1. ## GEM, c/sqrt(3)

This is not a Q, but it's related to several threads here -- feel free to move if you don't think it belongs.

In my sparring about Mach and GR, I ran across a commentary by B. Mashhoon and another about Mach. Masshoon has a number of papers on gravitomagnetism, relativistic rotation, etc stuff, and I was sort of smoking them over. Interesting stuff, at least for me. Anyway, this one caught my eye:

http://arxiv.org/PS_cache/astro-ph/pdf/0510/0510002.pdf

This is about GEM and what Mashhoon calls "critical speed", which is c/sqrt(3). Sound familiar? That is the speed above which an object "thrown down" radially from infinity in Schwarzchild will appear to *slow down*, rather than accelerate (it's coordinate speed as measured by the far away Schwarzschild observer, of course). You may recall a number of threads about this, one prompted by that Felber/Farber/Whatever guy who made such a big deal about it.

Well, Mashhoon discusses that in GEM context. There's a nice quick derivation of the GEM equations, and you'll note how the factor of 2/4 comes out. Mashhoon like to split the factor of 4 up, letting one go with the B_g field and the other go in the force equation:

F = mg + 2mv x B

He also prefers Gaussian-style units (like most EM high priests), so he has a v/c there, rather than plain v.

Anyway, Mashhoon thinks this "critical speed" may play an important role in what is *obversed* with relativistic flows of matter. Read on.

-Richard

2. Originally Posted by publius
Couldn't

3. Try

http://arxiv.org/abs/astro-ph/0510002

and click on the .pdf to the right. I think the above was a cached thing, which gets dumped after so long.

-Richard

4. Originally Posted by publius
This is about GEM and what Mashhoon calls "critical speed", which is c/sqrt(3). Sound familiar? That is the speed above which an object "thrown down" radially from infinity in Schwarzchild will appear to *slow down*, rather than accelerate (it's coordinate speed as measured by the far away Schwarzschild observer, of course).
What about the critical speed of light itself? How does that work out?

5. Well the speed of light is the speed of light. I noticed you posted something about this in your ATM thread. Remember this c/sqrt(3) is a *coordinate speed*. That is the speed using the ruler and clock of the far away, nearly flat observer. Observers deep down in the well using their own rulers and clock will say that maximum speed is different. They will all agree that things reach a maximum speed, then slow down, but they will disagree on just what it is.

And grav, keep in mind this is also a *velocity dependent* effect of gravity. It is not a gravitomagnetic effect, either, which seems strange to say, since magnetic-like effects are velocity dependent. It is simply a velocity dependent character of the "gravitoelectric" field, which is due to the more complex tensorial nature of gravity.

At first you might say, hey, doesn't that violate the equivalence principle, that all objects "fall at the same rate". Well it doesn't at all. All objects, *travelling along the same geodesic* fall at the same rate. Objects with different velocity vectors at a given point are different geodesics. In the weak field, that effect is small, but velocity itself enters into it no matter how weak the field is. At high v/c, what we think of as "g(r)" is actually very different, indeed, a g(r, v) function.

-Richard

6. I've forgotten exactly how it goes, but you can write a expression for the gravitational acceleration/force in a 3D vector form (and this, as always, using the ruler and clock of the far-away Schwarzschild observer, remember local observers doing whatever they are doing, would see things differently).

That expression has a normal radial component, but has another velocity dependent component that is directed along the velocity vector. Both the radial and the velocity vector part, well, depend on velocity themselves. And all that is still a pure "gravitoelectric" field for Schwarzschild, just the rather complex things a tensor field can do. So, in strong field Schwarzschild think of gravity has having a "drag" component, directed along the velocity vector itself, along with the regular central (radial) component.

When you have a pure radial trajectory, both of those, the velocity vector part, and the central part are directed along the same vector, radial. So the radial free fall critical speed comes from both of those, just looks like it is the same because they're both in the same direction. But for a trajectory with a tangential velocity component, those two will break apart differently.

You might think that would not conserve angular momentum, but it does, just looks like it wouldn't.

