1. Finally, some attempt to actually address a question!
Originally Posted by chinglu
You do not understand any of this.
You do certainly seem hard to understand.
z=0 is not necessary. Look at the y,

y = ±√(c²t² - [vtγ /( 1 + γ )]² )
I made z = 0 for simplicity. The actual requirement is

y² + z² = c²t² - [vtγ /( 1 + γ )]².
What is this equation supposed to be doing?
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.
This makes less sense, but let's move on.
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?
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.

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).
And, y = √(8/9) light seconds. So all is correct since z = 0 and x² + y² = 1

(SLW ) = spherical light wave
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.

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.

See, we have a problem with your reasoning and mine is correct. Events are light receptions and not light emissions.
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.
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.
Again, there is no such thing as "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.
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.
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.

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.
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.
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.
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.
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.)

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.
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.
Good for you, you worked through some similar math to what we did above.
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.
Well, it seems you didn't really pay attention.

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.

2. Originally Posted by chinglu

Now, as far as A and B not being simultaneous, I have proven that is false.

Each frame sees the strikes as simultaneous and claims the other does not see them as simultaneous.

If you can prove any of the very public math I presented is incorrect, then have at it.

I certainly have nothing to hide.
I fail to see your reasoning on that.

If A and B are at a location where the train observer would perceive them to be simultaneous, then they are at a location where the Embankment observer would perceive A to have preceded B.

If A and B are at a location where the embankment observer would perceive them to be simultaneous, then they are at a location where the train observer would perceive B to have preceded A.

I'm pretty sure could position the points strategically so that both observers observed them both to be simultaneous (haven't actually tried it), but that would require moving them up or down vertically as well as side to side horizontally. They can't be colinear with the embankment. The line formed by the two points A and B would have to be diagonal instead of horizontal.

3. Originally Posted by kojax
Originally Posted by chinglu

Now, as far as A and B not being simultaneous, I have proven that is false.

Each frame sees the strikes as simultaneous and claims the other does not see them as simultaneous.

If you can prove any of the very public math I presented is incorrect, then have at it.

I certainly have nothing to hide.
I fail to see your reasoning on that.

If A and B are at a location where the train observer would perceive them to be simultaneous, then they are at a location where the Embankment observer would perceive A to have preceded B.

If A and B are at a location where the embankment observer would perceive them to be simultaneous, then they are at a location where the train observer would perceive B to have preceded A.

I'm pretty sure could position the points strategically so that both observers observed them both to be simultaneous (haven't actually tried it), but that would require moving them up or down vertically as well as side to side horizontally. They can't be colinear with the embankment. The line formed by the two points A and B would have to be diagonal instead of horizontal.

Let's have a little exercise.

assume the below
M'---------------------------------------A'

Now assume when A is a distance d when the light is flashed and A' is at rest with M'.

Since light measure c, then t = d/c.

Now say A' is moving toward M'. When A' is a distance d, light is emitted.

What is t?

t is still d/c.

So, just because A, the lightning strike is a moving light source, that does not mean M' is moving toward the light in the view of the primed frame.

The primed frame is at rest and the light source is moving. So, the distance is measured when the light is emitted and has nothing to do with the motion of the light source.

In other words, M' never moves toward the light in its own view.

4. Originally Posted by chinglu
In other words, M' never moves toward the light in its own view.
This is true. Yet we know, in the Chapter 9 thought experiment, that the light from B reaches M' before the light from A. We know this because we already worked this out in the embankment frame and all frames must have the same event. The only explanation for this is that the light from B left earlier than the light from A.

This is called the relativity of simultaneity.

5. [quote="PhysBang"]

y = ±√(c²t² - [vtγ /( 1 + γ )]² )

Because c²t² = x² + y² with z = 0 for the SLW.

So, solve for y

y = ±√(c²t² -x² )

Then substitute x = vtγ /( 1 + γ ), hence

y = ±√(c²t² - [vtγ /( 1 + γ )]² )

I made z = 0 for simplicity. The actual requirement is

y² + z² = c²t² - [vtγ /( 1 + γ )]².
What is this equation supposed to be doing?

It is showing z does not need to be 0.

We have based on the light postulate

x² + y² + z² = c² t²*
Solve for y² + z²
y² + z² = c²t² - x²

Substitute x = vtγ /( 1 + γ )

Hence, y² + z² = c²t² - [vtγ /( 1 + γ )]²

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?
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.

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).
Wow, impressive, I did not even want to go into the x'=-x issue as part of all this, but you saw it anyway.

In fact, reciprocol length contraction implies t'=t.

Look in the literature, you will find frame to frame clock synchronization is not decidable under SR currently.

This method syncs clocks between the frames.

It does reveal some much worse though.

The question is along this y = ±√(c²t² - [vtγ /( 1 + γ )]² ) line,
how can we have t'=t on that line?

What is going on before t'=t and after t'=t.

OK, let y = K instead.

The SLW reaches the coordinate (0,K,0) in the unprimed frame. Now, use LT and you will find t' = Kγ/c.

But, t = K/c. So, t'>t.

Now, how can it be along the y=K line that eventually t'=t as the SLW expands?

In the view of the unprimed frame, time proceeds backwards along that y line until t'=t.

What else happens?

The SLW continues to expand along that y line until it reaches the primed coordinate (0,K,0).

At that place, t=Kγ/c and t'=K/c.

As we can see with these math tools, in order for SR to be true, the unprimed frame must conclude time moves backward in the primed frame.

But, the primed frame concludes time never moves backward along the line y=K.

6. Originally Posted by PhysBang
Originally Posted by chinglu
In other words, M' never moves toward the light in its own view.
This is true. Yet we know, in the Chapter 9 thought experiment, that the light from B reaches M' before the light from A. We know this because we already worked this out in the embankment frame and all frames must have the same event. The only explanation for this is that the light from B left earlier than the light from A.

This is called the relativity of simultaneity.
No, we don't know M' sees A before B. We were told that. We received instructions and submitted to a master without proof. We wanted to be non-scientific and accept a human God.

I, on the other hand, do not accept human Gods.

So, I am looking at this problem mathematically. Now, do you want to do this or submit to human Gods?

Reason your own way through this. Don't quote an authority, do it on your own.

7. Originally Posted by chinglu
Wow, impressive, I did not even want to go into the x'=-x issue as part of all this, but you saw it anyway.

In fact, reciprocol length contraction implies t'=t.
What reciprocal length contraction? There is no length contraction here, only a change of coordinates.
Look in the literature, you will find frame to frame clock synchronization is not decidable under SR currently.
This method syncs clocks between the frames.
No, it doesn't. This method picks out a location where the clocks happen to be synchronized. If one goes outside of this region, one finds that the clocks are no longer synchronized. Did you fail to notice that? Did you fail to notice that you had to assume very special conditions where x could only have certain values?
It does reveal some much worse though.

The question is along this y = ±√(c²t² - [vtγ /( 1 + γ )]² ) line,
how can we have t'=t on that line?
We can have t=t' on that line because you assumed that x = vtγ /( 1 + γ ) and this picks out the regions where t=t'.
What is going on before t'=t and after t'=t.
That makes no sense.
As we can see with these math tools, in order for SR to be true, the unprimed frame must conclude time moves backward in the primed frame.
Well, yes. This is the relativity of simultaneity.
But, the primed frame concludes time never moves backward along the line y=K.
Well, yes. This is the relativity of simultaneity.
What contradiction? SR shows us that time assigned by different frames differs with location and isn't constant across locations. Again, you simply show us that you do not understand the relativity of simultaneity.

8. Originally Posted by PhysBang
That's not what I wanted. I want you to prove that some point meets that criteria of x = vtγ /( 1 + γ ) and z = 0. .
The condition x = vtγ /( 1 + γ ) and z = 0 will be satisfied by a two-dimensional surface in spacetime. Clearly x=y=z=t=0 is one such point,

There are indeed spacetime points (this surface) for which the time coordinate is preserved by some fixed element of the Lorentz group. So there are in fact infinitely many such points.

The real question is "So what ?".

Restricting attention to points on that surface tells you nothing of interest.

The notion of a single event being sinultaneous in two reference frames is ludicrous. What can be simultaneous are two events in one reference frame.

Trying to argue logic with chinglu is going the be about as successful as trying to play chess with a pigeon.

9. Originally Posted by PhysBang
This method syncs clocks between the frames.
No, it doesn't. This method picks out a location where the clocks happen to be synchronized. If one goes outside of this region, one finds that the clocks are no longer synchronized. Did you fail to notice that? Did you fail to notice that you had to assume very special conditions where x could only have certain values?
Who cares about the special conditions?

You can stil sync the clocks between frames by using a light pulse and place observers at specific locations.

The mainstream says currently this is impossible under ALL circumstances.

As we can see, that is false.

It does reveal some much worse though.

I read the rest of you post.

Do some thinking and math on your own.

Assume the context of the unprimed frame and a light pulse is emitted.

Let the light pulse reach the coordinate (0,K,0).

Check with LT the times along the light y=K.

That is the partial derivative I produced earlier, dt'/dt<0.

Time proceeds backward for the primed frame in the view of the unprimed frame along the line y=K.

Do the numbers yourself.

10. Originally Posted by DrRocket
Originally Posted by PhysBang
That's not what I wanted. I want you to prove that some point meets that criteria of x = vtγ /( 1 + γ ) and z = 0. .
The condition x = vtγ /( 1 + γ ) and z = 0 will be satisfied by a two-dimensional surface in spacetime. Clearly x=y=z=t=0 is one such point,

There are indeed spacetime points (this surface) for which the time coordinate is preserved by some fixed element of the Lorentz group. So there are in fact infinitely many such points.

The real question is "So what ?".

Restricting attention to points on that surface tells you nothing of interest.

The notion of a single event being sinultaneous in two reference frames is ludicrous. What can be simultaneous are two events in one reference frame.

Trying to argue logic with chinglu is going the be about as successful as trying to play chess with a pigeon.
You are so intense.

All you people are claiming light emission is an event.

I tries to tell you folks that was false, but you all fell into it.

Now, you will need to confess you were wrong and when you do, I win the ROS false argument.

Otherwise, let the light emission be an event.

It is at t'=t=0.

Now, you have agreed at some t>0, t=t'.

We now have 2 simultaneous events betwen the frames for all t.

We have the light emission at t'=t=0 and the event where t'=t>0.

LOL.

11. Originally Posted by chinglu
Who cares about the special conditions?
Anyone who does a math proof wants to know when that proof holds. You seemingly ignore this part of math.
You can stil sync the clocks between frames by using a light pulse and place observers at specific locations.
This was pretty much the first thing that Einstein ruled out in 1905. So you're about 105 years behind the time.
The mainstream says currently this is impossible under ALL circumstances.
No, the mainstream says that t=t' at all points where x=-x (assuming a v in the positive x direction). One can see this quite clearly in the math of SR if one takes the time to look at it. Only someone obsessive takes one tiny fact about SR and uses it as a basis for ignoring the rest of it.
No, we don't know M' sees A before B. We were told that. We received instructions and submitted to a master without proof. We wanted to be non-scientific and accept a human God.
Again, you show us that you failed to actually read the very thought experiment that you brought up.

Here's what Einstein said:
"Just when the flashes of lightning occur, this point M' naturally coincides with the point M, but it moves towards the right in the diagram with the velocity v of the train. If an observer sitting in the position M’ in the train did not possess this velocity, then he would remain permanently at M, and the light rays emitted by the flashes of lightning A and B would reach him simultaneously, i.e. they would meet just where he is situated. Now in reality (considered with reference to the railway embankment) he is hastening towards the beam of light coming from B, whilst he is riding on ahead of the beam of light coming from A. Hence the observer will see the beam of light emitted from B earlier than he will see that emitted from A."

All we have to do is note that M' is moving in a certain direction to see that the light from B will hit M' before the light from A hits M'. Do you disagree with this?

If a car is moving towards another car, will the two cars hit each other sooner than if one of them was sitting still?
So, I am looking at this problem mathematically. Now, do you want to do this or submit to human Gods?
We looked at it mathematically, but you seem to want to take mathematical conclusions far beyond their scope.
Reason your own way through this. Don't quote an authority, do it on your own.
I'm not simply quoting an authority, I'm showing you the argument that you should have already read on your own.

12. Originally Posted by PhysBang
Originally Posted by chinglu
Who cares about the special conditions?
Anyone who does a math proof wants to know when that proof holds. You seemingly ignore this part of math.
Make your case with proof and not opinions.

