# Does an object orbiting close to a black hole appear to orbit slower due to time dilation?

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#### Unclear Engineer

The problem I have with thinking about an event horizon as a spherical surface in space where there is no passage of time is that that radius from a very large mass is not necessarily the outer radius of that mass in some solid form. For instance, the Schwarzschild radius for a mass with the density we think we observe in our universe is about 13.8 billion light years. And the masses inside are not even conglomerated into one single body. (Remember, I am not arguing (here) that our observable universe is a black hole, I am just arguing that there can be non-solid volume inside a black hole.)

So, if the event horizon is not a solid surface, for say a star that goes supernova and creates a black hole, what happens to additional matter that the black hole subsequently accretes? If it stops at the event horizon, it eventually would form a hollow shell. And, what matter would be strong enough not to collapse? But wait, it takes time to collapse. I see another paradox. Well the real problem is that I can't see it, so I'll use the other metaphor and say I "smell" it. (Yeah, I know that wouldn't actually work, either.)

I think what we are really saying is that the equations we have derived so far break down at the event horizon, and we don't have any observations to help us conceptualize more equations that "work" in that environment. Which is a roundabout way to say "We don't understand what happens there."

But, there are people who claim that the inside of a black hole appears to a local observer to be an expanding universe. See
.

In that video, note the animation for the space around a black hole's event horizon. It does not show a static place where time stands still, but rather a place where the flow of space into the black hole reaches the speed of light. That would answer my question about what a photon of light would do if emitted below a black hole's event horizon - it would still travel through space at the speed of light, but the space itself would be traveling even faster toward the center of the black hole, so the light goes "down" with the space, even though it is traveling against the flow.

The mind-bending part of that is that it seems as though an ever-increasing amount of space is accumulating inside a fixed volume. But, if we are willing to believe that an ever increasing amount of space can be added to our observed universe, why would we not be willing to believe that an ever increasing amount of space could be subtracted from the "interior" of a black hole?

We seem to believe that the sum of energy and matter cannot be increased or decreased, but we don't seem to believe that the "amount" of "space" itself must be constant, right? The Big Bang Theory requires that the measured volume of space increases with time, and with a speed not restricted to the speed of light.

#### billslugg

As a person falls farther and farther down the gravitational well their clock slows down and the EM waves we use to see them redshift. The person's clock runs slower and slower until the point where they reach the event horizon and they appear frozen in time. Except we can't see them since the EM waves they emit are of infinite length. Resolving power is inversely proportional to wavelength so we could not discern them.

#### Unclear Engineer

That seems entirely consistent with the light emitted as that person crosses the event horizon getting stretched from that point in time and (apparent) space "forever", as space continues to flow into the event horizon and the light continues to just keep up with the event horizon, at least as viewed from outside the EH. The light just outside the EH would need to keep increasing wave length as it was stretched between the stationary part at the EH and the part just outside the EH that is still moving outward in net velocity through the inflowing space. A distant observer outside the EH would "see" light emitted at the EH as zero wave length.

The time issue is another matter for the person transiting the event horizon and a distant observer outside the event horizon. As I understand General Relativity, if it were possible for us to retrieve a person from just outside the event horizon, that person would still be younger when brought back to a twin who only observed from a distance, due to "gravitational time dilation" affecting the one that went near the event horizon. So, at least coming from outside the event horizon, time does slow to zero at the EH to an outside observer. Does it effectively go "negative" on the inside, at least with respect to how the outside observer considers time passage? Does time speed up as the observer progresses "inward" from the event horizon? Would that look like the Hubble "constant" was increasing or decreasing as the observer went farther inside the event horizon? Would that observer be perceiving the "universe" that he/she can observe to be expanding at an increasing rate with the "passage" of time?

#### billslugg

The person falling through the event horizon can be completely unaware of what is going on. As they are falling through a gravitational field they are weightless. The only sensation they would have would arise from tidal forces. The larger the black hole, the smaller the tidal forces would be at the event horizon. I believe a multi billion solar mass black hole would have tidal forces so small they would not be noticed by the person falling in. I read this somewhere but don't have a source for it.

#### Unclear Engineer

I have read the same thing about extremely large black holes, but don't know what "extremely large" means in that context.

Here is a list of black hole masses computed to date: https://en.wikipedia.org/wiki/List_of_most_massive_black_holes

Note that "ours" here in the Milky Way is the smallest on that list. And, at "only"4.3 million solar masses, there might be a rough ride on the way in.

