"The Ghost In The Machine" wrote in message
...
: In sci.physics.relativity, Androcles
:
: wrote
: on Sun, 02 Sep 2007 11:10:04 GMT
: :
:
: "The Ghost In The Machine" wrote in
message
: ...
: : In sci.physics.relativity, Androcles
: :
: : wrote
: : on Sat, 01 Sep 2007 14:03:45 GMT
: : :
: :
: : "Paul B. Andersen" wrote in message
: : ...
: : : Androcles wrote:
: : : "Paul B. Andersen" wrote in
message
: : : ...
: : : : Androcles wrote:
: : : : "Paul B. Andersen" wrote in
: : message
: : : : ...
: : : : : Henri Wilson wrote:
: : : : : Explain the 90 deg phase lag then George.
: : : : :
: : : : : Not hard at all to explain why the curves are different.
: : : : : If the radius of the star didn't change, it is obvious
: : : : : from Planck's blackbody equation that the luminosity
: : : : : variation due to the changing temperature is much
: : : : : bigger in visible light (V-band) than it is in IR (K-band
: 2.2u).
: : : :
: : : : Everything is "obvious". Obviously you are a lunatic.
: : : :
: : : : I note with a yawn that Androcles doesn't find the obvious
: : : : consequence of Planck's black body radiation law to be obvious.
: : : :
: : : : Nothing is obvious in a haze, is it?
: : :
: : : ASSistant Professor "Paul B. Andersen" of
:
: : : Agder University College (HiA)
: : : Serviceboks 422, N-4604 Kristiansand, NORWAY Tel (+47) 38 14 10 00
: Fax
: : : (+47) 38 14 10 01
: : : has executed the biggest fumble ever seen in the history of
: : : sci.physics.relativity
: : : in message
: : : ...
: : :
: : : "The spectral class [of stars] is determined by the relative
: positions
: : : and intensities
: : : of the absorption lines, and these are unaffected by a Doppler
: shift."
: : :
: : : The all time classic:
: : :
: : : "That is, we can reverse the directions of the frames
: : : which is the same as interchanging the frames,
: : : which - as I have told you a LOT of times,
: : : OBVIOUSLY will lead to the transform:
: : : t = (tau-xi*v/c^2)/sqrt(1-v^2/c^2)
: : : x = (xi - v*tau)/sqrt(1-v^2/c^2)
: : : or:
: : : tau = (t+xv/c^2)/sqrt(1-v^2/c^2)
: : : xi = (x + vt)/sqrt(1-v^2/c^2)" -Paul B. Andersen
: : :
: : :
: : The faster you go the longer it takes to get there, OBVIOUSLY.
: : yawn
: :
: : Here's how I look at it.
:
: [snip wrong argument]
:
: ... reach the star, d-vt = 0.
:
: The Andersen Transforms are d+vt 0.
: The faster you go the longer it takes to get there, OBVIOUSLY.
: yawn
:
: Since I was assuming a different transform:
:
: xi = (x - vt) * g
: tau = (t - vx/c^2) * g
:
: this doesn't quite work.

: However, I'd have to dig through
: the posts to see what Paul was claiming and your objections
: thereto,

That's an old worn-out transform that doesn't quite work.
The Andersen Transforms date back to 1999 to 2000 and
are the symmetric form of the Cuckoo... err... sorry... Lorentz
Transforms. Once you realise that the station comes to the train
(i.e. the Principle of Relativity) and that the K-frame that was
the "stationary" frame becomes the "moving" frame as seen by
an observer on the train (we always have observers in relativity,
never passengers) then

"That is, we can reverse the directions of the frames
which is the same as interchanging the frames,
which - as I have told you a LOT of times,
OBVIOUSLY will lead to the transform:
t = (tau-xi*v/c^2)/sqrt(1-v^2/c^2)
x = (xi - v*tau)/sqrt(1-v^2/c^2)"

which is the cuckoo malformation as seen by the train observer
obviously to do. (I wrote it in the original Norwegian by adding
"to do" at the end and included "obviously".)
Obviously all that has happened is the names 't' and 'tau', 'x' and 'xi'
have been exchanged, quite, to do.