You can stil sync the clocks between frames by using a light pulse and place observers at specific locations.
This was pretty much the first thing that Einstein ruled out in 1905. So you're about 105 years behind the time.
Yea, well we have the math above that proves that is false. Even RocketMan agrees my math is correct.

Tough luck.

Again, you show us that you failed to actually read the very thought experiment that you brought up.

Here's what Einstein said:
"Just when the flashes of lightning occur, this point M' naturally coincides with the point M, but it moves towards the right in the diagram with the velocity v of the train. If an observer sitting in the position M’ in the train did not possess this velocity, then he would remain permanently at M, and the light rays emitted by the flashes of lightning A and B would reach him simultaneously, i.e. they would meet just where he is situated. Now in reality (considered with reference to the railway embankment) he is hastening towards the beam of light coming from B, whilst he is riding on ahead of the beam of light coming from A. Hence the observer will see the beam of light emitted from B earlier than he will see that emitted from A."

All we have to do is note that M' is moving in a certain direction to see that the light from B will hit M' before the light from A hits M'. Do you disagree with this?
How many times do I have to tell you, this is a farce.

M' does not move when it is stationary. Why are you have so many problems with this.

I want you to tell me M' moves toward the light when you are stationary with M'. How does a stationary frame move toward light?

Explain that under SR.

[/quote]

13. Debating creationists on the topic of evolution is rather like trying to play chess with a pigeon -- it knocks the pieces over, craps on the board, and flies back to its flock to claim victory." - Scott D. Weitzenhoffer

14. Originally Posted by chinglu
Yea, well we have the math above that proves that is false. Even RocketMan agrees my math is correct.
We agree that your math is correct, but everyone but you remembers that your proof requires that you only consider a restricted area of the entire spacetime. Everyone but you realizes that there are regions in spacetime where t=t' even though this is not a general property.
How many times do I have to tell you, this is a farce.

M' does not move when it is stationary. Why are you have so many problems with this.
Does the midpoint of a train move when the train moves? Why don't you answer my questions?
I want you to tell me M' moves toward the light when you are stationary with M'. How does a stationary frame move toward light?
It doesn't. As everyone has told you already and as you refuse to read for yourself, in the frame where M' is stationary, event B happens earlier than it does in the frame where M is moving. You can work this out for yourself with the Lorentz transformations.

Again, I will direct you to some unanswered questions:

All we have to do is note that M' is moving in a certain direction to see that the light from B will hit M' before the light from A hits M'. Do you disagree with this?

If a car is moving towards another car, will the two cars hit each other sooner than if one of them was sitting still?

15. Originally Posted by DrRocket
Debating creationists on the topic of evolution is rather like trying to play chess with a pigeon -- it knocks the pieces over, craps on the board, and flies back to its flock to claim victory." - Scott D. Weitzenhoffer

Anyway, since you have agreed t'=t>0 and you know the math but have not yet put it together, I can take any light reception place t1=t1'>0 with t2>t1 and I have simultaneity between the frames with different x coordinates.

That kills ROS.

What is it like being a flat earther?

16. Originally Posted by PhysBang
We agree that your math is correct, but everyone but you remembers that your proof requires that you only consider a restricted area of the entire spacetime. Everyone but you realizes that there are regions in spacetime where t=t' even though this is not a general property.
Oh, I have this part figured out. You see, this is absolute simultaneity between the frames for any time t. That refutes ROS.

You see, ROS must always work for different x coordinates or ROS is false.

I do not have to prove ROS is always false, like you think, I need only to prove it is false on some case.

When dealing with a flat earth GOD like ROS and SR, they must always be Godlike and true.

Well, i proved SR is a flat earth theory because it is the case SR is false.

Does the midpoint of a train move when the train moves? Why don't you answer my questions?
No problem, the answer is no.

I want you to tell me M' moves toward the light when you are stationary with M'. How does a stationary frame move toward light?
It doesn't. As everyone has told you already and as you refuse to read for yourself, in the frame where M' is stationary, event B happens earlier than it does in the frame where M is moving. You can work this out for yourself with the Lorentz transformations.
Yea, but I am working LT from the M' frame. SR does not have a preferred frame. Have you done that yet?

M' claims the strikes are simultaneous. prove with math I am wrong.

Good luck!

Again, I will direct you to some unanswered questions:

All we have to do is note that M' is moving in a certain direction to see that the light from B will hit M' before the light from A hits M'. Do you disagree with this?
Wait, if you are at rest with M', how is M' moving? Sure, from M you are correct. But, M is not a preferred frame. So, be at rest with M' and prove this.

[/quote]

17. Originally Posted by chinglu
Originally Posted by kojax
Originally Posted by chinglu

Now, as far as A and B not being simultaneous, I have proven that is false.

Each frame sees the strikes as simultaneous and claims the other does not see them as simultaneous.

If you can prove any of the very public math I presented is incorrect, then have at it.

I certainly have nothing to hide.
I fail to see your reasoning on that.

If A and B are at a location where the train observer would perceive them to be simultaneous, then they are at a location where the Embankment observer would perceive A to have preceded B.

If A and B are at a location where the embankment observer would perceive them to be simultaneous, then they are at a location where the train observer would perceive B to have preceded A.

I'm pretty sure could position the points strategically so that both observers observed them both to be simultaneous (haven't actually tried it), but that would require moving them up or down vertically as well as side to side horizontally. They can't be colinear with the embankment. The line formed by the two points A and B would have to be diagonal instead of horizontal.

Let's have a little exercise.

assume the below
M'---------------------------------------A'

Now assume when A is a distance d when the light is flashed and A' is at rest with M'.

Since light measure c, then t = d/c.

Now say A' is moving toward M'. When A' is a distance d, light is emitted.

What is t?

t is still d/c.

So, just because A, the lightning strike is a moving light source, that does not mean M' is moving toward the light in the view of the primed frame.

The primed frame is at rest and the light source is moving. So, the distance is measured when the light is emitted and has nothing to do with the motion of the light source.

In other words, M' never moves toward the light in its own view.

Let's extend this. Suppose

M'-------------------------A
N'-------------------------A

Let's say A is in motion toward M' and N', M' is stationary, and N' is moving to the left. N' and M' are originally the same distance from A when A (I'm treating both A's as one point) emits a spherical light wave.

Even though A's motion has no effect on the time at which the light will reach M' or N', their respective motions relative to each other ensures that M' will still receive the signal at a different time than N'.

18. Originally Posted by chinglu
Does the midpoint of a train move when the train moves? Why don't you answer my questions?
No problem, the answer is no.
OK, now you're definitely crazy. No more discussion.

19. Originally Posted by PhysBang
Originally Posted by chinglu
Does the midpoint of a train move when the train moves? Why don't you answer my questions?
No problem, the answer is no.
OK, now you're definitely crazy. No more discussion.
As a bed bug.

For those who are not candidates for a padded room, it is quite easy to see the relativity of simultaneity from the Lorentz transformations. (Edited to ad the word "not").

So what is "simultaneus". In the unprimed reference frame simultaneous events fill out a hyperplane, all points of the form where is fixed and vary over the real numbers.

So appplying a Lorentz transformation to such a point you get

The first (time) coordinate is not constant with respect to the spatial coordinates and so is not a hyperplane of simultaneity in the primed reference frame. Thus two events seen as simultaneous in the unprimed frame may not be simultaneous in the primed frame.

20. Originally Posted by PhysBang
Originally Posted by chinglu
Does the midpoint of a train move when the train moves? Why don't you answer my questions?
No problem, the answer is no.
OK, now you're definitely crazy. No more discussion.
Oh I see. We were talking from the context of the train. In the train frame, the midpoint does not move, but the earth moves.

From the bank, the train and midpoint of the train moves.

But, to understand the train, you must operate in the train. It does not move toward the light.

21. Originally Posted by kojax

Let's extend this. Suppose

M'-------------------------A
N'-------------------------A

Let's say A is in motion toward M' and N', M' is stationary, and N' is moving to the left. N' and M' are originally the same distance from A when A (I'm treating both A's as one point) emits a spherical light wave.

Even though A's motion has no effect on the time at which the light will reach M' or N', their respective motions relative to each other ensures that M' will still receive the signal at a different time than N'.
You are thinking logically. But you are not thinking SR and the T/E experiment is not SR.

Under SR, M' is moving toward A in the view of N'.

N' is moving away from A in the view of M'.

And, yes, buth will calculate a different time for light reception. That has never been in dispute.

But, if length contraction did not exist, under SR, they would calculate the same time because A is the same distance from N' and M' when light emission occured.

Then under SR, you take that d and have t = d/c.

22. Originally Posted by DrRocket
Originally Posted by PhysBang
Originally Posted by chinglu
Does the midpoint of a train move when the train moves? Why don't you answer my questions?
No problem, the answer is no.
OK, now you're definitely crazy. No more discussion.
As a bed bug.

For those who are candidates for a padded room, it is quite easy to see the relativity of simultaneity from the Lorentz transformations.

So what is "simultaneus". In the unprimed reference frame simultaneous events fill out a hyperplane, all points of the form where is fixed and vary over the real numbers.

So appplying a Lorentz transformation to such a point you get

The first (time) coordinate is not constant with respect to the spatial coordinates and so is not a hyperplane of simultaneity in the primed reference frame. Thus two events seen as simultaneous in the unprimed frame may not be simultaneous in the primed frame.
This post is mathematically accurate.

This is not in dispute.

Now, try to apply this to T/E because even though the phrase relativity of simultaneity has been applied to T/E it is not applicable.

In addition, have you learned yet to take the derivative using the SLW? I see that you confine yourself to static single light beams.

As we all know, SLW is dynamic and must be examined with the derivative.

I wonder why Einstein failed to complete that simple task and apply calculus to SLW.

Well, I have and time must proceed backward for the view of one frame while it does not for the view of the other frame.

Under SR, time must proceed backward for the primed frame while it does not proceed backward for the primed frame. that was proven with dt'/dt < 0 in the calculations of the unprimed frame. Yet, in the view of the primed frame, there are no conditions in which dt' < 0.

I can't imagine believing in such flat earth concepts.

23. You are simply crazy.

24. Originally Posted by PhysBang
You are simply crazy.
No, you simply resort to name calling when you are afraid because your flat earth theory has failed.

Flat earthers cling to their guns and Flat Earth bibles when valid ideas are mathematically presented that are different from their own failed ones.

Anyway, the math presented here is valid and final.

25. Sure. Everyone has made a simple error of overlooking the math of SR for 106 years. It's a good thing there are no applications of SR or GR anywhere.

26. Originally Posted by chinglu
This post is mathematically accurate.
yep

Originally Posted by chinglu
This is not in dispute.

Now, try to apply this to T/E because even though the phrase relativity of simultaneity has been applied to T/E it is not applicable.

In addition, have you learned yet to take the derivative using the SLW? I see that you confine yourself to static single light beams.
I applied the Lorentz transformation to spacetime. This has nothing to do with light beams.

What in the hell is a "static single light beam"? Under any reasonable interpretation that is an oxymoron.

27. Chinglu, it would help if you weren't so arrogant in your misunderstanding.

The bottom line here is that you are wrong. Your math is based on a misconception about simultaneity, which is what we all have been saying all along. We cannot refute your math, as there is nothing actually wrong with the math itself, it is the foundations you are building the math on that are misconceived - you are using terms in a nonsensical manner. If you were to submit your ideas to any mainstream journal for peer review, you would be rejected on that basis.

There is no point continually shouting "disprove the math!" if the math is built on misconceptions. What you have done is built a valid mathematical argument for an invalid situation. It is the situation as you are describing it that is messed up, not the math you use to describe it.

It is all laid out for you in Einsteins chapter on the train and embankment, so why don't you work through the actual problem, rather than putting in your own interpretation?

28. Originally Posted by chinglu
Originally Posted by kojax

Let's extend this. Suppose

M'-------------------------A
N'-------------------------A

Let's say A is in motion toward M' and N', M' is stationary, and N' is moving to the left. N' and M' are originally the same distance from A when A (I'm treating both A's as one point) emits a spherical light wave.

Even though A's motion has no effect on the time at which the light will reach M' or N', their respective motions relative to each other ensures that M' will still receive the signal at a different time than N'.
You are thinking logically. But you are not thinking SR and the T/E experiment is not SR.

Under SR, M' is moving toward A in the view of N'.

N' is moving away from A in the view of M'.

And, yes, buth will calculate a different time for light reception. That has never been in dispute.

But, if length contraction did not exist, under SR, they would calculate the same time because A is the same distance from N' and M' when light emission occured.