While you are looking at that link, note the comment about estimating the mass of black holes: "Another problem for this list is the method used in determining the mass. Such methods, such as broad emission-line reverberation mapping (BLRM), Doppler measurements, velocity dispersion, and the aforementioned M–sigma relation have not yet been well established. Most of the time, the masses derived from the given methods contradict each other's values. "

I would expect that whatever effects are causing the observed large masses of ionized gas in the vicinity of black holes would be hard on any solid observer going in. The largest one on that list is a quasar, so something extremely energetic must be happening around it.

Absolute zero

#### billslugg

Yes, for the sake of survivability there must be no matter accreting into the black hole. Accreting matter forms discs which rotate, but at different speeds depending on the radius. Very high shear forces cause collisions that emit vast quantities of heat, x-rays, gamma rays, etc. To survive a fall into a black hole's event horizon it must be a 'naked black hole' with no disc around it. The person's infall must be a straight shot, no orbiting.

Catastrophe

#### Absolute zero

The person falling through the event horizon can be completely unaware of what is going on. As they are falling through a gravitational field they are weightless. The only sensation they would have would arise from tidal forces. The larger the black hole, the smaller the tidal forces would be at the event horizon. I believe a multi billion solar mass black hole would have tidal forces so small they would not be noticed by the person falling in. I read this somewhere but don't have a source for it.
Unfortunately, the tidal forces are larger the larger the black hole. The statement of tidal forces getting smaller the larger the black hole comes from using Newtonian equations where GR use is mandatory.

#### Unclear Engineer

Unfortunately, the tidal forces are larger the larger the black hole. The statement of tidal forces getting smaller the larger the black hole comes from using Newtonian equations where GR use is mandatory.
Can you explain that better? "Tidal" forces implies to me some spin of an approaching object, which need not be the case.

For a straight-in trajectory of a non-rotating object, it seems to me that the "spaghettification" forces are what would be disruptive. And those would be calculated by the gradient in gravitational acceleration as a function of radial distance. If the acceleration of the leading part of the object is different enough from the acceleration of the trailing part of the object that it creates a force that exceeds the structural integrity of the object material, then the object stretches apart and breaks, right?

So, using GR theory, how does the gradient in gravitational acceleration at the event horizon change as black holes increase in mass and the event horizon radius increases, making the event horizon look "flatter" to an approaching object of a specified size, say 100 meters?

Let's ignore black hole spin for this calculation.

#### billslugg

Unfortunately, the tidal forces are larger the larger the black hole. The statement of tidal forces getting smaller the larger the black hole comes from using Newtonian equations where GR use is mandatory.
Here is an analysis appearing on the NASA website comparing the tidal forces of a stellar mass black hole and a 100 million stellar mass black hole. (Problems #3 & #4) Can you show us where their error is?

4Page33.pdf (nasa.gov)

Unclear Engineer

#### Absolute zero

Here is an analysis appearing on the NASA website comparing the tidal forces of a stellar mass black hole and a 100 million stellar mass black hole. (Problems #3 & #4) Can you show us where their error is?

4Page33.pdf (nasa.gov)
The error in problem three and four is the use of a newtonian equation near a black hole.

Take orbital velocity for example, at R=1.5Rs, which answer would you consider correct

V=1C or V~.58C

Orbital velocity is connected to gravitational acceleration. A higher orbital velocity at a given radius means a higher acceleration. At the event horizon, the Newton equation is only off by a magnitude to large to write down. (Trying to avoid infinite )

#### billslugg

We are not talking about orbital velocity, we are talking about tidal forces for a body falling straight into the black hole as it crosses the event horizon.

Saying they used the wrong equation is unconvincing unless you can provide the correct equation. Show us, for the above mentioned masses, what you believe the tidal forces would be, using your choice of relativistic equations.

Unclear Engineer

#### Unclear Engineer

As I understand it:

Tidal force near a black hole is M/r^3 and event horizon radius is 2GM/c^2.

So, tidal force at the event horizon is (c^6)/(8G^3 M^2)

So, tidal force at the event horizon should decrease as the mass increases and thus the event horizon radius increases.

#### billslugg

These are Newtonian equations. The Schwarzschild radius increases by the square while the tidal force decreases by the cube. The radius will always outrun the tidal force.

#### Unclear Engineer

Bill, you asked "Absolute zero" for his equations. I just chimed in with my equations. If you wade through my link, you will find the statement that the Newtonian derivation of the event horizon radius made by Schwarzchild does produce the same radius as the General Relativity derivation.

