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Adel Makram
Mar2-12, 01:17 PM
P: 163
Quote Quote by ghwellsjr View Post
Adel, I'm going to try to explain what time-like and space-like mean, and why if two events have one type of likeness in one frame, they have the same type of likeness in all frames. Again, I'm going to use compatible units where the value of c=1 and I'm going to align the first even at the origin. Then we can use the Lorentz transformation in a simple way to understand these terms.

First off, time-like simply means that the time coordinate is greater than the spatial coordinate. Space-like simply means that the spatial coordinate is greater than the time coordinate. Light-like simply means that both coordinates are equal.

So if we look at the simplified Lorentz transform, we see:

t' = γ(t-xβ)
x' = γ(x-tβ)

What we want to do is compare the values of these two equations for different values of β, where β can range from -1 to +1 (but not including those values). So we compare like this where the calculation of the new time coordinate, t', is on the left and the calculation of the new space coordinate, x', is on the right (the question mark can be replaced with =, <, or > depending on our evaluation):

γ(t-xβ) ? γ(x-tβ)

Now we note that γ is a function of β but for all values of β, γ is a positive number equal to 1 or greater than 1:

γ = 1/√(1-β2)

Therefore, we can remove γ from our conditional statement without effecting the evaluation of the condition:

t-xβ ? x-tβ

Next we rearrange the terms so that the t factors are on the left and the x factors are on the right:

t+tβ ? x+xβ

Now we factor out the common terms:

t(1+β) ? x(1+β)

Now since β can have a range of -1 to +1 but not including those numbers, the factor 1+β has a range of 0 to 2, not including those nuimbers.

Therefore we can divide out the (1+β) factor without changing the evaluation of the condition:

t ? x

What does this mean? It means what everyone has been telling you:

If t>x in one frame, t'>x' in any other frame, no matter what the value of β is.
If t=x in one frame, t'=x' in any other frame, no matter what the value of β is.
If t<x in one frame, t'<x' in any other frame, no matter what the value of β is.

So no amount of coming up with specific scenarios, especially ill-defined ones, can produce an example that will violate the conclusion that whatever likeness two events have in one frame, they have the same likeness in any other frame.

It's kind of a waste of time to try to decipher your scenarios and point out where your confusion is so that you will accept our critiques of what you are doing. My single most important advice to you is to learn what an event is, how to define a scenario in ONE frame, not two like you have been doing, and then use the Lorentz Transform to see what that scenario looks like in another frame.
Thank u for your care to clarify things for me. But my scenario is different from just assigning events to different coordinates in different frames. I posted here a clear detailed diagram and calculation based on my idea. Even from the diagram it is clear that A and B are still spacelike for the train observer because the source is put near to A than B which means that Signal must reaches B in no later than a signal would have taken to reach B coming from A. But how about the math! I see no contradictory to insert ∆t``= (AB`)/c and equating it with ∆t`= (AB`v/c2)/(1-(v2/c2)) can you? In other words, if A and B are spacelike for the train observer, why the math of LT allows the time difference between receiving signal at both ends to equal the time difference a light signal would take to reaches B from A?
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