Centrifugal Forces and Black Holes

[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"

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.

 

[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?

 

[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.

 

[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:<br /> <br /> L = r^2\frac{d\phi}{ds}<br /> <br /> 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.<br /> <br /> 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&lt;r&lt;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&lt;=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). <br /> <br /> 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.<br /> <br /> *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 <img src="https://cdn.jsdelivr.net/joypixels/assets/8.0/png/unicode/64/1f609.png" class="smilie smilie--emoji" loading="lazy" width="64" height="64" alt=":wink:" title="Wink :wink:" data-smilie="2"data-shortname=":wink:" /><br /> <br /> 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.

 

[Altabeh]

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.

That would be awesome if you also included in your post the proof procedure.

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.

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

 

[yuiop]

That would be awesome if you also included in your post the proof procedure.

AB

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 \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

 

[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 v_s into a_c=a_g 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 v_s 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.

 

[Altabeh]

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

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

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.

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.

Proof. Start with the following simple expansion,

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

 

[yuiop]

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

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.

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 v_s into a_c=a_g 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}}

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.

 

[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}

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

Start with the radial coordinate velocity of a free falling particle given earlier:

\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} = <br /> -\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&#039;) 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&#039; = \frac{d^2r&#039;}{dt&#039;\,^2} = \frac{d^2r}{dt\,^2}\gamma_g^3 - \frac{2M}{r^2}\left(\frac{dr}{dt}\right)^2 \gamma_g^5<br /> =-\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&#039; = \frac{d^2r&#039;}{dt&#039;\,^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_{*}&#039; 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_{*}&#039; = -\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_{*}&#039; = -\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_{*}&#039; = -\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_*&#039; 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.

 

[Tomsk]

As far as I know, the proofs of why these values are constant are non-trivial and involve Killing vectors or Noether's theorem

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?

 

[yuiop]

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?

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.

 

[starthaus]

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 \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.

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

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)}

Nope, there is a much cleaner expression:

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

Why would 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)

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}

 

[starthaus]

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:

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

 

[yuiop]

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)}

Nope, there is a much cleaner expression:

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

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

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)

Where did R come from?

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.

 

[starthaus]

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".

:lol:

 

[starthaus]

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 H (the conserved angular momentum) at that point.

But you don't know H .

 

[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".

:lol:

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.

 

[yuiop]

But you don't know H .

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

 

[starthaus]

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

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}

 

[yuiop]

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}

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

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?

 

[starthaus]

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}


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?

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.

 

[yuiop]

Why would dr/ds=0?

dr/ds is always zero at the apogee.

 

[starthaus]

dr/ds is always zero at the apogee.

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.

 

[yuiop]

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}

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.

 

[yuiop]

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.

It might be a hyperbola in certain cases. The point where dr/dt=0 might be an apogee or perigee (or more precisely an apoapsis or periapsis if we are not talking about the Earth). That is why I said in #6:

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)

To be more rigorous you would have to find the effective potential and determine if the particle is at maxima or minima. For a quick check if r<R and dr/dt is imaginary that is a pointer that (dr/dt=0 and R=r) might be the perigee and the particle is moving outwards (which happens if H and R are large). By entering a value of r>R a real solution for dr/dt is found in this case. This is one way of determining if the solution is a positive square root or a negative square root.

 

[yuiop]

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}

It is not simpler, because if you have a moving particle you do not know dt/ds unless you know its velocity which is defined by L (its angular momentum/rest mass) and dr/dt. K and L together define the trajectory.

 

[starthaus]

It might be a hyperbola in certain cases. The point where dr/dt=0 might be an apogee or perigee (or more precisely an apoapsis or periapsis if we are not talking about the Earth). That is why I said in #6:

Since you don't know, you can't use dr/ds=0. Just as simple.
Besides, you don't know R.
You don't know the equations of motion=> you don't know the trajectory.
You don't know the trajectory=> you don't know your "R", whatever that may be.

 

[starthaus]

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.

Yes, you are right. Come to think , I was the one showing you that

r^2\frac{d\phi}{ds}=constant from the Euler-Lagrange equations.

 

[yuiop]

Since you don't know, you can't use ds/dr=0. Just as simple.

Let us say we have problem expressed like this:

"An initially stationary particle is dropped from height R. What is its coordinate velocity at r?"

