On 1/29/2014 5:39 PM, LizR wrote:
On 30 January 2014 14:17, meekerdb <[email protected] <mailto:[email protected]>>
wrote:
On 1/29/2014 5:19 AM, Edgar L. Owen wrote:
Brent,
Here's another relativity question I'd like to get your explanation for
if I may...
In Thorne's 'Black Holes and Time Warps' he gives the following example.
Two observers A and B.
A leaves earth orbit to travel to the center of the galaxy, 30,100
light year
away, using a constant 1g acceleration to the midpoint and a constant 1g
decelleration on the second half of the journey to arrive stationary at
the
galactic center,
Thorne tells us that the 30,100 light year trip takes 30,102 years on
B's clock
back on earth but only 20 years on A's clock aboard the spaceship.
Now my question is what causes the extreme slowing of A's clock?
It can't be the acceleration as both A and B experience the exact same
1g
acceleration for the duration of the trip.
I can understand that during the trip B will observe A's clock to be
greatly
slowed due to the extreme relative motion, but since the motion IS
relative
wouldn't A also observe B's clock to be slowed by the same amount
during the trip?
And since the time dilation of relative motion is relative then how
does it
actually produce a real objective slowing of A's clock that both
observers can
agree upon?
You had said yesterday that "geometry doesn't cause clocks to slow" but
other
than the trivial 1g acceleration isn't all the rest just geometry in
this case?
What's the proper way to analyze this to get Thorne's result?
A rough way to see it is right is to note that c/g = 3e7sec ~ 1year <<
30,000yr. So
the spaceship spends essentially the whole flight at very near c. So the
trip takes
30,100+ years in the frame of the galaxy. But the proper time for the
spaceship is
very small; if it were actually at speed c, like a photon, its proper time
lapse
would be zero. Only, because it can't quite reach c, the time turns out to
be 20
years. To get the exact values you have to integrate the differential
equations:
dt/dtau = 1/gamma
dv/dtau = accel/gamma^2
dx/dtau = v/gamma
where gamma=sqrt(1-v^2)
The equivalence principle indicates that both A and B are in a 1g gravitational field
throughout the exercise, hence the time dilation experienced by A can't be
gravitational. All that leaves is the different distances they travel through space-time
to reach their final meeting, which is indeed down to "geometry" (in this case involving
curves rather the straight lines - but that is minor detail, and can be solved by
integrating the relevant equations, as indicated).
So I assume the overall geometry of their paths through space-time /is/ responsible for
the final mismatch between their clocks. I'm not sure whether that contradicts "geometry
doesn't cause clocks to slow" - probably not.
Exactly. The clocks faithfully measure the interval along their respective paths. It's
the difference in the paths, the geometry, that is the difference in duration.
PS I would instruct A to fly above the plane of the galaxy. There is a lot of stuff
between the Earth and the galactic centre and I suspect that even a dust grain would hit
a relativistic spacecraft like a nuclear bomb once it was near peak velocity, which
according to my calculations is 0.9999995c (or in any case p.d.q.)
Even without dust the intergalactic hydrogen atoms would make it similar to standing in
the LHC beam - but with a lot more luminosity.
Brent
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