Are Newton's Laws Still Valid Near Black Hole Orbits?

[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?

 

[Orodruin]

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.

 

[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.

 

[A.T.]

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.

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

 

[DParlevliet]

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

 

[Nugatory]

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?

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).

 

[DParlevliet]

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

 

[PWiz]

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

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

 

[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?

 

[A.T.]

What is the difference?

See description of r here:
https://en.wikipedia.org/wiki/Schwarzschild_metric#The_Schwarzschild_metric

 

[Nugatory]

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

Yes

and how?

See Orodruin's answer in post #2.

 

[pervect]

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?

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.

 

[PWiz]

Note that this implies that Newton's first law as usually written breaks down - there is not necessarily and equal and opposite reaction,

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.)

 

[Orodruin]

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.

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.

 

[pervect]

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.

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

However, if we choose to calculate the forces which must be applied by a distant observer at infinity (i.e by means of a long string), we find that this force differert from the local force by a factor of V.

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.

 

[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?

 

[pervect]

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.

 

[PeterDonis]

Does that mean that Newton laws are still valid at (or just outside) this radius?

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.

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

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.

 

[pervect]

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?

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:

<br /> 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}}<br />

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.

 

[DParlevliet]

The GR formula are not so simple,

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?

 

[Nugatory]

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?

I expect that by "the 'real' radius" you mean the straight-line distance measured with meter sticks, a micrometer, or whatever from the center to the horizon? If that's what you mean, then possible answers are "infinity", "undefined", or "That's like asking for the distance in kilometers between noon and Tokyo - I don't know how to answer".

When you hear that the Schwarzschild radius of something is 9mm, that tells you that the surface area of a spherical shell arbitrarily close to the event horizon will be arbitrarily close to $4\pi(.009^2)$, a formula that you will recognize as the surface area of a sphere. But that's about it - 9mm is not any distance that you'll find anywhere in or around the object.

 

[DParlevliet]

that the surface area of a spherical shell arbitrarily close to the event horizon will be arbitrarily close to $4\pi(.009^2)$,

That is what I mean (or the diameter 2r)
So in both formula r = 6G.M/v² (GT) and r = G.M/v² (Newton) r is (about) the same.
I suppose G and M and v are also the same in both formula? Then the general formula is r = a.G.M/v²
Closest to BH and v is close to c: a = 6
Large distance and low V: a = 1
How does a behave in between? Is it dependend on r or v? Or more scientific: what is the formula of a?

 

[Nugatory]

That is what I mean (or the diameter 2r)
So in both formula r = 6G.M/v² (GT) and r = G.M/v² (Newton) r is (about) the same.
I suppose G and M and v are also the same in both formula? Then the general formula is r = a.G.M/v²
Closest to BH and v is close to c: a = 6
Large distance and low V: a = 1
How does a behave in between? Is it dependend on r or v? Or more scientific: what is the formula of a?

G and M are constants and the same in both sets of formulas. However, the r that appears in the GR formulas has a completely different physical meaning than the r that appears in the Newtonian formulas - it's little more than a historical accident that we use the same letter for both, especially near and inside the event horizon. Thus, any attempt to define velocity and acceleration in terms of the Schwarzschild r is going to go seriously wrong somewhere.

 

[stevebd1]

The equation for a stable orbit in Schwarzschild space time (i.e. around a static black hole) is-

v_s=\sqrt{\frac{M}{r\left(1-\frac{2M}{r}\right)}}

where M=Gm/c^2 (note the answer would be a fraction of the speed of light, c). The above reduces to v_s=\sqrt(M/r), the Newtonian version, when at a great distance from the black hole. You can see that if you introduce 3M, which is the photon sphere (the radius at which light is predicted to orbit the BH), then v_s=1.

Calculating the marginally stable orbit (MSO) for an object of mass is more complicated though for a Schwarzschild BH, it is 6M, you also have the marginally bound orbit (MBO) at 4M.

On the side note of Newtonian equations appearing in GR equations, it's probably also worth pointing out that the equation for tidal forces (or gravity gradient) remains unchanged in GR (dg=dr\ 2M/r^3), even though gravity is infinite at the event horizon (2M).

