centrifugal acceleration because for a stationary observer in the Schwarzschild metric, the centrifugal force would be non zero. I suppose it worth mentioning that the only force we can directly measure in a gravitational field is the total force and the centrifugal force (however we define it) can only be calculated with knowledge of the tangential velocity. I guess there is a reluctance on the part of some of the authors to consider gravitational acceleration to be a function of velocity. However, Matsas derived an expression for the gravitational acceleration which is a function of velocity where he deliberately uses Rindler coordinates to eliminate centrifugal effects and he obtains:

$f_{grav} = \frac{-mg}{(1-v^2)}$

where the velocity dependent factor $1/(1-v^2)$ of the gravitational acceleration is exactly as given in the split of your equation in the previous post. See equation 17 of http://arxiv.org/pdf/gr-qc/0305106v1.pdf

Here is a question based on the mention of conserving fuel in your blog. Consider two rockets with identical fuel loads both at $r=3M$, but one is hovering and the other is orbiting at 0.8c. Which runs out of fuel first? they both experience the same proper acceleration but time passes differently for them in global coordinate terms.

10. Feb 18, 2014

### WannabeNewton

Sorry let me explain my confusion with more clarity as I have not conveyed it very well thus far. Take again the example of the rotating disk in flat space-time. In the inertial frame centered on the disk, there is a centripetal force acting on the observers circling around with the disk given by $\vec{a} = -\gamma^2 \omega^2 r \partial_r$. Now say we go to the frame, still centered on the disk, but rotating with the angular velocity $\omega$ of the disk. In this frame the observers on the disk are all at rest. As observed in this frame there should be a centrifugal force on these observers. Obviously in this frame we still have $\vec{a} = -\gamma^2 \omega^2 r \partial_r$ since there is no change in the radial coordinate when going from the inertial frame to the corotating frame but this is not the centrifugal acceleration in the Newtonian sense-it's still just the centripetal acceleration as measured in the original inertial frame. The paper however calls this the centrifugal acceleration which doesn't make sense to me-for starters the centrifugal acceleration on the observers as measured in the corotating frame should be outwards. So how does one extract the centrifugal force on these observers in the corotating frame akin to the Newtonian definition of centrifugal force which is $\vec{a}_c = -\vec{\omega}\times (\vec{\omega}\times \vec{x})$ where $\vec{x}$ is the position vector to the observer from the origin of the corotating frame?

Furthermore there are two different rotating objects at play here. There's the rotating frame centered at the disk relative to which the observers on the disk are all at rest and have exerted on them a centrifugal force as seen from this frame. There's also the rest frame of any given observer on the disk. This too rotates because of the Thomas precession. The observer to whom the rest frame belongs attributes a centrifugal force to all particles not at the origin of this rest frame.

Coming back to the case of spacecrafts in circular orbit at the photon radius in Schwarzschild space-time, in which sense is centrifugal force being used? Say we have a spacecraft in circular orbit at the photon radius with angular velocity $\omega$. We can perform a coordinate transformation on Schwarzschild coordinates by eliciting the change $\phi \rightarrow \phi - \omega t$ i.e. by rotating Schwarzschild coordinates with the angular velocity of the spacecraft orbit-this is just the corotating coordinate system of the orbit. This would bring us to a coordinate system in which the spacecraft would be at rest but at the photon radius and not at the origin of the coordinate system. If you recall the Rindler/Perlick paper referenced in previous threads, this is exactly the rotating coordinate transformation they employ. Will there be a centrifugal force on the spacecraft in this coordinate system? If not, why not?

On the other hand, we also have the rest frame of the spacecraft itself. According to proposition 3.2.5. in Malament's text, this frame is non-rotating relative to local gyroscopes no matter the angular velocity of the orbit because the vorticity of $\xi^{\mu} + \omega \eta^{\mu}$ vanishes for all values of $\omega$ at the photon radius (we take as usual $\omega$ to be constant throughout space-time). This would mean that the spacecraft attributes no centrifugal (or Coriolis) forces to particles in its rest frame. Is this the sense in which the first paper refers to the centrifugal force or is it rather in the sense outlined in the previous paragraph?

Thanks again.

11. Feb 18, 2014

### yuiop

Following WBN's objections and clarifications in #8 I revisited peter's equation and see that it can be reformulated to look like this:

$f_{total} = f_{grav}+f_{centrifugal} =-\frac{Mm}{r^2} \gamma + \frac{mv^2}{r}\left(\frac{1-3M/r}{1 - v^2}\right)\gamma$

or alternatively in terms of local angular velocity $\omega$

$f_{total} =-\frac{Mm}{r^2} \gamma + m\omega^2r\left(\frac{1-3M/r}{1 - \omega^2r^2}\right)\gamma$

Expressed like this the gravitational velocity is independent of velocity and the centrifugal force term does indeed go to zero for any velocity at $r=3M$ and reverses direction at that radius. This may be a more agreeable way of splitting the forces.

