Shouldn't objects entering black holes be instantly shredded?

[qwertyflatty]

Nothing can escape a black hole, so as an object enters the event horizon, the atoms currently inside the black hole wouldn't be able to bond with the atoms currently outside of the black hole, causing it to sever. But I've never heard of any physicist saying that this would happen, so I'm probably wrong.

Could someone explain why an object wouldn't get shredded immediately from entering a black hole's event horizon?

 

[ModusPwnd]

Something like this maybe?

http://en.wikipedia.org/wiki/Spaghettification

 

[qwertyflatty]

I mean instantly, as it's entering the event horizon. Spaghettification may happen sometime after passing through the event horizon.

 

[Integral]

Actually Spaghettification happens as you approach the event horizon due to the tremendous gravitational gradient present.

 

[Chronos]

For a supermassive black hole, you would not even notice you had entered its event horizon [unless you tried to escape]. Tidal forces are virtually insignificant. So, nothing would appear out of sorts for an object in free fall entering the event horizon of a supermassive black hole - until later.

 

[qwertyflatty]

I'm not talking about tidal forces.

Anything that crosses the event horizon can't get out. When an object crosses the event horizon, it doesn't all cross at the exact same time. Only parts of it cross at any instant. If nothing inside the event horizon can get out, then the atoms of the object currently inside the event horizon shouldn't be able to bond to the atoms of the objects outside the event horizon, causing it to be severed.

 

[phinds]

Actually Spaghettification happens as you approach the event horizon due to the tremendous gravitational gradient present.

No, as chronos pointed out, large black holes have such a small gravity gradient near the event horizon that the EH is a "non-event" horizon.

The EH has NOTHING to do with spahettification. It can happen before or after the EH depending on the size of the BH and for it to happen AT the EH is possible but extremely unlikely as it would require a precise confluence of BH size and particular strength of material entering.

 

[Chronos]

I'm not talking about tidal forces.

Anything that crosses the event horizon can't get out. When an object crosses the event horizon, it doesn't all cross at the exact same time. Only parts of it cross at any instant. If nothing inside the event horizon can get out, then the atoms of the object currently inside the event horizon shouldn't be able to bond to the atoms of the objects outside the event horizon, causing it to be severed.

Actually, this idea is very reminiscent of that proposed by Polchinski last July: Black Holes: Complementarity or Firewalls?, http://arxiv.org/abs/1207.3123. Obviously, there is no way to test such an idea.

 

[willem2]

I'm not talking about tidal forces.

Anything that crosses the event horizon can't get out. When an object crosses the event horizon, it doesn't all cross at the exact same time. Only parts of it cross at any instant. If nothing inside the event horizon can get out, then the atoms of the object currently inside the event horizon shouldn't be able to bond to the atoms of the objects outside the event horizon, causing it to be severed.

Objects just outside of the event horizon, will need a very large acceleration to get out. It is the force necessary for the acceleration that will tear the object apart, but the same force would also tear the object apart without a nearby black hole.

 

[phinds]

Objects just outside of the event horizon, will need a very large acceleration to get out. It is the force necessary for the acceleration that will tear the object apart, but the same force would also tear the object apart without a nearby black hole.

Huh? I don't get what you're saying

 

[Spourk]

Speaking of tidal forces, has anyone read the shortstory, "Neutron Star" by Larry Niven? It describes what it might be like if a pilot of an indestructable ship gets close enough to a Neutron star to experience tidal effects.

Poor Larry Niven though, he has been chastized into the floor by all kinds of students and physicists alike, especially for Ringworld.

Niven writes: "I keep meeting people who have done mathematical treatments of the problem raised in the short story 'Neutron Star', ... Alas and dammit, Shaeffer can't survive. It turns out that his ship leaves the star spinning, and keeps the spin."

 

[willem2]

Huh? I don't get what you're saying

Any object very close to the event horizon of a black hole needs to accelerate very quickly to escape from the black hole. It is the force needed for this acceleration that will tear an object apart that has gone halfway past the event horizon. If you don't try to accelerate the object it will fall in entirely and won't get torn apart. (as long as tidal forces aren't too big)

 

[WannabeNewton]

Any object very close to the event horizon of a black hole needs to accelerate very quickly to escape from the black hole. It is the force needed for this acceleration that will tear an object apart that has gone halfway past the event horizon. If you don't try to accelerate the object it will fall in entirely and won't get torn apart. (as long as tidal forces aren't too big)

The force that must be exerted at infinity to not even accelerate backwards but keep stationary a unit test mass in the limit as one approaches the event horizon is given by the surface gravity $\kappa$.

