Kim B wrote:

[snip]

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.

Thanks. Of course you are quite right about that, and I
apologize for my many mistakes here.

wrote:
Spoonfed wrote:

[snip]

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.

Oops, I goofed here. Kim B has corrected me on this
point.

Btw I agree with Kim that you should have written
"faster" instead of "slower" here. That is why the
clocks tick slower at the left side.

Kim B wrote:

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


Right.

wrote:
Kim B wrote:

[snip]

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.

Thanks. Of course you are quite right about that, and I
apologize for my many mistakes here.

I believe it when I hear it, but it's a little tricky to figure out.

It seems surprising that no matter what reference frame you are in, all
parts of the rod will pass v=0 at the same time. I'm not in the mood
to develop a proof, but it seems right.

On 6 Sep 2005 14:19:27 -0700, "Spoonfed"
wrote:


wrote:
Kim B wrote:

[snip]

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.

Thanks. Of course you are quite right about that, and I
apologize for my many mistakes here.

I believe it when I hear it, but it's a little tricky to figure out.

It seems surprising that no matter what reference frame you are in, all
parts of the rod will pass v=0 at the same time. I'm not in the mood
to develop a proof, but it seems right.

I've been working quite a lot with the acceleration problem and must
say it is a very good case to learn from ... and to get rid of some
wrong intuitive ideas :-) I was inspired by an old post from Ben
Rudiak-Gould tallking about the event horizon, and then I worked
through the math.

Kim

Spoonfed wrote:
wrote:
Kim B wrote:

[snip]

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.

Thanks. Of course you are quite right about that, and I
apologize for my many mistakes here.

I believe it when I hear it, but it's a little tricky to figure out.

It seems surprising that no matter what reference frame you are in, all
parts of the rod will pass v=0 at the same time. I'm not in the mood
to develop a proof, but it seems right.

I won't prove it here either; I'll just give some
further motivation.

If you imagine Born-rigid acceleration over a finite
time, and then ending, it must be the case that in the
end, no part of the rod is moving relative to any other
part (otherwise the rod would not be rigid). So, in
that sense it's not a surprising result at all.

On the other hand, given the complications that occur
with acceleration, I agree it's a bit surprising that
Born-rigid acceleration is possible at all, even in
theory.

You can see what's happening if you take Kim B's
diagram and, say, pick some point on the leftmost
hyperbola and draw the tangent there. Then draw
parallel tangents on the other hyperbolas, at whatever
points are determined by the parallel requirement.
Then notice that the points you have picked all lie
on the same straight line, tilted slightly upward to
the right. This is a line of simultaneity in the
frame that is moving at the speed given by the tangent
slope, and (recognizing that our diagram is drawn in
Euclidean rather than Minkowskian space) we see that
this line would actually be perpendicular to the tangent
if we redrew the diagram in the coordinates of that
frame. Thus, as Kim B says, the diagram looks the same
no matter what frame we draw it in.

On 7 Sep 2005 10:54:47 -0700, wrote:

Spoonfed wrote:
wrote:
Kim B wrote:

[snip]

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.

Thanks. Of course you are quite right about that, and I
apologize for my many mistakes here.

I believe it when I hear it, but it's a little tricky to figure out.

It seems surprising that no matter what reference frame you are in, all
parts of the rod will pass v=0 at the same time. I'm not in the mood
to develop a proof, but it seems right.

I won't prove it here either; I'll just give some
further motivation.

If you imagine Born-rigid acceleration over a finite
time, and then ending, it must be the case that in the
end, no part of the rod is moving relative to any other
part (otherwise the rod would not be rigid). So, in
that sense it's not a surprising result at all.

On the other hand, given the complications that occur
with acceleration, I agree it's a bit surprising that
Born-rigid acceleration is possible at all, even in
theory.

You can see what's happening if you take Kim B's
diagram and, say, pick some point on the leftmost
hyperbola and draw the tangent there. Then draw
parallel tangents on the other hyperbolas, at whatever
points are determined by the parallel requirement.
Then notice that the points you have picked all lie
on the same straight line, tilted slightly upward to
the right. This is a line of simultaneity in the
frame that is moving at the speed given by the tangent
slope, and (recognizing that our diagram is drawn in
Euclidean rather than Minkowskian space) we see that
this line would actually be perpendicular to the tangent
if we redrew the diagram in the coordinates of that
frame. Thus, as Kim B says, the diagram looks the same
no matter what frame we draw it in.

All lines of simultaneities in my diagram, drawn this way, pass
through (0,0).

Kim

Kim B wrote:
On 7 Sep 2005 10:54:47 -0700, wrote:

Spoonfed wrote:
wrote:
Kim B wrote:

[snip]

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.

