Re: The velocity of light going pass a moving train.
- From: "Sue..." <suzysewnshow@xxxxxxxxxxxx>
- Date: Thu, 21 Jun 2007 05:48:18 -0700
On Jun 21, 8:52 am, BZ <WQAHBGMXS...@xxxxxxxxxxxxx> wrote:
On Jun 21, 1:33 am, Dono <s...@xxxxxxxxxxx> wrote:
On Jun 20, 10:26 pm, "Jeckyl" <n...@xxxxxxxxxxx> wrote:
"Dono" <s...@xxxxxxxxxxx> wrote in message
news:1182387866.761051.23130@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
On Jun 20, 5:27 pm, "papar...@xxxxxxxxx" <papar...@xxxxxxxxx> wrote:
On 20 jun, 18:29, Dono <s...@xxxxxxxxxxx> wrote:
On Jun 19, 11:05 pm, "Jeckyl" <n...@xxxxxxxxxxx> wrote:
"Dono" <s...@xxxxxxxxxxx> wrote in message
news:1182303768.824683.175520@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
Yes, I kept trying to get dr.Jeckyll to understand it ......I
already
showed him the formula, to no avail (at least, so far).
You were simply misunderstanding the problem and so coming up with
the wrong
solution.
No, idiot. You simply don't understand aberration, that is all.
If you stopped talking and you started using math you would
understand. But since you avoid using math like the plague, you keep
repeating the same errors.
It helps when you are actually talking about the same problem as
everyone else (G, Harry, myself). As I said .. you were using the
right
formulas but misapplying it (as far as the problem the rest of us
were
talking about).
Again, no, idiot. The description of the problem in math terms is not
as ambigous as you keep making it to be.
Here it is, one more time, mr. Jackasss:
-In the traincar frame theta_car=pi/2
-In the track frame
cos(theta_track)=(cos(theta_car)-v/c)/(1-v/c*cos(theta_car))
So, can you calculate cos(theta_track)? I asked you 5 times, why are
you so shy about using a little math?
Since you don't get the math and you didn't get the "separation speed"
explanation, I will give you a third explanation: since in relativity
all frames are equivalent, instead of having the train moving Left to
Right with respect to the tracks, imagine that the tracks move Right
to Left while the light bounces vertically in the car frame. How is
the light inclined in the track frame? If you still don't get it, look
at these pictures:
http://www.fourmilab.ch/cship/aberration.html
But in those pictures, the observer is in the train frame and he sees
through the window as if the ground is moving from left to right and,
obviously the rain is falling with an angle that clearly is inclined
into the direction of the movement of the ground, again as seen from
the train frame. So those pictures actually contradict what you are
saying.
Miguel Rios
In both cases the light is inclined from right to left,
No .. if the traing goes left to right in the FoR of the tracks, then the
light that is vertical in the train goes left to right in the FoR of the
tracks.
You are using the correct formula .. but applying it incorrectly. BTW: Do
you even understand relativistic aberration works, and why you get a great
difference in angle when you take relativity into accout.
Yes, I do i****. Now check with the guy who was the first to derive
the formula:
http://www.fourmilab.ch/etexts/einstein/specrel/www/
i.e. it makes an angle greater than 90 degrees with the semipositive
x-axis.
Try understanding the relativistic aberration formula, would you?
How about you try it .. you seem to think the light goes the wrong way
Check with Einstein, you ignorant t***:
http://www.fourmilab.ch/etexts/einstein/specrel/www/
Do you understand that in the train iFoR, the vertical beam in the
light
clock hits both mirrors 'dead' center, all the time, because the
mirrors
are aligned in the trains iFoR to be parallel?
If you see that, then you should realize that a basketball player,
riding
on the train, sitting in his seat, dribbling the ball in the isle,
would
have the ball moving vertically in the trains iFoR.
If you agree with those, then you should realize that from ANY iFoR,
the
light must continue to be seen to hit the mirror in the center and the
ball
to hit the floor and the basketball players hand dead center.
Since the mirrors and the hand are traveling left to right, the light
and
the ball must travel left to right and follow the path that Jeckle
keeps
describing to you.
At the moment that the image of the ball [light beam] hitting the
floor [mirror] reaches the trackside observer, the objects causing the
images have moved even further to the right, but the track side
observer sees the ball following a path that looks like a series of
mmmmms. The light, on the other hand, zigzags to the right. Both
continue to hit the exact same spots.
Were that not true, the basketball player and the guy building the
light clock would both have ways of detecting and measuring ABSOLUTE
MOTION.
If you are going to argue for such an absolute motion detector, you
will need to abandon quoting Einstein, because he clearly said that it
could not be done, as have over a century of experiments.
That's a fair and square analysis. Whether you represent the mass
with a ball or the mirror's dielectric, the moving path is the one
that is effective. You can even prove it with QED.
Now! I nominate you to explain why we can't see light on
its journey from the sun to the moon and why raindrops
viewed from a moving car don't take longer to hit the ground
in spite of the diagonal path. >:-)
Sue...
Aside to Jeckle, and others: please keep your language here clean and
polite. Children may read the postings here. In fact, your great great
great grand children WILL almost certainly read what you have written.
You set a very poor example for them when you curse and call each
other idiot, etc. I understand your frustration when someone 'just
doesn't get it', but losing your temper just makes you look bad.
--
bz
please pardon my infinite ignorance, the set-of-things-I-do-not-know
is an
infinite set.
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