Environmental Engineering Reference
In-Depth Information
δv
2 πh/( 1day ) ) but that leads to a negligible correction. We can
eliminate the dependence upon the time measured in the inertial frame by
constructing the ratio:
=
τ plane
τ ground
v) 2
V 2
c 2
v 2
c 2
gh
c 2
1
2
(V
+
1
2
gh
c 2
1
2
Vv
c 2 .
1
+
+
1
+
c 2
Thus the clock on the airplane speeds up by a fractional amount gh/c 2 due
to the fact it is in a weaker gravitational field but it slows down in part due to
time dilation (by a factor v 2 /( 2 c 2 ) ) and in part due to the fact that the Earth
is rotating (the Vv/c 2
factor). Putting in the numbers: gh/c 2
10 12
×
1 . 1
v 2 /( 2 c 2 )
10 13
≈−
×
which corresponds to a speeding up of 3.9 ns/h;
4 . 3
10 12
which corresponds to a slowing down of -5.2 ns/h. The total journey time
for one circuit of the equator is just over 40 hours and thus the net effect is
that the clock on the airplane slows down by 0 . 11 µ s which is an effect large
enough to have been measured.
14.7 The tension in the rope would increase until it snaps. This is easiest to see
in the inertial frame in which the rockets are initially at rest. If the rockets
are initially a distance L 0 apart then they must remain a distance L 0 apart
in this frame since they undergo identical accelerations (they are identical
rockets). However the rope should suffer a length contraction (it is of length
L 0 only in its rest frame) but it is prevented from so doing since it is attached
to the rockets. Hence it snaps. Alternatively, we can looking at things from
the point of view of an observer on the rocket at the rear. Eq.(14.28) is
useful for it informs us that in order to keep the distance between the rockets
fixed, the rocket at the front must experience a reduced proper acceleration,
but since both rockets are identical this is not the case (observers on each
rocket experience the same acceleration) and so the rocket at the front has too
great an acceleration and the distance between the rockets increases, again
eventually causing the rope to snap.
Vv/c 2
≈−
×
which corresponds to a slowing down of -1.5 ns/h;
1 . 4
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