Centrifugal Forces and Black Holes

[starthaus]

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

You obtain this form after you put in dr/ds=0 by hand. So, it is normal that if you make r=R you recover dr/dt=0. This isn't a valid proof.

 

[starthaus]

I define R as a turning point where the free falling particle is at it its nearest approach (perigee) or at its greatest separation (apogee).
If I insert r=R into this equation:



then dr/dt=0 automatically falls out the equation.
How to determine if dr/dt=0 at R=r is an apogee or a perigee is described in the previous post by analysing the sign of the acceleration at that point.

You obtained this form (post 10) after you put in dr/ds=0 by hand. So, it is normal that if you make r=R you recover dr/dt=0. This isn't a valid proof.

 

[yuiop]

You obtained this form (post 10) after you put in dr/ds=0 by hand. So, it is normal that if you make r=R you recover dr/dt=0. This isn't a valid proof.

Ok, it is a bit circular but we are talking about orbits after all. To most people it is self evident that dr/dt=0 at the apogee or perigee. However, we can forget about about R being a turning point in the altitude of the particle. All we have to do is declare that the radius when dr/dt=0 is called R without needing to know anything about the point R other than it is a specific point in the orbit where dr/dt=0. Remember that the constants of motion can be defined at any arbitrary point in the trajectory. Now if we know the angular velocity at this point then we know everything we need to know about the trajectory. Sure there are trajectories where there is no point on the path where dr/dt=0 for r>2M but they are not relevant to the OP of this thread.

 

[Altabeh]

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?

Can you tell us what exactly you want here? I don't understand what you mean by "a way to get rid of the second derivatives of coordinates wrt $$s$$."

AB

 

[espen180]

Pointing out that your method doesn't work. You can't really put in stuff by hand, you need to be able to arrive to the equations of motion and solve them. This way, you will get to a discussion of the different possible trajectories. Putting in dr/ds=0 by hand will not do it.

Is it just me or has the thread derailed? Where did the discussion about centripetal acceleration disappear to? Now suddenly there are equations of motion and possible trjectories?

In order to study the OP's question you don't NEED a full orbit anyway, just an infinitesimal segment of one in which to study the acceleration, right?

As long as we are talking about orbits I see no problem assuming circular orbits. dr/dt is always 0 and r=R (assuming R is apogee radius), so we can study the centripetal acceleration without quarreling about whether the usage of Euler-Lagrange or whatever in order to obtain the equations hold, as long as the equations themselves are valid.

 

[Tomsk]

Can you tell us what exactly you want here? I don't understand what you mean by "a way to get rid of the second derivatives of coordinates wrt $$s$$."

AB

Kev posted an equation for dr/dt, I was wondering how to get to that from the Euler-Lagrange equations of motion. I can't figure out how to solve them to get dr/dt. The final equation of motion is
$$\frac{d}{ds}\frac{\partial L}{\partial \dot{r}} - \frac{\partial L}{\partial r} = 0 = \frac{d}{ds}\left( \frac{-\dot{r}}{\alpha L} \right) - \frac{1}{2L}\left( \frac{2m}{r^2}\dot{t}^2 + \frac{2m}{\alpha^2 r^2}\dot{r}^2 - 2r\dot{\phi}^2 \right) = 0$$

And I'm not sure how to solve it to get dr/dt - assuming that is what we want here?

Also I've gone and got confused about L=1. Is that really right, and is dL/ds = 0 or not? I'm not sure it is because if you work out $$d/ds (g_{\mu\nu}\dot{x}^\mu \dot{x}^\nu )$$ I can't see how it goes to zero.

 

[Jonathan Scott]

What do you think the answer to the OP is? Does effective reactive centrifugal force reverse (act inwards) when a particle is free falling below r=3m or not?

I haven't followed through the details, but I think the answer is "not".

The gravitational rate of change of momentum can be split into two parts: the part due to curvature of spacetime with respect to time, which affects all objects equally (regardless of speed) as seen in a local Minkowski space, and the part due to curvature of space, which affects objects according to their velocity through that space (creating an effect proportional to the square of the speed).

