Co-ordinate Time Vs. Real Time

**Mitchell**

"Franz Heymann" wrote in message ...
"Mitchell" wrote in message
om...
Bill Rowe wrote in message
...
In article ,
(Mitchell) wrote:

Bill Rowe wrote in message

...

big snip

The lack of a valid description (failure!) in a given
coordinate
system does not mean there can be no valid description in some
other
coordinate system. Nor does this lack in any way imply there
must be a
physical singularity.

No the invariance must be expressed through the coordinate
system Bill.
You must be able to transform from one to another without
changing the results(space-time curvature). That would spell
invariance.
You can't juggle them about willy nilly Bill.

It seems quite clear you've little understanding of the terms you
are
using. Forget relativity for the moment and consider coordinates
in a
Euclidean plane.

Start with some coordinate system and choose two points (X1, Y1)
and
(X2, Y2). Now rotate the coordinate axes by some angle theta, to
get
new axes.

In the new system, the coordinates previously called (X1, Y1) will
have
values ( X1 cos[theta] + Y1 sin[theta], -X1 sin[theta] + Y1
cos[theta]).. That is in the new coordinate system formed by
rotating
the axes, the coordinates have different values, are not
invariant.

But consider the distance between the points in both systems.

In the (x, y) system the distance between the points is:

sqrt[(X1-X2)^2 + (Y1-Y2)^2]

In the rotated system, the distance is given by

sqrt[
((X1 cos[theta] + Y1 sin[theta]) -(X2 cos[theta] +Y2
sin[theta]))^2 +
((-X1 sin[theta] + Y1 cos[theta]) - (-X2 sin[theta] + Y2
cos[theta]))^2
]

Expanding just the first term inside the square root function
results in:

(X1 cos[theta] + Y1 sin[theta] - X2 cos[theta] - Y2 sin[theta]) ^2

which is

(cos [theta]( X1 - X2) + sin[theta]( Y1- Y2))^2

which is

cos^2[theta](X1-X2)^2 + sin^2[theta](Y1-Y2)^2 +
2sin[theta]cos[theta](X1-X2)(Y1-Y2)

A similar expansion of the second term inside the square root
results in

sin^2[theta](X1-X2)^2 + cos^2[theta](Y1-Y2)^2 -
2sin[theta]cos[theta](X1-X2)(Y1-Y2)

adding these together results in

cos^2[theta](X1-X2)^2 + sin^2[theta](Y1-Y2)^2 +
sin^2[theta](X1-X2)^2 + cos^2[theta](Y1-Y2)^2

which is

(X1-X2)^2 (cos^2 + sin^2) + (Y1-Y2)^2 (cos^2 + sin^2)

But for any angle sin^2 + cos^2 = 1

So, the last result is nothing more than

(X1 - X2)^2 + (Y1 - Y2)^2. But this is the same expression inside
the
square root when computing distance in the original (x,y) system.
That
is distance is invariant.

The point of this exercise is to show something (distance)
computed in
two coordinate systems is invariant even though the coordinates
themselves are not invariant. In fact, the concept of invariance
applies
to something computed from coordinates not the coordinates
themselves.

Note in this example, the coordinates do not represent the
invariant
thing (distance). The coordinates are merely convenient labels put
on
the points to be used in the computation of distance. The same is
true
of invariance as used in relativity.

"The patterns of space-time curvature around mass are absolute."
Einstein

I would interpret the Einstein quote your provided as saying the
Einstein tensor (patters of spacetime curvature) is invariant
(absolute). And this is consistent with my comments about
coordinates
and coordinate systems. However, with a single quote taken out of
context it is quite difficult to know whether I've interpreted the
quote
correctly. In fact, quotes taken out of context are generally
meaningless and often prone to significant misinterpretation.

That is the invariance I am talking about.

Your previous comments indicate you really don't understand the
concept
of invariance or you have some totally non-standard definition in
mind.

I believe that the coordinate systems used to describe gravity
and black holes must be invariant.

That is a bloody silly belief for starters.
Go and learn something about coordinates and the concept of invariance
when converting from one set of coordinates and another.

Something has to give.
In one coordinate system (Schwarzschild) there is an end to
time(time coordinate singularity) and in another(Kruskal)
there isn't. Remember that matter is falling at light speed
there and that implies an end to time. Just like the schwarzschlid
time coordinate singularity describes.

