What happens when you enter a black hole

[GoodPR]

According to some members on this site and other facts I've found, it would be possible to travel towards Earth with such a velocity that Earth would perceive the traveller as a black hole.
Inversely at that same speed, the travelleler would see Earth as the black hole.

What would happen when the traveller entered the black hole.

My guess, is that space has a limit to how much mass it can contain. The size of the black hole compared to Earth, will be just large enough to contain all the relative mass of Earth without breaking that limit.
In other words, its center of gravity will be larger than Earth itself.
I believe what would happen is that the traveller would think he's going straight in, and would end up on the other side of the black hole. This seems to violate less things, because if he could crash into the planet, well, idunno.

 

[Pengwuino]

According to some members on this site and other facts I've found, it would be possible to travel towards Earth with such a velocity that Earth would perceive the traveller as a black hole.
Inversely at that same speed, the travelleler would see Earth as the black hole.

What? Who said that.

 

[vin300]

When you enter a black hole, you lose all hopes of exiting..

 

[ZikZak]

It's either a black hole or it isn't. Either light can escape from it or it can't. If you're not a black hole in your own reference frame, no one can "perceive" you to be one in theirs.

 

[fleem]

According to some members on this site and other facts I've found, it would be possible to travel towards Earth with such a velocity that Earth would perceive the traveller as a black hole.

They were wrong. It is the rest mass of a black hole (or of anything, for that matter) that gives it its gravitational field. Note that the location of kinetic energy is relative to the reference frame of the observer. For example, would the abovementioned traveller also see the Earth as a black hole but no change to himself? Which thing is a black hole and which thing isn't, is not a relative concept. As ZikZak said, you're either a black hole or you aren't. In this case, neither the Earth nor the traveler are black holes. However, if there is kinetic energy contained in a system, then that relativistic mass becomes rest mass, and there is a gravitational field gained from it. So if that traveler, when he hit the Earth, were trapped in a box, then the Earth-traveler system would become a black hole if there had been enough energy to begin with.

 

[Bob_for_short]

When you approach a black hole, you get burned by the Hawking's radiation.
Whent you get inside, you are burned by the light that could not escape from it.

 

[George Jones]

When you approach a black hole, you get burned by the Hawking's radiation.

No. According to the book Quantum Fields in Curved Space by Birrell and Davies, pages 268-269,

These consideration resolve an apparent paradox concerning the Hawking effect. The proper time for a freely-falling observer to reach the event horizon is finite, yet the free-fall time as measured at infinity is infinite. Ignoring back-reaction, the black hole will emit an infinite amount of radiation during the time that the falling observer is seen, from a distance to reach the event horizon. Hence it would appear that, in the falling frame, the observer should encounter an infinite amount of radiation in a finite time, and so be destroyed. On the other hand, the event horizon is a global construct, and has no local significance, so it is absurd to0 conclude that it acts as physical barrier to the falling observer.

The paradox is resolved when a careful distinction is made between particle number and energy density. When the observer approaches the horizon, the notion of a well-defined particle number loses its meaning at the wavelengths of interest in the Hawking radiation; the observer is 'inside' the particles. We need not, therefore, worry about the observer encountering an infinite number of particles. On the other hand, energy does have a local significance. In this case, however, although the Hawking flux does diverge as the horizon is approached, so does the static vacuum polarization, and the latter is negative. The falling observer cannot distinguish operationally between the energy flux due to oncoming Hawking radiation and that due to the fact that he is sweeping through the cloud of vacuumm polarization. The net result is to cancel the divergence on the event horizon, and yield a finite result, ..

This finite amount is negligible for observers freely falling into a black hole.

Whent you get inside, you are burned by the light that could not escape from it.

I assume that you're referring to the infinitely blue-shifted radiation at the Cauchy horizon inside a rotating black (or charged) hole, which creates a weak (not crushing) curvature singularity at the Cauchy horizon. Some researchers believe that this is a non-destructive singularity.

