black hole, singularity...

[zztop]

 

This is the part that's hanging you up. There's no canceling effect if the pull from every direction is towards a single direction, and if it is a singularity at the center, it's dimensionless.

 

Can someone clear up this conflict?

This question is unanswerable since no information can escape from beyond the EH. We cannot know what happens, we can only theorize.

[Strange]

This question is unanswerable since no information can escape from beyond the EH. We cannot know what happens, we can only theorize.

 

 

And theory says ... ?

 

I am assuming there is only one theoretical answer. (Based on the fact that, as far as I now, we only have one theory describing black holes: GR.) Maybe that is not the case.

 

Also, does the type of black hole matter? Is a Schwarzschild black hole different from a Kerr black hole in this respect?

[JohnLesser]

Beyond the event horizon of a singularity, all mass experiences isotropic force from the singularity. The affected mass has its own gravitation inertia holding the mass in its own orbit. The singularity then ripping apart the affected mass from the opposing force of the inertia orbit.

[J.C.MacSwell]

There are many paths inside the event horizon...they all lead to the centre, which is possibly a singularity.

There is a gradient in the gravitational field, both inside and outside the event horizon.

The smaller less massive the black hole the greater is the gradient. The gradient is there because gravity is stronger as you approach the centre.This gives rise to tidal forces. If the gradient is great enough you get "spaghettified"...if you are heading feet first your feet are pulled from your body, which is pulled from your head. If the black hole is massive enough this can take place well inside the event horizon, since though the gravity is strong the gradient can still be tolerable. If the black hole is small enough this can take place outside of the event horizon since though the gravity might not be quite as strong (still very strong) the gradient is stronger there than just inside the event horizon of a more massive black hole.

 

It is all about the gradient. With a good gradient all you need is the sauce, grated cheese, and perhaps some meatballs.

Edited March 28, 2017 by J.C.MacSwell

[MigL]

The event horizon is a mathematical construct. There is nothing there.

It is simply the radius at which escape velocity equals c .

Gravity acts no different on one side of it than the other.

 

A simple explanation for the 'direction' dilemma is to look at what happens to light cones as you cross the event horizon and approach the singularity. In effect, unless you can move faster than light, all POSSIBLE directions lead to the singularity.

 

Edit : Sorry cross posted with J.C.

Edited March 28, 2017 by MigL

[zztop]

 

 

And theory says ... ?

 

I am assuming there is only one theoretical answer. (Based on the fact that, as far as I now, we only have one theory describing black holes: GR.) Maybe that is not the case.

 

Also, does the type of black hole matter? Is a Schwarzschild black hole different from a Kerr black hole in this respect?

The theory gives us the "internal solution" for the EFEs in the form of metric that is quite different from the :external solution". The metric IS the "theory", it predicts how test probes would move inside the EH. We can only say that the predictions are quite different from the ones made by the "external" solution. We will never know because we will never be able to test it.

To your second question, I only know of the internal Schwarzschild solution, I do not know about the internal solutions for Kerr, Riessner, etc.

[Pugdaddy]

You might want to see this lecture by Leonard Susskind. Apparently Bob would not even know he crossed the event horizon until the tidal forces spaghetatized him, but Alice would see Bob slow down and never cross.

 

Edited March 29, 2017 by Pugdaddy

[zztop]

You might want to see this lecture by Leonard Susskind. Apparently Bob would not even know he crossed the event horizon until the tidal forces spaghetatized him, but Alice would see Bob slow down and never cross.

 

Actually, if you listened to him to the end , you would have found out that we can never find out what happened to Bob near or at the EH. Though ANY experiment. Which is exactly what I said earlier. What we DO know is that the tidal forces WAY OUT of a BH (far from the EH) spaghettify Bob. Tidal forces are REAL, they are not a coordinate "artifact".

Edited March 29, 2017 by zztop

[Pugdaddy]

Wouldn't it depend on the mass of the BH? The more massive the BH is, the larger the EH Radius is. Every bit of information that falls into the BH adds another Planck area to the surface area of the EH. So if the mass of BH is sufficiently large, the EH could be far enough away from the singularity, because the tidal force falls off with square of the distance, that Bob could pass through the EH without feeling the spaghetifing tidal force.

[zztop]

Wouldn't it depend on the mass of the BH? The more massive the BH is, the larger the EH Radius is. Every bit of information that falls into the BH adds another Planck area to the surface area of the EH. So if the mass of BH is sufficiently large, the EH could be far enough away from the singularity, because the tidal force falls off with square of the distance, that Bob could pass through the EH without feeling the spaghetifing tidal force.

