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yuiop

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In other threads the topic of how the proper time of a clock under acceleration has come up a number of times so I have decided to analyse this subject in a little more detail.

For this analysis we have these conditions:

a)Two clocks are initially at rest separated by a distance [itex]L_o[/itex] along the x axis.

b) Both clocks are simultaneously born-rigid accelerated along the x-axis (maintaining constant proper separation).

c) The experiment happens in flat space.

d) When the rear clock reaches velocity v relative to the initial frame it stops accelerating and maintains constant velocity v.

e) When the front clock reaches velocity v relative to the initial frame it stops accelerating and maintains constant velocity.

f) In the final state both clocks have constant and equal velocity and are separated by the proper distance [itex]L_o[/itex]

Referring to the attached diagram:

[itex] a_F [/itex] is the acceleration of the front clock proportional to [itex]c^2/x_F[/itex]

[itex] a_R [/itex] is the acceleration of the rear clock proportional to [itex]c^2/x_R[/itex]

[itex] t_F [/itex] is the coordinate time that the front clock clock takes to get to velocity v.

[itex] t_R [/itex] is the coordinate time that the rear clock clock takes to get to velocity v.

The dashed green line from the origin going through [itex]T_{R1}[/itex] and [itex]T_F[/itex] is a line of simultaneity in the accelerated clock frame.

The red lines are constant velocity phase after the acceleration phase.

When the front clock stops accelerating the proper time that has elapsed ([itex]t_f[/itex]) is given by the formula:

(Eq 1) [tex]T_F = { c \over a_F} \sinh^{-1} \left({a_F t_F \over c} \right)[/tex]

When the rear clock stops accelerating the proper time that has elapsed ([itex]t_R[/itex]) is given by the formula:

(Eq 2) [tex]T_{R1} = { c \over a_R} \sinh^{-1} \left({a_R t_R \over c} \right)[/tex]

The rear clock accelerates faster and gets to velocity v quicker than the front clock. The proper time that elapses from the coordinate time the rear clock stops accelerating to coordinate time that the front clock stops accelerating is simply the coordinate time difference adjusted by the Lorentz factor:

(Eq 3) [tex]T_{R2} - T_{R1} = (t_f - t_r)\sqrt{1-v^2/c^2}[/tex]

The difference in proper time between the front clock and the rear clock after accelerating is given by

(Eq 4) = (Eq1) - (Eq 2) - (Eq 3)

The following equations are generally known:

(Eq 5) [tex] a=\sqrt{\left({c^4 \over x^2-c^2 t^2} \right)}[/tex]

(Eq 6) [tex] v={a t \over \sqrt{\left(1+(a t /c)^2\right)}}[/tex]

and the acceleration of the front clock in terms of acceleration of the rear clock can easily be derived as :

(Eq 7) [tex] a_F = {a_R c^2 \over (L_o a_R +c^2) [/tex]

By substituting (Eq 5), (Eq 6) and (Eq 7) into (Eq 4) and solving the time difference between the front and rear clocks as seen in the initial frame is given by:

(Eq 8) [tex] T_F - T_{R2} = {-L_o v \over c^2} + {L_o \over c} \sinh^{-1} \left({ v \over c \sqrt{1-v^2/c^2}} \right)[/tex]

When the clocks are synchronised again it is well known that:

(Eq 9) [tex] T_F - T_{R2} = {-L_o v \over c^2} [/tex]

It is interesting to note that the difference in the proper times of the clocks after born-rigid acceleration and before synchronisation (Eq 8) is completely determined by the final velocity and the proper spatial separation and is independent of the rate of acceleration or time taken to transfer from one frame to another.

It is obvious from (Eq 8) that if constant proper separation is maintained the clocks will not remain naturally synchronised.

It is further conjectured that there is no acceleration scheme that can keep two spatially separated clocks synchronised when transferring the clocks from one inertial reference frame to another.*

I have left quite a few gaps in the derivation to try and keep it brief, but I will fill the gaps in, if anyone is really interested.