Then under SR, you take that d and have t = d/c.
It doesn't matter in SR when emission occurred, only when detection occurred.

29. Originally Posted by PhysBang
Sure. Everyone has made a simple error of overlooking the math of SR for 106 years. It's a good thing there are no applications of SR or GR anywhere.
This is an invalid argument when math has been clearly presented and can be inspected.

30. Originally Posted by DrRocket
Originally Posted by chinglu
This post is mathematically accurate.
yep

Originally Posted by chinglu
This is not in dispute.

Now, try to apply this to T/E because even though the phrase relativity of simultaneity has been applied to T/E it is not applicable.

In addition, have you learned yet to take the derivative using the SLW? I see that you confine yourself to static single light beams.
I applied the Lorentz transformation to spacetime. This has nothing to do with light beams.

What in the hell is a "static single light beam"? Under any reasonable interpretation that is an oxymoron.

What I am saying is you have not proven your space over continuity.

Have you ever considered that? No, you use the typical vector space translation without consideration for the fact that this space describes the motion of the spherical light wave (SLW).

Therefore, the vector space translation must be also consistent when analyzed by the derivative since the motion of the SLW is dynamic.

Since the space describes a continuous operation, then it must also be consistent under the derivative.

Now, you are left to argue the SLW does not move and hence the derivative is not applicable, but that is flat earth thinking.

But, as I showed with the proof, ∂t'/∂t < 0 for the SLW which is inconsistent with reality.

So, you have yet to address this, but you can't.

31. Originally Posted by SpeedFreek
Chinglu, it would help if you weren't so arrogant in your misunderstanding.

The bottom line here is that you are wrong. Your math is based on a misconception about simultaneity, which is what we all have been saying all along. We cannot refute your math, as there is nothing actually wrong with the math itself, it is the foundations you are building the math on that are misconceived - you are using terms in a nonsensical manner. If you were to submit your ideas to any mainstream journal for peer review, you would be rejected on that basis.

There is no point continually shouting "disprove the math!" if the math is built on misconceptions. What you have done is built a valid mathematical argument for an invalid situation. It is the situation as you are describing it that is messed up, not the math you use to describe it.

It is all laid out for you in Einsteins chapter on the train and embankment, so why don't you work through the actual problem, rather than putting in your own interpretation?
All you do is make accusations.

Take my math and prove the assumptions are false.

Here is your problem, I am well aware of your kind. I design the math on purpose such that if you try to refute it, you refute SR.

Now, point out the exact assumption I made that is invalid and I will show you exactly what I mean.

For example, I already routed all of you against trying to claim the light emission at A is not and agreed upon event. Some of you went this way, but that then invalidates Einstein's mirror experiment.

So, I preprogrammed my argument to handle that assertion along the mirror experiment.

Therefore, we are left to conclude

t = r/c

and

t' = r/(rγ )

Then, the same exact argument applies to B and so,

t = r/c

and

t' = r/(rγ )

Therefore, both frames will see the strikes as simultaneous and SR is therefore false.

All you do then is run to Einstein and beg him to help you. Do you know the answer you get from Einstein and the mainstream? The train agrees the train observer moves toward the light which is not SR. You don't get any math under SR to validate that assertion.

Now, if you and the mainstream are correct, show me your math. You have all your buddies out here to help.

I have presented clear math under the rules of SR without opinion but based on the postulates of SR.

I am sorry it makes SR false, but, that is the way it goes.

32. Originally Posted by chinglu
Originally Posted by DrRocket
Originally Posted by chinglu
This post is mathematically accurate.
yep

Originally Posted by chinglu
This is not in dispute.

Now, try to apply this to T/E because even though the phrase relativity of simultaneity has been applied to T/E it is not applicable.

In addition, have you learned yet to take the derivative using the SLW? I see that you confine yourself to static single light beams.
I applied the Lorentz transformation to spacetime. This has nothing to do with light beams.

What in the hell is a "static single light beam"? Under any reasonable interpretation that is an oxymoron.

What I am saying is you have not proven your space over continuity.

Have you ever considered that? No, you use the typical vector space translation without consideration for the fact that this space describes the motion of the spherical light wave (SLW).

Therefore, the vector space translation must be also consistent when analyzed by the derivative since the motion of the SLW is dynamic.

Since the space describes a continuous operation, then it must also be consistent under the derivative.

Now, you are left to argue the SLW does not move and hence the derivative is not applicable, but that is flat earth thinking.

But, as I showed with the proof, ∂t'/∂t < 0 for the SLW which is inconsistent with reality.

So, you have yet to address this, but you can't.
My puppy is looking better and better.

33. Originally Posted by kojax
Originally Posted by chinglu
Originally Posted by kojax

Let's extend this. Suppose

M'-------------------------A
N'-------------------------A

Let's say A is in motion toward M' and N', M' is stationary, and N' is moving to the left. N' and M' are originally the same distance from A when A (I'm treating both A's as one point) emits a spherical light wave.

Even though A's motion has no effect on the time at which the light will reach M' or N', their respective motions relative to each other ensures that M' will still receive the signal at a different time than N'.
You are thinking logically. But you are not thinking SR and the T/E experiment is not SR.

Under SR, M' is moving toward A in the view of N'.

N' is moving away from A in the view of M'.

And, yes, buth will calculate a different time for light reception. That has never been in dispute.

But, if length contraction did not exist, under SR, they would calculate the same time because A is the same distance from N' and M' when light emission occured.

Then under SR, you take that d and have t = d/c.
It doesn't matter in SR when emission occurred, only when detection occurred.
I mostly agree with your conclusion, except, light emission is an agreed upon event for all frames.

Now, based on that fact, try calculating t and t' for the T/E experiment and light emission at A.

Any scientific mind would want to do that. They want proof not some dogma.

34. Originally Posted by DrRocket
Originally Posted by chinglu
Originally Posted by DrRocket
Originally Posted by chinglu
This post is mathematically accurate.
yep

Originally Posted by chinglu
This is not in dispute.

Now, try to apply this to T/E because even though the phrase relativity of simultaneity has been applied to T/E it is not applicable.

In addition, have you learned yet to take the derivative using the SLW? I see that you confine yourself to static single light beams.
I applied the Lorentz transformation to spacetime. This has nothing to do with light beams.

What in the hell is a "static single light beam"? Under any reasonable interpretation that is an oxymoron.

What I am saying is you have not proven your space over continuity.

Have you ever considered that? No, you use the typical vector space translation without consideration for the fact that this space describes the motion of the spherical light wave (SLW).

Therefore, the vector space translation must be also consistent when analyzed by the derivative since the motion of the SLW is dynamic.

Since the space describes a continuous operation, then it must also be consistent under the derivative.

Now, you are left to argue the SLW does not move and hence the derivative is not applicable, but that is flat earth thinking.

But, as I showed with the proof, ∂t'/∂t < 0 for the SLW which is inconsistent with reality.

So, you have yet to address this, but you can't.
My puppy is looking better and better.
Do you get excited when you are outmatched?

Are you going to agree SR is based on dynamics and hence the vector space must be consistent under the derivative?

35. If we look at section 1 of Einstein's paper, "Definition of Simultaneity", we find given the entire paper it is impossible to establish simultaneity between frames.

http://www.fourmilab.ch/etexts/einstein/specrel/www/

Yet, with my proof of setting x = vtγ /(1+γ) , we are able to establish the concept of simultaneity between frames meaning there exists absolute simultaneity. More specifically, given the line I published, all frame to frame observers along that line measure the same time since light emission. therefore, there exists a notion of absolute time.

In addition, I proved ∂t'/∂t < 0 which means in order for SR to be true, time must proceed backward in the primed frame from the view of the unprimed frame where time does not ever proceed backward in the primed frame or the spherical light wave would collapse back to the light emission point.

36. Originally Posted by chinglu
Now, based on that fact, try calculating t and t' for the T/E experiment and light emission at A.

Any scientific mind would want to do that. They want proof not some dogma.
This just shows that you are crazy based on your own standards. You haven't been able to do thatcalculation!

37. Originally Posted by PhysBang
Originally Posted by chinglu
Now, based on that fact, try calculating t and t' for the T/E experiment and light emission at A.

Any scientific mind would want to do that. They want proof not some dogma.
This just shows that you are crazy based on your own standards. You haven't been able to do thatcalculation!

I have done this over and over and over.

Let r = the distance from M to A.

t = r/c

t'/ = r/(γc)

Same holds for B.

38. Originally Posted by PhysBang
Originally Posted by chinglu
Now, based on that fact, try calculating t and t' for the T/E experiment and light emission at A.

Any scientific mind would want to do that. They want proof not some dogma.
This just shows that you are crazy based on your own standards. You haven't been able to do thatcalculation!
Standards ? chinglu ?

chinglu is quite sane by his standards. That's how we knw that he is crazy.

39. Originally Posted by DrRocket
Originally Posted by PhysBang
Originally Posted by chinglu
Now, based on that fact, try calculating t and t' for the T/E experiment and light emission at A.

Any scientific mind would want to do that. They want proof not some dogma.
This just shows that you are crazy based on your own standards. You haven't been able to do thatcalculation!
Standards ? chinglu ?

chinglu is quite sane by his standards. That's how we knw that he is crazy.
Hey, I think folks would be quite amused if you and I tangled mathematically.

You see, they will only be convinced by your flat earth arguments if you prove your case with math.

That means, you need to use math to prove my derivative is wrong.

Otherwise, SR is proven false.

You have 2 ways I can think of, Noway and Nohow.

So, care to tangle?

40. I wanted to make sure RocketMan remembered the derivative and proof that refutes SR.

http://freepdfhosting.com/23a2424bf6.pdf

41. Originally Posted by chinglu

I mostly agree with your conclusion, except, light emission is an agreed upon event for all frames.

Now, based on that fact, try calculating t and t' for the T/E experiment and light emission at A.

Any scientific mind would want to do that. They want proof not some dogma.
No. The time of light emission is only agreed upon by observers in the frame from which it was emitted. Everyone outside that frame can only assume it was emitted distance/C seconds ago (which means observers in different frames are going to disagree).

Originally Posted by chinglu
Originally Posted by PhysBang
Originally Posted by chinglu
Now, based on that fact, try calculating t and t' for the T/E experiment and light emission at A.

Any scientific mind would want to do that. They want proof not some dogma.
This just shows that you are crazy based on your own standards. You haven't been able to do thatcalculation!

I have done this over and over and over.

Let r = the distance from M to A.

t = r/c

t'/ = r/(γc)

Same holds for B.
Why is there no r'? The distance changed in the interval between emission and reception of the light.

42. Originally Posted by kojax
Originally Posted by chinglu

I mostly agree with your conclusion, except, light emission is an agreed upon event for all frames.

Now, based on that fact, try calculating t and t' for the T/E experiment and light emission at A.

Any scientific mind would want to do that. They want proof not some dogma.
No. The time of light emission is only agreed upon by observers in the frame from which it was emitted. Everyone outside that frame can only assume it was emitted distance/C seconds ago (which means observers in different frames are going to disagree).
No sorry, SR does not work that way. Both frames call the light emission a start event and set their clocks to t'=t'=0. It is an agreed upoing event.

Originally Posted by kojax
Originally Posted by chinglu
Originally Posted by PhysBang
Originally Posted by chinglu
Now, based on that fact, try calculating t and t' for the T/E experiment and light emission at A.

Any scientific mind would want to do that. They want proof not some dogma.
This just shows that you are crazy based on your own standards. You haven't been able to do thatcalculation!

I have done this over and over and over.

Let r = the distance from M to A.

t = r/c

t'/ = r/(γc)

Same holds for B.
Why is there no r'? The distance changed in the interval between emission and reception of the light.
There is an r' it is above, r' = r/γ

43. Originally Posted by chinglu
No sorry, SR does not work that way. Both frames call the light emission a start event and set their clocks to t'=t'=0. It is an agreed upoing event.
That's crazy.

44. Originally Posted by PhysBang
Originally Posted by chinglu
No sorry, SR does not work that way. Both frames call the light emission a start event and set their clocks to t'=t'=0. It is an agreed upoing event.
That's crazy.

We now have to prove that any ray of light, measured in the moving system, is propagated with the velocity c, if, as we have assumed, this is the case in the stationary system; for we have not as yet furnished the proof that the principle of the constancy of the velocity of light is compatible with the principle of relativity.
At the time t = τ = 0 when the origin of the co-ordinates is common to the two systems, let a spherical wave be emitted therefrom, and be propagated with the velocity c in system K.

http://www.fourmilab.ch/etexts/einstein/specrel/www/

45. Someone has to be pretty stupid or pretty crazy to take one example and then assume that all examples have to have exactly the same variables with exactly the same values assigned to them.

