Any coordinate system in GR?
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From: Daryl McCullough on 28 Aug 2006 18:09 Koobee Wublee says... >JanPB wrote: > >> No, it doesn't. I have just CALCULATED it from YOUR OWN metric. If you >> want to claim otherwise, you have to find an error in my calculation >> (you can't). > >The metric indicates a distortion in space. Thus, after computing the >area in curved space, the radius does not necessarily follow the >computation in flat space. Your mistake is to mix curved space with >flat space. How about telling the truth, Koobee? You had no idea how to calculate area from the metric prior to Jan's explaining it to you. Rather than honestly saying "Thanks for explaining that to me, Jan", you instead pretended that *Jan* was the one who was confused. Jan didn't make a mistake---*YOU* did. -- Daryl McCullough Ithaca, NY
From: Koobee Wublee on 29 Aug 2006 00:46 After a few days of fastening these cows, it is time for slaughter. The cows are ** Mr. Bielawski, a known crackpot and a master Voodoo mathematician ** Mr. McCullogh, also a known crackpot and a Voodoo mathematician want-to-be Given the following metrics, ** 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 want-to-be suggested the following. "Daryl McCullough" <stevendaryl3016(a)yahoo.com> wrote in message news:ecs3oe0a7t(a)drn.newsguy.com... > > 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. The master smelled blood and came in for the kill by adding the following. "JanPB" <filmart(a)gmail.com> wrote in message news:1156719174.925230.246280(a)75g2000cwc.googlegroups.com... > > 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. After a little fastening up, the want-to-be boldly suggests the following. "Daryl McCullough" <stevendaryl3016(a)yahoo.com> wrote in message news:ecuq5k0j38(a)drn.newsguy.com... > > 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*. And after more fluttering, the want-to-be felt ever more confident also came in for the kill. "Daryl McCullough" <stevendaryl3016(a)yahoo.com> wrote in message news:ecvpjk0apq(a)drn.newsguy.com... > > How about telling the truth, Koobee? You had no idea > how to calculate area from the metric prior to Jan's > explaining it to you. Rather than honestly saying > "Thanks for explaining that to me, Jan", you instead > pretended that *Jan* was the one who was confused. > > Jan [the master] didn't make a mistake---*YOU* did. These two gentlemen fail to understand what the metric actually represents. It is thus necessary to explain what the metric really represents. Given the ubiquitously familiar Schwarzschild metric as described in the very second equation, dt, dr, and dO (longitudinal and latitudinal information) are the observed quantities. In GR, the observed quantities are what actually are. The role of the metric is to measure how much distortion in the observed quantities from the non-distorted ones. Thus, what the want-to-be suggested regarding the surface area of a sphere in a distorted space where (r = 0) resides right at the center of the gravitating body is totally wrong. The observed surface area of such sphere is still always (4 pi R^2) not (4 pi (R + K)^2) where R is the observed radius. Mr. McCullough needs to return his degree and ask for a complete refund for his tuition because of this embarrassment. Mr. Bielawski after receiving a somewhat higher degree in math should stay where he is that is watching second-rated films for a living. Minds practicing Voodoo math need to allocate wasteful time to vent their nonsense. In the meantime, the Voodoo mathematical minds still need to answer and address the following. "Koobee Wublee" <koobee.wublee(a)gmail.com> wrote in messa
From: Tom Roberts on 29 Aug 2006 01:55 I.Vecchi wrote: > One more consideration. A clock is, by its very nature, a massive, > extended object. Beyond the horizon no information can be tranferred > from the bottom to the top of the clock. Yes a clock has finite height. Let me assume it remains a clock as it passes the horizon (that is, it is not ripped apart by an attempt to hover the upper part of the clock above the horizon while slowly "dipping" the lower part into the horizon). As soon as the lower edge of the clock reaches the horizon, the horizon zooms past the clock at the speed of light, so there is no issue of transferring information from inside to outside the horizon -- the top will be inside as soon as any such information can reach it. The horizon always moves with speed c relative to ANY locally inertial frame valid at the horizon. All such frames are infalling of course. Yes, this means that in principle an outgoing light pulse could remain at the horizon forever; in practice this is an unstable equilibrium. > It is not clear to me that any > clock would keep working under such conditions. As long as the clock is not ripped apart, it is not possible to know when it passes the horizon by any local measurement; the operation of the clock is, of course, completely local. This is true for any freefalling clock, and for any clock that is accelerated within its specification. Tom Roberts
