Any coordinate system in GR?
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From: JanPB on 27 Aug 2006 15:40 I.Vecchi wrote: > Daryl McCullough wrote: > > I.Vecchi says... > > > > >The question is whether the requirement of geodesic completeness (which > > >in this setting is obtained by shifting trouble to infinity) is > > >appriopriate, i.e. physically relevant. > > > > The answer is definitely "yes". Consider an observer in freefall near the > > event horizon. Using his local coordinates, there is *nothing* to prevent > > him from reaching and passing the event horizon in a finite amount of > > proper time, because for him, spacetime near the event horizon is > > approximately *flat*. > > Yes, but the problem is to define what is the event horizont for him. > What does it mean PHYSICALLY that he has crossed the event horizon? > > Let me go back for a moment to the Agata and Bruno example ([1]). For > Agata, Bruno never crosses the horizon . She cannot associate a time > T_o to Bruno stepping over the horizon. For an external observer Bruno > NEVER crosses the horizon. Except Agata makes this conclusion based either on her visual (literally) observation - which is subject to light delay, or based on her Schwarzschild simultaneity notion - which we already know is very odd. > Let's consider Bruno perspective then. You are right that nothing > prevents him to reach the horizon, the problem is the the horizon's > position in his perspective becomes arbitrary. If his proper time at > the horizon had a physically determined value he should be able to > look at his clock just before talking the plunge (e.g. switching off > the rocket that keeps him over the horizon) calculate and say: "In ten > minutes I will have crossed the horizont", and that would mean that in > his proper time 10 minutes would pass and then he'd look at his clock > and know he's over . The problem is that in order to calculate that > proper time he needs to know on which KS chart he is. What do you mean by "on which KS chart he is"? (There is only one.) Do you mean Daryl's A and B time shifts? Bruno could calculate the time to reach the horizon using only the exterior Schwarzschild chart. If just before beginning his free fall he sits above the horizon at r=r0>2m, then: time to the horizon = lim_{r->2m+} integral_r0^r ds (integrate along the free-fall geodesic). Adding the constant A doesn't change this integral. > Now, as we have > seen ([2]), the correspondence between the KS chart and the > Schwarzschild chart is undetermined, so he will not be able to do > that. It is determined off the horizon and then one needs to prove that this diffeomorphism extends uniquely over the horizon. > The question is then, is there any chart independent way (i.e. > independent from the space-time measurement model that defines a chart) > to calculate Bruno's proper time across the horizon? The integral above is chart-independent. -- Jan Bielawski
From: Koobee Wublee on 27 Aug 2006 17:56 Daryl McCullough wrote: > Koobee Wublee says... > > ** 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 > > > Of course, r >=0. This is spherically symmetric polar coordinate. > > What about the first metric implies that the range of r is between > 0 and infinity? Are you thinking that the use of the name "r" implies > such a range? Yes. > You do realize, don't you, that the first metric, with range > r > -K is exactly the same as the second metric, with range r > 0? Yes. > Or, the other way around, the first metric with range r > 0 is > exactly the same as the second metric with range r > K>? Yes. Do you realize these two metrics are mutually independent of each other? Do you realize these two metrics are independent of each other? Do you realize both of these metrics have their origins at (r = 0)? Do you realize both of these two metrics are solutions to Einstein Field Equations in spherically symmetric vacuum?