The radius of the EH (also known as the Schwarzschild radius) is extremely small (for the Sun is 3km, as an example (the Earth is 9mm). The reason is the [math]c^2[/math] in the [math]r_s=\frac{2GM}{c^2}[/math]

The "G" is not helping either.

Edited March 29, 2017 by zztop

[Pugdaddy]

How massive would the BH have to be to make the tidal force at the EH not be noticed by Bob? Say M = 10^70. Would that make the EH radius about 10^ 47? Would the tidal force at that distance from the singularity be small enough to be unnoticed by Bob? Sorry my math is not great, but do I have the orders of magnitude right?

Edited March 29, 2017 by Pugdaddy

[zztop]

How massive would the BH have to be to make the tidal force at the EH not be noticed by Bob? Say M = 10^70. Would that make the EH radius about 10^ 47? Would the tidal force at that distance from the singularity be small enough to be unnoticed by Bob? Sorry my math is not great, but do I have the orders of magnitude right?

[math]\frac{G}{c^2}=10^{-27}[/math]

 

See the answer to problem 3

[MigL]

The answer to problem #3 is meaningless.

There are no such things as stellar sized BHs with the mass of the Sun, as they could not possibly undergo gravitational collapse.

They cannot even form neutron stars and are destined to be dwarfs in their senior years.

 

Most galactic center BHs consist of millions, if not billions, of solar masses, and you could easily pass right through the event horizon without even noticing it. The fact that an outside observer would never see this does not change your reality and future meeting with the ( possible ) singularity.

 

An excellent, non-mathematical reference is Kip Thorne's Black Holes and Time Warps.

[zztop]

The answer to problem #3 is meaningless.

There are no such things as stellar sized BHs with the mass of the Sun, as they could not possibly undergo gravitational collapse.

They cannot even form neutron stars and are destined to be dwarfs in their senior years.

 

Most galactic center BHs consist of millions, if not billions, of solar masses, and you could easily pass right through the event horizon without even noticing it. The fact that an outside observer would never see this does not change your reality and future meeting with the ( possible ) singularity.

 

An excellent, non-mathematical reference is Kip Thorne's Black Holes and Time Warps.

You completely miss the point, a BH of Sun's mass will spaghettify an object at 100 km distance.

[Strange]

I just noticed that the interior Schwarzschild solution is for a uniform density. I guess this means it is not a realistic description of what actually happens inside a black hole.

[Bender]

I just noticed that the interior Schwarzschild solution is for a uniform density. I guess this means it is not a realistic description of what actually happens inside a black hole.

It doesn't have to be uniform, as long as most of the mass is closer to the singularity than you are, you are fine (well, probably not, but the calculations work, as far as we can tell).

[Strange]

 

 

In Einstein's theory of general relativity, the interior Schwarzschild metric (also interior Schwarzschild solution or Schwarzschild fluid solution) is an exact solution for the gravitational field in the interior of a non-rotating spherical body which consists of an incompressible fluid (implying that density is constant throughout the body)

 

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

[Bender]

My mistake. In that case, what is the relevance? A homogeneous interiour sounds like a pretty poor hypothesis.

[Strange]

My mistake. In that case, what is the relevance? A homogeneous interiour sounds like a pretty poor hypothesis.

 

 

It was a simplifying assumption to allow a solution to the Einstein Field Equations to be found. It doesn't (just) apply to black holes, but to any spherical mass.

 

The external solution is more relevant to us (because we can observe what happens) but even that makes a number of simplifying assumptions: a non-rotating eternal unchanging black hole in an empty universe. But that is good enough for many purposes.

[Pugdaddy]

So as is it says in problem #5, if the BH is large enough, it is possible for Bob to cross the EH without realizing it? At least for a while.

[zztop]

I just noticed that the interior Schwarzschild solution is for a uniform density. I guess this means it is not a realistic description of what actually happens inside a black hole.

We don't know what happens inside the EH. We will NEVER know (because no information escapes from beyond the EH).

So as is it says in problem #5, if the BH is large enough, it is possible for Bob to cross the EH without realizing it? At least for a while.

Correct. To put it in math form, let's say the spaghettification occurs for an acceleration of [math]a=k*g[/math] where [math]k[/math] is a value greater than 100 and [math]g=\frac{GM}{r^2}[/math].

On the other hand, the EH is at [math]r_s=\frac{2GM}{c^2}[/math]

 

So, spaghettification happens at a distance [math]r[/math] from the singularity:

 

[math]k*g=\frac{r_s c^2}{2r^2}[/math]

 

i.e.

