Any coordinate system in GR?
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From: Daryl McCullough on 27 Aug 2006 19:04 Koobee Wublee says... > >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 >> 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, that's only true for *Euclidean* geometry. The formula for area is given by (in the special case of a diagonal metric) Integral from theta=0 to theta = pi Integral from phi=0 to phi=2pi square-root(|g_{theta,theta} g_{phi,phi}|) dtheta dphi In the case of the metric ds^2 = c^2 dt^2 / (1 + K / r) - (1 + K / r) dr^2 - (r + K)^2 dO^2 the important term is - (r + K)^2 dO^2 = - (r + K)^2 dtheta^2 - (r + K)^2 sin^2(theta) dphi^2 The metric component g_theta, theta is just the term multiplying dtheta^2, and the component g_phi,phi is the term multiplying dphi^2. So we see |g_theta,theta| = (r+K)^2 |g_phi,phi| = (r+K)^2 sin^2(theta) So the area is given by Integral from theta=0 to theta = pi Integral from phi=0 to phi=2pi (r+K)^2 sin(theta) dtheta dphi Integrating over phi just gives a factor of 2pi, so we have Area = (r+K)^2 2pi Integral from theta=0 to pi of sin(theta) dtheta Since the integral of sin(theta) is -cos(theta), we get Area = (r+K)^2 2pi [-cos(pi) - -cos(0)] = (r+K)^2 2pi [-(-1) - (-1)] = 4pi (r+K)^2 >How do you know which of the two is the first form? And why? Didn't you just ask that question? Didn't I just answer it? 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. -- Daryl McCullough Ithaca, NY
From: Daryl McCullough on 27 Aug 2006 19:06 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 >Do you realize these two metrics are mutually independent of each >other? No, they are not. If you can transform from one metric to another via a coordinate change, then that means that they are physically the *same* metric. -- Daryl McCullough Ithaca, NY
From: Daryl McCullough on 27 Aug 2006 20:26 I.Vecchi says... >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? Using Schwarzchild coordinates, you can compute the proper time for a freefalling observer as a function of his radius r. The exact computation depends on the initial conditions, but for one particular geodesic, the relationship is simple: r(tau) = 3/2 (2m)^{1/3} (tau_0 - tau))^{2/3} where tau_0 is a constant depending on the initial conditions. For this geodesic, the observer passes the event horizon at a proper time tau_1 given by tau_1 = tau_0 - (2/3)^{3/2} 2m and reaches r=0 at proper time tau = tau_0 So what it means physically for the observer to pass the event horizon is just that his proper time is greater than tau_1. As I said, from the point of view of the freefalling observer, there is no physical reason for his proper time to stop before reaching tau_1 (when he crosses the event horizon) and tau_0 (when he reaches r=0). >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. Yes, that shows that Agata's coordinate system is incomplete. There are physically meaningful events that are given no coordinate whatsoever. The same thing happens in flat spacetime with accelerated observers. If you are on board a rocket ship with constant proper acceleration g, then times and distances as measured by clocks and rulers on the ship will be related to times and distances as measured by inertial clocks that are at rest relative to the ship's initial reference frame as follows: x = X cosh(gT) t = X sinh(gT) where (x,t) are the coordinates as measured by the inertial observers, and (X,T) are the coordinates as measured by the accelerated observers. Note that the point (x=1 light year, t=1 year) is given no coordinates *at* *all* in the accelerated coordinate system: there is no values of X and T such that X cosh(gT) = 1 and X sinh(gT) = 1. From the point of view of the accelerated observer, the point (x=1, t=1) can be approached as T--> infinity, but it can never be reached. That just means that the accelerated coordinates are incomplete. The same thing is true of the Schwarzchild coordinates. They are incomplete in exactly the same way. There are valid points on the manifold that are given no coordinates whatsoever in the Schwarzchild coordinate system. >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. Yes, from his point of view, there is nothing special about the horizon. However, Agata can calculate the proper time tau_1 at which Bruno will hit the event horizon, and she can make the prediction that Bruno's clock will never advance past tau_1. Bruno can watch his clock and prove that Agata is wrong (although his message gloating about it will never reach Agata). >If his proper time at >the horizon had a physically determined value It does. Agata can calculate it. >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 horizon", Yes, and at 1 second past ten minutes, assuming he hasn't hit the singularity, he'll know that he has crossed the event horizon. >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. The calculation can be done using KS, or using Schwarzchild