Inverse Square Law for Black Holes

[dsaun777]

When you are calculating the gravitational force between two masses and one of them is a black hole, do you still use the distance to the center of mass as you would in Newtonian gravity to find the force? Or is the distance measured only to the event horizon? Is the inverse square law modified when working with black holes?

 

[phinds]

other than close to one, a black hole is just a mass like any other.

 

[PeterDonis]

When you are calculating the gravitational force between two masses

If you're doing that, you're not doing relativity, you're doing Newtonian physics. Gravity is not a force in GR.

In GR you compute the spacetime geometry and derive things like the behavior of geodesics (worldlines of freely falling objects) from that.

and one of them is a black hole, do you still use the distance to the center of mass

There is no such thing for a black hole.

is the distance measured only to the event horizon?

This is a meaningful distance but doesn't play any useful role in computing the spacetime geometry or its consequences.

Is the inverse square law modified when working with black holes?

Again, in GR gravity is not a force and there is no inverse square law.

If you are asking about the Newtonian approximation in GR, then black holes are irrelevant because they can't be analyzed within the Newtonian approximation. Perhaps that is a kind of answer to your question.

 

[Ibix]

When you are calculating the gravitational force between two masses and one of them is a black hole, do you still use the distance to the center of mass as you would in Newtonian gravity to find the force? Or is the distance measured only to the event horizon? Is the inverse square law modified when working with black holes?

If you are talking about forces you are using a Newtonian approximation, so a black hole is just another mass like any other. Such calculations will become inaccurate if the black holes are moving at significant fractions of the speed of light with respect to one another and/or if they get within a few Schwarzschild radii of each other. At that point "force" isn't really a helpful term, and you need general relativity and numerical solutions to the Einstein field equations to calculate trajectories.

 

[PeterDonis]

If you are talking about forces you are using a Newtonian approximation, so a black hole is just another mass like any other.

As long as you don't get too close to them and can approximate them as point masses (which means you're also ignoring issues like the fact that a black hole, strictly speaking, doesn't have a center of mass), or as "world tubes" whose interior structure and geometry is ignored.

 

[dsaun777]

other than close to one, a black hole is just a mass like any other.

So some particle at a large distance away and outside the black hole's horizon would just measure R to the center of mass, not the horizon?

 

[PeterDonis]

So some particle at a large distance away and outside the black hole's horizon would just measure R to the center of mass

No, because that's impossible; a black hole has no center of mass, as I've already said.

Far enough away from the hole, the "areal radius" $r$ (which is also the Schwarzschild radial coordinate) can be plugged into the equations as if it were a "distance from the center" with reasonable accuracy. That is basically what is done in cases where this level of approximation is sufficient.

 

[phinds]

So some particle at a large distance away and outside the black hole's horizon would just measure R to the center of mass, not the horizon?

Yes, within the caveats mentioned in the other posts above.

 

[dsaun777]

I know Newtonian physics is not well suited for black hole gravity, but I am trying to see if the distance within the Schwarzschild radius would be subtracted from the distance of a distant observer or particle. But according to posts above the distances within the black hole are accounted for in gravity.

 

[dsaun777]

In other words, the "force" wouldn't work like 1/(R-r)²
, where r is the radius and R is the distance from the center of mass and the particle.

 

[russ_watters]

I know Newtonian physics is not well suited for black hole gravity, but I am trying to see if the distance within the Schwarzschild radius would be subtracted from the distance of a distant observer or particle.

That distance is a rounding error, so I'd say no. E.G. for a solar mass black hole vs earth's distance, 3 km vs 150 million.

 

[PeterDonis]

I am trying to see if the distance within the Schwarzschild radius would be subtracted from the distance of a distant observer or particle.

This can't be done because there is no such thing as "the distance within the Schwarzschild radius" for a black hole. The interior of a black hole is not like ordinary space.

R is the distance from the center of mass

There is no such thing for a black hole. This has been repeated many times now. You need to take it into account and stop asking questions that presume the existence of this thing that doesn't exist.

What is done in practice is to estimate the areal radius $r$ and use that as the "distance" when the Newtonian approximation is being used.

That distance is a rounding error

No, it isn't, it's a nonexistent quantity. There is no such thing as "distance to the center" of a black hole.

I have described, above and in a previous post, what is actually done in practice.

 

[dsaun777]

areal radius

Is the areal radius something like sqrt{A/Pi}?

