# Crossing the Event Horizon

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#### skeptic

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<p class="MsoNormal"><span style="font-size:10pt;font-family:Arial"><span>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; </span>This is a discussion about possibility or impossibility of crossing the event horizon of a black hole and I&rsquo;m specifically soliciting the input of the moderators.<span>&nbsp; </span>To begin the discussion I reference the third paragraph of page 218 of Kip Thorn's Black Holes and Time Warps where, speaking of the implosion of a star into a black hole, he writes "...the implosion freezes forever as measured in the static, external frame but continues rapidly on past the freezing point as measured in the frame of the star's surface..."<span>&nbsp; </span>This statement is consistent with both the Lorentz Transformation and relativity.<span>&nbsp; </span></span></p> <p class="MsoNormal"><span style="font-size:10pt;font-family:Arial"><span>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; </span>The Lorentz Transformation describes how much space is contracted and time dilated by the velocity of a moving object according to the formula 1/sqrt(1-v^2/c^2).<span>&nbsp; </span>In a gravitational field the v term is replaced by escape velocity of the gravitational field.<span>&nbsp; </span>This explains why at the event horizon length has contracted to zero and time has stopped. A star and everything else must be frozen at the event horizon forever.<span>&nbsp; </span>The falling object however, doesn&rsquo;t experience this transformation but continues falling through the event horizon.<span>&nbsp; </span>How can the perception from the perspective of the infalling object be consistent with that of the external observer?</span></p> <p class="MsoNormal"><span style="font-size:10pt;font-family:Arial"><span>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; </span>As Woody Allen said &ldquo;Eternity is a long time, especially towards the end.&rdquo;, and being frozen at the event horizon forever is a long, long time, perhaps long enough for the black hole to evaporate by Hawking radiation.<span>&nbsp; </span>If this is the case, might the infalling object never cross the event horizon but see the event horizon shrink to nothing beneath it as it falls?</span></p> <p class="MsoNormal"><span style="font-size:10pt;font-family:Arial"><span>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; </span>It has been suggested that one means of crossing the event horizon is that objects don&rsquo;t have to fall through the horizon, the horizon will grow past the object as more matter falls into it. <span>&nbsp;</span>This doesn&rsquo;t really work.<span>&nbsp; </span>Just as two objects dropped an instant apart will gradually get farther and farther apart, matter falling behind an object will lag farther and farther behind the object and thus their mass will have less and less effect on the event horizon for the objects ahead of them. <span>&nbsp;</span>In effect each infalling object sees its own event horizon based on the time, position and speed at which it approaches.</span></p> <p class="MsoNormal"><span style="font-size:10pt;font-family:Arial"><span>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; </span>Another way an object might cross the event horizon is that when the object gets so close that the uncertainty of its position lies both outside and inside the event horizon, the particle may then tunnel into the black hole. <span>&nbsp;</span>Of course it may also tunnel its way out again.<span>&nbsp; </span>It may turn out that by the time a particle could get that close the event horizon the black hole will have already evaporated.</span></p> <p class="MsoNormal"><span style="font-size:10pt;font-family:Arial"><span>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; </span>Is the Lorentz Transformation valid inside the event horizon? <span>&nbsp;</span>If so does it require that both space and time become imaginary. <span>&nbsp;</span>If we try to calculate the velocity of a particle inside the event horizon, we end up dividing imaginary distance by imaginary time and get NEGATIVE velocity. <span>&nbsp;</span>Does that mean that instead of falling towards the center of the black hole, matter is repelled by it?<span>&nbsp; </span>If matter does continue falling towards the center, how fast does it fall? <span>&nbsp;</span>How is its velocity calculated?</span></p> <p class="MsoNormal"><span style="font-size:10pt;font-family:Arial"><span>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; </span>I&rsquo;m sure these interpretations are overly na&iuml;ve and I&rsquo;m hoping the experts in this forum can explain and clarify these issues.</span></p>