In particular, it can be shown that if $\xi^{a}$ is the killing vector field normal to the (null) killing horizon that represents the event horizon of a stationary black hole, then $\kappa^{2} = \lim \{\frac{-(\xi^{b}\nabla_{b}\xi^{c})(\xi^{a}\nabla_{a}\xi_{c})}{\xi^{d}\xi_{d}}\}$ where the limit is as one approaches the event horizon. It is easy to show that the local proper acceleration of an object stationary in this limit is just $a^{c} = \frac{(\xi^{b}\nabla_{b}\xi^{c})}{(-\xi^{d}\xi_{d})}$ so, letting $V = (-\xi^{a}\xi_{a})^{1/2}$, the above turns into $\kappa^{2} = \lim(Va)$.

For a static black hole, $V$ is nothing more than the redshift factor so the above limit just gives the force that must be exerted at infinity in order to keep a unit test particle in place as one approaches the event horizon. The reason why what you're saying, regarding the object being halfway into the black hole, makes no sense is because the local acceleration $a$ diverges to infinity at the event horizon whereas the redshift factor vanishes at the event horizon (the horizon is a null surface and the killing vector field is normal to it hence it is null on the horizon so it has zero length) meaning the force that must be exerted at infinity to keep the object in place once parts of it have gone through the event horizon isn't even well-defined, let alone a force that could accelerate it backwards.

 

[qwertyflatty]

I'm not talking about tidal forces, and I'm not talking about hovering near an event horizon.

When an object enters the event horizon, the entire object can't enter at the same instant. The same is true for a person passing through any point. Ex. A person diving into a swimming pool would enter hands first, then head, then the rest of his body.

Assume the black hole is massive enough that it's still intact as it reaches the event horizon. Any object that passes through the event horizon can't transmit information outside of the event horizon. Ergo, the atoms currently inside the event horizon can't bond with the atoms outside during the instances where some parts of object are inside. Ergo, the atoms inside should become severed from the outside.

Do you understand my question now?

 

[Borek]

To put it another way (using your pool analogy) - if you are half submerged in the black hole, you can't feel your legs, as the signal from the legs is not able to get to the brain.

It makes me think about different, related scenario. Imagine two oppositely charged point like objects, spinning around their center of mass (let's assume objects are large enough to be described by the classic mechanics). They approach the event horizon and one of them falls below before the other. What happens to the trajectory of the object that is not yet below the event horizon?

 

[Fredrik]

To put it another way (using your pool analogy) - if you are half submerged in the black hole, you can't feel your legs, as the signal from the legs is not able to get to the brain.

I think you would be going much faster than the nerve signal, so the rest of your body will be inside before the signal reaches your brain.

Now you're probably thinking about ways to go slowly through the event horizon. I wouldn't recommend it. Keep in mind that the force required to keep you at a fixed coordinate "altitude" goes to infinity as you approach the horizon. I haven't done the math on this, but I think it's safe to say that slowing you down to the speed of a nerve signal will require forces that are many orders of magnitude greater than what it would take to crush you.

 

[Borek]

Now you're probably thinking about ways to go slowly through the event horizon. I wouldn't recommend it. Keep in mind that the force required to keep you at a fixed coordinate "altitude" goes to infinity as you approach the horizon. I haven't done the math on this, but I think it's safe to say that slowing you down to the speed of a nerve signal will require forces that are many orders of magnitude greater than what it would take to crush you.

For a supermassive black hole, you would not even notice you had entered its event horizon [unless you tried to escape]. Tidal forces are virtually insignificant. So, nothing would appear out of sorts for an object in free fall entering the event horizon of a supermassive black hole - until later.

I feel like there is some kind of contradiction here.

 

[VantagePoint72]

I feel like there is some kind of contradiction here.

No, they are agreeing with each other. The surface gravity at an event horizon is huge, but you don't feel surface gravity when you're in free fall. You only feel tidal forces when in free fall, which are not large at an event horizon. To feel to the absolute value of the gravitational acceleration, you need to resist it and slow your descent—Frederik's point. I agree with him that this force would almost certainly be enough to crush you. On the other hand, if you start free falling towards the event horizon from a far enough distance that you could have maintained a fixed height without being crushed, then by the time you reach the horizon you will be going extremely fast. Much faster than the propagation of nerve signals in the pool analogy.