 

[PeterDonis]

Is the areal radius something like sqrt{A/Pi}?

It's $\sqrt{A / 4 \pi}$, where $A$ is the surface area of a 2-sphere surrounding the massive object.

 

[Ibix]

In other words, the "force" wouldn't work like 1/(R-r)²
, where r is the radius and R is the distance from the center of mass and the particle.

No. For example, if a black hole passes near another one and escapes I would expect the escape trajectory to depend on the rotation axes (both the magnitude and angle between) of the black holes.

In any situation where you can measure precisely enough to be worrying about the difference between $r$ and $r-R_S$ you shouldn't be using Newtonian gravity for a black hole, because at that level of precision the fact that the Schwarzschild $r$ isn't a distance in the naive sense will become important. Attempting to "correct" for $R_S$ in the way you are trying to do makes no sense.

 

[Nugatory]

Is the areal radius something like sqrt{A/Pi}?

The areal radius is $\sqrt{A/(4\pi)}$ and compare with the formula for the area of a sphere in Euclidean three-dimensional space.

And has been said several times above, you cannot interpret this quantity as the distance between the horizon and the center of the black hole - there is no center and the horizon isn’t what you’re thinking it is.

 

[dsaun777]

If there were two particles on the opposite sides of a supermassive black hole outside the event horizon, what would the distance between them be? Would you use geodesics going around the black hole similar to photon spheres?

 

[Ibix]

Is the areal radius something like sqrt{A/Pi}?

As others have noted, it's $\sqrt{A/4\pi}$. In a Euclidean space this would be just the radius, but in a non-Euclidean space you may find the distance across the center of the sphere to be more or less than $2\sqrt{A/4\pi}$. Or it may not be possible to define a path across the center of the sphere at all, as is the case in Schwarzschild spacetime because $r=0$ is something nearer a moment in time than a place in space and there is no way (even in principle) to draw a "diameter" through it.

 

[Ibix]

what would the distance between them be?

Undefined.

 

[Ibix]

Would you use geodesics going around the black hole similar to photon spheres?

You can measure distances along many different paths between two objects. But which one do you pick out and call "the" distance? In Euclidean space it's the straight line distance because that's a unique minimum length. But consider a sphere. In general there are two "interesting" paths along its surface between two points on its surface - the shortest path and the longest path, which together make the great circle through the two points. Which one is "the" distance? Why prefer one over the other?

Things get more complex in curved spacetime because there are usually infinite choices of geodesic paths between a pair of points. You could certainly find extrema of the interval along all possible geodesics, but like in the sphere case there will probably be more than one and why prefer one over the other? And extrema of interval may not correspond to extrema of spatial distance. And there's the fact that "space" doesn't have a unique definition in most spacetimes (isolated black holes yes, pairs of black holes no) so even if you pick out a path by some mathematically sound process the spatial distance along it is partially a matter of choice.

This stuff is complicated, I'm afraid. There will always be an answer to directly measurable questions ("how long until I receive the first radar echo from that object according to my clock"), although you may need a supercomputer to get it. But almost everything beyond that ("...so how far away is the object") is open to interpretation and definitional choices in a way that doesn't happen in Newtonian physics.

So yes, I can send a light signal round the outside of a black hole from one object to another. Probably in several different ways, in fact. But I can't convert that to a distance without (usually arbitrary) assumptions, and certainly can't construe it as "the" distance.

 

[Vanadium 50]

but I am trying to see if the distance within the Schwarzschild radius would be subtracted from the distance

The answer is "no"

Either the Newtonian approximation is good enough, or it isn't, but in neither case is this the correct procedure.

 

[phinds]

The answer is "no"

@dsaun777, you have asked the same question in posts 1, 6, 9, and 10 and have been give the same answer repeatedly. Do you get it now?

 

[vanhees71]

If you are asking about the Newtonian approximation in GR, then black holes are irrelevant because they can't be analyzed within the Newtonian approximation. Perhaps that is a kind of answer to your question.

If you are not too close to the event horizon, the Newtonian approximation is valid. If the Sun were a black hole, we could still describe the motion of the planets as we do with our real Sun with Kepler orbits a la Newton's theory are a very good approximation. It's just that for $2 r_S/r \ll 1$ the Newtonian approximation for the geodesic equation of a massive body in the Schwarzschild metric is a good approximation.

