Black Hole thought experiment

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barrykirk

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Let us take a Supermassive Black Hole with a very large Event horizon.<br /><br />For such a Black Hole, the tidal forces exerted outside of the event horizon would be fairly small.<br /><br />This is because of the following.<br /><br />Gravity falls off as 1/R^2 <br /><br />but Tidal forces fall off as 1/R^3 <br /><br />This means that for two objects with the same mass but different densities, the tides experienced on the surface are much greater for the denser object.<br /><br />The mean density inside the Schwarzschild radius decreases as the mass of the black hole increases, so while an earth-mass black hole would have a density of 2 × 1030 kg/m3, a supermassive black hole of 109 solar masses has a density of around 20 kg/m3, less than water.<br /><br />This also means that the more massive the black hole, the smaller the tidal forces for an object orbiting just outside the Schwarzschild radius.<br /><br />Let's take an arbitrarily large black hole such that an object in orbit outside of the event horizon is experiencing small tidal forces.<br /><br />From a spacecraft orbiting just outside the event horizon, if one were to lower an object using a cable from this orbit through the event horizon, what would happen as the object passed through the event horizon.<br /><br />Just before the event horizon, the forces on the object being lowered down would have a small and smooth function. Is their a discontinuity in the forces as the object travels through the event horizon?
 
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billslugg

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I don't believe there is a discontinuity in the forces affecting the object lowered through the Schwarzschild radius. Indeed, it could be pulled out again, thus rescuing someone trapped in the black hole. How can this be? Does it not violate the rule that nothing can leave a Black Hole? Not if you consider that the object, the cable and the spacecraft together are considered as one object. As long as any part of the object is outside, then it can rescue the rest.<br /><br />Then suppose you have a spacecraft that flies out from the center of the black hole to just shy of the Schwarzschild radius, then extends a probe outside the radius. The probe contains rockets that then rescue the spacecraft. Is this possible? <div class="Discussion_UserSignature"> <p> </p><p> </p> </div>
 
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yevaud

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Barry: no, no discontinuity. Remember, the Event Horizon isn't a physical barrier - it's merely that point where light itself cannot escape from the influence of the Singularity.<br /><br />Bill: no, it wouldn't work, as anything beneath the Event Horizon is forever trapped there. Even trying to raise that small extension through the EH would require a velocity in excess of "C." <div class="Discussion_UserSignature"> <p><em>Differential Diagnosis:  </em>"<strong><em>I am both amused and annoyed that you think I should be less stubborn than you are</em></strong>."<br /> </p> </div>
 
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barrykirk

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OK,<br /><br />I'm in orbit a few feet above the event horizon.<br /><br />I lower a flashlight to within an inch above the event horizon. Since the tidal forces are so low, the red-shift from the flashlight to me should be fairly low also?<br /><br />Now I lower the flashlight two more inches.... The red shift goes from almost zero to total? What am I missing?<br /><br />Yes, I agree the light from under the event horizon can't escape to inifinity, but I'm not understanding this situation.<br /><br />Unless, the space-time at that orbital point is really weird.
 
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yevaud

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Well, no. Tidal forces in this case are the gravitational gradient from weaker to stronger gravitational force as you approach the Event Horizon. In the case of a small BH, the gradient is quite steep, meaning that gravity increases hugely in a very short distance. That's what's meant by "Spaghettifaction" - the force on your feet is much stronger than your head. You are literally stretched out like taffy.<br /><br />With a larger mass BH, the gradient isn't so steep. But the Event Horizon is still the Event Horizon. So the light from that flashlight would be red-shifted very low indeed.<br /><br />Does that help? <div class="Discussion_UserSignature"> <p><em>Differential Diagnosis:  </em>"<strong><em>I am both amused and annoyed that you think I should be less stubborn than you are</em></strong>."<br /> </p> </div>
 
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Saiph

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yevauds got it. The light will be highly redshifted, but because of the low tidal forces, your flashlight will still be intact. If you put it over a stellar black hole, you're flashlight would be ripped to shreds, and it would still be highly redshifted. <br /><br />I'm not sure if it's the exact same redshift amount off hand, I'll stew on it, and get back to you. <div class="Discussion_UserSignature"> <p align="center"><font color="#c0c0c0"><br /></font></p><p align="center"><font color="#999999"><em><font size="1">--------</font></em></font><font color="#999999"><em><font size="1">--------</font></em></font><font color="#999999"><em><font size="1">----</font></em></font><font color="#666699">SaiphMOD@gmail.com </font><font color="#999999"><em><font size="1">-------------------</font></em></font></p><p><font color="#999999"><em><font size="1">"This is my Timey Wimey Detector.  Goes "bing" when there's stuff.  It also fries eggs at 30 paces, wether you want it to or not actually.  I've learned to stay away from hens: It's not pretty when they blow" -- </font></em></font><font size="1" color="#999999">The Tenth Doctor, "Blink"</font></p> </div>
 
