Re: The "thought" experiment of David Strich
- From: Sean McHugh <seanm@xxxxxxxxxxxxxx>
- Date: Sat, 06 Jun 2009 19:21:10 +1000
"Henry Wilson, DSc" wrote (86):
On Sun, 24 May 2009 20:14:00 +1000, Sean McHugh <seanm@xxxxxxxxxxxxxx> wrote:
"Henry Wilson, DSc" wrote:
On Sun, 24 May 2009 17:14:41 +1000, Sean McHugh ?seanm@xxxxxxxxxxxxxx? wrote:
?
?Androcles wrote (142):
??
?
?Observe that I didn't derive the pythagorean theorem either. Again we
?have seem to have the obstacle, English. The derivation of gamma was
?not the exercise. Read the question.
?
?I found where you wrote this:
?
?http://sci.tech-archive.net/Archive/sci.physics.relativity/2004-08/0296.html
?
?~ Hardly... I dunno where h comes from, since nobody ever derived
?~ sqrt(1-w^2/c^2) for me, but never mind, I'll go along with the joke.
?
?That's odd, given that there's a plethora of derivations out there,
?ranging from the dead easy to the diabolical. Anyway, I'll give it a
?go.
?
?Consider the X'Y'T' system, which, according to an inertial observer
?in XYT, is moving at v in the X direction. Within the X'Y'T' system,
?a light signal is bounced back and forth, at c, along Y', over the
?distance y', taking t' for each coverage of y'.
?
? v --?
? --- --- ------------------
? | | /\ /\
? | | / \ / \
? y = y' d \ / \
? | | / \ / \
? | | / \/ \
? --- --- ----------------------
? 0x---1x---2x---3x---4x
?
?According to XYT the light will be travelling at c in a sawtooth
?motion,
No it wont.
Oh for crying down the sink.
The light beam remains vertical but moves sideways.
Apart from you, who mentioned a 'beam'? Why not a 'signal' pulse that
is much shorter than ct'? How will XYT see it get to the corresponding
point on the other side, if he doesn't see it travel in both X and Y?
Even with your beam, think of its leading point.
Any pulse of light has a length.
And a very brief pulse can be very small compared to the distance it
traverses and can approximate a point. That oscillating point, in the
moving frame, will describe a series of diagonal lines as viewed by
the relatively stationary frame. In any case, any point along a
beam of light - the leading end, for example - can be regarded as a
point to which the above will apply.
How any more times do I have to explain this to the relativist morons here?
I'd say you have explained it to yourself far too many times. You need
to stop it before you go completely blind.
Look, I appreciate how difficult it is to explain logical science to ardent
supporters of Einstein because they are generally people of low intelligence
and high gullibility. But I will persevere...
Yeah, unlike newsgroup trolls, those notable physicists are notable
for their low intelligence. There but for the grace of God . . .
If you care to plot the paths of infinitesimal elements of a short light pulse
that moves vertically in its source frame, you will find that each follows a
different infinitely thin diagonal path in a relatively moving frame.
Yep.
These elements are not 'light'. They are merely points on a graph. Their speed
in the moving frame is sqrt(c^2+v^2).
No! The speed is c. That the speed of the source doesn't alter the
speed of the light. This was demonstrated in astronomy by observing
double stars:
http://tinyurl.com/lfmmpb
That source independence could be explained by the light being
transited at c trough a medium (aether) but the Michelson-Morely
experiment showed that either the aether did not exist or it had no
effect. In other words, it was undetectable. But this also meant that
either the source or the recipient of the light could be considered as
moving or stationary and so regardless of which system measures the
speed of the light, it will be measured as c.
According to the stationary or reference frame, the light takes a
diagonal path and takes longer to cover the same vertical distance
than according to the perspective of the moving frame. There is no
speeding up of the light to compensate and that is the whole point.
The velocity vector of the pulse, no
matter how short you make it, remains vertical in all frames.
Here you are talking more about the orientation of the 'element' than
the direction of travel and the distance to be covered. For the one
observing the 'moving' frame, the direction of travel for any
'element' (or part thereof) becomes diagonal and will take longer to
transit. The diagonal travel occurs even in your animation so I'm
stumped as to what is your problem with the model having the cycles
describing a sawtooth. The orientation of any element is actually
immaterial and necessitates no change in anything that I presented.
Better still, if you are capable of using a computer,
I get by.
you can actually see what happens by running my program:
www.users.bigpond.com/hewn/movingframe.exe
Understanding that a beam in Y'T', that follows its Y' (vertical)
axis, will be seen by YT to follow the moving frame's Y' axis, doesn't
really need moving pictures. Nevertheless, if Y'T' is moving with
respect to YT, the latter will be regarded as travelling diagonally.
To illustrate, if YT intercepted any photon from Y'T' with a tube,
desiring for the light to continue trough the tube uninterrupted, YT
would need to tilt that tube - roughly 45 degrees in your models. The
tube's bearing _must_ sensibly represent the light's bearing.
Best Regards,
Sean McHugh
.
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