Spaghettification near event horizon

[Stephanus]

Dear PF Forum,
Can we avoid spaghettification for some times, once we're inside event horizon?
I choose a rather massive black hole, so the tidal force won't be so big at EH.
Mass: 1 trillion solar mass
Schwarzschild radius: 2.950 trillion KM
What is the gravitational force at EH?
I calculate: $g = \frac{GM}{r^2} = 3 m/s^2$
But

You are doing the wrong calculation. That formula is not correct; the correct GR formula is
$$a = \frac{GM}{R^2 \sqrt{1 - \frac{2GM}{c^2 R}}}$$
According to this formula, $a$ increases without bound as $R \rightarrow 2GM/c^2$, i.e., as the horizon is approached.

So, once we're inside the event horizon minus 1 cm, acceleration increases without bound? But if acceleration is infinity then once we're inside EH, no matter how big massive the black hole is we'll arrive at the singularity at once (in the observer inside EH frame)?

 

[phinds]

So, once we're inside the event horizon minus 1 cm, acceleration increases without bound?

That doesn't seem to make sense. There is, as far as I am aware, nothing in particular that happens as you cross the event horizon of a supermassive black hole. You don't even notice it.

 

[PeterDonis]

Can we avoid spaghettification for some times, once we're inside event horizon?

Yes. Spaghettification only has to happen as you approach the singularity, not the horizon.

 

[Stephanus]

That doesn't seem to make sense. There is, as far as I am aware, nothing in particular that happens as you cross the event horizon of a supermassive black hole. You don't even notice it.

Don't even notice it?
But...

$$a = \frac{GM}{R^2 \sqrt{1 - \frac{2GM}{c^2 R}}}$$
According to this formula, $a$ increases without bound as $R \rightarrow 2GM/c^2$, i.e., as the horizon is approached.

I'm sorry I can't imagine it.
Is the acceleration force approaching EH very high? It's not the speed that I ask. The acceleration.
Thanks for explain it to me.

 

[Stephanus]

Yes. Spaghettification only has to happen as you approach the singularity, not the horizon.

But you said before, near EH the acceleration is very, very high. I cannot picture that. And object can withstand high velocity (even at 99% speed of light, right?) But how can object withstand high acceleration?

 

[Jorrie]

A person free-falling through an EH does not feel the acceleration, just tidal acceleration (difference in acceleration between body parts). This is small over human sizes when the BH is large.

 

[Stephanus]

A person free-falling through an EH does not feel the acceleration, just tidal acceleration (difference in acceleration between body parts). The is small over human sizes when the BH is large.

Thanks for the answer Jorrie.
Actually I want to know more about this thread
https://www.physicsforums.com/threa...on-of-a-large-black-hole.827183/#post-5200300
But there things that I have to learn before I can understand that thread. And rather than interfere with someone's thread, I think it's better that I create my own
So I think I study acceleration inside EH. About the spaghettification, I agree with you.

- A person free-falling through an EH does not feel the acceleration,
- just tidal acceleration
- (difference in acceleration between body parts) this is small over human sizes when the BH is large

And I remember something a mentor posted months ago in my own thread.
And this is the FIRST time I argue about a mentor post. Is the acceleration inside EH is extremely big?

$a = \frac{GM}{R^2\sqrt{1-\frac{2GM}{c^2R}}}$

 

[Jorrie]

Is the acceleration inside EH is extremely big?

It is problematic talking about "acceleration inside the EH", because relative to what shall one specify it?
The equation that you have quoted is acceleration relative to an observer momentarily at constant radial parameter R, but there cannot be such observers at or inside the EH - everything is traveling towards the central singularity (whatever that may be).

Do you already know the Schwarzschild geometry? If not, you will have to master that before thinking about the 'inside'. A good introduction with the various embedding diagrams is available here: http://casa.colorado.edu/~ajsh/schwp.html

 

[PeterDonis]

Is the acceleration force approaching EH very high?

If you're hovering at a constant altitude above the horizon, yes. But that doesn't cause spaghettification. It just causes you to have to withstand a higher and higher g force, until finally your body is crushed by it.

If you're free-falling into the hole, you feel no acceleration at all.

 

[A.T.]

If you're hovering at a constant altitude above the horizon, yes. But that doesn't cause spaghettification.

Depending on how the force is applied to you during hover, you might experience spaghettification or pizzafication.

 

[PAllen]

Depending on how the force is applied to you during hover, you might experience spaghettification or pizzafication.

