Any coordinate system in GR?
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From: Daryl McCullough on 28 Aug 2006 08:34 Tom S. says... >"Daryl McCullough" <stevendaryl3016(a)yahoo.com> wrote >> 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). That's true. I think, though, that if e1 and e2 are very close together in time, then the difference between the result of an inertial clock and a noninertial clock becomes negligible. -- Daryl McCullough Ithaca, NY
From: Daryl McCullough on 28 Aug 2006 09:13 Koobee Wublee says... >I do. Your assumption is very faulty in the very start. The surface >area of a sphere remains (4 pi R^2). In *Euclidean* geometry, but not in curved space. Distances have to be computed using the metric. Areas are computed using distances. In curved 2-D, if you have two coordinates x and y (assumed to be orthogonal) and you look at the rectangle with corners (x,y) (x+delta_x,y) (x+delta_x,y+delta_y) (x,y+delta_y) what is the area of the rectangle? Well, the length of the sides are *not* delta_x and delta_y. To compute lengths in curved space, you have to use the metric. The length of the sides are square-root(|g_xx|) delta_x and square-root(|g_yy| delta_y). The area of that rectangle is square-root(|g_xx| |g_yy|) delta_x delta_y The area of a sphere is calculated using exactly this principle. In the case of a sphere, instead of using x and y, you use theta and phi. So the area of a tiny patch on the surface of a sphere is given by square-root(|g_theta,theta| |g_phi,phi|) delta_theta delta_theta In Euclidean geometry with the usual coordinates, g_theta,theta = R^2 g_phi,phi = R^2 sin^2(theta) So the area of the little patch is square-root(R^2 * R^2 sin^2(theta)) delta_theta delta_theta = R^2 sin(theta) delta_theta delta_phi >Your calculation is based on something that only applies to >velocity terms You are confused about the meaning of "metric". The metric tells how to measure proper distances and proper times. It doesn't have anything to do with velocity. Once again, look at the surface of a sphere. Suppose you have two points on the sphere. The first point has coordinates (theta=pi/4, phi=0). The second point has coordinates (theta=pi/4, phi=pi). What is the distance between these two points along the curve theta=pi/4? To compute distances, you use the metric: ds^2 = g_uv dx^u dx^v Since the only coordinate that is varying is phi, this simplifies to ds^2 = g_phi,phi dphi^2 So ds = square-root(g_phi,phi) dphi g_phi,phi for spherical coordinates is R^2 sin^2(theta). So ds = R sin(theta) dphi In our case, theta = pi/4, so sin(theta) = 1/square-root(2). So ds= R/square-root(2) dphi Integrating from phi=0 to phi=pi gives s = R/square-root(2) pi The metric is primarily about calculating *distances* and *times*. -- Daryl McCullough Ithaca, NY
From: Daryl McCullough on 28 Aug 2006 09:20 Koobee Wublee says... >Daryl McCullough wrote: >> >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. > >No, as you have written the spacetime is its incremental discrete form, >you are actually referring to its change of line element of spacetime >which translates to velocity. This (r + K) term only affect the >velocity. Nice try but no cigar. We're not talking about velocity. Do you understand what a metric is? It doesn't have anything to do with velocity. The metric is used for computing *distances*. To compute the area, you have to know distances. To compute distances, you have to use the metric. So you have to use the metric to compute areas. >Is there any reason to avoid my comment other than this puny >introduction from my previous post? I answered several times: The area of a sphere using your metric is given by 4pi(r+K)^2, thus showing that the relevant radius is r+K, not r. -- Daryl McCullough Ithaca, NY
From: Daryl McCullough on 28 Aug 2006 11:16 Koobee Wublee says... >JanPB wrote: >> > The surface area of a sphere in this case still >> > remains to be (4 pi R^2). >> >> It's 4 pi (R+K)^2 - I have just CALCULATED it from YOUR OWN metric. Do >> you know calculus? > >I do. Your assumption is very faulty in the very start. The surface >area of a sphere remains (4 pi R^2). At this point, the disagreement has nothing to do with physics or relativity, but is a purely mathematical question. Given a two-dimensional surface described by coordinates u and v, what is the area enclosed by the region 0 < u < pi 0 < v < 2 pi Can you answer that question without knowing anything about the coordinates u and v? Of course not. If u and v are cartesian coordinates, then the area will be 2 pi^2 If the surface is a flat 2-D plane, and u is the radial coordinate and v is the angle (in radians) then the area will be pi^2 If the surface is the surface of a sphere of radius R in flat Euclidean 3D space, and u and v are the colatitude and azimuth, respectively, then the area is 4 pi R^2 If the surface is the surface of a cylinder of radius R, and u is the distance along the axis and v is the angle around the circumference, then the area is 2 pi^2 You can't compute the area without knowing more information about the coordinates u and v. Where does that additional information come from? In Riemannian geometry, that additional information is found in the *metric*. The distinction between spherical coordinates and rectangular coordinates is given by the metric coefficients. You seem to think that the information is conveyed by the *names* of the coordinate. If the names are r and theta, then we are using polar coordinates, and if the names are theta and phi, then we are using spherical coordinates, and if the names are x and y, then we are using Cartesian coordinates. However, that's just a convention. You can name a coordinate anything you like. You have to look at the geometry (which is described by the metric). -- Daryl McCullough Ithaca, NY
From: Ken S. Tucker on 28 Aug 2006 12:55
JanPB wrote: > 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... That's the basis of the Schwarz. Solution, Weinberg's "Grav&Cosmo" pg. 181. It is how energy is *stored* in the spacetime field. > > > My point was that at this stage this is really the ODE theory, hence > > > the uniqueness. > > > > What is ODE? > > Ordinary differential equations. Thanks Ken |