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. See, for example,

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

particularly pages 15 (starting at "Consequence to the curvature singularity at the IH: (IH = Inner Horizon)), 16, 19 (for physical implications of the weak singularity), and 24.

 

[chi22]

What would happen when the traveller entered the black hole.

I know this might be a rough explanation... but if your trying to get into a BH, you'll need to get pass the gravitational gaps and the huge amounts of disintegrating space junk traveling close to c first

 

[GoodPR]

Well, thanks you guys, it seems from your counter explanations, that what I have in mind may be possible. As vin said, you cannot escape a black hole, and as Zikzak said, you would be looking at yourself from your own reference frame and therefor only have rest mass.
So, although Earth see's the traveller as a black hole, the traveller sees the Earth as a black hole.
Remember that this mass is only relativistic and just like the light that is blue shifted in front and red shifted behind ,the mass and therefor gravitational field would be the same.

As the traveller began passing through the black hole, he would look behind him and not see a black hole, because relativisticly his "gravity" has redshifted behind him and would not see a black hole. In that possible example, exiting the other side of this black hole would be easy, because it doesn't exist from the other direction.

This would be more like a wormhole than a conventional black hole I suppose, and although it may be pointless to argue at this point, I think because of the relativistic gravity, relative faster than light speed could be established in this manner.

 

[Dale]

it would be possible to travel towards Earth with such a velocity that Earth would perceive the traveller as a black hole.
Inversely at that same speed, the travelleler would see Earth as the black hole.

This is not correct. See http://math.ucr.edu/home/baez/physics/Relativity/BlackHoles/black_fast.html

 

[GoodPR]

That web page does not seem like it contradicts this argument.
It uses the fact that a black hole would not form around its rest frame as evidence it would not form in any other frame
"since its mass and volume haven't changed in its rest frame, it should not form a black hole in that frame--and therefore not in any other frame either"
Just because it doesn't exist from its rest frame doesn't mean it doesn't exist in another frame.
Clearly from Earths frame, light is different, blue shifted, in accordance with its velocities. So what your visually experiencing when on Earth is definitely changing.

What types of particles give off Gamma Radiation? Do you think that if you could visually see the actual particles they would appear normal? Just a normal heated element in a flash light, giving off Gamma Radiation?, probably not. If you could actually see the element it would probably look so condensed, so dense that a gamma ray would be expected from the object.

It would take very powerful releases of energy to create Gamma Rays, usually coming from very dense sources, our perception of something relies soley on the Energy that it emits or reflects. Since we know that our perception of this rocket will change based on its velocity, why can't it be a black hole, in realitly all that would change would be how it looks.

 

[Dale]

You missed the critical point:

"this is based on a particular static solution to the Einstein field equations of general relativity, and ignores momentum and angular momentum as well as the dynamics of spacetime itself. In general relativity, gravity does not only couple to mass as it does in the Newtonian theory of gravity. Gravity also couples to momentum and momentum flow; the gravitational field is even coupled to itself".

The source of gravity in GR is not mass, but the stress-energy tensor. When you have a moving mass then the stress energy tensor not only has a relativistic mass term but also has momentum and momentum flow terms. These violate the assumptions of the Schwarzschild solution and prevent an event horizon from forming.

Again, you do not turn into a black hole by going fast, nor does something else turn into a black hole by going fast in your frame. If there is no event horizon in the rest frame then no event horizon forms in any other frame. It is always possible to send light signals and they always have finite Doppler shift.

 

[GoodPR]

Ok, So, I agreed with you, for the sake of the argument that gravity is not affected by your relative mass.

However, that does not mean it will not necessarily form a black hole, because, that's assuming that gravity is a necessary component of a black hole. That may not be true, nothing would be violated necessarily if a visual black hole was created with the same rest mass, or gravity it always had.
Light would still reach you as it always would, with it's finite doppler shift, it just will seem to come from around the event horizon instead of directly from the flashlight.