46. Originally Posted by chinglu
Originally Posted by kojax
Originally Posted by chinglu

I mostly agree with your conclusion, except, light emission is an agreed upon event for all frames.

Now, based on that fact, try calculating t and t' for the T/E experiment and light emission at A.

Any scientific mind would want to do that. They want proof not some dogma.
No. The time of light emission is only agreed upon by observers in the frame from which it was emitted. Everyone outside that frame can only assume it was emitted distance/C seconds ago (which means observers in different frames are going to disagree).
No sorry, SR does not work that way. Both frames call the light emission a start event and set their clocks to t'=t'=0. It is an agreed upoing event.
If that were true, Chinglu, then observers in different frames would have to disagree as to how fast C is.

Every observer must believe that the beams of light they see arriving right now were emitted distance/C seconds ago. If one observer is subject to length contraction (due to his/her/its own motion relative to the light source), and another isn't, then the first observer must therefore believe the beam of light was emitted a shorter time ago than the second observer does.

47. Originally Posted by kojax
If that were true, Chinglu, then observers in different frames would have to disagree as to how fast C is.
Come on, it's crazier than that!

If what chinglu says is true, then there was only ever one time in the entire history of the universe when light was emitted. There can be no other time, since we would have to have two different light emission events at two different time and this is impossible.

You see how crazy this guy is?

48. Originally Posted by PhysBang
Someone has to be pretty stupid or pretty crazy to take one example and then assume that all examples have to have exactly the same variables with exactly the same values assigned to them.
The point that I made that all you can do is agree is whenever there is a light emission, the time in the respective frames is set to 0. That is the logic of Einstein and I showed you his reasoning.

Now, given a light emission at A, we have t'=t=0.

The next issue to decide is how far is A from M and M'.

For M, we assumed, so that is r. For M', based on length contraction, the diatance is r/γ.

Then, based on light being measured c in each frame, we have

t = r/c for the light to reach M.

t' = r/(γc) for the light to reach M'.

The same reasoing holds for B.

Hence, t'A = t'B and tA = tB.

I can't seem to get anyone to offer any other solution and so anyone that believes ROS does that without a basis in math and logic.

From the above factual math, ROS is false.

49. Originally Posted by kojax
Originally Posted by chinglu
Originally Posted by kojax
Originally Posted by chinglu

I mostly agree with your conclusion, except, light emission is an agreed upon event for all frames.

Now, based on that fact, try calculating t and t' for the T/E experiment and light emission at A.

Any scientific mind would want to do that. They want proof not some dogma.
No. The time of light emission is only agreed upon by observers in the frame from which it was emitted. Everyone outside that frame can only assume it was emitted distance/C seconds ago (which means observers in different frames are going to disagree).
No sorry, SR does not work that way. Both frames call the light emission a start event and set their clocks to t'=t'=0. It is an agreed upoing event.
If that were true, Chinglu, then observers in different frames would have to disagree as to how fast C is.

Every observer must believe that the beams of light they see arriving right now were emitted distance/C seconds ago. If one observer is subject to length contraction (due to his/her/its own motion relative to the light source), and another isn't, then the first observer must therefore believe the beam of light was emitted a shorter time ago than the second observer does.

Your reasoning is wrong. I showed

t = r/c and

t' = r'/c.

Both frames measure c. That is SR.

Since r is measured in the frame of M, then M' would see this distance as length contracted based on SR, hence, r' = r/γ.

This is all standard SR.

But, what is not standard is the fact, ushing math and SR, M' will see the strikes as simultaneous.

50. Originally Posted by PhysBang
Originally Posted by kojax
If that were true, Chinglu, then observers in different frames would have to disagree as to how fast C is.
Come on, it's crazier than that!

If what chinglu says is true, then there was only ever one time in the entire history of the universe when light was emitted. There can be no other time, since we would have to have two different light emission events at two different time and this is impossible.

You see how crazy this guy is?
I am surprised you understand how to reach this place.

I never said anything like this. I simply quoted Einstein such that for any light emission the frames agree t'=t=0 at the light emission point.

This is not actually true and you clearly do not understand this is a convention. All the clocks in a frame cannot be set to 0. In reality, the light emission occurs when
t=t0
and t'=t0'.

Then, instead of using t/t' in the LT equations, you would use

(t1-t0) for t
and

(t1'-t0') for t'

That is cumberson and not needed, so for convention, they are set such t=t'=0.

Read Einstein's paper and you will see I am correct.

51. Originally Posted by chinglu
I never said anything like this. I simply quoted Einstein such that for any light emission the frames agree t'=t=0 at the light emission point.
Yeah, you quoted a section where he begins one mathematical proof with an assumption that one can make without any loss of generality. He could get the same results in the proof if he used t=1 and t'=17.5.

You could equally well quote a mathematics text that says that "x=4" to claim that x is always equal to four.

Your approach is crazy, plain and simple. You cannot see this because you are mentally ill.
This is not actually true and you clearly do not understand this is a convention. All the clocks in a frame cannot be set to 0. In reality, the light emission occurs when
t=t0
and t'=t0'.

Then, instead of using t/t' in the LT equations, you would use

(t1-t0) for t
and

(t1'-t0') for t'

That is cumberson and not needed, so for convention, they are set such t=t'=0.

Read Einstein's paper and you will see I am correct.
Now you seem to be changing your tune. Perhaps you are a little less mentally ill or you have occasional bouts of clarity.

We do not forget that you wrote, "No sorry, SR does not work that way. Both frames call the light emission a start event and set their clocks to t'=t'=0. It is an agreed upoing event. " This is crazy talk.

That you might be temporarily getting better does not help. As we have seen in this thread, you can't actually produce a mathematical description of the train experiment in Chapter 9. If you could have done so, you would have.

52. Originally Posted by chinglu
Originally Posted by PhysBang
Someone has to be pretty stupid or pretty crazy to take one example and then assume that all examples have to have exactly the same variables with exactly the same values assigned to them.
The point that I made that all you can do is agree is whenever there is a light emission, the time in the respective frames is set to 0. That is the logic of Einstein and I showed you his reasoning.

Now, given a light emission at A, we have t'=t=0.

The next issue to decide is how far is A from M and M'.

For M, we assumed, so that is r. For M', based on length contraction, the diatance is r/γ.

Then, based on light being measured c in each frame, we have

t = r/c for the light to reach M.

t' = r/(γc) for the light to reach M'.

The same reasoing holds for B.

Hence, t'A = t'B and tA = tB.

I can't seem to get anyone to offer any other solution and so anyone that believes ROS does that without a basis in math and logic.

From the above factual math, ROS is false.
Now I see why so many other posters are giving up on this thread.

Suppose as in the picture you gave us that:

A -------- M' ----------B
A---------M------------B

Suppose M' is moving to the right, and M is also moving to the right but at a slower speed.

If M receives the light signal from A and B simultaneously, then it's impossible for M' to also receive the light signal from A and B simultaneously. Are we in agreement so far?

53. Originally Posted by PhysBang
Originally Posted by chinglu
I never said anything like this. I simply quoted Einstein such that for any light emission the frames agree t'=t=0 at the light emission point.
Yeah, you quoted a section where he begins one mathematical proof with an assumption that one can make without any loss of generality. He could get the same results in the proof if he used t=1 and t'=17.5.

You could equally well quote a mathematics text that says that "x=4" to claim that x is always equal to four.

Your approach is crazy, plain and simple. You cannot see this because you are mentally ill.
I have already explained all this in another post.

Either way, this has no impact on the math I presented. I set t'=t=0 just like Einstein, the start could be anything.

And, the reason you call out names is bacuse you are forced to agree with the math since it is correct.

That causes you frustration and so you call out names. It's sad really that you cannot stay on task with the math.

This is not actually true and you clearly do not understand this is a convention. All the clocks in a frame cannot be set to 0. In reality, the light emission occurs when
t=t0
and t'=t0'.

Then, instead of using t/t' in the LT equations, you would use

(t1-t0) for t
and

(t1'-t0') for t'

That is cumberson and not needed, so for convention, they are set such t=t'=0.

Read Einstein's paper and you will see I am correct.
Now you seem to be changing your tune. Perhaps you are a little less mentally ill or you have occasional bouts of clarity.

We do not forget that you wrote, "No sorry, SR does not work that way. Both frames call the light emission a start event and set their clocks to t'=t'=0. It is an agreed upoing event. " This is crazy talk.

That you might be temporarily getting better does not help. As we have seen in this thread, you can't actually produce a mathematical description of the train experiment in Chapter 9. If you could have done so, you would have.
I'm not changing my tune. I simply did as Einstein did for ease of use.

Here, let me show you again since you are so confused.

At the time t = τ = 0 when the origin of the co-ordinates is common to the two systems, let a spherical wave be emitted therefrom,

http://www.fourmilab.ch/etexts/einstein/specrel/www/

Now do you understand I used the same convention as Einstein? This seems to have gotten you very confused.

54. Originally Posted by kojax
Originally Posted by chinglu
Originally Posted by PhysBang
Someone has to be pretty stupid or pretty crazy to take one example and then assume that all examples have to have exactly the same variables with exactly the same values assigned to them.
The point that I made that all you can do is agree is whenever there is a light emission, the time in the respective frames is set to 0. That is the logic of Einstein and I showed you his reasoning.

Now, given a light emission at A, we have t'=t=0.

The next issue to decide is how far is A from M and M'.

For M, we assumed, so that is r. For M', based on length contraction, the diatance is r/γ.

Then, based on light being measured c in each frame, we have

t = r/c for the light to reach M.

t' = r/(γc) for the light to reach M'.

The same reasoing holds for B.

Hence, t'A = t'B and tA = tB.

I can't seem to get anyone to offer any other solution and so anyone that believes ROS does that without a basis in math and logic.

From the above factual math, ROS is false.
Now I see why so many other posters are giving up on this thread.

Suppose as in the picture you gave us that:

A -------- M' ----------B
A---------M------------B

Suppose M' is moving to the right, and M is also moving to the right but at a slower speed.

If M receives the light signal from A and B simultaneously, then it's impossible for M' to also receive the light signal from A and B simultaneously. Are we in agreement so far?
They have given up on the thread because they cannot stop the math. This is not an opinion section.

You are struggling with this.

Instead of repeating books, let's see exactly your timing for light to travel from A to M' and the time for light to travel from B to M'.

Folks give up on the thread because they struggle with this simple calculation. Instead, they cling to the T/E scripture without having any understanding how to prove if the conclusions are valid.

Let anyone come here with the math and prove the conclusions of the T/E experiment.

55. Nobody has problems with the math. Indeed, everyone follows and agrees with each step of your proofs. They have problems with your insanity.

You constantly take mathematical claims out of context. First you claim that events where light is emitted have to be assigned t=0. Then, when even you realize how crazy this is, you are forced to change your tune. However, you leave all the crazy in place. You keep saying that all events where light is emitted must take place at the same time. That's crazy.

Another example is that you said that t always has to equal t' because you found some places where t=t'. That's crazy.

You never answer direct questions, since part of you knows that you don't have a good answer and you are shielding yourself from that. That's crazy.

56. Originally Posted by PhysBang
Nobody has problems with the math. Indeed, everyone follows and agrees with each step of your proofs. They have problems with your insanity.

You constantly take mathematical claims out of context. First you claim that events where light is emitted have to be assigned t=0. Then, when even you realize how crazy this is, you are forced to change your tune. However, you leave all the crazy in place. You keep saying that all events where light is emitted must take place at the same time. That's crazy.

Another example is that you said that t always has to equal t' because you found some places where t=t'. That's crazy.

You never answer direct questions, since part of you knows that you don't have a good answer and you are shielding yourself from that. That's crazy.
As I have demonstrated to you, Einstein set the convention of t'=t=0. You will need to take that up with your master. You lose on this point.

Another example is that you said that t always has to equal t' because you found some places where t=t'. That's crazy.
I said clearly along this line, t'=t. I did not say under SR, it is the case that t'=t. So, you are just confused.

Finally, I answer all questions and so you are a liar.

In fact, I presented clear math out here.

Take you choice on the math I presented, submit to it or refute it.

You have chosen this in-between world of crying in frustration.

Get your emotions under control and let's deal with the math.

I notice RocketMan loves to correct people on their math. Yet, he has run from this thread knowing for a fact he cannot refute any of the math.

