[SOLVED] Proper Time in General Relativity

Ken S. Tucker

Murat Ozer wrote:
[...]
> Do you really think that I have not done that? There cannot be and
> there is not a flaw in my derivation. This is because my derivation
> is the step by step operational derivation and it cannot fail.
> The (non operational) definition dTau = ds/c works in all the cases
> but the turntable case. And one failure is enough to declare that
> the definition dTau = ds/c in GR is flawed...
>
> Murat Ozer

Mr. Ozer,
I'd hesitate to call GRT flawed, but your
good problem is challenging.

The approach I would use is begin with

ds^2 = g_uv dx^u dx^v , {u,v=0,1,2,3}.


Set the turntable to rotate in the x,y plane,
then only 9 metrics are needed,

g00 g01 g02
g10 g11 g12
g20 g21 g22

Usually an algorithm aka a field equation is
used to solve the metrics above, however I
don't have one off-hand, maybe someone else
knows, so I'll wing it pending improvement.

I'll set radius parallel to x, then only V(y)
exists because V(x)=0 and x=r.

Then I'll use the following values in place of
the 9 metrics above,

1 0 -V
0 1 -V
-V V 1

where g12=-g21 provides the rotation direction,
for example,

g12 = V(x)*(y/r)- V(y)*(x/r).

Solving that determinant g = |g_uv| using
these relations and cofactors gives,

g = g00 G(00)+ g01 G(01) + g02 G(02)

= 1+V^2 +V^2 = 1 + 2*V^2.

Then use ds = sqrt(g)*dt.

Colloquially, we find a clock moving out on
a turn table *slows down*, but a clock moving
up in gravitational field *speeds up*.

Regards
Ken S. Tucker

Ken S. Tucker

Murat Ozer wrote:
[...]
> Do you really think that I have not done that? There cannot be and
> there is not a flaw in my derivation. This is because my derivation
> is the step by step operational derivation and it cannot fail.
> The (non operational) definition dTau = ds/c works in all the cases
> but the turntable case. And one failure is enough to declare that
> the definition dTau = ds/c in GR is flawed...
>
> Murat Ozer

Mr. Ozer,
I'd hesitate to call GRT flawed, but your
good problem is challenging.

The approach I would use is begin with

ds^2 = g_uv dx^u dx^v , {u,v=0,1,2,3}.


Set the turntable to rotate in the x,y plane,
then only 9 metrics are needed,

g00 g01 g02
g10 g11 g12
g20 g21 g22

Usually an algorithm aka a field equation is
used to solve the metrics above, however I
don't have one off-hand, maybe someone else
knows, so I'll wing it pending improvement.

I'll set radius parallel to x, then only V(y)
exists because V(x)=0 and x=r.

Then I'll use the following values in place of
the 9 metrics above,

1 0 -V
0 1 -V
-V V 1

where g12=-g21 provides the rotation direction,
for example,

g12 = V(x)*(y/r)- V(y)*(x/r).

Solving that determinant g = |g_uv| using
these relations and cofactors gives,

g = g00 G(00)+ g01 G(01) + g02 G(02)

= 1+V^2 +V^2 = 1 + 2*V^2.

Then use ds = sqrt(g)*dt.

Colloquially, we find a clock moving out on
a turn table *slows down*, but a clock moving
up in gravitational field *speeds up*.

Regards
Ken S. Tucker

Ken S. Tucker

Murat Ozer wrote:
[...]
> Do you really think that I have not done that? There cannot be and
> there is not a flaw in my derivation. This is because my derivation
> is the step by step operational derivation and it cannot fail.
> The (non operational) definition dTau = ds/c works in all the cases
> but the turntable case. And one failure is enough to declare that
> the definition dTau = ds/c in GR is flawed...
>
> Murat Ozer

Mr. Ozer,
I'd hesitate to call GRT flawed, but your
good problem is challenging.

The approach I would use is begin with

ds^2 = g_uv dx^u dx^v , {u,v=0,1,2,3}.


Set the turntable to rotate in the x,y plane,
then only 9 metrics are needed,

g00 g01 g02
g10 g11 g12
g20 g21 g22

Usually an algorithm aka a field equation is
used to solve the metrics above, however I
don't have one off-hand, maybe someone else
knows, so I'll wing it pending improvement.

I'll set radius parallel to x, then only V(y)
exists because V(x)=0 and x=r.

Then I'll use the following values in place of
the 9 metrics above,

1 0 -V
0 1 -V
-V V 1

where g12=-g21 provides the rotation direction,
for example,

g12 = V(x)*(y/r)- V(y)*(x/r).

Solving that determinant g = |g_uv| using
these relations and cofactors gives,

g = g00 G(00)+ g01 G(01) + g02 G(02)

= 1+V^2 +V^2 = 1 + 2*V^2.

Then use ds = sqrt(g)*dt.

Colloquially, we find a clock moving out on
a turn table *slows down*, but a clock moving
up in gravitational field *speeds up*.

Regards
Ken S. Tucker

Ken S. Tucker

Murat Ozer wrote:
[...]
> Do you really think that I have not done that? There cannot be and
> there is not a flaw in my derivation. This is because my derivation
> is the step by step operational derivation and it cannot fail.
> The (non operational) definition dTau = ds/c works in all the cases
> but the turntable case. And one failure is enough to declare that
> the definition dTau = ds/c in GR is flawed...
>
> Murat Ozer

Mr. Ozer,
I'd hesitate to call GRT flawed, but your
good problem is challenging.

The approach I would use is begin with

ds^2 = g_uv dx^u dx^v , {u,v=0,1,2,3}.


Set the turntable to rotate in the x,y plane,
then only 9 metrics are needed,

g00 g01 g02
g10 g11 g12
g20 g21 g22

Usually an algorithm aka a field equation is
used to solve the metrics above, however I
don't have one off-hand, maybe someone else
knows, so I'll wing it pending improvement.

I'll set radius parallel to x, then only V(y)
exists because V(x)=0 and x=r.

Then I'll use the following values in place of
the 9 metrics above,

1 0 -V
0 1 -V
-V V 1

where g12=-g21 provides the rotation direction,
for example,

g12 = V(x)*(y/r)- V(y)*(x/r).

Solving that determinant g = |g_uv| using
these relations and cofactors gives,

g = g00 G(00)+ g01 G(01) + g02 G(02)

= 1+V^2 +V^2 = 1 + 2*V^2.

