Re: Accelerating rocket/light beam question

David Rutherford says...

According to your setup, this should be the arrangement of the rockets
at rest and in motion.

B A
Rockets at rest: ---------> --------->
--------->
C

B A
Rockets in motion: -----> ----->
----->
C

Yes, that's right.

The contractions in this post are only meant to show the qualitative
relationship between the rockets at rest and in motion.

If there's a contraction in the rocket lengths, then there should also
be a contraction in the space between the rockets.

Yes, if all *three* rockets are rigidly connected, then
the space between the rockets will remain constant. In
that case, the front rocket accelerates *less* than the
rear rocket. This is easy to see:

Let x_0 be the location of the rear of rocket B at time t=0.
Let x_1 be the location of the rear of rocket B at time t=T.
Let x_2 be the location of the rear of rocket C at time t=0.
Let x_3 be the location of the rear of rocket C at time t=T.

Now, suppose that both rockets B and C are undergoing Lorentz
contraction. So at time 0, the length of rocket B is L, and
at time T, the length is L/gamma, where gamma > 1. Then if
C stays connected to B, then

x_2 = x_0 + L
x_3 = x_1 + L/gamma

So the total distance traveled by rocket B in time T is
x_1 - x_0.

The total distance traveled by rocket C is
x_3 - x_2 = x_1 + L/gamma - (x_0 + L) = (x_1 - x_0) - L (1 - 1/gamma)

So the distance traveled by C is *less* than the distance traveled
by B. Since distance is the integral of velocity, it follows that
at each moment, C has a lower velocity than B. Since velocity is
the integral of acceleration, it follows that at each moment,
C has a lower acceleration than B.

So we're just getting back to the two options that I mentioned:
(1) If the rockets are rigidly connected, then the front rocket
travels with smaller acceleration than the rear rocket.
(2) If the rockets are traveling at the same acceleration, then
the rockets cannot remain connected; the rear of rocket C will
separate from the front of rocket B.

As you can see, the space between rockets A and B contracts along with
the rockets (and the space they occupy). So, the top of rocket B won't
recede from the bottom of rocket C, and the top of rocket C won't recede
from the bottom of rocket A. If there was a string connecting the bottom
of rocket A to the top of rocket B, it would also contract by the same
amount as the space between the rockets, so it wouldn't break.

Right, but that's *not* a situation in which all three rockets
are traveling at the same acceleration. Rocket B is accelerating
more than rocket C, which is accelerating more than rocket A.

You are confusing two different relationships between accelerating
rockets:

1. Rockets A, B, and C are "rigidly connected".

The rockets are not connected to each other, in any way.

Instead of "connected" read "touching". Option 1 has
the top of rocket B touching the bottom of rocket C,
and the top of rocket C touching the bottom of rocket A.

That *is* the situation you are talking about. In that
situation, rocket B accelerates more than rocket C, which
accelerates more than rocket A.

This
means that they are accelerating in a straight line in
such a way that the distance between them remains constant (as
measured by a comoving inertial reference frame).

2. Rockets A, B, and C are undergoing identical proper accelerations.

These are *not* the same properties. If (1) is true, then (2) is *false*,
and vice-versa.

(1) is not true.

My definition of "rigidly connected" is just that the
distance between the top of rocket B and the bottom of
rocket C does not increase with time. By this definition,
you *are* talking about rigidly connected rockets.

If, on the other hand, the rockets
are not connected, but are following the same "flight plan"
(blast off at such and such a time, with rocket engines set
to such and such thrust level), then the rockets will *not*
accelerate rigidly. Instead, the distance between the
rockets will increase (as measured in a comoving inertial
frame).

As I showed, above, if the rocket's lengths contract, then I think the
distance between the rockets will decrease.

Not if the rockets are undergoing identical accelerations.

Let F be the inertial frame in which the rockets are initially
at rest. Let L be the initial distance between the rockets,
as measured in frame F.

Let e_1 be some event (location in space and time) taking place
at the rear rocket after it has begun accelerating.

Let F' be the inertial frame in which the rear rocket is
momentarily at rest at event e_1.

