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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DrRocket

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<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 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.&nbsp; 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..."&nbsp; This statement is consistent with both the Lorentz Transformation and relativity.&nbsp; &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 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).&nbsp; In a gravitational field the v term is replaced by escape velocity of the gravitational field.&nbsp; 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.&nbsp; The falling object however, doesn&rsquo;t experience this transformation but continues falling through the event horizon.&nbsp; How can the perception from the perspective of the infalling object be consistent with that of the external observer? &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 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.&nbsp; 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? &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 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. &nbsp;This doesn&rsquo;t really work.&nbsp; 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. &nbsp;In effect each infalling object sees its own event horizon based on the time, position and speed at which it approaches. &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 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. &nbsp;Of course it may also tunnel its way out again.&nbsp; It may turn out that by the time a particle could get that close the event horizon the black hole will have already evaporated. &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; Is the Lorentz Transformation valid inside the event horizon? &nbsp;If so does it require that both space and time become imaginary. &nbsp;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. &nbsp;Does that mean that instead of falling towards the center of the black hole, matter is repelled by it?&nbsp; If matter does continue falling towards the center, how fast does it fall? &nbsp;How is its velocity calculated? &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 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. <br />Posted by skeptic</DIV></p><p>There are two answers to your question, depending on what that question is, as you have left a thing or two unsaid.</p><p>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.</p><p>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.</p><p>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.</p><p>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 <em>A Brief History of Time.&nbsp; </em>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.</p><p>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.</p> <div class="Discussion_UserSignature"> </div>
 
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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'> Dr. Rocket, Thank you for responding.&nbsp; The intention of my post was not so much to find out what happens as how or why it happens.&nbsp; 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.&nbsp; 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.&nbsp; The result I get is that when an object reaches the event horizon its infalling velocity is equal to the speed of light.&nbsp; 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.&nbsp; &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 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.&nbsp; I would appreciate it very much if you could give me the correct formula. &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 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.&nbsp; It is not possible for an object to cross the event horizon in one inertial frame and not cross it in another.&nbsp; So we come back to the original question, can an object cross the event horizon and if so how? &nbsp; (1)&nbsp;&nbsp; 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. &nbsp;&nbsp; <br />Posted by skeptic</DIV></p><p>Your formula for the velocity as a function of radius is classical -- derived from Newtonian mechanics and Newton's law of universal gravitation.&nbsp; It relies on F=ma strongly to relate velocity to kineteic energy.&nbsp; In special relativity the kinetic energy is not 1/2 mv^2&nbsp; but is (m-m0)c^2 where m0 is rest mass and m0 is the total mass, which depends on speed.&nbsp; You are mixing classical mechanics with relativity.&nbsp; It is the classical mechanics that results in your prediction of velocity, and classically there is no limit to velocity.&nbsp; So the problem with your conclusion is that you are using classical mechanics to compute a velocity which results in an impossibility under the rules of relativity.&nbsp; But we know that classical mechanics does not apply when velocities are near the speed of light, that is why it was necessary to replace it with relativity.</p><p>An object can cross the event horizon with no trouble.&nbsp; It will in fact be pulled in by the gravity of the black hole, exactly as an object approaching the earth is caused to fall towards the center of our planet.&nbsp; </p><p>I don't have a simple formula for you to calculate the kinetic energy or speed as you move from infinity to the Schwarzchild radius, or to describe time dilation effects as function of position relative to the event horizon..&nbsp; There may be one but I don't have it off the top of my head, and what is needed is to consider the effect of a body accelerating under and force that obeys the inverse sqare law within relativity.&nbsp; You might try looking at a text on general relativity, for instance <em>Gravitation</em> by Misner, Thorne and Wheeler.</p><p>You need to be very careful about the inertial reference frames that are being used.&nbsp; Classical mechanics and special relativity work with global inertial reference frames, and assume that such things exist.&nbsp; Special relativity does not apply under circumstances in which gravity is a significant factor -- and it is a VERY significant factor when you are talking about black holes.&nbsp; In general relativity, which is our best available theory&nbsp;of gravity, &nbsp;an inertial reference frame is one that is attached to a freely falling body, unrestrained.&nbsp; Moreover, the inertial reference frame is local, not global.</p><p>I think that a good deal of the confusion arises from mixing classical mechanics, special relativity and general relativity together inappropriately.&nbsp; However, I am not an expert in general relativity or black holes.&nbsp; You may need to find a real general relativist.&nbsp; There are such people, but you don't find one on every street corner.&nbsp; Thorne, who wrote your book (and was one of three who wrote <em>Gravitation</em>) is one of he best.</p><p>There is guy who goes by the&nbsp;forum name Publius, who inhabits the Bad Astronomy forum who seems to&nbsp;specialize in general relativity.&nbsp; He might be able to shed some more light on your questions.&nbsp; <br />&nbsp;</p> <div class="Discussion_UserSignature"> </div>
 
