ydoaPs Posted November 29, 2005 Share Posted November 29, 2005 ok, i was REALLY bored. i decided to try to derive the newtonian analogue of GR with absolute space and time-no space-time. i started like this: you have a piece of matter. this piece is the center of an imaginary sphere(not in the i sense, but in the sense of not existing). the gravity would cause a volume reduction. this could be seen by having the edge of the sphere be composed of matter. [math]V=\frac{4}{3}{\pi}r^3[/math] [math]\frac{dV}{dt}=4{\pi}r^2{\frac{dr}{dt}}[/math] [math]a{\frac{dV}{dt}}=4{\pi}GM\frac{dr}{dt}[/math] now, i know acceleration has something to do with curvature, but i'm not sure how to express it. the acceleration causing curvature can be seen by means of the first law of motion. how do i get rid of the [math]\frac{dV}{dt}[/math] and the [math]\frac{dr}{dt}[/math]? how do i use the a to make a term for curvature? i am pretty sure i need an r term in there as well. would [math]\frac{dr}{dt}[/math] be the same as v=at? Link to comment Share on other sites More sharing options...
timo Posted November 29, 2005 Share Posted November 29, 2005 I wouldn´t want to stop anyone from trying to fiddle around with equations but in this case I must tell you that I don´t really know what you´re up to. Furthermore, I don´t have the feeling that you´re knowing what you´re doing either. That´s not nessecarily bad. Sometimes it´s interesting just to do some stuff and critically review what you actually did afterwards - it sometimes yields a very personal learning experience. Are you aware that "curvature" (whatever you think it is) is usually connected to some second derivative, for example? Well, one point where I can help you out a bit (at least that´s what I hope): >> Would dr/dt be the same as v=at? If r is a position, then the total derivative dr/dt (t being time) is called velocity, in other words: dr/dt=v. In your case r was supposed to be a radius of a sphere, so it´s more or less up to you to decide to what extend you could still talk about a velocity (one might think about things like "expansion velocity of the sphere"). Link to comment Share on other sites More sharing options...
ydoaPs Posted December 3, 2005 Author Share Posted December 3, 2005 I must tell you that I don´t really know what you´re up to.i told you what i was trying to do. if you don't understand, make specific questions and i'll try to explain it better. Are you aware that "curvature" (whatever you think it is) is usually connected to some second derivative, for example?i know that concavity is connected to the second derivative, but i don't know anything about curvature, eccept that it should involve the acceleration and radius. Link to comment Share on other sites More sharing options...
□h=-16πT Posted January 10, 2006 Share Posted January 10, 2006 Newtonian mechanics is set in flat space-time, curvature is a tensor that describes the degree to which parrallelism is lost. Are you thinking of radius of curvature? Link to comment Share on other sites More sharing options...
CanadaAotS Posted January 11, 2006 Share Posted January 11, 2006 how come [math]r^2[/math] suddenly turns into [math]GM[/math]? Link to comment Share on other sites More sharing options...
timo Posted January 11, 2006 Share Posted January 11, 2006 Purely equation-wise it´s because of F = ma = G*M*m/r² => a = G*M/r². Then, he multiplied the left-hand side of the equation with a and the right-hand one with GM/r². Link to comment Share on other sites More sharing options...
□h=-16πT Posted January 12, 2006 Share Posted January 12, 2006 Frames freely falling in a gravitational field are inertial. If you were going to try and create a Newtonian analogue of GR you'd have to include the weak equivalence principle, which would remove the acceleration from your equations. Einstein arrived at the idea of inertial motion over a curved space-time being gravity because of this equivalence principle. Link to comment Share on other sites More sharing options...
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