From: throopw@sheol.org (Wayne Throop)
Newsgroups: sci.physics.relativity
Subject: Re: SRT and LET Improved
Distribution: world
Organization: sheol
Message-ID: <883187931@sheol.org>
Date: Sat, 27 Dec 1997 01:58:51 GMT
References: <3497ec17.2274334@news.erinet.com> <01bd0b17$49229b40$34297dc2@dbyr.iol.ie> <349952dc.2833361@news.erinet.com> <01bd0c01$af994500$162a7dc2@dbyr.iol.ie> <3499d2e4.3734625@news.erinet.com> <01bd0c73$28f10b60$c0297dc2@dbyr.iol.ie> <349d3497.1084439@news.erinet.com> <01bd0e47$fff735a0$302a7dc2@dbyr.iol.ie> <349de868.9333570@news.erinet.com> <01bd0ec7$ceb21520$15297dc2@dbyr.iol.ie> <349fd202.2346890@news.erinet.com> <01bd0fd9$45c06b00$04297dc2@dbyr.iol.ie> <34a07654.8905594@news.erinet.com> <882943039@sheol.org> <34a11e05.5692811@news.erinet.com> <882995442@sheol.org> <34a3f90d.1817852@news.erinet.com>
:: throopw@sheol.org (Wayne Throop)
:: He is, in essence, forgetting that the "contracted" rod is in motion,
: kenseto@erinet.com (Ken H. Seto)
: No I did not forget. In fact that's the main point of my argument.
: It will take a longer time for light to traverse a rod that is in
: motion than a rod that is at rest. The longer duration of B's second
: must be accompany by a longer rod length to give a constant math ratio
: for light speed.
Well, there he goes again, with his favorite ploy; the rephrasing the
problem into vague English, rather than precise terms or formal statements,
so that he can sweep his errors under the rug.
In this case, just as I said, he's completely ignoring the fact that,
since the endpoints of the rods are in motion, they must be *shorter*
to give the correct answer, not *longer*. A fact I demonstrated in
the simplest possible terms: arithmetic on a very simple example.
: Here's what A sees of B:
: 1. B's proper time and proper length will measure light speed to be c
: --- 299,792,485 m in B's frame/9,192,631,770 periods of Cs atom in B's
: frame (this is also a proper second in B's frame)
Note that kenseto is setting up the same problem I did, but
he does it in a clumsier notation, easier to make mistakes in.
And he does indeed make mistakes. One of minor importance,
and one complete conceptual snafu.
: 2. From A's point of view, each of A's proper second is worth 0.4
: proper second of B's clock according to the following calculation:
: Ta=1 proper second in A's frame
: V=0.8 c
: Tb= = 0.4 seconds proper second in B's
: frame. This value is confirmed experimentally. Also, this means that it
: takes 2.5 of A's seconds to equal to the duration of one of B's second.
Wrong, kenseto. Evaluate 1*Sqrt{(1- (0.8 c)^2/c^2)} a bit more
carefully: we cancel the c^2, and drop the times 1, and we have
sqrt(1-(0.8)^2). Which is 0.6, not 0.4. This is the minor error.
It takes 1 and (2/3) seconds of A to equal one second of B, in A's frame.
( Of course, kenseto also drops that final, and required, "in A's frame",
and pretends it's an unqualified truth, a "duration", rather than a
frame-relative relationship. Yes, he really misunderstands coordinates
in general and SR in particular that badly. )
: 3. Also, from A's point of view, B's rod length must be apparently
: 2.5 times longer than A's rod. La=1 proper length (or 299,792,485 m)
: V=0.8c Lb=La/Sqrt(1-V^2/c^2) =1/Sqrt{(1- (.8c)^2/c^2)} = 2.5 times of
: the proper length of A
No. You can see right there: kenseto didn't account for the fact
that *from* *A's* *POV*, B's rod is in motion at .8c.
He treated it as if the rod were *standing still* as light went
from one end to the other, adjusted the length by the inverse
of the time for light to span the endpoints.
But light isn't traveling the length of a B rod in those 5/3 A seconds.
It's traveling the length of a B rod, plus the distance the rod moved
in the 5/3 seconds.
This is kenseto's major error; an error to which, for some
unaccountable reason, kenseto is completely blind.
Kenseto simply refuses to accurately trace the endpoints of the
rod in both coordinate systems, in order to determine the
needed length of that unit rod in B when reckoned in A,
to get lightspeed to come out c.
It's obvious that the formula he uses isn't right,
for the reasons I've just discussed above.
And he didn't even *try* to point out any error in
my calculation; he just asserted there was one:
: The rest of your post is erroneous and irrelevant.
But of course, kenseto doesn't say *why* it's erronious,
and exactly therefore, it's obviously relevant.
Let's go through it again, quite slowly, and with diagrams
of what's going in each frame. First, in coordinate system B,
we have a light pulse traversing a light-second-long rod.
That's events B(t=0,x=0) and B(t=0,x=1) for the endpoints of
the rod at the start, and B(t=1,x=0), B(t=1,x=1) for the
endpoints of the rod at the end, when the ray reaches
the other end.