 

[dsaun777]

As long as you don't get too close to them and can approximate them as point masses (which means you're also ignoring issues like the fact that a black hole, strictly speaking, doesn't have a center of mass), or as "world tubes" whose interior structure and geometry is ignored.

For a world tube, what you are only calculating is the surface area and neglecting the internal distance?

 

[Ibix]

For a world tube, what you are only calculating is the surface area and neglecting the internal distance?

No. A worldtube is a term for a collection of worldlines. A worldline is a 0d point extended in time. If I'm modelling an object as a point in space then I'm modelling it as a worldline. A 1d object forms a worldsheet - i.e. a section of a plane with one spacelike direction and one timelike one. A 2 or 3d object forms a worldtube, with two or three spacelike dimensions and one timelike one.

 

[dsaun777]

No. A worldtube is a term for a collection of worldlines. A worldline is a 0d point extended in time. If I'm modelling an object as a point in space then I'm modelling it as a worldline. A 1d object forms a worldsheet - i.e. a section of a plane with one spacelike direction and one timelike one. A 2 or 3d object forms a worldtube, with two or three spacelike dimensions and one timelike one.

How can an object be truly 1 dimensional in this universe? Is that where strings come in?

 

[Ibix]

How can an object be truly 1 dimensional in this universe?

It can't. Point-like, line-like and plane-like objects are useful models for things that are small in one or more dimensions compared to the scale you are modelling.

Is that where strings come in?

This has nothing to do with string theory.

 

[PeterDonis]

For a world tube, what you are only calculating is the surface area and neglecting the internal distance?

For the case of a black hole, yes; you're pretending that what is inside the tube is just an ordinary mass, not a black hole, and ignoring all of the internal structure inside the tube.

A worldtube is a term for a collection of worldlines.

Normally, yes, but for the case of a black hole this doesn't work; a black hole cannot be modeled as a congruence of worldlines. You have to just deal with the surface of the world tube and ignore its interior. See above.

 

[dsaun777]

For the case of a black hole, yes; you're pretending that what is inside the tube is just an ordinary mass, not a black hole, and ignoring all of the internal structure inside the tube.Normally, yes, but for the case of a black hole this doesn't work; a black hole cannot be modeled as a congruence of worldlines. You have to just deal with the surface of the world tube and ignore its interior. See above.

So in this context, the distance of a point to the world tube would be the distance from the surface of the tube to the point and not the areal radius of the black hole?

 

[PeterDonis]

So in this context, the distance of a point to the world tube would be the distance from the surface of the tube to the point and not the areal radius of the black hole?

The relevant distance would be the areal radius of the point (not of the hole's horizon) relative to the hole.

 

[pervect]

When you are calculating the gravitational force between two masses and one of them is a black hole, do you still use the distance to the center of mass as you would in Newtonian gravity to find the force? Or is the distance measured only to the event horizon? Is the inverse square law modified when working with black holes?

What you do in GR rather than calculate the force at some distance d is to calculate the proper acceleration of a static observer at some particular space-time coordinates. Then to get the equivalent of a "force" holding the object in place, you mutiply this proper acceleration of the static observer by the mass of the test particle. The mass of the test particle is assumed to be small, so that it doesn't affect the overall space-time geometry.

To carry out the calculation you need to specify exactly what coordinates you are using,. Typically one would uses the Schwarzschild coordinates associated with the Schwarzschild metric. On occasion, though, one might want to use isotropic coordiantes associated with a different line element.

If we assume the use of Schwarzschild coordiantes, and let the radial coordinate of the test mass be "r", the proper acceleration of the test particle is:

$$ \frac{GM}{r^2 \sqrt{1-\frac{2GM}{r c^2}}}$$

This is mentioned somewhere in Wald, I used an old PF post of mine, https://www.physicsforums.com/threa...rcular-orbits-in-symmetric-spacetimes.686147/

Note that 2GM / c^2 is just r_s, r_s being the Schwarzschild radius of the black hole. So we can re-write this as:

$$ \frac{GM}{r^2 \sqrt{1-\frac{r_s}{r}}}$$

Thus we see that it would require an infinite proper acceleration, and hence an infinite force, at r=r_s. Note also that this is the force as measured by a local scale co-located with the test mass.

 

[PeterDonis]

Thread closed for moderation.

 

[PeterDonis]

After moderator review, the thread will remain closed. The substantive question in the OP has been answered.