Then use ds = sqrt(g)*dt.

Colloquially, we find a clock moving out on
a turn table *slows down*, but a clock moving
up in gravitational field *speeds up*.

Regards
Ken S. Tucker

Ken S. Tucker

Murat Ozer wrote:
[...]
> Do you really think that I have not done that? There cannot be and
> there is not a flaw in my derivation. This is because my derivation
> is the step by step operational derivation and it cannot fail.
> The (non operational) definition dTau = ds/c works in all the cases
> but the turntable case. And one failure is enough to declare that
> the definition dTau = ds/c in GR is flawed...
>
> Murat Ozer

Mr. Ozer,
I'd hesitate to call GRT flawed, but your
good problem is challenging.

The approach I would use is begin with

ds^2 = g_uv dx^u dx^v , {u,v=0,1,2,3}.


Set the turntable to rotate in the x,y plane,
then only 9 metrics are needed,

g00 g01 g02
g10 g11 g12
g20 g21 g22

Usually an algorithm aka a field equation is
used to solve the metrics above, however I
don't have one off-hand, maybe someone else
knows, so I'll wing it pending improvement.

I'll set radius parallel to x, then only V(y)
exists because V(x)=0 and x=r.

Then I'll use the following values in place of
the 9 metrics above,

1 0 -V
0 1 -V
-V V 1

where g12=-g21 provides the rotation direction,
for example,

g12 = V(x)*(y/r)- V(y)*(x/r).

Solving that determinant g = |g_uv| using
these relations and cofactors gives,

g = g00 G(00)+ g01 G(01) + g02 G(02)

= 1+V^2 +V^2 = 1 + 2*V^2.

Then use ds = sqrt(g)*dt.

Colloquially, we find a clock moving out on
a turn table *slows down*, but a clock moving
up in gravitational field *speeds up*.

Regards
Ken S. Tucker

Ken S. Tucker

Murat Ozer wrote:
[...]
> Do you really think that I have not done that? There cannot be and
> there is not a flaw in my derivation. This is because my derivation
> is the step by step operational derivation and it cannot fail.
> The (non operational) definition dTau = ds/c works in all the cases
> but the turntable case. And one failure is enough to declare that
> the definition dTau = ds/c in GR is flawed...
>
> Murat Ozer

Mr. Ozer,
I'd hesitate to call GRT flawed, but your
good problem is challenging.

The approach I would use is begin with

ds^2 = g_uv dx^u dx^v , {u,v=0,1,2,3}.


Set the turntable to rotate in the x,y plane,
then only 9 metrics are needed,

g00 g01 g02
g10 g11 g12
g20 g21 g22

Usually an algorithm aka a field equation is
used to solve the metrics above, however I
don't have one off-hand, maybe someone else
knows, so I'll wing it pending improvement.

I'll set radius parallel to x, then only V(y)
exists because V(x)=0 and x=r.

Then I'll use the following values in place of
the 9 metrics above,

1 0 -V
0 1 -V
-V V 1

where g12=-g21 provides the rotation direction,
for example,

g12 = V(x)*(y/r)- V(y)*(x/r).

Solving that determinant g = |g_uv| using
these relations and cofactors gives,

g = g00 G(00)+ g01 G(01) + g02 G(02)

= 1+V^2 +V^2 = 1 + 2*V^2.

Then use ds = sqrt(g)*dt.

Colloquially, we find a clock moving out on
a turn table *slows down*, but a clock moving
up in gravitational field *speeds up*.

Regards
Ken S. Tucker

Ken S. Tucker

Murat Ozer wrote:
[...]
> Do you really think that I have not done that? There cannot be and
> there is not a flaw in my derivation. This is because my derivation
> is the step by step operational derivation and it cannot fail.
> The (non operational) definition dTau = ds/c works in all the cases
> but the turntable case. And one failure is enough to declare that
> the definition dTau = ds/c in GR is flawed...
>
> Murat Ozer

Mr. Ozer,
I'd hesitate to call GRT flawed, but your
good problem is challenging.

The approach I would use is begin with

ds^2 = g_uv dx^u dx^v , {u,v=0,1,2,3}.


Set the turntable to rotate in the x,y plane,
then only 9 metrics are needed,

g00 g01 g02
g10 g11 g12
g20 g21 g22

Usually an algorithm aka a field equation is
used to solve the metrics above, however I
don't have one off-hand, maybe someone else
knows, so I'll wing it pending improvement.

I'll set radius parallel to x, then only V(y)
exists because V(x)=0 and x=r.

Then I'll use the following values in place of
the 9 metrics above,

1 0 -V
0 1 -V
-V V 1

where g12=-g21 provides the rotation direction,
for example,

g12 = V(x)*(y/r)- V(y)*(x/r).

Solving that determinant g = |g_uv| using
these relations and cofactors gives,

g = g00 G(00)+ g01 G(01) + g02 G(02)

= 1+V^2 +V^2 = 1 + 2*V^2.

Then use ds = sqrt(g)*dt.

Colloquially, we find a clock moving out on
a turn table *slows down*, but a clock moving
up in gravitational field *speeds up*.

Regards
Ken S. Tucker

Ken S. Tucker

Murat Ozer wrote:
[...]
> Do you really think that I have not done that? There cannot be and
> there is not a flaw in my derivation. This is because my derivation
> is the step by step operational derivation and it cannot fail.
> The (non operational) definition dTau = ds/c works in all the cases
> but the turntable case. And one failure is enough to declare that
> the definition dTau = ds/c in GR is flawed...
>
> Murat Ozer

Mr. Ozer,
I'd hesitate to call GRT flawed, but your
good problem is challenging.

The approach I would use is begin with

ds^2 = g_uv dx^u dx^v , {u,v=0,1,2,3}.


Set the turntable to rotate in the x,y plane,
then only 9 metrics are needed,

g00 g01 g02
g10 g11 g12
g20 g21 g22

Usually an algorithm aka a field equation is
used to solve the metrics above, however I
don't have one off-hand, maybe someone else
knows, so I'll wing it pending improvement.

I'll set radius parallel to x, then only V(y)
exists because V(x)=0 and x=r.

Then I'll use the following values in place of
the 9 metrics above,

1 0 -V
0 1 -V
-V V 1

where g12=-g21 provides the rotation direction,
for example,

g12 = V(x)*(y/r)- V(y)*(x/r).