From: Sorcerer on 29 Aug 2006 04:42 "Koobee Wublee" <koobee.wublee(a)gmail.com> wrote in message news:1156826811.550173.60410(a)b28g2000cwb.googlegroups.com... | After a few days of fastening these cows, it is time for slaughter. ROFL! According to Bielawski, Relativity works because An error in Relativity "would be like Stephen Hawking dividing by zero or something equally trivial." -- Bielawski. It's WAY too simple-minded.-- Bielawski. "would have been caught immediately by the AdP reviewer." -- Bielawski. It is boring to find out how far Bielawski can be pushed into irrelevancies and nonsense. news:1127968741.095065.269300(a)o13g2000cwo.googlegroups.com Actually, it is like Albert Einstein dividing by zero: http://www.androcles01.pwp.blueyonder.co.uk/DominoEffect.GIF Androcles. | The cows are | | ** Mr. Bielawski, a known crackpot and a master Voodoo mathematician | | ** Mr. McCullogh, also a known crackpot and a Voodoo mathematician | want-to-be | | Given the following metrics, | | ** 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 want-to-be suggested the following. | | "Daryl McCullough" <stevendaryl3016(a)yahoo.com> wrote in message | news:ecs3oe0a7t(a)drn.newsguy.com... | > | > 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. | | The master smelled blood and came in for the kill by adding the | following. | | "JanPB" <filmart(a)gmail.com> wrote in message | news:1156719174.925230.246280(a)75g2000cwc.googlegroups.com... | > | > 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. | | After a little fastening up, the want-to-be boldly suggests the | following. | | "Daryl McCullough" <stevendaryl3016(a)yahoo.com> wrote in message | news:ecuq5k0j38(a)drn.newsguy.com... | > | > 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*. | | And after more fluttering, the want-to-be felt ever more confident also | came in for the kill. | | "Daryl McCullough" <stevendaryl3016(a)yahoo.com> wrote in message | news:ecvpjk0apq(a)drn.newsguy.com... | > | > How about telling the truth, Koobee? You had no idea | > how to calculate area from the metric prior to Jan's | > explaining it to you. Rather than honestly saying | > "Thanks for explaining that to me, Jan", you instead | > pretended that *Jan* was the one who was confused. | > | > Jan [the master] didn't make a mistake---*YOU* did. | | These two gentlemen fail to understand what the metric actually | represents. It is thus necessary to explain what the metric really | represents. Given the ubiquitously familiar Schwarzschild metric as | described in the very second equation, dt, dr, and
From: Daryl McCullough on 29 Aug 2006 10:50
Koobee Wublee says... [stuff deleted] I'm sorry if what I wrote went over your head. I tried to make things as simple as possible, but I see that you don't understand what a "metric" is, so we should back up a little. If x and y are Cartesian coordinates, then we can easily compute the length of a line segment in terms of the coordinates of its endpoints as follows: If the line segment runs from (x0,y0) to (x1,y1), then the length is given by the Pythagorean theorem as follows: L = square-root(L_x^2 + L_y^2) where L_x = (x1 - x0) and L_y = (y1 - y0). Now, suppose that instead of a straight line, we are trying to compute the length of a *curve*. We can approximate the length of the curve by breaking it up into tiny segments, using Pythagoras to approximate the length of each segment, and then summing the lengths to get the length of the entire curve. In the limit as the number of segments becomes very large, the result will be the length of the curve. Let (x_i,y_i) be the starting point of segment number i. The endpoint of segment number i will be the starting point of segment number i+1: (x_{i+1}, y_{i+1}). The length of each segment will be approximated by L_i = square-root(delta-x_i^2 + delta-y_i^2) where delta-x_i = (x_{i+1} - x_i) and delta-y_i = (y_{i+1} - y_i). So we can approximate the length of the entire curve by the sum L = sum over i of L_i = sum over i of square-root(delta-x_i^2 + delta-y_i^2) In the limit as the number of segments gets very large, this becomes an integral: L = integral along the curve of square-root(dx^2 + dy^2) The combination dx^2 + dy^2 is called the "line element" and is usually written ds^2. So for Euclidean geometry with cartesian coordinates, we have ds^2 = dx^2 + dy^2 Now, what happens when we use non-Cartesian coordinates? For example, what if, instead of using x and y, we use r and theta, defined indirectly via x = r cos(theta) y = r sin(theta) Well, what we need to do is to rewrite the line element ds^2 in terms of new coordinates r and theta. We can easily compute: dx = dr cos(theta) - r sin(theta) dtheta dy = dr sin(theta) + r cos(theta) dtheta ds^2 = dx^2 + dy^2 = dr^2 cos^2(theta) - 2 dr dtheta cos(theta) sin(theta) + r^2 sin^2(theta) dtheta^2 + dr^2 sin^2(theta) + 2 dr dtheta cos(theta) sin(theta) + r^2 cos^2(theta) dtheta^2 = (dr^2 + r^2 dtheta^2) (sin^2(theta) + cos^2(theta)) = dr^2 + r^2 dtheta^2 So in order to compute the length of a curve using r and theta, we use: L = Integral along the curve of square-root(dr^2 + r^2 dtheta^2) If we want to compute L using any coordinates whatsoever, then we can use the metric g to write it as L = Integral along the curve of square-root(sum over j, k of g_jk dx^j dx^k) That's the *fundamental* use of the metric: to compute lengths of curves. The metric doesn't have anything, a priori, to do with gravity or relativity or velocity. -- Daryl McCullough Ithaca, NY |