Something else I'll ETA: Above, you'll note I talking about gravity as a "field", ie complex tensorial field character. Then in other threads, you'll see me saying gravity is just geometry. Well, those two are not in conflict, just different ways to say the same thing. Whichever one you use sort of depends on the point you're trying to make.

-Richard
Last edited by publius; 2007-Apr-06 at 03:44 AM.

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Originally Posted by publius
Try

http://arxiv.org/abs/astro-ph/0510002

and click on the .pdf to the right.

-Richard
Good find Richard; I really like Mashhoon....he always seems to bring out new insights others seem to pass over; and this report is no exception.

In this case, I like the fact he provides new insight into the General Relativistic 'mechanism' for the Shapiro time delay (of light in a gravitational field)...as well as making the comparative case for the critical velocity for MASSIVE particles in the field.

It is well for us all to have more exposure to these paradoxical post Newtonian gravitational effects, which is particularly evident with respect to the existence of a velocity dependent inflection point for massive particles, (a point which I believe you tried to bring out in previous posts).

The point to be made here (as you indicated) is the (non-intuitive) GR prediction that....

....1. the gravitational field exerts no force upon a massive particle that comes in radially at the critical velocity (of about .57c),

......and 2. for velocities above V(c), gravity becomes 'repulsive'.
(I'm sure that will provoke a lively discussion among the 'gravity is a repulsive force' people ).

Both are evidenced by the GEM equation of motion.....dv/dt = (GMx/r^3)(1-3v^2).

Although the reduction in the coordinate speed of light through a gravity field has much experimental verification, it seems that physical evidence of the similar retardation of relativistic massive particles will be much harder to come by.

Having said that however, if neutrinos prove to be massive (as recent data seems to indicate), then such evidence is already at hand in the data from SN1987A.

Let me expain:
It can be calculated that the Shapiro time delay of the SN1987A photons (due to the gravity from the galactic center) amounts to about 500 days. See here:...
.http://prola.aps.org/abstract/PRL/v60/i3/p173_1

Thus the (almost) simultaneous arrival of SN1987A photons and neutrinos (if shown to be massive) verifies the Gen Rel. prediction of an equivalent gravitational 'slow-down' in ultra-relativistic massive particles (and extends further exp. verification of EP to massive particles) at intergalactic distances, (at least to the .2% level).

Furthermore, terrestrial tests might prove feasible if Felber is correct that the repulsion amounts to -8(Gamma) times Newtonian gravity....although I haven't had time to look closely at it,....you may want to review it here: http://arxiv.org/ftp/gr-qc/papers/0505/0505098.pdf

Mashhoon is interested obviously in the application to the dynamics of astrophysical 'jets', . My first thought on this post Newtonian gravitational effect was its relevance to the ORIGIN of ultra-relativistic cosmic protons (as for example, providing the acceleration mechanism for these enigmatic visitors).

.....Thanks for bringing up such an intriguing topic and, in particular, the report; guys like me aren't always able to go fast enough to dilate the clocks sufficienty to find extra research time.

Gsquare
Last edited by Gsquare; 2007-Apr-14 at 01:21 AM.

8. Originally Posted by Gsquare
Both are evidenced by the GEM equation of motion.....dv/dt = (GMx/r^3)(1-3v^2).
This is interesting. I have recently found that ax=(GMx/d^3)(1+3v^2/c^2) and ay=(GMy/d^3)(1+3v^2/c^2) gives the precession of the planets in a computer simulation program, where M is the mass of our sun. Multiplying by (1-3v^2/c^2) instead would give the same precession but opposite the line of travel of the orbit. First of all, which way should the precession proceed, in the line of travel of the orbit, right? Second, how does this compare to the formula for precession according to GR and with what is being discussed here?

9. Grav,

The (relatively) simple forms of the acceleration quoted above, and in my own various ramblings here are for *radial* trajectories, not the full case with tangential velocity. That gets a bit more involved. Fooling around with radial geodesics is fairly simple, but going "orbital" gets involved.

Mashhoon's paper has the GEM form of the full vector "force" on a test particle. While that is not the full, non-linear Schwarzchild form, that should give the precession of Mercury to first order as it is.