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skeptic

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<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>Your formula for the velocity as a function of radius is classical -- derived from Newtonian mechanics and Newton's law of universal gravitation.&nbsp; It relies on F=ma strongly to relate velocity to kineteic energy.&nbsp; In special relativity the kinetic energy is not 1/2 mv^2&nbsp; but is (m-m0)c^2 where m0 is rest mass and m0 is the total mass, which depends on speed.&nbsp; You are mixing classical mechanics with relativity.&nbsp; It is the classical mechanics that results in your prediction of velocity, and classically there is no limit to velocity.&nbsp; So the problem with your conclusion is that you are using classical mechanics to compute a velocity which results in an impossibility under the rules of relativity.&nbsp; But we know that classical mechanics does not apply when velocities are near the speed of light, that is why it was necessary to replace it with relativity.An object can cross the event horizon with no trouble.&nbsp; It will in fact be pulled in by the gravity of the black hole, exactly as an object approaching the earth is caused to fall towards the center of our planet.&nbsp; I don't have a simple formula for you to calculate the kinetic energy or speed as you move from infinity to the Schwarzchild radius, or to describe time dilation effects as function of position relative to the event horizon..&nbsp; There may be one but I don't have it off the top of my head, and what is needed is to consider the effect of a body accelerating under and force that obeys the inverse sqare law within relativity.&nbsp; You might try looking at a text on general relativity, for instance Gravitation by Misner, Thorne and Wheeler.You need to be very careful about the inertial reference frames that are being used.&nbsp; Classical mechanics and special relativity work with global inertial reference frames, and assume that such things exist.&nbsp; Special relativity does not apply under circumstances in which gravity is a significant factor -- and it is a VERY significant factor when you are talking about black holes.&nbsp; In general relativity, which is our best available theory&nbsp;of gravity, &nbsp;an inertial reference frame is one that is attached to a freely falling body, unrestrained.&nbsp; Moreover, the inertial reference frame is local, not global.I think that a good deal of the confusion arises from mixing classical mechanics, special relativity and general relativity together inappropriately.&nbsp; However, I am not an expert in general relativity or black holes.&nbsp; You may need to find a real general relativist.&nbsp; There are such people, but you don't find one on every street corner.&nbsp; Thorne, who wrote your book (and was one of three who wrote Gravitation) is one of he best.There is guy who goes by the&nbsp;forum name Publius, who inhabits the Bad Astronomy forum who seems to&nbsp;specialize in general relativity.&nbsp; He might be able to shed some more light on your questions.&nbsp; &nbsp; <br /> Posted by DrRocket</DIV></p><p>&nbsp;</p><p> </p><p style="text-indent:0.5in" class="MsoNormal"><span style="font-size:10pt;font-family:'Arial','sans-serif'">Thank you for clarifying the issue with kinetic energy in special relativity.<span>&nbsp; </span>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.<span>&nbsp; </span>Likewise I didn&rsquo;t realize that a freely falling object gains mass with speed.<span>&nbsp; </span>If I understand your point correctly, as a freely falling object gains speed, its mass also increases.<span>&nbsp; </span>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.<span>&nbsp; </span>Is this a correct interpretation of your point? </span></p> <p style="text-indent:0.5in" class="MsoNormal"><span style="font-size:10pt;font-family:'Arial','sans-serif'">I appreciate your informing me of how I have confused classical Newtonian mechanics with relativity.<span>&nbsp; </span>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.<span>&nbsp; </span>Despite that, I wonder if you could help me with the calculation of the radius of the 10 solar mass black hole described above.<span>&nbsp; </span>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.<span>&nbsp; </span>The value I got was very close to 29,540 meters. Could you describe how the Schwarzschild radius would be different using general relativity?</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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skeptic

