# Orbits around black holes

1. Jun 29, 2015

### DParlevliet

Classical calculated with Newton the escape speed is r = 2G.M/v2. With v = c that would be the same as the Scharzschild radius calculated with GR. Does that mean that Newton laws are still valid at (or just outside) this radius?

2. Jun 29, 2015

### Orodruin

Staff Emeritus
This depends on what exactly you mean by "Newton's" laws. Newton's gravitational law is definitely not valid, for example, there are no circular orbits with radius less than 3R/2, where R is the Schwarzschild radius.

3. Jun 29, 2015

### DParlevliet

The classical escape speed is based on Newton's law g = G.M/r2. If GM results in the same formula, does that mean that this Newton law is valid there (supposing a simple BH, sferical, no charge, no rotation)?

Indeed that gives a conflict with minimal circular orbits, but that is the second step.

4. Jun 29, 2015

### A.T.

Only if you interpret the radial Schwarzschild coordinate, which isn’t the physical radial distance, as equivalent to the classical radial distance.

5. Jun 29, 2015

### DParlevliet

What is the difference? If sizes of BK are mentioned (Schwarzschildradius), is that physical or not?

6. Jun 29, 2015

### Staff: Mentor

No. Mercury orbits well outside of the Sun's Schwarzschild radius, and its orbit is not completely and accurately predicted by Newton's laws. This anomaly was noticed early in the eighteenth century, but it wasn't explained until GR came along two centuries later.

(The other planets will have a similar anomaly, but it is much smaller so no one notices).

7. Jun 29, 2015

### DParlevliet

But the formula of escape speed is classical the same as in GM, near the BH

8. Jun 29, 2015

### PWiz

That is a mere coincidence. Most other coordinate values, such as coordinate acceleration, have "correction" factors which make them different from their Newtonian counterparts.

9. Jun 29, 2015

### DParlevliet

Allright. Now the formula of a simple orbit r = G.M/v2 is based on:
1. g = G.M/r2
2. s = 0.5g.t2
3. The curve of the time space dimensions as a circle y = x2/2r (x ≈ 0), which in above is: s = v2.t2/2r.
Is any of those items changing near a black hole according GR and how?

10. Jun 29, 2015

### A.T.

11. Jun 29, 2015

### Staff: Mentor

Yes
See Orodruin's answer in post #2.

12. Jun 29, 2015

### pervect

Staff Emeritus
No, it means there are some formal similarities between the results of Newton's laws, and the result of GR calculations. Interpreting the formal similarities as identities will tend to lead to confusion - if you want to learn about general Relativity, there's no quick and easy substitute for learning it. Most introductory approches to learning GR, such as "Exploring black holes", for instance, of "General Relativity from A to B", will not rely on Newton's laws. Furthermore, it would be advisible to study some about special relativity first before trying to learn GR.

Some of the differences between Newton's laws and GR have already been mentioned, but without the background it may be difficult to appreciate the significance of the differences. But I'll repeat the cautions - "r" is the radial distance in a Newtonian formulation, but it's just a coordinate in GR, and a change in "r" does not correspond to a change in distance.

Other differences between Newton's laws and GR's laws are more severe, for instance if you suspend a weight by a light-weight cable (which takes several pages of text or some rather advanced math to describe precisely and unambiguously what we mean by this), the tension at the lower end of the cable will be different than the tension at the upper end. So even talking about "the force" is ambiguous, you can consider the idea of the force to be equal and opposite to the tension on the cable, but where you measure the tension matters as to what result you get.

Note that this implies that Newton's first law as usually written breaks down - there is not necessarily and equal and opposite reaction, i.e. force, on two ends of a cable in GR, if both ends are at different gravitational potentials.