 

[DParlevliet]

In your previous answer #21 you told "When you hear that the Schwarzschild radius of something is 9mm, that tells you that the surface area of a spherical shell arbitrarily close to the event horizon will be arbitrarily close to 4π(.009²)"
In Newton formula the spherical area is also 4π (.009²)

 

[PeterDonis]

In Newton formula the spherical area is also 4π (.0092)

Yes, that's because the radial coordinate $r$ is defined so that the surface area of a 2-sphere at radial coordinate $r$ is $4 \pi r^2$.

The difference is that, in ordinary Euclidean geometry (which is really where the formula $4 \pi r^2$ comes from, not Newton), the quantity $r$ is also the physical distance from the center of the 2-sphere to its surface. In Schwarzschild spacetime, it isn't; in fact, there is no such thing as "the physical distance from the center of the 2-sphere to its surface", because there is no such thing as "the center of the 2-sphere". The geometry of Schwarzschild spacetime does not work that way.

 

[DParlevliet]

Allright, then change the formule in D = 12G.M/v² (GT) and D = 2G.M/v² (Newton)
Is D (diameter) in both formula the same?

 

[PeterDonis]

D = 12G.M/v2 (GT)

Where does this come from?

Is D (diameter) in both formula the same?

A Schwarzschild black hole does not have a meaningful diameter, any more than it has a meaningful radius.

 

[DParlevliet]

A Schwarzschild black hole does not have a meaningful diameter, any more than it has a meaningful radius.

Suppose there is a BH with the mass of the sun with r_s = 3 km. Suppose we are able to see this BH with a telescope. What would we see, with what size?

 

[Nugatory]

Suppose there is a BH with the mass of the sun with r_s = 3 km. Suppose we are able to see this BH with a telescope. What would we see, with what size?

We would, if we get the light and other conditions right and don't confuse things with accretion disks and other stuff falling in and orbiting, see a circular black area that covers an area of the sky. We could then find the number $r$ such that a massless spherical shell of surface area $4\pi{r}^2$ would cover the same amount of sky - but that number would have no more physical significance than that. For example, if we had two such shells, one with surface area $4\pi{R_1}^2$ and the other with surface area $4\pi{R_2}^2$ where $R_1 > r$ and $R_2 = R_1+\Delta{r}$ the gap between them would be not be $\Delta{r}$.

 

[pervect]

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?

I don't understand where this formula came from. If we take the limit of large L in the formula I gave above, so that (L/c) >> (GM/c^2), we find that the term in the square root is essentially equal to L/c. Note that physically GM/c^2 is just the Schwarzschild radius. Then we can write that $R = L^2 / GM$. This is the same as the Newtonian formula. In the limit of large R -> large L, we can say that there are no significant time dilation effects, so that we can replace $d\phi / d\tau$ with $\omega$. Substituting in the definition of L, our formula reduces to $R = R^4 \omega^2 / GM$ . We can re-write this to solve for $\omega$ and we get $\omega^2 = GM/R^3$.

The lowest possible value of L for a stable circular orbit is when $\frac{L^2}{c^2} = 12 \left(\frac{GM}{c^2}\right)^2$, because the term in the square root must be positive for the formula to apply. Solving for R, the term in the square root is zero, and $R = \left(\frac{L}{c}\right)^2 / \left(2GM/c^2\right)$ To find the value of R at which this occurs, we substitute $L^2/c^2 = 12 G^2 M^2 / c^4$ and get $R = 6GM / c^2$.

There are circular orbits inside this radius, but they are unstable. The smallest possible unstable orbit occurs when $R = 3 \frac{GM}{c^2}$, which is the photon sphere. Inside this, no circular orbits exist at all.

The web-page gives the differential equation and a simulator for what happens for the orbit of a particle with a given L and a given initial radius, along with the "effective potential" diagram. The behavior isn't intuitive, but it follows from the differential equations of motion.

Perhaps the exposition in terms of L is less satisfying than one in terms of "velocity", but it's much easier to define. It would be possible to further expound on the nature of what we mean by "velocity", but that would take a lot of writing, and I'm not sure that it's not a digression from the topic as well as being a lot of work.

 

[DParlevliet]

a circular black area that covers an area of the sky. We could then find the number $r$ such that a massless spherical shell of radius $r$ would cover the same amount of sky

That is what I mean and that is the radius Newton would have used.