Solving that determinant g = |g_uv| using
these relations and cofactors gives,

g = g00 G(00)+ g01 G(01) + g02 G(02)

= 1+V^2 +V^2 = 1 + 2*V^2.

Then use ds = sqrt(g)*dt.

Colloquially, we find a clock moving out on
a turn table *slows down*, but a clock moving
up in gravitational field *speeds up*.

Regards
Ken S. Tucker

Ken S. Tucker

Murat Ozer wrote:
[...]
> Do you really think that I have not done that? There cannot be and
> there is not a flaw in my derivation. This is because my derivation
> is the step by step operational derivation and it cannot fail.
> The (non operational) definition dTau = ds/c works in all the cases
> but the turntable case. And one failure is enough to declare that
> the definition dTau = ds/c in GR is flawed...
>
> Murat Ozer

Mr. Ozer,
I'd hesitate to call GRT flawed, but your
good problem is challenging.

The approach I would use is begin with

ds^2 = g_uv dx^u dx^v , {u,v=0,1,2,3}.


Set the turntable to rotate in the x,y plane,
then only 9 metrics are needed,

g00 g01 g02
g10 g11 g12
g20 g21 g22

Usually an algorithm aka a field equation is
used to solve the metrics above, however I
don't have one off-hand, maybe someone else
knows, so I'll wing it pending improvement.

I'll set radius parallel to x, then only V(y)
exists because V(x)=0 and x=r.

Then I'll use the following values in place of
the 9 metrics above,

1 0 -V
0 1 -V
-V V 1

where g12=-g21 provides the rotation direction,
for example,

g12 = V(x)*(y/r)- V(y)*(x/r).

Solving that determinant g = |g_uv| using
these relations and cofactors gives,

g = g00 G(00)+ g01 G(01) + g02 G(02)

= 1+V^2 +V^2 = 1 + 2*V^2.

Then use ds = sqrt(g)*dt.

Colloquially, we find a clock moving out on
a turn table *slows down*, but a clock moving
up in gravitational field *speeds up*.

Regards
Ken S. Tucker

Ken S. Tucker

Murat Ozer wrote:
[...]
> Do you really think that I have not done that? There cannot be and
> there is not a flaw in my derivation. This is because my derivation
> is the step by step operational derivation and it cannot fail.
> The (non operational) definition dTau = ds/c works in all the cases
> but the turntable case. And one failure is enough to declare that
> the definition dTau = ds/c in GR is flawed...
>
> Murat Ozer

Mr. Ozer,
I'd hesitate to call GRT flawed, but your
good problem is challenging.

The approach I would use is begin with

ds^2 = g_uv dx^u dx^v , {u,v=0,1,2,3}.


Set the turntable to rotate in the x,y plane,
then only 9 metrics are needed,

g00 g01 g02
g10 g11 g12
g20 g21 g22

Usually an algorithm aka a field equation is
used to solve the metrics above, however I
don't have one off-hand, maybe someone else
knows, so I'll wing it pending improvement.

I'll set radius parallel to x, then only V(y)
exists because V(x)=0 and x=r.

Then I'll use the following values in place of
the 9 metrics above,

1 0 -V
0 1 -V
-V V 1

where g12=-g21 provides the rotation direction,
for example,

g12 = V(x)*(y/r)- V(y)*(x/r).

Solving that determinant g = |g_uv| using
these relations and cofactors gives,

g = g00 G(00)+ g01 G(01) + g02 G(02)

= 1+V^2 +V^2 = 1 + 2*V^2.

Then use ds = sqrt(g)*dt.

Colloquially, we find a clock moving out on
a turn table *slows down*, but a clock moving
up in gravitational field *speeds up*.

Regards
Ken S. Tucker

Ken S. Tucker

Murat Ozer wrote:
[...]
> Do you really think that I have not done that? There cannot be and
> there is not a flaw in my derivation. This is because my derivation
> is the step by step operational derivation and it cannot fail.
> The (non operational) definition dTau = ds/c works in all the cases
> but the turntable case. And one failure is enough to declare that
> the definition dTau = ds/c in GR is flawed...
>
> Murat Ozer

Mr. Ozer,
I'd hesitate to call GRT flawed, but your
good problem is challenging.

The approach I would use is begin with

ds^2 = g_uv dx^u dx^v , {u,v=0,1,2,3}.


Set the turntable to rotate in the x,y plane,
then only 9 metrics are needed,

g00 g01 g02
g10 g11 g12
g20 g21 g22

Usually an algorithm aka a field equation is
used to solve the metrics above, however I
don't have one off-hand, maybe someone else
knows, so I'll wing it pending improvement.

I'll set radius parallel to x, then only V(y)
exists because V(x)=0 and x=r.

Then I'll use the following values in place of
the 9 metrics above,

1 0 -V
0 1 -V
-V V 1

where g12=-g21 provides the rotation direction,
for example,

g12 = V(x)*(y/r)- V(y)*(x/r).

Solving that determinant g = |g_uv| using
these relations and cofactors gives,

g = g00 G(00)+ g01 G(01) + g02 G(02)

= 1+V^2 +V^2 = 1 + 2*V^2.

Then use ds = sqrt(g)*dt.

Colloquially, we find a clock moving out on
a turn table *slows down*, but a clock moving
up in gravitational field *speeds up*.

Regards
Ken S. Tucker

Ken S. Tucker

Murat Ozer wrote:
[...]
> Do you really think that I have not done that? There cannot be and
> there is not a flaw in my derivation. This is because my derivation
> is the step by step operational derivation and it cannot fail.
> The (non operational) definition dTau = ds/c works in all the cases
> but the turntable case. And one failure is enough to declare that
> the definition dTau = ds/c in GR is flawed...
>
> Murat Ozer

Mr. Ozer,
I'd hesitate to call GRT flawed, but your
good problem is challenging.

The approach I would use is begin with

ds^2 = g_uv dx^u dx^v , {u,v=0,1,2,3}.


Set the turntable to rotate in the x,y plane,
then only 9 metrics are needed,

g00 g01 g02
g10 g11 g12
g20 g21 g22

Usually an algorithm aka a field equation is
used to solve the metrics above, however I
don't have one off-hand, maybe someone else
knows, so I'll wing it pending improvement.

I'll set radius parallel to x, then only V(y)
exists because V(x)=0 and x=r.

Then I'll use the following values in place of
the 9 metrics above,

1 0 -V
0 1 -V
-V V 1

where g12=-g21 provides the rotation direction,
for example,

g12 = V(x)*(y/r)- V(y)*(x/r).