This seems like it is possible without violating any laws does it not?
The question remains however, would the "visual" black hole be represented according to relative mass? Or would the size of the black hole be the size of the rockets matter compressed to shwartzwald density? I think it would make more sense if the second were true.

Also, I guess what I'm saying essentially, the rocket as it approaches definitely appears condensed, normally we think of this as being a 1 dimensional compression of depth, maybe its a 2 dimensional compression of height and width (if the rocket is heading directly towards you) and a 1 dimensional expansion of depth. Which, if were true would mean you could travel fast enough that you could go straight through a planet without hitting a particle (if you were going fast enough).

 

[Dale]

that does not mean it will not necessarily form a black hole ... Light would still reach you as it always would, with it's finite doppler shift

Then it is not a black hole. End of story.

 

[GoodPR]

Then it is not a black hole. End of story.

Wow, you paraphrased me so badly you changed the meaning of what I said.
Lets see if I can paraphrase a different part of it and get a different answer

Quote myself " Light would still reach you ...it just will ...come from around the event horizon instead of directly from the...rocket"

It must be a black hole, end of story.
Hmm now it's different

 

[Dale]

In your frame you can send a light signal to the earth, therefore in the Earth's frame you can also. In the Earth's frame the Earth can send a light signal to you, therefore in your frame the Earth can also. There is no event horizon for the light to go around. No event horizon in either frame so no black hole in either frame.

Let's see how you randomly assert that your idea is still tenable

PS I avoided the "around the event horizon" part because there is no event horizon as the other part indicated. Your post was self-contradictory.

 

[Naty1]

...thats assuming that gravity is a necessary component of a black hole. That may not be true,

gravity is what causes a black hole to form. Dalespam stated the situation as clearly as possible...

 

[jtbell]

I know this might be a rough explanation... but if your trying to get into a BH, you'll need to get pass the gravitational gaps and the huge amounts of disintegrating space junk traveling close to c first

Gravitational gaps?

 

[chi22]

Gravitational gaps?

As your traveling further and further into the gravitational well, there must be huge gravitational gaps among each millimeter taken, especially as your getting closer to the event horizon

 

[GoodPR]

In your frame you can send a light signal to the earth, therefore in the Earth's frame you can also. .

Not true, your comparing two different frames.
From the rockets frame it can send light to Earth, and from Earths frame it can send light to the rocket. That is what you meant, that is the comparable frame. What your saying is like comparing one time frame with a distance frame from a different location.

In the Earth's frame the Earth can send a light signal to you, therefore in your frame the Earth can also.

More of the same.

PS I avoided the "around the event horizon" part because there is no event horizon as the other part indicated. Your post was self-contradictory.

Avoiding part of the argument may make it easier for you to feel right, that doesn't mean you are.
Obviously experimental data shows that the event horizon is comparable to the mass of the black hole only, but how many fast moving black holes has anyone witnessed? Would be hard to take notice of, considering its mass wouldn't change, only its size.

 

[Dale]

Not true, your comparing two different frames.
From the rockets frame it can send light to Earth, and from Earths frame it can send light to the rocket. That is what you meant, that is the comparable frame. What your saying is like comparing one time frame with a distance frame from a different location.

OK, here is a thought experiment for you to help you see how your idea is self-contradictory.

Dr. Doom is inertially flying by Earth at relativistic speeds (time dilation factor 1E6) in the SS Doomsday. In the ship's frame the ship is 1E3 Schwarzschild radii in size, and in the Earth's frame the Earth is 7E8 Schwarzschild radii in size. Thus, according to you, the ship is a black hole in the Earth's frame, but the Earth is not a black hole in the ship's frame. Dr. Doom sends a signal to his Doomsday Device (located on the Earth and equipped with a sufficiently broadband reciever) which will destroy earth. In the ship's frame the signal reaches the Earth and the Earth blows up. (According to you) in the Earth's frame the signal cannot cross the event horizon and so the Earth does not blow up. Your idea thus leads to a paradox.

