# Centrifugal Forces and Black Holes

1. Jun 4, 2010

### stevebd1

I thought this may be an isolated idea but on doing a search on the web, there seems to be a common interest in the idea that centrifugal force reverses near a black hole. Below are a couple of links-

http://www.npl.washington.edu/av/altvw55.html" [Broken]

http://arxiv.org/abs/0903.1113v1" [Broken]

The same subject was mentioned in https://www.physicsforums.com/showthread.php?t=10369" (post #4).

According to most sources, it appears that the reactive centrifuge becomes zero at the photon sphere, my question is, how does this fit into the centripetal acceleration equation? I had a look the relativistic equation for the tangential velocity required for a stable orbit in Kerr metric and reduced it for a Schwarzschild solution (see https://www.physicsforums.com/showthread.php?t=354583"). Based on ac=ag (where ac is centripetal acceleration and ag is gravity), the only way the equations would work is if centripetal acceleration reduced in accordance with the redshift, becoming zero at the event horizon and negative beyond the EH.

This works also with the Kerr metric where frame dragging increases exponentially within the ergoregion, without ac being reduced, it would appear that objects would tend to be thrown out of the ergoregion before crossing the EH but if ac reduces in accordance with the redshift, then the object is overcome by gravity regardless of it's tangential velocity (relative to infinity) and crosses the event horizon.

Last edited by a moderator: May 4, 2017
2. Jun 6, 2010

### yuiop

I think this is an ineresting question worth exploring. I have seen this claim too several times while browsing around the subject. The intuitive heuristic picture is that a particle requires to be orbiting at the speed of light at the photon sphere in order not to lose altitude, and so at a radius of less than 3M it is impossible for a freely orbiting particle to maintain altitude because it would have to have a tangential velocity greater than the speed of light.

I think it would be interesting to try and work out if a purely radially falling particle with zero angular momentum is dropped from r=3M, would it arrive at the event horizon after a second particle released from the same point at the same time, that has non-zero angular momentum?

Last edited: Jun 7, 2010
3. Jun 6, 2010

### Naty1

I am skeptical so far because Leonard Susskind's THE BLACK HOLE WAR and Kip Thorne's BLACK HOLES AND TIME WARPS, both recent publications, give many explanations of black hole effects and never mention such a reversal.

It's so unexpected that I would think at least one of those authors would have mentioned it because they do discuss a number of other "strange" phenomena, like thermal vs virtual radiation, time dilation, the string nature of black holes, and a number of other "unexpected" phenomena....

I look forward to reading more here.

4. Jun 7, 2010

### yuiop

I have done a rough calculation using the Lagrangian equations of motion and the Schwarzschild metric and obtained this equation:

$$\frac{dr}{dt} = (1-2M/r)\sqrt{1-\frac{(1-2M/r)}{(1-2M/R)}\frac{(1+L^2/r^2)}{(1+L^2/R^2)}}$$