Solving that determinant g = |g_uv| using
these relations and cofactors gives,

g = g00 G(00)+ g01 G(01) + g02 G(02)

= 1+V^2 +V^2 = 1 + 2*V^2.

Then use ds = sqrt(g)*dt.

Colloquially, we find a clock moving out on
a turn table *slows down*, but a clock moving
up in gravitational field *speeds up*.

Regards
Ken S. Tucker

Ken S. Tucker

Murat Ozer wrote:
[...]
> Do you really think that I have not done that? There cannot be and
> there is not a flaw in my derivation. This is because my derivation
> is the step by step operational derivation and it cannot fail.
> The (non operational) definition dTau = ds/c works in all the cases
> but the turntable case. And one failure is enough to declare that
> the definition dTau = ds/c in GR is flawed...
>
> Murat Ozer

Mr. Ozer,
I'd hesitate to call GRT flawed, but your
good problem is challenging.

The approach I would use is begin with

ds^2 = g_uv dx^u dx^v , {u,v=0,1,2,3}.


Set the turntable to rotate in the x,y plane,
then only 9 metrics are needed,

g00 g01 g02
g10 g11 g12
g20 g21 g22

Usually an algorithm aka a field equation is
used to solve the metrics above, however I
don't have one off-hand, maybe someone else
knows, so I'll wing it pending improvement.

I'll set radius parallel to x, then only V(y)
exists because V(x)=0 and x=r.

Then I'll use the following values in place of
the 9 metrics above,

1 0 -V
0 1 -V
-V V 1

where g12=-g21 provides the rotation direction,
for example,

g12 = V(x)*(y/r)- V(y)*(x/r).

Solving that determinant g = |g_uv| using
these relations and cofactors gives,

g = g00 G(00)+ g01 G(01) + g02 G(02)

= 1+V^2 +V^2 = 1 + 2*V^2.

Then use ds = sqrt(g)*dt.

Colloquially, we find a clock moving out on
a turn table *slows down*, but a clock moving
up in gravitational field *speeds up*.

Regards
Ken S. Tucker

Ken S. Tucker

Murat Ozer wrote:
[...]
> Do you really think that I have not done that? There cannot be and
> there is not a flaw in my derivation. This is because my derivation
> is the step by step operational derivation and it cannot fail.
> The (non operational) definition dTau = ds/c works in all the cases
> but the turntable case. And one failure is enough to declare that
> the definition dTau = ds/c in GR is flawed...
>
> Murat Ozer

Mr. Ozer,
I'd hesitate to call GRT flawed, but your
good problem is challenging.

The approach I would use is begin with

ds^2 = g_uv dx^u dx^v , {u,v=0,1,2,3}.


Set the turntable to rotate in the x,y plane,
then only 9 metrics are needed,

g00 g01 g02
g10 g11 g12
g20 g21 g22

Usually an algorithm aka a field equation is
used to solve the metrics above, however I
don't have one off-hand, maybe someone else
knows, so I'll wing it pending improvement.

I'll set radius parallel to x, then only V(y)
exists because V(x)=0 and x=r.

Then I'll use the following values in place of
the 9 metrics above,

1 0 -V
0 1 -V
-V V 1

where g12=-g21 provides the rotation direction,
for example,

g12 = V(x)*(y/r)- V(y)*(x/r).

Solving that determinant g = |g_uv| using
these relations and cofactors gives,

g = g00 G(00)+ g01 G(01) + g02 G(02)

= 1+V^2 +V^2 = 1 + 2*V^2.

Then use ds = sqrt(g)*dt.

Colloquially, we find a clock moving out on
a turn table *slows down*, but a clock moving
up in gravitational field *speeds up*.

Regards
Ken S. Tucker

Murat Ozer

Igor Khavkine wrote:
> On 2005-09-10, Murat Ozer <Murat.H.Ozer@gmail.com> wrote:
> > Igor Khavkine wrote:
> >> On 2005-09-07, Murat Ozer <Murat.H.Ozer@gmail.com> wrote:
> >> > Igor Khavkine wrote:
> >> >> On 2005-09-01, Murat <Murat.H.Ozer@gmail.com> wrote:
> >>
> >> >> >using the fact that dt' = dTau when dx' = 0=dy'=dz'
> >> >> > we get
> >> >> >
> >> >> > dt = gamma dTau, or dTau = Sqrt(1-v^2/c^2)dt. (8)
> >> >>
> >> >
> >> >> I don't see what you mean here. You obtain the right answer in this
> >> >> case,
> >> >
> >> > What I mean is very clear. This is actually how proper time is derived
> >> > in some textbooks. (See, for example, p.468 of An Introduction to
> >> > Mechanics, by Kleppner and Kolenkow, Mc Graw Hill)Furthermore, that
> >> > dt' = dTau when dx' = dy' = dz' =0, namely the time read by a clock
> >> > that is at rest is THE definition of proper time and ALWAYS gives
> >> > the CORRECT expression for it..
> >>
> >> I'm sorry, but I still don't understand what you mean. What do you mean
> >> by a clock "at rest"? As you should know, this statement is meaningless
> >> unless you specify an inertial frame with respect to which it holds.
> >
> > Again, what I mean is very clear because I am using the standard
> > terminology used by every body else. I am not introducing new
> > terminology here. When someone drives his car, for example, the car
> > has a certain velocity relative to the road but the driver's velocity
> > relative to the car is zero. This is because the driver is comoving
> > with the car and is "at rest" in the car. Furthermore, one does not
> > need an inertial frame to define the rest frame of a clock. After all,
> > does not a clock in a gravitational field have a rest frame in which
> > it is momentarily at rest?
>
> My confusion rests on the fact that you said "read by a clock at rest",
> but did not specify what it was at rest with respect to. Compare to your
> example, where you did specify that the driver is at rest with respect
> to the car. I find that perfectly clear.
>
> Hmm... "one does not need an inertial frame to define the rest frame of
> a clock" you say. See below.

You seem to have failed to include the argument. Or is it due to the
failure of the Google editor?


>

It's true that one does not need an inertial frame to define the rest
frame of a clock. Inertial or not, a frame with respect to which the
clock is not moving is the rest frame of the clock. If we consider the
rotating turntable again, it's the rest frame of all the nonmoving
clocks located at any point r_0, where 0<= r_0 =< R with R being the
radius of the turntable.