Last edited: Feb 18, 2014
12. Feb 18, 2014

### WannabeNewton

Thanks yuiop! How would you interpret the term $f_{\text{centrifugal}} \propto 1 - \frac{3M}{r}$ you derived with regards to my question(s?) in post #10?

EDIT: Hopefully I didn't make any mistakes but I wanted to compute the expression for the centrifugal force given in the Nayak paper, expression (40), so as to compare it with yuiop's expression above. All of the following is on the equatorial plane so $\theta = \frac{\pi}{2}$. We have $u^{\mu} = e^{\psi}(\xi^{\mu} + \omega \eta^{\mu})$ where $e^{\psi} = (1 - \frac{2M}{r} - \omega^2 r^2)^{-\frac{1}{2}}$ and $e^{\alpha} = r$.

Furthermore $\eta^{\mu}\eta_{\mu} = r^2$ and $\xi^{\mu}\xi_{\mu} = -(1 - \frac{2M}{r})$ so $$f^{\text{centrifugal}}_{\mu} = \frac{1}{2}\omega^2 \frac{1 - \frac{2M}{r}}{1 - \frac{2M}{r} - \omega^2 r^2}\nabla_{\mu}[r^2(1 - \frac{2M}{r})^{-1}] = \omega^2 r (1 - \frac{2M}{r} - \omega^2 r^2)^{-1}(\frac{1 - \frac{3M}{r}}{1 - \frac{2M}{r}})(dr)_{\mu}$$ thus $f^{\mu}_{\text{centrifugal}} = \omega^2 r(1 - \frac{2M}{r} - \omega^2 r^2)^{-1}(1 - \frac{3M}{r})(\partial_r)^{\mu}$

If we evaluate this for a circular geodesic orbit then we get $f^{\mu}_{\text{centrifugal}} = \frac{M}{r^2}(\partial_r)^{\mu}$ which is exactly the Newtonian expression for the centrifugal force on the circular geodesic orbit as observed in the coordinate system centered on the symmetry axis corotating with the orbit.

At $r = 3M$ we get $f^{\mu}_{\text{centrifugal}} = 0$ independently of $\omega$ but is this signifying the vanishing of the centrifugal force as observed in the coordinate system corotating with a spacecraft in circular orbit at the photon radius but with center on the symmetry axis or is it signifying the vanishing of the centrifugal force in the rest frame of the spacecraft itself due to the non-rotation of the spacecraft rest frame as a consequence of vanishing vorticity?

This is basically what I was asking in post #10 and I'm still confused in this regard.

Last edited: Feb 18, 2014
13. Feb 18, 2014

### Mentz114

For the congruence $u^\mu=\dot{t}\partial_t + \dot{\phi}\partial_\phi$ in the Schwarzschild coords ( equatorial) I find the acceleration $u^\nu\nabla_\nu u_\mu$ has a component in the $r$-direction which can be decomposed into two terms thus $\frac{m}{{r}^{2}-2\,m\,r}-\frac{{\dot{\phi}}^{2}\,r\,\left( r-3\,m\right) }{r-2\,m}$ which seems to support the previous calculations.

14. Feb 18, 2014

### Staff: Mentor

No, this is not obvious. You're confusing coordinate acceleration with proper acceleration, and what counts if you're talking about centrifugal force as a frame-dependent concept (i.e., it appears in rotating frames but not in inertial frames) is coordinate acceleration. In the rotating frame, the coordinate acceleration of an observer rotating with the disk is zero, since such an observer is at rest in the rotating frame. That's why you have to introduce centrifugal force in this frame: to explain why the observer at rest in the rotating frame remains at rest, even though he is being subjected to a centripetal force. The centripetal force is exactly balanced, in the rotating frame, by the centrifugal force, so the net force on the observer at rest in the frame is zero, and hence he has no (coordinate) acceleration.

Of course on this view, the centrifugal force does not produce any proper acceleration, only coordinate acceleration; that's why such forces are often called "fictitious" or "pseudo" forces. But if you're going to use terms like "centrifugal force" at all, you've already accepted this sort of weirdness.