A more succinct way to say this is that if there exists a null geodesic from A to B in one reference frame then there exists a null geodesic from A' to B' in any other reference frame. Again, the bottom line is that you do not become a black hole if you travel fast enough. Full stop. None of your justifications are sound.

 

[GoodPR]

Though I appreciate your well thought out experiment, in it's beginning you claimed I said something, which I did not.

I am not claiming the ship is a black hole in Earths frame but the Earth is not a black hole in the ships frame,
I am claiming the Earth is a black hole from the ships frame, and the ship is a black hole from the Earth's frame.

So in your example, the signal sent from the ship would make it as far as the event horizon as far as you could tell, and from Earth the signal would not escape the event horizon of the ship, atleast until the two collided.

 

[Dale]

I am not claiming the ship is a black hole in Earths frame but the Earth is not a black hole in the ships frame,
I am claiming the Earth is a black hole from the ships frame, and the ship is a black hole from the Earth's frame.

No, your claim is that relative velocity can cause an object to turn into a black hole in some frames due to the increase in relativistic mass. Since the relativistic mass increase is symmetric and since the formation of a black hole depends on density if two objects have different proper densities then it is always possible to formulate such a scenario. In any case, my more succinct phrasing (if there exists a null geodesic from A to B in one reference frame then there exists a null geodesic from A' to B' in any other reference frame) works across event horizons so it includes the case of equal proper density objects.

You still have nothing supporting your argument. If you wish to proceed further then simply solve the EFE for a fast moving point mass and show the formation of this supposed event horizon.

 

[ZikZak]

Well, thanks you guys, it seems from your counter explanations, that what I have in mind may be possible. As vin said, you cannot escape a black hole, and as Zikzak said, you would be looking at yourself from your own reference frame and therefor only have rest mass.
So, although Earth see's the traveller as a black hole, the traveller sees the Earth as a black hole.

Incorrect. No one sees anyone as a black hole. Black holes are regions of spacetime from which light cannot escape to arbitrary distances. I can send a light ray out from me to arbitrary distance, and no matter what frame you are in, that photon recedes an arbitrary distance from me. Therefore, since the photon recedes an arbitrary distance from in in all frames, I am not a black hole in any of them.

It makes no difference where the light flash "appears" to come from, and in any case, there is an event where the light flash is adjacent to my body that can be observed by anyone and everyone. Thus the light flash appears to come from me, it does in fact come from me, and does recede to arbitrary distance. I am not a black hole. Period.

Remember that this mass is only relativistic and just like the light that is blue shifted in front and red shifted behind ,the mass and therefor gravitational field would be the same.

Same as what? Mass does not redshift. It does not blueshift. Even the "relativistic mass" does not do these things. You seem to be confused by the concept of Relativistic Mass, which is a very confusing and therefore deprecated concept. You CANNOT just plug in the Relativistic Mass into, say, Newton's theory of gravity and expect get any kind of physical result. Newtonian Gravity is Wrong. Relativity is not GMM/r^2 with relativistic mass. In Relativity, the source of gravity is stress-energy, a tensor, and the "relativistic mass" doesn't enter into it. What goes into it for the purposes of this problem is mass and energy flux, which gravitate in different ways. Yes, as seen in a frame in which the body moves, its gravity is different than the frame in which it is at rest, but not simply stronger. The gravity induced by the kinetic energy is not purely attractive (in fact, it is mostly skew to the line of velocity, like magnetism) and NEVER, EVER generates black holes.

That web page does not seem like it contradicts this argument.
It uses the fact that a black hole would not form around its rest frame as evidence it would not form in any other frame...
Just because it doesn't exist from its rest frame doesn't mean it doesn't exist in another frame.

YES IT DOES. Because if a photon can escape from the body in its rest frame, then it is observed to escape in ANY frame. If a photon can escape, it is not a black hole.