 

[Borek]

OK, so the analogy was poor. Still, the analogy was there only to explain what the original question was about.

What about trajectory of the charged object?

 

[phinds]

... You only feel tidal forces when in free fall, which are not large at an event horizon

Absolutely not true as a generality. Tidal forces are so huge at the EH of a small BH that you would be ripped apart WAY before you even get to the EH.

 

[Fredrik]

I typed this before I saw that LastOneStanding has already said pretty much the same thing.

I feel like there is some kind of contradiction here.

I don't see a contradiction. Chronos and I are talking about two different things.

The world line of a particle in free fall is a geodesic. But if you fall through the event horizon of a typical black hole, and all the particles in your body move as described by geodesics, you will die quickly because the distance between your feet and your head will be increasing rapidly.

Acceleration is a measure of the deviation from geodesic motion, and force is mass times acceleration, so it takes a force to prevent your head and feet from moving as described by geodesics, while some part in the middle of your body is moving as described by a geodesic. The only forces at play here are the internal forces in your body. So your survival of a "free fall" through an event horizon depends on whether those forces are strong enough to keep your body parts at constant distances from each other.

In the case of a very massive black hole, the natural elasticity of your body will generate enough force to keep you intact. In the case of a small black hole, those forces won't be anywhere near enough.

This is what Chronos was talking about. I was talking about the external force (from say the rocket that you're in) that it would take to prevent your speed as you cross the horizon from being much greater than that of a nerve signal. If you fall freely from a great distance, you will obviously accelerate (in Schwartzschild coordinates) to an enormous speed before you get to the horizon. And if you use your rocket engines to descend slowly so that you can jump in from a low altitude, you will be crushed against the floor, because of the rocket's enormous acceleration.

 

[VantagePoint72]

Absolutely not true as a generality. Tidal forces are so huge at the EH of a small BH that you would be ripped apart WAY before you even get to the EH.

The conversation I was referring to was specifically about supermassive BHs; but, it's worthwhile to highlight this point so thank you.

 

[Borek]

That leaves us again with the original unanswered question

 

[VantagePoint72]

That leaves us again with the original unanswered question

Well—putting aside these recent proposals about black hole firewalls we've been hearing about—the equivalence principle implies that nothing wonky happens as you cross the event horizon, and so the answer to your 'what happens to the orbiting particles?' question is 'nothing' and the answer to the original question is 'no'. Of course, picking out the hole in the reasoning that says something untoward should happen is trickier. For one thing, as far as your particles orbiting thought experiment goes, there's no reason why the particle behind the event horizon should have any trouble feeling the force from the one that has not yet passed through and should continue moving normally (even after the time for disturbances in the EM field to propagate out is taken into account); however, I may be mistaken, but I don't believe "particle A crosses the event horizon before particle B" is a coordinate invariant statement.

This, to me, suggests that the apparent paradox (that the equivalence principle says nothing happens to electromagnetic bonds, but such bonds constituting a causal connection suggest otherwise) is—as with most of the apparent paradoxes with relativity—a result of using bad coordinates. I'll leave it to someone more competent with black hole GR to explain how the whole thing looks in good coordinates. Whatever that analysis is, though, I believe the result will be that objects are not shredded by a severing of chemical bonds as they pass through the event horizon.

 

[Chronos]

Before the firewall paper, most of us were probably smugly confident that passing through an event horizon would be unremarkable, or even unnoticeable providing the black hole was sufficiently massive. I am considerably less confident than I was before that Polchinski guy crashed into the apple cart. A naive calculation implies you should be traveling at the speed of light once you reach the EH, which is presumably impossible for a massive particle. The logical answer is you would be converted to massless particles at that point. A proper theory of quantum gravity is probably necessary to say with any assurance how, when, or what happens when you approach an EH.

 

[VantagePoint72]

A proper theory of quantum gravity is probably necessary to say with any assurance how, when, or what happens when you approach an EH.

Well, yes, that may well be true. But it's still very worthwhile to ask, "What is the answer to this question according to general relativity," even if the aims are purely pedagogical. There is clearly some insight to be gained into how motion near black holes—even as classical objects—works. qwerty's and Borek's thought experiments are perfectly well-posed questions about what happens as you cross the event horizon of a classical black hole; I don't see the possibility of quantum gravity changing the answer as a reason not to come to a clear conclusion about what GR says. We can, for the sake argument, treat the electromagnetic field as a classical object too and ask what happens to electrodynamically-bound objects orbiting each other as one of them crosses the event horizon.

