Re: The velocity of light going pass a moving train.

On Jun 21, 10:50 am, BZ <WQAHBGMXS...@xxxxxxxxxxxxx> wrote:
On Jun 21, 9:00 am, Dono <s...@xxxxxxxxxxx> wrote:



On Jun 21, 4:52 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?

Correct.

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.

I hope you also agreed with this. The ball, of course, moves much more
sllllooowwwly than the light.





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.

Incorrect. In the track frame, while the light travelled from the
floor to the ceiling, the train moved from LEFT to RIGHT.
In mathematical terms:

1. In the train frame the light makes an angle pi/2 with the
semipositive x axis, so cos(theta_car)=0

right, for the train's frame.

cos(pi) = -1.

The beam misses the mirror, in which direction?
cos(0) = 1.

The beam misses the mirror, in which direction?


In the track frame the beam hits the mirror a little behind the
center, in the direction opposite to the train movement.


2. The track frame, according to the relativistic aberration formula

cos(theta_track)=(cos(theta_car)-v/c)/(1-v/c*cos(theta_car))

so

cos(theta_track)=-v/c

meaning that , in the track frame, the light wave vector k, makes an
andgle LARGER than pi/2 with the semipositive x-axis

We KNOW that the beam HITS the mirror dead center, no matter which FoR
the observer is in.

We KNOW that the ceiling mirror (and the floor mirror or laser) are
moving toward our right.
We KNOW that the basketball hits the basketball player's hand, no
matter which FoR the observer is in.

We KNOW that the basketball player, his hand and the ball are moving
toward the right.

RIGHT???

And we are talking about the trip from the floor toward the ceiling,
or basketball players hand, right?

Assuming the bounce starts just as the train passes, going 20 MPH.
Assume that the ball bounces at 20 MPH (strong basket ball player).

The ball will be seen to be rising, along a path that is almost
straight and at 45 degrees [off of the vertical AND off of the floor],
as seen by the trackside observer, through the picture window in the
side of the car.

We need not worry about Einstein at all at 20 MPH. Right?

And we are even dealing with something that can be seen from the side
[a beam of light could only be seen from the side if there were
something in its path to scatter the light].

By similar reasoning, a beam of light, in a light clock, aboard a
rapidly moving train, would STILL have to travel along a path that
slants to the right, in order to hit the center to the mirror [and we
KNOW that is where it does hit because the guy riding on the train
told us so.]



In other words, contrary to what you are saying, the light beam in
inclined towards the REAR of the car as viewed from the frame of the
tracks.
This makes perfect sense, since, as pointed out earlier, the car has
moved from the LEFT to the RIGHT while the light travelled from the
floor to the ceiling.

The mirror is ALSO traveling from left to right and the beam MUST hit
the mirror dead center. All observers will see that happen.



Now, the above system would not allow for dtection of absolute motion
because, inside the closed car the light moves perfectly vertically
regardless of the car speed v. So, an observer closed inside the car
would not be able to detect any "absolute motion"

They WOULD if the beam did NOT hit the center of the mirror for some
observers while it does hit dead center for other observers!

-bz-

Correct, the beam hits the center for the train observer but does not
for the track observer. The reason is that the mirror has moved to the
right while the beam travelled from the floor to the ceiling.


.

Relevant Pages

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