This equation is defined for $\theta = \pi/2$ and $d\theta = 0[/tex] to simplify things. R is the height of the path apogee and L is the angular momentum per unit rest mass defined as: $$L = r^2\frac{d\phi}{ds}$$ L is a constant and since it is defined in terms of the proper time of the orbiting particle, L can take any value between zero and infinity. Checking the equation numerically, a large value of L at large R and r can produce negative* dr/dt which seems reasonable. When R=3M (the photon sphere radius) and 2M<r<3M it seems that any arbitrarily large value of L can not produce a negative fall rate dr/dt, which is also in agreement with our expectations. However a particle with non-zero angular momentum falls slower than a particle with zero angular momentum even for R<=3M. This implies that centrifugal reactive force does not reverse below r=3M. Non-zero angular momentum does not increase the rate of fall. Rather, it slows the descent between r=3M and r=2M but the acceleration due to gravity is always greater than the reactive centrifugal acceleration below r=3M, for any value of L(assuming my equation is correct). In the literature most of the equations for the motion of a particle near a black hole are in terms of dr/ds rather dr/dt as I have done here. The trouble with using dr/ds is that a particle with horizontal motion approaching the speed of light experiences significant time dilation due to its horizontal motion, making the vertical fall rate appear to increase. Using the proper time of the particle also makes it difficult to compare the fall rates of two different particles with different trajectories. Using coordinate velocity dr/dt allows a direct comparison of the fall rate of two particles. *If anyone is curious how a real negative fall rate can fall out the equation when it seems to produce an imaginary result, just ask Disclaimer: This equation is based on a non rigorous derivation with plenty of room for error and is offered in good faith as a starting point for discussion. Last edited: Jun 7, 2010 5. Jun 7, 2010 ### Altabeh That would be awesome if you also included in your post the proof procedure. The first part is not a reason for why L is a constant from a rigorous viewpoint. If the last of geodesic equations in Schwarzschild metric is integrated, then L appears to be the constant of integration and from the conservation law for the angular momentum of the particle, it is obvious that the angular momentum is a conserved quantity (no matter whatever value between zero and infinity it has) and this is backed when L is a constant. . AB 6. Jun 7, 2010 ### yuiop I would not call this a proof, just my thought process. I might be way off base. Starting with Schwarzschild metric and assuming orbital motion in a plane about the equator such that [itex]\theta = \pi/2$ and $d\theta = 0$
(Due to the spherical symmetry of the metric we can define the equator to be wherever we like, for convenience.)

$$ds^2=\alpha dt^2-\frac{1}{\alpha}dr^2-r^2d\phi^2$$

$$\alpha=1-\frac{2m}{r}$$

Solve for dr/dt:

$$\frac{dr}{dt} = \sqrt{ \alpha^2 - \alpha (\frac{rd\phi}{dt})^2 - \alpha(\frac{ds}{dt})^2}$$

The equations of motion where K and L are defined as constants are given as:

$$K = \alpha \frac{dt}{ds}$$

$$L = r^2 \frac{d\phi}{ds}$$

As far as I know, the proofs of why these values are constant are non-trivial and involve Killing vectors or Noether's theorem, which I do not pretend to understand.
If it is trivial and someone cares to elucidate, then please feel free to do so

Substitute these constants into the equation for dr/dt given above:

$$\frac{dr}{dt} = \sqrt{ \alpha^2 - \frac{\alpha^3}{K^2} \frac{L^2}{r^2} - \frac{\alpha^3}{K^2}} =\alpha \sqrt{ 1- \frac{\alpha}{K^2}(1+L^2/r^2)}$$

Now obtain an alternative formulation of K by solving the original Schwarzschild metric for $\alpha^2dt^2/ds^2$:

$$\alpha^2 (\frac{dt}{ds})^2 = K^2 = \alpha + (\frac{dr}{ds})^2 + \alpha (\frac{rd\phi}{ds})^2$$

Now K is a constant and I "fix" its value at the apogee (or perigree) where dr/ds=0 so that the alternative form of K is defined as:

$$K^2 = (1-2M/r ) + (1-2M/r) (\frac{rd\phi}{ds})^2 = (1-2M/R ) + (1-2M/R)(L^2/R^2)$$

$$K^2 = (1-2M/R ) (1 + L^2/R^2)$$

This alternative form of K^2 takes into account angular motion and is substituted into the equation for dr/dt above:

$$\frac{dr}{dt} = (1-2M/r)\sqrt{1-\frac{(1-2M/r)}{(1-2M/R)}\frac{(1+L^2/r^2)}{(1+L^2/R^2)}}$$

which is the equation I gave earlier. I am sure some people will consider it a bit hacky