Ah, yes, if you want to be pizza, don't sit on a grate.

 

[Stephanus]

If you're hovering at a constant altitude above the horizon, yes. But that doesn't cause spaghettification. It just causes you to have to withstand a higher and higher g force, until finally your body is crushed by it.

If you're free-falling into the hole, you feel no acceleration at all.

Yes, we won't feel acceleration at all. The spaghettification will be felt near singularity. How near? Should have calculated it.
But before I read about spaghettification, I would to know about the speed between EH and the singularity.
For example there's a massive black hole, out side earth.
An object fall just inside EH. The acceleration is very high. So the instant (more likely before entering EH) is the speed of the object will be very high if it free falling?
For example: If an object is as far as the moon from earth, it will be attracted to Earth at 2GM/r^2 = 0.00276m/s². Rather slow, but the velocity acceleration is accelerating toward the earth, right.?
Okay, anyway according to your equation. If something is hovering at Event Horizon then suddenly we release it. It doesn't crawl at a couple of meter per seconds.
Not at $a = \frac{GM}{R^2}$ but at $\frac{a}{\sqrt{1-\frac{2GM}{R^2}}}$
No, wait. It still has some value in it. It's not undefined.
$R_s = \frac{2GM}{c^2}$,
so
$a = \frac{GM}{{R_s}^2\sqrt{1-\frac{2GM}{{R_s}^2}}}$
Let me simplify this.
$a = \frac{GM}{{{\frac{2GM}{c^2}}}^2\sqrt{1-\frac{2GM}{{\frac{2GM}{c^2}}^2}}}$
$a = \frac{c^2}{2GM\sqrt{1-\frac{c^4}{2GM}}}$
I tought the speed is undefined. But is this the simplified formula for acceleration in black hole (or any object after adjusting with relativity)?

 

[PeterDonis]

Depending on how the force is applied to you during hover

You don't have a choice about that; if you're hovering, the force is being applied radially outward.

 

[stevebd1]

The spaghettification will be felt near singularity. How near?

The following is taken from this thread-

Regarding tidal forces, Edward Wheeler came up with the 'ouch' radius which is derived from the equation for gravity gradient- $dg=(2Gm/r^3)dr$ which can be rewritten in respect of a change of 1g $(g_E)$ from head to toe (i.e. dr=2m) which is considered a limit we could tolerate-

$$r_{ouch}=\left(\frac{4Gm}{g_E}\right)^{1/3}$$​

this ouch radius will occur well outside the event horizon of small black holes and inside the event horizon for very large black holes such as SMBHs.

You might also find these two link of interest-

Spacetime Geometry Inside a Black Hole
http://www.jimhaldenwang.com/black_hole.htm

Falling Into and Hovering Near A Black Hole
http://mathpages.com/rr/s7-03/7-03.htm

 

[A.T.]

You don't have a choice about that;

https://www.physicsforums.com/threads/spaghettification-near-event-horizon.827980/#post-5201229

 

[PeterDonis]

I would to know about the speed between EH and the singularity.

There is no invariant way to assign a "speed" to an object inside the horizon. Outside the horizon, you can use observers "hovering" at constant altitude above the horizon to define an invariant notion of "being at rest", and define "speed" as speed relative to those observers. By this definition, the speed of an object free-falling into the hole approaches the speed of light as the object approaches the horizon (but never quite reaches $c$ as long as the object is above the horizon).

At and inside the horizon, however, there are no "hovering" observers, and there is no other family of observers that can be viewed as "being at rest", so there's no fixed standard from which to define "speed".

If something is hovering at Event Horizon then suddenly we release it.

Nothing can hover at the event horizon; it would have to be moving outward at the speed of light to do so.

is this the simplified formula for acceleration in black hole (or any object after adjusting with relativity)?

You've done your algebra wrong. It's a lot easier to adopt "geometric" units, in which $G = c = 1$; then your equations don't get cluttered with a lot of extra coefficients.

Start with

$$
a = \frac{M}{r^2 \sqrt{1 - 2M / r}}
$$

Now plug in $r = 2M$. It should be obvious that the factor under the square root in the denominator goes to zero, making $a$ undefined, but let's do the algebra anyway:

$$
a = \frac{M}{r \sqrt{r^2 - 2M r}}
$$

So at the horizon, we have

$$
a_{EH} = \frac{M}{2M \sqrt{4M^2 - 4M^2}} = \frac{1}{2 \sqrt{4M^2 - 4M^2}} = \frac{1}{4M \sqrt{1 - 1}}
$$

This is obviously undefined.