coordinates. It's not hard using Schwarzchild coordinates. From the metric and the geodesic equation, you can prove that there are two constants of motion for radial geodesics: (1-2m/r) V^t = P (1-2m/r) (V^t)^2 - 1/(1-2m/r) (V^r)^2 = E where V^t = dt/dtau and V^r = dr/dtau. Using the first equation in the second, we find P/(1-2m/r) - (V^r)^2/(1-2m/r) = E So V^r = - square-root(P - E(1-2m/r)) dr/dtau = - square-root(P - E(1-2m/r)) (where I chose the minus sign because the observer is falling, so dr/dtau is negative). This gives tau as a function of r: dtau/dr = - 1/square-root(P - E(1-2m/r)) tau = tau_0 - integral of 1/square-root(P - E(1-2m/r)) dr The integral is a perfectly ordinary integral, with no singularities except at r=0. So the value of tau at which r=2m is computable. >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. The relationship between tau and r is perfectly well determined. The relationship between tau and t is not completely determined but that's not relevant to the question of what the value of tau is when the infalling observer hits the event 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? Yes. I just gave it to you. If you know Bruno's initial radius and the initial time on Bruno's clock, then you can easily compute the time on Bruno's clock when he reaches the event horizon. >>Anything other than demanding geodesic completeness >> would violate the equivalence principle, I think. > >Would it? I would like to inspect a rigorous proof that it would, I don't know what a rigorous proof would look like, but suppose that we compute (as shown above) that Bruno will hit the event horizon a
From: Tom S. on 27 Aug 2006 23:12 "Daryl McCullough" <stevendaryl3016(a)yahoo.com> wrote in message news:ecs7a70i7a(a)drn.newsguy.com... > Let e1 and e2 be two successive ticks of a clock (assuming that > you have an old-fashioned clock that ticks, if they make those > anymore). There are at least three different ways to compute the > "time between ticks": (1) Use the metric, and compute > Integral from e1 to e2 of square-root |g_ij dx^i dx^j| > along a geodesic path connecting e1 and e2. (2) Look at > the elapsed time shown on the clock. (3) Use a particular > coordinate system, and use t2 - t1 as the time between > ticks (where t2 is the time of e2 in that coordinate > system, and t1 is the time of e1). > > The usual assumption is that methods (1) and (2) will > give the same answer. Maybe I'm misinterpreting what you're saying. But I don't think that (1) and (2) are usually assumed to yield the same result. For example, let e1 and e2 be ticks of a clock sitting at rest on the surface of a neutron star. Then the elapsed time between e1 and e2 as shown on this clock is your method (2). Let another clock be launched radially outward from the surface of the star coincident with the fixed clock at event e1 with an initial speed chosen such that it freefalls out and back, returning to the surface at event e2. This clock has traveled along a geodesic between e1 and e2, so it's elapsed time corresponds to method (1). But (1) will not agree with (2). In fact, you can arrange a scenario where two different geodesic paths connect the same two events e1 and e2 such that the ''lengths'' of the two geodesics between e1 and e2 are different. So, method (1) doesn't necessarily lead to a unique result. Tom
From: Bill Hobba on 27 Aug 2006 23:47
"Edward Green" <spamspamspam3(a)netzero.com> wrote in message news:1156695713.871863.29260(a)m73g2000cwd.googlegroups.com... > Bill Hobba wrote: > >> "Edward Green" <spamspamspam3(a)netzero.com> wrote in message >> news:1156542691.425027.214860(a)75g2000cwc.googlegroups.com... > >> > You are merely being insulting now. >> >> If you thought that then I apologize. You have done nothing to deserve >> insults. > > Thank you. > >> > >> > Obviously a manifold is a mathematical abstraction. Obviously any word >> > normally applied to material objects tentatively when applied to a >> > mathematical abstraction should be understood to refer to a putative >> > property of the abstraction analogous to the named material property. >> >> That is not so obvious to me. > > Possibly it should be. If I ask you if a manifold can be twisted, does > it mean I can't distinguish mathematical from physical objects, or does > it mean I want to know whether any property of the manifold is > analogous to twisting a material body? No because twist is a well defined topological property with a well accepted meaning in physics and mathematics. Stress however does not have such a generaly accepted interpretation. Physics is not a game of definitions. Unless you are uber experienced then in any endeavor it is best to stick to the standard usage of words before venturing your own. Thanks Bill > As a courtesy, please assume > enough background on my part to suppose the latter. > >> Have you considered your musings are better suited to a philosophy >> forum? > > Sigh. I guess it just comes naturally then. > |