On 30 Aug 2005 09:11:50 -0700, "Spoonfed"
wrote:


Kim B wrote:

In the space-time diagram, any uniformly accelering object will have a
45° asymptote. Any events on the other side of this asymptote will
never be visible to the object


Any event that can be seen by an inertial observer should also be seen
by a uniformly accelerating observer sitting right next to him, passing
him, or coming to a momentary rest adjacent to him.

If that conflicts with whatever you just said, you should let me know.

Take a look at http://micki.dk/diverse/accel.jpg - it shows 4 points
accelerating, which all could belong to the same rigid rod (David, you
may have a look too!).

The event at the red dot is never seen by any of these points.

Kim

bz wrote:
wrote in news:430b11f5.38471559@news-
server.austin.rr.com:

We see the optical equivalent of a thunder roll, as the sound from
different portions of the path reach our ears.

We recognize that sight and sound can be missleading and that no wrapping
actually occurs.
Wrapping actually does occur. In the rest frame of the rod and of the
rotating cylinder, we are attaching one end of the rod to the cylinder
a second before we attach the other end. The cylinder rotates 10
times before the other end is attached. Since the speeds are everyday
values can actually do this experiment over a much shorter length.
The short rod will attach to the short segment of the rotating disk,
and if we attach one end at a different time then we attach the other
end (the stationary frame view), the attached rod will spiral about
the cylinder. In the problem I posted because of the lengths involved
the spiral and wrapping occur over large distances, but it still
occurs.
David


Not if the attachment is done simultaniously all along the length using
sync'd clocks.

Bz is again talking out of ignorance. The attachment is
synched in the *moving* frame. So, in the stationary frame
the attachment is not simultaneous, and the rod wraps of
necessity.


We could sync it another way:
At a great distance, we have a signal source. The signal from that source
approximates a plane wave all along the length of our cylinder.

BTW, at that distance, we have a powerful telescope and we watch the
joining take place. No twisting is observed.

None is observed in the frame for which the attachment is
simultaneous. But enough of bz -- my post is mainly about
something else, to wit:

I think dseppala has actually managed to ask an interesting
question here, although as usual he's made it more complicated
than it needs to be. Also I believe he errs (along with most
of his respondents here) in thinking that this can be treated
as a 1D problem when in fact all 3 spatial dimensions must be
considered.

Let me recast it in simpler terms: Imagine an ordinary steel
corkscrew (actually any suitably asymmetric object would do)
rotating along its long axis; and suppose for simplicity that
it has exactly an integral number of turns so that it doesn't
wobble as it rotates. The C.O.M. is fixed; 3-momentum is
conserved in this frame.

Now, look at it from a different frame, one that is moving
along the screw axis. We find that the number of turns in
the corkscrew has *changed* so that there is more mass on one
side of the axis than the other. What happened to momentum
conservation?

Note that in some frame the corkscrew will even be *straight*.
(This is the extreme case that dseppala was considering.) And
yet it will still be going round and round its axis as it zips
along. 3-momentum definitely not conserved.

You can't just say this is impossible; I'm talking here about
an *ordinary* corkscrew, not some 300000km long one. The
inescapable conclusion is that 3-momentum is not conserved.

Kim B wrote:
On 30 Aug 2005 09:11:50 -0700, "Spoonfed"
wrote:


Kim B wrote:

In the space-time diagram, any uniformly accelering object will have a
45° asymptote. Any events on the other side of this asymptote will
never be visible to the object


Any event that can be seen by an inertial observer should also be seen
by a uniformly accelerating observer sitting right next to him, passing
him, or coming to a momentary rest adjacent to him.

If that conflicts with whatever you just said, you should let me know.

Take a look at http://micki.dk/diverse/accel.jpg - it shows 4 points
accelerating, which all could belong to the same rigid rod (David, you
may have a look too!).

The event at the red dot is never seen by any of these points.

Kim

Thanks, Kim

Your diagram at http://micki.dk/diverse/accel.jpg looks very similar to
the hyperbolic arcs in the upper right quadrant of this animation:

http://www.spoonfedrelativity.com/files/world-lines.gif

While I don't know for sure that the two diagrams are derived from the
same place, I do know what the same arcs represent in my diagram, so
forgive me if this seems off-the-mark.

Take a set of simultaneous events, occurring at
(x,t) = (1,0),
(2,0),
(3,0),
(4,0)

Perform Lorentz transformations around the origin to find points
(x',t')

t' = gamma*t - v*gamma*t/c^2
x' = gamma x - v*gamma*t

(where gamma=1/sqrt(1-(v/c)^2)

By varying the velocity and plotting each new set of points, map out a
locus of points which form hyperbolic arcs which are similar to the
ones in your diagram.

Though we are varying the velocity, this is not acceleration.
Acceleration involves the passage of time. This animation applies
repeated transformations at time t=0, whereas an acceleration should
apply transformations at incrementally increased times.

