DParlevliet Posted June 26, 2015 Share Posted June 26, 2015 Classical calculated with Newton the escape speed is r = 2G.M/v2. With black holes and v = c that would be r = 2G.M/c2. That is the same as the Scharzschild radius calculated with relativity. Does that mean that Newton laws are still valid at this radius? Link to comment Share on other sites More sharing options...
Robittybob1 Posted June 26, 2015 Share Posted June 26, 2015 Classical calculated with Newton the escape speed is r = 2G.M/v2. With black holes and v = c that would be r = 2G.M/c2. That is the same as the Scharzschild radius calculated with relativity. Does that mean that Newton laws are still valid at this radius? Can you get a mass to go at the speed of light? I thought the answer might be "no", therefore the equation r = 2G.M/v^2 will not work at light speed. Link to comment Share on other sites More sharing options...
pavelcherepan Posted June 26, 2015 Share Posted June 26, 2015 Can you get a mass to go at the speed of light? I thought the answer might be "no", therefore the equation r = 2G.M/v^2 will not work at light speed. The formula gives the escape velocity, not the velocity of a massive body. Escape velocity can be higher than the speed of light, no problem. Classical calculated with Newton the escape speed is r = 2G.M/v2. With black holes and v = c that would be r = 2G.M/c2. That is the same as the Scharzschild radius calculated with relativity. Does that mean that Newton laws are still valid at this radius? I guess, only to some extent. As object approaches the event horizon of a black hole it would be expected to be moving at a very high velocity where you just can't ignore relativistic effects. The formula for Scharzschild radius was derived from GR so IMO it's more of a remarkable coincidink than the fact that Newtonian mechanics can be used for objects near black holes. Link to comment Share on other sites More sharing options...
Endy0816 Posted June 27, 2015 Share Posted June 27, 2015 Escape velocity set to the speed of light did play a role in the origin of the earlier concept for a Black hole. https://en.wikipedia.org/wiki/Dark_star_%28Newtonian_mechanics%29#John_Michell_and_dark_stars Funny how many people plug c into the escape velocity equation and see the same thing. Understanding that Relativity is a refinement of Newtonian mechanics is the key here. At slower speeds Newtonian mechanics is accurate enough, at higher speeds Relativity is the way to go. Sometimes both describe roughly the same scenario and the math agrees. Link to comment Share on other sites More sharing options...
md65536 Posted June 27, 2015 Share Posted June 27, 2015 Does Newtonian and GR escape velocity agree at all radii? Classical calculated with Newton the escape speed is r = 2G.M/v2. With black holes and v = c that would be r = 2G.M/c2. That is the same as the Scharzschild radius calculated with relativity. Does that mean that Newton laws are still valid at this radius? All the laws? I think the answer is 'no' partly because Newtonian and GR are not describing the same thing here. With Newtonian laws, an object with a velocity of v=c pointed away from the mass at that radius would leave it at a velocity of c, and escape to infinity while its velocity approaches 0. With GR, that escape velocity would only allow it to hover at the event horizon, and the object would never truly escape, never climb anywhere out of the gravity well to where escape is easier. But I don't fully know what that means. It is only at the event horizon that escape is impossible, but does that mean that only at that radius does escape velocity no longer have the same meaning as outside the radius? If we instead considered a point just outside the radius, is the Newtonian and GR escape velocity the same? And in GR does escape velocity allow it to coast out to infinity? Link to comment Share on other sites More sharing options...
DParlevliet Posted June 27, 2015 Author Share Posted June 27, 2015 The formula of a simple orbit r = G.M/v2 is based on: g = G.M/r2 s = 0.5g.t2 The curve of the time space dimensions as a circle y = x2/2r (x ≈ 0), which in above is: s = v2.t2/2r. So which item is changing near a black hole according GR and how? The smallest radius for a photon is 3G.M/c2 of for mass 6G.M/c2. Again a simple formula resembling Newton formula but with whole numbers. I suppose that simple formula has simple explanations. Link to comment Share on other sites More sharing options...
MigL Posted June 29, 2015 Share Posted June 29, 2015 For a simple ( non-rotating and non-charged ) BH with a spherical event horizon ( probably not too many, if any, of those ), the mathematical radius which describes the event horizon is the distance at which the escape velocity is c. As massive particles cannot move at c, anything which reaches the mathematical radius describing the event horizon will be forced to continue to the possible singularity. This is perfectly consistent with Newtonian gravity. Massive particles from just outside the event horizon however, can escape to 'infinity' ( the definition of escape velocity ), by sacrificing their partner virtual particle to the BH and thereby 'stealing' energy from it. This is called Hawking radiation. Link to comment Share on other sites More sharing options...