When I get around to it, I'll post the full Schwarzschild equation of motion (the differential equation, not particular solutions) for you. I think I remember a good page showing the derivation of Mercury's precession from those full equations too. The approximate form of that takes a very simple form, and again, it should agree with the linear GEM formulation above.

Anyway, understand that the GEM stuff above, while going beyond Newton, is still an approximation to the full GR in the general case. However, Schwarzschild does have exact solutions. Well, let me be careful there. The metric is exact, an exact solution of the EFE, and the geodesic equations can be written. Now, particular solutions of those geodesic equations, especially for the full angular momentum/orbital cases, may not all have closed form solutions.

Now that I'm thinking about it, in some thread a while back you and I were discussing this, and I think I did post that first precessional result. IIRC, there's a 3 coefficient ( ) and turns out to depend (to the order of approximation) on the "semilatus rectum" of the orbital ellipse.

-Richard

10. Originally Posted by publius
Grav,

The (relatively) simple forms of the acceleration quoted above, and in my own various ramblings here are for *radial* trajectories, not the full case with tangential velocity. That gets a bit more involved. Fooling around with radial geodesics is fairly simple, but going "orbital" gets involved.

Mashhoon's paper has the GEM form of the full vector "force" on a test particle. While that is not the full, non-linear Schwarzchild form, that should give the precession of Mercury to first order as it is.

When I get around to it, I'll post the full Schwarzschild equation of motion (the differential equation, not particular solutions) for you. I think I remember a good page showing the derivation of Mercury's precession from those full equations too. The approximate form of that takes a very simple form, and again, it should agree with the linear GEM formulation above.

Anyway, understand that the GEM stuff above, while going beyond Newton, is still an approximation to the full GR in the general case. However, Schwarzschild does have exact solutions. Well, let me be careful there. The metric is exact, an exact solution of the EFE, and the geodesic equations can be written. Now, particular solutions of those geodesic equations, especially for the full angular momentum/orbital cases, may not all have closed form solutions.

Now that I'm thinking about it, in some thread a while back you and I were discussing this, and I think I did post that first precessional result. IIRC, there's a 3 coefficient ( ) and turns out to depend (to the order of approximation) on the "semilatus rectum" of the orbital ellipse.

-Richard
Thanks, publius. If this is for the radial trajectory, it actually agrees with what I found for gravitational lensing in order to comply with the tangial component for precession. The full formulation for this would then be a gravitational acceleration of a=(GM/d^2)(1+3[(sin L)^2-(cos L)^2](v/c)^2). There are other possibilities too, but this is the most obvious and promising that provides for both the precession and a bending for light that is twice the classical value. The tangial component of orbit, then, would provide for the precession of the planets when (sin L) is close to 1 and (cos L) is close to zero, as with an orbit. For the radial, (sin L)=0 and (cos L)=1, so it becomes a=(GM/d^2)(1-3(v/c)^2), exactly as is shown. If you have any more knowledge on this (no doubt you do ), or if there are any more terms in an extended formulation, it would be much appreciated.

11. Grav,

http://www.astro.ku.dk/~cramer/RelVi...web/node3.html

Here is a page on the Schwarzschild geodesics. The math may be a little heavy, but it covers the equations of motion. Note that Schwarzschild is done in polar coordinates (r, theta) and the equations of motion are therefore in those. Polar coordinates are used in Newtonian orbital mechanics, and familiarity with that would be helpful.

http://www.mathpages.com/rr/s6-02/6-02.htm

Here is a derivation of the perhelion precession. The lowest order approximation can be written simply as

"radians per orbit" = 6*pi(R/L), where L is the "semilatus rectum" of the ellipse -- semimajor/average radius will do fine for 'L' for near circular orbits -- and R is the Schwarzschild radis, 2GM/c^2.

Now, divide by 2pi, to get the "circles per orbit" or "orbits per orbit" figure, and that is simply 3R/L. This the fraction of circle the periapsis will shift with each orbit.

-Richard

12. Grav,

I happened to see your ATM thread, and I want to stress that the GEM equation of motion (radial only), ie

dv/dt = GM/r^2 *(1 - 3(v/c)^2)

which says the acceleration would be zero the whole way down when the initial speed was v = c/sqrt(3), is only the GEM linear approximation.