Let e_2 be the event such that (1) e_1 and e_2 are
simultaneous, according to frame F, and (2) the spatial
distance between e_1 and e_2 is L, according to frame F.

Let e_3 be the event such that (1) e_1 and e_3 are
simultaneous, according to frame F', and (2) the spatial
distance between e_1 and e_3 is L, according to frame F'.

You have a choice: (1) If the front rocket travels with
constant acceleration so that it passes through event e_2,
then it will *not* pass through event e_3. So if the rockets
maintain a constant distance apart, as measured in frame F,
then their distance apart as measured in frame F' will be
*larger* than L at the time of event e_1.

In choice (1), assume that the lengths of all three rockets and the
distance between rockets A and B _expand_, in frame F', like this,

B A
Rockets at rest: -------> ------->
----------------------->
C

B A
Rockets in motion: ---------> --------->
----------------------------->
C

What mechanism is supposed to be causing the rockets to expand?
If the rockets are expanding, that means that the front of the
rocket is accelerating *more* than the rear of the rocket. What
could cause that to happen?

A string between rockets A and B would also expand by the same amount as
the interval between rockets A and B.

Why?

If it's possible for rockets and intervals (and strings) to contract,
why isn't it possible for them to expand? Contraction of lengths, in SR,
can be thought of as a result of the fact that, in SR, clocks at forward
positions, in the direction of motion, run behind clocks at rear
positions. Consequentially, a clock at a forward position is,
essentially, where it was at an earlier time, making it look a little
further toward the rear than when at rest. This makes lengths appear
contracted.

The same kind of thing might happen in reverse for accelerated frames.
Clocks in forward positions, for motion in the direction of the
acceleration, run ahead of, and faster than, clocks in rear positions.

How in the world could that happen? If the two clocks (one in the
rear, and one in the front) have the same trajectory, then at any
given time in the initial reference frame, the front clock will show
the same time as the rear clock.

Same as above, but lengths are contracted, in frame F. But, the string
still doesn't break.

So, basically, the string doesn't break, no matter how you look at it,
period.

Let me try one more attempt. Suppose that at time t=0, we have
the following situation: (Let's switch from rockets to cars
to make it more concrete)

1. There is a car in the rear traveling at velocity -v
(so it's traveling in the negative x-direction).

2. There is a car in the front traveling at velocity +v
(so it's traveling in the positive x-direction).

3. There is a string connecting the two cars. We'll
assume that the center of the string is at rest, while
the front of the string is being pulled in the +x direction,
and the rear of the string is being pulled in the -x direction.

Wouldn't you expect that the string would be stressed, in such
a situation? This doesn't have to be a purely theoretical issue.
You can try it out yourself: Take two cars, and have one car
traveling at speed 10 miles per hour along a road, and have
the other car traveling at speed 10 miles per hour in the
opposite direction. Try it yourself: the string will break.

Now, let's put some acceleration in. Suppose that both
cars are undergoing identical acceleration. The front car
is accelerating in the +x direction, so that its speed is
increasing, while the rear car is accelerating in the +x
direction so that its speed is decreasing. Their rates
of acceleration are the same (so that the relative
velocity remains 2v). In that case, do you think the
string will break? Try it for yourself. It will.

So after you've tried it out, let's assume the
following fact about cars and strings: Let
F be the frame of your cars.

1. Fact: If in frame F, two cars have equal
and opposite velocities, and they are accelerating
in such a way that their separation speed remains
constant, then a string stretched between them will
eventually break.

2. Assumption: Applying the relativity principle, we
conclude that for *any* inertial frame F', if
two cars have equal and opposite velocities, and
there is a string stretched between them, then the
string will eventually break.

3. Fact: According to the Lorentz transforms, if two
cars are undergoing identical acceleration from
rest in frame F, then at a later time, let F' be
some frame in which the point on the string
halfway between the two cars is momentarily
at rest. Then in frame F', the two cars are
traveling in opposite directions. So by
assumption 2, the string will break in frame
F'.

--
Daryl McCullough
Ithaca, NY

.

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