I will try to interpret the red dot in this schema. Assuming there is
an observer at the origin, the red dot could represent an event in the
observer's future.

The observer accelerates until the red dot is directly above the
origin. As time goes by, the observer will see each of the four events
well before he reaches the red dot. The light from those events
travels at a 45 degree angle up and to the left, and when it crosses
the path of the observer, he will see it.

On 31 Aug 2005 05:04:35 -0700, "Spoonfed"
wrote:

Kim B wrote:
On 30 Aug 2005 09:11:50 -0700, "Spoonfed"
wrote:


Kim B wrote:

In the space-time diagram, any uniformly accelering object will have a
45° asymptote. Any events on the other side of this asymptote will
never be visible to the object


Any event that can be seen by an inertial observer should also be seen
by a uniformly accelerating observer sitting right next to him, passing
him, or coming to a momentary rest adjacent to him.

If that conflicts with whatever you just said, you should let me know.

Take a look at http://micki.dk/diverse/accel.jpg - it shows 4 points
accelerating, which all could belong to the same rigid rod (David, you
may have a look too!).

The event at the red dot is never seen by any of these points.

Kim

Thanks, Kim

Your diagram at http://micki.dk/diverse/accel.jpg looks very similar to
the hyperbolic arcs in the upper right quadrant of this animation:

http://www.spoonfedrelativity.com/files/world-lines.gif

While I don't know for sure that the two diagrams are derived from the
same place, I do know what the same arcs represent in my diagram, so
forgive me if this seems off-the-mark.

Take a set of simultaneous events, occurring at
(x,t) = (1,0),
(2,0),
(3,0),
(4,0)

Perform Lorentz transformations around the origin to find points
(x',t')

t' = gamma*t - v*gamma*t/c^2
x' = gamma x - v*gamma*t

(where gamma=1/sqrt(1-(v/c)^2)

By varying the velocity and plotting each new set of points, map out a
locus of points which form hyperbolic arcs which are similar to the
ones in your diagram.

Though we are varying the velocity, this is not acceleration.
Acceleration involves the passage of time. This animation applies
repeated transformations at time t=0, whereas an acceleration should
apply transformations at incrementally increased times.

I will try to interpret the red dot in this schema. Assuming there is
an observer at the origin, the red dot could represent an event in the
observer's future.

The observer accelerates until the red dot is directly above the
origin. As time goes by, the observer will see each of the four events
well before he reaches the red dot. The light from those events
travels at a 45 degree angle up and to the left, and when it crosses
the path of the observer, he will see it.

The hyperbolic lines in my diagram ARE the worldline of 4 observers on
a uniformly accelerating rod -- and none of these will ever see the
light from the read dot.

Kim

All parts of the rod are traveling at same speed ... but time passes
not synchronously along the rod. Clocks run slower at the back end

Kim



All parts of the rod are traveling at the same speed according to an
observer standing on the accelerating rod.

To an inertial observer, who watches the rod slow down, stop, and
accelerate to the right again, it will always appear that the left end
of the rod is moving slower than the right end, (except for the moment
that the entire rod is stopped.)

To the inertial observer the difference in the speed of the clocks at
the front and back end is accounted for entirely by the difference in
time dilation caused by the difference in velocity between the ends.

If I ever have the time to finish my demo, I'll show that in more
detail.

Spoonfed wrote:

All parts of the rod are traveling at same speed ... but time passes
not synchronously along the rod. Clocks run slower at the back end

I don't see in what sense your first statement could
be true; it certainly isn't true in the original inertial
frame, where at all t0 the rear of the rod is moving a
tiny bit faster than the front. Indeed that is *required*
for your second stmt to be true -- which btw it is. It's
speed, not acceleration, that determines clock rates
relative to clocks in an inertial frame.


Kim



All parts of the rod are traveling at the same speed according to an
observer standing on the accelerating rod.

Well, to say that, you would need to specify what
coordinates the accelerated observer is using. There
is really no obvious choice in this example, AFAIK.

I was aware of this complication, but didn't mention it,
when I wrote earlier about the rod maintaining its original
proper length. Such a statement sweeps a lot of important
details under the carpet. I'm not an expert in this area,
and that fact that such a distinguished name (as Born) is
attached to it suggests that it is not simple.

One thing I do know is true, is that in the original frame
the rod will always have length L/gamma, where L is the
original proper length. But this isn't true in any other
frame, unless I err somewhere in my thinking. However, it
would be possible to turn off the thrusters on some exactly
timed schedule in the original frame (*not* simultaneously
there) such that the rod would continue to coast inertially
with its proper length unchanged; and thereafter of course
it would have length L/gamma in all inertial frames.