The equation for a stable orbit in Schwarzschild space time (i.e. around a static black hole) is

and that is what I was looking for.

 

[Nugatory]

The equation for a stable orbit in Schwarzschild space time (i.e. around a static black hole) is-
...

That is what I mean and that is the radius Newton would have used.
and that is what I was looking for.

As long as you understand that those orbits look nothing like the orbits that Newton would have calculatde by plugging r into his formulas, and that if two objects are both in circular orbits of radius $R_1$ and $R_2$ the distance between their orbits is not $R_2-R_1$... Yes, I suppose you could say that.

Personally, I really seriously doubt that Newton would have accepted a definition of "radius" that had that property, but he's not around to ask.

 

[DParlevliet]

that if two objects are both in circular orbits of radius $R_1$ and $R_2$ the distance between their orbits is not $R_2-R_1$

Perhaps, but through our telescope we would see two black circles with differ $R_2-R_1$ in radius.

 

[Nugatory]

Perhaps, but through our telescope we would see two black circles with differ $R_2-R_1$ in radius.

We would see two black circles, one with radius $R_2$ and one with radius $R_1$ so are different in that one has radius $R_2$ and the other radius $R_1$. However, they do not differ in radius by $R_2-R_1$ - you won't find that quantity anywhere in the setup, it's just something that you might write down on a piece of paper. It's like saying that two airplanes are "two inches apart" because you lined one end of a ruler up with the horizon and then the three-inch mark happened to line up with one of the airplanes and the five-inch mark happened to line up with the other.

 

[Ibix]

Note: I'm still learning GR. Don't assume the following is a valid analogy until Peter, pervect, Nugatory et al haven't shot it full of holes.

If you look at the rings on a dartboard, the radius of each circle is r, and this is equal to the distance to the centre of the board.

If you look at latitude lines on the Earth, each one is a circle with a radius r which is not simply related to the distance to the pole.

Newtonian theory assumes a flat, Euclidean background (like the dartboard), where the circumference of a circle is simply related to the distance from the bullseye. GR allows more complex geometries where the distance to the the singularity (analogous to the pole of our sphere) is not simply related to the circumference of a circle around it (analogous to a line of latitude).

So, Newtonian calculations tell you that the escape velocity at some distance from the bullseye is something. GR calculations tell you that the escape velocity at some distance from the rotation axis of the Earth is the same as the Newtonian calculation. But the distance to the rotation axis of the Earth is not the distance from the pole and, in the real space-time case rather than the sphere analogy, isn't a distance with any real meaning - just a convenient mathematical concept. Which is why everyone is telling you that it is a coincidence that the expressions are the same.

 

[DParlevliet]

It's like saying that two airplanes are "two inches apart" because you lined one end of a ruler up with the horizon and then the three-inch mark happened to line up with one of the airplanes and the five-inch mark happened to line up with the other.

No, I measure the diameters of the black holes with the telescope and subtract them. By definition of the GR formula I know that the radius is half that size. No reference needed.
I do the same with objects orbeting the BH on large distances where Newton can be applied.

 

[DParlevliet]

The equation for a stable orbit in Schwarzschild space time (i.e. around a static black hole) is-
v_s=\sqrt{\frac{M}{r\left(1-\frac{2M}{r}\right)}} with : M=Gm/c^2 and : v_s=v/c .

If I transfer this to a more familiar form it becomes: (r - r_s) = G.m/v², so standard Newton, only he would think that in a BH all mass is concentrated in a shell with radius r_s (instead of a singularity in the center).

For photons (v = c): r = G.m/c² + r_s = G.m/c² + 2G.m/c² = 3G.m/c²

Are these calculations right?

 

[stevebd1]

Your rearranging of the equation at first glance looks correct though as you've already pointed out, it's unlikely Newton would have totally understood the significance of r_s (or 2M).

On a side note, as it's also already been pointed out, if you have two separate orbits at say, 6M and 4M, the 'proper' distance between the two orbits wouldn't be 2M (which would be the coordinate distance), it would actually be slightly greater. Sample problem 2 on page 2-28 from the following link shows how to calculate the proper distance between 2 orbits near a static BH-

http://www.eftaylor.com/pub/chapter2.pdf

 

[PeterDonis]

I measure the diameters of the black holes with the telescope and subtract them. By definition of the GR formula I know that the radius is half that size.