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nexium

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I agree with barrykirk. The flash light would be infinately redshifted with respect to a distant observer, but hardly any with respect to the spacecraft a few inches (or miles) away, as the beam is passing though only a tiny percentage of the gravity well of the supermassive black hole. The space craft could easily retreve the flashlight, and the Spacecraft could escape the black hole, if it had sufficiently energetic fuel.<br />In fact, I'm of the opinion that the space craft can fly a few feet inside the super massive black hole and escape, if it has sufficient fuel, without help from outside the event horizon. Light cannot escape to a distant observer. A ballistic projectal can escape, but it must return after it reaches the top of it's trajectory. The space craft however can be powered continiously or be multistaged, either of which gives it a big advantage over a light beam, I think. This assumes the super massive black hole has a puny accreation disk, and managable tidal gradient near the event horizon. Neil
 
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Saiph

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<blockquote><font class="small">In reply to:</font><hr /><p>I agree with barrykirk. The flash light would be infinately redshifted with respect to a distant observer, but hardly any with respect to the spacecraft a few feet away, as the beam is passing though only a tiny percentage of the gravity well of the supermassive black hole. The space craft could easily retreve the flashlight, and the Spacecraft could escape the black hole, is it had sufficiently energetic fuel. <p><hr /></p></p></blockquote><br /><br />So far, so good.<br /><br /><blockquote><font class="small">In reply to:</font><hr /><p>In fact, I'm of the opinion that the space craft can fly a few feet inside the super massive black hole and escape, if it has sufficient fuel, without help from outside the event horizon.<p><hr /></p></p></blockquote> <br /><br />Unfortunately, there's a problem here. Inside the event horizon, there is <i>no</i> path that leads out. The <i>only</i> way to get out, is to go faster than C, but then you're not actually using the sheer speed through space to get out, but the entire "going backwards in time" thing (think tachyons here) to do so. <div class="Discussion_UserSignature"> <p align="center"><font color="#c0c0c0"><br /></font></p><p align="center"><font color="#999999"><em><font size="1">--------</font></em></font><font color="#999999"><em><font size="1">--------</font></em></font><font color="#999999"><em><font size="1">----</font></em></font><font color="#666699">SaiphMOD@gmail.com </font><font color="#999999"><em><font size="1">-------------------</font></em></font></p><p><font color="#999999"><em><font size="1">"This is my Timey Wimey Detector.  Goes "bing" when there's stuff.  It also fries eggs at 30 paces, wether you want it to or not actually.  I've learned to stay away from hens: It's not pretty when they blow" -- </font></em></font><font size="1" color="#999999">The Tenth Doctor, "Blink"</font></p> </div>
 
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doubletruncation

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<blockquote><font class="small">In reply to:</font><hr /><p>but Tidal forces fall off as 1/R^3<p><hr /></p></p></blockquote><br /><br />I've seen this calculated assuming you are on a freely falling radial trajectory into the BH. Does anyone know if this applies if you were actually on an orbit around the black hole as well? While I think that intution from cartesian space suggests that it should, I don't really trust my intuition in strong gravity environments. <div class="Discussion_UserSignature"> </div>
 
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barrykirk

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Again, from the Newtonian point of view, you should be able to pull the flashlight back into the spacecraft, even if it goes below the event horizon.<br /><br />Einstein says different....<br /><br />Obviously, the fabric of space-time near the event horizon is highly warped, and my intuition for this location is probably way off.
 
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doubletruncation

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I've been thinking about this a little bit, and I think what would happen is as follows:<br /><br />The 1/R^3 law basically gives a rule for how much particles will tend to drift apart as they fall into the singularity - so if you were to fall into a supermassive black hole you might not stretch apart too much as you cross the event horizon, but as you actually fall into the singularity you'd be ripped apart. Now say you actually wanted to keep from falling into the black hole, you can calculate the amount of energy at any given radius and intial velocity that you'd need to go from falling across the event horizon to not falling across it. That energy asymptotes to infinity at the event horizon regardless of your velocity. So if your ship were falling into the black hole with the probe hanging out in front you wouldn't be tidally stretched too much, but when you put on the breaks you would essentially feel a much stronger tidal force, and if your probe had already crossed the event horizon it would be ripped off (assuming that you were actually able to break). So I think hovering outside the black hole with your thrusters on full, you will feel a much larger tidal force (or gradient in the amount of force necessary to keep each bit of you from falling into the BH) than you would feel if you were falling in. I think a similar thing must apply if you were orbiting the black hole rather than being stationary with your thrusters on. If you release the probe you would need to apply an energy that asymptotes to infinity as it approaches the event horizon to keep it moving around the BH at your orbital velocity and not crossing the event horizon. <div class="Discussion_UserSignature"> </div>
 
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barrykirk

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Actually the 1/R^3 applies to actual forces applied, not<br />the drifting apart is due to the accelerations caused by<br />those forces.<br /><br />Please note that the 1/R^3 law is a Newtonian law.<br />It does NOT take General Relativity into account.
 
