Re: A friendly request to Ken Seto from Uncle Ben

Correct.  They have no effect on the transit times.

But they have a great effect on the starting times of the light
travelling to the front and to the back.

That's the whole point. Simultaneity is relative.

No you are the clueless one. x1 and x2 are coordinates of the ends of
the train. But these coordinates are irrelavent....they have no effect
on the transit times needed for the light fronts from the original
points of the flashes to arrive at M'.

Ken Seto

[...]

You are really clueless (eventhough we gav you the clues). In the LT, x1 is
NOT a distance; similarly, x2 is NOT a distance!

Harald

Helloooo....the points of the flashes are at the ends of the train
when M and M' are coincide with each other. So how does that mean that
the train is a point train? x1 is the distance of M to the front of
the train and x2 is the distance of M to the rear of the train since M
is at the middle of the train then x1=x2.

Ken Seto

Ken I'm afraid you missed that clarification - you would be wise to
contemplate about it. In this one-dimensional problem (as y1=y2 and z1=z2),
if x1=x2, this can only mean that the two points are the same point. Do you
think that the strikes were said to occur at the same point? And, as Ben
said, that the train is a point train?

Indeed.

Harald

NO, NO, NO....x1 and x2 are points in space where the strikes occur
simultaneously and these points are at equal distance from the M and
M' observers. The locations of the ends of the train is irrelevant.

Ken Seto

you for showing some Lorentz Transformation equations.

But if you examine the statement of the problem cafrefully you will
see that x1 is at the caboose of the train when the lightning strikes
it and x2 is at the engine of the train when the lightning strikes
it.

So if you want to say that x1 = x2, you are saying that the length of
the train w.r.t. the track is ZERO.  In other words, the train is a
point train!

I don't think so.

And if you realize that x2 > x1, for a real train, you will see that
t'1 > t'2, and the simultaneity is, indeed, GONE.

Uncle Ben

Event 1  t'1 = gamma(t1 - vx1/c*c)    Lorentz Transformation, track ->
train
Event 2  t'2 = gamma(t2 - vx2/c*c)    Lorentz Transformation, track -

When M and M' coincide with each other t1 = t2 =0
Therefore:
Event 1 arrive at M':
t'1 = -vx1/c^2
Event 2 arrive at M':
t'2 = -vx2/c^2

Since x1 = x2  therefore the light fronts from events 1 and 2 will
arrive at M simultaneously.

Ken Seto

- Hide quoted text -

What you say may or may not be true, but I don't see any Lorentz
Transformation yet.  What is the Lorentz Transformation?

When M and M' are coincide with each other:
The transit time for the light fronts from +x and -x arrive at M
simultaneously is L/c from both directions.
The light path length for the light fronts from +x and -x to arrive at
M' is gamma*L
Therefore the transit time for the light fronts from +x and -x to
arrive at M' simultaneously is gamma*L/c.

Ken Seto

Dear Ken,

For weeks now you have been telling me, as we discuss elementary
problems in Specialk Relativiy, that I have been executing the Lorentz
Transformation incorrectly.

Maybe you are right, Ken.  Educate me a little.  Tell me, what is the
Lorentz Transformation?  In particular,

1. What is it for?  It is a tool, I know, but what is it used for?

2. Show me a very simple example of how it is used for any problem you
like.

If it turns out that I have bveen misusing the Lorentz Transformation
for all the years I have taught relativity, I will be very forever
grateful to you for correcting me.

Uncle Ben