Any coordinate system in GR?
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From: Koobee Wublee on 28 Aug 2006 01:14 Daryl McCullough wrote: > 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. 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. Is there any reason to avoid my comment other than this puny introduction from my previous post? In case if you have a short memory, here it is .... 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 28 Aug 2006 01:22 JanPB wrote: > 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. > > [...] The metric specifies how the 2nd derivative of spatial components are going to behave in which there is no definitive saying on the integrated result. The surface area of a sphere in this case still remains to be (4 pi R^2). However, the trajectory is a different matter. This is not funny! This is serious stuff. You clowns trying to make this issue into funny matter to confuse everyone else is absolutely sinful.
From: JanPB on 28 Aug 2006 01:46 Koobee Wublee wrote: > JanPB wrote: > > 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. > > > > [...] > > The metric specifies how the 2nd derivative of spatial components are > going to behave in which there is no definitive saying on the > integrated result. Words words words. > 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? -- Jan Bielawski
From: Koobee Wublee on 28 Aug 2006 02:28 JanPB wrote: > Koobee Wublee wrote: > > JanPB wrote: > > > 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. > > > > > > [...] > > > > The metric specifies how the 2nd derivative of spatial components are > > going to behave in which there is no definitive saying on the > > integrated result. > > Words words words. And mathematics between the words. > > 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). Your calculation is based on something that only applies to velocity terms which is grossly inapproppriate and faulty in nature. So, please act like an adult for a change and stop whining about spilled milk.
From: JanPB on 28 Aug 2006 03:04
Koobee Wublee wrote: > JanPB wrote: > > Koobee Wublee wrote: > > > JanPB wrote: > > > > 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. > > > > > > > > [...] > > > > > > The metric specifies how the 2nd derivative of spatial components are > > > going to behave in which there is no definitive saying on the > > > integrated result. > > > > Words words words. > > And mathematics between the words. What "mathematics"? After I CALCULATED the sphere area you said "The metric specifies how the 2nd derivative of spatial components are going to behave in which there is no definitive saying on the integrated result" - which is poetry and has nothing to do with the area of the sphere in question. > > > 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). 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). > Your calculation is based on > something that only applies to velocity terms which is grossly > inapproppriate and faulty in nature. There is no velocity there. There is just a metric you yourself wrote - all I've done is calculated the area of the sphere r=const. according to your metric. > So, please act like an adult for a change and stop whining about spilled milk. I prefer facts to rethoric. -- Jan Bielawski |