The extra area on the Sun is caused by its curved extension. In other
words its extenison is in curved space. It has an extra area of about
the size of the USA.

On Dec 13, 1:03 am, BURT wrote:
The extra area on the Sun is caused by its curved extension. In other
words its extenison is in curved space. It has an extra area of about
the size of the USA.

Geeze. This isn't even wrong. Extra area relative to what?
Curved extension? Make sense there Skippy.
Socks

On Dec 12, 10:03 pm, BURT wrote:

The extra area on the Sun is caused by its curved extension. In other
words its extenison is in curved space. It has an extra area of about
the size of the USA.

The surface area of the sun relative to an observer using the common
spherically symmetric polar coordinate system always would be the
following regardless how much space or spacetime is curved around the
sun.

A = 4 pi R^2

Where

** A = Observed surface area of the sun
** R = Observed radius of the sun

This is because when an observer does an observation, he would always
default the metric that would result in flat geometry. If the
geometry is "curved", then using this flat metric would result in
distorted observation. However, if you know exactly how space or
spacetime is curved around the sun, then you can use the metric that
would result in distortion-free observation. Also, remember that
looking into the curved space or spacetime, there is no way in hell
that you can determine directly what the curvature is.

I hope you can understand this because this very simple concept still
eludes all the PhDs in physics. shrug

On Dec 13, 11:04 pm, Koobee Wublee wrote:
On Dec 12, 10:03 pm, BURT wrote:

The extra area on the Sun is caused by its curved extension. In other
words its extenison is in curved space. It has an extra area of about
the size of the USA.

The surface area of the sun relative to an observer using the common
spherically symmetric polar coordinate system always would be the
following regardless how much space or spacetime is curved around the
sun.

A = 4 pi R^2

Where

** A = Observed surface area of the sun
** R = Observed radius of the sun

This is because when an observer does an observation, he would always
default the metric that would result in flat geometry. If the
geometry is "curved", then using this flat metric would result in
distorted observation. However, if you know exactly how space or
spacetime is curved around the sun, then you can use the metric that
would result in distortion-free observation. Also, remember that
looking into the curved space or spacetime, there is no way in hell
that you can determine directly what the curvature is.

I hope you can understand this because this very simple concept still
eludes all the PhDs in physics. shrug

Yet you still can't prove that the area of a sphere is 4piR^2...

On Dec 14, 3:04 am, Koobee Wublee wrote:
On Dec 12, 10:03 pm, BURT wrote:

The extra area on the Sun is caused by its curved extension. In other
words its extenison is in curved space. It has an extra area of about
the size of the USA.

The surface area of the sun relative to an observer using the common
spherically symmetric polar coordinate system always would be the
following regardless how much space or spacetime is curved around the
sun.

A = 4 pi R^2

Where

** A = Observed surface area of the sun
** R = Observed radius of the sun

This is because when an observer does an observation, he would always
default the metric that would result in flat geometry. If the
geometry is "curved", then using this flat metric would result in
distorted observation. However, if you know exactly how space or
spacetime is curved around the sun, then you can use the metric that
would result in distortion-free observation. Also, remember that
looking into the curved space or spacetime, there is no way in hell
that you can determine directly what the curvature is.

I hope you can understand this because this very simple concept still
eludes all the PhDs in physics. shrug

So, you have a ready estimate of the total effect of the curvature
of spacetime around the sun on the area of the sun, yes?

For example, you can tell me the radius of curvature near the
surface of the sun, yes?
Socks

On Dec 14, 8:22 am, Puppet_Sock wrote:
On Dec 14, 3:04 am, Koobee Wublee wrote:





On Dec 12, 10:03 pm, BURT wrote:

The extra area on the Sun is caused by its curved extension. In other
words its extenison is in curved space. It has an extra area of about
the size of the USA.

The surface area of the sun relative to an observer using the common
spherically symmetric polar coordinate system always would be the
following regardless how much space or spacetime is curved around the
sun.

