Re: Accelerating rocket/light beam question
- From: David Rutherford <drutherford@xxxxxxxxxxx>
- Date: Mon, 6 Apr 2009 17:52:45 +0200 (CEST)
Daryl McCullough wrote:
[...]
I hope you don't mind my snipping all of the above, but I think
answering what you have below will clarify my point.
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.
Of course, but that has nothing to do with the situation we're discussing.
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.
Sure, but this doesn't have anything to do with the situation we're
discussing, either.
Think, for a moment, about a system consisting of two rockets B and C
with a string S connecting them. If the system undergoes a Lorentz
contraction, due to a Lorentz transformation, it might look something
like this.
B S C
Rockets+string at rest: --------->..........--------->
B S C
Rockets+string in motion: ----->......----->
Neither the rockets nor the string are stressed. The same is true if the
system undergoes an _expansion_, due to a coordinate transformation,
like this.
B S C
Rockets+string at rest: ----->......----->
B S C
Rockets+string in motion: --------->..........--------->
Again, neither the rockets nor the string are stressed.
You can kind of look at it like this. Make a drawing of two rockets
connected by a string. Hold the plane of the drawing perpendicular to
your line of sight, with the rockets pointing horizontally. That's the
system of rockets+string, at rest. Now rotate the drawing around a
vertical axis, so that it makes an acute angle with your line of sight.
This perspective view of the rockets+string system would represent the
contracted system in motion. Does the contraction result in stress to
the rockets or string? Of course not.
Now imagine that the drawing, rotated at the acute angle, is the system
at rest. Turn it back to its original position, so that it is again
perpendicular to your line of sight. It now represents the expanded
rockets+string system in motion. Does that mean that the rockets or
string are stressed, or that the string breaks? Again, the answer is no.
Although this is not an exact analogy, I think it emphasizes that
contractions or expansions, due to coordinate transformations, are
closer to an optical illusion than a physical compressing or stretching
due to a force (which would result in stresses).
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'.
Your cars+string system is not equivalent to the rockets+string system.
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.
--
Dave Rutherford
"New Transformation Equations and the Electric Field Four-vector"
http://www.softcom.net/users/der555
Applications:
"4/3 Problem Resolution"
"Action-reaction Paradox Resolution"
"Energy Density Correction"
"Proposed Quantum Mechanical Connection"
"Biot-Savart's Companion"
"Horizon Problem Resolution"
.
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