DParlevliet Posted June 29, 2015 Author Share Posted June 29, 2015 (edited) This is perfectly consistent with Newtonian gravity. Do you mean that (just) outside the BH Newton laws are still valid (in the case you mentioned)? Edited June 29, 2015 by DParlevliet Link to comment Share on other sites More sharing options...
Mordred Posted June 29, 2015 Share Posted June 29, 2015 (edited) No Newtonian physics is strictly Euclidean. Does not include curvature. While Newtons laws works great for everyday science, it doesn't have the lorentz coordinate transformations. Here is a good article specifically on Bh metrics https://www.google.ca/url?sa=t&source=web&cd=1&ved=0CBsQFjAA&url=http%3A%2F%2Fwww.phys.uu.nl%2F~thooft%2Flectures%2Fblackholes%2FBH_lecturenotes.pdf&rct=j&q=black%20holes%20pdf&ei=QkOQVf6BFIWrNtSRgaAN&usg=AFQjCNERag-FH9DCbw66GsxObohS8wEq9A&sig2=3_JOCHB5wqHkhJ__D5-ATQ Key note this doesn't mean Newtonian physics change, it means that the Newtonian physics doesn't cover spacetime curvature. Newtons three laws are still valid including escape velocity but you need to account for the coordinate changes. Edited June 29, 2015 by Mordred Link to comment Share on other sites More sharing options...
DParlevliet Posted June 29, 2015 Author Share Posted June 29, 2015 Newtons three laws are still valid including escape velocity but you need to account for the coordinate changes. That is what I mean But what coordinate is changing? Radius r is the same Link to comment Share on other sites More sharing options...
Mordred Posted June 29, 2015 Share Posted June 29, 2015 (edited) Not really here Ned Wright's page shows the coordinate transformation http://ned.ipac.caltech.edu/level5/March01/Carroll3/Carroll7.html You should note there is numerous calculations for R and r becomes infinite Should note that page covers the Schwartzchild Metric. The lecture notes link I provided shows other solutions such as the Kruskal coordinates ( these type of coordinate transformations isn't my strongest area, despite how long I've studied them)you'd be better off with the references than my attempts Edited June 29, 2015 by Mordred 1 Link to comment Share on other sites More sharing options...
DParlevliet Posted June 30, 2015 Author Share Posted June 30, 2015 Not really here Ned Wright's page shows the coordinate transformation If I was able to understand GR calculations I would not need to ask a forum. To start again: the classic orbits r = G.M/v2 is based on: g = G.M/r2 , s = 0.5g.t2 en the curve of the time space dimensions as a circle y = x2/2r (x ≈ 0), which in above is: s = v2.t2/2r. Those are simple formula. Simple formula has simple explanations. Around the BH for mass the minimun orbit is r = 6G.M/v2. So what happens. Is r 6 times larger? G smaller? Is v root 6 c? Link to comment Share on other sites More sharing options...
Endy0816 Posted June 30, 2015 Share Posted June 30, 2015 That radius would be 6 times larger. Link to comment Share on other sites More sharing options...
DParlevliet Posted June 30, 2015 Author Share Posted June 30, 2015 That radius would be 6 times larger. So in general r = a.G.M/v2 At small distance and speed close to c: a = 6 At large distance and low speed: a = 1 What is it in between? Does it depend on distance or speed. Or more scientific: what is the formula of a? Link to comment Share on other sites More sharing options...
DParlevliet Posted July 13, 2015 Author Share Posted July 13, 2015 In another forum I got the general formula for orbits: If I transfer this to a more familiar form it becomes: (r - rs) = G.m/v2 And then indeed for photons (v = c): r = G.m/c2 + rs = G.m/c2 + 2G.m/c2 = 3G.m/c2 Link to comment Share on other sites More sharing options...
DParlevliet Posted July 20, 2015 Author Share Posted July 20, 2015 (edited) Then there is a problem. With the formula s = 0.5g.t2 (acceleration) and y = x2/2r = v2t2/2r (circle at x ≈ 0) one gets: 0.5g.t2 = v2t2/2r, resulting in: g = v2/r In classic the orbit formula is v2 = G.m/r with gives in above: g = G.m/r2 In GR the orbit formula is v2 = G.m/(r-rs) (see above) with gives: g = G.m/r(r-rs) = G.m/r2(1-rs/r) However, according a writer on the above mentioned forum, the formula is: If this is true, then s = 0.5g.t2 (acceleration) and y = x2/2r is also changing close to the BH. Does someone knows what chances in the formula? Edited July 20, 2015 by DParlevliet Link to comment Share on other sites More sharing options...
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