It does not hold in the strong field. Indeed, there, in the limit of a black hole, everything, light included is going to slow to down to zero (coordinate speed) at the horizon.

As I said in that other thread, in the weak field, it holds pretty good, and something "thrown down" radially at the earth or sun at critical speed would pretty much maintain constant speed all the way until it splattered against the surface.

-Richard

13. G^2,

Glad you found that paper as interesting as I did. Yes, I would agree that more people need to appreciate the "paradoxical", or at least very
non-Newtonian behavior. There is a tendency to think that in a weak field, things are nearly Newtonian, but besides the "weak field" (R/r << 1, or
Phi/c^2 << 1), there is also the *low velocity* (v/c << 1) caveat as well. For high v/c, things are very non-Newtonian even in the weak field, and I think that's an underappreciate point.

About it being "repulsive". That really should be considered "only" a coordinate effect -- basically gravity still "pulls things toward it", it just grabs a hold of things going too fast and brakes them while it pulls them in. If it's not going fast enough, it speeds them up, but it is still attractive in the broad sense.

And you're right indeed -- the reverse of this, throwing something "up" out of the well has quite the opposite effect too: A highly relativistic particle thrown up will *speed up* as it climbs, not slow down!

-Richard

14. Thanks again, publius. It sounds like you're saying that is just a first order approximation that needs to be expanded further. I'll keep working on it. The links you provided should be helpful once I've studied them enough. I'll keep looking through them and working with the Schwarzschild radius to see what else might be determined or if what I am working on tends to deviate in some way.

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## Provocative

Originally Posted by publius

About it being "repulsive". That really should be considered "only" a coordinate effect -- ....
I know Richard; I shouldn't think shock & awe is necessary to awaken the dead heads; I guess I'm just a myth buster disguised as an intentionally provocative trouble maker.
However, the gravitational 'repulsion' is still observed by the local observer, no? ...since the locally measured kinetic energy of the incoming particle deviates from the Newtonian.
Oops; ... there I go again.

And you're right indeed -- the reverse of this, throwing something "up" out of the well has quite the opposite effect too: A highly relativistic particle thrown up will *speed up* as it climbs, not slow down!
Now who's using shock treatment ! ? Let me moderate that so anyone who chooses can ignore the full impact of your statement....Lets see..."If you are REALLY determined to get off the planet, Schwarzchild will assist you".
If you're not careful......I fully expect the 'Daily Planet" will try to contact you about writing (provocative) science articles for their newspaper. he., he

If you are not too offended by my warped GR humor, I do have a somewhat 'intelligent looking' question for you Richard....but it will have to wait till I get back from Church.

G^2

16. I am never offended by warped senses of humor, GR or otherwise, and probably do more offending, than being offended, myself.

Back to being serious, but still "warped" in the GR sense. While the coordinate speed (of something thrown down radially fast enough) does decrease (and in full Schwarzchild, everything is eventually going to slow down), the kinetic energy does increase!

And going backwards, throwing something up fast enough, the kinetic does decrease even though the coordinate speed increases. And, in all cases, the total (relativistic) energy remains constant.

There ain't no such thing as gravitational potential energy -- something that falls (or orbits, or escapes) just "sits there" and follows its geodesic in space-time. All that stuff about accelerating (or not) is just a matter of the projection of that "sitting there" on some observer's coordinate system, his local ruler and clock.

There is no energy associated with the gravitational field at all, it is simply geometry. (Except when there is, such as with radiation, but that gets too complicated for me; seriously, in a space-time that is static, there is no field energy at all; well, one can adopt that description. Now, when mass gets to radiating gravitationally, the field does carry energy [and momentum] away, or can be interepreted to do so. This business gets tricky, and I'm not sure even the high priests are in agreement on the "pedagogy" of this)

If you're interested, and once I get warmed up, we can go through the above odd statements. And for radial motion, we can do it exactly with Schwarzschild, the full "strong field" all the way down and all. It's absolutely elegant, and really very simple, once you get in the necessary "warped" frame of mind.