 

[Stephanus]

Thanks for any responses and answer.
I think I'm ready to my original question.
Okay, now after I gather (or rather read) the facts, there's an interesting problem at the other thread.
https://www.physicsforums.com/threa...ast-the-horizon-of-a-large-black-hole.827183/
Here let me quote some post at those thread that indirectly address my issue here.. I create this "spaghetti" thread, because I focus on the acceleration. Not trying to derail that thread.
If this violates PF rule, please let me now.

Can one create a black hole when already inside the black hole ?

Intriguing thought. If a star was in the process of collapsing to start a supernova as it passed the event horizon of a large black hole, what would cause it to cease being able to collapse? Tidal forces pulling it apart? If the large black hole is really large, tidal forces would be fairly small. So I'm not sure about that.

All paths inside the event horizon lead towards the singularity. So a particle trying to gravitationally collapse still must move towards the singularity, not towards the center of mass of the region trying to collapse.

So if the black hole is big enough, say 1 trillion solar mass. (10 times the mass of milky way?) Is it possible for a star in iron burning process inside the event horizon to become a black hole before it gets "spaghettificated"?
No wait, once it become a black hole itself, it can't get streamed, right?
But is it possible for an object to become a black hole inside the event horizon of a black hole?

 

[Stephanus]

@stevebd1
He did call that the "ouch" radius for 1 g different from about 1.7 m? Talk about some fancy names like Chandrasekhar limit, Planck length, Hubble volume, this one just "ouch"?

 

[Stephanus]

You've done your algebra wrong. It's a lot easier to adopt "geometric" units, in which $G = c = 1$; then your equations don't get cluttered with a lot of extra coefficients.

I'm sorry what is G here? $G = c = 1$
Is it Newton G: $6.674×10_{−11} N⋅\frac{m^2}{kg^2}$ or
g as acceleration 9.m/s²
But looking at your equation G is speed of light.
Schwarzschild radius is $R_s = \frac{2GM}{c^2}$
Is G in $G = c = 1$ the same as
G in $R_s = \frac{2GM}{c^2}$ ?

 

[PeterDonis]

is it possible for an object to become a black hole inside the event horizon of a black hole?

The short answer is "no", but there are some complications. There was a thread on this some time ago:

https://www.physicsforums.com/threads/observing-black-holes-inside-black-holes.780809/

 

[George Jones]

Consider an observer who freely falls radially from rest at a great distance from a spherical black hole. Where "ouch" occurs, inside or outside the event horizon, does depend on the size of the black hole, but the observer wristwatch time from "ouch" to "arrival" at the singularity (i.e., the limit of tau as the observer's r-value approaches zero) is independent of the size of the black hole.

 

[Stephanus]

Consider an observer who freely falls radially from rest at a great distance from a spherical black hole. Where "ouch" occurs, inside or outside the event horizon, does depend on the size of the black hole, but the observer wristwatch time from "ouch" to "arrival" at the singularity (i.e., the limit of tau as the observer's r-value approaches zero) is independent of the size of the black hole.

Time (after adjusted to relativiy effect) is independent of the size of the black hole? That's interesting, I like that.
But I have trouble understanding this "ouch"
What is "ouch" radius? Acceleration difference 1 g at 1.7 meters? The height of a person? Why 1 g? Why 1.7m?
That is very human sentris, right.
I imagine like in "Contact" movie, the frequency for finding alien is the frequency of the vibration of hydrogen atom (times Pi, irrational number, so it will not resonance with other frequency) eliminating human factor such as the frequency of SSFM radio channel is 100Mhz. Human frequency.

 

[DrGreg]

I'm sorry what is G here? $G = c = 1$
Is it Newton G: $6.674×10^{−11} N⋅\frac{m^2}{kg^2}$

Yes.

But looking at your equation G is speed of light.

"$G = c = 1$" is maybe open to misunderstanding. It means "$G = 1$ and $c = 1$".

"$c = 1$" means "choose units to make $c = 1$", e.g. 1 light-second per second, or 1 light-year per year, or 1 light-nanosecond per nanosecond.

Similarly "$G = 1$" means "choose units to make $G = 1$".

Is G in $G = c = 1$ the same as
G in $R_s = \frac{2GM}{c^2}$ ?

Yes.