Thanks. Of course you are quite right about that, and I
apologize for my many mistakes here.

I believe it when I hear it, but it's a little tricky to figure out.

It seems surprising that no matter what reference frame you are in, all
parts of the rod will pass v=0 at the same time. I'm not in the mood
to develop a proof, but it seems right.

I won't prove it here either; I'll just give some
further motivation.

If you imagine Born-rigid acceleration over a finite
time, and then ending, it must be the case that in the
end, no part of the rod is moving relative to any other
part (otherwise the rod would not be rigid). So, in
that sense it's not a surprising result at all.

On the other hand, given the complications that occur
with acceleration, I agree it's a bit surprising that
Born-rigid acceleration is possible at all, even in
theory.

You can see what's happening if you take Kim B's
diagram and, say, pick some point on the leftmost
hyperbola and draw the tangent there. Then draw
parallel tangents on the other hyperbolas, at whatever
points are determined by the parallel requirement.
Then notice that the points you have picked all lie
on the same straight line, tilted slightly upward to
the right. This is a line of simultaneity in the
frame that is moving at the speed given by the tangent
slope, and (recognizing that our diagram is drawn in
Euclidean rather than Minkowskian space) we see that
this line would actually be perpendicular to the tangent
if we redrew the diagram in the coordinates of that
frame. Thus, as Kim B says, the diagram looks the same
no matter what frame we draw it in.

All lines of simultaneities in my diagram, drawn this way, pass
through (0,0).

Thanks, yes, that helps with the visualization. It agrees
with our intuition that the one special event (the location
of the bear when he starts chasing us) can't depend on how
we draw the diagram.

The 45-degree line through (0,0) is a degenerate hyperbola
where *all* of the change in slope happens, as it were,
exactly at (0,0). So, no matter what slope one chooses
for the tangent, (0,0) must be on the line of simultaneity
determined by that choice.

On 7 Sep 2005 11:51:46 -0700, wrote:

Kim B wrote:
On 7 Sep 2005 10:54:47 -0700, wrote:

Spoonfed wrote:
wrote:
Kim B wrote:

[snip]

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.

Thanks. Of course you are quite right about that, and I
apologize for my many mistakes here.

I believe it when I hear it, but it's a little tricky to figure out.

It seems surprising that no matter what reference frame you are in, all
parts of the rod will pass v=0 at the same time. I'm not in the mood
to develop a proof, but it seems right.

I won't prove it here either; I'll just give some
further motivation.

If you imagine Born-rigid acceleration over a finite
time, and then ending, it must be the case that in the
end, no part of the rod is moving relative to any other
part (otherwise the rod would not be rigid). So, in
that sense it's not a surprising result at all.

On the other hand, given the complications that occur
with acceleration, I agree it's a bit surprising that
Born-rigid acceleration is possible at all, even in
theory.

You can see what's happening if you take Kim B's
diagram and, say, pick some point on the leftmost
hyperbola and draw the tangent there. Then draw
parallel tangents on the other hyperbolas, at whatever
points are determined by the parallel requirement.
Then notice that the points you have picked all lie
on the same straight line, tilted slightly upward to
the right. This is a line of simultaneity in the
frame that is moving at the speed given by the tangent
slope, and (recognizing that our diagram is drawn in
Euclidean rather than Minkowskian space) we see that
this line would actually be perpendicular to the tangent
if we redrew the diagram in the coordinates of that
frame. Thus, as Kim B says, the diagram looks the same
no matter what frame we draw it in.

All lines of simultaneities in my diagram, drawn this way, pass
through (0,0).

Thanks, yes, that helps with the visualization. It agrees
with our intuition that the one special event (the location
of the bear when he starts chasing us) can't depend on how
we draw the diagram.

The 45-degree line through (0,0) is a degenerate hyperbola
where *all* of the change in slope happens, as it were,
exactly at (0,0). So, no matter what slope one chooses
for the tangent, (0,0) must be on the line of simultaneity
determined by that choice.

Right. And it is also the limit for how long we can stretch the rod to
the left.

Kim

[I just happened to see this.]

wrote:
On the other hand, given the complications that occur
with acceleration, I agree it's a bit surprising that
Born-rigid acceleration is possible at all, even in
theory.

This is not "surprising", it is natural:

Born rigid motion is the natural way a solid object accelerates, as long
as it is pushed or pulled from a single point, and the acceleration is
small enough that the inter-molecular bonds of the object remain on
average unstressed (i.e. the object does not break or deform). Those
bonds will try to keep the inter-molecular spacings constant, making
this as rigid as possible for an atomic object, and that's just what
Born rigid motion is.


Tom Roberts