As I said, I think it's a pretty clear consequence of the equivalence principle that nothing unusual should occur. However, an explicit analysis—which I'm unable to do—showing confirming this would be illuminating. It would be interesting if someone more capable did such an analysis.

 

[WannabeNewton]

Well if we had just one charged particle falling in radially, once it enters the Schwarzschild black hole the black simply takes on the charge of the particle and becomes an Reissner-Nordstrom black hole with a radial electric field emanating from $r = 0$. Now if we have another charged particle electrically bound to the first when the first falls into the black hole, I don't know if the act of the black hole acquiring the charge of the first and consequently having an electric field emanate from $r = 0$ changes the trajectory of the other charge or fails to affect it at all. I'm trying to find a paper on it but am having no luck. Hopefully someone else can find a paper on the matter.

 

[Nugatory]

To put it another way (using your pool analogy) - if you are half submerged in the black hole, you can't feel your legs, as the signal from the legs is not able to get to the brain.

You can't be "half-submerged" unless part of your body is stationary (constant Schwarzschild r coordinate) at and inside the event horizon. That's impossible because the horizon is a lightlike surface and r is timelike inside the horizon. So we're moving through the horizon, not suspended with half our body on each side (and a good thing too, because being torn apart as one half of our body heads down into the interior of the black hole while some enormous force holds the other half outside doesn't sound like fun to me). And if we're moving through the horizon in free fall, there's no problem: the nerve impulse leaves my toes before my head has passed through the horizon, reaches my head after it's passed through the horizon. That's not even strange or weird; the nerve impulse isn't instantaneous so we always expect to arrive at the brain some time after it's sent from the toe.

It makes me think about different, related scenario. Imagine two oppositely charged point like objects, spinning around their center of mass (let's assume objects are large enough to be described by the classic mechanics). They approach the event horizon and one of them falls below before the other. What happens to the trajectory of the object that is not yet below the event horizon?

We can simplify a bit by assuming that the two particles are not charged but are connected by a string that keeps them orbiting around one another (assume the string's tension varies with the square of its length if you really care about the exact trajectory). Now we don't have to involve Reissner and Nordstrom. I'm still not going to try calculating the trajectory in Schwarzschild coordinates, but I'd expect that (as long as the distance between the two particles is small enough that we can ignore tidal effects) in the inertial center of mass frame of the infalling pair they both fall happily through the horizon without noticing.

 

[Borek]

We can simplify a bit by assuming that the two particles are not charged but are connected by a string that keeps them orbiting around one another.

I have intentionally avoided the string as it magically carries the information "we are connected" both ways through the EH. I hoped assuming just classic electrostatic interaction will make the analysis easier to understand (if there is anything easy to understand in places like EH).

 

[Nugatory]

I have intentionally avoided the string as it magically carries the information "we are connected" both ways through the EH. I hoped assuming just classic electrostatic interaction will make the analysis easier to understand (if there is anything easy to understand in places like EH).

Ah - right - I see why you'd prefer describing the problem using the Coulomb attraction to tie the two particles together.

But whether we tie the particles together with a string or their mutual electrical attraction, the "we are connected" information does not propagate instantaneously. In the electrical case, changes in the electrical field propagate at the speed of light, so when the first particle drops below the horizon its electrical field outside the horizon does not immediately disappear.
In the string case, the "we are connected" information propagates with a delay as described in this entry from the relativity sub-forum FAQ.

Either way, as long as the tidal forces are small enough to ignore and we're in free fall, the second particle makes it through the horizon before the "OMG, my partner is on the other side of the event horizon" information can reach it and there's no reason to expect the pair to be torn apart.

(As with any problem involving falling through the event horizon, the Schwarzschild coordinates are not a good choice; the coordinate singularity at the horizon means that they won't properly describe a system that's "divided" by the horizon)

 

[skeptic2]

A naive calculation implies you should be traveling at the speed of light once you reach the EH, which is presumably impossible for a massive particle. The logical answer is you would be converted to massless particles at that point.

Please excuse my naivete Chronos, but why is it [presumably] impossible for a massive to reach the speed of light at the horizon? The free falling massive particle is not being accelerated in the usual sense but is at rest.

Why would being converted to massless particles be the logical answer?