Last edited: Jun 7, 2010
7. Jun 8, 2010

### stevebd1

The common census seems to be that centrifuge reduces to zero at the photon sphere but there also seems to be some evidence that reactive centrifugal force reduces to zero at the event horizon synonymous with the redshift. If we look at the Kepler equation for a stable orbit in Schwarzschild metric-

$$v_s=\frac{\pm\sqrt{M}\,r^2}{\sqrt{r^2-2Mr}\,r^{3/2}}$$

If we incorporate the results for vs into ac=ag where $a_c=f(r)v_s^2/r$ where f(r) is a function relative to r, and $a_g=M/(r^2 \sqrt(1-2M/r))$ which is the equation for gravity in Schwarzschild metric, we get-

$$f(r)\frac{v_s^2}{r}=\frac{M}{r^2 \sqrt{1-2M/r}}$$

substituting for vs and rearranging the equation, we get-

$$f(r)=\frac{(r^2-2Mr)}{r^2 \sqrt{1-2M/r}}=\sqrt{1-2M/r}$$

Using this function, we can also demonstrate that the max tangential velocity of 1 (i.e. c) is at the photon sphere (3M)-

$$\sqrt{1-2M/r}\,\frac{v_s^2}{r}=\frac{M}{r^2 \sqrt{1-2M/r}}$$

which when r=3M reduces to-

$$v_s=\sqrt{\frac{M}{r\,(1-2M/r)}}=1$$

gravitational acceleration is understood in Schwarzschild metric and the above tells us at the very least that something is going on with centripetal acceleration in extreme gravity fields.

Last edited: Jun 8, 2010
8. Jun 8, 2010

### Altabeh

I checked every line of it carefully and I have to say it is 100% flawless!

For this, first we have to prove the following theorem about Killing vectors. I take into account that you're a little bit familiar with Killing vectors.

Theorem. Let $$\xi$$ be a Killing vector field i.e. it satisfies the Killing equation$$\nabla_a\xi_b+\nabla_b\xi_a=0.$$ Assume that $$S$$ is a geodesic with tangent vector $$u^a.$$ Then $$\xi_au^a$$ is constant along S.

$$u^b\nabla_b(\xi_au^a) = u^au^b\nabla_b\xi_a+\xi_au^b\nabla_b u^a.$$

From the geodesic equation one knows that the second equation is clearly zero, and by multiplying the Killing equation by $$u^au^b$$ we can also obtain

$$u^au^b\nabla_b\xi_a+u^au^b\nabla_a\xi_b=2u^au^b\nabla_a\xi_b=0$$

which is the first term on the right-hand side of the above expansion. And we are done.

Now if we calculate the Killing vectors of the Schwarzschild metric, it can be well-understood that the only nun-null vectors are

$$(\partial/ \partial \phi)^a=(0,0,0,1)$$

and

$$(\partial/ \partial t)^a=(1,0,0,0).$$

This is so because the Schwarzschild metric is independent of $$t$$ and $$\phi$$ and this follows that the Lie derivative of metric tensor would be zero if $$\xi^a=(\partial/ \partial t)^a$$ or $$\xi^a=(\partial/ \partial {\phi})^a.$$ Hence the above theorem yields

$$L=\xi_au^a=\xi^bg_{ab}u^a=\xi^bg_{ab}u^a=r^2\sin^(\theta)\dot{\phi},$$
$$E=\xi_au^a=\xi^bg_{ab}u^a=\xi^bg_{ab}u^a=-(1-2m/r)\dot {t}.$$