This measurement doesn't tell you the physical distance from the horizon of a black hole to its center. It only tells you how much of your field of vision the black hole takes up. They're not the same thing. They would be if the black hole were just an ordinary object sitting in Euclidean space; but it isn't.

 

[PeterDonis]

If I transfer this to a more familiar form it becomes: (r - rs) = G.m/v2, so standard Newton

No, it is not "standard Newton". Standard Newton has $r$, not $r - r_s$. You can't just handwave that away.

only he would think that in a BH all mass is concentrated in a shell with radius rs (instead of a singularity in the center).

No, he would think the result was inconsistent, because it is telling you that the "center" is at $r - r_s$, not $r$; but the surface area of a 2-sphere at $r - r_s$ is not zero, so it can't be the "center". So Newton would say you've simply made a mistake and the correct formula should just have $r$.

Newton would be wrong, but not because the mass is concentrated in a shell at radius $r_s$. He would be wrong because his formulas for gravity are wrong; gravity is correctly described by GR, not Newton's laws.

 

[PWiz]

He would be wrong because his formulas for gravity are wrong;

I think a better restatement of that would be "He would be wrong because his formulas for gravity are approximations based on empirical evidence".

 

[PeterDonis]

I think a better restatement of that would be "He would be wrong because his formulas for gravity are approximations based on empirical evidence".

In general I would agree; but the regime we are discussing in this thread, close to a black hole, is precisely the regime in which Newton's formulas are not good approximations at all. So for this thread I chose to emphasize the fact that Newton's formulas are not the correct ones; the GR formulas are.

 

[PWiz]

In general I would agree; but the regime we are discussing in this thread, close to a black hole, is precisely the regime in which Newton's formulas are not good approximations at all. So for this thread I chose to emphasize the fact that Newton's formulas are not the correct ones; the GR formulas are.

I see your point. I didn't see the explicit mentioning of the context in your previous post however, which is why I suggested a restatement.

 

[DParlevliet]

Your rearranging of the equation at first glance looks correct though as you've already pointed out, it's unlikely Newton would have totally understood the significance of r_s (or 2M)

Indeed, although he would be please when informed that information in a BH is proportional with the surface of the sphere r_s in stead of its content (but I know, still for the wrong reason)

But now about mass: is your formule the same (r - r_s) = G.m/v²? Then in the smallest stable orbit v is smaller then c.
Or is the formula different (for instance (r - 5r_s) = G.m/c²) and the mass speed is about c.

 

[PeterDonis]

in the smallest stable orbit v is smaller then c.

That's correct; in the smallest stable orbit, which is at three times the horizon radius, the orbital velocity is $c / 2$. At the smallest orbit of any kind (which is unstable), at 3/2 the horizon radius, the orbital velocity is $c$; this is called the "photon sphere" because photons can orbit the hole at this radius.

 

[PeterDonis]

he would be please when informed that information in a BH is proportional with the surface of the sphere rs

Why would that have pleased Newton?

 

[DParlevliet]

So summing up: the formula outside a BH is (r - r_s) = G.m/v², which means that g = G.m/(r - r_s)². This applies to both photons and mass, although for mass r cannot be smaller then 6r_s (v = c/2), because of reasons not explained here the orbit becomes unstable.

 

[PeterDonis]

the formula outside a BH is (r - r_s) = G.m/v²

Only for $r \ge \frac{3}{2} r_s$. For $r$ smaller than that there are no free-fall orbits possible.

which means that g = G.m/(r - rs)²

If by "g" you mean "acceleration due to gravity", no, this is not correct. The correct formula is

$$
g = \frac{Gm}{r^2 \sqrt{1 - \frac{r_s}{r}}}
$$

 

[DParlevliet]

Only for $r \ge \frac{3}{2} r_s$. For $r$ smaller than that there are no free-fall orbits possible.

That is right.

If by "g" you mean "acceleration due to gravity", no, this is not correct. The correct formula is

My formula was not correct indeed, but when I calculate it I get (and is until r_s):
$$
g = \frac{Gm}{r^2 (1 - \frac{r_s}{r})}
$$
If that is not the case, then other formula does change.