## [SOLVED] Proper Time in General Relativity

Ken S. Tucker wrote:
> 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}.
>
>

Mr. Tucker,
Since the turntable is rotating relative to some stationary
frame, even though it's OK as long as there is no confusion,
I would have labeled the coordinates of the table with primes,
and would have reserved the unprimed coordinates for the
stationary frame relative to which the turntable is rotating.

> 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

,where you have dropped the primes on the coordinates.

>
> 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.

Unfortunately, you seem to have made a mistake. This determinant
is equal to 1, not 1 + 2*V^2.

>
> Then use ds = sqrt(g)*dt.
>

Since g = 1, this expression would give ds = dt.
Furthermore, this example has g_0'0' = 1 and the
expression ds = cdTau is OK. My claim is, if you recall,
ds = cdTau fails whenever g_0'0' is NOT 1...
Incidentally, I have never seen the expression
ds = sqrt(g)*dt in the literature. I would have appreciated
a reference.

Regards,

Murat Ozer

> 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 wrote: > 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}. > > Mr. Tucker, Since the turntable is rotating relative to some stationary frame, even though it's OK as long as there is no confusion, I would have labeled the coordinates of the table with primes, and would have reserved the unprimed coordinates for the stationary frame relative to which the turntable is rotating. > 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 ,where you have dropped the primes on the coordinates. > > 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. Unfortunately, you seem to have made a mistake. This determinant is equal to 1, not 1 + 2*V^2. > > Then use ds = sqrt(g)*dt. > Since g = 1, this expression would give ds = dt. Furthermore, this example has g_0'0' = 1 and the expression ds = cdTau is OK. My claim is, if you recall, ds = cdTau fails whenever g_0'0' is NOT 1... Incidentally, I have never seen the expression ds = sqrt(g)*dt in the literature. I would have appreciated a reference. Regards, Murat Ozer > 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 wrote: > 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}. > > Mr. Tucker, Since the turntable is rotating relative to some stationary frame, even though it's OK as long as there is no confusion, I would have labeled the coordinates of the table with primes, and would have reserved the unprimed coordinates for the stationary frame relative to which the turntable is rotating. > 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 ,where you have dropped the primes on the coordinates. > > 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. Unfortunately, you seem to have made a mistake. This determinant is equal to 1, not 1 + 2*V^2. > > Then use ds = sqrt(g)*dt. > Since g = 1, this expression would give ds = dt. Furthermore, this example has g_0'0' = 1 and the expression ds = cdTau is OK. My claim is, if you recall, ds = cdTau fails whenever g_0'0' is NOT 1... Incidentally, I have never seen the expression ds = sqrt(g)*dt in the literature. I would have appreciated a reference. Regards, Murat Ozer > 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 wrote: > 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}. > > Mr. Tucker, Since the turntable is rotating relative to some stationary frame, even though it's OK as long as there is no confusion, I would have labeled the coordinates of the table with primes, and would have reserved the unprimed coordinates for the stationary frame relative to which the turntable is rotating. > 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 ,where you have dropped the primes on the coordinates. > > 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. Unfortunately, you seem to have made a mistake. This determinant is equal to 1, not 1 + 2*V^2. > > Then use ds = sqrt(g)*dt. > Since g = 1, this expression would give ds = dt. Furthermore, this example has g_0'0' = 1 and the expression ds = cdTau is OK. My claim is, if you recall, ds = cdTau fails whenever g_0'0' is NOT 1... Incidentally, I have never seen the expression ds = sqrt(g)*dt in the literature. I would have appreciated a reference. Regards, Murat Ozer > 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 wrote: > 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}. > > Mr. Tucker, Since the turntable is rotating relative to some stationary frame, even though it's OK as long as there is no confusion, I would have labeled the coordinates of the table with primes, and would have reserved the unprimed coordinates for the stationary frame relative to which the turntable is rotating. > 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 ,where you have dropped the primes on the coordinates. > > 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. Unfortunately, you seem to have made a mistake. This determinant is equal to 1, not 1 + 2*V^2. > > Then use ds = sqrt(g)*dt. > Since g = 1, this expression would give ds = dt. Furthermore, this example has g_0'0' = 1 and the expression ds = cdTau is OK. My claim is, if you recall, ds = cdTau fails whenever g_0'0' is NOT 1... Incidentally, I have never seen the expression ds = sqrt(g)*dt in the literature. I would have appreciated a reference. Regards, Murat Ozer > 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 wrote: > 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}. > > Mr. Tucker, Since the turntable is rotating relative to some stationary frame, even though it's OK as long as there is no confusion, I would have labeled the coordinates of the table with primes, and would have reserved the unprimed coordinates for the stationary frame relative to which the turntable is rotating. > 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 ,where you have dropped the primes on the coordinates. > > 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. Unfortunately, you seem to have made a mistake. This determinant is equal to 1, not 1 + 2*V^2. > > Then use ds = sqrt(g)*dt. > Since g = 1, this expression would give ds = dt. Furthermore, this example has g_0'0' = 1 and the expression ds = cdTau is OK. My claim is, if you recall, ds = cdTau fails whenever g_0'0' is NOT 1... Incidentally, I have never seen the expression ds = sqrt(g)*dt in the literature. I would have appreciated a reference. Regards, Murat Ozer > 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