It is motion through space which creates the effect of centrifugal force and I reckon that centrifugal force would be expected to be negative whenever the curvature of space had a smaller radius than a circular tangential path.

At r=3m, the total curvature of a tangential light beam matches the circular tangential path, but this is equal to the sum of the two effects. This means that the curvature of space on its own is less than the curvature of the circular tangental path.

The curvature of space itself would become equal to the curvature of the circle at r=2m, the Schwarzschild radius, as can be seen from Flamm's Paraboloid (as pointed out recently by DrGreg).

[Edit: I'd missed DrGreg retracting part of what he previously said in the other thread, which may mean I got something wrong too]

[Edit 2: No, after checking that more carefully, I think that what I said here stands]

 

[starthaus]

Kev posted an equation for dr/dt, I was wondering how to get to that from the Euler-Lagrange equations of motion. I can't figure out how to solve them to get dr/dt. The final equation of motion is
$$\frac{d}{ds}\frac{\partial L}{\partial \dot{r}} - \frac{\partial L}{\partial r} = 0 = \frac{d}{ds}\left( \frac{-\dot{r}}{\alpha L} \right) - \frac{1}{2L}\left( \frac{2m}{r^2}\dot{t}^2 + \frac{2m}{\alpha^2 r^2}\dot{r}^2 - 2r\dot{\phi}^2 \right) = 0$$

And I'm not sure how to solve it to get dr/dt - assuming that is what we want here?

The above is the third Euler-Lagrange equation. It has several proprties:

-it is dependent on the other two , so
- it is devoid of any extra information (only n-1 Euler equations are independent)
-it isn't solvable

Also I've gone and got confused about L=1. Is that really right, and is dL/ds = 0 or not? I'm not sure it is because if you work out $$d/ds (g_{\mu\nu}\dot{x}^\mu \dot{x}^\nu )$$ I can't see how it goes to zero.

Don't worry about kev's hacks. You already have the three correct Euler-Lagrange equations.

 

[starthaus]

Ok, it is a bit circular but we are talking about orbits after all.

They are circular. Hacks don't constitute valid solutions.

 

[yuiop]

Don't worry about kev's hacks. You already have the three correct Euler-Lagrange equations.

The point of this thread is not about obtaining the correct Euler-Langrange equations. it is about determing whether or not reactive centrifugal acceleration reverses in a real sense at r<3m. As far as I can tell using my equations, the answer is no.

So instead of making this thread a continued personal attack on me, why not be constructive and say what you think the answer is using whatever method suits you. If our conclusions differ then will try and figure out why that is.

Hacks don't constitute valid solutions.

Why don't you produce what you think is a valid solution?

 

[yuiop]

Here is a different approach in terms of proper time that clearly illustrates where the idea of reversal of the reactive centrifugal acceleration comes from.

First directly solve the truncated Schwarzschild metric for dr/ds:

$$\frac{dr}{ds} = \sqrt{ \alpha^2\frac{dt^2}{ds^2} - \alpha \left(1 + r^2\frac{d\phi^2}{ds^2}\right) }$$

Substitute the constants of motion in and expand $$\alpha$$:

$$\frac{dr}{ds} = \sqrt{ K^2 - \left(1-\frac{2M}{r}\right) \left(1 + \frac{H^2}{r^2} \right) }$$

Differentiate wrt ds to obtain the acceleration:

$$\frac{d^2r}{ds^2} = -\frac{M}{r^2} + \frac{H^2(r-3M)}{r^4}$$

From the above it easy to see that when r<3m the last term on the right changes sign for any real value of H and this is where the conclusion that reactive centrifugal acceleration reverses below r=3M originates from.

However it should be noted that above equation of acceleration is expressed in terms of a mixture of proper time of the particle and coordinate length. It is not what any single observer directly measures. The first term on the right is supposedly the gravitational acceleration but it easy to see that it does not correspond with the proper acceleration of particle at rest at that point. The equation I gave earlier for acceleration in terms of $$d^2r'/dt'^2$$ is a more realistic physical interpretation of what is happening.

In terms of local instantaneous tangential orbital velocity, the above equation can be written as:

$$\frac{d^2r}{ds^2} = -\frac{M}{r^2} + \frac{v_s^2}{ r }\frac{(1-3M/r)}{(1-v_s^2/c^2)}$$

 

[Altabeh]

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.