 

[stevebd1]

According to some members on this site and other facts I've found, it would be possible to travel towards Earth with such a velocity that Earth would perceive the traveller as a black hole.

One way you might look at this is by starting with a simple form of the GR equation for gravity-

g=\rho+\frac{1}{c^2}(P_x+P_y+P_z)

where g is gravity, \rho is density and P is pressure.

Pressure can be described as confined kinetic energy (in fact, the units for P and KE can be described as synonymous) so P in the above equation can also loosely represent KE. If we apply the above to an object that is moving in the direction of x, the P_x would be positive, P_y would be negative (i.e. negative pressure behind the moving object) and P_z would be zero. This means while the object increases in relativistic mass, its gravity might remain unchanged.

While pressure and kinetic energy can only be described as synonymous when considering the non-relativistic equation for kinetic energy (E_k=\frac{1}{2} m v^2), the same might apply in principle at relativistic speeds.

 

[GoodPR]

ZikZak the statements your making assume that you know what happens when you flash a light while your in a black hole, I find that hard to believe.
Just because you see the light leaving your frame how can you say that another frame would see the same thing. You can't your just making predictions because you don't like this idea in general because it goes against your high school physics lessons. There is no math anywhere that states if you shine a light in your frame that it is visible in any other frame. Your just assuming that.

 

[bm0p700f]

I am no expert on GR. I do understand a enough of SR to be dangerous though. Please correct me.

If a mass (observer) is moving you get a relatalvistic mass increase which is easy to calculate. The length contracts as apparent to a stationary observer. I suppose to the moving observer the length in the direction of movement appears to be the same as it was when it was stationary.

So for a moving observer there is a density increase. Density and the gravitational field strength that results are linked. So a black hole could form in that respect.

However I am uncomforatbele with that because for the moving observer the length in the direction of motion remains unchanged so both observers would disagree on their calculations of their density. However density increases still by this logic.

Also with regards to the above equation;

<br /> g=\rho+\frac{1}{c^2}(P_x+P_y+P_z)<br /> &lt;br /&gt; &lt;br /&gt; &amp;quot;If we apply the above to an object that is moving in the direction of x, the Px would be positive, Py would be negative (i.e. negative pressure behind the moving object) and Pz would be zero.&amp;quot;&lt;br /&gt; &lt;br /&gt; Py surley is not behind Px, it should be a right angles to it so no negative pressure&lt;br /&gt; &lt;br /&gt; So if pressure is energy density (energy divided by volume) then if the momentum increases the relativistic energy increases E = sqrt(p^2c^2+m^2c^4) where m is rest mass. So the energy density surely rises. So g increases. If it increases enough a black could result. This I know is not right so what is this flaw in the way I am reading this.&lt;br /&gt; &lt;br /&gt; I can accept that no black hole forms as both observers can communicate but the my confused logic is not helping me. Can each of my confused ideas be put straight.

 

[Ich]

Just because you see the light leaving your frame how can you say that another frame would see the same thing.

Instead of accusing other people of narrow-mindedness, you better learn some basic concepts. Like: what is a reference frame.
As long as you have diffenent frames "seeing" different things, you're certainly not in a position to ramble about other people's high scool education.

So the energy density surely rises.

Energy density is one of ten components of the stress-energy tensor. The latter is the source of gravity, not energy density.

 

[bm0p700f]

I have just been thinking. If the relativistic mass is relative to the observer. The mass increases as measured by the stationary observer but for the moving observer his mass remains the same. This would stop a BH forming for the moving observer.

If the mass is the relativistic mass in dependent of the observer then for the moving observer he can continue to accelerate past c. I told you I enough about SR to be dangerous. I am getting a bit confused. I answer one question but another pops up.

Although the one regarding my flawed interpretation of g equation in GR needs addressing.