57. Originally Posted by chinglu
Originally Posted by kojax
Originally Posted by chinglu
Originally Posted by PhysBang
Someone has to be pretty stupid or pretty crazy to take one example and then assume that all examples have to have exactly the same variables with exactly the same values assigned to them.
The point that I made that all you can do is agree is whenever there is a light emission, the time in the respective frames is set to 0. That is the logic of Einstein and I showed you his reasoning.

Now, given a light emission at A, we have t'=t=0.

The next issue to decide is how far is A from M and M'.

For M, we assumed, so that is r. For M', based on length contraction, the diatance is r/γ.

Then, based on light being measured c in each frame, we have

t = r/c for the light to reach M.

t' = r/(γc) for the light to reach M'.

The same reasoing holds for B.

Hence, t'A = t'B and tA = tB.

I can't seem to get anyone to offer any other solution and so anyone that believes ROS does that without a basis in math and logic.

From the above factual math, ROS is false.
Now I see why so many other posters are giving up on this thread.

Suppose as in the picture you gave us that:

A -------- M' ----------B
A---------M------------B

Suppose M' is moving to the right, and M is also moving to the right but at a slower speed.

If M receives the light signal from A and B simultaneously, then it's impossible for M' to also receive the light signal from A and B simultaneously. Are we in agreement so far?
They have given up on the thread because they cannot stop the math. This is not an opinion section.

You are struggling with this.

Instead of repeating books, let's see exactly your timing for light to travel from A to M' and the time for light to travel from B to M'.

Folks give up on the thread because they struggle with this simple calculation. Instead, they cling to the T/E scripture without having any understanding how to prove if the conclusions are valid.

Let anyone come here with the math and prove the conclusions of the T/E experiment.
All well and good, but do we agree, that the time of receiving the light signals cannot be simultaneous for M', if it is simultaneous for M

58. Originally Posted by chinglu
Originally Posted by PhysBang
Nobody has problems with the math. Indeed, everyone follows and agrees with each step of your proofs. They have problems with your insanity.

You constantly take mathematical claims out of context. First you claim that events where light is emitted have to be assigned t=0. Then, when even you realize how crazy this is, you are forced to change your tune. However, you leave all the crazy in place. You keep saying that all events where light is emitted must take place at the same time. That's crazy.

Another example is that you said that t always has to equal t' because you found some places where t=t'. That's crazy.

You never answer direct questions, since part of you knows that you don't have a good answer and you are shielding yourself from that. That's crazy.
As I have demonstrated to you, Einstein set the convention of t'=t=0. You will need to take that up with your master. You lose on this point.

Another example is that you said that t always has to equal t' because you found some places where t=t'. That's crazy.
I said clearly along this line, t'=t. I did not say under SR, it is always the case that t'=t. So, you are just confused.

Finally, I answer all questions and so you are a liar.

In fact, I presented clear math out here.

Take you choice on the math I presented, submit to it or refute it.

You have chosen this in-between world of crying in frustration.

Get your emotions under control and let's deal with the math.

I notice RocketMan loves to correct people on their math. Yet, he has run from this thread knowing for a fact he cannot refute any of the math.

59. Originally Posted by chinglu
Finally, I answer all questions and so you are a liar.
We can all see you ducking kojax' question right now. We're not crazy. We can see that you keep ducking questions. Perhaps you can't see it, because you're crazy, but we all can.

60. Originally Posted by chinglu
Originally Posted by chinglu
Originally Posted by PhysBang
Nobody has problems with the math. Indeed, everyone follows and agrees with each step of your proofs. They have problems with your insanity.

You constantly take mathematical claims out of context. First you claim that events where light is emitted have to be assigned t=0. Then, when even you realize how crazy this is, you are forced to change your tune. However, you leave all the crazy in place. You keep saying that all events where light is emitted must take place at the same time. That's crazy.

Another example is that you said that t always has to equal t' because you found some places where t=t'. That's crazy.

You never answer direct questions, since part of you knows that you don't have a good answer and you are shielding yourself from that. That's crazy.
As I have demonstrated to you, Einstein set the convention of t'=t=0. You will need to take that up with your master. You lose on this point.

Another example is that you said that t always has to equal t' because you found some places where t=t'. That's crazy.
I said clearly along this line, t'=t. I did not say under SR, it is always the case that t'=t. So, you are just confused.

Finally, I answer all questions and so you are a liar.

In fact, I presented clear math out here.

Take you choice on the math I presented, submit to it or refute it.

You have chosen this in-between world of crying in frustration.

Get your emotions under control and let's deal with the math.

I notice RocketMan loves to correct people on their math. Yet, he has run from this thread knowing for a fact he cannot refute any of the math.
No, we cannot agree.

I want to see your t'A and t'B.

Always back up your assertions with math.

Wait, come to think about it, Einstein provided lots of math in his paper to back up his assertions.

Yet, he did not prove with math the strikes on the train will not be simultaneous.

Why is that?

Do you have the math?

It seems like a simple task and yet all the flat earthers in physics do not have a proof that t'A > t'B in the view of the M' frame.

Do you have said proof?

61. Originally Posted by PhysBang
Originally Posted by chinglu
Finally, I answer all questions and so you are a liar.
We can all see you ducking kojax' question right now. We're not crazy. We can see that you keep ducking questions. Perhaps you can't see it, because you're crazy, but we all can.
What?

I proved with math t'A = t'B.

What am I ducking?
I have done my job of refuting SR.

Otherwise prove t'A ≠ t'B in the context and view of the M' frame.

I want to see that.

62. Sadly, I cannot find a good reference with numbers for the train experiment. Since this is a good example for use in class, I've done some preliminary work.

Let's assign coordinates to Chapter 9. http://www.bartleby.com/173/9.html

We'll talk about an embankment frame with events A and B and we'll just use the time coordinate and the x coordinate (t,x)

The coordinates for A and B are (0,0) and (0,10). The midpoint in space between the spacial locations of these events, in this frame, will thus be x=5.

If we set c=1, then light from the events reaches this midpoint above at (5,5).

Let's set the velocity of the train as v=0.5. This means that the train is moving in a direction towards B and away from A, so to speak.

We can identify a point on the train that is next to the midpoint above at t=0. We will call this the midpoint of the train (relative to the flashes), M'. M' moves relative to the embankment frame, so we know that it will take some time for the light from A to catch up. At t=5, M' will be at (5, 7.5).

The motion of light is described by x=t; the motion of the midpoint M' is described by x=5+0.5t. We solve for t, taking t=5+0.5t . This means that these lines intersect at

t-0.5t=5
0.5t=5
t=10

Now we can ask what happens with the light from B. The motion of this light is x=10-t. Again we solve for t, taking 10-t=5+0.5t. This means that these lines intersect at

0.5t+t=10-5
1.5t=5
t=3+1/3

So we know that, according to the standard picture of movement in the frame of the embankment, there are two events where light from these flashes hits the midpoint of the train. This is a fact of this example. That is, given the starting assumptions (i.e., that the flashes were simultaneous in the embankment frame) we deduce that there are two events that correspond with the light reaching M'.

According to the frame of the train, the midpoint M' is stationary. We should be reminded of the definition of simultaneity used by Einstein from the previous chapter:

"After thinking the matter over for some time you then offer the following suggestion with which to test simultaneity. By measuring along the rails, the connecting line AB should be measured up and an observer placed at the mid-point M of the distance AB. This observer should be supplied with an arrangement (e.g. two mirrors inclined at 90°) which allows him visually to observe both places A and B at the same time. If the observer perceives the two flashes of lightning at the same time, then they are simultaneous."

Assuming that this train is set up like the one in Chapter 9, then by this definition the lightning flashes were not simultaneous, since the light from these flashes did not get to the midpoint at the same time.

63. Originally Posted by PhysBang
Sadly, I cannot find a good reference with numbers for the train experiment. Since this is a good example for use in class, I've done some preliminary work.

Let's assign coordinates to Chapter 9. http://www.bartleby.com/173/9.html

We'll talk about an embankment frame with events A and B and we'll just use the time coordinate and the x coordinate (t,x)

The coordinates for A and B are (0,0) and (0,10). The midpoint in space between the spacial locations of these events, in this frame, will thus be x=5.

If we set c=1, then light from the events reaches this midpoint above at (5,5).

Let's set the velocity of the train as v=0.5. This means that the train is moving in a direction towards B and away from A, so to speak.

We can identify a point on the train that is next to the midpoint above at t=0. We will call this the midpoint of the train (relative to the flashes), M'. M' moves relative to the embankment frame, so we know that it will take some time for the light from A to catch up. At t=5, M' will be at (5, 7.5).

The motion of light is described by x=t; the motion of the midpoint M' is described by x=5+0.5t. We solve for t, taking t=5+0.5t . This means that these lines intersect at

t-0.5t=5
0.5t=5
t=10

Now we can ask what happens with the light from B. The motion of this light is x=10-t. Again we solve for t, taking 10-t=5+0.5t. This means that these lines intersect at

0.5t+t=10-5
1.5t=5
t=3+1/3

So we know that, according to the standard picture of movement in the frame of the embankment, there are two events where light from these flashes hits the midpoint of the train. This is a fact of this example. That is, given the starting assumptions (i.e., that the flashes were simultaneous in the embankment frame) we deduce that there are two events that correspond with the light reaching M'.

According to the frame of the train, the midpoint M' is stationary. We should be reminded of the definition of simultaneity used by Einstein from the previous chapter:

"After thinking the matter over for some time you then offer the following suggestion with which to test simultaneity. By measuring along the rails, the connecting line AB should be measured up and an observer placed at the mid-point M of the distance AB. This observer should be supplied with an arrangement (e.g. two mirrors inclined at 90°) which allows him visually to observe both places A and B at the same time. If the observer perceives the two flashes of lightning at the same time, then they are simultaneous."

Assuming that this train is set up like the one in Chapter 9, then by this definition the lightning flashes were not simultaneous, since the light from these flashes did not get to the midpoint at the same time.

We can identify a point on the train that is next to the midpoint above at t=0. We will call this the midpoint of the train (relative to the flashes), M'. M' moves relative to the embankment frame, so we know that it will take some time for the light from A to catch up. At t=5, M' will be at (5, 7.5).

This is the basis on the error of your thinking.

Let's look at the context of the M' frame.

First, M' is not moving, the earth is and so is M.

Now, when A strikes, we apply the light postulate.

The motion of the light source does not impact the timing of the light to strike M or M'.

So, you said, M' moves relative to the embankment frame, so we know that it will take some time for the light from A to catch up.

Under the light postulate, light does not catch up with a rest frame, M'. It is c period regardless of the motion of A.

So, in the frame of M', A emits at some distance and yes A moves away from M', but no frame moves away from light.

Do you understand yet?

64. Originally Posted by chinglu
This is the basis on the error of your thinking.

Let's look at the context of the M' frame.

First, M' is not moving, the earth is and so is M.

Now, when A strikes, we apply the light postulate.

The motion of the light source does not impact the timing of the light to strike M or M'.

So, you said, M' moves relative to the embankment frame, so we know that it will take some time for the light from A to catch up.

Under the light postulate, light does not catch up with a rest frame, M'. It is c period regardless of the motion of A.

You didn't read my post, just like you didn't read Einstein's chapter. This is because you are crazy. You are making the same crazy person error that many crazy people make. It's not relativity theory that makes them crazy, it is that they are crazy, make a mistake when trying to understand relativity theory, and they use this to feed their insanity.

For those who are sane, let me explain. If I were describing the relationship between the flash of lightning and the midpoint of the train in frame of the train, I would use the light postulate. But at no point did I do this. The only time I invoked the frame of the train was when I discussed the criterion for simulteneity. If I were to discuss this example in detail using the reference frame of the train, I would have to introduce specific coordinates within that frame.

chinglu wants to say that it is wrong for anyone to claim that M' moved away from A. In order to do this, he has to be crazy, because we know losts of real-world examples where people move away from light sources like car headlights and flashlights. So we know that it must be possible to describe M' moving away from A. Note that chinglu never brings up any problem with the actual calculations performed, just like he never answers direct questions, just like he never does any actual analysis of this train example. This is because he is protecting his cherished fantasies.
So, in the frame of M', A emits at some distance and yes A moves away from M', but no frame moves away from light.

Do you understand yet?
Nobody could understand that, because frames do not move away from things.

65. Originally Posted by PhysBang
Originally Posted by chinglu
This is the basis on the error of your thinking.

Let's look at the context of the M' frame.

First, M' is not moving, the earth is and so is M.

