[SOLVED] Proper Time in General Relativity

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

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

Murat Ozer

Igor Khavkine wrote:
> 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?

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.

No...

> For example, such coordinate
> systems are obtained by both Lorentzian and Galilean boosts.

No!...The rest frame is imposed upon you by the specific problem
given to you. For example, the (global) rest frame of a clock in a
gravitational field is not a Lorentzian rest frame. It is definitely
noninertial. (Of course, we are not talking locally here.)
As for Galilean versus Lorentzian boosts for a clock in uniform linear
(as well as circular) motion there is a different kind of problem.
If you take a look at my "Is a uniformly rotating system an
inertial frame?" posting, to which nobody has made any remarks so far,
you will see that I have raised a very important issue there.


> Quite
> clearly, the only the Lorentzian boost gives the rigth choice of
> coordinates for the particle's rest frame.

Again, the particle's rest frame is not defined by choice, it is
defined by the space given to you. If the space is flat, the rest
frame of the particle is Lorentzian. If not, the rest frame is simply
not Lorentzian, it is noninertial.



> 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?

That g_0'0' is not equal to 1, and there is a mixing term, namely
g_0'1'.


>
> > (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.
>

You are missing what I am trying to say here. These two moving systems
that are related to the stationary system by Galilean and Lorentz
transformations are two different systems. My point is, the operational
definition gives the correct answer for the Lorentzian system.
Therefore,there is no reason why it should not give the correct
answer for the Galilean system. These two systems are not
identical; one is flat and the other one is curved. This issue
becomes clearer for a rotating system. One can use the Galilean
transformations to obtain a curved system, or the Lorentz
transformations to obtain a seemingly inertial system. These
are completely different systems whose metrics are different,
especially whose g_0'0's are different. The proper times for
these two nonidentical systems must be different. The proper time
obtained by the operational definition applied to the metric and
the proper time obtained from the transformation equations are
the same for each system. Though ds = cdTau gives the same answer
as the transformation equation for the Lorentzian system, it does
not give the same answer as the transformation equations for the
Galilean system. Therefore, I conclude, the definition ds = c dTau
fails whenever g_0'0' is not equal to 1...

> 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.


That's your claim which you have not supported by any physical or
mathematical reason or fact so far. Claims must be supported by reason
or fact as you know.


>
> I hope it's clear by now that everything revolves around inertia.

I do not agree. I believe we need the opinion of the GR experts.
Where are they? Why are they so silent?

Murat Ozer


>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

Murat Ozer

Igor Khavkine wrote:
> 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?

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.

No...

> For example, such coordinate
> systems are obtained by both Lorentzian and Galilean boosts.

No!...The rest frame is imposed upon you by the specific problem
given to you. For example, the (global) rest frame of a clock in a
gravitational field is not a Lorentzian rest frame. It is definitely
noninertial. (Of course, we are not talking locally here.)
As for Galilean versus Lorentzian boosts for a clock in uniform linear
(as well as circular) motion there is a different kind of problem.
If you take a look at my "Is a uniformly rotating system an
inertial frame?" posting, to which nobody has made any remarks so far,
you will see that I have raised a very important issue there.


> Quite
> clearly, the only the Lorentzian boost gives the rigth choice of
> coordinates for the particle's rest frame.

Again, the particle's rest frame is not defined by choice, it is
defined by the space given to you. If the space is flat, the rest
frame of the particle is Lorentzian. If not, the rest frame is simply
not Lorentzian, it is noninertial.



> 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?

That g_0'0' is not equal to 1, and there is a mixing term, namely
g_0'1'.


