# Insights Centrifugal Force Reversal Near A Black Hole - Comments

1. Aug 19, 2015

### Staff: Mentor

2. Aug 19, 2015

### A.T.

3. Aug 19, 2015

### Staff: Mentor

Hmm. I hadn't thought of that, but it might work, yes.

4. Aug 19, 2015

### WannabeNewton

5. Aug 20, 2015

### Staff: Mentor

Nice post!

Out of curiosity, some acceleration values at r=3M (they scale with 1/M):
Sun as BH: 7*1012 m/s2
Sagittarius A*: 1.6*106 m/s2
S5 0014+813 (largest one known, about 40 billion solar masses): 1.7*102 m/s2 or 17g. There are some rockets that can deliver this for a few seconds.

6. Aug 20, 2015

### bcrowell

Staff Emeritus

Here's another possible way of approaching this. Say we have three people:

* h = hovering observer
* s = spaceship pilot circling the black hole
* o = radially free-falling observer who is instantaneously at rest relative to h

Let

* $a^s$ = proper acceleration of s
* $a^h$ = proper acceleration of h
* $a_o^s$ = three-acceleration of s as measured by o
* $a_o^h$ = three-acceleration of h as measured by o

Then assuming I have my math right, we have

$\frac{a^s}{a^h} = \gamma_s^2 \frac{a_o^s}{a_o^h}$

The second factor on the right is $<1$, which makes us intuitively expect that the result will be $<1$. In the Newtonian case, if s is in a circular orbit with the engines turned off, then $a_o^s$ can even be zero. However, for $r<3M$, an inertial circular orbit is not possible (discussed here: https://en.wikipedia.org/wiki/Schwarzschild_geodesics#Circular_orbits_and_their_stability ), so there is some minimum positive value for the ratio $a_o^s/a_o^h$. I think what must happen is that the factor of $\gamma_s^2$ dominates, and the result is $>1$. Obviously there are more details that I haven't worked out, but I think the two analyses should give the same answer.

7. Aug 20, 2015

### Staff: Mentor

We must always have $a^h = a_o^h$, so this reduces to $a^s = \gamma_s^2 a_o^s$, which looks ok.

I may be missing something obvious, but I don't understand why this would have to always be the case. I understand why it's the case in the Newtonian limit (weak field, very far away from the source), but I don't understand why it must always be the case, even for $r < 3M$.

8. Aug 20, 2015

### bcrowell

Staff Emeritus
I'm starting a second glass of wine, so I may be getting this wrong, but let the world-lines of s, h, and o all coincide at some time. Near that point of intersection, introduce a local Minkowski coordinate system representing o's frame, with y the radial coordinate and x azimuthal. Then for times soon after that, don't we have $y_o<y_s<y_h$?

9. Aug 21, 2015

### Staff: Mentor

I was assuming that that event was the origin of the local inertial coordinates, yes.

Ah, I see now. I should have just used the formula in my article to compute $a^s_o / a^h_o$. Here's what that gives; the ratio of proper accelerations is:

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

But we know that $a^h = a^h_o$ and $a^s = \gamma^2 a^s_o$, so we can substitute those into the above to get:

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

You're correct that this will always be less than 1, so the key is how much less, i.e., whether the extra factor of $\gamma^2$ in front is enough to overcome the above effect. Evidently it will be for $r < 3M$, but not for $r > 3M$ (at $r = 3M$ the two effects cancel and the ratio of proper accelerations is 1).

10. Aug 21, 2015

### bcrowell

Staff Emeritus
I'm not sure who's making a mistake here, you or me. This doesn't seem right to me for the following reason. In the Newtonian limit, we have $r\gg M$, so this becomes $a^s_0/a^h_o = rv^2/M$. Reinserting constants for SI units makes this $rv^2/GM$, which for a Newtonian circular orbit equals 1 identically. But in the Newtonian case, $a^s_o=0$ exactly, regardless of $r$, while $a^h_o=g\ne0$.

11. Aug 21, 2015

### Staff: Mentor

No, it doesn't. It becomes

$$\frac{a^s_o}{a^h_o} = 1 - \frac{r}{M} v^2$$

You can't ignore the $1$, because if $v$ is small enough the second term will be smaller than the first even if $r \gg M$. (In fact, regardless of how large $r / M$ is, there will be some $v$ for which the two terms cancel and $a^s_o = 0$--this will be the free-fall orbital velocity, $v = \sqrt{M / r}$, or $v = \sqrt{GM / r}$ in conventional units.)

12. Aug 21, 2015

### bcrowell

Staff Emeritus
Ah, I see. That makes sense.

13. Aug 21, 2015

### bcrowell

Staff Emeritus
Given that (in my notation) $a^s>a^h$, but $a_o^s<a_o^h$, it's not obvious to me that one should really interpret this verbally as a reversal of the centrifugal force. At best, the term "centrifugal force" is vague as applied here.

Is this related to the so-called Hilbert repulsion? Hilbert repulsion seems to be a good example of the danger of applying these verbal interpretations to things. You get kooks like Loinger who say, "OMG, this has earthshattering implications!"

14. Aug 21, 2015

### Staff: Mentor

Yes, particularly since your analysis is done in a local inertial frame, in which there is, by definition, no "centrifugal force" (or any other "fictitious" force). The only reason I used the term "centrifugal force reversal" in the title of the article is that many sources (including some in particular that were cited in PF threads a while back) use that term, and I wanted to show that there was, indeed, a "real" phenomenon involved, it wasn't all just an artifact of coordinate choices.

The unquestionably "real" phenomenon, though, is the behavior of the ratio $a^s / a^h$, i.e., the proper accelerations, not the coordinate accelerations. If we want to try to construct an interpretation in which this behavior corresponds to the behavior of "centrifugal force", the best way I can see to do it would be to work in the rotating frame in which the "orbiting" spaceship "s" is at rest. In this frame, there is a "force of gravity" from the black hole, which always "points downward" and requires an upward thrust to counter in order to stay at rest (the "h" part in your labeling). Assuming the "s" ship is not hovering, that it has some nonzero velocity relative to the hovering "h" observer, then there will, in general, also be a "centrifugal force" in this frame, but it won't always point in the "usual" direction, i.e., opposite to the direction of the "force of gravity". For $r > 3M$, the centrifugal force points in the "usual" direction, upwards; at $r = 3M$, the centrifugal force is absent; and for $r < 3M$, the centrifugal force points downwards, in the same direction as gravity, hence the term "reversal". But this is a frame-dependent interpretation (which is why some of those past PF threads got all bogged down in the question of whether it was "real"), whereas the behavior of the proper acceleration ratio $a^s / a^h$ is not.

If you mean what's discussed in this paper, I think it's somewhat related, in the sense that calling the phenomenon "repulsion" is, IMO, a confusion based on looking at coordinate-dependent quantities instead of invariants.

15. Aug 22, 2015

### stevebd1

FWIW, using the equations from page 10 to 12 from the following paper 'Geometric transport along circular orbits in stationary axisymmetric spacetimes' http://arxiv.org/abs/gr-qc/0407004 , I removed the spin properties $(a)$ to get the gravity field for an object with tangential velocity around a static BH and got the following equation-

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

Which gives the exact same results and displays the same phenomena near the photon sphere as the equation in PeterDonis's Insight post.

16. Aug 22, 2015

### Staff: Mentor

Yes, with some algebra this equation is the same as the one I gave in the article (in fact I obtained this one as an intermediate step in computing the one I gave in the article).