Sam Yes Even a car going at 60,but no relativistic effects can be
measured. Bert

"The Ghost In The Machine" <ewill@sirius.athghost7038suus.net> wrote in message news:dg0lf2-49r.ln1@sirius.athghost7038suus.net...

In sci.physics, Dirk Van de moortel
<dirkvandemoortel@ThankS-NO-SperM.hotmail.com>
wrote
on Wed, 2 Mar 2005 16:49:39 +0100

[snip]

To make it easier...

Adam, Eve1 and Eve2 are hanging there, far from
any gravitational influences...
At a certain moment Eve1 and Eve 2 go away and
return together after t years, as seen by Adam.

Eve1 gets acceleration a = g for t/4 years. Then
a = -g for t/2 years, and finally again g for t/4 years,
so she makes the smoothest dreamable re-docking
with Adam.

Eve2 is teletransported to Earth, feeling 1g all the
time. After t years as seen by Adam, Eve2 is
transported back.

How much have the Eves aged?
Try with t = 10 years.
Easy, but you have to find the two equations :-)

Dirk Vdm

Not quite that simple. One issue is that Eve1's acceleration
is g *relative to her reference frame only*; Adam will see
Eve1's acceleration to be initially g, but it will decrease
as Eve1's velocity increases relative to Adam. (Assuming, of
course, that Adam is employing the naive calculation a = dv/dt.)

It is not simple indeed, unless you have the solution ;-)
See below...

As for Eve2, that's easy; Old Man's formula (I don't know
offhand where he got it from) perverts into

dt(r) / dt(inf) = sqrt(1 - 2*G*M/(r*c^2))

where r is the planet's radius and M is its mass, and G
is the universal gravitational constant. Since
g = GM/r^2, one can also rewrite this as

dt(r) / dt(inf) = sqrt(1 - 2*g*r/c^2)

Note the r in this formula; this makes for some difficulties
when comparing this GR formula with Einstein's "infinite elevator"
(or, in this problem, Eve1's impossible spacecraft).

For r = 6.378*10^6 m, c = 3 * 10^8 m/s, and g = 9.805 N/kg,
one gets a correction factor of (1 - 6.9583*10^-10), which
is about right. (Note that this is not the actual
factor used for GPS calculation, as one has to factor
in the height of the satellite and the fact that it's moving.)

Indeed. Ignoring the rotation, multiplying your factor
by 10 gives you something like 9.999999993 years.
Not much of a difference with Adam :-)

In Eve1's spacecraft, she is moving at velocity v. Her
craft accelerates at a constant rate a for a short time dt'.
After this time, her new rate will be approximately
(v + a * dt') / (1 + v*a*dt'/c^2), as derivable from
the Lorentz.

Subtracting, one gets

dv = (v + a * dt') / (1 + v*a*dt'/c^2) - v
= (v + a * dt' - v - v^2*a*dt'/c^2) / (1 + v*a*dt'/c^2)
= a*(1-v^2/c^2)*dt' / (1+v*a*dt'/c^2)

or dv/dt' = a*(1-v^2/c^2) / (1+v*a*dt'/c^2).

In this form, one can easily let dt'->0 and get

dv/dt' = a*(1-v^2/c^2).

Admittedly, this isn't quite right, as v is also a function of t'.
Therefore, this is a differential equation. However, it's
easily reversed:

dt'/dv = 1/(a*(1-v^2/c^2))
= 1/(a*(1-v/c)*(1+v/c))
= c/(2*a*v) *(1/(1-v/c) - 1/(1+v/c))

[Sign error alert on second term]

= c/(2*a) *(1/(c-v) - 1/(c+c))

and therefore t' = c/(2*a) * (log(c-v) - log(c+v))
= c/(2*a) * (log( (c-v)/(c+v))

With the sign error corrected:
dt'/dv = c/(2*a*v) *(1/(1-v/c) + 1/(1+v/c))
= c/(2*a) *(1/(c-v) + 1/(c+v))
and therefore (correcting another sign error):
t' = c/(2*a) * ( -log(c-v) + log(c+v) )
= c/(2*a) * (log( (c+v)/(c-v))

How one expresses v in terms of t', I've no clue at this point,
though an infinite series is of course always possible.

And then there's the t' => t conversion issue. Oy vey.

Exponentiating the expression for t' gives and pulling v out gives:
v(t') = c ( exp(2 a t'/c) - 1 ) / ( exp(2 a t'/c ) + 1 )
= c ( exp(a t'/c) - exp(-a t'/c) ) / ( exp(a t'/c) + exp(-a t'/c) )
= c sinh(a t'/c) / cosh(a t'/c)
= c tanh(a t'/c)

So your solution with the double sign error would have
given v(t') = - c tanh(a t'/c). You were so very close :-)

You find the complete derivation (and a pointer to the FAQ entry) on
http://users.pandora.be/vdmoortel/dirk/Physics/Acceleration.html
The equations are all together at the bottom. My T is your t'.

I have given Eve1 a constant proper acceleration a = g
during 1/4 of the Adam-time of her trip.
Adam can then use the equation
T(t) = c/a argsinh( a/c t )
where t = 10/4 years is Adam's perception of the time
of the quarter trip, and T(t) is Eve1's perception of
the time.

For symmetry reasons (seen on a spacetime diagram) the total trip
takes 4 times as much for both parties, so Eve1's final age will
increase with T = 4 T(10/4 years).
When taking 365.25*24*60*60 seconds per year, I get something like
6.5 years
which is much less than her sister's :-)

Dirk Vdm