>
> > (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.
>

You are missing what I am trying to say here. These two moving systems
that are related to the stationary system by Galilean and Lorentz
transformations are two different systems. My point is, the operational
definition gives the correct answer for the Lorentzian system.
Therefore,there is no reason why it should not give the correct
answer for the Galilean system. These two systems are not
identical; one is flat and the other one is curved. This issue
becomes clearer for a rotating system. One can use the Galilean
transformations to obtain a curved system, or the Lorentz
transformations to obtain a seemingly inertial system. These
are completely different systems whose metrics are different,
especially whose g_0'0's are different. The proper times for
these two nonidentical systems must be different. The proper time
obtained by the operational definition applied to the metric and
the proper time obtained from the transformation equations are
the same for each system. Though ds = cdTau gives the same answer
as the transformation equation for the Lorentzian system, it does
not give the same answer as the transformation equations for the
Galilean system. Therefore, I conclude, the definition ds = c dTau
fails whenever g_0'0' is not equal to 1...

> 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.


That's your claim which you have not supported by any physical or
mathematical reason or fact so far. Claims must be supported by reason
or fact as you know.


>
> I hope it's clear by now that everything revolves around inertia.

I do not agree. I believe we need the opinion of the GR experts.
Where are they? Why are they so silent?

Murat Ozer


>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

Murat Ozer

Igor Khavkine wrote:
> 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?

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.

No...

> For example, such coordinate
> systems are obtained by both Lorentzian and Galilean boosts.

No!...The rest frame is imposed upon you by the specific problem
given to you. For example, the (global) rest frame of a clock in a
gravitational field is not a Lorentzian rest frame. It is definitely
noninertial. (Of course, we are not talking locally here.)
As for Galilean versus Lorentzian boosts for a clock in uniform linear
(as well as circular) motion there is a different kind of problem.
If you take a look at my "Is a uniformly rotating system an
inertial frame?" posting, to which nobody has made any remarks so far,
you will see that I have raised a very important issue there.


> Quite
> clearly, the only the Lorentzian boost gives the rigth choice of
> coordinates for the particle's rest frame.

Again, the particle's rest frame is not defined by choice, it is
defined by the space given to you. If the space is flat, the rest
frame of the particle is Lorentzian. If not, the rest frame is simply
not Lorentzian, it is noninertial.



> 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?

That g_0'0' is not equal to 1, and there is a mixing term, namely
g_0'1'.


>
> > (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.
>

You are missing what I am trying to say here. These two moving systems
that are related to the stationary system by Galilean and Lorentz
transformations are two different systems. My point is, the operational
definition gives the correct answer for the Lorentzian system.
Therefore,there is no reason why it should not give the correct
answer for the Galilean system. These two systems are not
identical; one is flat and the other one is curved. This issue
becomes clearer for a rotating system. One can use the Galilean
transformations to obtain a curved system, or the Lorentz
transformations to obtain a seemingly inertial system. These
are completely different systems whose metrics are different,
especially whose g_0'0's are different. The proper times for
these two nonidentical systems must be different. The proper time
obtained by the operational definition applied to the metric and
the proper time obtained from the transformation equations are
the same for each system. Though ds = cdTau gives the same answer
as the transformation equation for the Lorentzian system, it does
not give the same answer as the transformation equations for the
Galilean system. Therefore, I conclude, the definition ds = c dTau
fails whenever g_0'0' is not equal to 1...

> 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.


That's your claim which you have not supported by any physical or
mathematical reason or fact so far. Claims must be supported by reason
or fact as you know.


>
> I hope it's clear by now that everything revolves around inertia.

I do not agree. I believe we need the opinion of the GR experts.
Where are they? Why are they so silent?

Murat Ozer


>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

Murat Ozer

Igor Khavkine wrote:
> 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?

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.

No...

> For example, such coordinate
> systems are obtained by both Lorentzian and Galilean boosts.

No!...The rest frame is imposed upon you by the specific problem
given to you. For example, the (global) rest frame of a clock in a
gravitational field is not a Lorentzian rest frame. It is definitely
noninertial. (Of course, we are not talking locally here.)
As for Galilean versus Lorentzian boosts for a clock in uniform linear
(as well as circular) motion there is a different kind of problem.
If you take a look at my "Is a uniformly rotating system an
inertial frame?" posting, to which nobody has made any remarks so far,
you will see that I have raised a very important issue there.


> Quite
> clearly, the only the Lorentzian boost gives the rigth choice of
> coordinates for the particle's rest frame.

