At 12:11 AM 4/2/2010, Stephen A. Lawrence wrote:
On 04/01/2010 07:28 PM, Abd ul-Rahman Lomax wrote:
> Re: [Vo]:checking my understanding of Lorentz contraction
>
> At 02:04 PM 4/1/2010, Abd ul-Rahman Lomax wrote:
>
>> But there is a limit suggested by this, that the maximum flattening is
>> by one-half.
This is false.
I figured that much out. But I'm trying to get an intuitive grasp of
this and citing formulas doesn't do that. Sure, formulas can be used
to determine the application of theory to a situation, and perhaps an
intuitive grasp isn't possible. But I'm not giving up on that yet.
I'm finding this quite useful in terms of my own learning, and
perhaps it will ultimately benefit someone else, because if I'm
confused, I'm sure there are others who are as well, and appeal to
authority doesn't clear up confusion at all.
Stephen recommended I pick up a textbook. Perhaps I should, it occurs
to me that I never did get a copy of Feynman's text, even though I
sat there for much of it. Not enough, by the way, my interests were
diverging already. (Would kick self if inclined to self-kick.)
It seems that I'm not alone in having some difficulty with this.
Lorenz himself seems to have gotten it wrong, and it's claimed that
Einstein, in his first paper, gave the same impression. This is, I
think, the paper that lent the name to Terrell rotation.
http://www.guspepper.net/electro/Segundo%20semestre/Seminarios/Funez.pdf
So, I'm proposing this:
A flat circular disc is moving sideways (along a diameter) at
velocity v. This flat circle passes just above a film sensitive to
gamma rays. A burst of gammas from a source at some distance impinges
upon the circle and film, leaving a shadow image on the film. What is
the shape of this shadow? Does it change as v approaches c?
What does "length" mean for a moving object? If I'm correct in my
understanding of Terrell's paper, which is shallow because I have not
studied all of it in depth, the image would be circular, not
flattened. (Under these conditions, Terrell rotation would not
appear, AFAIK. Note that Terrell rotation, which appears at a
distance, due to the time-of-flight for light, would flatten a shadow
in the direction of motion. But I haven't followed this yet.)
My "camera" is a shadow camera that gets rid of the complications of
solid angles, etc. Simultaneity of determination of the front and
trailing edge of the disc is assured (in the "stationary frame, not
in the moving frame.)
Now imagine that thre is a gamma source at the front of the disc and
at the back, and on the two sides, a burst is emitted simultaneously
(in the moving frame), which will mark the film. How far apart are the marks?
This is a quite different question! It reduces to a question of the
timing of the bursts. If they are simultaneous in the stationary
frame, we would have the same situation as with the shadow photo, and
there would be no reduction in distance. What do we mean by "simultaneous"?
Okay, in the center of the disc there is a light source that sends a
signal that triggers the burst. The signal initiates when the gamma
burst arrives that creates the shadow image. This signal travels to
the four gamma emitters so that the bursts, in the moving (disc)
frame, are simultaneous.
Let's look, then, at that timing signal as it appears in our
stationary frame. In our frame, the signal arrives sooner at the back
than at the front, because the back "comes up to meet it." Closing
velocity for the signal to the back is c + v. Toward the front, it's
reversed, the front is receding, so the closing velocity is c - v.
Therefore the burst from the back is emitted before the front. The
side "markers" are active simultaneously in our frame. The "image" is
elongated as the disc approaches c!
If the disc diameter is d, then the time to the back is d/2(c+v), and
the time to the front is d/2(c-v). The delay for the front is then
d/2(c-v) - d/2(c+v), simplified to dv/(c^2-v^2). In that time, the
disc would move on a distance of d/(c^2-v^2). Which yields a distance
between marks of d + d(c^2 - v^2). The marks are further apart in the
direction of motion by a factor of 1 + 1/(c^2-v^2). No change at zero
velocity, infinite change as v approaches c.
Under one concept of length (what a camera would see!) there is no
flattening. Under what concept of length is it shortened? That's what
I haven't found so far.
Reading the Terrell paper again after having written this, I see that
Terrell follows a similar argument as I. He writes this:
There is, however, a clear distinction between observing and seeing.
An observation of the shape of a fast-moving object involves
simultaneous measurement of the position of a number of points on
the object. If done by means of light, all the quanta should leave
the surface simultaneously, as determined in the observer's system,
but will arrive at the observer's position at different times.
The shadow camera eliminates the "different times" of arrival, by
sourcing the gammas in the observer's system. At the time of the
photograph, the gammas have already traveled the same distance from
the source, are obscured or not obscured by the disc, and then travel
the same distance to the film (very small).
In seeing the object, on the other hand, or photographing it, all
the light quanta arrive simultaneously at the eye (or shutter)
having departed from the object at various earlier times. Clearly
this should make a difference in the contracted shape which is in
principle observable and the actual visual appearance of a fast-moving object.
Terrill, I assume, is so familiar with the Lorenz contraction that he
does not explain the contraction itself, and he assumes this of the
reader. The shadow camera is a method of measuring the length of the
disc. It will show no change in length with velocity. It will agree
with the measurements made in the moving frame. What is the "Lorenz"
method of measuring length that does not behave in this manner?
I'll come back later, I do have some ideas, having to do with
simultaneity. The simultaneous snapshot of the front and back edge,
in the shadow camera, is not simultaneous in the moving frame.
