Any coordinate system in GR?
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From: Koobee Wublee on 27 Aug 2006 02:56 JanPB wrote: > Koobee Wublee wrote: >> ** ds^2 = c^2 dt^2 / (1 + K / r) - (1 + K / r) dr^2 - (r + K)^2 dO^2 >> ** ds^2 = c^2 (1 - K / r) dt^2 - dr^2 / (1 - K / r) - r^2 dO^2 >> >> K is an integration constant chosen to fit Newtonian result. It is >> >> ** K = 2 G M / c^2 >> >> How do you know which of the two is the first form? And why? > > If one starts - as is usual - with a spherically symmetric > (t,r,theta,phi)-space with r>0, etc., then any spherically symmetric > metric can be locally written in the two forms we had before (not the > two forms you wrote above - the two forms with the sign switch). One > solves the Einstein equations which describe a metric (uniquely, > because the equations happen to be ODEs in this case) described by > components which happen to blow up at r=2m in that basis. This is your > second equation. Of course, r >=0. This is spherically symmetric polar coordinate. > If one started instead with a spherically symmetric > (t,r,theta,phi)-space with labelling set so that r>-K (K - some number) > then you'd end up with the solution that looks like your first > equation. Same thing, different chart. Is there any reason for you to avoid my simple question? > > > ...where you are recycling the letter "r" in the first form where I > > > have used "u". Look at the formula r=u+K. It's patently obvious that > > > when r changes between 0 and infinity then u changes between -K and > > > infinity. Tell me, is this not true? > > > > No. > > Are you sure? This is aritmetic now. One more time: assuming r=u+K if r > has values between 0 and infinity then u has values between what and > what? Very sure. In both equations, r >= 0.
From: Ken S. Tucker on 27 Aug 2006 02:59 JanPB wrote: > Ken S. Tucker wrote: > > JanPB wrote: > > > Ken S. Tucker wrote: > > > > > > > > Jan, I think you should study the original paper > > > > Ed Green cited, there is no such thing as an > > > > event horizon or Black-Hole's. BTW, Dr. Loinger > > > > and I discussed this at length. In short, the > > > > original Schwarzschild Solution has been > > > > bastardized and mis-understood for simplicity. > > > > > > No, the "original" solution is the only one. This follows from basic > > > ODE theory and the definition of tensor. > > > > > > > The bastardized version became popularized > > > > and embraced by astronomers who now see > > > > BH's under their beds. > > > > > > There is no other version, bastardized or not. How many solutions do > > > you have to the following ODE: > > > > > > f'(x) = f(x) > > > > > > ...given the initial condition f(0) = 1 ? I can see f(t) = exp(t). Is > > > there any other? > > > > > > How about another ODE, just slightly more complicated: > > > > > > -2 f(x) f'(x) + 1/x * (1 - f(x)^2) = 0 > > > > > > ...given the initial condition f(1) = 0 ? > > > > > > -- > > > Jan Bielawski > > > > What Newton assumed was, > > > > r= sqrt(x^2 + y^2 +z^2) > > > > but in GR the "real" R = r - m > > where m =1.47 kms in the case of the Sun. > > > > Most treatise begin by defining "r" that way. > > Schwarzschild emphasized that difference and obtained > > a singularity only at the very center of a mass, and that > > would assume an infinite density, which is physically > > impossible. But it's a good exercise. > > Not sure what you're alluding to. > > For the record, the second equation I wrote: > > -2 f(x) f'(x) + 1/x * (1 - f(x)^2) = 0 > > ...is what the Einstein field equations reduce to (in the case that > turns out to be the exterior). > > The solution is the familiar: > > f(x) = sqrt(1 + C/x) > > I set up the initial condition at random where normally one chooses the > constant C so that the metric is Newtonian for large x, namely C = -2m. I perfer other approaches but I'll use your's here. f(x) ~ (1-m/x) , x*f(x) = x - m. In more standard notation, that is written, R = r - m as I posted. Where Newton says the light particle would be at "r" Einstein predicted it would be at "R", with the caveat that we're in 3D. When translating to 4D the actual quantity becomes R(4D) = r - 2m. > My point was that at this stage this is really the ODE theory, hence > the uniqueness. What is ODE?