> >> >> but I don't follow your reasoning.
> >> >
> >> > So, please consult an elementeray book on the subject.
> >>
> >> I don't have the particular book you cited, but I have consulted various
> >> books on relativity in the past. Still, I would not give the same
> >> definition of proper time. Since I think this is the key point in the
> >> misunderstanding, could you explain what you mean in the above paragraph
> >> in more detail?
> >
> > I don't know if I can do better than that. But instead let me quote
> > from another book, Landau & Lifshitz, The Classical Theory of Fields,
> > p.7:
> >
> > " Suppose that in a certain inertial reference system we observe
> > clocks which are moving relative to us in an arbitrary manner. At each
> > different moment of time this motion can be considered uniform. Thus at
> > each moment of time we can introduce a coordinate system rigidly linked
> > to the moving clocks, which with the clocks constitutes an inertial
> [ emphasis added ] ^^^^^^^^
> > reference system.
>
> Compare this to what you said above. Now, suppose we ignore the
> requirement of the reference frame being inertial. Additionally, let us
> identify the choice of frame with a choice of coordinate system.
>
> If there is a clock moving with velocity v in the x direction. I can
> choose a coordinate system where the position of the clock is always at
> the origin. Usually, this choice is done with a Lorentz transformation.
> However, I'm feeling fancy, so I'll use a Galilean transformation
> instead: x' = x - vt, y' = y, z' = z, t' = t. This is a perfectly valid
> choice of coordinate system, and since the coordinates of the clock in
> it are x' = y' = z' = 0, according to you I it describes a frame of
> reference in which the clock is at rest. Then again, according to you,
> the proper time will be just: dTau = dt' = dt. Obviously, this is
> wrong. And you've already found what the right answer is.

Yes, according to the operational definition of proper time which is
also my position in regard to this issue, the proper time is
dTau = dt' = dt. This is not wrong as I have shown before. For the sake
of completenes let us treat your example carefully:
Let us consider two frames,a stationary frame S and another one, S',
moving at velocity v relative to S. Assume, for simplicity, that
the metric in S is given by

ds^2 = c^2dt^2 - dx^2 - dy^2 - dz^2. (1)

As for the transformations that relate S to S', let us consider two
sets of transformations, Galilean transformations and Lorentz
transformations.
(I) Galilean transformations:

x = x' + vt, t = t', y = y', z = z',

under which the metric in S' is

ds'^2 = (1-v^2/c^2)c^2dt'^2-2vdx'dt' - dx'^2 - dy'^2 - dz'^2. (2)

Obviously, under these transformations S' is not an inertial frame.

(II) Lorentz Transformations:

x = gamma(x' + vt'), t = gamma(t' + v/c^2x'), y = y', z = z'

under which the metric in S' is the usual one

ds'^2 = c^2dt'^2 - dx'^2 - dy'^2 - dz'^2. (3)

Now, let us follow Landau & Lifshitz step by step:

In a system of coordinates linked to the moving clocks, the latter
are at rest, i.e., dx'=dy'=dz'=0. Because of the invariance of
intervals

ds^2 = c^2dt^2 -dx^2-dy^2-dz^2

ds^2 = (1-v^2/c^2)c^2dt'^2, from (2) (Galilean Transformations),

ds^2 = c^2dt'^2, from (3) (Lorentz Transformations),

from which,

Sqrt(1-v^2/c^2)dt' = dtSqrt(1 -[dx^2-dy^2-dz^2]/(c^2dt^2))

= dtSqrt(1-v^2/c^2),

and hence dt' = dTau = dt (for Galilean Transformations),

and
dt' = dTau = dtSqrt(1-v^2/c^2) (for Lorentz Transformations).

As you see,firstly, the operational definition gives the correct
answer for the Lorentz Transformations, and therefore it must give
the correct answer for the Galilean transformations too. Secondly,
the same results are obtained from a direct use of the transformation
equations. Thirdly and more importantly, the general relativistic
definition ds = cdTau gives the SAME result, namely
dTau = Sqrt(1-v^2/c^2)dt for these completely different Galilean
and Lorentzian systems.
Now, are you going to object by saying that I have obtained this
result because the system of the clocks S' is not an inertial system
in the case of the Galilean transformations?


>
> The gap in reasoning comes from the neglect of the condition that the
> choice of reference frame be inertial.

You do not choose the reference frame; it is given to you. The
relativistic Galilean turntable, or the relativistic Galilean
linear frame whose metrics are given above are imposed upon you. They
are defined via the Galilean transformations and the ensuing metrics,
and they are noninertial frames. These frames are the rest frames
of clocks that are nonmoving in them. There is no reason why the same
"operational definition" employed in SR cannot be used for these
systems.
Saying that a noninertial system to which a clock is attached is not
the rest frame of the clock is equivalent to saying that the proper
time for that clock cannot be defined...The definition ds = cdTau is
not an operational definition. You need to define Tau operationally.
How are you going to do that in a noninertial frame in a way different
than mine and obtain the result ds = cdTau?

> If the coordinate system is
> chosen through a Lorentz transformation, it is inertial and as you've
> already shown the correct answer is obtained.
>
> To really understand what this means, you have to be able to answer the
> following questions:
>
> 1. What is an inertial frame? How will you know it when you see one?
>
> And to apply this understanding to the case of a rotating plate, these
> ones:
>
> 2. Is the "co-rotating" coordinate system that you described for the
> rotating plate inertial?
> 3. Can you find a coordinate system that is inertial for at all points
> on the plate?
> 4. If not, can you find a coordinate system that is inertial for a given
> point on the plate?
> 5. In this frame, do you find correct expression for the proper time
> using the definition you tried to apply previously?
>
> >> >> Learning to spot one's own mistakes is a very valuable skill.
> >> >
> >> > I cannot dispute this remark. I am very much humbled by it.
> >> > But unfortunately, you have not spotted any mistake(s) in my treatment.
> >>
> >> Lets just say we differ in opinion here. However, my remark was aimed at
> >> something that you already noticed yourself. Namely, that you did not
> >> obtain the corrected expression for the proper time in your example.
> >
> >> Your first instinct should be to find a flaw in your derivation and
> >> understanding, not a flaw in the definition of proper time.
> >
> > Do you really think that I have not done that? There cannot be and
> > there is not a flaw in my derivation.
>
> You should also at some point realize that it is very hard to defend
> absolutes.
>
> > This is because my derivation
> > is the step by step operational derivation and it cannot fail.
> > The (non operational) definition dTau = ds/c works in all the cases
> > but the turntable case. And one failure is enough to declare that
> > the definition dTau = ds/c in GR is flawed...
>
> Let me throw another maxim into the mix. Pride cometh before a fall.

