Finally, some attempt to actually address a question!

You do certainly seem hard to understand.Originally Posted bychinglu

Why is this an equation for y? You've not been clear about this.z=0 is not necessary. Look at the y,

y = ±√(c²t² - [vtγ /( 1 + γ )]² )

What is this equation supposed to be doing?I made z = 0 for simplicity. The actual requirement is

y² + z² = c²t² - [vtγ /( 1 + γ )]².

This makes less sense, but let's move on.In fact, that implies for each t, there exists an infinite number of space

coordinates such that t'=t.

But, if z=0, then there exists only two space coordinates in which t'=t.

OK, so you've found an equation to show how, at certain points within the light cone of the origin of a coordinate system, t=t' at that point. Not only is this point within the light cone of the origin, it is also a place where x'<0.Since you are having trouble with the theoritical math, I will give you an example.

Let v = (3/5)c. t = 1 second. Then gamma(γ) = 5/4

Let's calculate x.

x = vtγ/( 1 + γ ) = (3/5)(1)(5/4)/(1+5/4) = 1/3 light seconds.

Now, lets calculated t'

t' = ( t - vx/c² )γ = ( 1 - (3/5)(1/3)/c² )(5/4) = ( 1 - 1/5)(5/4) = (4/5)(5/4) = 1

Therefore, t = t' = 1.

See how simple and factual that is?

We can verify this by noting that

x' = γ(x-vt) = 5/4(1/3-(3/5)1) = 5/4(1/3-3/5) = 5/4(5/15-9/15) = 1/4(-4/3) = -1/3

Note that at a slightly higher x at t=1, t' would be slightly higher and at a slightly lower x at t=1, t' would be slightly lower.

What happens for this special equation at t=2? Then x=2/3, x'=-2/3.

What happens at v=4/5 and t=1? Then γ=5/3, x= 1/2, x'= -1/2

So we've really learned is how to look for the point where x=-x' and there we find that t=t'.

Is this really so amazing?

Well, it is if one doesn't understand the relativity of simultaneity. Because as we go backwards along the x axis, we are going to a point later along the t' axis (assuming a frame associated with a positive velocity). These are things that one figures out as one plays around with the mathematics of relativity. The planes of simultaneity that are associated with x'=q (where q is some constant), are at an angle relative to the plane x=r (where r is some constant).

I assume, from your ramblings, that you are trying to do something with light looked at from the reference frame co-moving with v=3/5.And, y = √(8/9) light seconds. So all is correct since z = 0 and x² + y² = 1

(SLW ) = spherical light wave

You're going to have to provide a little more information to make clear what you're going on about, though. So far it doesn't seem so amazing.

That just seems to be crazy talk. Events are anything that happens at a location and at a time, whether there is light emitted, light received, or no light involved whatsoever.See, we have a problem with your reasoning and mine is correct. Events are light receptions and not light emissions.

Again, there is no such thing as "a simultaneous event between the frames."But, let's go with your reasoning. Since M and M' are co-located, then you apply ROS and must claim the light emission event at A is not a simultaneous event between the frames.

It is not true in all frames that the event where M and M' are adjacent is simultaneous with event A.

All frames will agree that all events occur. They may disagree on what time these events occur. An observer, regardless of where they are, using the train frame to determine time will claim that M and M' are not next to each other when event A occurs, even though they will eventually agree that A occurs and that the light from A and B reaches M at the same time.That means if an unprimed observer is at A and light strikes there and there exists a primed observer co-located with at at the strike, then A will claim light is there with A and A' but A" will not. That is logically absurd.

So far, you've made a lot of crazy claims about Einstein's thought experiment but not backed them up. Show us your calculations for this supposed A and A' scenario of yours.

They might if it's useful, but it's not always so, especially when one has to consider two spatially separated events where light is emitted.Next, light emission is considered a start event if you want to call it an event. Now, most folks set t=t'=0 at the place when the light emission occurs, A in this case.

We have to be careful, since in the reference frame of the train, which I'm assuming is associated with t', we do not know for certain that event A happens when M and M' are at the same place. (Indeed, by the end of Chapter 9 we should be sure that these events aren't simultaneous.)Now, in the context of each frame, there is but one time.

So let's operate from the context of each frame. At t=0, what is the distance of M/M' to A? We already said it is r.

At t'=0, what is the distance of M/M' to A? By length contracition, that is r/γ.

Do you agree or disagree with the above.

The length at t'=0 from x=0 to the location of M' is actually rγ, since the proper length along the train is the length in the rest frame and this length is contracted in the embankment frame.

If we take our insights about relativity of simultaneity from our mathematics above, we should realize that at t'=0, M' has already moved past M and B happened even earlier.

Good for you, you worked through some similar math to what we did above.Let's take this math further.

So far we have for t'=t, when

x = vtγ/( 1 + γ )

and

y = ±√(c²t² - x² )

Let's solve for t

x = vtγ/( 1 + γ )

t = x( 1 + γ )/vγ

Plug this into y = ±√(c²t² - x² )

y = ±√(c²[x( 1 + γ )/vγ]² - x² )

y = ±√(c²x²( 1 + γ )²/)(v²γ²) - x² )

y = ±x√(c²( 1 + γ )²/)(v²γ²) - 1 )

Since v and γ are constants in SR, set K = √(c²( 1 + γ )²/)(v²γ²) - 1 )

Then, we have y = ±Kx.

Therefore, the set of points such that t'=t for positive y form a line. The same is true for negative y.

Hence, all space coordinates that satisfy y = ±x√(c²( 1 + γ )²/)(v²γ²) - 1 ), we have

t'=t.

Well, it seems you didn't really pay attention.Here is the bad part. Since t is not involved in the equation y = ±x√(c²( 1 + γ )²/)(v²γ²)

- 1 ) for deciding t'=t, meaning no information about time has value, then this line has no requirement for time between the frames.

This kills the concept of space-time.

Again, you proved that you can find specific regions where t=t'. This is something that people who study SR should already know. Your proof has nothing to say about what goes on in other regions. Indeed, your proof relies on equations that demonstrate that t<>t' at other locations.