The diagram of this is at http://sheol.org/throopw/length-of-rod-01.gif
Now, plot those events in A-coordinates, using the lorentz transforms
A(t=(Tb-v*Xb)/sqrt(1-v^2),x=(Xb-v*Tb)/sqrt(1-v^2))
We get A(t=0,x=0), A(t=-4/3,x=5/3) for the endpoints at the start,
and A(t=5/3,x=-4/3), A(t=1/3,x=1/3) for the endpoints at the end.
The diagram of this is at http://sheol.org/throopw/length-of-rod-02.gif
Now. The length of the red line segment (the A(t=0) line) from the
origin to its intersection with the right end of the rod is the length
of that rod in A coordinates. What kenseto does is take the length from
the origin to the perpendicular dropped through x=5/3, which is an easy
distance to find, but physically meaningless; the diagram shows just
*why* kenseto's calculation is physically meaningless. Again: the
endpoint of the rod has moved away from where he's calculated a length.
We know that the slope (or velocity) of that blue line is 0.8,
and it goes through the point A(t=1/3,x=5/3) (which we just proved above).
So we take x=(1/3)-.8*(t-(1/3)), set t=0, and get x=0.6.
The rod is contracted.
Again, kenseto's error is that he refuses to track the endpoints
of the rod explicitly, and account for their movement, but rather
prefers to confuse himself and others by ignoring the movement
of the endpoints of the rod during the light transit.
Even a naive approach of supposing that Ta*c = La + Ta*0.8*c (hence La =
0.2*Ta*c) yeilds a rod contraction. Of course, it yields the wrong
number; it gives 1/3 length (0.2*5/3=(1/3)). I leave as an excersize
for the reader why this is 5/9ths of the correct answer; I consider it
obvious from the diagram. Hint: the diagram shows 5/3 time above the
red t=0 line, and 4/3 below it. Hence there are nine
duration-one-third-second periods of time in A coordinates; ponder which
of those the 1/3 wrong answer corresponds to in the diagram.
But no matter how you look at it, kenseto's approach is obviously wrong,
once you remember that the ends of the rod are in motion during the
period of time Ta. Contrary to the givens of the scenario,
kenseto's approach assumes they are not.
--
Wayne Throop throopw@sheol.org http://sheol.org/throopw
From: throopw@sheol.org (Wayne Throop)
Newsgroups: sci.physics.relativity
Subject: Re: SRT and LET Improved
Distribution: world
Organization: sheol
Message-ID: <883244642@sheol.org>
Date: Sat, 27 Dec 1997 17:44:02 GMT
References: <3497ec17.2274334@news.erinet.com> <01bd0b17$49229b40$34297dc2@dbyr.iol.ie> <349952dc.2833361@news.erinet.com> <01bd0c01$af994500$162a7dc2@dbyr.iol.ie> <3499d2e4.3734625@news.erinet.com> <01bd0c73$28f10b60$c0297dc2@dbyr.iol.ie> <349d3497.1084439@news.erinet.com> <01bd0e47$fff735a0$302a7dc2@dbyr.iol.ie> <349de868.9333570@news.erinet.com> <01bd0ec7$ceb21520$15297dc2@dbyr.iol.ie> <349fd202.2346890@news.erinet.com> <01bd0fd9$45c06b00$04297dc2@dbyr.iol.ie> <34a07654.8905594@news.erinet.com> <882943039@sheol.org> <34a11e05.5692811@news.erinet.com> <882995442@sheol.org> <34a3f90d.1817852@news.erinet.com> <883187931@sheol.org>
: throopw@sheol.org (Wayne Throop)
: Let's go through it again, quite slowly, and with diagrams of what's
: going in each frame.
Oops; I note I diagramed the rod moving towards the lightsource;
it's more likely kenseto meant a case witht he rod moving away
from the lightsource. It doesn't change the facts of the matter
any, and it doesn't change the main point, which is that kenseto
neglects the motion of the rod's endpoints in A coordinates in
trying to determine its length. But it's probably best to diagram
the sitation with the rod moving the other way as well.
The diagram of light traversing the rod in B coordinates
is at http://sheol.org/throopw/length-of-rod-01.gif
The diagram of the same situation in "A" coordinates moving
to the right WRT those B coordinates is at
http://sheol.org/throopw/length-of-rod-02.gif
This post refers to the new diagram of the same situation, but
in "A" coordinates moving to the left WRT the same B coordinates.
It's at http://sheol.org/throopw/length-of-rod-03.gif
As before, we take the B coordinates of the endpoints of the unit
rod before and after light traverses it
left endpoint right endpoint
before B(t=0,x=0) B(t=0,x=1)
after B(t=1,x=0) B(t=1,x=1)
This time, we lorentz-transform them with
A(t=(Tb-v*Xb)/sqrt(1-v^2),x=(Xb-v*Tb)/sqrt(1-v^2))
and v=-0.8 in stead off v=+0.8.
left endpoint right endpoint
before A(t=0,x=0) A(t=4/3,x=5/3)
after A(t=5/3,x=4/3) A(t=3,x=3)
This also makes the calculation of the rod length more clear.
It's 3-0.8*3. The A-time-coordinate where the the light hits the other
end of the rod is inflated above the 1+2/3 time dilation because the rod
moved, but when you subtract it back out, the rod itself
MUST BE CONTRACTED.
And the bottom line is the same as it's always been: kenseto
simply doesn't account for the movement of the rod.
--
Wayne Throop throopw@sheol.org http://sheol.org/throopw