-Richard

17. Let's do a little warm-up. I'm something of an electrician and shade-tree eee-lectric power engineer (engineer in the sense a garbage man is a sanitation engineer, of course), and one of my profound observations is that to do 3-phase engineering, all one needs to know is the square root of 3, and when to multiply and when to divide by it.

And so imagine my surprise at the coincidence that "critical speed" involved the square root of three, too. Yep, "it's all connected somehow". But, the main point is I have a similiar observation about Schwarzschild. All one needs to know is the factor 1 - R/r, and what powers to use and when to multiply and divide by it.

To that end, the relativisitc energy, E, of test particle of mass 'm' is given by the following in Schwarzchild:

E = mc^2 *sqrt(1 - R/r) * gamma_coordinate

In general, getting a bit fancier, we can write:

E/mc^2 = sqrt(g_00) * gamma_coordinate, and call this ratio the "specific energy". Now, a word about gamma_coordinate.

We can make this look very much like the familiar SR gamma:

gamma^2 = 1/[1 - (v/c)^2].

However, the 'c' has to be the coordinate speed of light, or 'v' has to be the *local speed* (as measured by an observer stationary in our coordinate system). Either way that ratio gets modified by terms involving the metric. So, in GR, the v/c ratio in gamma becomes the ratio of the speed to the coordinate speed of light. In general, the coordinate speed of light is directionally dependent as well as dependent on just position. And Schwarzschild is no exception, which is what makes orbits get more complex.

And that shows you how something crazy, like kinetic energy increasing even though speed was decreasing, can happen. If the coordinate speed of light in the denominator is decreasing faster than v is decreasing, the ratio increases.

And finally, this says the rest energy is actually mc^2*sqrt(g_00). What of that? Well, the g_00 factor is the clock rate. If the clock of something stationary is ticking different than our reference rate, then the rest energy is different as well. "Rest energy" is just the "kinetic energy of motion through time". So, if something's clock is slower, it's moving through time slower (relative to us) and so it has less energy (according to us) than we do. And if its clock is faster, it has more energy.

Now, with the above in mind, when something follows a geodesic, it is doing nothing but moving through its own sweet proper time, experiencing no force, no interaction with anything. It's specific energy remains constant. So let's simply hold that specific energy constant:

(1 - R/r)/[1 - (v/c)_coord^2] = E_0^2, where E_0 is now the initial specific energy. If we drop something from infinity with zero initial velocity, E_0 is just 1. mc^2/mc^2. Now, knowing that pure radial paths are actually geodesics, we can restrict ourselves to pure radial paths. The radial coordinate speed of light is just c*(1 - R/r). Plug that in, and after a little algebra, you'll get:

v(r) = sqrt(2GM/r) * (1 - R/r).

That is indeed the radial coordinate free-fall speed as a function of r. And that is exact. Not how that goes to zero at r = R -- that is the "freezing" at the event horizon. Now, we're getting warmed up...... You will see that has a maximum, and that maximum occurs at r = 3/2 *R, the "photon sphere", where circular orbital speed = c (coordinate).

-Richard

-Richard

18. Now that we're warmed up, let's get in second gear, and look at the radial energy relation for any initial specific energy:

1 - R/r = E^2 *[1 - (v/[c(1 - R/r)])^2].

After a little algebra of getting the v on one side, one gets this:

v = c(1 - R/r) *sqrt[1 - (1- R/r)/E^2]

{Side Note: If you consider tangetial velocity, the factor on the right with E gets another term involving L, the angular momentum, on it, and that sucker is the general form of this}

You will note that for light, or massless particles, the specific energy E is *infinite* (finite total energy over zero rest energy). Let E go to infinity there, and one recovers the radial coordinate speed of light. If one plugs in
E = 1, then one gets the previous expression for dropping something from infinity with zero initial velocity.

Now that we're in second gear, let's shift into third, and get the (coordinate) acceleration. That is dv/dt. But the above gives us a v(r), velocity vs position, not velocity vs time. We could integrate the above (messy, messy) then take two time derivatives. But we can do a little trick.

What is d(v^2)/dr? That is 2v dv/dr. But what is v? That is dr/dt. So, that is just 2 dr/dt * dv/dr = 2dv/dt. So taking the derivative of v^2 above with respect to r gives us the twice the acceleration as a function of r.