From: FrediFizzx on 27 Aug 2006 03:04 "Tom Roberts" <tjroberts137(a)sbcglobal.net> wrote in message news:Sg9Ig.20724$gY6.6996(a)newssvr11.news.prodigy.com... > Edward Green wrote: > > Now, I can rephrase my question. Suppose we replaced "t" with t' = > > a(t)t , a( ) a positive increasing function. This can be interpreted > > as changing the units of time from "natural clock seconds", and not in > > a uniform way. How does the explicit form of the metric handle this? > > The metric itself is of course unaffected. > > How the metric components behave is given by the usual coordinate > transform of the components of a rank-2 covariant tensor. > > > > In this case, a possible answer seems trivial: we simply insert the > > inverse of the positive factor "a(t)"... better, a^-1(t') ... in the > > formal metric, and everything is fine. > > No. that is not correct. > > g_i'j' = (dx^i/dx^i') (dx^j/dx^j') g_ij > > Here primed indices are for the primed coordinates, and unprimed are for > unprimed coords. Basically you forgot to account for how the basis > vectors of the coordinate system change because of the transform. > > [Note that a^-1(t') will occur in dx^t/dx^t'.] > > > > OK, Ken, here is my current postulation for an answer: you _can't_ > > reconstruct a physical spacetime simply from a set of variable labels > > and an explicit form of a metric on such labels: there is an element > > missing. That element is a perscription of the relationship between > > the local differentials of the variables and standard local physical > > processes -- "so many wavelengths of a certain spectral transition", > > and so forth. > > Yes. Coordinates by themselves are essentially useless in physics. One > needs a metric, and in particular the metric components in terms of > those coordinates. That permits one to compute distances between points > labeled by the coordinates, and from that everything else follows. > > > > If this perscription is the same everywhere, then it can > > simply be stated as "choice of units", as in, centimeters and seconds; > > if the perscription varies, then the additional information will be > > more complicated. But there is additional structure required to > > complete the model. > > Yes. This "additional structure" is called a metric. > > > > I'm thinking however that the units must be stated, which > > is sufficiently trivial as to be invisible, and that if we allow > > arbitrary point to point fluctuations in the units, we may need some > > additional non-trivial statements to complete the specification, beyond > > the explicit form of the metric. > > The metric components and the coordinate differentials must include > units. It is a matter of personal choice whether the units are put into > the metric components or into the coordinate differentials; in any case > ds^2 must have units (length)^2. Tom, please don't get sloppy with terminology. It is ds^2 must have *dimensions* of length^2. However when c = 1, length = time so it can also have dimensions of time^2. FrediFizzx Quantum Vacuum Charge papers; http://www.vacuum-physics.com/QVC/quantum_vacuum_charge.pdf or postscript http://www.vacuum-physics.com/QVC/quantum_vacuum_charge.ps http://www.arxiv.org/abs/physics/0601110 http://www.vacuum-physics.com
From: JanPB on 27 Aug 2006 03:53 Sorcerer wrote: > "JanPB" <filmart(a)gmail.com> wrote in message > news:1156659448.039855.139290(a)m73g2000cwd.googlegroups.com... > | Koobee Wublee wrote: > | > > | > No. > | > | Are you sure? This is aritmetic now. One more time: assuming r=u+K if r > | has values between 0 and infinity then u has values between what and > | what? > | > > err... -1 and 3? I used ter be gud at ritmatec 'n' speling. That's only if K=42. -- Jan Bielawski
From: JanPB on 27 Aug 2006 03:58
Ken S. Tucker wrote: > JanPB wrote: > > Ken S. Tucker wrote: > > > JanPB wrote: > > > > Ken S. Tucker wrote: > > > > > > > > > > Jan, I think you should study the original paper > > > > > Ed Green cited, there is no such thing as an > > > > > event horizon or Black-Hole's. BTW, Dr. Loinger > > > > > and I discussed this at length. In short, the > > > > > original Schwarzschild Solution has been > > > > > bastardized and mis-understood for simplicity. > > > > > > > > No, the "original" solution is the only one. This follows from basic > > > > ODE theory and the definition of tensor. > > > > > > > > > The bastardized version became popularized > > > > > and embraced by astronomers who now see > > > > > BH's under their beds. > > > > > > > > There is no other version, bastardized or not. How many solutions do > > > > you have to the following ODE: > > > > > > > > f'(x) = f(x) > > > > > > > > ...given the initial condition f(0) = 1 ? I can see f(t) = exp(t). Is > > > > there any other? > > > > > > > > How about another ODE, just slightly more complicated: > > > > > > > > -2 f(x) f'(x) + 1/x * (1 - f(x)^2) = 0 > > > > > > > > ...given the initial condition f(1) = 0 ? > > > > > > > > -- > > > > Jan Bielawski > > > > > > What Newton assumed was, > > > > > > r= sqrt(x^2 + y^2 +z^2) > > > > > > but in GR the "real" R = r - m > > > where m =1.47 kms in the case of the Sun. > > > > > > Most treatise begin by defining "r" that way. > > > Schwarzschild emphasized that difference and obtained > > > a singularity only at the very center of a mass, and that > > > would assume an infinite density, which is physically > > > impossible. But it's a good exercise. > > > > Not sure what you're alluding to. > > > > For the record, the second equation I wrote: > > > > -2 f(x) f'(x) + 1/x * (1 - f(x)^2) = 0 > > > > ...is what the Einstein field equations reduce to (in the case that > > turns out to be the exterior). > > > > The solution is the familiar: > > > > f(x) = sqrt(1 + C/x) > > > > I set up the initial condition at random where normally one chooses the > > constant C so that the metric is Newtonian for large x, namely C = -2m. > > I perfer other approaches but I'll use your's here. > > f(x) ~ (1-m/x) , x*f(x) = x - m. > > In more standard notation, that is written, > > R = r - m > > as I posted. Where Newton says the light particle > would be at "r" Einstein predicted it would be at "R", > with the caveat that we're in 3D. When translating > to 4D the actual quantity becomes R(4D) = r - 2m. Not sure what the Einstein equation has to do with R = r - m at all... > > My point was that at this stage this is really the ODE theory, hence > > the uniqueness. > > What is ODE? Ordinary differential equations. -- Jan Bielawski |