And, since this does support your (and my) belief that the event horizons of extremely large masses could be crossed without spaghettification, I just provided my understanding of why that is so.

If "Absolute zero" can come up with different equations that are from GR theory and show that they produce an opposite result (i.e., spaghettification is greater for larger black holes - as he posted before), then we might both learn something.

But, no such showing by Mr. zero to this point in time/space. My post was not intended to take him off the hook that you posted, just to demonstrate your position, which is also mine.

Catastrophe

#### billslugg

Sorry about that! I lose track of who says what.

#### Absolute zero

We are not talking about orbital velocity, we are talking about tidal forces for a body falling straight into the black hole as it crosses the event horizon.

Saying they used the wrong equation is unconvincing unless you can provide the correct equation. Show us, for the above mentioned masses, what you believe the tidal forces would be, using your choice of relativistic equations.
Do you want just equations or are you interested in a logical reasoning to why RsC²/r² should not be used at R=Rs? Manipulating the wrong equation at the start will most often result in an incorrect equation as a result.

Here is a link to how the equation in question was derived. Note: It starts with acceleration.

As does the formulation of orbital velocity in the following link.

And there is this link on deriving the expression for acceleration in relativity.

Note that Rc is much less than r .

Catastrophe

#### billslugg

I want to know your evaluation of the tidal force experienced by a 2 meter tall person at the event horizon of a 100 million solar mass black hole who is not orbiting but simply falling into it. Here are the equations used by NASA:

Formulas:
Event horizon distance: Rs = 2 * G * M / c^2
Tidal force: a = 2 * G * M * d / R^3

Variables:
a = tidal force in m/s^2
G = gravitational constant = 6.67e-11 N m^2 / kg^2
M = mass of black hole = 100 million times mass of Sun (2e30 kg) or 2e38 kg
d = height of human = 2 meters
Rs = event horizon distance = 295 million km or 2.95e11 meters
c = 3e8 m/s

Event horizon distance Rs is:
Rs = 2 * 6.67e-11 * 2e38 / (3e8)^2
Rs = 2.95e11 meters

Tidal force is:
a = 2 * G * M *a / R^3
a = 2 * 6.67e-11 * 2e38 * 2 / (2.95e11)^3
= 2e-6 m/s^2
This amount of tidal force would be undetectable by a person.

Now show me your calculations in the same format and show that a person would be torn apart as you assert.

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#### Unclear Engineer

I think we are going to want the equations derived from General Relativity, if Absolute zero is telling us that the equations we are using are giving the wrong answer.

The link he provided in his latest post goes to a discussion where the participants failed to derive the equations he thinks are required. For their purpose in that discussion, they do make some simplifying assumptions that they are far out from the event horizon and the radial component of the velocity is much less than the speed of light. So they end up with equations similar to what we are already using here. But, they do show the complexity of the full GR math.

So, that link is not helpful for addressing what the General Relativity equations are for things just outside the event horizon. That is not a big surprise, since GRT "breaks down" at the event horizon. So, trying to get an answer extremely close to it would involve dealing with all of those parts of the mathematical expressions that are discarded as "too small to matter" when far enough away from the event horizon to use Newtonian physics, anyway.

If Absolute zero wants to spend the time and mental effort on this, I suggest that he not try to solve the equations to closed form expressions, but rather focus on showing whether the partial derivative of the radial acceleration with respect to mass is greater or less than zero at the Schwarzchild radius. I am not sure even that is possible with GR equations, but maybe approaching it as a limit of those equations would give some insights.

Absolute zero

#### Absolute zero

If you take the time to understand how the equation was derived, maybe the realization of the equation NASA used is indistinguishable from Newtonian. Also I am trying to show the equation was derived from

RsC²/r² or is exact equivalent in Newton's theory of gravity.

For Unclear Engineer : you are very close but at this point, any Newtonian equation I introduce correction factors into most likely will be rejected because "it's not NASA'S ". The use of orbital velocity is to show what a correction factor would look like.

Newton : RsC²/ R

Correction factor for the change in the way velocities are added in General Relativity : 1/(1-Rs/R) or R/(R-Rs)

Multiplying the two : RsC²R/R(R-Rs) or RsC²/(R-Rs)

The final result is exactly the same as GR.

What happens to
Tidal force: a = 2 * G * M * d / R^3
If even one R is replaced with (R-Rs), Tidal force at Rs goes to infinity.

#### billslugg

I want you to back up your assertion you made in post #32:

"Unfortunately, the tidal forces are larger the larger the black hole. The statement of tidal forces getting smaller the larger the black hole comes from using Newtonian equations where GR use is mandatory." - Absolute Zero in post #32.