13. Jun 29, 2015

### PWiz

Isn't that Newton's 3rd law?
And to just add to what pervect said, Newton's 1st law does hold in GR, but in a modified form. The non-relativistic version of the law states that particles continue to move in straight lines unless and until forces act on them. In GR, the modified version becomes "particles continue to move along geodesics unless and until forces act on them". (Remember that in GR, gravity is not a 'real' force, just a fictitious one.)

14. Jun 30, 2015

### Orodruin

Staff Emeritus
But this is certainly true also in Newtonian mechanics. Those forces were never a third law pair. The third law pairs would be the internal forces at any given point on the rope. These act at the same point in space-time and are equal and opposite.

15. Jun 30, 2015

### pervect

Staff Emeritus
I don't see why you say this. It's pretty common in freshman physics to have masses suspended by (lightweight) ropes, and the tension at both ends of the rope is constant.

If we have to get into the mathematical details, a lot of the appeal of the simplicity of the argument to the layperson is lost. But I suppose it's worth exploring to make sure there isn't some error.

I believe the formal non-relativistic version of what I'm saying would involve the conservation form of the cauchy momentum equation

https://en.wikipedia.org/w/index.php?title=Cauchy_momentum_equation&oldid=667601617

$$j = \rho u \quad F = \rho u \times u + \sigma \quad s = \rho g$$
$$\frac{\partial j}{\partial t} + \nabla \cdot F = \rho g$$

In our application, the velocity u is zero which implies that the momentum density j is zero, and the body forces $\rho g$ are presumed negligible because we've assumed that the rope is "lightweight" so its own weight doesn't contribute to the stress. These are the same assumptions as in freshman physics, we don't account for the ropes own weight, we assume it's negligible. Then we can write

$\nabla \cdot \sigma = 0$

the vanishing of the divergence of the classical stress-energy tensor. If we assume a cartesian basis and a corresponding constant-width rope, then the stress-energy tensor in the rope will be

$\left( \begin{array}{ccc}T & 0 & 0 \\0 & 0 & 0 \\ 0 & 0 & 0 \end{array} \right)$

i.e. there will be tension along it's length and no stresses in the other directions. Then the vanishing of the divergence is just $\frac{\partial T}{\partial x} = 0$, which imples the tension in the rope is constant, i.e. it does not depend on length.

The relativistic version is similar in concepet, but different in details. The important difference is we need to take into account the covariant derivative of T. A short summary would be that we replace $\sigma$ with the stress-energy tensor T, and the continuity equation with $\nabla_a T^{ab}=0$. Rather than attempt to muddle through the relativistic version, I'll refer you to Wald , "General Relativity", pg 288

Rather than quote the details, I'll just explain that V here is the redshift factor, so that $V^2 = -\xi^a \xi_a$, the length of a timelike Killing vector, which I would describe in lay language as the time dilation factor, though the later term is coordinate dependent and the Killing vector form isn't.

A much shorter but non-rigorous and potentially argumentative way of saying the same thing is that the conservation of momentum means that a certain amount of it flows through the rope in one coordinate second, and that because of gravitational time dilation, the conservation of this flow of momentum means that the rate of flow with respect to proper time, which is what you can actually measure with a force gauge, varies.

In the Newtonian case, the conservation of the flow of momentum and the static nature of the suspended object means that the flow of momentum, i.e. the force, is constant along the length of the rope.

Constancy of flow of mometum with time was assumed in the Newtonian case, in the GR case we instead talk about a stationary metric and it's associated Killing vector.

16. Jun 30, 2015

### DParlevliet

Let us go back to basics. I know that GR in not the same at Newton and gives strange effect in some situations. But I choose a simple simplified situation:

the classic orbits r = G.M/v2 is based on: g = G.M/r2, s = 0.5g.t2 and the curve of the time space dimensions as a circle y = x2/2r (x ≈ 0), which in above is: s = v2.t2/2r.

Those are simple formula. Simple formula has simple explanations. Around the BH for mass the minimun orbit is r = 6G.M/v2. So what happens. Is r 6 times larger? G smaller? Is v root 6 c?