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doubletruncation

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A similar 1/R^3 law can actually be derived in GR as well. In that case, you use the equation of geodesic deviation to calculate the acceleration of the space-like vector connecting two particles in free fall into the BH. You find that the acceleration of the radial component of this vector goes as 2*m*n/r^3 where n is the radial component of the vector and m is the mass in units where G=c=1. Similarly you'll see that there is an acceleration in the tangential directions that tends to shorten the length of the tangetial component of the connecting vector - so you're stretched radially and squeezed tangentially. (See for example, chapter 16 of D'Inverno). So I was wrong to say it is the amount by which they drift apart, it's really the acceleration in their drift apart and you have to apply a force by that amount to keep the particles together. But it does work out in GR as well as in Newtonian physics. <div class="Discussion_UserSignature"> </div>
 
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barrykirk

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Thanks for that info.... But if that's the case,<br />that the particles have a small seperation force,<br />then why does light suddenly "switch off" at the event horizon?
 
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yevaud

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Because that's the point where the velocity limit of C is achieved. <div class="Discussion_UserSignature"> <p><em>Differential Diagnosis:  </em>"<strong><em>I am both amused and annoyed that you think I should be less stubborn than you are</em></strong>."<br /> </p> </div>
 
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doubletruncation

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There are some neat animations showing what you would see as you fell across the event horizon at:<br />http://casa.colorado.edu/~ajsh/schw.shtml<br /><br />In particular see question 3 of the following quiz:<br />http://casa.colorado.edu/~ajsh/quiz.html#quiz<br /><br />The question is basically the one that you asked (what would happen if you were sitting stationary outside the event horizon and dropped a fishing line in). The answer to the question (given at http://casa.colorado.edu/~ajsh/singularity.html ) is:<br />"To remain at rest just above the horizon, you would have to accelerate like crazy just to stay put. Radial distances measured in your frame are greatly stretched compared to radial distances measured by observers who go with the flow, so you'd have to pay out your fishing line a long way before it hit the horizon. The acceleration goes to infinity for something that attempts to remain at rest at the horizon, so the fishing line that passed the horizon would either break, or drag you in. "<br /> <div class="Discussion_UserSignature"> </div>
 
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barrykirk

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Radial Distances are stretched....<br /><br />And that is the part that I missed. I was thinking in Newtonian terms.<br /><br />I'm very familiar with Special Relativity and I understand<br />some of the concepts of General Relativity, but I don't<br />have the math for GR.<br /><br />I knew that time gets stretched in a gravity field, but I<br />didn't know, but should have known that space gets<br />stretched too.<br /><br />That is the missing piece of the puzzle.<br /><br />So correct me if I'm wrong, but.<br /><br />From the point of view of a distant observer the tidal<br />forces produce a minimal force, but from the point of<br />view of the observer in orbit just above the event<br />horizon. Space is stretched radially.<br /><br />Is this correct, or did I miss something?
 
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doubletruncation

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<blockquote><font class="small">In reply to:</font><hr /><p>From the point of view of a distant observer the tidal<br />forces produce a minimal force, but from the point of<br />view of the observer in orbit just above the event<br />horizon. Space is stretched radially.<br /><br />Is this correct, or did I miss something? <p><hr /></p></p></blockquote><br /><br />I think so, also if you were freely falling into the black hole (i.e. not accelerating, or I guess orbiting) then radial distances are not stretched so much either. <div class="Discussion_UserSignature"> </div>
 
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skeptic

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> Space is stretched radially. <br /><br />I may be wrong but thought that space was shortened in the radial direction close to a BH the same way it is shortened in the direction of motion of accelerated objects. At the event horizon, isn't space "compressed" into zero radial distance. How would this affect the gravitational gradient?
 
