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

[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.

 

[PeterDonis]

when I calculate it I get (and is until rs):

$$
g = \frac{Gm}{r^2 \left( 1 - \frac{r_s}{r} \right)}
$$

I don't know how you're calculating this, but your answer is wrong. The factor $\left( 1 - \frac{r_s}{r} \right)$ should be under a square root sign.

If that is not the case, then other formula does change.

What other formula?

 

[DParlevliet]

The classic calculation of an orbit is based on: s = 0.5g.t² and y = x²/2r (circle at x ≈ 0) which is : s = v²t^2/2r
together: 0.5g.t² = v²t^2/2r, resulting in: g = v²/r

In classic the orbit formula is v² = G.m/r with gives in above: g = G.m/r²

In GR the orbit formula is v² = G.m/(r-r_s) with gives: g = G.m/r(r-r_s) = G.m/r²(1-r_s/r)

If this is not right, then one of the two classic formula on top is changing too in GR. Which one?

 

[PeterDonis]

The classic calculation of an orbit is based on

The classic calculation of an orbit in Newtonian physics. To repeat once more Newtonian physics is wrong. It's not a matter of which one of the Newtonian formulas has to change. They are all built on the wrong conceptual foundation. The correct procedure is to start over with the right conceptual foundation, that of GR, and derive the formulas for orbits (and everything else) from first principles using GR. Any GR textbook will show you how to do that. I'm pretty sure Carroll's online lecture notes on GR do it too, which is nice because they're online and free.

with gives: g = G.m/r(r-rs) = G.m/r2(1-rs/r)

No, it doesn't. The (1 - r_s / r) factor is under a square root sign, as I've posted before. You say you've calculated it, but unless I see your calculation I can't tell where it's going wrong. I suspect it's because you are continuing to assume that it's just a matter of tweaking some Newtonian formulas and you just need to know which ones. It isn't. See above.

then one of the two classic formula on top is changing too in GR. Which one?

All of them, because the entire "classic" conceptual foundation is wrong. See above.

 

[DParlevliet]

Someone else who can follow my simple math calculations?
Or must I write it in small steps?

 

[PeterDonis]

Someone else who can follow my simple math calculations?

It's not a matter of following them; of course the equations you're using are simple. They're just the wrong ones. It's not a matter of which ones "change" in GR; GR doesn't even use the conceptual foundation you're using. You can't just take your Newtonian formulas and replace them with "GR versions". You have to rework your whole approach.

Or must I write it in small steps?

No, you need to go back and start from the right conceptual foundation.

 

[Drakkith]

Someone else who can follow my simple math calculations?
Or must I write it in small steps?

As Peter has repeatedly told you, Newtonian physics is not correct here, so you need to use GR. GR is not a simple modification of the former.

 

[DParlevliet]

...You say you've calculated it, but unless I see your calculation I can't tell where it's going wrong.

That is what you wrote. That math is there. Then you are not interested or not familiar with this math

There are three formula which together give a wrong result. Then one of the formula must wrong. Unless math also changes in GR...
If you don't know the answer, please wait for someone who does instead of repeating the same clincher.

 

[Nugatory]

There are three formula which together give a wrong result. Then one of the formula must wrong.

No. All three of them are wrong or do not apply to the Schwarzschild spacetime or both.

 

[PeterDonis]

The OP's question has been answered, and the discussion has run its course. Thread closed.