From: Koobee Wublee on 27 Aug 2006 18:17 Daryl McCullough wrote: > Koobee Wublee says... >> ** 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 > >> 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 you compute the area of a sphere of radius r and constant t using > the top metric, you will find that it is > > 4 pi (r+K)^2 > > which implies that the effective radius is r+K, not r. Wrong. The surface area of a sphere is always (4 pi r^2) where r is the radius of the sphere. 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. > > Are you saying that it's not true that if we start with > > ds^2 = c^2 (1 - K / r) dt^2 - dr^2 / (1 - K / r) - r^2 dO^2 > > and change variables to u = r-K, then we obtain > > ds^2 = c^2 dt^2 / (1 + K / u) - (1 + K / u) du^2 - (u + K)^2 dO^2 Yes, this is true. You can also start with the following trial equation (Safartti's term). ds^2 = c^2 (1 - K / u) dt^2 - du^2 / (1 - K / u) - u^2 dO^2 Setting (u = r + K), you get ds^2 = c^2 dt^2 / (1 + K / r) - (1 + K / r) dr^2 - (r + K)^2 dO^2 You can also start with the following trial equation. ds^2 = c^2 dt^2 / (1 + K / u) - (1 + K / u) du^2 - (u + K)^2 dO^2 Setting (u = r - K), you get ds^2 = c^2 (1 - K / r) dt^2 - dr^2 / (1 - K / r) - r^2 dO^2 You can also start with the following trial equation. ds^2 = c^2 dt^2 / (1 + K / u) - (1 + K / u) du^2 - (u + K)^2 dO^2 Setting (u = K^2 / r), you get ds^2 = c^2 dt^2 / (1 + r / K) - (1 + r / K) (K^4 / r^4) dr^2 - (K^4 / r^2) (1 + r / K)^2 dO^2 If the following is a valid metric, ds^2 = c^2 A(r) dt^2 - B(r) dr^2 - C(r)^2 dO^2 Then, the following is also a valid metric. ds^2 = c^2 A(u) dt^2 - B(u) (du/dr)^2 dr^2 - C(r)^2 dO^2
From: Koobee Wublee on 27 Aug 2006 18:27 JanPB wrote: > 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 > > >> > > >> 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. > > Yes, although there is nothing magical about 0. It's just a convention > everyone uses by analogy with the planar polar coordinates. One can > assign the number 42 to the centre of symmetry. With the following spacetime, ds^2 = c^2 dt^2 / (1 + r / K) - (1 + r / K) (K^4 / r^4) dr^2 - (K^4 / r^2) (1 + r / K)^2 dO^2 How do you specify a distance 2 AU from the center of the sun? > > > 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? > > I though I'd just answered it. I assume it was "How do you know which > of the two is the first form?" Did you mean something else? I assumed > you meant "how does one know to which form r>0 applies to?". You did not answer my question. I am expecting a 'first' or 'second' answer following by an expalantion on why the answer is 'first' or 'second'.
From: JanPB on 27 Aug 2006 18:52
Koobee Wublee wrote: > Daryl McCullough wrote: > > Koobee Wublee says... > > >> ** 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 > > > >> 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 you compute the area of a sphere of radius r and constant t using > > the top metric, you will find that it is > > > > 4 pi (r+K)^2 > > > > which implies that the effective radius is r+K, not r. > > Wrong. The surface area of a sphere is always (4 pi r^2) where r is > the radius of the sphere. No. (This is too funny.) The surface area is determined by the metric. Let's compute then this area precisely using the top metric. Fix r = C = const. and t = D = const. - this describes the sphere corresponding to the fixed value of your "r" coordinate equal to C. Using your top form and inserting r=C and t=D, we get dr = dt = 0, and: dsigma^2 = -(C+K)^2 dtheta^2 - (C+K)^2 sin^2(theta) dphi^2 ....so the volume 2-form on the sphere is: w = sqrt(g) dtheta /\ dphi ....where: [ -(C+K)^2 0 ] g = det [ ] = (C+K)^4 sin^2(theta) [ 0 -(C+K)^2 sin^2(theta) ] ....hence: w = (C+K)^2 sin(theta) dtheta /\ dphi ....and we integrate over the sphere: Area = integral(volume form) = integral_{-pi}^{pi} integral_0^{pi} (C+K)^2 sin(theta) dtheta dphi = = (C+K)^2 integral_{-pi}^{pi} [-cos(theta)]_0^{pi} dphi = = (C+K)^2 * 2 * integral_{-pi}^{pi} dphi = = (C+K)^2 * 2 * 2pi = 4 pi (C+K)^2. QED. You can change and substitute every which way and you'll never change the tensor. That's part of what it means to be a tensor. Plain vectors are the same way. -- Jan Bielawski |