Logic constrained under well defined precise principles and procedures
knows nothing about pride.

Murat Ozer
>
> Igor

Murat Ozer

Igor Khavkine wrote:
> On 2005-09-10, Murat Ozer <Murat.H.Ozer@gmail.com> wrote:
> > Igor Khavkine wrote:
> >> On 2005-09-07, Murat Ozer <Murat.H.Ozer@gmail.com> wrote:
> >> > Igor Khavkine wrote:
> >> >> On 2005-09-01, Murat <Murat.H.Ozer@gmail.com> wrote:
> >>
> >> >> >using the fact that dt' = dTau when dx' = 0=dy'=dz'
> >> >> > we get
> >> >> >
> >> >> > dt = gamma dTau, or dTau = Sqrt(1-v^2/c^2)dt. (8)
> >> >>
> >> >
> >> >> I don't see what you mean here. You obtain the right answer in this
> >> >> case,
> >> >
> >> > What I mean is very clear. This is actually how proper time is derived
> >> > in some textbooks. (See, for example, p.468 of An Introduction to
> >> > Mechanics, by Kleppner and Kolenkow, Mc Graw Hill)Furthermore, that
> >> > dt' = dTau when dx' = dy' = dz' =0, namely the time read by a clock
> >> > that is at rest is THE definition of proper time and ALWAYS gives
> >> > the CORRECT expression for it..
> >>
> >> I'm sorry, but I still don't understand what you mean. What do you mean
> >> by a clock "at rest"? As you should know, this statement is meaningless
> >> unless you specify an inertial frame with respect to which it holds.
> >
> > Again, what I mean is very clear because I am using the standard
> > terminology used by every body else. I am not introducing new
> > terminology here. When someone drives his car, for example, the car
> > has a certain velocity relative to the road but the driver's velocity
> > relative to the car is zero. This is because the driver is comoving
> > with the car and is "at rest" in the car. Furthermore, one does not
> > need an inertial frame to define the rest frame of a clock. After all,
> > does not a clock in a gravitational field have a rest frame in which
> > it is momentarily at rest?
>
> My confusion rests on the fact that you said "read by a clock at rest",
> but did not specify what it was at rest with respect to. Compare to your
> example, where you did specify that the driver is at rest with respect
> to the car. I find that perfectly clear.
>
> Hmm... "one does not need an inertial frame to define the rest frame of
> a clock" you say. See below.

You seem to have failed to include the argument. Or is it due to the
failure of the Google editor?


>

It's true that one does not need an inertial frame to define the rest
frame of a clock. Inertial or not, a frame with respect to which the
clock is not moving is the rest frame of the clock. If we consider the
rotating turntable again, it's the rest frame of all the nonmoving
clocks located at any point r_0, where 0<= r_0 =< R with R being the
radius of the turntable.

> >> >> but I don't follow your reasoning.
> >> >
> >> > So, please consult an elementeray book on the subject.
> >>
> >> I don't have the particular book you cited, but I have consulted various
> >> books on relativity in the past. Still, I would not give the same
> >> definition of proper time. Since I think this is the key point in the
> >> misunderstanding, could you explain what you mean in the above paragraph
> >> in more detail?
> >
> > I don't know if I can do better than that. But instead let me quote
> > from another book, Landau & Lifshitz, The Classical Theory of Fields,
> > p.7:
> >
> > " Suppose that in a certain inertial reference system we observe
> > clocks which are moving relative to us in an arbitrary manner. At each
> > different moment of time this motion can be considered uniform. Thus at
> > each moment of time we can introduce a coordinate system rigidly linked
> > to the moving clocks, which with the clocks constitutes an inertial
> [ emphasis added ] ^^^^^^^^
> > reference system.
>
> Compare this to what you said above. Now, suppose we ignore the
> requirement of the reference frame being inertial. Additionally, let us
> identify the choice of frame with a choice of coordinate system.
>
> If there is a clock moving with velocity v in the x direction. I can
> choose a coordinate system where the position of the clock is always at
> the origin. Usually, this choice is done with a Lorentz transformation.
> However, I'm feeling fancy, so I'll use a Galilean transformation
> instead: x' = x - vt, y' = y, z' = z, t' = t. This is a perfectly valid
> choice of coordinate system, and since the coordinates of the clock in
> it are x' = y' = z' = 0, according to you I it describes a frame of
> reference in which the clock is at rest. Then again, according to you,
> the proper time will be just: dTau = dt' = dt. Obviously, this is
> wrong. And you've already found what the right answer is.

Yes, according to the operational definition of proper time which is
also my position in regard to this issue, the proper time is
dTau = dt' = dt. This is not wrong as I have shown before. For the sake
of completenes let us treat your example carefully:
Let us consider two frames,a stationary frame S and another one, S',
moving at velocity v relative to S. Assume, for simplicity, that
the metric in S is given by

ds^2 = c^2dt^2 - dx^2 - dy^2 - dz^2. (1)

As for the transformations that relate S to S', let us consider two
sets of transformations, Galilean transformations and Lorentz
transformations.
(I) Galilean transformations:

x = x' + vt, t = t', y = y', z = z',

under which the metric in S' is

ds'^2 = (1-v^2/c^2)c^2dt'^2-2vdx'dt' - dx'^2 - dy'^2 - dz'^2. (2)

Obviously, under these transformations S' is not an inertial frame.

(II) Lorentz Transformations:

x = gamma(x' + vt'), t = gamma(t' + v/c^2x'), y = y', z = z'

under which the metric in S' is the usual one

ds'^2 = c^2dt'^2 - dx'^2 - dy'^2 - dz'^2. (3)

Now, let us follow Landau & Lifshitz step by step:

In a system of coordinates linked to the moving clocks, the latter
are at rest, i.e., dx'=dy'=dz'=0. Because of the invariance of
intervals

ds^2 = c^2dt^2 -dx^2-dy^2-dz^2

ds^2 = (1-v^2/c^2)c^2dt'^2, from (2) (Galilean Transformations),

ds^2 = c^2dt'^2, from (3) (Lorentz Transformations),

from which,

Sqrt(1-v^2/c^2)dt' = dtSqrt(1 -[dx^2-dy^2-dz^2]/(c^2dt^2))

= dtSqrt(1-v^2/c^2),

and hence dt' = dTau = dt (for Galilean Transformations),

and
dt' = dTau = dtSqrt(1-v^2/c^2) (for Lorentz Transformations).