17. Jun 30, 2015

### pervect

Staff Emeritus
After writing the above, I realized I glossed over a few things. To solve the continuity equation, i.e.$\nabla \cdot \sigma = 0$ or the relativistic equivalent, we have a partial differential equation. To solve this equation, we also need the boundary conditions. The boundary conditions are that the force normal to the rope vanish at the edge of the rope. With a constant with rope this justifies our assumption that $\sigma_xx = T, \sigma_yy = 0, sigma_zz = 0$. If we had a non-constant width rope, we'd have to revisit the boundary conditions in more detail. I'm sure we'd still get the same result, that the continuity equations demanded that the tension in the rope multiplied by the cross section of the rope would be constant, but I haven't waded through the details. I suppose we would also need to look at the existence and uniqueness of the solution to the partial differential equation if we are being extremely thorough, my recollection is that given the boundary conditions the solutions exist and are unique. But I haven't double-checked.

18. Jun 30, 2015

### Staff: Mentor

The underlying question you seem to be interested in is whether there are circular free-fall orbits around a black hole just outside its event horizon. The answer to that question is an emphatic "no". There are no circular free-fall orbits at all inside a radius of 3/2 of the horizon radius, and the circular orbits from 3/2 of the horizon radius to 3 times the horizon radius are unstable to small perturbations. So it would seem that the answer to your question is "no"; you can't use Newton's laws, even as approximations, to understand orbits close to a black hole.

It's formally the same, but the meanings of the symbols are different (in particular the symbol $r$), so it's not the same in any meaningful sense. As what I said just above should make clear, trying to understand trajectories close to a black hole based on formal similarities with Newtonian formulas is not going to be a fruitful strategy.

19. Jun 30, 2015

### pervect

Staff Emeritus
The GR formula are not so simple, you can find them online at "orbits in strongly curves spacetime", http://www.fourmilab.ch/gravitation/orbits/. If you want a textbook reference with the same formula, it will be discussed is Misner, Thorne, Wheeler's "Gravitation".

The details of the derivation would be too complex for a post, and actually you'd be better off with a different approach and book that the "effective potential" approach used in the article, perhaps book such as "Exploring black holes". Not knowing you're background, I don't know if you're familiar with the concept of an effective potential, I will pessimiestically assume the worst.

Be that as me way, we can summarize the results from the webpage and convert them back to standard units.

$L = r^2 \frac{d\phi}{d\tau}$, where L is the angular momentum per unit mass, which has units of r x v = meters*meters/second

This is rather like $r^2 \omega$, but note that the symbols mean something different. R is "just a coordinate", though it may not be too misleading to think of it as a distance. You can think of $2 \pi R$ as a distance, that's defined as the circumference at the R coordinate value of R. Also note that $\tau$ isn't a time coordinate, but is proper time, the time you'd read on a wriswatch on the orbiting oberver.

The webpage gives us the R coordinate at which orbits occur in geometric units. Converting back to standard units, we note that L/c has units of meters, and GM/c^2 has units of meters. And in geometric units, c, and G have values of 1. So we can write:

$$R =\frac{ \left[ \frac{L}{c} \, \left( \frac{L}{c} + \sqrt{\left(\frac{L}{c}\right)^2 - 12 \frac {G^2 M^2}{c^4} } \right) \right] } {\frac{2GM}{c^2}}$$

where I've chosen only to write the equation for the stable orbits. I'm sure the equations can be simplifed considerably, my goal was only to conveert the geometric units on the webpage in as simple a manner as possible.

20. Jul 1, 2015

### DParlevliet

I understand, but the result is simple.
GT is r = 6G.M/v2 and Newton r = G.M/v2. Where is 6 coming from? Is one of the variables 6 times smaller or larger?
It is often suggested that r is not the physical distance we know. If the earth would be a black hole its Scharzschild radius would be 9 mm. Then what is the "real" radius here?