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border_ruffian

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if you were falling into a black hole you get spaghettified.<br /><br />could somebody explain this to me.... doesn't the light from the flashlight being pointed INTO the black hole get blueshifted instead of redshifted because of the pull of gravity? .....<br /><br />as oppose to....<br /><br />somebody at the event horizon pointing the flashlight OUT..... then the light gets redshifted as it tries to escape the pull of gravity??????<br /><br />
 
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doubletruncation

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<blockquote><font class="small">In reply to:</font><hr /><p>if you were falling into a black hole you get spaghettified.<br />could somebody explain this to me.... doesn't the light from the flashlight being pointed INTO the black hole get blueshifted instead of redshifted because of the pull of gravity? .....<br />as oppose to....<br />somebody at the event horizon pointing the flashlight OUT..... then the light gets redshifted as it tries to escape the pull of gravity??????<p><hr /></p></p></blockquote><br /><br />When talking about general relativity you have to be very careful about reference frames, even more so than in special relativity because there is no such thing as a global Lorentz frame in general relativity. So when someone says that you're spaghettified what they mean is that if every part of you is freely falling into the black hole, then your head, which let's assume is closer to the black hole than your feet, will appear to accelerate away from your feet and towards the black hole from your feet's point of view. As a result, if you didn't do anything about it, you would be stretched out thin like a spaghetti noodle. Keep in mind that when your head and feet are both freely falling neither of them are locally in accelerating frames (that is, special relativity applies to them locally), in GR gravity is not a force and if you are freely falling then you are not accelerating even though from your point of view another freely falling object will be accelerating... hence space is curved. To avoid stretching like that you need to supply a force to accelerate your head towards your feet, and the force that you need approaches infinity as you approach the black hole. Now let's think about the blue-shift/red-shift issue. First of all, this is usually depicted assuming one observer is very far away from the black hole, and the other observer is nearer to the black hole and at rest with respect to the distant observer. Now here's a thing to keep in mind, if you are stayin <div class="Discussion_UserSignature"> </div>
 
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doubletruncation

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<blockquote><font class="small">In reply to:</font><hr /><p>I may be wrong but thought that space was shortened in the radial direction close to a BH the same way it is shortened in the direction of motion of accelerated objects. At the event horizon, isn't space "compressed" into zero radial distance. How would this affect the gravitational gradient?<p><hr /></p></p></blockquote><br /><br />Again you have to be careful about whose point of view you're talking about. From the point of view of a distant observer far from the black hole space is compressed into zero radial distance at the event horizon. Now suppose you're at rest with respect to the black hole but near the event horizon - you are now in an accelerating frame, you will still see that objects become inifinitely compressed at the event horizon but the distance that you think have to go to the event horizon becomes longer. <div class="Discussion_UserSignature"> </div>
 
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border_ruffian

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yes i understand it's relative to whose point of view we're talking about.... but i want to understand this....<br /><br />*observer A at the event horizon holding a flashlight out into space....<br /><br />*observer B far away from the black hole pointing another flashlight in the general direction of A...<br /><br />B will see the light from A redshifted, correct?<br /><br />and<br /><br />A will see the light from B blueshifted as it gets pulled by gravity?<br /><br />isn't this what the blue sheets (blue wall) at the event horizon all about???<br /><br />
 
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skeptic

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Thanks. You correctly interrpreted what I was trying to say but put it into better words. I'm still curious as to how the coontracted space close to the event horizon affects the gravitaional gradient as seen by a distant observer. Surely the gradient would appear to be much steeper than 1/r^2 close to the EH and would affect the apparent (to a distant observer) orbits of objects around the BH.
 
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doubletruncation

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<blockquote><font class="small">In reply to:</font><hr /><p>*observer A at the event horizon holding a flashlight out into space.... <p><hr /></p></p></blockquote><br /><br />Observer A, just outside the event horizon, would need to be **at rest** with respect to the black hole (and hence accelerating against the pull of the black hole) to see Observer B blueshifted, if he were falling he would not see an infinite blueshift as he falls across the event horizon. In fact, he could see Observer B to be redshifted rather than blueshifted. Regardless of whether Observer A is falling or at rest, Observer B will see him redshifted. Keep in mind that unless you are actually going the speed of light you could not be at rest with respect to the black hole when you are at the event horizon, and once you are inside the event horizon even light cannot avoid falling into the singularity. Even if you are at rest with respect to the black hole (and see the observer at infinity blueshifted) you will still experience a very strong tidal force which would tend to rip you apart (I think the tidal force and redshift/blueshift issues are not really related here). Say your front side is facing away from the black hole, then you will need to put a much larger force on your back side then you would have to put on your front side to keep all of you from falling into the black hole. If you aren't able to supply the correct forces you will be stretched apart. Say you can just supply the right amount of force to keep your front side from falling into the black hole but not quite enough for your backside, then your backside will start accelerating toward the black hole while your front side stays at a fixed distance from it. The gradient in the force that you would need will go to infinity as you actually cross the event horizon, this is just another way of saying that you can't actually keep yourself at rest with respect to the black hole when you're at the event horizon. If you don't try to stay at rest <div class="Discussion_UserSignature"> </div>
 
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