simple problem in length contraction;
we have a double booster rocketship assembly consisting
of two identical 100 meter rocket ships connected by a
1 millimeter fiber optic communications cable of 100 meters.
100meter 100 meter
rocket X rocket Y
|========>>----------|========>>
100 meter
fiber optic cable
assuming uniform acceleration from a
launch pad on planet Xanadu to (0.99476 c)
for the rocketship assembly, what will the length
of the fiber optic cable be at this speed as
calculated by a mission control supervisor
resting on planet Xanadu.
and the answer is "c"
no, not -that- "c"
the multiple choice "c"
a. 0.24795 meters
b. 1500 meters
c. 10.227 meters <---
d. 1.3 meters
L' = L [1- v^2/c^2]^0.5
100 m [ 1 - 0.
9895474576 ]^0.5
10.227 meters
and, in another imaginative construct,
we simply demand that the space between the two rocketships,
with -no- string attached, must undergo -no- lengthening nor
contraction, inasmuch as it is, perceived as vacuum.
100meter rocketA 100 meter rocket B
_________ _________
|<100 meter gap>|
===>[near light speed travel]
1meter rocketA 1 meter rocket B
_ _
|<100 meter gap>|
_______ _______ _______ _______
|<10m>| |<10 m>| |<10 m>|
__ __ __ __
|<10m>| |<10 m>| |<10 m>|
but still, when the string is attached
we still don't say that the string remains 10m
_______......_______......_______......_______
__...__...__...__
it's difficult to say;
"empty space lengthens while filled space contracts"
but the other kid wants us to take a 10 m section of 'space'
and accelerate -that- to near light speed.
sort of like saying;
we take two projectiles which are constructed
of 100 decaliter evacuated glass tubes
and as light speed is approached, the glass contracts,
but the 'space' remains the same, and so, will the
glass tubing break?
100meter rocketA 100 meter rocket B
_________ _________
|<100 meter gap>|
[[ Launch pad ]] <---stationary observer
that's you, the inert observer
doing your calculations on
rocket A and rocket B assuming
light speed travel.
you calculate a length contraction on the TWO rockets
and -no- length contraction on
the gap between the two rockets,
and suppose that the space between
the two rockets has really increased.
1meter rocketA 1 meter rocket B
_ _
|<---198 meter gap-->|
and so, you say, the gap between
the two rockets enlarged, and, if you attached
a string between the two and did the same sort
of calculation, by gum the string will snap.
but, now, you take the same rocket ships
and attach a string that is the same
length as the gap
100meter rocketA 100 meter rocket B
_________................._________
100meter string
and -now- you do your calculation
on the entire 300 meter 'frame'
and find that the entire 'frame'
has contracted to 3 meters
as observed from the launch pad.
1meter rocketA 1 meter rocket B
_..._
1 meter string
the string remains in place and does not pull away.
the material design of the rockets and how
they manage to project themselves to light
speed is wholly irrelevant,
the relevant datum, is the length contraction
that is =calculated= by an inert observer, -you-
the strength of the rope is irrelevant.
the rope does not hold the rockets
together by virtue of its strength
the rope converts the =frame=
from two 100 meter rocketc
to ONE 300 meter rocket-string-rocket assembly
and =that= is the length that
contracts in the second case.
i'm telling you Major Matt Mason =proved= this already...