Re: Finally, Special Relativity Is Proven False
- From: Strich 9 <Strich.9.2c29301@xxxxxxxxxxxxxxxxx>
- Date: Tue, 15 Jul 2008 13:31:18 +0100
paparios@xxxxxxxxx;1187405 Wrote:
On 14 jul, 09:26, Strich 9 Strich.9.2c14...@xxxxxxxxxxxxxxxxx wrote:-
papar...@xxxxxxxxx;1186354 Wrote:
-
On 13 jul, 09:44, Strich 9 Strich.9.2bff...@xxxxxxxxxxxxxxxxx
wrote:-
-It is indeed amazing how a century of indoctrination dulls the
intellect.--
-
Do moving rods contract? Do moving clocks run slow?-
-
*The DAVID Arguments*:-
-
Let there be two frames M and N with relative velocity zero. A rod D
measuring 1 meter in M is equal to a rod D' measuring 1 meter in N.-
-
These 2 frames are also moving at 0.995c relative to an arbitrary
observer O. This does not alter the measurements noted above.-
-
Now M slows down to zero relative to O. Now had an actual length
contraction occurred when rod D was speeding at 0.995c relative to
O,
then when it comes to rest with O it un-contracts to a length of
10meters. Also, as M is now moving at -0.995c relative to N, N
observes
this rod as measuring 1 meter. This is the same length observed by N
when M was at rest relative to itself. A contradiction.
--
-
As you have been told several times, your argument is wrong. The
moving observers in relativity do not observe length contraction or
time dilation on their comoving objects (we could say all the moving
stuff contracts the same), so M and N measuring a rod as 1m on their
respective frames means the rod proper length at v=0 is exactly 1m.
Now observer O should measure those moving rods to be 0.1m, since
with
respect to this observer both M and N are experiencing length
contraction. When M stops, his rod will be 1m long both in M and O
frames and N will measure that rod to be 0.1m. So there is no
contradiction at all.
-
_The_same_argument_applies_for_moving_clocks._-
-
Let there be two frames M and N with relative velocity zero. An
atomic
clock A ticks 1 second in M synchronously with a 1 second tick in an
atomic clock A' in N.-
-
These 2 frames are also moving at 0.995c relative to an arbitrary
observer O. This does not alter the measurements noted above.-
-
Now M slows down to zero relative to O. Now had an actual time
dilation
occurred when A was speeding at 0.995c relative to O, then when it
comes
to rest with O it un-dilates and ticks 10 times faster. Also, as M
is
now moving at -0.995c relative to N, N observes A to be running slow
by
a factor of 10, so A' ticks 100 seconds for each tick of A in M. A
contradiction again.
--
-
Again, and for the same reason given in the length contraction case,
you are wrong. Moving observers will not detect time dilation of
their
comoving clocks, so all your reasoning is flawed.
--
-
--
Strich 9---
-
-
Miguel Rios-
Did you just say the rod was *-1 meter-* from M viewpoint when M was
going at *-0.995c-*, and was -*1 meter*- from M viewpoint when it was
going at at *-0c-*? Implying that the rod in M remained
-*invariant*-
at 0c and 0.995c? Have you not just shown to yourself and your ilk
that the premise of proposition 5 in this thread is true (moving rods
and clocks keep same length and time)?
Well I have proved that already, but you will believe it if you
derive
it yourself.
By the way, do a complete reanalyses of the atomic clocks as well to
complete your proof. Do not just say 'for the same reason' as it is
not exactly the same analyses. Note also you have applied SR using
the
21SR interpretation rather than EISR which is why you arrived at -no
contradiction-. See my elegant post:http://tinyurl.com/5grdvm
--
Strich 9-
You have to be careful in what terms you use. If v=0.995c, sqrt(1-v^2/
c^2) = 0.099875, gamma = 10.0125
delta t' = delta t / 0.099875 ; L = Lo * 0.099875
If I'm aboard a space ship M that is sitting still with respect to an
near earth inertial frame of reference E (Earth Space Station), so as
the relative speed is 0, then a 1m rod will be measured as 1m in both
frames. If then M speeds up to 0.995c, when that M space ship is
moving inertially at that speed with respect to E, that same rod
measured within the moving frame M will still be 1m long. From the
point of view of one observer at the E frame, the rod length will be
contracted to 0.1m. On the other hand the moving M clock will still
appears to a co-moving observer to tick at the same rate it did before
the trip. However for the observer in the E frame the clock at M will
appear to slow with respect to the rate it had before the trip, all
this following the Lorentz equations.
If the trip last for 10 years, on the E frame, the space ship will be
9.95 light years away at the end of that period. The space ship clock
will, however, show its Captain that the trip took only one year
(approx.).
There are no contradictions!
Miguel Rios
The proposition of the Argument assumes that time dilation and length
contraction represent -*real physical changes*-. This leads to a
-*contradiction*-.
When you remove the reality of the relativistic changes and assume that
time dilation and length contraction are -*illusory *-, then the
*-contradiction disappears-*.
You have proved EXACTLY my point that time dilation and length
contraction are only apparent changes and do not represent reality,
which is what the Argument is supposed to prove. Notice that your rods
and clocks kept EXACTLY the same measure and time whether they were in
motion or not.
What, you need another century to get this point?
--
Strich 9
.
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