[SOLVED] Proper Time in General Relativity

Murat Ozer

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

Murat Ozer

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

Murat Ozer

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

Murat Ozer

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

Murat Ozer

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

Murat Ozer

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

Igor Khavkine

On 2005-09-11, Murat Ozer <Murat.H.Ozer@gmail.com> wrote:
> Igor Khavkine wrote:

> 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.

Consider an even simpler example. A particle moving with a uniform
velocity. There should be only one rest frame for this particle, right?
But if we take what you are saying at face value, then any coordinate
system where the coordinates of the particle's position stay constant,
can be considered as its rest frame. For example, such coordinate
systems are obtained by both Lorentzian and Galilean boosts. Quite
clearly, the only the Lorentzian boost gives the rigth choice of
coordinates for the particle's rest frame. Something makes this choice
special. Do you know what that something is?

>> 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.

Ok, and what is the precise reason that makes it obvious?

> (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.

There is only one proper time, which is independent of the choice of
coodinate system, and measurable (just by looking at the clock). If you
obtain different expressions for it, then both of them cannot be corect.

Your operational definition gives the right answer for one choice of
frame, but not for another. You must identify what makes the frame where
you get the right answer special and incorporate this condition as part
of your operational definition.

> 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?

That's precisely what makes your "operational definition" give an
incorrect result in this case.

I hope it's clear by now that everything revolves around inertia. You
are very close to getting the "operational definition" of proper time to
work. You just need to incorporate inertia into it. BTW, do you know how
to identify physicall, or operationally if you will, whether a given
coordinate system describes an inertial frame of reference?


>> 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.

Only as long as it is correct.

Igor

Igor Khavkine

On 2005-09-11, Murat Ozer <Murat.H.Ozer@gmail.com> wrote:
> Igor Khavkine wrote:

> 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.

Consider an even simpler example. A particle moving with a uniform
velocity. There should be only one rest frame for this particle, right?
But if we take what you are saying at face value, then any coordinate
system where the coordinates of the particle's position stay constant,
can be considered as its rest frame. For example, such coordinate
systems are obtained by both Lorentzian and Galilean boosts. Quite
clearly, the only the Lorentzian boost gives the rigth choice of
coordinates for the particle's rest frame. Something makes this choice
special. Do you know what that something is?

>> 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.

Ok, and what is the precise reason that makes it obvious?

> (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.

There is only one proper time, which is independent of the choice of
coodinate system, and measurable (just by looking at the clock). If you
obtain different expressions for it, then both of them cannot be corect.

Your operational definition gives the right answer for one choice of
frame, but not for another. You must identify what makes the frame where
you get the right answer special and incorporate this condition as part
of your operational definition.

> 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?

That's precisely what makes your "operational definition" give an
incorrect result in this case.

I hope it's clear by now that everything revolves around inertia. You
are very close to getting the "operational definition" of proper time to
work. You just need to incorporate inertia into it. BTW, do you know how
to identify physicall, or operationally if you will, whether a given
coordinate system describes an inertial frame of reference?


>> 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.

Only as long as it is correct.

Igor

Igor Khavkine

On 2005-09-11, Murat Ozer <Murat.H.Ozer@gmail.com> wrote:
> Igor Khavkine wrote:

> 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.

Consider an even simpler example. A particle moving with a uniform
velocity. There should be only one rest frame for this particle, right?
But if we take what you are saying at face value, then any coordinate
system where the coordinates of the particle's position stay constant,
can be considered as its rest frame. For example, such coordinate
systems are obtained by both Lorentzian and Galilean boosts. Quite
clearly, the only the Lorentzian boost gives the rigth choice of
coordinates for the particle's rest frame. Something makes this choice
special. Do you know what that something is?

>> 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.

Ok, and what is the precise reason that makes it obvious?

> (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.

There is only one proper time, which is independent of the choice of
coodinate system, and measurable (just by looking at the clock). If you
obtain different expressions for it, then both of them cannot be corect.

