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

On Jun 15, 1:45 am, "harry" <harald.vanlintelButNotT...@xxxxxxx>
wrote:
"G" <gehan_ameresek...@xxxxxxxxxxx> wrote in message

news:1181841217.718007.143680@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx



On 14 Jun, 20:53, "harry" <harald.vanlintelButNotT...@xxxxxxx> wrote:
"G" <gehan_ameresek...@xxxxxxxxxxx> wrote in message

news:1181832996.422174.142370@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx> On 14 Jun,
15:15, "harry" <harald.vanlintelButNotT...@xxxxxxx> wrote:
"G" <gehan_ameresek...@xxxxxxxxxxx> wrote in message

news:1181750413.104128.37500@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx> On Jun
13,
4:52 pm, "harry" <harald.vanlintelButNotT...@xxxxxxx>
wrote:
<luke.s...@xxxxxxxxx> wrote in message

[...]

For example: a light beam shot upwards in a moving train moves
relative to the source only.

A light beam is measured to propagate at c relative to the used
coordinate
system (the "rest frame"). I hope you don't overlook that the speed of
light
is independent of the speed of the source. For example, it doesn't
matter
if
your light source is attached to the railway track instead of to the
train:

If the light source is attached to the track the light goes directly
upwards in the track ref frame

*Only* if the light is directed straight up in the frame of the track!

-> Probably it should be pointed out that most people here use the word
"speed" to indicate the absolute value of the velocity, and *not* the
direction. And "velocity" is nowadays often used to indicate the vector
speed (thus absolute value and direction), while that was not the case 100
years ago when SRT was formulated.
Indeed, the *direction* of the light is a function of the velocity of the
source.
Does that solve the problem for you?

If the light source is on the train, the light goes up at an angle
according to AE and other texts

SO how come it does not matter?

OK, fine, for purposes of description it is to visualize a virtual
ether in the reference frame of the observer
that is, the platform reference frame.

If you choose so. It also works if you choose the train as reference
frame.

Choose the platfrom as the reference frame since we are on the
platform

It doesn't matter what frame you choose; as long as you stick to discussing
one example and not two examples that differ.
Here we have the example of a light ray that is bouncing between a point on
the ceiling and a point at the floor of the train.
That is seen in the train as going straight up and down (relative to the
train's restframe).
Now try to depict the same reality with the platform as reference frame. You
will see (if you can do that!) that the textbooks and the drawing that you
refer to below are correct.

[...]

Therefore, unlike the illustration given by in some texts (http://
abyss.uoregon.edu/~js/ast123/lectures/lec08.html
photon clock for example), to an observer on the train, the light beam
will be seen moving diagonally in the opposite direction to the
direction of motion.

No! The illustration in the middle is correct. If the ray goes straight
up
relative to the train,
it necessarily goes diagonally relative to the track.

You wrote "if": indeed, it's useless to mix examples - except if you *want*
to be confused.

But why should it go straigh up relative to the train when seen from
the track reference frame?

As we see it, we see a source emit a light and it does not matter what
the source is doing. Or does it?

You can point a laser pointer straight up in the train so that the laser
beam will propagate like this:

¦ t2
¦
¦
¦ t1

Inside the pointer, the light will go straight up *as measured in the
train*, because in the train the exit end of the pointer does not move
between time t1 and time t2.

Can you draw the trajectory of the light inside the pointer as measured on
the tracks, relative to the frame of the tracks? Necessarily the light
remains inside the pointer during that time; thus you must take into account
where in the frame of the tracks the bottom of the pointer is at t1, and
where the top of the pointer is at t2.

You can see that for yourself if you make the drawings yourself: also
according to an observer on the track, the light must remain bouncing
inside
the train between the same points on the floor and the ceiling!

Why? If the source is on the track, the light enters the train and
goes bouncing directly up and down in the train? Why
If it does not matter if the source is on the track or in the train
why does the light behave differently depending if
it is emitted by a source moving relative to the track or not?

See above: first work out one example, and then when you understand that,
you will rapidly understand other cases. :-)

I hope this helped; if not, perhaps someone else wants to try...

Regards,
Harald



Actually , you got it wrong, Harry.
And it is such a simple problem. "G" is indeed correct, how would the
light go diagonally in the track frame? Moreover, why would the light
ray get inclined forward in the direction of the train motion?
I'll give you a couple of hints, Harry.

1. Think about the separation (closing) speed (actually velocity)
between the ray of light in the vertical mirrors. If this is too
difficult, start with the calculation imagining that the "light clock"
is set horiziontally, on its side.

2. If you manage to solve question 1, then tell us where will the
light strike the upper mirror? (hint: it is not in the center)

3. Now, you can look this problem from a different "angle" (excuse the
pun). If you use Einstein light aberration formula :

cos (Theta_track)=(cos(Theta_train)-v/c)/(1-v/c*cos(Theta_train))

and you set cos(Theta_train)=pi/2 what do you get?
(hint: cos (Theta_track)=-v/c)

.

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