Now, when A strikes, we apply the light postulate.

The motion of the light source does not impact the timing of the light to strike M or M'.

So, you said, M' moves relative to the embankment frame, so we know that it will take some time for the light from A to catch up.

Under the light postulate, light does not catch up with a rest frame, M'. It is c period regardless of the motion of A.

You didn't read my post, just like you didn't read Einstein's chapter. This is because you are crazy. You are making the same crazy person error that many crazy people make. It's not relativity theory that makes them crazy, it is that they are crazy, make a mistake when trying to understand relativity theory, and they use this to feed their insanity.

For those who are sane, let me explain. If I were describing the relationship between the flash of lightning and the midpoint of the train in frame of the train, I would use the light postulate. But at no point did I do this. The only time I invoked the frame of the train was when I discussed the criterion for simulteneity.

chinglu wants to say that it is wrong for anyone to claim that M' moved away from A. In order to do this, he has to be crazy, because we know losts of real-world examples where people move away from light sources like car headlights and flashlights. So we know that it must be possible to describe M' moving away from A. Note that chinglu never brings up any problem with the actual calculations performed, just like he never answers direct questions. This is because he is protecting his cherished fantasies.
So, in the frame of M', A emits at some distance and yes A moves away from M', but no frame moves away from light.

Do you understand yet?
Nobody could understand that, because frames do not move away from things.
If you are at rest and a light source is moving, use the light postulate to prove an observer at rest moves toward/away from the light.

Here is the LP.

Any ray of light moves in the “stationary” system of co-ordinates with the determined velocity c, whether the ray be emitted by a stationary or by a moving body

http://www.fourmilab.ch/etexts/einstein/specrel/www/

66. Originally Posted by chinglu
If you are at rest and a light source is moving, use the light postulate to prove an observer at rest moves toward/away from the light.

Here is the LP.

Any ray of light moves in the “stationary” system of co-ordinates with the determined velocity c, whether the ray be emitted by a stationary or by a moving body

http://www.fourmilab.ch/etexts/einstein/specrel/www/
This is a common way that crazy people like to misinterpret the light postulate. Einstein is very clear what he means. Crazy people like chinglu like to quote the first part of the light postulate but not the last part.

Einstein continues right after crazy chinglu cuts him off:
"Hence

velocity = lightpath / time interval

where time interval is to be taken in the sense of the definition in §1."

And what is the time interval in §1? It's an interval between clocks that are at rest relative to the so-called stationary system that are at specific points of the system of coordinates. So when we describe the motion of light, we have to describe it relative to the system of coordinates, not relative to any specific object, whether or not that object is moving.

67. Originally Posted by PhysBang
Originally Posted by chinglu
If you are at rest and a light source is moving, use the light postulate to prove an observer at rest moves toward/away from the light.

Here is the LP.

Any ray of light moves in the “stationary” system of co-ordinates with the determined velocity c, whether the ray be emitted by a stationary or by a moving body

http://www.fourmilab.ch/etexts/einstein/specrel/www/
This is a common way that crazy people like to misinterpret the light postulate. [/
I think that you and Scott have had similar experiences:

"Debating creationists on the topic of evolution is rather like trying to play chess with a pigeon -- it knocks the pieces over, craps on the board, and flies back to its flock to claim victory." - Scott D. Weitzenhoffer

68. Originally Posted by PhysBang
Originally Posted by chinglu
If you are at rest and a light source is moving, use the light postulate to prove an observer at rest moves toward/away from the light.

Here is the LP.

Any ray of light moves in the “stationary” system of co-ordinates with the determined velocity c, whether the ray be emitted by a stationary or by a moving body

http://www.fourmilab.ch/etexts/einstein/specrel/www/
This is a common way that crazy people like to misinterpret the light postulate. Einstein is very clear what he means. Crazy people like chinglu like to quote the first part of the light postulate but not the last part.

Einstein continues right after crazy chinglu cuts him off:
"Hence

velocity = lightpath / time interval

where time interval is to be taken in the sense of the definition in §1."

And what is the time interval in §1? It's an interval between clocks that are at rest relative to the so-called stationary system that are at specific points of the system of coordinates. So when we describe the motion of light, we have to describe it relative to the system of coordinates, not relative to any specific object, whether or not that object is moving.

Good, you have finally understood the logic, c = lightpath / time interval.

Now, let's see if you can calculate t'.

Note, t' = lightpath /c.

Now, what is the light path?

69. Originally Posted by DrRocket
Originally Posted by PhysBang
Originally Posted by chinglu
If you are at rest and a light source is moving, use the light postulate to prove an observer at rest moves toward/away from the light.

Here is the LP.

Any ray of light moves in the “stationary” system of co-ordinates with the determined velocity c, whether the ray be emitted by a stationary or by a moving body

http://www.fourmilab.ch/etexts/einstein/specrel/www/
This is a common way that crazy people like to misinterpret the light postulate. [/
I think that you and Scott have had similar experiences:

"Debating creationists on the topic of evolution is rather like trying to play chess with a pigeon -- it knocks the pieces over, craps on the board, and flies back to its flock to claim victory." - Scott D. Weitzenhoffer
I have a derivative outstanding.

Since you cannot refute it, you therefore, submit to it.

Further, do you have the math to prove ROS? I want to see the t' calculations for A and B.

Since you refuse to supply these calculations, you submit to my results.

It is time for you to confess SR is false.

70. Originally Posted by chinglu

Now, what is the light path?

If you weren't crazy, you'd see that was already explicitly answered. But since you missed that, you are more obviously crazy.

71. Originally Posted by chinglu
It is time for you to confess SR is false.
Try telling that to the people who have been testing SR for the past century.

http://math.ucr.edu/home/baez/physic...periments.html

72. Originally Posted by PhysBang
Originally Posted by chinglu

Now, what is the light path?

If you weren't crazy, you'd see that was already explicitly answered. But since you missed that, you are more obviously crazy.
Really? You claimed the train frame when considered at rest viewed light as measured c+v and c-v. Do you imderstand this is not SR.

Under SR, light is measured c to the frame.

Beginners do not understand light is measured c when the frame is considered at rest.

Anyway, no one has been able to refute that

t' = r/(γc) for A and B

and

t = r/c.

No math has yet refuted this. If you present math contrary to this, put it out here so I may correct it.

73. Originally Posted by SpeedFreek
Originally Posted by chinglu
It is time for you to confess SR is false.
Try telling that to the people who have been testing SR for the past century.

http://math.ucr.edu/home/baez/physic...periments.html
Curious. I can't find anything prpving the T/E experiment.

Further, I can't find anything refuting my calclus.

I would be surprised if SR could refute calculus. But, that is not the point, both calculus and SR cannot be true.

RocketMan has not been able to refute the calculus. No one has been able to refute my calculated times for the T/E experiment.

Do you have the calcululated times or do you quit like everyone else?

74. SpeedFreek has implicitly issued a nice challenge. Can you explain why the experimental evidence to date seems to agree with SR? Can you suggest a practical test which would uniquely support your position?

Refuting the math becomes sort of irrelevant if your ideas make unique and observable predictions about the universe.

75. chinglu, you have convinced me. Your understanding of these issues apparently excedes that of not only the professionals on this forum, but the greatest minds of the last century and a half. It is an indescribable honour to be alive at the same time as you.

I suspect your intellect extends beyond the realms of physics and cosmology. I imagine you can grasp the nature of complex problems and provide valid, insightful solutions with minimal effort. So I come to you for help, as a last resort, confident you will provide a solution to a problem that has vexed me for the last several days.

I am quite certain that chinglu and ignoramus are anagrams of each other, but try as I may I can't seem to rearrange the letters in a way that works. What am I doing wrong?

76. Originally Posted by chinglu
Originally Posted by SpeedFreek
Originally Posted by chinglu
It is time for you to confess SR is false.
Try telling that to the people who have been testing SR for the past century.

http://math.ucr.edu/home/baez/physic...periments.html
Curious. I can't find anything prpving the T/E experiment.

Further, I can't find anything refuting my calclus.

I would be surprised if SR could refute calculus. But, that is not the point, both calculus and SR cannot be true.

RocketMan has not been able to refute the calculus. No one has been able to refute my calculated times for the T/E experiment.

Do you have the calcululated times or do you quit like everyone else?
I have already said I cannot refute mathematics based on misconceptions. It doesn't have to be the maths that's wrong, its what you are using it to do.

You have already been shown the correct way to do it, which you continue to ignore. The reason you cannot find anything to refute your calculus is because you are getting the right answer to the wrong question.

You say that SR is false, but you will find no experiment where SR is applicable that conflicts with the predictions of SR.

It is not SR you are arguing with, it is your interpretation of what SR implies.

Let me test this...
I see a train moving across my view at 75% of c. It flashes a beacon that is attached to it, and the light from that beacon propagates in a spherical wave centred on the position, relative to me, that the beacon was when it flashed. The train is therefore moving through the spherical light wave as it propagates.

From the point of view of an observer on the train, the light propagates at c in all directions from the beacon, and therefore the spherical wave remains centred on the train. The light to the rear of the train propagates away at the same speed as the light to the front of the train. The occupants therefore see a spherical light wave.

This is SR. Do you agree with the above scenario?

77. Originally Posted by chinglu
Really? You claimed the train frame when considered at rest viewed light as measured c+v and c-v.
No, I didn't. You might have hallucinated this, but anyone who is not crazy can see that I never said that. I told you again and again that frames don't move. Neither can the be considered to be at rest. They are the context in which things are measured. I did not give any descriptions in the train frame, I only gave them in the frame in which the embankment is at rest. You might have your on fantasies about this, but that doesn't really matter because you are crazy.
Do you imderstand this is not SR.
And yet I agree with everyone else here and you are crazy.
Anyway, no one has been able to refute that

t' = r/(γc) for A and B

and

t = r/c.

No math has yet refuted this. If you present math contrary to this, put it out here so I may correct it.
No math refutes it because nobody can tell what you mean when you say this. You are too crazy to use the mathematics properly, even if you can follow a proof.

78. Originally Posted by TheBiologista
SpeedFreek has implicitly issued a nice challenge. Can you explain why the experimental evidence to date seems to agree with SR? Can you suggest a practical test which would uniquely support your position?

Refuting the math becomes sort of irrelevant if your ideas make unique and observable predictions about the universe.
No, it is not a challenge.

As I said, there does not exists a single experiment that refutes my math. Therefore, that would imply I found a loophole.

79. Originally Posted by Ophiolite
chinglu, you have convinced me. Your understanding of these issues apparently excedes that of not only the professionals on this forum, but the greatest minds of the last century and a half. It is an indescribable honour to be alive at the same time as you.

I suspect your intellect extends beyond the realms of physics and cosmology. I imagine you can grasp the nature of complex problems and provide valid, insightful solutions with minimal effort. So I come to you for help, as a last resort, confident you will provide a solution to a problem that has vexed me for the last several days.

I am quite certain that chinglu and ignoramus are anagrams of each other, but try as I may I can't seem to rearrange the letters in a way that works. What am I doing wrong?
All you need to is address the math.

Everyone is running around here like chickens with their heads cut off and none have addressed the math.

Now, can you produce the math for your SR for the T/E experiment? It would seem after 106 years you people would have this math down.

So, let's have it.

Then, after you failed at that task, you may address my derivative and refute that since you are certain in your views.

80. Originally Posted by SpeedFreek
Originally Posted by chinglu
Originally Posted by SpeedFreek
Originally Posted by chinglu
It is time for you to confess SR is false.
Try telling that to the people who have been testing SR for the past century.

http://math.ucr.edu/home/baez/physic...periments.html
Curious. I can't find anything prpving the T/E experiment.

Further, I can't find anything refuting my calclus.

I would be surprised if SR could refute calculus. But, that is not the point, both calculus and SR cannot be true.

RocketMan has not been able to refute the calculus. No one has been able to refute my calculated times for the T/E experiment.

Do you have the calcululated times or do you quit like everyone else?
I have already said I cannot refute mathematics based on misconceptions. It doesn't have to be the maths that's wrong, its what you are using it to do.

You have already been shown the correct way to do it, which you continue to ignore. The reason you cannot find anything to refute your calculus is because you are getting the right answer to the wrong question.

You say that SR is false, but you will find no experiment where SR is applicable that conflicts with the predictions of SR.

It is not SR you are arguing with, it is your interpretation of what SR implies.

Let me test this...
I see a train moving across my view at 75% of c. It flashes a beacon that is attached to it, and the light from that beacon propagates in a spherical wave centred on the position, relative to me, that the beacon was when it flashed. The train is therefore moving through the spherical light wave as it propagates.