Yep, there are all kinds of cans of worms we can open with a scenario like this. But the comments I made above still apply; the "centrifugal force" in this frame doesn't cause any proper acceleration, only coordinate acceleration. Any proper acceleration (such as the proper acceleration of the observer at rest at the origin in this frame) must have some other cause.

I would say no, but this opens yet another can of worms. First consider a "hovering" observer, with zero angular velocity (relative to infinity). This observer is at rest in the standard Schwarzschild coordinate chart, but he has nonzero proper acceleration; i.e., he is subject to an outward force, say from his rocket engine. What balances this force to keep him at rest? The inward force of "gravity". Of course this means that gravity itself is a "fictitious" or "pseudo" force, since it doesn't cause any proper acceleration; only the outward force of the observer's rocket engine does.

Now suppose our observer is rotating about the hole, at the photon radius, with nonzero angular velocity. We switch to a rotating chart such that the observer is at rest. What is his "force balance" now? It is *unchanged* from the "hovering" case; his rocket engine is exerting an outward force which is exactly balanced by the inward force of gravity. (This is because, as we've seen, the outward force required to maintain altitude is independent of angular velocity at the photon radius, for any timelike orbit.) So there's no extra "centrifugal force" required to make things balance.

Yes.

I think the first paper is using the term in the sense of the previous paragraph. However, I agree that they could be clearer about it.

15. Feb 18, 2014

### WannabeNewton

Awesome, thanks Peter that helped clarify a lot. I still have one confusion left however and it relates to the Rindler/Perlick paper, linked here by yuiop: https://www.physicsforums.com/showthread.php?t=733916&page=4

Expression (11) in the paper writes down the metric in the coordinate system corotating with a given circular orbit of angular velocity $\omega$ in Schwarzschild space-time. The observer in the circular orbit has the 4-velocity $u^{\mu} = (1 - \frac{2M}{r} - \omega^2 r^2)^{-1/2} \delta^{\mu}_t$ i.e. the observer is of course at rest in the corotating coordinates at some fixed radius. Now the 4-acceleration on the observer is given by $$a^{\mu} = \frac{1}{2}\nabla^{\mu}\ln (1 - \frac{2M}{r} - \omega^2 r^2)= \frac{1}{2}g^{rr}\partial_{r}\ln (1 - \frac{2M}{r} - \omega^2 r^2) \delta^{\mu}_r \\= (1 -\frac{2M}{r} )(1 - \frac{2M}{r} - \omega^2 r^2)^{-1}(\frac{M}{r^2} - \omega^2 r)\delta^{\mu}_r$$

At the bottom of p.3 of the paper, in the 2nd to last paragraph, the authors state that $a^{\mu}$ is the centrifugal acceleration on the observer in the circular orbit as measured in the corotating coordinate system. Actually they say "centrifugal or centripetal force" which is confusing because they make it seem like the two are interchangeable-mathematically in the case of the circular orbit they would be opposites of each other in sign but same in magnitude and they certainly aren't the same conceptually. Regardless, does it make sense to call $a^{\mu}$ the centrifugal acceleration?

Consider the case of a circular geodesic orbit. Then $a^{\mu} = 0$ but the centrifugal force definitely doesn't vanish on an observer in a circular geodesic orbit as measured in the corotating coordinate system. In fact in the corotating coordinate system the centrifugal force on a circular geodesic should be $a^{\mu}_{\text{centrifugal}} = \frac{M}{r^2}\delta^{\mu}_r$ shouldn't it, just like in Newtonian mechanics?

Therefore shouldn't there be a way to decompose $a^{\mu}$ into a gravitational part and a centrifugal part such that the two cancel each other out for circular geodesics and such that the centrifugal part vanishes identically for all circular orbits at the photon radius?

Thanks again.

16. Feb 18, 2014

### Staff: Mentor

I assume you mean "gravitational force" or "gravitational acceleration" here. However, those things are not independent of velocity as you've split them up here, because $\gamma$ depends on the velocity. There is no general way that I can see to split the acceleration (or force) up into two terms one of which is independent of velocity; the only case in which that works is the Newtonian limit, where the velocity is small enough that $\gamma \approx 1$ so that factor drops out and the gravitational force can be viewed as velocity-independent (but for this to work at orbital velocity you also need to have $r >> 2M$ so the orbital velocity is also small).

17. Feb 19, 2014

### Staff: Mentor

I'm not entirely sure that's what they mean. Consider the regime $r > 3M$, where there are timelike geodesic orbits. Suppose we pick an angular velocity that is *faster* than the orbital velocity at a given radius. Then, as I noted in my blog post, an *inward* (i.e., centripetal) proper acceleration will be required to maintain altitude. It's possible that this possibility is why the term "centrifugal or centripetal" was used.