 

[math]r=c \sqrt{\frac{r_s}{2k*g}}[/math]

 

It is possible to have [math]r<r_s[/math] if :

 

[math]r_s>\frac{c^2}{2k*g}[/math]

 

The above condition can be fulfilled for supermassive BHs. For BH with [math]r_s<\frac{c^2}{2k*g}[/math] the spaghettification occurs way outside the EH. The main lesson from this thread is that the language of physics is MATH. We need to learn to express ourselves in mathematical terms.

Edited March 29, 2017 by zztop

[Bender]

We don't know what happens inside the EH. We will NEVER know (because no information escapes from beyond the EH).

 

Correct. To put it in math form, let's say the spaghettification occurs for an acceleration of [math]a=k*g[/math] where [math]k[/math] is a value greater than 100 and [math]g=\frac{GM}{r^2}[/math].

On the other hand, the EH is at [math]r_s=\frac{2GM}{c^2}[/math]

 

So, spaghettification happens at a distance [math]r[/math] from the singularity:

 

[math]k*g=\frac{r_s c^2}{2r^2}[/math]

 

i.e.

 

[math]r=c \sqrt{\frac{r_s}{2k*g}}[/math]

 

It is possible to have [math]r<r_s[/math] if :

 

[math]r_s>\frac{c^2}{2k*g}[/math]

 

The above condition can be fulfilled for supermassive BHs. For BH with [math]r_s<\frac{c^2}{2k*g}[/math] the spaghettification occurs way outside the EH. The main lesson from this thread is that the language of physics is MATH. We need to learn to express ourselves in mathematical terms.

For spaghettification, shouldn't you look at the gradient da/dr rather than at the magnitude a?

[Pugdaddy]

(Correct. To put it in math form, let's say the spaghettification occurs for an acceleration of where is a value greater than 100 and .

On the other hand, the EH is at

 

So, spaghettification happens at a distance from the singularity:

 

 

i.e.

 

 

It is possible to have if :

 

)

 

 

 

The above condition can be fulfilled for supermassive BHs)

 

Thank You for the math. It would have taken me years.

 

All paths lead to the singularity once the EH is passed. Either directly or asymptotically along the EH? That is what it looks like from Professor's Susskind's drawing. There would not be any cancelling of forces. Wasn't that the original question?

[swansont]

The "ripping apart" happens BEFORE the object crosses the event horizon. A picture will help.

 

 

Depends on the size of the black hole. Big ones have larger Schwarzschild radii, and since tidal forces nominally depend on r^3, these are relatively weak at the event horizon. It's not the acceleration, it's the gradient, as Bender has noted.

 

https://en.wikipedia.org/wiki/Spaghettification#Inside_or_outside_the_event_horizon

 

There's a discussion of the tensile strength needed for a few different scenarios in the link.

[zztop]

We don't know what happens inside the EH. We will NEVER know (because no information escapes from beyond the EH).

Correct. To put it in math form, let's say the spaghettification occurs for an acceleration of [math]a=k*g[/math] where [math]k[/math] is a value greater than 100 and [math]g=\frac{GM}{r^2}[/math].

On the other hand, the EH is at [math]r_s=\frac{2GM}{c^2}[/math]

 

So, spaghettification happens at a distance [math]r[/math] from the singularity:

 

[math]k*g=\frac{r_s c^2}{2r^2}[/math]

 

i.e.

 

[math]r=c \sqrt{\frac{r_s}{2k*g}}[/math]

 

It is possible to have [math]r<r_s[/math] if :

 

[math]r_s>\frac{c^2}{2k*g}[/math]

 

The above condition can be fulfilled for supermassive BHs. For BH with [math]r_s<\frac{c^2}{2k*g}[/math] the spaghettification occurs way outside the EH. The main lesson from this thread is that the language of physics is MATH. We need to learn to express ourselves in mathematical terms.

I was trying to keep things simple, since some of you argued for using the gravitational acceleration GRADIENT instead of the acceleration proper (acceleration proper makes you into a "pancake", not a "spaghetti"), here is the more complex treatment for "spaghetti":

 

[math]a=\frac{2GMd}{r^3}[/math]

 

where d is the length of the object and [math]a[/math] is the acceleration gradient over the distance d. Re-doiing the calculations:

 

[math]r=(\frac{r_sc^2d}{k*g})^{1/3}[/math]

 

The points I made stand, depending on the characteristics of the BH spaghettification can occur inside the EH or way outside it, you need to compare [math](\frac{r_sc^2d}{k*g})^{1/3}[/math] with [math]r_s[/math]

.

 

There would not be any cancelling of forces.

correct

Edited March 29, 2017 by zztop