From the point of view of an observer on the train, the light propagates at c in all directions from the beacon, and therefore the spherical wave remains centred on the train. The light to the rear of the train propagates away at the same speed as the light to the front of the train. The occupants therefore see a spherical light wave.

This is SR. Do you agree with the above scenario?
So, if you cannot address the math because of my misconceptions, then you can show the math where I am wrong.

That means you can take the T/E experiment and produce t' for the A and B flashes in the context of M'.

How long will this take?

I already produced my math. I have no fear.

81. Originally Posted by PhysBang
Originally Posted by chinglu
Really? You claimed the train frame when considered at rest viewed light as measured c+v and c-v.
No, I didn't. You might have hallucinated this, but anyone who is not crazy can see that I never said that. I told you again and again that frames don't move. Neither can the be considered to be at rest. They are the context in which things are measured. I did not give any descriptions in the train frame, I only gave them in the frame in which the embankment is at rest. You might have your on fantasies about this, but that doesn't really matter because you are crazy.
Do you imderstand this is not SR.
And yet I agree with everyone else here and you are crazy.
Anyway, no one has been able to refute that

t' = r/(γc) for A and B

and

t = r/c.

No math has yet refuted this. If you present math contrary to this, put it out here so I may correct it.
No math refutes it because nobody can tell what you mean when you say this. You are too crazy to use the mathematics properly, even if you can follow a proof.

Good, let's see your t' for A and B for the at rest train frame.

How long will this take?

82. OK, I am going to be gone for a while.

But, while I am away, I expect those SR supporters to refute the calculus and also provide t' for A and B with the T/E experiment.

Have a nice many days.

83. Originally Posted by chinglu
Good, let's see your t' for A and B for the at rest train frame.
Let's assume, without loss of generality, that the origin of the frame in which the embankment is at rest is the same at that frame in which the train is at rest.

v = 0.5 in the positive x direction. Therefore γ = 1.15470.

We know that t for the events A and B is 0 and that x for these events is 0 and 10.

Therefore we simply plug in the Lorentz transformations, t' = γ ( t - v x ).

For A:

t' = 1.15470 ( 0 - 0.5 * 0) = 0

For B:

t' = 1.15470 (0 - 0.5 * 10) = 1.15470(-5) = -5.7735

Thus we know, in the frame in which the train is at rest, that B happened well before A.

Now, where is M'?

At t=0, M' is at x=5, so we use the appropriate transformation.

x'= γ (x - vt) = 1.15470 (5 - 0.5 * 0) = 5.7735

At t=10, M' is at x=10 and this is when the light from A reaches M'.

t' = γ (t - vx) = 1.15470 (10 - 0.5 * 10) = 5.7735
x'= γ (x - vt) = 1.15470 (10 - 0.5 * 10) = 5.7735

Of course, this indicates that the length of the train in the embankment frame is 10 and the length of the train in the train frame is 11.547. But this is to be expected, since the maximum length of any object is found in that frame in which that object is at rest. Thus the length train must be contracted in the embankment frame.

At t= 3 + 1/3, M' is at x = 6 + 2/3 and this is when the light from B reaches M'.

x' = γ (x - vt) ~ 1.15470 (6.6666 - 0.5 * 3.3333) ~ 5.7735
t' = γ (t - vx) ~ 1.15470 (3.333 - 0.5 * 6.6666) = 0

So, unsurprisingly, the midpoint of the train doesn't move in this frame. We note that in this frame, the light from B reaches the midpoint of the train at the same time that the A occurs.

(Additionally, in the embankment frame B occurs at t=0 and x=10, so

x'= γ (x - vt) = 1.15470 (10 - 0.5 * 0) = 11.547 = 2 * 5.7735 .

We expect this given the calculation of the train length done above.)

Note that, given the spatial coordinate locations of A and B in the train frame, this shows that the light from A and B is indeed moving at c (which we set at 1 for convenience).

(NB: That B happens when A takes place is an artefact of the initial conditions used and is not generic to every choice of velocity for the train and every choice of train length.)
How long will this take?
About five minutes. I was slowed down by having to find my original example and the numbers I used there and right this out with explanations so that others could follow.

84. Originally Posted by chinglu
This is the basis on the error of your thinking.

Let's look at the context of the M' frame.

First, M' is not moving, the earth is and so is M.

Now, when A strikes, we apply the light postulate.

The motion of the light source does not impact the timing of the light to strike M or M'.
The light source's motion doesn't affect it, but the motions of M and M' prime certainly does affect it.

So, you said, M' moves relative to the embankment frame, so we know that it will take some time for the light from A to catch up.

Under the light postulate, light does not catch up with a rest frame, M'. It is c period regardless of the motion of A.
If M' is at rest, then M is moving in the <---- direction. What's the difference between a beam of light taking the normal amount of time to reach M', and a shorter than normal amount of time to reach M vs. a beam of light taking a longer than normal amount of time to reach M' and the normal amount of time to reach M?

None. There is no difference.

So, in the frame of M', A emits at some distance and yes A moves away from M', but no frame moves away from light.

Do you understand yet?
Yes. We've established that the motion of a light source doesn't affect anything. For simplicity, it's not a bad idea to just treat light sources as being perfectly stationary in the absolute sense. You'll confuse yourself a lot less if you do that.

Originally Posted by kojax
Originally Posted by chinglu
Originally Posted by kojax
Originally Posted by chinglu
Originally Posted by PhysBang
Someone has to be pretty stupid or pretty crazy to take one example and then assume that all examples have to have exactly the same variables with exactly the same values assigned to them.
The point that I made that all you can do is agree is whenever there is a light emission, the time in the respective frames is set to 0. That is the logic of Einstein and I showed you his reasoning.

Now, given a light emission at A, we have t'=t=0.

The next issue to decide is how far is A from M and M'.

For M, we assumed, so that is r. For M', based on length contraction, the diatance is r/γ.

Then, based on light being measured c in each frame, we have

t = r/c for the light to reach M.

t' = r/(γc) for the light to reach M'.

The same reasoing holds for B.

Hence, t'A = t'B and tA = tB.

I can't seem to get anyone to offer any other solution and so anyone that believes ROS does that without a basis in math and logic.

From the above factual math, ROS is false.
Now I see why so many other posters are giving up on this thread.

Suppose as in the picture you gave us that:

A -------- M' ----------B
A---------M------------B

Suppose M' is moving to the right, and M is also moving to the right but at a slower speed.

If M receives the light signal from A and B simultaneously, then it's impossible for M' to also receive the light signal from A and B simultaneously. Are we in agreement so far?
They have given up on the thread because they cannot stop the math. This is not an opinion section.

You are struggling with this.

Instead of repeating books, let's see exactly your timing for light to travel from A to M' and the time for light to travel from B to M'.

Folks give up on the thread because they struggle with this simple calculation. Instead, they cling to the T/E scripture without having any understanding how to prove if the conclusions are valid.

Let anyone come here with the math and prove the conclusions of the T/E experiment.
All well and good, but do we agree, that the time of receiving the light signals cannot be simultaneous for M', if it is simultaneous for M
I'd still like an answer to this question.

85.

86. It was the same at BAUT, too.

87. It's almost as if Chinglu's wrong...

88.

89. Originally Posted by PhysBang
Originally Posted by chinglu
Good, let's see your t' for A and B for the at rest train frame.
Let's assume, without loss of genreality, that the origin of the frame in which the embankment is at rest is the same at that frame in which the train is at rest.

v = 0.5 in the positive x direction. Therefore γ = 1.15470.

We know that t for the events A and B is 0 and that x for these events is 0 and 10.

Therefore we simply plug in the Lorentz transformations, t' = γ ( t - v x ).

For A:

t' = 1.15470 ( 0 - 0.5 * 0) = 0

For B:

t' = 1.15470 (0 - 0.5 * 10) = 1.15470(-5) = -5.7735

Thus we know, in the frame in which the train is at rest, that B happened well before A.

Now, where is M'?

At t=0, M' is at x=5, so we use the appropriate transformation.

x'= γ (x - vt) = 1.15470 (5 - 0.5 * 0) = 5.7735

At t=10, M' is at x=10 and this is when the light from A reaches M'.

t' = γ (t - vx) = 1.15470 (10 - 0.5 * 10) = 5.7735
x'= γ (x - vt) = 1.15470 (10 - 0.5 * 10) = 5.7735

Of course, this indicates that the length of the train in the embankment frame is 10 and the length of the train in the train frame is 11.547. But this is to be expected, since the maximum length of any object is found in that frame in which that object is at rest. Thus the length train must be contracted in the embankment frame.

At t= 3 + 1/3, M' is at x = 6 + 2/3 and this is when the light from B reaches M'.

x' = γ (x - vt) ~ 1.15470 (6.6666 - 0.5 * 3.3333) ~ 5.7735
t' = γ (t - vx) ~ 1.15470 (3.333 - 0.5 * 6.6666) = 0

So, unsurprisingly, the midpoint of the train doesn't move in this frame. We note that in this frame, the light from B reaches the midpoint of the train at the same time that the A occurs.

(Additionally, in the embankment frame B occurs at t=0 and x=10, so

x'= γ (x - vt) = 1.15470 (10 - 0.5 * 0) = 11.547 = 2 * 5.7735 .

We expect this given the calculation of the train length done above.)

Note that, given the spatial coordinate locations of A and B in the train frame, this shows that the light from A and B is indeed moving at c (which we set at 1 for convenience).

(NB: That B happens when A takes place is an artefact of the initial conditions used and is not generic to every choice of velocity for the train and every choice of train length.)
How long will this take?
About five minutes. I was slowed down by having to find my original example and the numbers I used there and right this out wiht explanations so that others could follow.
You have failed to apply the LT equations correctly. They can be quite difficult.

You must remember, A is the light emission point in both frames. You reversed the light emission point and light reception point./

Now, the the light source A moves from the view of the train. But that has no impact on the results.

In order to calculate t, you mist use the distance to the light source at light emissiom and divide that by c.

You can find that in the origin paper with the light postulate.

90. Originally Posted by kojax
Originally Posted by chinglu
This is the basis on the error of your thinking.

Let's look at the context of the M' frame.

First, M' is not moving, the earth is and so is M.

Now, when A strikes, we apply the light postulate.

The motion of the light source does not impact the timing of the light to strike M or M'.
The light source's motion doesn't affect it, but the motions of M and M' prime certainly does affect it.
The motion of the train is 0 for the view of the train. You must take the distance to the light emission point in the frame / c.

That will give you t'. You did not do this and thus you are in error.

So, you said, M' moves relative to the embankment frame, so we know that it will take some time for the light from A to catch up.

Under the light postulate, light does not catch up with a rest frame, M'. It is c period regardless of the motion of A.
If M' is at rest, then M is moving in the <---- direction. What's the difference between a beam of light taking the normal amount of time to reach M', and a shorter than normal amount of time to reach M vs. a beam of light taking a longer than normal amount of time to reach M' and the normal amount of time to reach M?

None. There is no difference.

So, in the frame of M', A emits at some distance and yes A moves away from M', but no frame moves away from light.

Do you understand yet?
Yes. We've established that the motion of a light source doesn't affect anything. For simplicity, it's not a bad idea to just treat light sources as being perfectly stationary in the absolute sense. You'll confuse yourself a lot less if you do that.

Originally Posted by kojax
Originally Posted by chinglu
Originally Posted by kojax
Originally Posted by chinglu
Originally Posted by PhysBang
Someone has to be pretty stupid or pretty crazy to take one example and then assume that all examples have to have exactly the same variables with exactly the same values assigned to them.
The point that I made that all you can do is agree is whenever there is a light emission, the time in the respective frames is set to 0. That is the logic of Einstein and I showed you his reasoning.

Now, given a light emission at A, we have t'=t=0.

The next issue to decide is how far is A from M and M'.

For M, we assumed, so that is r. For M', based on length contraction, the diatance is r/γ.

Then, based on light being measured c in each frame, we have

t = r/c for the light to reach M.

t' = r/(γc) for the light to reach M'.

The same reasoing holds for B.

Hence, t'A = t'B and tA = tB.

I can't seem to get anyone to offer any other solution and so anyone that believes ROS does that without a basis in math and logic.

From the above factual math, ROS is false.
Now I see why so many other posters are giving up on this thread.

Suppose as in the picture you gave us that:

A -------- M' ----------B
A---------M------------B

Suppose M' is moving to the right, and M is also moving to the right but at a slower speed.