Now, everyone, go to town doing that, keeping E in there -- it's just a constant, don't worry. And remember, this is *exact*, so one is rewarded for one's differentation efforts with the full, strong field form of the gravitational acceleration.

-Richard

19. Well shoot. I was hoping someone would've taken the hint and done that differentation for me. And an on top of that, I just got through doing the delicate self-surgical operation of getting a fish bone out of the gum under a bridge in my mouth. And it was supposed to have been boneless.......

Let's see what have we got:

v = c(1 - R/r) *sqrt[1 - (1- R/r)/E^2]. We want the d/dr of the square of that sucker. Well, square that, and make the substituion u = (1 - R/r), our magic factor that's everywhere:

v^2 = c^2 *u^2[1 - u/E^2] -- >

v^2 = c^2[u^2 - u^3/E^2]

Now, d/dr = d/u*du/r. d/du of v^2 is just:

d(v^2)/du = c^2*[2u - 3u^2/E^2]. Now, du/dr is just R/r^2. Recalling that this derivative is just 2a, where a is the coordinate acceleration:

a = c^2*R/2r^2 *[2u - 3(u/E)^2]. But recall that R = 2GM/c^2. So that becomes:

a = GM/r^2 *[2u - 3(u/E)^2]. Note how that factor of 2 on the v^2 trick came in handy to cancel out the 2 in R and gives us our inverse square term on the front. No accident that -- "it's all connected".

But wait a minute, have I forgotten a minus sign? Well, let's see. For light E --> infinity, so that second term in brackets goes to zero, and so this says the coordinate acceleration of a photon is:

a_photon = 2GM/r^2 *(1 - R/r). Far away, for R/r ~ 0, this says that is
twice the Newtonian value in the opposite direction. Have I lost my marbles? And that goes to zero at r = R, agreeing that light freezes at r = R (to our ruler and clock).

The astute lurker will remember the odd result with gravitational lensing that one can show the acceleration (of curve paths with a tangential component) in the weak field limit is twice the Newtonian value. The above odd result is how that factor of two plays out for a pure radial light trajectory. It *decelerates* at twice the Newtonian 'g(r)'!

I'm getting tired of doing all this derivatating and ciphering, so I'm gonna require some of you guys plug E = 1 in there and see what happens. Does that recover Newton in the weak field, low velocity regime? Or did I screw up something and all that nonsense about light slowing down at twice the Newtonian value is crazy? .........................

-Richard

20. And, before I go to bed, I want to point out what just give me the warm and fuzzies about all this. We've had all this fun, derived all these fancy, and *exact* results for strong field Schwarzschild by doing nothing more than conserving energy (and momentum, angular in particular, but that is not explicit in our simple radial excursion here because L = 0). Not a Christoffel symbol in sight (but oh how we stand on the shoulders of such things, they're just hiding in what we've been doing in a more simple form).

We just said that if we drop something, nothing happens, essentially. A test mass simply follows its geodesic, its inertial path, in its own sweet proper time. All this stuff about acceleration, movement, yada yada, is just *a projection* on our own ruler and clock. We're sitting out there stationary watching all this stuff, and all the action is nothing but the "image" of something doing nothing being projected on our local coordinate system.

The only physics there was what we started with, 1 - R/r, which comes from the EFE. All that cipherin' above was simply geometry.

-Richard

21. Originally Posted by publius
I'm getting tired of doing all this derivatating and ciphering, so I'm gonna require some of you guys plug E = 1 in there and see what happens. Does that recover Newton in the weak field, low velocity regime? Or did I screw up something and all that nonsense about light slowing down at twice the Newtonian value is crazy? .........................
Well, even though I've already forgotten much of what you've shown me about derivatives (it's one thing to learn it, another to retain it ), I'll go ahead and try to answer this one, seeing as how you practically gave it away and all, and just to show I am at least attempting to read through this stuff. It seems like the more I learn and figure out on this forum, the more amateurish what I already did know seems to become.

