Re: Spaceship Question
- From: The Ghost In The Machine <ewill@xxxxxxxxxxxxxxxxxxxxxxx>
- Date: Sat, 04 Feb 2006 03:00:29 GMT
In sci.physics.relativity, PD
<TheDraperFamily@xxxxxxxxx>
wrote
on 3 Feb 2006 07:52:34 -0800
<1138981954.556418.146260@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>:
The Ghost In The Machine wrote:
In sci.physics.relativity, PD
<TheDraperFamily@xxxxxxxxx>
wrote
on 2 Feb 2006 07:57:48 -0800
<1138895868.339436.248970@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>:
The Ghost In The Machine wrote:
In sci.physics.relativity, Randy Poe
<poespam-trap@xxxxxxxxx>
wrote
on 1 Feb 2006 11:05:17 -0800
<1138820717.097263.297860@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>:
kenseto wrote:
"Randy Poe" <poespam-trap@xxxxxxxxx> wrote in message
news:1138721521.463219.61170@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
kenseto wrote:
"Randy Poe" <poespam-trap@xxxxxxxxx> wrote in message
news:1138636003.832078.155550@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
plane....so you
PD wrote:
kenseto wrote:
You make the claim that a radar can measure the length of a
need to back up your claim with references.
Crap. I can also claim that live, moving images can
be transmitted through the air and rendered in a box,
and you can claim not, and I can say "TV", and you can
say "Back up your claim with references," and I can
laugh and walk away from that stupid demand. If you want
to know how something *commonplace* works, then go to
the library and read a book.
Piggybacking:
Measuring the time difference between nose and tail reflections
and multiplying by c gives you a number equal to aircraft length
within the precision of any radar.
In what way does that *NOT* qualify as a length measurement
with radar? If the aircraft is 10 m, you'll get a number of
10 m. If the aircraft is 13.6 m, you'll get a value
of 13.6 m. What does Ken see as a problem with this
length measurement?
But you and PD claimed that length contraction is real.
Yes.
Now you say that
radar measures the same length as if the plane was at
rest in your frame. That's the problem.
"... within the precision of any radar."
I have gone into a fair amount of detail on exactly what
the difference between measured length and rest length
will be. The differences are very tiny. If you are measuring
a quantity, and you come up with a number which is
very close to that quantity, in what sense is that not
a measurement.
Is this all too confusing to you? Let me go back to the
numbers. A plane has a rest length of 30.0 m. I measure
the time difference between nose and tail reflections. Those
differences happen a distance of 29.99988 apart, and
my radar, being accurate down to 0.1 m, decides that
the length is in the 30.0+-0.05 bin rather than the
29.9 +- 0.05 or 30.1 +- 0.05 bin.
My radar tells you that the length is 30 m. Do you
disagree? In what sense is this not a measurement of
the length?
You are taking the position that there's no way at all
to do a length measurement of something moving. That's
nonsense. This is a very easy measurement to make.
The up shot is: Your so called "measurement" comes up with
the same answer that the plane is the same length as though
it is at rest in your frame. So it is not supporting your
claim that length contraction is real.
Wrong, since if length contraction is real, the prediction is
*also* that the measurement will fall in the 30.0 +- 0.05 bin.
I should note that SR length contraction is not real; the
contraction is part measurement artifact, part lightspeed
delay caused. If anything, the rod is getting *longer*, though
because of the delay we can't see it getting longer.
This is a *highly* misleading statement.
Probably! :-) But it's a highly misleading measurement to attempt
to determine the length of a moving rod, even without SR's
complications.
As you might see from my other post, the length measured
depends on the technique, and the simplest technique --
using a radar gun -- will result in shrinkage, but if
one can set up a "simultaneous detector array" by very
careful arrangement of signal delays and bathe the entire
experiment continuously in photons (e.g., from a set of
fluorescent tubes), the rod is measured as being longer.
Or one can have the rod flash its endpoints "simultaneously",
and then the detector grid measures the flashes, except
that the detector grid won't see them as being simultaneously.
Right, and this is where Russell corrected you. If an observer is using
a tape to measure a moving rod, it's important to mark the ends of the
rod against the tape at the *same time*.
In which frame? Simultaneity in SR is a big problem.
(Obviously, if you mark the
back end of the rod first, and then the front end, you'll get a length
measurement than is longer than intended.) This simultaneity has to be
set by the *observer*, not by the rod. If you let the rod determine the
simultaneity ("flashing its endpoints"), this will not be simultaneous
in the observer's frame, and in fact you'll get precisely the problem
referred to before: in the observer's frame, the back end will flash
first and the front end will flash a little later.
Ah, OK.
It is certainly possible to measure the length of a moving rod, but the
simultaneity has to be determined by the system of synchronized clocks
tied to the tape -- that is, NOT moving along with the rod. This is the
only sensible definition of length measurement.
Of course, when you do that, and the rod's clocks mark the time of the
simultaneous flashes from the tape, the rod's clocks will disagree that
the flashes were simultaneous. This is true even if there is a rod
clock *immediately* adjacent to the tape flasher. (It is NOT an
illusion based on propagation of light to the center of the rod or
anything like that.) And this is why the rod will say that the
measurement done by the observer with the tape is not the right
measurement of the length.
I think we can agree here, though there are definitely problems
in measuring a moving rod no matter how one slices it. :-)
This does not invalidate SR any more than Galileo's attempts
to measure lightspeed disprove SR. :-)
But here's a couple of issues, even with a "light-photo"
(simultaneity of the measurer) mentality:
[1] Rod is moving away, straight line. Assuming both endpoints
are always visible (one can take a limit if need be, getting
closer and closer to the rod's travel line), the rod is
mismeasured with the ratio sqrt(1-v/c)/sqrt(1+v/c). In
other words, it has shrunk, as far as the observer is concerned.
[2] Now move the rod towards you. All this does is reverse v's
sign; the rod is now too *long* by the ratio
sqrt(1+v/c)/sqrt(1-v/c).
[3] If one takes the measurement between two certain times
-- probably the times at which the endpoints move
under the measurer's feet -- the rod can have any
length at all between these two extremes, if not even
wackier results (since one endpoint is moving towards
the observer while the other is moving away).
I'd have to work out the "angle problem" -- an observer offset some
distance along the Y axis measuring the rod by eyeball as it moves
on the line x=0. The main issue there is that a photon from the
rod's endpoint now has to travel the distance sqrt(x^2+y^2) instead
of just plain abs(x), translating into a time distortion of
sqrt(x^2+y^2)/c and modifying various Lorentz results appropriately.
It should lessen the shrinkage or lengthenage (is that a word?)
slightly.
[rest snipped]
--
#191, ewill3@xxxxxxxxxxxxx
It's still legal to go .sigless.
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- From: Randy Poe
- Re: Spaceship Question
- From: The Ghost In The Machine
- Re: Spaceship Question
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