Each of these corresponds to a special conservation law; but how? Let me first give a more intuitive form of these equations so the rest of discussion could be better pinned down. Here I want to compute the $$t-$$ and $$\phi-$$ components of covariant 4-velocity along the geodesics of the Schwarzschild metric. To do so, I simply follow the general formula

$$u_a=g_{ab}u^b.$$

Therefore

$$u_0=g_{00}u^0=-(1-2m/r)\dot{t}=const.$$
$$u_3=g_{33}u^3=r^2\dot{\phi}=const.$$

where I made use of the previous forms of L and E and also put $$\theta=\pi/2$$ without loss of generality. These versions of equations of L and E are easily seen to be compatible with the following form of the geodesic equation,

$$\dot{u}_a=\frac{1}{2}\partial_ag_{bc}u^bu^c$$

where the over-dot represents the differentiation wrt the parameter of geodesics, e.g. $$s$$. Introducing $$a=0$$ and $$3$$ into this equation, respectively, gives

$$\dot{u}_0=0,$$
$$\dot{u}_3=0.$$

Which verifies our newly obtained equations for $$u_3$$ and $$u_0$$ above. Actually the first equation is related to the conservation of the energy of a test particle moving in a time-independent field. To wit, the energy of the particle with a rest mass $$m_0$$ is given by $$m_0u_0$$ and thus by being a constant it is invariant under time translations which inspires the conservation law of energy. On the other hand the angular momentum in GR is defined in any plane about the equator to be

$$m_0r^2\dot{\phi}.$$

Since this is a constant, by construction, we are led to believe that its value is invariant under $$\phi$$-angle translations, confirming the conservation law of angular momentum.

Hope this helps.

AB

Last edited: Jun 8, 2010
9. Jun 8, 2010

### yuiop

Thanks for checking it out Altabeh Thanks also for showing the proof for the constants of motion using Killing vectors. I am only just starting to learn about tensors so the explanation is little above my pay grade, but I will refer back to it as I learn more.

Hi Steve, I see what you are trying to do here, but unfortunately the term on the right for gravitational acceleration is only valid for a stationary particle and the velocity due to orbital motion changes things. I will post the (hopefully) correct term in the next post.

10. Jun 8, 2010

### yuiop

[EDIT] To avoid confusion with the L symbol for the Lagragian, I have relabelled the conserved angular momentum per unit rest mass as H so that the constant is now:

$$H = r^2 \frac{d\phi}{ds}$$

===========================

$$\frac{dr}{dt} = (1-2M/r)\sqrt{1-\frac{(1-2M/r)}{(1-2M/R)}\frac{(1+H^2/r^2)}{(1+H^2/R^2)}} \:\: (eq1)$$

Differentiate the above to obtain the coordinate radial acceleration of a free falling particle:

$$a = \frac{d^2r}{dt^2} = \frac{d(dr/dt)}{dr}\frac{dr}{dt} = -\frac{M}{r^2}(1-M/r)\left(\frac{(1-2M/r)}{(1-2M/R)}\left(3\frac{(1+H^2/r^2)}{(1+H^2/R^2)} - \frac{H^2}{rM}\frac{(1-2M/r)}{(1+H^2/R^2)}\right) - 2 \right)\:\: (eq2)$$

Not too pretty eh, but as a quick check, if H is set to zero we get the purely coordinate acceleration given by mathpages here: http://www.mathpages.com/rr/s6-07/6-07.htm

If we set R=r and H=0 the expected coordinate acceleration for a particle at apogee is obtained.

For your purposes we need the local acceleration ($a_g'$) and this can be obtained by using the method shown in this post in a different thread https://www.physicsforums.com/showpost.php?p=2747788&postcount=345

The local acceleration a' using the notation $\gamma_g = 1/\sqrt{(1-2M/r)}$ is:

$$a' = \frac{d^2r'}{dt'\,^2} = \frac{d^2r}{dt\,^2}\gamma_g^3 - \frac{2M}{r^2}\left(\frac{dr}{dt}\right)^2 \gamma_g^5 =-\frac{M}{r^2}\gamma_g \left(\frac{(1-2M/r)}{(1-2M/R)}\left(3\frac{(1+H^2/r^2)}{(1+H^2/R^2)} - \frac{H^2}{rM}\frac{(1-2M/r)}{(1+H^2/R^2)}\right) - 2 \right) - \frac{2M}{r^2}\left(\frac{dr}{dt}\right)^2 \gamma_g^5 \;\; (eq3)$$