Your operational definition gives the right answer for one choice of
frame, but not for another. You must identify what makes the frame where
you get the right answer special and incorporate this condition as part
of your operational definition.

> 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?

That's precisely what makes your "operational definition" give an
incorrect result in this case.

I hope it's clear by now that everything revolves around inertia. You
are very close to getting the "operational definition" of proper time to
work. You just need to incorporate inertia into it. BTW, do you know how
to identify physicall, or operationally if you will, whether a given
coordinate system describes an inertial frame of reference?


>> 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.

Only as long as it is correct.

Igor

Igor Khavkine

On 2005-09-11, Murat Ozer <Murat.H.Ozer@gmail.com> wrote:
> Igor Khavkine wrote:

> 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.

Consider an even simpler example. A particle moving with a uniform
velocity. There should be only one rest frame for this particle, right?
But if we take what you are saying at face value, then any coordinate
system where the coordinates of the particle's position stay constant,
can be considered as its rest frame. For example, such coordinate
systems are obtained by both Lorentzian and Galilean boosts. Quite
clearly, the only the Lorentzian boost gives the rigth choice of
coordinates for the particle's rest frame. Something makes this choice
special. Do you know what that something is?

>> 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.

Ok, and what is the precise reason that makes it obvious?

> (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.

There is only one proper time, which is independent of the choice of
coodinate system, and measurable (just by looking at the clock). If you
obtain different expressions for it, then both of them cannot be corect.

Your operational definition gives the right answer for one choice of
frame, but not for another. You must identify what makes the frame where
you get the right answer special and incorporate this condition as part
of your operational definition.

> 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?

That's precisely what makes your "operational definition" give an
incorrect result in this case.

I hope it's clear by now that everything revolves around inertia. You
are very close to getting the "operational definition" of proper time to
work. You just need to incorporate inertia into it. BTW, do you know how
to identify physicall, or operationally if you will, whether a given
coordinate system describes an inertial frame of reference?


>> 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.

Only as long as it is correct.

Igor

Igor Khavkine

On 2005-09-11, Murat Ozer <Murat.H.Ozer@gmail.com> wrote:
> Igor Khavkine wrote:

> 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.

Consider an even simpler example. A particle moving with a uniform
velocity. There should be only one rest frame for this particle, right?
But if we take what you are saying at face value, then any coordinate
system where the coordinates of the particle's position stay constant,
can be considered as its rest frame. For example, such coordinate
systems are obtained by both Lorentzian and Galilean boosts. Quite
clearly, the only the Lorentzian boost gives the rigth choice of
coordinates for the particle's rest frame. Something makes this choice
special. Do you know what that something is?

>> 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.

Ok, and what is the precise reason that makes it obvious?

> (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.

There is only one proper time, which is independent of the choice of
coodinate system, and measurable (just by looking at the clock). If you
obtain different expressions for it, then both of them cannot be corect.

Your operational definition gives the right answer for one choice of
frame, but not for another. You must identify what makes the frame where
you get the right answer special and incorporate this condition as part
of your operational definition.

> 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?

That's precisely what makes your "operational definition" give an
incorrect result in this case.

I hope it's clear by now that everything revolves around inertia. You
are very close to getting the "operational definition" of proper time to
work. You just need to incorporate inertia into it. BTW, do you know how
to identify physicall, or operationally if you will, whether a given
coordinate system describes an inertial frame of reference?


>> 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.

Only as long as it is correct.

Igor

Igor Khavkine

On 2005-09-11, Murat Ozer <Murat.H.Ozer@gmail.com> wrote:
> Igor Khavkine wrote:

> 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.

Consider an even simpler example. A particle moving with a uniform
velocity. There should be only one rest frame for this particle, right?
But if we take what you are saying at face value, then any coordinate
system where the coordinates of the particle's position stay constant,
can be considered as its rest frame. For example, such coordinate
systems are obtained by both Lorentzian and Galilean boosts. Quite
clearly, the only the Lorentzian boost gives the rigth choice of
coordinates for the particle's rest frame. Something makes this choice
special. Do you know what that something is?