To my knowledge, considering a radial motion, only when $$r=2m$$ the coordinate velocity $$dr/dt$$ is zero and I don't know if your "apogee" points at where particle hits the surface on which the coordinate singularity occurs but putting $$dr/dt =0$$ is not by itself self-consistent and in essence calls for some good reasoning. But since you first assume $$dr/ds=0$$ for which "apogee" could happen in a way completely different from how the coordinate velocity vanishes, I see no reason to back starthaus's position on the problem. To wit, $$dr/ds=0$$ makes the first geodesic equation get the following form (in any plane with $$\theta=\pi/2$$):

$$\ddot{t}=0,$$, or by integrating this, $$\dot{t}=k=const.$$

So that the radial geodesic equations now gives

$$\ddot{r}+\frac{m}{r^2}k^2-r\dot{\phi}^2=0.$$

But since $$\ddot{r}=0,$$ this reduces to

$$L=r^2\dot{\phi}^2= \frac{m}{r}k^2.$$

To this point everything looks fine because we can put any value in front of $$r$$ to get a constant L by looking at the last geodesic equation,

$$\ddot{\phi}=0$$

which if got integrated, would give $$\dot{\phi}=const.$$ Introducing this into the above equation for $$L$$ yields the fact that $$r$$ is not necessary to be $$2m$$ in order for $$dr/ds=0$$ to hold in general. Thus we just proved the non-necessariness of $$r=R=2m$$ if assuming a vanishing $$dr/ds$$ and this shows the "apogee" of trajectories of freely falling particles at least in the planes with a constant $$\theta$$ can be any point i.e. the the particle is instantaneously at rest but definitely this has nothing to ruin the validity of the proof until the condition $$dr/dt=0$$ has not been imposed and fortunately you haven't done so. In contrary to this, if I assume by "apogee" getting a vanishing coordinate velocity, then it is undoubtedly clear that we must put $$r=2m$$ in the formula given by you.

AB

 

[starthaus]

$$\frac{d^2r}{ds^2} = -\frac{M}{r^2} + \frac{H^2(r-3M)}{r^4}$$

From the above it easy to see that when r<3m the last term on the right changes sign for any real value of H and this is where the conclusion that reactive centrifugal acceleration reverses below r=3M originates from.

No, it doesn't. The correct analysis is to write:

$$\frac{d^2r}{ds^2}= -\frac{M}{r^4}(r^2-\frac{H^2}{M}r+3H^2)/\sqrt{...}$$

You can then do an elementary discussion based on the discriminant:

$$\Delta=\frac{H^4}{M^2}-12H^2=H^2(\frac{H^2}{M^2}-12)$$

Obviously, the roots are not only a function of $$M$$ but also of $$H$$, this is why the "$$r<3M$$" above is meaningless.

In addition, in order for the problem to make sense, one needs to add the condition:

$$K^2>(1-\frac{2M}{r})(1+\frac{H^2}{r^2})$$ for any r.

 

[yuiop]

To my knowledge, considering a radial motion, only when $$r=2m$$ the coordinate velocity $$dr/dt$$ is zero ...

I was thinking of it like this. Consider a particle fired directly upwards with less than escape velocity. It will eventually reach a height where it slows down to a stop and reverses direction and falls back down again even if the motion is purely radial. At the turning point where the particle changes from upward motion to downward motion, dr/dt=0 and d^2/rdt^2 is negative. For an eliptical orbit when the particle reaches the point of nearest approach the motion reverses from inwards to outwards and at that point dr/dt also equals zero but d^2r/dt^2 is positive. That is my informal interpretation but it agrees with your rigorous analysis.

Curiously in Schwarzschild coordinate terms, a particle with less than escape velocity at a given height, has dr/dt at its apogee and accelerates downwards but as it gets near the event horizon it starts deccelerating and approaches dr/dt=0 again.

 

[yuiop]

$$\frac{d^2r}{ds^2} = -\frac{M}{r^2} + \frac{H^2(r-3M)}{r^4}$$

From the above it easy to see that when r<3m the last term on the right changes sign for any real value of H and this is where the conclusion that reactive centrifugal acceleration reverses below r=3M originates from.

