Hi Kees,

According to Stephen Hawking, particles traveling faster than c do,
indeed, travel backwards in time, so you're right on that one.

You're wrong on the second part, though. SR applies to inertial
non-accelerating) frames of reference. In the Twin Paradox, B
accelerates rapidly up to c/2, travels at c/2 for 20 Earth years,
delerates rapidly to rest, accelerates rapidly up to c/2 back towards
Earth, travels at a steady c/2 for another 20 Earth years, and finally
decelerates rapidly down to rest.

According to General Relativity, the forces caused by
acceleration/deceleration are indistinguishable from those caused by a
gravitational field. Further, clocks in high gravitational fields
definitely do run slower than those in lower fields. If we assume that
B's accelerations/decelerations were considerably higher than g, then
both A and B will agree that B has aged less than A when they are
together again.

Incidentally, travelling at c/2 does not cause the traveler's time rate
to halve. According to SR:

t(moving) = t(stationary)*sqrt(1 + v^2/c^2)

Hope this helps,

Cheers,

Pete

"Brit" wrote in message ups.com...
Hi Kees,

According to Stephen Hawking, particles traveling faster than c do,
indeed, travel backwards in time, so you're right on that one.

You're wrong on the second part, though. SR applies to inertial
non-accelerating) frames of reference.

But see
http://math.ucr.edu/home/baez/physic...eleration.html

In the Twin Paradox, B
accelerates rapidly up to c/2, travels at c/2 for 20 Earth years,
delerates rapidly to rest, accelerates rapidly up to c/2 back towards
Earth, travels at a steady c/2 for another 20 Earth years, and finally
decelerates rapidly down to rest.

According to General Relativity, the forces caused by
acceleration/deceleration are indistinguishable from those caused by a
gravitational field. Further, clocks in high gravitational fields
definitely do run slower than those in lower fields. If we assume that
B's accelerations/decelerations were considerably higher than g, then
both A and B will agree that B has aged less than A when they are
together again.

Incidentally, travelling at c/2 does not cause the traveler's time rate
to halve. According to SR:

t(moving) = t(stationary)*sqrt(1 + v^2/c^2)

Better write:
t(stationary) = t(moving) / sqrt(1 - v^2/c^2)
where
t(moving) is the time as seen in the moving frame
between two colocal events of that same frame, and
t(stationary) is the time as seen in the stationary frame
between those two same events.

But just as well:
t(moving) = t(stationary) / sqrt(1 - v^2/c^2)
where
t(stationary) is the time as seen in the stationary frame
between two colocal events of that same frame, and
t(moving) is the time as seen in the moving frame
between those two same events.

Dirk Vdm


Hope this helps,

Cheers,

Pete