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#### skeptic

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<p><font size="1"><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>There are two answers to your question, depending on what that question is, as you have left a thing or two unsaid.You can easily cross the event from the outside towards the singularity.&nbsp; The black hole has a gravitational field and you can simply fall into it, crossing the event horizon in the process.You cannot cross the event horizon from the inside.&nbsp; Once matter or light has entered the sphere bounded by the event horizon it does not and cannot come back out.There is also a time warpage effect (recall that a black hole is ultimately an extreme warpage of space-time which is very highly curved in the vicinity of the black hole and the singularity is called a singularity becaue the curvature at that point is predicted to be infinite by general relativity).&nbsp; That warpage affects what would be seen by an outside observer.When&nbsp; a star collapses to form&nbsp; a black hole, there is an extreme warpage of space-time that occurs.&nbsp;&nbsp; Light from the surface of the star continues to be emitted, but time at the surface of the star is different from time measured by an observer outside the event horizon.&nbsp; And in the final moments of the collapse time at the surface of the collapsing star is stretched infintely with respect to the outside observer, so light continues to reach that observer forever.&nbsp; But it is also red-shifted in the extreme so that it quickly becomes very very dark.&nbsp; To see an good explanation of this I refer you to pages 87 and 88 in Stephen Hawking's A Brief History of Time.&nbsp; I think this is what Thorne is trying to get across in the passage that you cited.&nbsp; Also you should study pages 249 to 257 of Thorne's book where covers this material as well.I am not sure why you are asking about the Lorentz transformation.&nbsp; The Lorentz transformation applies to special relativity, and special relativity specifically assumes that gravity can be neglected and that space-time can be taken as flat.&nbsp; That is certainly not the case near a black hole.&nbsp; Quite the contrary. <br /> Posted by DrRocket</DIV></font></p><p> </p><p><span style="font-size:10pt;font-family:'Arial','sans-serif'">Dr. Rocket,</span></p> <p style="text-indent:0.5in"><span style="font-size:10pt;font-family:'Arial','sans-serif'">Thank you for responding.<span>&nbsp; </span>The intention of my post was not so much to find out what happens as how or why it happens.<span>&nbsp; </span>For instance, let&rsquo;s consider the case of a 10 solar mass black hole and calculate the velocity profile for an infalling object from rest beginning effectively at infinity.<span>&nbsp; </span>The formula I use is sqrt(2*G*M/r) where G = 6.67428*10^-11 m^3/kg/s^2, M = 10 times the solar mass or 1.98892*10^31 kg and r is the distance from the center of the black hole in meters.<span>&nbsp; </span>The result I get is that when an object reaches the event horizon its infalling velocity is equal to the speed of light.<span>&nbsp; </span>Just as we agree that nothing can escape a black hole because in doing so its velocity would have to exceed c, it seems that nothing can fall in for the same reason.<span>&nbsp; </span></span></p> <p><span style="font-size:10pt;font-family:'Arial','sans-serif'"><span>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; </span>I understand the point about time at the surface of the collapsing star being stretched (sometimes called time dilation) infinitely at the Swarzschild radius (1) but was looking for more a quantitative expression or formula to describe how much time is dilated at various distances from the event horizon.<span>&nbsp; </span>I would appreciate it very much if you could give me the correct formula.</span></p> <p class="MsoNormal"><span style="font-size:10pt;font-family:'Arial','sans-serif'"><span>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; </span>It&rsquo;s my understanding that two observers in different inertial frames may disagree about when something happens or even if event A occurs before event B (as long as there is no cause and effect relationship between the two), but they cannot disagree over whether or not event A ever happens.<span>&nbsp; </span>It is not possible for an object to cross the event horizon in one inertial frame and not cross it in another.<span>&nbsp; </span>So we come back to the original question, can an object cross the event horizon and if so how?</span></p> <p class="MsoNormal"><span style="font-size:10pt;font-family:'Arial','sans-serif'">&nbsp;</span></p> <p style="margin-left:0.5in;text-indent:-0.25in" class="MsoNormal"><span style="font-size:10pt;font-family:'Arial','sans-serif'"><span>(1)<span style="font-family:'TimesNewRoman';font-style:normal;font-variant:normal;font-weight:normal;font-size:7pt;line-height:normal;font-size-adjust:none;font-stretch:normal">&nbsp;&nbsp; </span></span></span><span style="font-size:10pt;font-family:'Arial','sans-serif'">I particularly like this term because although it refers to Karl Swarzschild, in German it means &lsquo;black shield&rsquo; which is in fact what it is.</span></p> &nbsp;<p>&nbsp;</p>

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#### DrRocket

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<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>&nbsp; Thank you for clarifying the issue with kinetic energy in special relativity.&nbsp; Frankly, I was not aware that it was equal to (m-m0)c^2 where m0 is rest mass and m0 is the total mass.&nbsp; Likewise I didn&rsquo;t realize that a freely falling object gains mass with speed.&nbsp; If I understand your point correctly, as a freely falling object gains speed, its mass also increases.&nbsp; Since the black hole&rsquo;s gravity cannot accelerate more massive objects as rapidly as less massive ones, its speed, when it reaches the event horizon, falls short of c.&nbsp; Is this a correct interpretation of your point? I appreciate your informing me of how I have confused classical Newtonian mechanics with relativity.&nbsp; One never realizes the limits of one's understanding until confronted by someone with superior knowledge. Your knowledge of relativity is all the more remarkable as it seems to be unencumbered by obscure mathematical expressions.&nbsp; Despite that, I wonder if you could help me with the calculation of the radius of the 10 solar mass black hole described above.&nbsp; I&rsquo;m afraid I erred by using the strictly Newtonian method of calculating the radius at which the escape velocity is equal to c.&nbsp; The value I got was very close to 29,540 meters. Could you describe how the Schwarzschild radius would be different using general relativity? &nbsp;&nbsp; <br />Posted by skeptic</DIV></p><p>The formula for the Schwarzchild radius per se, is based on Newtonian gravity.&nbsp; But the physics questions that you have raised are not amenable to analysis on the basis of Newtonian mechanics.</p><p>http://en.wikipedia.org/wiki/Schwarzschild_radius</p><p>http://en.wikipedia.org/wiki/Black_hole</p><p>http://en.wikipedia.org/wiki/Event_horizon<br /></p> <div class="Discussion_UserSignature"> </div>

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