 

[PeterDonis]

what is G here? G=c=1G = c = 1
Is it Newton G: $6.674×10_{−11} N⋅\frac{m^2}{kg^2}$ or

Yes.

 

[PeterDonis]

the observer wristwatch time from "ouch" to "arrival" at the singularity (i.e., the limit of tau as the observer's r-value approaches zero) is independent of the size of the black hole.

I don't think this is correct; the time from horizon to singularity should increase with the mass of the hole. The "normalized" time, i.e., the time divided by the mass of the hole (in geometric units) is constant, but the actual proper time itself is not.

The formulas as I understand them are, for proper time from a large finite radius $R$ to the singularity (for an object released from rest at infinity, so it has some nonzero inward velocity at $R$, but we can make $R$ large enough for that velocity to be negligible):

$$
\frac{\tau_S}{2M} = \frac{2}{3} \left( \frac{R}{2M} \right)^{\frac{3}{2}}
$$

And for the proper time from $R$ to the horizon:

$$
\frac{\tau_{EH}}{2M} = \frac{2}{3} \left[ \left( \frac{R}{2M} \right)^{\frac{3}{2}} - 1 \right]
$$

Subtracting the second from the first and multiplying by $2M$ gives

$$
\tau_{EH-S} = \frac{4}{3} M
$$

 

[Stephanus]

[..]So at the horizon, we have

$$
a_{EH} = \frac{M}{2M \sqrt{4M^2 - 4M^2}} = \frac{1}{2 \sqrt{4M^2 - 4M^2}} = \frac{1}{4M \sqrt{1 - 1}}
$$

This is obviously undefined.

Using "my" algebra, I come up with this, actually there was an error in my previous calculation.
$a = \frac{c^4}{{4GM}\sqrt{1-c^2}}$
But I don't see why we can replace G with 1 and C with 1. Altough in SR forum, we often replace C = 1 and V = V/C.
I'll contemplate it.

 

[George Jones]

Subtracting the second from the first and multiplying by $2M$ gives

$$
\tau_{EH-S} = \frac{4}{3} M
$$

The time from event horizon to singularity is not the same as the time from "ouch" to singularity. For a small black hole, "ouch" occurs outside the event horizon, while for a large black hole, "ouch" occurs inside the event horizon, and this happens in such that the time from "ouch" to singularity is the same. Here, "ouch" is defined to occur when relative tidal acceleration equals g. If you have access, see the box on page B-21 of the book "Exploring Black Holes" by Taylor and Wheeler.

 

[PAllen]

And to add a terminological historical note, the phrase "ouch time", in a different context, arises from the ill fated Star Wars Episode I:

 

[PeterDonis]

The time from event horizon to singularity is not the same as the time from "ouch" to singularity.

Ah, sorry, I was misinterpreting what the "ouch" meant.

 

[PeterDonis]

Using "my" algebra, I come up with this

Still wrong. If you're getting any answer other than the one I gave, you're doing it wrong. This calculation has been done thousands of times by physicists and students in GR. Please consult a textbook.

I don't see why we can replace G with 1 and C with 1.

Because that just amounts to a choice of units. You already understand what it means to set $c = 1$; it just means we're measuring speeds in units of $c$, so the speed of light is $1$. Setting $G = 1$ means we are measuring mass in units of length; $G / c^2$ is just a conversion factor from ordinary units of mass to mass in length units. So the mass of the Earth, for example, instead of being about $6 \times 10^{24}$ kg, is about 4.5 millimeters in these units (called "geometric units").

If you insist on seeing the calculation I did with the factors of $G$ and $c$ included, here it is:

$$
a = \frac{GM}{r^2 \sqrt{1 - \frac{2GM}{c^2 r}}} = \frac{GM}{r \sqrt{r^2 - \frac{2GM}{c^2} r}} = \frac{G M c}{r \sqrt{c^2 r^2 - 2 G M r}}
$$

Plugging in $r = 2GM/c^2$ then gives

$$
a = \frac{G M c}{\left( \frac{2GM}{c^2} \right) \sqrt{c^2 \left( \frac{2GM}{c^2} \right)^2 - 2 G M \left( \frac{2GM}{c^2} \right)}} = \frac{G M c^3}{2 G M \sqrt{ \frac{4 G^2 M^2}{c^2} - \frac{4 G^2 M^2}{c^2} }} = \frac{c^4}{4 G M \sqrt{1 - 1}}
$$

Note that if we choose units so $c = 1$ and $G = 1$, this is identical to what I gave before. Also, of course, it is undefined.