No, it doesn't. The correct analysis is to write:

$$\frac{d^2r}{ds^2}= -\frac{M}{r^4}(r^2-H^2r+3MH^2)$$

Nope, that should be:

$$\frac{d^2r}{ds^2}= -\frac{M}{r^4}(r^2-\frac{H^2r}{M}+3H^2)$$

 

[starthaus]

Nope, that should be:

$$\frac{d^2r}{ds^2}= -\frac{M}{r^4}(r^2-\frac{H^2r}{M}+3H^2)$$

Same difference, you now know how to do the analysis right.

 

[stevebd1]

It seems there is some agreement that the relativistic equation for centripetal acceleration in Schwarzschild metric is-

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

which is supported by the rearranging the Kepler equation for stable orbit and Kev's derivation, (what's slightly odd is that while the equation for gravity in Schwarzschild metric is considerably well known, you literally have to dig to find the equivalent equation for centripetal acceleration). The whole notion of centrifuge becoming negative at the photon sphere may be nothing more than a mixing of coordinates.

Another equation I found that I thought was of interest is the redshift for an orbiting object, which in Schwarzschild metric is-

$$A=\sqrt{g_{tt} - \Omega^2 g_{\phi \phi}}$$

where $g_{tt}=(1-2M/r)$, $g_{\phi \phi}=r^2$ and $\Omega$ is angular velocity (the equation is reduced from a more complete version which is relative to Kerr metric- $A=\sqrt{(g_{tt} + 2\Omega g_{\phi t}-\Omega^2 g_{\phi \phi})}$).

The equation for a stable orbit (in Schwarzschild metric) for $\Omega$ is-

$$\Omega_s=\frac{\sqrt{M}}{r^{3/2}}$$

If $\Omega_s$ is used in the equation for A then $A\equiv\sqrt{(1-3M/r)}$ and becomes zero at the photon sphere.

The relativistic equation for a_c in Schwarzschild metric can also be derived using $\Omega_s$ where-

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

which again, is reduced from a more complete equation relative to Kerr metric.

 

[stevebd1]

According to a number of sources, the equation for specific angular momentum of an object in stable orbit is-

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

which has been reduced from an equation relative Kerr metric (http://www.tat.physik.uni-tuebingen.de/~kokkotas/Teaching/Relativistic_Astrophysics_files/GTR2009_4.pdf" page 21). This equation can be rewritten-

$$L/m=\frac{\Omega_s\,g_{\phi \phi}}{A}=\frac{\sqrt{M}\,r^2}{r^{3/2}\sqrt{1-3M/r}}$$

where $\Omega_s,\ A$ and $g_{\phi \phi}$ are as the equations in post #52. The equation can also be written as-

$$L/m=\frac{v_s r}{\sqrt{(1-(v_s/c)^2)}}$$

where $v_s$ is as the equation in post #7

In all cases, the equations are equivalent and approach infinity at 3M.


Another equation I see a lot for specific angular momentum is-

$$l=-\frac{\Omega\,g_{\phi \phi}}{g_{tt}}$$

where $g_{tt}=(1-2M/r)$ which again is reduced from an equation relative to Kerr metric (http://luth2.obspm.fr/fichiers/seminaires/straub-seminar.pdf" page 13-14). When $\Omega=\Omega_s$, this isn't equivalent to the other equations and approaches infinity at 2M. It's possible there is a reduction factor that needs to be included for, possibly relative to $\Omega$, in order to make it match. I'd be interested to hear other peoples opinions.