A = 4 pi R^2

Where

** A = Observed surface area of the sun
** R = Observed radius of the sun

This is because when an observer does an observation, he would always
default the metric that would result in flat geometry. If the
geometry is "curved", then using this flat metric would result in
distorted observation. However, if you know exactly how space or
spacetime is curved around the sun, then you can use the metric that
would result in distortion-free observation. Also, remember that
looking into the curved space or spacetime, there is no way in hell
that you can determine directly what the curvature is.

I hope you can understand this because this very simple concept still
eludes all the PhDs in physics. shrug

So, you have a ready estimate of the total effect of the curvature
of spacetime around the sun on the area of the sun, yes?

For example, you can tell me the radius of curvature near the
surface of the sun, yes?
Socks- Hide quoted text -

- Show quoted text -

Spherical Curved extension of matter.

Mitch Raemsch

On Dec 14, 8:22 am, Puppet_Sock wrote:
On Dec 14, 3:04 am, Koobee Wublee wrote:

The surface area of the sun relative to an observer using the common
spherically symmetric polar coordinate system always would be the
following regardless how much space or spacetime is curved around the
sun.

A = 4 pi R^2

Where

** A = Observed surface area of the sun
** R = Observed radius of the sun

This is because when an observer does an observation, he would always
default the metric that would result in flat geometry. If the
geometry is "curved", then using this flat metric would result in
distorted observation. However, if you know exactly how space or
spacetime is curved around the sun, then you can use the metric that
would result in distortion-free observation. Also, remember that
looking into the curved space or spacetime, there is no way in hell
that you can determine directly what the curvature is.

I hope you can understand this because this very simple concept still
eludes all the PhDs in physics. shrug

So, you have a ready estimate of the total effect of the curvature
of spacetime around the sun on the area of the sun, yes?

This is absolutely true if I know the actual metric to observe the
geometry without any distortion through the coordinate system I have
chosen well in advance. shrug

For example, you can tell me the radius of curvature near the
surface of the sun, yes?

Again, with my chosen coordinate system, I need to know the metric to
compute the actual geometry. Without it, I cannot determine what the
actual metric is by doing a direct observation. Thus, I would tend to
use the metric only applied in flat space that I know so well. This
should be a very fundamental concept, no?

On Fri, 14 Dec 2007 23:31:21 -0800 (PST), Koobee Wublee
wrote:

[...]

Again, with my chosen coordinate system, I need to know the metric to
compute the actual geometry. Without it, I cannot determine what the
actual metric is by doing a direct observation. Thus, I would tend to
use the metric only applied in flat space that I know so well. This
should be a very fundamental concept, no?

If you know it so well, how come you are incapable of computing
anything with it?

On Dec 15, 1:02 am, Eric Gisse wrote:
On Fri, 14 Dec 2007, Koobee Wublee wrote:

Again, with my chosen coordinate system, I need to know the metric to
compute the actual geometry. Without it, I cannot determine what the
actual metric is by doing a direct observation. Thus, I would tend to
use the metric only applied in flat space that I know so well. This
should be a very fundamental concept, no?

If you know it so well, how come you are incapable of computing
anything with it?

You just don't understand the basics, and that is handicapping your
claims. shrug

On Sat, 15 Dec 2007 22:50:16 -0800 (PST), Koobee Wublee
wrote:

On Dec 15, 1:02 am, Eric Gisse wrote:
On Fri, 14 Dec 2007, Koobee Wublee wrote:

Again, with my chosen coordinate system, I need to know the metric to
compute the actual geometry. Without it, I cannot determine what the
actual metric is by doing a direct observation. Thus, I would tend to
use the metric only applied in flat space that I know so well. This
should be a very fundamental concept, no?

If you know it so well, how come you are incapable of computing
anything with it?

You just don't understand the basics, and that is handicapping your
claims. shrug

Ok.

Show us how to _compute_ an area in a manifold given the metric.