Re: Accelerating rocket/light beam question

David Rutherford says...

Daryl McCullough wrote:

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.

Of course, but that has nothing to do with the situation we're discussing.

It's *exactly* the situation we are discussing, only described
from the point of view of a frame in which the center of the
string is momentarily at rest. This is explained in more
detail below.

Your cars+string system is not equivalent to the rockets+string system.

Yes, it is. This is explained below.

The fundamental difference between your system and the rockets+string
system is that, in your system, no comoving inertial frame can be found,
in which your system is momentarily at rest.

That's exactly right. If two rockets are undergoing constant
equal accelerations, then after they have been accelerating
for a while then there *IS* no comoving inertial frame
in which both rockets are at rest. This is just a simple
application of the Lorentz transforms.

Let F be the frame in which the two rockets are initially
at rest. Let the two rockets be connected by a string.

Pick a velocity v. Let e_1 be the event at which the
front rocket reaches speed v relative to frame F. Before
e_1, the front rocket has speed < v. After e_1, it has
speed > v.

Let e_2 be the event at which the rear rocket reaches
speed v. Before e_2, the rear rocket has speed < v.
After e_2, the rear rocket has speed > v.

Let e_mid be the event at which the midpoint of the string
connecting the two rockets reaches speed v. This event
is approximately halfway between e_1 and e_2.

In frame F, all three events e_1, e_mid, and e_2 are simultaneous.

Let F' be a frame that is moving at speed v relative to
frame F along the positive x-axis.

The way velocity transforms between frames tells
us that if an object is traveling at a speed less than v
in frame F, then it is traveling in the negative x-direction
in frame F'. If it is traveling at speed v in frame F,
then it is traveling at speed 0 in frame F'. If it is
traveling at a speed greater than v in frame F, then
it is traveling in the positive x-direction in frame F'.

So in frame F':

1. The front rocket is traveling in the negative x-direction
prior to e_1, and in the positive x-direction afterwards.
2. The midpoint is traveling in the negative x-direction
prior to e_mid, and in the positive x-direction afterwards.
3. The rear rocket is traveling in the negative x-direction
prior to e_2, and in the positive x-direction afterwards.

Now, here's the key point: In frame F', event e_1 happens
*before* event e_mid, which happens before e_2. To see this,
look at the Lorentz transforms:

t_1' = gamma (t_1 - v/c^2 x_1)
t_mid' = gamma (t_mid - v/c^2 x_2)
t_2' = gamma (t_2 - v/c^2 x_2)

Since x_1 > x_mid > x_2, and t_1 = t_mid = t_2,
it follows that t_1' < t_mid' < t_2'.

So in frame F', we have the following situation:

1. Prior to t_1', both rockets and the midpoint are
traveling in the negative x direction.

2. After time t_1', the front rocket is traveling
in the positive x-direction, but the midpoint and
the rear rocket are traveling in the negative
x-direction.

3. At time t_mid', the front rocket is traveling
in the positive x-direction, the midpoint has speed 0,
and the rear rocket is traveling in the negative
x-direction.

4. After time t_2', both rockets and the midpoint are
traveling in the positive x-direction.

So at time t_mid', the front rocket is traveling in
one direction, and the rear rocket is traveling in
the opposite direction. So in F', at time t_mid',
we're in the same situation that you agreed would
break the string: the rockets are pulling on the
string in opposite directions.

--
Daryl McCullough
Ithaca, NY

.

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