> > > I agree. 
> > > The correct answer is that in this system GR must be applied. 

> > No, SR can deal with the question of the proper time recorded by an 
> > accelerating clock by considering its path as seen in an inertial frame. 
> > Just integrate [squareroot(1 - v(t)^2/c^2) dt] between two times t1 and 
> > t2 in your frame, with v(t) being the clock's instantaneous velocity at 
> > time t in your frame; this will give the correct answer for the time 
> > elapsed on the accelerating clock during the interval (t1, t2) in your 
> > frame 

> You can see this in action on 
>   https://home.deds.nl/~dvdm/dirk/Physics/Acceleration.html 

[snip] 

Good think I forgot to KILLFILE you Donkey. The ecerpt from the above 
imbecile's link: 

"Observer I can parametrize the worldline of A with the same 
proper time T, so he will see infinitesimally consecutive velocities 
of A as v(T) and v(T+dT). 

Since v(T+dT) is the standard SR composition ('addition') of the 
velocities v(T) and dV(T), we can write (using c=1): 
        v(T+dT) = [ v(T) + dV(T) ] / [ 1 + v(T) dV(T) ] 
which, using the expression for dV(T), becomes: 
        v(T+dT) = [ v(T) + a(T) dT ] / [ 1 + v(T) a(T) dT ] 

To calculate dv(T)/dT we use 
        [ v(T+dT) - v(T) ] / dT = a(T) [1-v^2(T)] / [ 1 + v(T) a(T) dT ] 
Take the limit dT --> 0 
        dv(T)/dT = a(T) [ 1 - v^2(T) ]" 

Now, you think you can fool people with this stupid derivation of 
yours. You ain't going to fool me Dinky-Donkey. 

The error in the above derivation is basic. Only a Dinky-Donkey could 
err that way. 

Are you so stupid Dink-Donk? 

Hint: maybe mixing Lorentz and Galilean relativity? 

Mike 

 See also

http://groups.google.com/group/sci.physics/browse_frm/thread/a27832a9a1e8c6f7