If M receives the light signal from A and B simultaneously, then it's impossible for M' to also receive the light signal from A and B simultaneously. Are we in agreement so far?
They have given up on the thread because they cannot stop the math. This is not an opinion section.

You are struggling with this.

Instead of repeating books, let's see exactly your timing for light to travel from A to M' and the time for light to travel from B to M'.

Folks give up on the thread because they struggle with this simple calculation. Instead, they cling to the T/E scripture without having any understanding how to prove if the conclusions are valid.

Let anyone come here with the math and prove the conclusions of the T/E experiment.
All well and good, but do we agree, that the time of receiving the light signals cannot be simultaneous for M', if it is simultaneous for M
I'd still like an answer to this question.[/quote]

As far as M and M' being simultaneous, that is the view of the frame when at rest with the frame.

So, at rest with M', A and B are simultaneous and the same hold for M.

From the view of M', they are not simultaneous in M.

That is basic SR.

The question you must resolve for yourself and refuse to do this is what is t' in the view of M'.

I still see no person here has answered the question as to why one would believe the T/E experiment conclusions of Einstein when he did not even provide the math to back up his failed conclusions.

Why would a science discipline accept an "opinion" from Einstein without even questioing the math?

91. Originally Posted by MeteorWayne
I proved dt'/dt < 0 under SR for a specific example. This cannot and has not been refuted.

Since SR requires time to proceed backward in order to be true and nature cannot possibly supply this logic, I think you have perfectly figured out SR.

92. Originally Posted by PhysBang
I can prove this logic as well. You will note RPenner could not refute the math as he tried and I corrected him.

I wonder why all you folks lack the math to back up your positions whereas I provide all math for all to see and clearly refute if possible.

Yet, noone can refute my math. Gee, I wonder why that is.

Note that for every position I take, I provide clear math. I am simply not into opinions.

93. What would Einstein conclude about the Granville train crash , 'the day of the roses' in about 1978, in which a carriage was crushed by the concrete-bridge slab falling on it.
The man says he saw the carriage being crushed as if in slow motion -- how?
In my life I became very energy depleted on day and was playing tennis. I saw the served ball coming in slow motion.
What would Albert say about such things?
Was the observer in each case seeing faster than the speed of light? Or merely processing the information faster?

As to the simultaneity of the lightning flash, why cant the observer on the bank and in the train simply agree to see the flashes happen at the same time?

On another level, sight perception actually works like radar, where the brain sends out a signal that bounces back to the eyes and is then interpreted by the brain according to how it changed?
In the case of the train crash, the man sent out light faster than light and received it back faster than he usually would and hence the carriage was in effect slowed down?
If this is true, then simultaneity of lightning flashes could be arranged or not arranged according to how the observers wanted it to be?

If two bank observers are looking at a train, and the guy on the left processes faster, the train appears to slow down relative to the other observer -- what if the train was a beam of light? That means reality depends on the observer: the observer is king of reality, rather than reality dictating terms to each observer.
Einsteins fixed speed of light c, is a dictator in disguise?

94. Originally Posted by chinglu
You have failed to apply the LT equations correctly. They can be quite difficult.

You must remember, A is the light emission point in both frames. You reversed the light emission point and light reception point./
No, crazy person, I did not. It's refreshing to see just how deluded you can be. We know that these delusions are a symptom of your mental illness. We can only hope that you get some professional help. You might not need medication.
Now, the the light source A moves from the view of the train. But that has no impact on the results.
A is an event, not a location. A never moves in any frame.
In order to calculate t, you mist use the distance to the light source at light emissiom and divide that by c.

You can find that in the origin paper with the light postulate.
No, in order to find t we must simply use the description given.
I still see no person here has answered the question as to why one would believe the T/E experiment conclusions of Einstein when he did not even provide the math to back up his failed conclusions.
Einstein did give all the math that was needed. If one reads chapter 8 and 9, and if one is not crazy like you, one finds all the math one needs. The results are so incredibly general that no specifics are needed.

We note that when given specifics, you simply have delusions like, "You reversed the light emission point and light reception point." You don't give any reasoning for this statement because it is merely a delusion.
I can prove this logic as well. You will note RPenner could not refute the math as he tried and I corrected him.
He, like everyone else, merely noted that you don't know what the math says. Just like here, you made basic mistakes about the scope of a mathematical proof and about what SR actually says.
I wonder why all you folks lack the math to back up your positions whereas I provide all math for all to see and clearly refute if possible.
Again, the mathematics is up there and you haven't been able to find a problem with it. This shows, again, that you are crazy.
Note that for every position I take, I provide clear math. I am simply not into opinions.
You don't get what mathematics is. You don't provide clear math, at least in the sense that you change your interpretation of math as your insanity hits you. For example, first you said that all clocks were coordinated in all reference frames, then you said that clocks run backwards in some reference frames. This is crazy.

95. Originally Posted by PhysBang
Now, the the light source A moves from the view of the train. But that has no impact on the results.
A is an event, not a location. A never moves in any frame.
You may call A whatever you want.

Yet, you claim A does not move in any frame. Can you explain this? That is absolute rest.

In order to calculate t, you mist use the distance to the light source at light emissiom and divide that by c.

You can find that in the origin paper with the light postulate.
No, in order to find t we must simply use the description given.
So, what is t'? Please be specific.

Don't worry, if you get it wrong, I will correct you in front of everyone.

I still see no person here has answered the question as to why one would believe the T/E experiment conclusions of Einstein when he did not even provide the math to back up his failed conclusions.
Einstein did give all the math that was needed. If one reads chapter 8 and 9, and if one is not crazy like you, one finds all the math one needs. The results are so incredibly general that no specifics are needed.
Good, let's see the math for the T/E experiment. How long will this take?

We note that when given specifics, you simply have delusions like, "You reversed the light emission point and light reception point." You don't give any reasoning for this statement because it is merely a delusion.
It is a shame that you do not understand light emission and light reception. LT only applies to the light emission point as applied to the light reception point. You applied LT to the light reception point as the light emission point. That is the error.

Go ask RPenner or ask him to come here. RPenner will refuse. It is odd that folks who really believe in SR cannot supply the math. It is like religion.

I can prove this logic as well. You will note RPenner could not refute the math as he tried and I corrected him.
He, like everyone else, merely noted that you don't know what the math says. Just like here, you made basic mistakes about the scope of a mathematical proof and about what SR actually says.
That's false. Read the thread and I provided the math he could not handle. Otherwise, take his position here since you understand it and I will correct you also as I did him.

You don't get what mathematics is. You don't provide clear math, at least in the sense that you change your interpretation of math as your insanity hits you. For example, first you said that all clocks were coordinated in all reference frames, then you said that clocks run backwards in some reference frames. This is crazy.
I will assume you can prove your statements above using math.

Everyone here wants to see it. I know I do. If it deviates from mine, then you will be corrected. So, show the math.

Note, I have no fear.

96. Originally Posted by chinglu
You may call A whatever you want.
No, I can't if I want to do physics. A crazy person like you can call things whatever they want.
Yet, you claim A does not move in any frame. Can you explain this? That is absolute rest.
The location of an event does not change once assigned in a given reference frame, it has a fixed location in space and time in each reference frame. You might as well say that the location of the origin changes within a reference frame, since you are crazy.
In order to calculate t, you mist use the distance to the light source at light emissiom and divide that by c.

You can find that in the origin paper with the light postulate.
No, in order to find t we must simply use the description given.
So, what is t'? Please be specific.
Now you are trying to pretend that you didn't ask the question of a crazy person. You asked about "t", now you are trying to change the subject.
Don't worry, if you get it wrong, I will correct you in front of everyone.
Sure you will. Regardless, I have given the value of t' for many, many events above. You have yet to actually address any of that. But this is OK, because you are crazy.
Good, let's see the math for the T/E experiment. How long will this take?
a) Read that reference that you gave.
b) Look above.

It is a shame that you do not understand light emission and light reception. LT only applies to the light emission point as applied to the light reception point.
That is crazy talk. Sane people know that the Lorentz transformations work for any spacetime point.
You applied LT to the light reception point as the light emission point. That is the error.
You keep saying that, but you fail to demonstrate it. Since you are crazy, these things exist only in your head, which is why you can't point to anything specific.
Go ask RPenner or ask him to come here. RPenner will refuse. It is odd that folks who really believe in SR cannot supply the math. It is like religion.
RPenner wasted enough time on your crazy talk. He also clearly went over exactly where you were wrong so that anyone who is not crazy, like you, can see.

You don't get what mathematics is. You don't provide clear math, at least in the sense that you change your interpretation of math as your insanity hits you. For example, first you said that all clocks were coordinated in all reference frames, then you said that clocks run backwards in some reference frames. This is crazy.
I will assume you can prove your statements above using math.
You prove my statements about you being crazy by saying crazy things. No math is needed for people to read what you write and conclude that you are crazy.
Everyone here wants to see it. I know I do. If it deviates from mine, then you will be corrected. So, show the math.

Note, I have no fear.
You also have no math. You asked for the train experiment with math, I gave it, and you simply spew crazy talk rather than "correct" it. Because you are crazy.

That shame that you bring on your family for not actually bothering to learn this stuff and correct your mistakes is great.

97. Originally Posted by chinglu
Originally Posted by PhysBang
Now, the the light source A moves from the view of the train. But that has no impact on the results.
A is an event, not a location. A never moves in any frame.
You may call A whatever you want.

Yet, you claim A does not move in any frame. Can you explain this? That is absolute rest.

An event is, by definition, a point in spacetime. An event cannot "move". It has fixed time and spatial coordinates in any given inertial frame. The coordinates will be different for different inertial frames -- that is what Lorentz transformations are all about.

Now, go take your medication and study some special relativity.

98. Originally Posted by DrRocket
Originally Posted by chinglu
Originally Posted by PhysBang
Now, the the light source A moves from the view of the train. But that has no impact on the results.
A is an event, not a location. A never moves in any frame.
You may call A whatever you want.

Yet, you claim A does not move in any frame. Can you explain this? That is absolute rest.

An event is, by definition, a point in spacetime. An event cannot "move". It has fixed time and spatial coordinates in any given inertial frame. The coordinates will be different for different inertial frames -- that is what Lorentz transformations are all about.

Now, go take your medication and study some special relativity.
The context of the conversation was whether the light source A moves or not and the impact that motion has on the light reception at M and M'.

The poster claimed the light source A did not move relative to any frame.

So, your comments are not based in reality. Nice try though. It must get frustrating.

Have you figured out yet that dt'/dt < 0, as I proved, demonstrates SR is logically inconsistent?

Have you figured out why?

I will try to teach you.

t or t' is a measure of the distance light has traveled from the origin. Hence, dt'/dt < 0 with t increasing implies t' decreases.

That means although light beams measure c, LT does not preserve the propagation of light away from the origin for the primed frame in all directions as required by the consistency of SR.

Tough luck.

99. Originally Posted by PhysBang
Originally Posted by chinglu
You may call A whatever you want.
No, I can't if I want to do physics. A crazy person like you can call things whatever they want.
Yet, you claim A does not move in any frame. Can you explain this? That is absolute rest.
The location of an event does not change once assigned in a given reference frame, it has a fixed location in space and time in each reference frame. You might as well say that the location of the origin changes within a reference frame, since you are crazy.
Good, you are beginning to understand the light postulate. However, the context of my statement was based on whether the light source A moves or not. So, you are off task.

Now, to complete your lessons, what is the distance from M' when A emits?

If r is the distance from A to M, then r/γ is the distance of A to M' at light emission when t'=t=x=x'=0 at A.

Now, the same logic applies to B.

Now, since t'=0 at B and t'=0 at A for the light emission and the distance of A and B to M' is r/γ when t' = 0, then t'=r/γ for A and t'=r/γ for B, hence the train observer will view the light flashes as simultaneous.

It is a shame Einstein did not provide you with the math as a guide, but I did,

100. Originally Posted by PhysBang
Sane people know that the Lorentz transformations work for any spacetime point.
OK Mr Newton.

LT works based on an agreed upon start event which is light emission. See Einstein's original paper and the derivation of LT. LT was derived based on the light mirror experiment.

So, the light emission was at A. Any sane person can understand that. You on the other hand, tried to claim M and M' is the start event. In order for you to make that claim, you are not using the frame origins as the light emission point.

you can find math on line to correct your mistake when light emission is not at the origin.

Now, have you yet figured out that the distance light must travel to M' from A is r/γ given the distance from A to M is r?

That is your first step in understanding this problem.

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