As you see,firstly, the operational definition gives the correct
answer for the Lorentz Transformations, and therefore it must give
the correct answer for the Galilean transformations too. Secondly,
the same results are obtained from a direct use of the transformation
equations. Thirdly and more importantly, the general relativistic
definition ds = cdTau gives the SAME result, namely
dTau = Sqrt(1-v^2/c^2)dt for these completely different Galilean
and Lorentzian systems.
Now, are you going to object by saying that I have obtained this
result because the system of the clocks S' is not an inertial system
in the case of the Galilean transformations?


>
> The gap in reasoning comes from the neglect of the condition that the
> choice of reference frame be inertial.

You do not choose the reference frame; it is given to you. The
relativistic Galilean turntable, or the relativistic Galilean
linear frame whose metrics are given above are imposed upon you. They
are defined via the Galilean transformations and the ensuing metrics,
and they are noninertial frames. These frames are the rest frames
of clocks that are nonmoving in them. There is no reason why the same
"operational definition" employed in SR cannot be used for these
systems.
Saying that a noninertial system to which a clock is attached is not
the rest frame of the clock is equivalent to saying that the proper
time for that clock cannot be defined...The definition ds = cdTau is
not an operational definition. You need to define Tau operationally.
How are you going to do that in a noninertial frame in a way different
than mine and obtain the result ds = cdTau?

> If the coordinate system is
> chosen through a Lorentz transformation, it is inertial and as you've
> already shown the correct answer is obtained.
>
> To really understand what this means, you have to be able to answer the
> following questions:
>
> 1. What is an inertial frame? How will you know it when you see one?
>
> And to apply this understanding to the case of a rotating plate, these
> ones:
>
> 2. Is the "co-rotating" coordinate system that you described for the
> rotating plate inertial?
> 3. Can you find a coordinate system that is inertial for at all points
> on the plate?
> 4. If not, can you find a coordinate system that is inertial for a given
> point on the plate?
> 5. In this frame, do you find correct expression for the proper time
> using the definition you tried to apply previously?
>
> >> >> Learning to spot one's own mistakes is a very valuable skill.
> >> >
> >> > I cannot dispute this remark. I am very much humbled by it.
> >> > But unfortunately, you have not spotted any mistake(s) in my treatment.
> >>
> >> Lets just say we differ in opinion here. However, my remark was aimed at
> >> something that you already noticed yourself. Namely, that you did not
> >> obtain the corrected expression for the proper time in your example.
> >
> >> Your first instinct should be to find a flaw in your derivation and
> >> understanding, not a flaw in the definition of proper time.
> >
> > Do you really think that I have not done that? There cannot be and
> > there is not a flaw in my derivation.
>
> You should also at some point realize that it is very hard to defend
> absolutes.
>
> > This is because my derivation
> > is the step by step operational derivation and it cannot fail.
> > The (non operational) definition dTau = ds/c works in all the cases
> > but the turntable case. And one failure is enough to declare that
> > the definition dTau = ds/c in GR is flawed...
>
> Let me throw another maxim into the mix. Pride cometh before a fall.

Logic constrained under well defined precise principles and procedures
knows nothing about pride.

Murat Ozer
>
> Igor

Murat Ozer

Igor Khavkine wrote:
> On 2005-09-10, Murat Ozer <Murat.H.Ozer@gmail.com> wrote:
> > Igor Khavkine wrote:
> >> On 2005-09-07, Murat Ozer <Murat.H.Ozer@gmail.com> wrote:
> >> > Igor Khavkine wrote:
> >> >> On 2005-09-01, Murat <Murat.H.Ozer@gmail.com> wrote:
> >>
> >> >> >using the fact that dt' = dTau when dx' = 0=dy'=dz'
> >> >> > we get
> >> >> >
> >> >> > dt = gamma dTau, or dTau = Sqrt(1-v^2/c^2)dt. (8)
> >> >>
> >> >
> >> >> I don't see what you mean here. You obtain the right answer in this
> >> >> case,
> >> >
> >> > What I mean is very clear. This is actually how proper time is derived
> >> > in some textbooks. (See, for example, p.468 of An Introduction to
> >> > Mechanics, by Kleppner and Kolenkow, Mc Graw Hill)Furthermore, that
> >> > dt' = dTau when dx' = dy' = dz' =0, namely the time read by a clock
> >> > that is at rest is THE definition of proper time and ALWAYS gives
> >> > the CORRECT expression for it..
> >>
> >> I'm sorry, but I still don't understand what you mean. What do you mean
> >> by a clock "at rest"? As you should know, this statement is meaningless
> >> unless you specify an inertial frame with respect to which it holds.
> >
> > Again, what I mean is very clear because I am using the standard
> > terminology used by every body else. I am not introducing new
> > terminology here. When someone drives his car, for example, the car
> > has a certain velocity relative to the road but the driver's velocity
> > relative to the car is zero. This is because the driver is comoving
> > with the car and is "at rest" in the car. Furthermore, one does not
> > need an inertial frame to define the rest frame of a clock. After all,
> > does not a clock in a gravitational field have a rest frame in which
> > it is momentarily at rest?
>
> My confusion rests on the fact that you said "read by a clock at rest",
> but did not specify what it was at rest with respect to. Compare to your
> example, where you did specify that the driver is at rest with respect
> to the car. I find that perfectly clear.
>
> Hmm... "one does not need an inertial frame to define the rest frame of
> a clock" you say. See below.

You seem to have failed to include the argument. Or is it due to the
failure of the Google editor?


>

It's true that one does not need an inertial frame to define the rest
frame of a clock. Inertial or not, a frame with respect to which the
clock is not moving is the rest frame of the clock. If we consider the
rotating turntable again, it's the rest frame of all the nonmoving
clocks located at any point r_0, where 0<= r_0 =< R with R being the
radius of the turntable.