If E=1, I'm assuming for normal matter, and R/r approaches zero for far away coordinates, then u=1 also and we get the Newtonian acceleration of gravity toward the body. For R approaches r, so that R/r is approximately 1, the gravitation is twice that of Newtonian, but repulsive. The gravity would become zero at 2u-3u^2=0, or u=2/3, so at R/r=1/3, hence r=3R, is that right? And this is not an actual repulsion, but a coordinate thing that gives the appearance of repulsion, right?
Last edited by grav; 2007-Apr-15 at 03:28 PM.

22. Originally Posted by grav
If E=1, I'm assuming for normal matter, and R/r approaches zero for far away coordinates, then u=1 also and we get the Newtonian acceleration of gravity toward the body. For R approaches r, so that R/r is approximately 1, the gravitation is twice that of Newtonian, but repulsive. The gravity would become zero at 2u-3u^2=0, or u=2/3, so at R/r=1/3, hence r=3R, is that right? And this is not an actual repulsion, but a coordinate thing that gives the appearance of repulsion, right?
Grav,

Exactly! In the low velocity (E ~ 1), weak field (u ~ 1) regime, the term in brackets becomes simple 2 - 3 = -1, and we get our minus sign on GM/r^2 and we've recovered Newton.

And, very good, for E =1, the acceleration does go to zero at r = 3R/2, the "photon sphere". And that's the E ~ 1 inflection point, where the coordinate velocity begins to decrease. If we drop something radially from a long distance away, it accelerates all the way down to there, the very quickly slows down and freezes at the horizon.

Consider a solar mass Schwarzschild black hole. R is about 1.5km (IIRC -- do 2GM/c^2 for M = 1sol), which means the diameter is about 3km -- that's familiar, but I forget if it's really radius or diameter. Anyway, the diameter of the photon sphere would then be 4.5km.

So just imagine that. We drop something, and it accelerates all the way down to about that 4.5km diameter boundary, then rapidly slows down over that little 0.75km, freezing at the horizon.

And finally, we can get "critical speed" by asking what value of E makes the term in brackets equal to zero for u =1:

2 - 3/E^2 = 0 --> E^2 = 3/2. For u = 1, out there far away, E is just the SR gamma, so E^2 = 1/(1 - (v/c)^2), so

1 - (v/c)^2 = 2/3 --> (v/c)^2 = 1 - 2/3 = 1/3

v/c = 1/sqrt(3)

And there we have it. Anything moving faster than that will be "repelled", or deccelerated, rather than accelerated.

And again, I stress that all this is *exact* for Schwarzschild, we used to full Monty, not any linear approximations, but you can certainly how the linear GEM approximations hold pretty good until R/r becomes significant.

-Richard

23. Originally Posted by publius
And, very good, for E =1, the acceleration does go to zero at r = 3R/2, the "photon sphere". And that's the E ~ 1 inflection point, where the coordinate velocity begins to decrease. If we drop something radially from a long distance away, it accelerates all the way down to there, the very quickly slows down and freezes at the horizon.
I was wondering about that. I thought it should probably be r=3R/2, for the photon sphere, but I got r=3R instead. If 2u-3u^2=0 and u=1-R/r, then u=2/3, so R/r=1-2/3=1/3, and r=3R. It would work out for E^2=1/2, or E=1/sqrt(2), though.

24. Originally Posted by grav
I was wondering about that. I thought it should probably be r=3R/2, for the photon sphere, but I got r=3R instead. If 2u-3u^2=0 and u=1-R/r, then u=2/3, so R/r=1-2/3=1/3, and r=3R. It would work out for E^2=1/2, or E=1/sqrt(2), though.
Grav,

You are exactly right, and that is indeed the correct inflection point,
r = 3R. I had it my head it was the photon sphere from somewhere (obviously messed up a factor of 2, saying u = 1/3, rather than 2/3), and I have been repeating it.

But, no, the inflection point is indeed 3R, not 3R/2. And that's even more interesting actually. 3R is the last *stable* circular orbit radius. 1.5R is where circular speed equals light speed, but orbits in between 3R and there are not stable, the slightest perturbation will send them flying out, or flying in to oblivion depending.

I suspect these two are related. At some point, circular orbit speed becomes greater than free fall speed, which is another odd non-Newtonian behavior, and I wonder if the inflection point is related to that some way.