 

[rustytxrx]

I think the problem is the Special theory of Relativity. Einstein predicts if a particle reached the speed of light the mass and energy would be infinite. There seems to be a problem with that.

 

[skeptic2]

I think the problem is the Special theory of Relativity. Einstein predicts if a particle reached the speed of light the mass and energy would be infinite. There seems to be a problem with that.

That is true if the particle is accelerated, but in the case of free fall, the particle is at rest. It is not being accelerated. If you were in a spaceship, free falling towards the event horizon, you would be weightless which is not possible if you were being accelerated.

 

[WannabeNewton]

Please excuse my naivete Chronos, but why is it [presumably] impossible for a massive to reach the speed of light at the horizon? The free falling massive particle is not being accelerated in the usual sense but is at rest.

A freely falling particle is only locally inertial; it is not "at rest" given the strict definition of the term: if the space-time is stationary, a notion of being "at rest" with respect to the space-time exists and this is defined as following an orbit of the time-like killing vector field of the space-time. This is what people mean when they say "what is the force needed at infinity in order to keep stationary a unit test mass near the event horizon of a static black hole". A particle stationary ("at rest") in the above sense has non-vanishing proper acceleration (whereas a freely falling particle has vanishing proper acceleration). In summary, don't confuse free fall with being "at rest" in the above sense (which is roughly what you intuitively think of as being at rest with respect to the space-time).

 

[Aryianna]

There is a physicist by the name of Amos Ori whose research is about this subject matter. I have met him many years ago at Caltech, and he did explain to me that it is possible to escape the event horizon. I had a copy of his paper, and he has worked with Kip Thorne on this subject matter. I'll look it up.

 

[George Jones]

There is a physicist by the name of Amos Ori whose research is about this subject matter. I have met him many years ago at Caltech, and he did explain to me that it is possible to escape the event horizon. I had a copy of his paper, and he has worked with Kip Thorne on this subject matter. I'll look it up.

Do you mean "escape the event horizon" or "survive the curvature singularity singularity at inner (Cauchy) horizon"?

There is a weak curvature singularity at the Cauchy inner horizon (IH) of a rotating black hole. Shredding is caused by tidal force, and tidal force is caused by spacetime curvature, so isn't shedding guaranteed by a curvature singularity?

Seminal work on this was done by Poisson and Israel, and this work was continued by Ori. See

http://physics.technion.ac.il/~school/Amos_Ori.pdf .

Ori writes "Consequence to the curvature tensor at the IH ... However, the IH-singularity is weak (namely, tidally non-destructive.)"

Roughly, if components of g (the metric) are continuous but "pointy" (like the absolute value function), then first derivatives of g have step diiscontinuities (like the Heaviside step function), and second derivatives of g (used in the curvature tensor) are like Dirac delta functions. If a curvature singularity blows up like a Dirac delta function, then integration produces only a finite contribution to the tidal deformation of an object, which, if the object is robust enough, it can withstand.

 

[Fredrik]

Do you mean "escape the event horizon" or "survive the curvature singularity singularity at inner (Cauchy) horizon"?

There is a weak curvature singularity at the Cauchy inner horizon (IH) of a rotating black hole. Shredding is caused by tidal force, and tidal force is caused by spacetime curvature, so isn't shedding guaranteed by a curvature singularity?

Chronos posted a link to a paper earlier in the thread (post #8) that says that you may get shredded at the horizon even when the tidal forces are negligible. The argument is obviously not based on GR only, but on some assumptions about how GR and QM work together.

I have no idea if that's the sort of thing Aryianna was referring to. Edit: I think you're right that this must be about missing the singularity, not about passing through the event horizon in one piece.

 

[rustytxrx]

cartoons anyone?

cartoon = 10 chalk boards of calculus :)

 

[Naty1]

Chronos posts:

[Before the firewall paper, most of us were probably smugly confident that passing through an event horizon would be unremarkable, or even unnoticeable providing the black hole was sufficiently massive. I am considerably less confident than I was before that Polchinski guy crashed into the apple cart. /QUOTE]

I had not heard about that...Interesting.

The adjacent paper in ARXIV is from Leonard Susskind who so far disagrees...
should be fun to see how it plays out...


http://arxiv.org/abs/1301.4505

Black Hole Complementarity and the Harlow-Hayden Conjecture


Conclusion

[Amps is the Amheiri-Marolf-Polchinski-Sully firewall argument referenced by Chronos.]