Again, the particle's rest frame is not defined by choice, it is
defined by the space given to you. If the space is flat, the rest
frame of the particle is Lorentzian. If not, the rest frame is simply
not Lorentzian, it is noninertial.



> 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?

That g_0'0' is not equal to 1, and there is a mixing term, namely
g_0'1'.


>
> > (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.
>

You are missing what I am trying to say here. These two moving systems
that are related to the stationary system by Galilean and Lorentz
transformations are two different systems. My point is, the operational
definition gives the correct answer for the Lorentzian system.
Therefore,there is no reason why it should not give the correct
answer for the Galilean system. These two systems are not
identical; one is flat and the other one is curved. This issue
becomes clearer for a rotating system. One can use the Galilean
transformations to obtain a curved system, or the Lorentz
transformations to obtain a seemingly inertial system. These
are completely different systems whose metrics are different,
especially whose g_0'0's are different. The proper times for
these two nonidentical systems must be different. The proper time
obtained by the operational definition applied to the metric and
the proper time obtained from the transformation equations are
the same for each system. Though ds = cdTau gives the same answer
as the transformation equation for the Lorentzian system, it does
not give the same answer as the transformation equations for the
Galilean system. Therefore, I conclude, the definition ds = c dTau
fails whenever g_0'0' is not equal to 1...

> 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.


That's your claim which you have not supported by any physical or
mathematical reason or fact so far. Claims must be supported by reason
or fact as you know.


>
> I hope it's clear by now that everything revolves around inertia.

I do not agree. I believe we need the opinion of the GR experts.
Where are they? Why are they so silent?

Murat Ozer


>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

Murat Ozer

Igor Khavkine wrote:
> 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?

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.

No...

> For example, such coordinate
> systems are obtained by both Lorentzian and Galilean boosts.

No!...The rest frame is imposed upon you by the specific problem
given to you. For example, the (global) rest frame of a clock in a
gravitational field is not a Lorentzian rest frame. It is definitely
noninertial. (Of course, we are not talking locally here.)
As for Galilean versus Lorentzian boosts for a clock in uniform linear
(as well as circular) motion there is a different kind of problem.
If you take a look at my "Is a uniformly rotating system an
inertial frame?" posting, to which nobody has made any remarks so far,
you will see that I have raised a very important issue there.


> Quite
> clearly, the only the Lorentzian boost gives the rigth choice of
> coordinates for the particle's rest frame.

Again, the particle's rest frame is not defined by choice, it is
defined by the space given to you. If the space is flat, the rest
frame of the particle is Lorentzian. If not, the rest frame is simply
not Lorentzian, it is noninertial.



> 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?

That g_0'0' is not equal to 1, and there is a mixing term, namely
g_0'1'.


>
> > (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.
>

You are missing what I am trying to say here. These two moving systems
that are related to the stationary system by Galilean and Lorentz
transformations are two different systems. My point is, the operational
definition gives the correct answer for the Lorentzian system.
Therefore,there is no reason why it should not give the correct
answer for the Galilean system. These two systems are not
identical; one is flat and the other one is curved. This issue
becomes clearer for a rotating system. One can use the Galilean
transformations to obtain a curved system, or the Lorentz
transformations to obtain a seemingly inertial system. These
are completely different systems whose metrics are different,
especially whose g_0'0's are different. The proper times for
these two nonidentical systems must be different. The proper time
obtained by the operational definition applied to the metric and
the proper time obtained from the transformation equations are
the same for each system. Though ds = cdTau gives the same answer
as the transformation equation for the Lorentzian system, it does
not give the same answer as the transformation equations for the
Galilean system. Therefore, I conclude, the definition ds = c dTau
fails whenever g_0'0' is not equal to 1...

> 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.


That's your claim which you have not supported by any physical or
mathematical reason or fact so far. Claims must be supported by reason
or fact as you know.


>
> I hope it's clear by now that everything revolves around inertia.

I do not agree. I believe we need the opinion of the GR experts.
Where are they? Why are they so silent?

Murat Ozer


>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