-Richard

25. Originally Posted by publius
Grav,

You are exactly right, and that is indeed the correct inflection point,
r = 3R. I had it my head it was the photon sphere from somewhere (obviously messed up a factor of 2, saying u = 1/3, rather than 2/3), and I have been repeating it.

But, no, the inflection point is indeed 3R, not 3R/2. And that's even more interesting actually. 3R is the last *stable* circular orbit radius. 1.5R is where circular speed equals light speed, but orbits in between 3R and there are not stable, the slightest perturbation will send them flying out, or flying in to oblivion depending.

I suspect these two are related. At some point, circular orbit speed becomes greater than free fall speed, which is another odd non-Newtonian behavior, and I wonder if the inflection point is related to that some way.

-Richard
Oh, okay. Those factors of two can be troublesome sometimes for some reason, can't they? Maybe easily overlooked. I find myself checking these days to see if the value I think something is shouldn't really be twice as great or half as much for one reason or another.

26. Grav,

I don't know what I did, I can't remember. Note that if I got screwed up and forget 'u' was (1 - R/r), but said it was R/r, I would've got the 3/2 for u = 2/3.

That may have been what I did, because going over it in the more simple form (without E, just letting E = 1), I can't see any obvious place to make the error other than doing the above.

-Richard

27. Speaking of factors of two, the formula I came up with for precession seems to be off by a factor of two from GR for some reason. Maybe you can tell me why. The formula is GM/r^2(1+3*(v/c)^2), and GM/r^2=v^2/r gives the regular Newtonian orbit, where (GM/r^2)(r/v^2)=1 is the number of orbits per orbit travelled according to Newton, so (GM/r^2)(r/v^2)*3(v/c)^2 is the extra amount of the orbit per orbit for the precession. This should be equal to the precession according to GR, which is 6(GM/r)/c^2. But instead I get (GM/r^2)(r/v^2)*3(v/c)^2=3((GM/r)/v^2)(v/c)^2=3(GM/r)/c^2, which is only half as great. What did I do wrong? It can't be time dilation because these are the factors we actually measure for this and v is no where relativistic.

28. Grav,

I'm not exactly clear on your interpretation of "orbits per orbit" -- a better way to say it would "fractions of circle per orbit" of periapsis shift. For a pure circular orbit, this would "degenerate" into a simple difference between the Newtonian orbital speed and the GR orbital speed at the same radius, I believe.

And Grav, remember the GR precession formula is actually ~3R/l, where 'l' is not the radius or semi-major axis, but is actually the semilatus rectum of the ellipse. For a long, skinny ellipse, 'l' is much less than the semi-major axis.

In one of those links above is the derivation of that. Other than this, I'm not going to much more help, because I'm just too lazy right now to get into Schwarzschild orbits. It gets, well, more difficult.........

But a word to the wise. The best way to do orbits is to use polar coordinates, not cartesians, at least for exact work. Doing numerical stuff, cartesians are probably as good any, even more straightfoward.

The above link uses polar coordiantes, that's what Schwarzschild is anyway. In any text on classical mechanics, you quickly switch to polar coordinates to solve Newtonian orbits. I would urge you to get such a text and see how it's done.

In polar coordinates, you're after two functions of time, r(t) and O(t), where O is theta, the angular coordinate. For a circular orbit, r(t) = constant, and O(t) just goes around at some w. Now, for an ellipse, r(t) become a variable. It turns out to be a periodic function, going from a minumum (perispasis) to a maximum (apoapsis, aopogee, whatever).

You then find out that r(t) and O(t) there are "commensurate". r(t) returns to the minimum at the same O, and you eliminate 't' and get r(O) and say, "Aha, that's an ellipse". Now, add in GR's correction. What happens there is that r(t) returns to the minimum a little before (or after). r(t) and O(t) are no longer exactly commensurate, and the effect is the angular position of the periapsis just "walks" along a bit with each orbit.

-Richard

29. Oh, sorry. I'm just figuring 'l' as the radius of a circular orbit, so an approximation to the first order is all we would need, in "revs per rev" or "fractions of a circle per orbit", as you say.

30. bump.........

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