 

[Stephanus]

Still wrong. If you're getting any answer other than the one I gave, you're doing it wrong. This calculation has been done thousands of times by physicists and students in GR. Please consult a textbook.
[..]
$$
a = \frac{GM}{r^2 \sqrt{1 - \frac{2GM}{c^2 r}}} = \frac{GM}{r \sqrt{r^2 - \frac{2GM}{c^2} r}} = \frac{G M c}{r \sqrt{c^2 r^2 - 2 G M r}}
$$

Plugging in $r = 2GM/c^2$ then gives

$$
a = \frac{G M c}{\left( \frac{2GM}{c^2} \right) \sqrt{c^2 \left( \frac{2GM}{c^2} \right)^2 - 2 G M \left( \frac{2GM}{c^2} \right)}} = \frac{G M c^3}{2 G M \sqrt{ \frac{4 G^2 M^2}{c^2} - \frac{4 G^2 M^2}{c^2} }} = \frac{c^4}{4 G M \sqrt{1 - 1}}
$$

Note that if we choose units so $c = 1$ and $G = 1$, this is identical to what I gave before. Also, of course, it is undefined.

Come on PeterDonis.
Is this wrong?
$\frac{c^4}{4GM\sqrt{1-c^2}}$
and yours
$\frac{c^4}{4GM\sqrt{1-1}}$
But I didn't change c to 1. Still needs time to understand why we can substitute c with 1 (or even G with 1).
I made some error in calculating 2GM^2.
$2GM^2 = 2G^2M^2$ it should be $4G^2M^2$ that's why I make a wrong calculation.
And it's not $2GM^2$ it's $(2GM)^2$.

 

[A.T.]

Still needs time to understand why we can substitute c with 1 (or even G with 1).

https://en.wikipedia.org/wiki/Geometrized_unit_system

 

[Stephanus]

https://en.wikipedia.org/wiki/Geometrized_unit_system

Wow, thanks for the link

 

[Stephanus]

[..]Because that just amounts to a choice of units. You already understand what it means to set $c = 1$; it just means we're measuring speeds in units of $c$, so the speed of light is $1$. Setting $G = 1$ means we are measuring mass in units of length; $G / c^2$ is just a conversion factor from ordinary units of mass to mass in length units. So the mass of the Earth, for example, instead of being about $6 \times 10^{24}$ kg, is about 4.5 millimeters in these units (called "geometric units").[..]

Okay, let's see if I can change G =1 and c = 1
$G = 6.674×10−11 N⋅m2/kg2$
in unit
$G = (kg⋅\frac{m}{s^2}) ⋅m2/kg2$
$G = \frac{m^3}{s^2⋅kg}$

$c = 300,000,000m/s$
in unit
$c = m/s$
$\frac{G}{c^2} = \frac{6.674⋅10^{-11}\frac{m^3}{s^2⋅kg}}{\frac{9⋅10^16m^2}{s^2}}$

$\frac{G}{c^2} = \frac{6.674⋅10^{-11}m}{9⋅10^{16}⋅kg}$

$\frac{G}{c^2} = \frac{7.41556⋅10^{-28}m}{kg}$

Earth mass = 6 x 10²⁴ Kg
I don't even know why we should multiply or divide Earth mass with G/c2
Okay, I choose multiply
$6 * 10^{24} Kg * \frac{7.41556⋅10^{-28}m}{kg}$
$6 * 10^{24} * 7.41556⋅10^{-28}m$
$44.49 * 10^{-4}m = 4.449 mm$ Okay...
I imagine we divide mass of the Earth with this unit, not multiply it.
Perhaps it's the same if we have a car that can runs 100km/s
And we say, what is 20 seconds according to you.
Then we multiply 20 seconds with Car_speed, then we'll say 20 seconds for my car is 200 km.
Or perhaps
There are two towns, 500 km apart. What is the distance of the towns.
Then we divide 500 km with Car_speed, then we'll say the distance of the towns is 5 seconds for my car.
I have to think it over, still don't get it fully.

 

[Stephanus]

I think G/c^2 is half of Schwarzschild radius.
If we multiply the mass of an (planetary) object with this unit, we'll get acceleration at that distance. But distance must in the format of
Distance/300000000
if We multiply the mass of that object with twice of this unit, we'll get its schwarzschild radius
I'll try to understand it deeper.