 

[stevebd1]

Py surley is not behind Px, it should be a right angles to it so no negative pressure

I made an error in my explanation. The above GR equation takes into account pressure pushing equally in all three spatial dimension, if we rewrite the equation to take into account the opposite directions (x_-, y_- and z_-), another way of viewing Einsteins equation would be-

g=\rho+\frac{1}{c^2}\left(\frac{(P_x+P_{x_-})+(P_y+P_{y_-})+(P_z+P_{z_-})}{2}\right)

While the above is a bit choppy, it still equals the original equation in post 25 but takes into account six directions instead of three. If an object was to travel in the direction of x, then x_- would be negative and y, y_- and z, z_- would remain zero.

This is talked about to some degree in the 'Gravity & Relativistic Mass' section in the Talk section of wikipedia for 'Mass in special relativity'-

http://en.wikipedia.org/wiki/Talk:Relativistic_mass#Gravity_.26_Relativistic_Mass

 

[Ich]

If the mass is the relativistic mass in dependent of the observer then for the moving observer he can continue to accelerate past c.

He can continue to accelerate. Of course, as he can always claim to be "at rest", so what should stop him? But his velocity relative to someone else will never be greater than c.

Although the one regarding my flawed interpretation of g equation in GR needs addressing.

I don't think it makes sense for you to delve into GR before getting the basics of SR.

 

[bm0p700f]

While I was in the shower I was thinking. A stationary observer (A) see a moving observer (B) increase in mass by m = gamma*rest mass. the moving observer disagrees, he (B) says he is stationary and A is moving. So each observer sees a mass increase, they just disagree who ate all the pies. In this interpreation the mass increase in relative and still mean a velocities must be less than c.

Also given stevebd1 GR equation, would a volume of gas in a closed box at the temperature T1 and pressure P1 see an increase in the gravitational field it creates when the pressure and temperature are increased to T2 and P2? I would say not as the temperature and pressure are directly related to the K.E of each particle. As already discussed increases in K.E do not increase it gravitational field strength. Is this right? Am I getting there, where ever there is?

 

[stevebd1]

As already discussed increases in K.E do not increase it gravitational field strength. Is this right? Am I getting there, where ever there is?

KE confined in a frame would contribute to the gravity of an object. For example, for a rapidly rotating neutron star, pressure and rotational kinetic energy (K_{rot}=\frac{1}{2}I \omega^2) would contribute, both have a place in the stress energy tensor, but if the KE is not confined (such as in an object moving in one direction) then it's contribution is negligable. As stated in the wiki talk section, there may be some kind of effect where the gravity field becomes uneven, maybe there's an increase in the field to the front and a decrease at the back (the integral of which shows no increase in g as a whole). Based on the gravity field being uneven, you can also consider the zeroth law of black hole thermodynamics that for a black hole to exist, the surface gravity (κ) must be constant over the black holes event horizon.

 

[ZikZak]

ZikZak the statements your making assume that you know what happens when you flash a light while your in a black hole, I find that hard to believe.

For any useful definition of the word "know," I absolutely do know what happens when I flash a light inside a black hole. The answer is given to me by the theory of General Relativity, the most well-supported theory in all of science.

Just because you see the light leaving your frame how can you say that another frame would see the same thing. You can't your just making predictions because you don't like this idea in general because it goes against your high school physics lessons. There is no math anywhere that states if you shine a light in your frame that it is visible in any other frame. Your just assuming that.

You misunderstand the role of math in science. There is nothing magical about math that makes things true. And in any case, there is certainly "math" that shows that no light flash in a black hole ever gets out. There is certainly "math" that shows that you do not become a black hole at high speeds. The "math" is called the Einstein Field Equations.

This is, of course, because the makers of General Relativity DID make an assumption. It's the radical, closed-minded assumption that there is an external reality on which all observers agree. In fact, ALL those crazy, closed-minded scientists seem to assume this. What the hell is up with that?? But the result of the assumption of external reality is: that if the photon gets out of the black hole, then it gets out of the black hole. It makes no difference what reference frame you are in.