No, but as I noted previously, in the co-rotating coordinates the centrifugal force is balanced by the inward "force of gravity". For a geodesic orbit, with zero proper acceleration, we therefore have two "fictitious" forces balancing each other.

Actually, I mis-stated things a bit just now. Conceptually, for a general circular orbit, not necessarily geodesic, we actually have *three* "forces" in play in the co-rotating coordinates:

* The inward "force of gravity" (fictitious), which we'll call $f_g$;

* The outward "centrifugal force" (fictitious), which we'll call $f_c$;

* The outward force exerted by the observer's rocket engine (not fictitious), which we'll call $f_e$.

So (at least for velocities less than or equal to orbital velocity--see below) we have $f_g$ acting inward, balanced by $f_c + f_e$ acting outward. The geodesic orbit, of course, is the special case where $f_e = 0$ and $f_g = f_c$.

For velocities greater than orbital velocity, we have $f_e$ acting inward, not outward, as I noted above; so we have $f_g + f_e$ acting inward, balanced by $f_c$ acting outward.

Of course, this conceptual scheme views $f_g$ as independent of velocity (it depends only on radius), whereas $f_c$ and $f_e$ are velocity-dependent. So I'm not sure how it matches up with the formulas we've been looking at (see below).

As I said in my previous post responding to yuiop, I don't see any general way to split $a^{\mu}$ into two additive terms such that one term (the "gravitational" term) is independent of velocity.

18. Feb 19, 2014

### yuiop

Yes, I meant gravitational force. I was using $\gamma$ as shorthand for $1/\sqrt{1-2M/r}$ as stated earlier in post #5 but I should have made that clear. Sorry for the confusion. The split equation in full is:

$a_{total} =-\frac{M}{r^2\sqrt{1-2M/r}} + \frac{\omega^2r}{\sqrt{1-2M/r}}\left(\frac{1-3M/r}{1 - \omega^2r^2}\right)$

This expression is mathematically identical to the one in your blog (except for change of sign convention). Since my expression is nothing more than an algebraic manipulation of your expression its correctness depends on the correctness of your blog expression. it should be clear that at least in terms of your expression the gravitational acceleration term can be stated in a velocity independent form.

This paper http://www.dmf.unisalento.it/~giordano/allow_listing/AJP000936.pdf linked earlier by WBN states that the centrifugal acceleration is zero at $r=3M$ and that the proper acceleration of the rockets required to maintain a circular orbit is independent of their orbital velocity, which implies that the gravitational acceleration must also be independent of velocity.

What is bothering me is that your expression does not appear to agree with the results from WBN or Mentz.

Your expression is in terms of local velocity measured by a stationary observer at r. In terms of coordinate angular velocity $\omega_{\infty}$ as measured by an observer at infinty, where $\omega^2 = \omega_{\infty}^2/(1-2M/r)$, the expression becomes:

$a_{total} = \frac{-M}{r^2\sqrt{1-2M/r}} + \frac{\omega_{\infty}^2r}{\sqrt{1-2M/r}}\left(\frac{1-3M/r}{1 - 2M/r -\omega_{\infty}^2r^2}\right)$

The above expression is now the total proper acceleration in terms of coordinate angular velocity.

The last term on the right representing the centrifugal acceleration is almost identical to the centrifugal acceleration given by WBN except for a factor of $1/\sqrt{1-2M//r}$.

While we might all be using different coordinate systems, we should at least all be able to agree on what the total proper acceleration is, but so far no luck.

Last edited: Feb 19, 2014
19. Feb 19, 2014

### Mentz114

These conditions are met by the non-zero component of $u^\nu\nabla_\nu u_\mu$, which is $\frac{m}{{r}^{2}-2\,m\,r}-\frac{{\dot{\phi}}^{2}\,r\,\left( r-3\,m\right) }{r-2\,m}$. Setting this zero and solving for $\dot{\phi}$ gives the geodesic. This is just the value where the two components cancel.

Using the Schwarzschild metric, written in the new $\phi$ cordinate from the Rindler et al. paper, I find the acceleration of the stationary frame has a radial component $({{r}^{3}\,{w}^{2}-m})/({{r}^{4}\,{w}^{2}-{r}^{2}+2\,m\,r})$. The geodesic therefore requires $\omega^2=m/r^3$.

The last acceleration can be factored into 2 parts $\frac{m}{r\,\left( r-2\,m\right) }$ and $\frac{{r}^{2}\,\left( r-3\,m\right) \,{w}^{2}}{\left( r-2\,m\right) \,\left( {r}^{3}\,{w}^{2}-r+2\,m\right) }$. The second part has a zero at $r=3m$.