The equation for dr/dt is given by (eq1) and can be inserted into (eq3) to obtain:

$$a' = \frac{d^2r'}{dt'\,^2} = -\frac{M}{r^2} \frac{\sqrt{1-2M/r}}{(1-2M/R)}\left(\frac{(1+H^2/r^2)}{(1+H^2/R^2)} - \frac{H^2}{rM}\frac{(1-2M/r)}{(1+H^2/R^2)}\right) \;\; (eq4)$$

(eq4) is the fully general local acceleration due to gravity of a particle at r that has a trajectory with its apogee at R and takes vertical and orbital motion into account. The only limitation is that the plane of the motion remains on the equatorial plane so that $\theta=\pi/2$ and $d\theta =0$.

To keep things fairly simple we can analyse the acceleration only at the turning point of the trajectory, so that dr/dt=0 and R=r and use (eq3) to obtain the local acceleration for this limited case $a_{*}'$ where the asterix in the subscript serves as a reminder that this equation is only valid at the apogee or perigee of a trajectory:

$$a_{*}' = -\frac{M}{r^2 \sqrt{1-2M/r}}+ \frac{H^2}{r^3}\frac{\sqrt{1-2M/r}}{(1+H^2/r^2)} \:\: (eq5)$$

$$a_{*}' = -\frac{M}{r^2 \sqrt{1-2M/r}} + \frac{(rd\phi/d\tau)^2\sqrt{1-2M/r}}{r(1+(rd\phi/d\tau)^2)}\right) \:\;\: (eq6)$$

If we call the local instantaneous tangential orbital velocity $v_s$ such that $v_s^2/(1-v_s^2/c^2) = (rd\phi/d\tau)^2$ then (eq6) can be written as:

$$a_{*}' = -\frac{M}{r^2 \sqrt{1-2M/r}} + \frac{ v_s^2\sqrt{1-2M/r}}{r} \:\: (eq7)$$

[Note] The above has been edited to better define $$a_*'$$ as the acceleration at the turning point rather than the acceleration of a circular orbit, as I incorrectly stated the first time around. A circular orbit is defined by dr/dt=0 and d^2r/dt^2=0 both being true.

Last edited: Jun 9, 2010
11. Jun 8, 2010

### Tomsk

I think they just come from the lagrangian equations of motion for the geodesic. The lagrangian is
$$L = \sqrt{ g_{\mu\nu} \dot{x}^\mu \dot{x}^\nu } = \sqrt{\alpha \dot{t}^2 - \alpha^{-1}\dot{r}^2 - r^2 \dot{\phi}^2}$$
which is just the magnitude of the 4-velocity, which is 1, so it's constant along the geodesic.

The first two equations of motion are:
$$\frac{\partial L}{\partial \dot{t}} = \frac{\alpha \dot{t}}{L} = K$$
and
$$\frac{\partial L}{\partial \dot{\phi}} = -\frac{r^2 \dot{\phi}}{L} = H$$
where H is a constant the negative of your L (my L is my lagrangian!) Since the lagrangian is constant along the geodesic the constants you gave are constant.

I'm trying to continue the derivation using the lagrangian approach but it doesn't seem to be working- I can't find a way to get rid of the second derivatives $$\ddot{t}, \ddot{r}, \ddot{\phi}$$. Anyone know if it's possible? Or is the whole point you end up with the accelerations?

12. Jun 8, 2010

### yuiop

If we take the (truncated) Schwarzschild metric as:

$$d\tau = \sqrt{\alpha {dt}^2 - \alpha^{-1}{dr}^2 - r^2 {d\phi}^2}$$

and divide both sides by $d\tau$ we get:

$$1 = \sqrt{\alpha \dot{t}^2 - \alpha^{-1}\dot{r}^2 - r^2 \dot{\phi}^2}$$

so it seems reasonable to equate L with unity and leave L out of the equations so that the first two equations of motion are:

$$\frac{\partial L}{\partial \dot{t}} = \alpha \dot{t} = K$$

and

$$\frac{\partial L}{\partial \dot{\phi}} = r^2 \dot{\phi} = H$$

As for the problem with the second derivatives, I probably can't help, but I am sure someone here can if you post where they are popping up.