>> 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.

Ok, and what is the precise reason that makes it obvious?

> (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.

There is only one proper time, which is independent of the choice of
coodinate system, and measurable (just by looking at the clock). If you
obtain different expressions for it, then both of them cannot be corect.

Your operational definition gives the right answer for one choice of
frame, but not for another. You must identify what makes the frame where
you get the right answer special and incorporate this condition as part
of your operational definition.

> 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?

That's precisely what makes your "operational definition" give an
incorrect result in this case.

I hope it's clear by now that everything revolves around inertia. You
are very close to getting the "operational definition" of proper time to
work. You just need to incorporate inertia into it. BTW, do you know how
to identify physicall, or operationally if you will, whether a given
coordinate system describes an inertial frame of reference?


>> 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.

Only as long as it is correct.

Igor

Igor Khavkine

On 2005-09-11, Murat Ozer <Murat.H.Ozer@gmail.com> wrote:
> Igor Khavkine wrote:

> 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.

Consider an even simpler example. A particle moving with a uniform
velocity. There should be only one rest frame for this particle, right?
But if we take what you are saying at face value, then any coordinate
system where the coordinates of the particle's position stay constant,
can be considered as its rest frame. For example, such coordinate
systems are obtained by both Lorentzian and Galilean boosts. Quite
clearly, the only the Lorentzian boost gives the rigth choice of
coordinates for the particle's rest frame. Something makes this choice
special. Do you know what that something is?

>> 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.

Ok, and what is the precise reason that makes it obvious?

> (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.

There is only one proper time, which is independent of the choice of
coodinate system, and measurable (just by looking at the clock). If you
obtain different expressions for it, then both of them cannot be corect.

Your operational definition gives the right answer for one choice of
frame, but not for another. You must identify what makes the frame where
you get the right answer special and incorporate this condition as part
of your operational definition.

> 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?

That's precisely what makes your "operational definition" give an
incorrect result in this case.

I hope it's clear by now that everything revolves around inertia. You
are very close to getting the "operational definition" of proper time to
work. You just need to incorporate inertia into it. BTW, do you know how
to identify physicall, or operationally if you will, whether a given
coordinate system describes an inertial frame of reference?


>> 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.

Only as long as it is correct.

Igor

Igor Khavkine

On 2005-09-11, Murat Ozer <Murat.H.Ozer@gmail.com> wrote:
> Igor Khavkine wrote:

> 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.

Consider an even simpler example. A particle moving with a uniform
velocity. There should be only one rest frame for this particle, right?
But if we take what you are saying at face value, then any coordinate
system where the coordinates of the particle's position stay constant,
can be considered as its rest frame. For example, such coordinate
systems are obtained by both Lorentzian and Galilean boosts. Quite
clearly, the only the Lorentzian boost gives the rigth choice of
coordinates for the particle's rest frame. Something makes this choice
special. Do you know what that something is?

>> 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.

Ok, and what is the precise reason that makes it obvious?

> (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.

There is only one proper time, which is independent of the choice of
coodinate system, and measurable (just by looking at the clock). If you
obtain different expressions for it, then both of them cannot be corect.

Your operational definition gives the right answer for one choice of
frame, but not for another. You must identify what makes the frame where
you get the right answer special and incorporate this condition as part
of your operational definition.

> 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?

That's precisely what makes your "operational definition" give an
incorrect result in this case.

I hope it's clear by now that everything revolves around inertia. You
are very close to getting the "operational definition" of proper time to
work. You just need to incorporate inertia into it. BTW, do you know how
to identify physicall, or operationally if you will, whether a given
coordinate system describes an inertial frame of reference?


>> 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.

Only as long as it is correct.

Igor

Igor Khavkine

On 2005-09-11, Murat Ozer <Murat.H.Ozer@gmail.com> wrote:
> Igor Khavkine wrote:

> 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.