[EDIT] *I found a counter proof to this conjecture in post #9. (This does not change the validity of (Eq 8) in the above context of born-rigid transportation.

For this analysis we have these conditions:

a)Two clocks are initially at rest separated by a distance [itex]L_o[/itex] along the x axis.

b) Both clocks are simultaneously born-rigid accelerated along the x-axis (maintaining constant proper separation).

c) The experiment happens in flat space.

d) When the rear clock reaches velocity v relative to the initial frame it stops accelerating and maintains constant velocity v.

e) When the front clock reaches velocity v relative to the initial frame it stops accelerating and maintains constant velocity.

f) In the final state both clocks have constant and equal velocity and are separated by the proper distance [itex]L_o[/itex]

Referring to the attached diagram:

[itex] a_F [/itex] is the acceleration of the front clock proportional to [itex]c^2/x_F[/itex]

[itex] a_R [/itex] is the acceleration of the rear clock proportional to [itex]c^2/x_R[/itex]

[itex] t_F [/itex] is the coordinate time that the front clock clock takes to get to velocity v.

[itex] t_R [/itex] is the coordinate time that the rear clock clock takes to get to velocity v.

The dashed green line from the origin going through [itex]T_{R1}[/itex] and [itex]T_F[/itex] is a line of simultaneity in the accelerated clock frame.

The red lines are constant velocity phase after the acceleration phase.

When the front clock stops accelerating the proper time that has elapsed ([itex]t_f[/itex]) is given by the formula:

(Eq 1) [tex]T_F = { c \over a_F} \sinh^{-1} \left({a_F t_F \over c} \right)[/tex]

When the rear clock stops accelerating the proper time that has elapsed ([itex]t_R[/itex]) is given by the formula:

(Eq 2) [tex]T_{R1} = { c \over a_R} \sinh^{-1} \left({a_R t_R \over c} \right)[/tex]

The rear clock accelerates faster and gets to velocity v quicker than the front clock. The proper time that elapses from the coordinate time the rear clock stops accelerating to coordinate time that the front clock stops accelerating is simply the coordinate time difference adjusted by the Lorentz factor:

(Eq 3) [tex]T_{R2} - T_{R1} = (t_f - t_r)\sqrt{1-v^2/c^2}[/tex]

The difference in proper time between the front clock and the rear clock after accelerating is given by

(Eq 4) = (Eq1) - (Eq 2) - (Eq 3)

The following equations are generally known:

(Eq 5) [tex] a=\sqrt{\left({c^4 \over x^2-c^2 t^2} \right)}[/tex]

(Eq 6) [tex] v={a t \over \sqrt{\left(1+(a t /c)^2\right)}}[/tex]

and the acceleration of the front clock in terms of acceleration of the rear clock can easily be derived as :

(Eq 7) [tex] a_F = {a_R c^2 \over (L_o a_R +c^2) [/tex]

By substituting (Eq 5), (Eq 6) and (Eq 7) into (Eq 4) and solving the time difference between the front and rear clocks as seen in the initial frame is given by:

(Eq 8) [tex] T_F - T_{R2} = {-L_o v \over c^2} + {L_o \over c} \sinh^{-1} \left({ v \over c \sqrt{1-v^2/c^2}} \right)[/tex]

When the clocks are synchronised again it is well known that:

(Eq 9) [tex] T_F - T_{R2} = {-L_o v \over c^2} [/tex]

It is interesting to note that the difference in the proper times of the clocks after born-rigid acceleration and before synchronisation (Eq 8) is completely determined by the final velocity and the proper spatial separation and is independent of the rate of acceleration or time taken to transfer from one frame to another.

It is obvious from (Eq 8) that if constant proper separation is maintained the clocks will not remain naturally synchronised.

It is further conjectured that there is no acceleration scheme that can keep two spatially separated clocks synchronised when transferring the clocks from one inertial reference frame to another.*

I have left quite a few gaps in the derivation to try and keep it brief, but I will fill the gaps in, if anyone is really interested.

[EDIT] *I found a counter proof to this conjecture in post #9. (This does not change the validity of (Eq 8) in the above context of born-rigid transportation.

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