> >> >> but I don't follow your reasoning.
> >> >
> >> > So, please consult an elementeray book on the subject.
> >>
> >> I don't have the particular book you cited, but I have consulted various
> >> books on relativity in the past. Still, I would not give the same
> >> definition of proper time. Since I think this is the key point in the
> >> misunderstanding, could you explain what you mean in the above paragraph
> >> in more detail?
> >
> > I don't know if I can do better than that. But instead let me quote
> > from another book, Landau & Lifshitz, The Classical Theory of Fields,
> > p.7:
> >
> > " Suppose that in a certain inertial reference system we observe
> > clocks which are moving relative to us in an arbitrary manner. At each
> > different moment of time this motion can be considered uniform. Thus at
> > each moment of time we can introduce a coordinate system rigidly linked
> > to the moving clocks, which with the clocks constitutes an inertial
> [ emphasis added ] ^^^^^^^^
> > reference system.
>
> Compare this to what you said above. Now, suppose we ignore the
> requirement of the reference frame being inertial. Additionally, let us
> identify the choice of frame with a choice of coordinate system.
>
> If there is a clock moving with velocity v in the x direction. I can
> choose a coordinate system where the position of the clock is always at
> the origin. Usually, this choice is done with a Lorentz transformation.
> However, I'm feeling fancy, so I'll use a Galilean transformation
> instead: x' = x - vt, y' = y, z' = z, t' = t. This is a perfectly valid
> choice of coordinate system, and since the coordinates of the clock in
> it are x' = y' = z' = 0, according to you I it describes a frame of
> reference in which the clock is at rest. Then again, according to you,
> the proper time will be just: dTau = dt' = dt. Obviously, this is
> wrong. And you've already found what the right answer is.

Yes, according to the operational definition of proper time which is
also my position in regard to this issue, the proper time is
dTau = dt' = dt. This is not wrong as I have shown before. For the sake
of completenes let us treat your example carefully:
Let us consider two frames,a stationary frame S and another one, S',
moving at velocity v relative to S. Assume, for simplicity, that
the metric in S is given by

ds^2 = c^2dt^2 - dx^2 - dy^2 - dz^2. (1)

As for the transformations that relate S to S', let us consider two
sets of transformations, Galilean transformations and Lorentz
transformations.
(I) Galilean transformations:

x = x' + vt, t = t', y = y', z = z',

under which the metric in S' is

ds'^2 = (1-v^2/c^2)c^2dt'^2-2vdx'dt' - dx'^2 - dy'^2 - dz'^2. (2)

Obviously, under these transformations S' is not an inertial frame.

(II) Lorentz Transformations:

x = gamma(x' + vt'), t = gamma(t' + v/c^2x'), y = y', z = z'

under which the metric in S' is the usual one

ds'^2 = c^2dt'^2 - dx'^2 - dy'^2 - dz'^2. (3)

Now, let us follow Landau & Lifshitz step by step:

In a system of coordinates linked to the moving clocks, the latter
are at rest, i.e., dx'=dy'=dz'=0. Because of the invariance of
intervals

ds^2 = c^2dt^2 -dx^2-dy^2-dz^2

ds^2 = (1-v^2/c^2)c^2dt'^2, from (2) (Galilean Transformations),

ds^2 = c^2dt'^2, from (3) (Lorentz Transformations),

from which,

Sqrt(1-v^2/c^2)dt' = dtSqrt(1 -[dx^2-dy^2-dz^2]/(c^2dt^2))

= dtSqrt(1-v^2/c^2),

and hence dt' = dTau = dt (for Galilean Transformations),

and
dt' = dTau = dtSqrt(1-v^2/c^2) (for Lorentz Transformations).

As you see,firstly, the operational definition gives the correct
answer for the Lorentz Transformations, and therefore it must give
the correct answer for the Galilean transformations too. Secondly,
the same results are obtained from a direct use of the transformation
equations. Thirdly and more importantly, the general relativistic
definition ds = cdTau gives the SAME result, namely
dTau = Sqrt(1-v^2/c^2)dt for these completely different Galilean
and Lorentzian systems.
Now, are you going to object by saying that I have obtained this
result because the system of the clocks S' is not an inertial system
in the case of the Galilean transformations?


>
> The gap in reasoning comes from the neglect of the condition that the
> choice of reference frame be inertial.

You do not choose the reference frame; it is given to you. The
relativistic Galilean turntable, or the relativistic Galilean
linear frame whose metrics are given above are imposed upon you. They
are defined via the Galilean transformations and the ensuing metrics,
and they are noninertial frames. These frames are the rest frames
of clocks that are nonmoving in them. There is no reason why the same
"operational definition" employed in SR cannot be used for these
systems.
Saying that a noninertial system to which a clock is attached is not
the rest frame of the clock is equivalent to saying that the proper
time for that clock cannot be defined...The definition ds = cdTau is
not an operational definition. You need to define Tau operationally.
How are you going to do that in a noninertial frame in a way different
than mine and obtain the result ds = cdTau?

> If the coordinate system is
> chosen through a Lorentz transformation, it is inertial and as you've
> already shown the correct answer is obtained.
>
> To really understand what this means, you have to be able to answer the
> following questions:
>
> 1. What is an inertial frame? How will you know it when you see one?
>
> And to apply this understanding to the case of a rotating plate, these
> ones:
>
> 2. Is the "co-rotating" coordinate system that you described for the
> rotating plate inertial?
> 3. Can you find a coordinate system that is inertial for at all points
> on the plate?
> 4. If not, can you find a coordinate system that is inertial for a given
> point on the plate?
> 5. In this frame, do you find correct expression for the proper time
> using the definition you tried to apply previously?
>
> >> >> Learning to spot one's own mistakes is a very valuable skill.
> >> >
> >> > I cannot dispute this remark. I am very much humbled by it.
> >> > But unfortunately, you have not spotted any mistake(s) in my treatment.
> >>
> >> Lets just say we differ in opinion here. However, my remark was aimed at
> >> something that you already noticed yourself. Namely, that you did not
> >> obtain the corrected expression for the proper time in your example.
> >
> >> Your first instinct should be to find a flaw in your derivation and
> >> understanding, not a flaw in the definition of proper time.
> >
> > Do you really think that I have not done that? There cannot be and
> > there is not a flaw in my derivation.
>
> You should also at some point realize that it is very hard to defend
> absolutes.
>
> > This is because my derivation
> > is the step by step operational derivation and it cannot fail.
> > The (non operational) definition dTau = ds/c works in all the cases
> > but the turntable case. And one failure is enough to declare that
> > the definition dTau = ds/c in GR is flawed...
>
> Let me throw another maxim into the mix. Pride cometh before a fall.

Logic constrained under well defined precise principles and procedures
knows nothing about pride.

Murat Ozer
>
> Igor