Any coordinate system in GR?
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From: Eric Gisse on 1 Sep 2006 00:19 JanPB wrote: [...] I wonder if he has ever even taken a calculus course that covers multiple variables. Seeing him deny, deny, deny is just FUN.
From: I.Vecchi on 1 Sep 2006 03:28 Daryl McCullough ha scritto: > 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. Yes, very nice. > > 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. > Yes, but this assumes that the parameter r corresponds to to the the radius of the black hole, right? When you write that tau_0 "depends on the initial conditons" you are making some assumptions on the time it will take to reach the singularity at r=0. This sounds circular, but you are welcome to expand on your argument and/or to provide a reference. I don't think that an infalling observer can measure the inner radius, since in order to do that he would have to reach the singularity at r=0, where the metric blows up for good. More in general, apparently your argument assumes that to an infalling observer the inner domain appears as a sphere of radius r, as it appears from the outside to external observers, who however have no access to space-time events beyond the horizon. Is there any reason to assume that
From: I.Vecchi on 1 Sep 2006 11:06 I am correcting some typos. The argument however stays the same. Daryl McCullough ha scritto: > 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. Yes, but this assumes that the parameter r corresponds to to the the radial distance inside the black hole, right? When you write that tau_0 "depends on the initial conditons" you are making some assumptions on the time it will take to reach the singularity at r=0. This sounds circular, but you are welcome to expand on your argument and/or to provide a reference. I don't think that an infalling observer can measure the inner radius, since in order to do that he would have to reach the singularity at r=0, where the metric blows up for good. If the metric becomes singular at r=0, how come the radius is finite? So, as far as I understand, your argument assumes that to an infalling observer the inner domain appears as a sphere of radius 2m, as it appears from the outsi
From: Daryl McCullough on 1 Sep 2006 12:57 I.Vecchi says... >Yes, but this assumes that the parameter r corresponds to to the the >radius of the black hole, right? When you write that tau_0 "depends on >the initial conditons" you are making some assumptions on the time it >will take to reach the singularity at r=0. No, not really. Let me switch from tau to s, because s is easier to write. The geodesic equation implies the following constraints on the motion 1. (1-2m/r) (dt/ds)^2 - 1/(1-2m/r) (dr/ds)^2 = 1 2. (1-2m/r) (dt/ds) = some constant k Using 2, we can simplify 1 to get k^2/(1-2m/r) - (dr/ds)^2/(1-2m/r) = 1 or 1'. (dr/ds)^2 + 2m/r + (1-k^2) = 0 To compute k, you need to know the initial conditions. In particular, let's suppose that the observer is initially at rest according to the Schwarzchild coordinates. That means that when s=0, we have: dr/ds = 0. In that case, letting r0 be the value of r when s=0 Then we have, at proper time s=0: (1-2m/r0) (dt/ds)^2 = 1 from equation 1 and (1-2m/r0) (dt/ds) = k from equation 2. These two together imply k = square-root(1-2m/r0) So our equation 1' above becomes (dr/ds)^2 + 2m/r + 2m/r0 = 0 This equation gives us r as a function of s and r0. >I don't think that an infalling observer can measure the >inner radius, since in order to do that he would have to reach the >singularity at r=0, where the metric blows up for good. Yes, you're right that r is not directly measurable. However, using the above equation, you can *compute* r from your proper time s. >More in general, apparently your argument assumes that to an >infalling observer the inner domain appears as a sphere of radius r, No, I didn't assume that. I just assumed that r and t are two coordinate such that the metric expressed in terms of r and t is ds^2 = -(1-2m/r) dt^2 + 1/(1-2m/r) dr^2 + angular part >Consider the following alternative scenario. Take the complete Kruskal >solution and paste together the horizons of the white and of the black >hole. I'm not sure what this means. Look at the Kruskal diagram as shown in http://io.uwinnipeg.ca/~vincent/4500.6-001/Cosmology/Black_Holes_files/image011.gif In that picture, the Kruskal "time" coordinate is v, whose axis is shown oriented vertically. The Kruskal "space" coordinate is u, whose axis is shown oriented horizontally. In terms of compass directions: The black hole singularity is to the north, the white hole singularity is to the south, the black hole exterior is to the east, and the mirror exterior is to the west. The black hole event horizon is the V-shaped boundary of the northern quadrant. It runs diagonally from the northwest corner to the center (v=0, u=0), and then turns a right angle and runs diagonally to the northeast corner. The white hole event horizon is the inverted V-shaped boundary of the southern quadrant. It runs from diagonally from the southwest corner to the center, and then diagonally from the center to the southeast corner. So which portions of the horizon are you proposing to paste together? >This is still a well-behaved extension. The infalling >observers, raising his eyes after checking on his clock that he's >passed the horizon of the black hole, would just find himself gushing >out of a white hole horizon. After browsing around I think this is >called an Einstein-Rosen bridge. As I understand it, the bridge connects the exterior in the east to the mirror exterior in the west. I don't see how it is accurate to describe someone crossing the bridge as "gushing out of the white hole horizon". -- Daryl McCullough Ithaca, NY
From: JanPB on 1 Sep 2006 15:55
I.Vecchi wrote: > > Consider the following alternative scenario. Take the complete Kruskal > solution and paste together the horizons of the white and of the black > hole. This is still a well-behaved extension. I don't see why the metric should remain smooth (at tne horizon) after such regluing. > The infalling > observers, raising his eyes after checking on his clock that he's > passed the horizon of the black hole, would just find himself gushing > out of a white hole horizon. After browsing around I think this is > called an Einstein-Rosen bridge. The Einstein-Rosen bridge is a spacelike surface (really a 3D manifold) corresponding to the lines T=const. (horizontal) on the K-S diagram. As you can see from the K-S it can be traversed only if you move faster than light. > I surmise this may be the same as a > blackhole were the singularity at r=0 is blown up to infinity > (together with tau_0) through a coordinate change for the interior > domain 2m<r<0. Repeating the above question, how do we know that this > is not the "right" chart (i.e, the chart corresponding to space-time > measurements of an infalling observer)? Not sure what you mean... Proper measurements don't change when coordinates change. -- Jan Bielawski |