Any coordinate system in GR?
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From: Tom Roberts on 27 Aug 2006 00:33 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. BTW Ken Tucker in the past has been particularly confused about the relationship among coordinates, units, and metric components.... Tom Roberts
From: Koobee Wublee on 27 Aug 2006 01:55 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 > > At r = -2GM/c^2. Your restriction r>0 is wrong, it should read > r>-2GM/c^2. > > The reason for it this is that the first form of the metric is obtained > from the second (for which r>0) by means of the coordinate change: How do you know which of the two is the first form? And why? > ...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.
From: JanPB on 27 Aug 2006 02:17 Koobee Wublee wrote: > 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 > > > > At r = -2GM/c^2. Your restriction r>0 is wrong, it should read > > r>-2GM/c^2. > > > > The reason for it this is that the first form of the metric is obtained > > from the second (for which r>0) by means of the coordinate change: > > 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. 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. > > ...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? -- Jan Bielawski
From: Sorcerer on 27 Aug 2006 02:33 "JanPB" <filmart(a)gmail.com> wrote in message news:1156659448.039855.139290(a)m73g2000cwd.googlegroups.com... | Koobee Wublee wrote: | > 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 | > > | > > At r = -2GM/c^2. Your restriction r>0 is wrong, it should read | > > r>-2GM/c^2. | > > | > > The reason for it this is that the first form of the metric is obtained | > > from the second (for which r>0) by means of the coordinate change: | > | > 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. | | 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. | | > > ...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? | err... -1 and 3? I used ter be gud at ritmatec 'n' speling. | -- | Jan Bielawski |
From: JanPB on 27 Aug 2006 02:33
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. My point was that at this stage this is really the ODE theory, hence the uniqueness. -- Jan Bielawski |