Perhaps you meant to say the tidal forces are greater, at a given radius, the larger the black hole.

If this is not the case then present your GR equation for the tidal force on a 2 meter human at the event horizon of a 100 million solar mass black hole and show us it is larger than 2e-6 m/s^2.

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#### Unclear Engineer

Bill and Absolute zero,

I don't think it is unlikely that tidal forces could be a bit higher when calculated with Relativistic equations than with Newtonian equations, but that is not the point.

The question is whether the tidal forces at the event horizon increase or decrease as the black hole mass increases and the event horizon increases.

With the radius of a black hole's event horizon increasing as a direct function of the mass, and the tidal force due to gravity decreasing as the inverse cube of the radius in Newtonian equations, the net effect for Newtonian calculations is that the tidal force at the event horizon will decrease as the inverse square of the black hole mass.

That is a very strong effect. And, it seems intuitively reasonable, because the change in radius of a fixed length object (such as a 2 meter tall astronaut) becomes a smaller and smaller fraction of the radial distance from the center of the black hole to the radius of the event horizon as that gets bigger and bigger. So, all of those calculations we see about stable orbits, etc. being multiples of the event horizon radius seem to indicate that a fixed length astronaut would see smaller tidal forces because he is a smaller fraction of the radius to the center of the black hole as that radius becomes larger, with larger mass.

To show that it wrong, and that properly-done relativistic calculations would somehow get the mass of the black hole out of the denominator and into the numerator for the tidal force results, is what we need to see to support Absolute zero's assertion.

#### billslugg

Very well put. At a given distance the tidal force goes up with the black hole mass, but at the event horizon it goes down since the event horizon recedes by the square and the tidal force recedes by the cube.

#### Absolute zero

I will amend my statement to " I believe a person would be torn apart at the event horizon of a very large black hole and the larger the black hole, the larger the tidal forces at the same distance above the black hole," It should never been stated in the "been proven mode" .

I had assumed the acceleration equation used in both Newtonian and GR was derived at to recover the former from the later in R<<Rs and v<<c conditions.

What I have a hard time believing is any object being accelerated to C with a finite and likely fixed amount of gravitational force equivalent.

#### Harry Costas

At the event Horizon and beyond your body atoms will break down to atoms and then Neutrons and protons which will take on an electron and form Neutrons. Neutrons will compact to 10 ^17.
If there is enough mass in the core than Neutrons will break down to Quarks with compaction over 10 ^ 20 and may break down to Partonic matter and may break down to the theoretical particle Axion Gluon Matter and compaction extreme.

#### Unclear Engineer

At the event Horizon and beyond your body atoms will break down to atoms and then Neutrons and protons which will take on an electron and form Neutrons. Neutrons will compact to 10 ^17.
If there is enough mass in the core than Neutrons will break down to Quarks with compaction over 10 ^ 20 and may break down to Partonic matter and may break down to the theoretical particle Axion Gluon Matter and compaction extreme.
Why would this happen at the event horizon? Something in free fall is not seeing any forces other than tidal forces, which tend to pull it apart, rather than compact it.

I tend to think of what is inside a black hole by extending what I think about neutron stars. At least in theory, a neutron star could accrete additional matter until it's surface escape velocity reaches the speed of light, and it would become a small black hole. I don't know of any reason why the material in the neutron star would suddenly change at that point in the accretion process, so I think of a small black hole as being a neutron star inside of an event horizon. I do agree that the material in the neutron star compacts as it accretes more material, and might change phase at a subatomic level to become the mass of quarks or whatever at some point in its mass increase. But, that would make the surface of the star become a smaller and smaller fraction of the space inside its event horizon.

So, I would expect some radial distance between the event horizon and and star surface inside. Working again from what is thought about neutron stars, it is when the accreting matter hits the star's surface that fusion reactions occur and make the events our astronomers see.

Once the event horizon exceeds the star's surface radius, I would still expect that accreting material fuses or whatever when it reaches the star's surface, but we would no longer be able to see that happen from outside, because no light or matter would escape past the event horizon.

If there are other factors involved that change that expectation, please educate me.

And, yes, I do know that neutron stars usually have complicating features like extreme spin rates and extreme magnetic fields, but I don't see how those change the thinking about the event horizon itself not being the location of nuclear fusion or disassembly into quarks. Those features might even tear things apart outside the event horizon and prevent some of the accretions, as we think we see happening around black holes.