13. Jun 8, 2010

### starthaus

No, they are simply the trivial solutions of the Euler-Lagrange equations.

Nope, there is a much cleaner expression:

$$\frac{dr}{dt}=\alpha\sqrt{1-\frac{L^2}{\alpha r^2}-\frac{\alpha}{K^2}}$$

Why would dr/ds=0?

Where did $$R$$ come from? Besides, expressing K as a function of L doesn't solve anything since you don't know L.

It is much simpler to observe that $$K=(1-2m/r_0)\frac{dt}{ds}|_{s=0}$$

Last edited: Jun 8, 2010
14. Jun 8, 2010

### starthaus

This is not how it's done, you need to construct the Euler-Lagrange equations.

15. Jun 8, 2010

### yuiop

I prefer the correct version:

$$\frac{dr}{dt}=\alpha\sqrt{1-\frac{\alpha L^2}{K^2 r^2}-\frac{\alpha}{K^2}}$$

to the incorrect "cleaner version".

My version works out nice later

K is a constant and its value can be chosen anywhere along the trajectory. I choose to fix it at the apogee where dr/dt =0 for convenience by defining it in terms of constants R (the height of the apogee) and L (the conserved angular momentum) at that point.

16. Jun 8, 2010

### starthaus

:lol:

17. Jun 8, 2010

### starthaus

But you don't know H .

18. Jun 8, 2010

### yuiop

Is that your way of saying you do not think my version is correct?

Please try and be constructive. We do want not this thread getting locked like the other ones.

Last edited: Jun 8, 2010
19. Jun 8, 2010

### yuiop

No, you define it. You don't know R either. Both are defined initial conditions.

20. Jun 8, 2010

### starthaus

Nope:

$$(\frac{dr}{dt})^2=\alpha^2-\alpha r^2(\frac{d\phi}{dt})^2-\alpha(\frac{ds}{dt})^2=\alpha^2-\alpha(\frac{L}{r})^2-\frac{\alpha^3}{K^2}$$

21. Jun 8, 2010

### yuiop

I was just making the observation that:

$$1 = \sqrt{\alpha \dot{t}^2 - \alpha^{-1}\dot{r}^2 - r^2 \dot{\phi}^2}$$

derived from the Schwarzschild metric looks a lot like:

$$L = \sqrt{\alpha \dot{t}^2 - \alpha^{-1}\dot{r}^2 - r^2 \dot{\phi}^2}$$

derived by Tomsk. It suggests to me that:

$$L = \sqrt{\alpha \dot{t}^2 - \alpha^{-1}\dot{r}^2 - r^2 \dot{\phi}^2} = 1$$

but maybe that is just a coincidence?

22. Jun 8, 2010

### starthaus

It just points that you don't know the Euler-Lagrange formalism that produces the equations of motion. We've been over this several times already.

23. Jun 8, 2010

### yuiop

dr/ds is always zero at the apogee.

24. Jun 8, 2010

### starthaus

How do you know you have an apogee? You haven't derived the equations of motion, let alone solved them, therefore you don't know the trajectory. The trajectory might be (and is, in certain cases) a hyperbola.

Last edited: Jun 8, 2010
25. Jun 8, 2010

### yuiop

You have substituted $L^2$ for $$r^4\frac{d\phi^2}{dt^2}$$

but that is a mistake because

$$L^2 = r^4\frac{d\phi^2}{ds^2}$$

so your objection is wrong. I can see this is going to be another long thread.

Last edited: Jun 8, 2010