Compare with yuiop's reduction of PeterDonis' result $a_{total} = \frac{-M}{r^2\sqrt{1-2M/r}} + \frac{\omega_{\infty}^2r}{\sqrt{1-2M/r}}\left(\frac{1-3M/r}{1 - 2M/r -\omega_{\infty}^2r^2}\right)$

Last edited: Feb 19, 2014
20. Feb 19, 2014

### WannabeNewton

Right but that still means the centrifugal force shouldn't vanish in the corotating coordinates. However the Rindler/Perlick paper calls the 4-acceleration $a^{\mu} = \nabla^{\mu}\psi$ itself the centrifugal force, where $g_{tt} = -e^{2 \psi}$ in the corotating coordinates. But then if we consider a geodesic orbit $a^{\mu} = 0$ regardless of the coordinate system so calling it the centrifugal force doesn't make sense to me because there is still a centrifugal force acting outwards on the geodesic orbit in the corotating coordinates that exactly cancels the inwards gravitational force.

I agree with all this but in light of the above doesn't this make the Rindler/Perlick definition of $a^{\mu} = \nabla^{\mu}\psi$ as the centrifugal force a confusing one since as stated the centrifugal force doesn't vanish for geodesic orbits whereas $a^{\mu} = 0$ for geodesic orbits?

But this is exactly what the Nayak paper does. However I found the Nayak paper linked in post #1 to lack explanatory detail so I found the original paper by Abramowicz which has much more clarity: http://iopscience.iop.org/0264-9381/10/10/001 Now my university lets me access the paper for free so I don't know if you're blocked behind a paywall and so I'll just try to summarize the relevant parts of the paper. The notations and definitions are the exact same as in the Nayak paper.

Given a 4-velocity field $u^{\mu}$ and a background irrotational 4-velocity field $n^{\mu}$ the authors decompose the orthogonal projection of the 4-force on $u^{\mu}$ relative to $n^{\mu}$ as $f^{\perp}_{\mu} = m h_{\mu}{}{}^{\nu}u^{\gamma}\nabla_{\gamma}u_{\nu} = G_{\mu} + Z_{\mu} + C_{\mu} + E_{\mu}$. As per their decomposition, $G_{\mu} = -m\nabla_{\mu}\Phi$ where $g^{tt} = -e^{2\Phi}$ so this is entirely independent of the velocity of $u^{\mu}$ relative to $n^{\mu}$-it depends only on the gravitational field. Next the authors note that the Coriolis force $C_{\mu}$ has no Newtonian counterpart so that answers one of the previous questions.

Most importantly they state that all of the above inertial forces are defined or measured in the rest frames of the observers following orbits of $u^{\mu}$ which is what you stated in an earlier post. But note this means that as per this decomposition, for a circular orbit in Schwarzschild space-time this centrifugal force $Z_{\mu}$ is as measured in the rest frame of the observer in the circular orbit; I calculated $Z_{\mu}$ in post #12. In particular, the vanishing of $Z_{\mu}$ at the photon orbit for all angular velocities means the vanishing of the centrifugal force in the rest frame of the observer in the orbit at the photon radius. At face value, this is definitely different from what one would call the centrifugal force on the observer in Newtonian mechanics, which is the outwards radial force on the observer as measured in the frame or coordinate system corotating with the orbiting observer but with origin fixed on the symmetry axis of the orbit.

I know you answered this before but just for added clarification, now with regards to the paper http://www.dmf.unisalento.it/~giordano/allow_listing/AJP000936.pdf, which for the record is by the same author and originally talked about centrifugal forces on observers in circular orbits in Schwarzschild space-time, do you still think the author is talking about centrifugal force on the orbit in this Newtonian sense or rather in the sense defined in the author's paper liked above as that measured in the rest frame of the observer in the orbit? Or does it not make a difference in light of my reply to Mentz below?

Alright good this is the same expression I got for $Z_{\mu}$ in post #12 so unsurprisingly the 4-acceleration $a^{\mu} = \nabla^{\mu}\psi$ in the corotating frame itself isn't the centrifugal acceleration, even though the Rindler/Perlick paper defined it as such for some reason, but rather we have to further decompose $a^{\mu}$ into a centrifugal part in the corotating coordinates. Thankfully this centrifugal part agrees with $Z_{\mu}$ which is the centrifugal force as measured in the rest frame of the observer in the circular orbit.

Thanks Mentz!

Last edited: Feb 19, 2014