Re: Help with apparent paradox please?


dunxuk@xxxxxxx wrote:
Thanks Martin. I am really getting confused with this 'speed of force
through matter' aspect; if the carriages are 'springy' because one end
can't keep up with the other how would this affect things or is this
not a key issue?

At the end of the day though does acceleration matter? If they both had
clocks and stopped them when the light arrived and left the brakes
alone would the clocks perfectly match?

Many thanks Martin!

Probably nothing will happen to the conductor, because at those
speeds, the breaks will break(fail). Seriously though, the front and
rear brakes cannot act at the same rate if the train is to stay
together.

By braking, your train is actually accelerating in the negative
direction.

Here's something I posted on accelerating rockets earlier, just
substitute train, accelerating the NEGATIVE direction, so the rear of
the train is equivalent to the front of the rocketship.

If the FRONT of a 1 light year long train accelerated at
1 light year per year per year, it would never receive a signal from
the rear of the train. Since rods and trains maintain their
structural
integrity through molecular forces, with a limiting speed of c, and
since
the molecules at the rear of the train whose forward end is
accelerating at
1 light year per year per year cannot interact with the molecules at
the
front of the train, there's no net molecular force holding the front
and
rear together.

Since there's no net molecular force holding the front and rear
together, the slightest bit of turbulence will separate the ends.
The train MUST come apart, per the pole- barn paradox.

Of course, this limit on acceleration is only a very high theoretical
upper bound, and there are other bounds on the size of the train, such
as the force
of gravity, and tensile strength of the material, which would place
much lower
bounds on the force of acceleraton a material body could withstand.


The clocks on the train won't be synchronized during acceleration.
Both the observer at the front (rear in braking train) of the train and
the observer at the
back of the train will measure the clock at the back of the train as
running 1/2 (aL/(c^2)) slower than the clock at the front, where a is
the acceleration of the train, L is the distance between the front
clock and rear clock.

Strictly speaking, there's no such thing as a rigid body, but as long
as the length of the train is small compared to the acceleration and
speed of light, so the (aL)/(c^2) figure is negligible, the
following formula gives distance and time for a given acceleration:


If a body has constant acceleration in its own rest frame, the



velocity at time T, with respect to the initial rest frame, will be

V= aT
_____________
SQRT( 1 + ((aT/c)^2))



the body's time, T' = c/a ln (aT/c + SQRT(1 +aaTT/cc).

An equivalent formula, easier to work with on your calculator, is
T= sinh aT' , X = cosh aT'


where a is the acceleration in light years/year,
T is earth time, and X-1 is distance traveled from earth's
persepctive.


sinh is the hyperbolic sine, cosh is the hyperbolic cosine.


0.97g is about 1 light year/year, so letting a rocket accelerate at 1
light year/year for 1 "rocket year,
T'= sinh 1 = 1.175 earth years.
The rocket will have traveled cosh(1) -1 = 1.54-1 =0.54 light years.
The initial rest frame's time will be sinh (1) = 1.175 years


The velocity will be 1*1)/sqrt(1 + (1)^2)
or 0.7017 c. From the Fitzgerald-Lorentz contraction, the length of
the ship will have contracted to 0.7071 of its "at rest" length. The
rear of the vessel(front of the decelerating train) will have to have
advanced .543 light years to
keep up,
plus have gained an additional 0.2929 light years to keep the
spaceship (train)
from stretching apart, for a total of 0.8359 light years in a year.


To do so, it will have to accelerate at an average of between
1.21 and 1.22 light years per year per year for the first year.
Compute the acceleration of the rear for the first 2 years, and
you will get something between 1.07 and 1.08. To stay even with
the front end of the ship, the relative acceleration of the rear
of the ship(train) is gradually reduced.


Notice that with an acceleration of 1 light year per year per year,
(about 0.97g), the ship will be able to outrun a light ray if it has
a one year head start. Likewise, a ship with an acceleration of
1/2 light year per year per year will be able to outrun a light ray
if it has a 2 year head start, and in general, a ship with an
acceleration
1/n light year per year per year will be able to outrun a light ray
if it has an n year head start. That automatically limits the length
of
a material body to a length of less than n if it undergoes an
acceleration of
1/n.
- A. McIntire

.

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