 

[stevebd1]

It appears the factor that brings the equation for $l$ in post #53 in line with the other equations for angular momentum is the equation for conserved energy per unit mass for an object in stable orbit, which for Schwarzschild metric is-

$$E/m=\frac{r^2-2Mr}{r\sqrt{r^2-3Mr}}$$

http://www.tat.physik.uni-tuebingen.de/~kokkotas/Teaching/Relativistic_Astrophysics_files/GTR2009_4.pdf" page 21

The equation for angular momentum for an object in stable orbit can be written as-

$$L/m \equiv (E/m)\frac{\Omega_s\,g_{\phi \phi}}{g_{tt}}$$

_____________________________________________________________________

Incidentally E/m in Schwarzschild metric is equivalent to the product of the gravitational redshift and the Lorentz factor relative to the orbiting object-

$$E/m \equiv \frac{\sqrt{1-2M/r}}{\sqrt{1-(v_s/c)^2}}$$

and A is equivalent to the gravitational redshift divided by the Lorentz factor-

$$A=\sqrt{1-2M/r}\cdot \sqrt{1-(v_s/c)^2}$$The above only applies to an object in a stable orbit.

 

[yuiop]

Your trajectory will not be a straight line but an ellipse, a parabola or a hyperbola. You don't know which until you find out and solve the equations of motion.

.. you need to be able to arrive to the equations of motion and solve them. This way, you will get to a discussion of the different possible trajectories. Putting in dr/ds=0 by hand will not do it.

No, it doesn't. The correct analysis is to write:

$$\frac{d^2r}{ds^2}= -\frac{M}{r^4}(r^2-\frac{H^2}{M}r+3H^2)/\sqrt{...}$$

You can then do an elementary discussion based on the discriminant:

$$\Delta=\frac{H^4}{M^2}-12H^2=H^2(\frac{H^2}{M^2}-12)$$

Are you suggesting that we can tell if the path is elliptical, circular, parabolic or hyperbolic simply by analysing this discriminant? If so I think you are barking
up the wrong tree.

 

[starthaus]

Are you suggesting that we can tell if the path is elliptical, circular, parabolic or hyperbolic simply by analysing this discriminant? If so I think you are barking
up the wrong tree.

No, I was simply pointing out that in addition to your making calculus mistakes you are also making glaring elementary algebra mistakes. That's all.

 

[yuiop]

No, I was simply pointing out that in addition to your making calculus mistakes you are also making glaring elementary algebra mistakes. That's all.

This is just a smear campaign with no substance. Let's have a look at your algebra.

$$\frac{d^2r}{ds^2} = -\frac{M}{r^2} + \frac{H^2(r-3M)}{r^4}$$

From the above it easy to see that when r<3m the last term on the right changes sign for any real value of H and this is where the conclusion that reactive centrifugal acceleration reverses below r=3M originates from. ...

No, it doesn't...

If H and r are positive or negative non-zero real numbers and M is a positive non-zero real number, then the term:

$$+ \frac{H^2(r-3M)}{r^4}$$

is positive when r>3M and negative when R<3M. If you believe otherwise (and your disagreement indicates you do) then it is you that is making a glaring elementary algebra mistake.

I also demonstrated that you made some very basic algebra blunders in https://www.physicsforums.com/showpost.php?p=2754412&postcount=50" of this thread.

Ever heard the expression "people who live in glass houses don't throw stones"?

 

[starthaus]

This is just a smear campaign with no substance. Let's have a look at your algebra.If H and r are positive or negative non-zero real numbers and M is a positive non-zero real number, then the term:

$$+ \frac{H^2(r-3M)}{r^4}$$

is positive when when r>3M and negative when R<3M. If you believe otherwise (and your disagreement indicates you do)

No, I don't believe otherwise, I simply see that you are unable to analyze the sign of the full algebraic expression correctly, that's all. I really don't understand why you keep arguing, I gave you the correct expression and sign analysis long ago, in post 48.

 

[hcgpi]

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"

http://articles.adsabs.harvard.edu//full/1990MNRAS.245..720A/0000720.000.html"

http://arxiv.org/abs/0903.1113v1"

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 a_c=a_g (where a_c is centripetal acceleration and a_g 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 a_c being reduced, it would appear that objects would tend to be thrown out of the ergoregion before crossing the EH but if a_c 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.

I wanted to comment on stevebd1's post of aug13-08 but can't get back to it. Can stevebd1tell me if there is any credible theory to support the notion that an initial Planck sphere of Planck radius at Planck time could have contained the sum of all components of the observable universe or say, simply, the universe? His '08 discussion of Planck density (to which there were no comments) has intrigued me for many months, but I sense I may have missed the point.