Consider an even simpler example. A particle moving with a uniform
velocity. There should be only one rest frame for this particle, right?
But if we take what you are saying at face value, then any coordinate
system where the coordinates of the particle's position stay constant,
can be considered as its rest frame. For example, such coordinate
systems are obtained by both Lorentzian and Galilean boosts. Quite
clearly, the only the Lorentzian boost gives the rigth choice of
coordinates for the particle's rest frame. Something makes this choice
special. Do you know what that something is?

>> 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.

Ok, and what is the precise reason that makes it obvious?

> (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.

There is only one proper time, which is independent of the choice of
coodinate system, and measurable (just by looking at the clock). If you
obtain different expressions for it, then both of them cannot be corect.

Your operational definition gives the right answer for one choice of
frame, but not for another. You must identify what makes the frame where
you get the right answer special and incorporate this condition as part
of your operational definition.

> 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?

That's precisely what makes your "operational definition" give an
incorrect result in this case.

I hope it's clear by now that everything revolves around inertia. You
are very close to getting the "operational definition" of proper time to
work. You just need to incorporate inertia into it. BTW, do you know how
to identify physicall, or operationally if you will, whether a given
coordinate system describes an inertial frame of reference?


>> 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.

Only as long as it is correct.

Igor

Igor Khavkine

On 2005-09-11, Murat Ozer <Murat.H.Ozer@gmail.com> wrote:
> Igor Khavkine wrote:

> 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.

Consider an even simpler example. A particle moving with a uniform
velocity. There should be only one rest frame for this particle, right?
But if we take what you are saying at face value, then any coordinate
system where the coordinates of the particle's position stay constant,
can be considered as its rest frame. For example, such coordinate
systems are obtained by both Lorentzian and Galilean boosts. Quite
clearly, the only the Lorentzian boost gives the rigth choice of
coordinates for the particle's rest frame. Something makes this choice
special. Do you know what that something is?

>> 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.

Ok, and what is the precise reason that makes it obvious?

> (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.

There is only one proper time, which is independent of the choice of
coodinate system, and measurable (just by looking at the clock). If you
obtain different expressions for it, then both of them cannot be corect.

Your operational definition gives the right answer for one choice of
frame, but not for another. You must identify what makes the frame where
you get the right answer special and incorporate this condition as part
of your operational definition.

> 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?

That's precisely what makes your "operational definition" give an
incorrect result in this case.

I hope it's clear by now that everything revolves around inertia. You
are very close to getting the "operational definition" of proper time to
work. You just need to incorporate inertia into it. BTW, do you know how
to identify physicall, or operationally if you will, whether a given
coordinate system describes an inertial frame of reference?


>> 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.

Only as long as it is correct.

Igor

Igor Khavkine

On 2005-09-11, Murat Ozer <Murat.H.Ozer@gmail.com> wrote:
> Igor Khavkine wrote:

> 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.

Consider an even simpler example. A particle moving with a uniform
velocity. There should be only one rest frame for this particle, right?
But if we take what you are saying at face value, then any coordinate
system where the coordinates of the particle's position stay constant,
can be considered as its rest frame. For example, such coordinate
systems are obtained by both Lorentzian and Galilean boosts. Quite
clearly, the only the Lorentzian boost gives the rigth choice of
coordinates for the particle's rest frame. Something makes this choice
special. Do you know what that something is?

>> 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.

Ok, and what is the precise reason that makes it obvious?

> (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.

There is only one proper time, which is independent of the choice of
coodinate system, and measurable (just by looking at the clock). If you
obtain different expressions for it, then both of them cannot be corect.

Your operational definition gives the right answer for one choice of
frame, but not for another. You must identify what makes the frame where
you get the right answer special and incorporate this condition as part
of your operational definition.

> 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?

That's precisely what makes your "operational definition" give an
incorrect result in this case.

I hope it's clear by now that everything revolves around inertia. You
are very close to getting the "operational definition" of proper time to
work. You just need to incorporate inertia into it. BTW, do you know how
to identify physicall, or operationally if you will, whether a given
coordinate system describes an inertial frame of reference?


>> 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.

Only as long as it is correct.

Igor