How do we know what is right if coordiante systems are arbitrary
and yield different results?

If two coordinate systems are not distinguishable by virtue of the
fact that theyassign different values to some physical quantities,
they must be identical coordinate systems.

You know sweet fanny adams.

Franz

Time can't be one of them Franz. It has to be invariant in gravity.

"The patterns of curved space-time around mass are absolute." Albert Einstein

Mitch Raemsch
-- Light Falls --

**Mitchell**

Bill Rowe wrote in message ...
In article ,
(Mitchell) wrote:

I believe that the coordinate systems used to describe gravity
and black holes must be invariant.

A coordinate system is an arbitrary system invented by the analyst to
describe something such as gravity. It makes no sense to describe a
coordinate system as invariant.
That is the problem Bill. If time coordinates yield different
results then they are not up to the job. We can longer be sure we know
anything if the coordinates don't represent time invariance.
Coordinate time(representing gravity) and proper time must be the
same.
Proper time in gravity is absolute.

In one coordinate system (Schwarzschild) there is an end to
time(time coordinate singularity) and in another(Kruskal)
there isn't.

So? The fact there is a singularity in one coordinate system and not
another simply means the singularity isn't physical. It is simply an
artifact of the choice of coordinates.

Wrong. Proper time and coordinate time coincide. Otherwise coordinate
time would be useless; with no certainty.

How do we know what is right if coordiante systems are arbitrary
and yield different results?

If the computation is done for an invariant quantity and it is done
correctly, the results cannot and will not be different. If the
computation is done correctly and you get different results, then the
thing being computed isn't invariant. In this case both results are
correct *in the specific coordinate system* the computation was made.

You can't have it both ways.

**Bill Rowe**

In article ,
(Mitchell) wrote:

snip

We are going in circles. It is quite clear you've little if any
understanding of the terms you are using. Or you are using them in some
non-standard way which you've not shared.

It also seems clear, my attempts to explain why coordinate singularities
are of no specific significance are inadequate to improve your
understanding. Possibly, I simply lack the ability of providing an
explanation you can understand. But since we do seem to be going in
circles with no apparent increase in your understanding, it seems rather
pointless to continue.

--
To reply via email subtract one hundred nine

**Franz Heymann**

"Mitchell" wrote in message
om...
Bill Rowe wrote in message
...
In article ,
(Mitchell) wrote:

I believe that the coordinate systems used to describe gravity
and black holes must be invariant.

A coordinate system is an arbitrary system invented by the analyst
to
describe something such as gravity. It makes no sense to describe
a
coordinate system as invariant.

In one coordinate system (Schwarzschild) there is an end to
time(time coordinate singularity) and in another(Kruskal)
there isn't.

So? The fact there is a singularity in one coordinate system and
not
another simply means the singularity isn't physical. It is simply
an
artifact of the choice of coordinates.

That is two different results representing time.
And it is not just the coordinate singularity where they differ.
They differ on the way to the singualrity.
The description of time is completely different in both systems.

How do we know what is right if coordiante systems are arbitrary
and yield different results?

If the computation is done for an invariant quantity and it is
done
correctly, the results cannot and will not be different.
If the
computation is done correctly and you get different results, then
the
thing being computed isn't invariant. In this case both results
are
correct *in the specific coordinate system* the computation was
made.

They can't both be right.
Balls.

If I walk northwards at 1 m.p.h. on a train which is travelling
Eastwards at 1 m.p.h., you, standing on the ground will see me
walking NE with a speed of 1.414 m.p.h. So, even in terms of common
garden Galilean relativity, you are shooting your mouth off.

Twerp.

**Franz Heymann**

"Mitchell" wrote in message
om...
"Franz Heymann" wrote in message
...
"Mitchell" wrote in message
om...
Bill Rowe wrote in message

...
In article ,
(Mitchell) wrote:

Bill Rowe wrote in
message

...

big snip

The lack of a valid description (failure!) in a given
coordinate
system does not mean there can be no valid description in
some
other
coordinate system. Nor does this lack in any way imply
there
must be a
physical singularity.

No the invariance must be expressed through the coordinate
system Bill.
You must be able to transform from one to another without
changing the results(space-time curvature). That would spell
invariance.
You can't juggle them about willy nilly Bill.