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<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>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.http://en.wikipedia.org/wiki/Schwarzschild_radiushttp://en.wikipedia.org/wiki/Black_holehttp://en.wikipedia.org/wiki/Event_horizon <br /> Posted by DrRocket</DIV></p><p> </p><p style="text-indent:0.5in" class="MsoNormal">Well at least we can agree on the Schwarzschild radius and escape velocity.<span>&nbsp; </span>One way of describing the escape velocity is the muzzle velocity needed to fire a projectile from the specified radius so that it just coasts to a stop at infinity.<span>&nbsp; </span>We would find not unexpectedly, that as the projectile ascends, its velocity is always equal to the escape velocity at whatever radius along the trajectory. <span>&nbsp;</span>One would also expect that if an object started at infinity and fell down the same trajectory as the projectile, its speed would be the same as the projectile&rsquo;s, albeit in the opposite direction, at all points along the trajectory.<span>&nbsp; </span>Thus an object that starts from infinity (more or less) and falls towards the event horizon of a black hole will approach the event horizon at very nearly the speed of light.<span>&nbsp; </span>To suggest otherwise requires an explanation of why the velocity of an infalling object should be different than that of the projectile.<span>&nbsp; </span><span>&nbsp;</span>Note that the calculation of the Schwarzschild radius at the event horizon where gravity and velocities are the most extreme is done with Newtonian physics so it should be valid to use the same method at all points along the trajectory.</p> <p style="text-indent:0.5in" class="MsoNormal">We all have been taught that as an object is accelerated, the energy of acceleration has mass which is added to the mass of the object, making it harder and harder to accelerate the object further.<span>&nbsp; </span>If this principle applies to free falling matter approaching the event horizon at nearly the speed of light, that matter may well have more mass than the black hole itself.<span>&nbsp; </span>Where could that much energy come from, certainly not from the black hole.</p> <p style="text-indent:0.5in" class="MsoNormal">If a space ship is floating free in space we say it is at rest even though it may be moving at considerable velocity.<span>&nbsp; </span>An astronaut in that space ship would be weightless because the ship is not accelerating.<span>&nbsp; </span>But what about a free falling space ship.<span>&nbsp; </span>Even though the ship is accelerating, the astronauts are still weightless inside.<span>&nbsp; </span>There is no test an astronaut could do, other than looking outside, that would tell him whether the ship is at rest or freely accelerating in a gravitational field. <span>&nbsp;</span>A freely falling ship is considered to be at rest and because it is, it doesn&rsquo;t gain mass as it falls towards the black hole.</p> <p style="text-indent:0.5in" class="MsoNormal">I suspect that not only does the Lorentz transformation apply near a black hole but it applies in two ways. <span>&nbsp;</span>The first is that it foreshortens the length and dilates the time of the infalling matter based on its velocity the same way it would away from a gravitational field.&nbsp; With the other way, space near the black hole is contracted and time dilated because of the intense gravitational field. <span>&nbsp;</span>My guess is the two effects are combined in the same way relativistic velocities are added by the formula <span>&nbsp;</span>(v + u)/(1 + (v/c)(u/c)) where v represents the infalling velocity of the object and u represents the escape velocity at the object&rsquo;s radius.<span>&nbsp; </span>Can anyone with real knowledge in this area confirm this or provide an alternate explanation?</p> &nbsp;<p>&nbsp;</p>
 