 

[Dmitry67]

But the result of the assumption of external reality is: that if the photon gets out of the black hole, then it gets out of the black hole. It makes no difference what reference frame you are in.

Different ACCELERATED frames do not agree on the number of real (non-virtual) particles.
Virtual particle in one frame can become a real one in another accelerated frame and vs
AFAIK the Hawking radiation is based on that fact.

 

[Dale]

So for a moving observer there is a density increase. Density and the gravitational field strength that results are linked. So a black hole could form in that respect.

A http://math.ucr.edu/home/baez/physics/Relativity/BlackHoles/black_fast.html" . GoodPR has yet to provide any evidence, logic, or math in support of his position, instead relying entirely on stubborn determination to ignore and misunderstand everything anyone else says. It is simply wrong, it has been refuted by reputable sources, by theoretical considerations, and by experimental observation (high energy cosmic rays do not form black holes).

 

[stevebd1]

I just want to add to post #33 the following extracted from wiki talk-

'The only time kinetic energy shows up as increased gravitational field is when it's the kinetic energy in a SYSTEM where the sum of momenta is zero. Otherwise, kinetic energy shows up.. ..as increased momentum, but not as gravity.'

which I think shed more light on the subject of KE as gravity or momentum.

While it's already been proved that objects at high speed increase in mass/energy (as in the LHC, the sum mass of the particles produced is approx. equal to the KE + rest mass of the particles collided), it would be interesting to know how the gravitational field of these objects is affected at high speed.

 

[stevebd1]

Some interesting recent threads on kinetic energy and gravity (all Jan 09)-

https://www.physicsforums.com/showthread.php?t=288327"

https://www.physicsforums.com/showthread.php?t=285014"

https://www.physicsforums.com/showthread.php?t=287998"


Older posts-

https://www.physicsforums.com/showthread.php?t=87991" (Sept 05)

https://www.physicsforums.com/showthread.php?t=83157" (July 05)

https://www.physicsforums.com/showthread.php?t=57680" (Dec 04)


Related-

https://www.physicsforums.com/showthread.php?t=300630" (Mar 09)

https://www.physicsforums.com/showthread.php?t=288819" (Jan 09)

https://www.physicsforums.com/showthread.php?t=252623" (Aug 08)

https://www.physicsforums.com/showthread.php?t=82546" (July 05)

 

[Naty1]

Different ACCELERATED frames do not agree on the number of real (non-virtual) particles. Virtual particle in one frame can become a real one in another accelerated frame and vs AFAIK the Hawking radiation is based on that fact.

Virtual particles are not observeable to a stationary observer. A free falling observer will pass the horizon of a black hole as virtual particles invisibly flash in and out of momentary existence harmlessly around him; a stationary observer outside the black hole horizon will be fried by those former virtual particles that now appear as REAL particles exhibiting thermal energy.

 

[dream431ca]

When you enter a black hole, you get ripped apart by the shear force of the gravitational field. Also, black holes could be considered not part of the rest of the universe because space-time is altered so much that time literary stops theoretically.

Here is an interesting point. Let's say you are just outside the black hole, before the event horizon so you don't get pulled in and you throw a rock into the black hole. What you will see is the rock heading towards the black hole then eventually slow down and come to a stop. You will see the rock just "floating" in the black hole. Because space-time is skewed so much, you will see the rock frozen in time while the rock is actually ripping apart in the black hole. Good old theory of relativity.

 

[Dmitry67]

When you enter a black hole, you get ripped apart by the shear force of the gravitational field. Also, black holes could be considered not part of the rest of the universe because space-time is altered so much that time literary stops theoretically.

What a nice collection of misconseptions about black holes!
Tidal forces are not infinite at the horizon. For very heavy black holes these forces can be quite small so humans can survive
Also, GR insists that BH *is* a part of our space time (and for that very re4ason it can calculate what is inside)
"times stops" is WRONG. I recommend checking space time diagrams to really understand what is inside.