Right away the question has defined H=0 because it's falling radially and dr/ds=0 at r=R because it was initially stationary at R.

 

[starthaus]

Let us say we have problem expressed like this:

"An initially stationary particle is dropped from height R. What is its coordinate velocity at r?"

Right away the question has defined H=0 because it's falling radially and dr/ds=0 at r=R because it was initially stationary at R.

But, as I explained to you eons ago, this is not the case you are trying to treat. There is the other pesky coordinate \phi that you need to consider.
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.

 

[yuiop]

But, as I explained to you eons ago, this is not the case you are trying to treat. There is the other pesky coordinate \phi that you need to consider.

I am sure you will agree I have considered coordinate \phi in great detail in posts #6 and #10. It is not difficult to observe the path of a particle and measure where its apogee or perigee is and at that point dr/dt=0. I am sure there are cases where for example a new comet is spotted and its perigee is unkown and determining its future trajectory will much more complicated, but that is going way beyond the remit of the OP of this thread.

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 have gone some way towards answering that question by determining the coordinate velocity and coordinate and local acceleration of the particle when the particle has angular momentum. What is your contribution?

 

[yuiop]

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_{*}&#039; = -\frac{M}{r^2 \sqrt{1-2M/r}} + \frac{ v_s^2\sqrt{1-2M/r}}{r} \:\: (eq7)

I have just noticed that (eq7) is in agreement with the equations given by Steve, if the a_*&#039; [/itex] is interpreted to be the total acceleration acting on the particle and not just the gravitational acceleration. I should have spotted that earlier but an unfortunate typo in (eq7) (that has now been corrected) prevented me making the obvious connection. In other words the term:<br /> <br /> -\frac{M}{r^2 \sqrt{1-2M/r}}<br /> <br /> is the gravitational acceleration a_g&amp;#039; acting inwards and<br /> <br /> + \frac{ v_s^2\sqrt{1-2M/r}}{r}<br /> <br /> is the effective reactive centrifugal acceleration a_c&amp;#039; acting outwards so that:<br /> <br /> a_*&amp;#039; = -\frac{M}{r^2 \sqrt{1-2M/r}} + \frac{ v_s^2\sqrt{1-2M/r}}{r} = a_g&amp;#039; + a_c&amp;#039;<br /> <br /> a_*&amp;#039; is the resultant acceleration that is actually measured by a stationary local observer using the proper length and proper time of his local rulers and clocks. This particular form of the equation is valid only to the apogee or perigee of a given orbital path and does not just apply to circular orbits as I said ealier. <br /> <br /> If a_*&amp;#039; is set to zero then the above equation is in agreement with Steve&#039;s claim that:<br /> <br /> <blockquote data-attributes="" data-quote="stevebd1" data-source="post: 2752124" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-title"> stevebd1 said: </div> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> \sqrt{1-2M/r}\,\frac{v_s^2}{r}=\frac{M}{r^2 \sqrt{1-2M/r}} </div> </div> </blockquote><br /> although the form of Steve&#039;s equation is only valid for circular orbits and can not be used when 2M&lt;R&lt;3M. The equation for a_*&amp;#039; can still usefully be applied to evaluate what happens between 2M&lt;R&lt;3M. A positive value of a_*&amp;#039; means that the point where dr/dt=0 is a perigee (point of nearest approach). With the constraint that v_s&lt;=1, it is not possible to have a perigee with R&lt;3M because total acceleration a_*&amp;#039; is always negative in this region.<br /> <br /> The equation I gave earlier for dr/dt seems to confirm that if two particles start at R=3M and one has angular velocity and the other does not, the one with angular velocity arrives at the event horizon later. Therefore, the effective reactive centrifugal acceleration is still slowing the rate of descent of the particle even below R=3M. What happens below R=2M is another matter.

 

[yuiop]

Since you don't know, you can't use dr/ds=0. Just as simple.
Besides, you don't know R.
You don't know the equations of motion=> you don't know the trajectory.
You don't know the trajectory=> you don't know your "R", whatever that may be.

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:

... the radial coordinate velocity of a free falling particle given earlier:

\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)

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.

 

[starthaus]

I have gone some way towards answering that question by determining the coordinate velocity and coordinate and local acceleration of the particle when the particle has angular momentum. What is your contribution?

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.

 

[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&#039;/dt&#039;^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&lt;3M" above is meaningless.

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

K^2&gt;(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)