To an inertial observer, who watches the rod slow down, stop, and
accelerate to the right again,

Wait, the acceleration is *always* in one direction (let's
say to the right).

it will always appear that the left end
of the rod is moving slower than the right end, (except for the moment
that the entire rod is stopped.)

There is *no* moment when the entire rod is stopped, in such
a frame. During the acceleration there is only one frame
that has all of the rod in it simultaneously, and that is
the original frame -- and only at time t=0.


To the inertial observer the difference in the speed of the clocks at
the front and back end is accounted for entirely by the difference in
time dilation caused by the difference in velocity between the ends.

This is true. Also the various inertial observers see a
different *constant* offset in the clocks due to RoS.


If I ever have the time to finish my demo, I'll show that in more
detail.

I think you may need to work a bit more on the theory,
from what you say above.

On 6 Sep 2005 09:10:42 -0700, "Spoonfed"
wrote:




All parts of the rod are traveling at same speed ... but time passes
not synchronously along the rod. Clocks run slower at the back end

Kim



All parts of the rod are traveling at the same speed according to an
observer standing on the accelerating rod.

That's what I meant.


To an inertial observer, who watches the rod slow down, stop, and
accelerate to the right again, it will always appear that the left end
of the rod is moving slower than the right end, (except for the moment
that the entire rod is stopped.)

Ehh .. assuming the rod is accelerating to the right, the left end is
the one moving faster, right?

Kim


To the inertial observer the difference in the speed of the clocks at
the front and back end is accounted for entirely by the difference in
time dilation caused by the difference in velocity between the ends.

If I ever have the time to finish my demo, I'll show that in more
detail.

wrote:
Spoonfed wrote:

All parts of the rod are traveling at same speed ... but time passes
not synchronously along the rod. Clocks run slower at the back end

I don't see in what sense your first statement could
be true; it certainly isn't true in the original inertial
frame, where at all t0 the rear of the rod is moving a
tiny bit faster than the front.

By "your" I meant "Kim B's". That is, I knew this was
Kim B's writing but didn't notice that Spoonfed had
provided no attribution. So I am following up, to
clarify.

And also, to make the following correction [after a snip]:

One thing I do know is true, is that in the original frame
the rod will always have length L/gamma, where L is the
original proper length.

Er, this was rather silly of me since I just finished
saying that the rod is moving at different speeds
in this frame -- so which gamma do we use? It seems
there are pitfalls at every turn, in talking about
this.

(Sure, if the acceleration is not great, the statement
is more or less true no matter which gamma we pick,
since they are all about the same. But I think I'd
better stop saying such categorical things about a
problem that I don't even know how to specify exactly.
Indeed it seems like the specification is really the
only hard part here, yet I don't think anyone in this
discussion has seriously addressed that part of it yet.)

wrote:

[snip]

Indeed it seems like the specification is really the
only hard part here, yet I don't think anyone in this
discussion has seriously addressed that part of it yet.)

Duh, me again -- I suppose one could say that the
exact specification *is* Kim B's spacetime diagram;
so in that sense Kim did address the specification
seriously. However, the pitfalls that I wrote about
are nevertheless there when one *talks* about this
spacetime diagram. And as I pointed out, I do not
think even Kim is getting that talk 100% right.

On 6 Sep 2005 12:33:42 -0700, wrote:

wrote:
Spoonfed wrote:

All parts of the rod are traveling at same speed ... but time passes
not synchronously along the rod. Clocks run slower at the back end

I don't see in what sense your first statement could
be true; it certainly isn't true in the original inertial
frame, where at all t0 the rear of the rod is moving a
tiny bit faster than the front.

By "your" I meant "Kim B's". That is, I knew this was
Kim B's writing but didn't notice that Spoonfed had
provided no attribution. So I am following up, to
clarify.

If you choose a point on the rod a use its current speed as your FOR,
the the rest of the rod will fit nicely in this FOR (along the FOR's
line of simultaneity) ... with the same speed all along and the
correct proper length, exactly as it fits in our "rest" frame at the
base line ... all frames are equal, assuming the rod has accelerated
and will accelerate forever.


And also, to make the following correction [after a snip]:

One thing I do know is true, is that in the original frame
the rod will always have length L/gamma, where L is the
original proper length.

The rod's length in the "rest frame" is not _that_ simple to calculate
(it's the horizontal distance between the hyperbolas of the ends).

Kim

Er, this was rather silly of me since I just finished
saying that the rod is moving at different speeds
in this frame -- so which gamma do we use? It seems
there are pitfalls at every turn, in talking about
this.

(Sure, if the acceleration is not great, the statement
is more or less true no matter which gamma we pick,
since they are all about the same. But I think I'd
better stop saying such categorical things about a
problem that I don't even know how to specify exactly.
Indeed it seems like the specification is really the
only hard part here, yet I don't think anyone in this
discussion has seriously addressed that part of it yet.)