It seems quite clear you've little understanding of the terms
you
are
using. Forget relativity for the moment and consider
coordinates
in a
Euclidean plane.

Start with some coordinate system and choose two points (X1,
Y1)
and
(X2, Y2). Now rotate the coordinate axes by some angle theta,
to
get
new axes.

In the new system, the coordinates previously called (X1, Y1)
will
have
values ( X1 cos[theta] + Y1 sin[theta], -X1 sin[theta] + Y1
cos[theta]).. That is in the new coordinate system formed by
rotating
the axes, the coordinates have different values, are not
invariant.

But consider the distance between the points in both systems.

In the (x, y) system the distance between the points is:

sqrt[(X1-X2)^2 + (Y1-Y2)^2]

In the rotated system, the distance is given by

sqrt[
((X1 cos[theta] + Y1 sin[theta]) -(X2 cos[theta] +Y2
sin[theta]))^2 +
((-X1 sin[theta] + Y1 cos[theta]) - (-X2 sin[theta] + Y2
cos[theta]))^2
]

Expanding just the first term inside the square root function
results in:

(X1 cos[theta] + Y1 sin[theta] - X2 cos[theta] - Y2
sin[theta]) ^2

which is

(cos [theta]( X1 - X2) + sin[theta]( Y1- Y2))^2

which is

cos^2[theta](X1-X2)^2 + sin^2[theta](Y1-Y2)^2 +
2sin[theta]cos[theta](X1-X2)(Y1-Y2)

A similar expansion of the second term inside the square root
results in

sin^2[theta](X1-X2)^2 + cos^2[theta](Y1-Y2)^2 -
2sin[theta]cos[theta](X1-X2)(Y1-Y2)

adding these together results in

cos^2[theta](X1-X2)^2 + sin^2[theta](Y1-Y2)^2 +
sin^2[theta](X1-X2)^2 + cos^2[theta](Y1-Y2)^2

which is

(X1-X2)^2 (cos^2 + sin^2) + (Y1-Y2)^2 (cos^2 + sin^2)

But for any angle sin^2 + cos^2 = 1

So, the last result is nothing more than

(X1 - X2)^2 + (Y1 - Y2)^2. But this is the same expression
inside
the
square root when computing distance in the original (x,y)
system.
That
is distance is invariant.

The point of this exercise is to show something (distance)
computed in
two coordinate systems is invariant even though the
coordinates
themselves are not invariant. In fact, the concept of
invariance
applies
to something computed from coordinates not the coordinates
themselves.

Note in this example, the coordinates do not represent the
invariant
thing (distance). The coordinates are merely convenient labels
put
on
the points to be used in the computation of distance. The same
is
true
of invariance as used in relativity.

"The patterns of space-time curvature around mass are
absolute."
Einstein

I would interpret the Einstein quote your provided as saying
the
Einstein tensor (patters of spacetime curvature) is invariant
(absolute). And this is consistent with my comments about
coordinates
and coordinate systems. However, with a single quote taken out
of
context it is quite difficult to know whether I've interpreted
the
quote
correctly. In fact, quotes taken out of context are generally
meaningless and often prone to significant misinterpretation.

That is the invariance I am talking about.

Your previous comments indicate you really don't understand
the
concept
of invariance or you have some totally non-standard definition
in
mind.

I believe that the coordinate systems used to describe gravity
and black holes must be invariant.

That is a bloody silly belief for starters.
Go and learn something about coordinates and the concept of
invariance
when converting from one set of coordinates and another.

Something has to give.
In one coordinate system (Schwarzschild) there is an end to
time(time coordinate singularity) and in another(Kruskal)
there isn't. Remember that matter is falling at light speed
there and that implies an end to time. Just like the
schwarzschlid
time coordinate singularity describes.

How do we know what is right if coordiante systems are arbitrary
and yield different results?

If two coordinate systems are not distinguishable by virtue of the
fact that theyassign different values to some physical quantities,
they must be identical coordinate systems.

You know sweet fanny adams.

Franz

Time can't be one of them Franz. It has to be invariant in gravity.

"The patterns of curved space-time around mass are absolute." Albert
Einstein

Itg is my considered opinion that you are ineducable, so let's call it
a day.

Franz

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