Now comes the miraculous part, to do.

"or:
tau = (t+xv/c^2)/sqrt(1-v^2/c^2)
xi = (x + vt)/sqrt(1-v^2/c^2)" -Paul B. Andersen

Obviously, quite, to do.



: However, I'd have to dig through
: the posts to see what Paul was claiming and your objections
: thereto, and I'd highly prefer a different nomenclature anyway,
: something along the following lines.

: If E is a symbol for the coordinate frame for Earth and S
: the symbol for the coordinate frame for the spacecraft,


Aha, we are off into spacetime, how exciting! Choo-choo
trains are definitely old hat these days.



: and the spacecraft is traveling at a uniform velocity
: v away from Earth in a straight line, and with
: coincident origins, one can establish the usual Lorentz transform
:
: x_S = (x_E - v * t_E) / sqrt(1-v^2/c^2)
: t_S = (t_E - v * x_E / c^2) / sqrt(1-v^2/c^2)
:
: This transform is invertable, with a little work,
: resulting in:
:
: x_E = (x_S + v * t_S) / sqrt(1-v^2/c^2)
: t_E = (t_S + v * x_S / c^2) / sqrt(1-v^2/c^2)

There ya go, the Andersen Transforms. Well done!

Now, keeping the origo (that's Norwegian for origin and not an
oreo cookie or cuckoo with cream) of the E-frame and the S-frame
coincident as you say above, v = 0.

So with a little more work,
x_E = (x_S + 0 * t_S) / sqrt(1-0^2/c^2)
t_E = (t_S + 0 * x_S / c^2) / sqrt(1-0^2/c^2)

x_E =x_S
t_E = t_S.

:
: I can also express these using the notation
:
: (x_E,t_E)_E = ( (x_E - vt_E)*g, (t_E - v*x_E/c^2)*g)_S
: (x_S,t_S)_S = ( (x_S + vt_S)*g, (t_S + v*x_S/c^2)*g)_E
:
: where _S and _E establish which frame I'm using for the
: given coordinate point, and g = 1/sqrt(1-v^2/c^2) the
: usual gamma. I'm not sure if this is a proper tensor,
: but it does establish a mapping from E-frame to S-frame
: (and vice versa; the transform is bijective for any v in
: the open interval (-1, +1)),

As long as the spacecraft is traveling at a uniform velocity
zero away from Earth in a straight line with coincident origins.

: and I for one think

No, surely not! Can you provide any evidence to support that
absurb claim?



it helps
: clarify things, as one doesn't have to ask such questions
: as "which frame is this relevant in?".

Of course. Keeping the origins coincident really does
clarify things.



: Now one can set up various problems: the most obvious
: one is when does SR predict that the spacecraft reaches
: the star?


It doesn't, the origins coincide. You'll need to bring the
star to the coincident origins.




: More specifically, what values of x_E, t_E,
: x_S, and t_S are such that
: (x_E, t_E)_E = (d,t_E)_E [Earth sees it reaching the star][*]
: and
: (x_S, t_S)_S = (0,t_S)_S ? [The spacecraft sees itself reaching the
star]
:
: The first equality should be obvious, but the second one might not be.
: The general issue is that an astronaut cannot leave his ship, and since
: the ship defines the ship-frame, x_S = 0 for most of the interesting
: measurements, unless one postulates the astronaut holding a rod of
: a certain length (which is a slightly different problem).
:
: So now we know x_E = d, x_S = 0. Since x_S = g * (x_E - v*t_E) = 0,
: t_E = x_E / v, or x_E = v * t_E. Since t_S = (t_E - v * x_E / c^2) * g,
: t_S = (t_E - v^2 * t_E / c^2) * g = t_E / g.
:
: Since t_S t_E, x_E - v * t_S 0 -- this may be what Paul meant.
: Unfortunately, the expression x_E - v * t_S isn't all that meaningful
: an expression, as the units are not consistent.
:
:[*] with this expression, we are eliding the issue of when the light of
: the spacecraft reaching the star actually gets to Earth, and
: assuming that Earth -- somehow -- sees things instantaneously.
:
Did you want further comment on coincident origins, the value of v
or your ability to think?