## [SOLVED] Proper Time in General Relativity

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 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 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 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 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 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 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 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 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 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 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 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 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 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