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DrRocket

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<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'> Well at least we can agree on the Schwarzschild radius and escape velocity.&nbsp; One way of describing the escape velocity is the muzzle velocity needed to fire a projectile from the specified radius so that it just coasts to a stop at infinity.&nbsp; We would find not unexpectedly, that as the projectile ascends, its velocity is always equal to the escape velocity at whatever radius along the trajectory. &nbsp;One would also expect that if an object started at infinity and fell down the same trajectory as the projectile, its speed would be the same as the projectile&rsquo;s, albeit in the opposite direction, at all points along the trajectory.&nbsp; Thus an object that starts from infinity (more or less) and falls towards the event horizon of a black hole will approach the event horizon at very nearly the speed of light.&nbsp; To suggest otherwise requires an explanation of why the velocity of an infalling object should be different than that of the projectile.&nbsp; &nbsp;Note that the calculation of the Schwarzschild radius at the event horizon where gravity and velocities are the most extreme is done with Newtonian physics so it should be valid to use the same method at all points along the trajectory.</DIV></p><p>You had better be careful.&nbsp; And read the material at the links that I provided.&nbsp; Because while the formula for the Schwarzchild radius appears to be classical, the meaning is not.&nbsp; Yes, you calculate it as though the escape velocity starting from the Schwarzchild radius is c, and it is.&nbsp; So one would indeed expect, simply from conservation of energy that an object falling to the black hole from infinity would have a velocity of c, and I think that is correct, although I am not positive.&nbsp; There are problems with energy and conservation of energy in general relativity, so the logic may be flawed. http://math.ucr.edu/home/baez/physics/Relativity/GR/energy_gr.htmlIn any case&nbsp;it is&nbsp;definitely not classical that from a radius less than the Schwarzchild radius, there is no escape.&nbsp; Not to infinity, and not even one centimeter beyond the Schwarzchild radius -- and that is not classical and is explainable with Newtonian mechanics.</p><p>Bottom line:&nbsp; You are asking questions to which I am not sure of the answer.&nbsp; But I am sure that the sort of reasong that you are using, while valid in non-relativistic physics, is on very shaky ground in the world of general relativity and black holes.&nbsp; </p><p>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>We all have been taught that as an object is accelerated, the energy of acceleration has mass which is added to the mass of the object, making it harder and harder to accelerate the object further.&nbsp; If this principle applies to free falling matter approaching the event horizon at nearly the speed of light, that matter may well have more mass than the black hole itself.&nbsp; Where could that much energy come from, certainly not from the black hole.</DIV></p><p>We are again confronting the problems with using non-relativistic physics in a situation that is driven by general relativity.&nbsp; Intuition breaks down here.&nbsp; So do arguments based on Newtonian mechanics.&nbsp; </p><p>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>&nbsp;&nbsp;If a space ship is floating free in space we say it is at rest even though it may be moving at considerable velocity.&nbsp; An astronaut in that space ship would be weightless because the ship is not accelerating.&nbsp; But what about a free falling space ship.&nbsp; Even though the ship is accelerating, the astronauts are still weightless inside.&nbsp; There is no test an astronaut could do, other than looking outside, that would tell him whether the ship is at rest or freely accelerating in a gravitational field. &nbsp;A freely falling ship is considered to be at rest and because it is, it doesn&rsquo;t gain mass as it falls towards the black hole.</DIV></p><p>No, a freely falling object is not at rest.&nbsp; It is simply an inertial reference frame for the purpose of general relativity.&nbsp; It might be considered at rest, if that is the reference frame in which you choose to work.&nbsp; But the notion of relativistic mass increase only applies to a object in one inertial reference frame when viewed from the point of view of another inertial reference frame with respect to which the object is moving.&nbsp; The concepts of mass and energy are dependent on the reference frame in which they are measured.&nbsp; In the reference frame of the object its mass never changes and it has no kinetic energy.</p><p>&nbsp;</p><p>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>I suspect that not only does the Lorentz transformation apply near a black hole but it applies in two ways. &nbsp;The first is that it foreshortens the length and dilates the time of the infalling matter based on its velocity the same way it would away from a gravitational field.&nbsp; With the other way, space near the black hole is contracted and time dilated because of the intense gravitational field. &nbsp;My guess is the two effects are combined in the same way relativistic velocities are added by the formula &nbsp;(v + u)/(1 + (v/c)(u/c)) where v represents the infalling velocity of the object and u represents the escape velocity at the object&rsquo;s radius.&nbsp; Can anyone with real knowledge in this area confirm this or provide an alternate explanation? &nbsp;&nbsp; <br />Posted by skeptic</DIV></p><p>You have to be careful about what is meant by a reference frame here.&nbsp; In special relativity&nbsp;you work with global reference frames and the Lorentz transformation lets you translate between those reference frames.&nbsp; In general relativity you have only local reference frames.&nbsp;To go much further you are probably going to need a specialist in general relativity.&nbsp; I don't know of any who participate in this forum.&nbsp; I can tell you that specific calculations in general relativity are difficult even for specialists.&nbsp; <br /></p> <div class="Discussion_UserSignature"> </div>
 
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