| Topic: |
Science > Physics |
| User: |
"BURT" |
| Date: |
13 Dec 2007 12:03:11 AM |
| Object: |
Area on the Sun |
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.
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| User: "Koobee Wublee" |
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| Title: Re: Area on the Sun |
14 Dec 2007 02:04:45 AM |
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On Dec 12, 10:03 pm, BURT <macromi...@yahoo.com> 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>
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| User: "Eric Gisse" |
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| Title: Re: Area on the Sun |
14 Dec 2007 03:22:06 AM |
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On Dec 13, 11:04 pm, Koobee Wublee <koobee.wub...@gmail.com> wrote:
On Dec 12, 10:03 pm, BURT <macromi...@yahoo.com> 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...
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| User: "Puppet_Sock" |
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| Title: Re: Area on the Sun |
14 Dec 2007 10:22:28 AM |
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On Dec 14, 3:04 am, Koobee Wublee <koobee.wub...@gmail.com> wrote:
On Dec 12, 10:03 pm, BURT <macromi...@yahoo.com> 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
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| User: "BURT" |
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| Title: Re: Area on the Sun |
14 Dec 2007 07:55:48 PM |
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On Dec 14, 8:22 am, Puppet_Sock <puppet_s...@hotmail.com> wrote:
On Dec 14, 3:04 am, Koobee Wublee <koobee.wub...@gmail.com> wrote:
On Dec 12, 10:03 pm, BURT <macromi...@yahoo.com> 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
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| User: "Koobee Wublee" |
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| Title: Re: Area on the Sun |
15 Dec 2007 01:31:21 AM |
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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?
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| User: "Eric Gisse" |
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| Title: Re: Area on the Sun |
15 Dec 2007 03:02:05 AM |
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On Fri, 14 Dec 2007 23:31:21 -0800 (PST), Koobee Wublee
<koobee.wublee@gmail.com> 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?
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| User: "Koobee Wublee" |
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| Title: Re: Area on the Sun |
16 Dec 2007 12:50:16 AM |
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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>
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| User: "Puppet_Sock" |
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| Title: Re: Area on the Sun |
17 Dec 2007 09:40:51 AM |
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On Dec 16, 1:50 am, Koobee Wublee <koobee.wub...@gmail.com> 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>
Who is it that does not understand the basics? I got A+ in all
of the classes I took in differential geometry. Plus the classes
I took in relativity. And then there's the classes I *taught* in
relativity.
If you think you can't determine the metric by observation, then
you don't understand what a metric is.
It should be a fundamental concept. It seems to be eluding you.
Socks
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| User: "Koobee Wublee" |
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| Title: Re: Area on the Sun |
17 Dec 2007 02:44:06 PM |
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On Dec 17, 7:40 am, Puppet_Sock wrote:
Who is it that does not understand the basics? I got A+ in all
of the classes I took in differential geometry. Plus the classes
I took in relativity. And then there's the classes I *taught* in
relativity.
If you think you can't determine the metric by observation, then
you don't understand what a metric is.
Oh, dear. We have another nut case here who thinks he knows how space
is curving by directly looking into that curved space. This nut case
is also bragging about getting exceptional high grade in differential
geometry. <shrug>
It should be a fundamental concept. It seems to be eluding you.
As a nut case, the only audience would probably be squirrels and
rodents in your backyard. Yes, they understand relativity very well
--- better than you do. <shrug>
Why the subject of relativity always brings out these nuts? Where did
they come from?
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| User: "Androcles" |
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| Title: Re: Area on the Sun |
17 Dec 2007 03:13:04 PM |
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"Koobee Wublee" <koobee.wublee@gmail.com> wrote in message
news:01e04c30-82c9-49c8-b693-3a32eaa1344d@e6g2000prf.googlegroups.com...
: On Dec 17, 7:40 am, Puppet_Sock wrote:
:
: > Who is it that does not understand the basics? I got A+ in all
: > of the classes I took in differential geometry. Plus the classes
: > I took in relativity. And then there's the classes I *taught* in
: > relativity.
: >
: > If you think you can't determine the metric by observation, then
: > you don't understand what a metric is.
:
: Oh, dear. We have another nut case here
Who, you?
We know that, you are a stupid aetherialist, just as nuts as any relativist.
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| User: "Eric Gisse" |
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| Title: Re: Area on the Sun |
16 Dec 2007 02:15:30 AM |
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On Sat, 15 Dec 2007 22:50:16 -0800 (PST), Koobee Wublee
<koobee.wublee@gmail.com> 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.
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| User: "Puppet_Sock" |
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| Title: Re: Area on the Sun |
13 Dec 2007 12:34:28 PM |
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On Dec 13, 1:03 am, BURT <macromi...@yahoo.com> 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
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