The AMPS paradox is currently forcing a rethinking of how, and where, information is
stored in quantum gravity. The possible answers range from more or less conventional
localization (proximity and firewalls) to the radical delocalization of A = RB:
The argument in favor of the proximity postulate assumes the possibility of an Alice
experiment", which in some respects resembles time travel to the past. From this per-
spective, the firewall would function as a chronology protection agent, but at the cost of
the destruction of the interior of the black hole. The Harlow-Hayden conjecture opens an
entirely new perspective on chronology protection. It is based on the extreme fine-grained
character of information that Alice needs to distill before returning to the present."
Fine-grained information is something that has never been of much use in the past,
given how hard it is to extract. But there is clearly a whole world of fine-grained data
stored in the massive entanglements of scrambled pure states. That world is normally
inaccessible to us, but if BHC is correct, it is accessible to an observer who passes through
the horizon of a black hole. Thus complementarity implies a duality:
Ordinary coarse-grained information in an infalling frame is dual to the fine-grained
information in the exterior description...

It's obviously premature to declare the paradox resolved, but the validity of the HH
conjecture would allow the strong complementarity of Bousso and Harlow [14] to be con-
sistent, without the need for firewalls. For these reasons I believe that BHC, as originally
envisioned by Preskill, 't Hooft, and Susskind-Thorlacius-Uglum, is still alive and kicking.

 

[Aryianna]

Do you mean "escape the event horizon" or "survive the curvature singularity singularity at inner (Cauchy) horizon"?

There is a weak curvature singularity at the Cauchy inner horizon (IH) of a rotating black hole. Shredding is caused by tidal force, and tidal force is caused by spacetime curvature, so isn't shedding guaranteed by a curvature singularity?

Seminal work on this was done by Poisson and Israel, and this work was continued by Ori. See

http://physics.technion.ac.il/~school/Amos_Ori.pdf .

Ori writes "Consequence to the curvature tensor at the IH ... However, the IH-singularity is weak (namely, tidally non-destructive.)"

Roughly, if components of g (the metric) are continuous but "pointy" (like the absolute value function), then first derivatives of g have step diiscontinuities (like the Heaviside step function), and second derivatives of g (used in the curvature tensor) are like Dirac delta functions. If a curvature singularity blows up like a Dirac delta function, then integration produces only a finite contribution to the tidal deformation of an object, which, if the object is robust enough, it can withstand.

George, sorry for the very late reply (four months later), but yes I was referring to the probability of traveling through the singularity, not the event horizon. The interesting study is where the traveler ends up upon entering the singularity. My apologies for not being able to contribute or explain myself quantitatively. It has been years since I have been active in Physics.

 

[Low-Q]

I have learned that if a spaceship approach a black hole, the gravity that is stronger at the front end than the rear end will stretch the spaceship into pieces. Is this theory taking into account that the stronger gravity at the front end also slows down time more than the less gravity at the space ships rear end? Will the spaceship actually "feel" that it is stretched into pieces? Will it be stretched at all - relatively speaking? And will it ever reach the event horizon as the time finally stops (from what I have learned about time and gravity)?

Vidar

 

[Algr]

As you pass the event horizon there are still a wide variety of speeds that an object might be moving. A particle doesn't need to move away from the black hole in order to interact with other particles, you can have a state where some particles move faster than others.

It's rather like the path of the moon, relative to the sun. When viewed from the Earth, the Moon seems to move in an ellipse. But from the sun's POV, the moon never moves in a retrograde direction. (As moons of the outer planets do)

 

[Nugatory]

I have learned that if a spaceship approach a black hole, the gravity that is stronger at the front end than the rear end will stretch the spaceship into pieces. Is this theory taking into account that the stronger gravity at the front end also slows down time more than the less gravity at the space ships rear end? Will the spaceship actually "feel" that it is stretched into pieces? Will it be stretched at all - relatively speaking?

yes, yes, and yes - if the tidal forces are strong enough.
Somewhat counterintuitively, the tidal forces are less intense for larger black holes; we're looking at the difference between the force at the nose and the tail, not the absolute strength of the force. For a sufficiently large black hole, not only will you not be stretched, you may not even notice when you pass through the event horizon.

And will it ever reach the event horizon as the time finally stops (from what I have learned about time and gravity)?

Time doesn't stop, or even slow down, for the observer free-falling into the black hole. An observer hovering far away from the black hole may never see the free-faller reach the event horizon, but plenty of other observers will.