1. In the T/E experiment, both observers are equidistant to the light sources when they are fired.

Also, Einstein suggested in the train frame, the train observer moves toward the light.

Let's take a look.

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. Observers who take the railway train as their reference-body must therefore come to the conclusion that the lightning flash B took place earlier than the lightning flash A. We thus arrive at the important result:

http://www.bartleby.com/173/9.html

Let's look at the "Now in reality". There is no reality that can be applied from the E to the T as suggested in Einstein's passage above. That would imply an absolute frame. But, carefully, Einstein claims the E frame imposes reality onto the T frame. That is not SR.

2.

3. Yet again, you make assertions about the incorrectness of SR because you fail to apply the postulates of SR.

It is reality, in the Embankment frame, that the flashes were simultaneous.

It is reality, in the Train frame, that the flashes were not simultaneous.

That's the relativity of simultaneity for you. SR tells us there is no absolute simultaneity between events separated by space.

4. Originally Posted by SpeedFreek
Yet again, you make assertions about the incorrectness of SR because you fail to apply the postulates of SR.

It is reality, in the Embankment frame, that the flashes were simultaneous.

It is reality, in the Train frame, that the flashes were not simultaneous.

That's the relativity of simultaneity for you. SR tells us there is no absolute simultaneity between events separated by space.
Good, so you agree the E frame cannot impose its frame logic on the T frame as suggested by Einstein.

Very good.

Now, both frames are equidistant to the flashes A and B when they occur, is that correct or not correct?

5. Originally Posted by chinglu
Now, both frames are equidistant to the flashes A and B when they occur, is that correct or not correct?
That makes no sense. Frames are not equidistant from anything, they give the context of distance.

6. Originally Posted by PhysBang
Originally Posted by chinglu
Now, both frames are equidistant to the flashes A and B when they occur, is that correct or not correct?
That makes no sense. Frames are not equidistant from anything, they give the context of distance.
I am talking about M and M' obviously.

7. Originally Posted by chinglu
I am talking about M and M' obviously.
And what are you saying about them? You haven't said anything about them that has made sense yet.

8. Originally Posted by PhysBang
Originally Posted by chinglu
I am talking about M and M' obviously.
And what are you saying about them? You haven't said anything about them that has made sense yet.
Of course I have. You do not understand the experiment.

Let M' be the mid-point of the distance A —> B on the travelling train. Just when the flashes 1 of lightning occur, this point M' naturally coincides with the point M,

http://www.bartleby.com/173/9.html

9. Originally Posted by PhysBang
Originally Posted by chinglu
I am talking about M and M' obviously.
And what are you saying about them? You haven't said anything about them that has made sense yet.
chinglu has never made sense. There is no reason to expect that record to suddenly be violated.

This discussion should be taken to a separate thread, in Pseudoscience, and not derail the topic, which is discussion of the primer on special relativity in the "sticky" thread.

10. Originally Posted by DrRocket
Originally Posted by PhysBang
Originally Posted by chinglu
I am talking about M and M' obviously.
And what are you saying about them? You haven't said anything about them that has made sense yet.
chinglu has never made sense. There is no reason to expect that record to suddenly be violated.

This discussion should be taken to a separate thread, in Pseudoscience, and not derail the topic, which is discussion of the primer on special relativity in the "sticky" thread.
I have made two simple statements. Can you explain specifically how these observations are Pseudoscience. Since when is it Pseudoscience to state the problem specifically as given by Einstein?

Now, can you explain why the train frame can impose its view onto the interpretation of the train frame? Then, can you explain why a frame imposing its view on another frame, commonly called frame mixing, is relativity?

Then can you explain the light postulate. Does it suggest the speed of light is c to the train observer regardless of the motion of the light source? Technically, aren't A and B moving light sources?

Finally, if r is the distance from A to M' when A emits light, what is t for the problem. Is it not t = r/c based on the light postulate?

In the T/E experiment, both observers are equidistant to the light sources when they are fired.

Also, Einstein suggested in the train frame, the train observer moves toward the light.

Let's take a look.

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. Observers who take the railway train as their reference-body must therefore come to the conclusion that the lightning flash B took place earlier than the lightning flash A. We thus arrive at the important result:

http://www.bartleby.com/173/9.html

Let M' be the mid-point of the distance A —> B on the travelling train. Just when the flashes 1 of lightning occur, this point M' naturally coincides with the point M,

http://www.bartleby.com/173/9.html

11. Originally Posted by chinglu
I have made two simple statements. Can you explain specifically how these observations are Pseudoscience. Since when is it Pseudoscience to state the problem specifically as given by Einstein?
It is pseudoscience when you have demonstrated in the past that you do not understand the science and when you demonstrate in this thread that you cannot write a proper English sentence about the relationship of frames and distance. The reason that everyone that reads your screeds disagrees with you is not a problem on their end, the problem is on your end.
Now, can you explain why the train frame can impose its view onto the interpretation of the train frame? Then, can you explain why a frame imposing its view on another frame, commonly called frame mixing, is relativity?
All frames impose things on other frames: if an event happens in one frame then it has to happen in all frames, if a spacetime interval between two events is given in one frame then the spacetime interval is the same in every frame (equivalently, one must apply the proper mathematical transformations to space and time distances).
Then can you explain the light postulate. Does it suggest the speed of light is c to the train observer regardless of the motion of the light source? Technically, aren't A and B moving light sources?
The light postulate doesn't do this. It says that the speed of light is c relative to points of space that are fixed in a frame.

12. Originally Posted by chinglu
I have made two simple statements. Can you explain specifically how these observations are Pseudoscience. Since when is it Pseudoscience to state the problem specifically as given by Einstein?
They can't, chinglu. So out comes the old thought-police you're-talking-nonsense abuse in lieu of a considered civil discussion. If you read what Einstein said carefully, you can see that there's an implicit absolute frame in there:

Originally Posted by Einstein
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...
Now have a read of The Other Meaning of Special Relativity to get to the bottom of this. Einstein knew about the wave theory of matter, so I don't know why he didn't come up with this. But such is life.

13. Originally Posted by Farsight
They can't, chinglu. So out comes the old thought-police you're-talking-nonsense abuse in lieu of a considered civil discussion.
Really? You've been attacking people whenever they've given substantive comments on your work since at least 2006. You have no idea what civil discussion is.

Perhaps you would like to explain how "Now, both frames are equidistant to the flashes A and B when they occur, is that correct or not correct?" makes any sense?
If you read what Einstein said carefully, you can see that there's an implicit absolute frame in there:
Sure, if you ignore everything else that Einstein ever wrote. Like always, all you have to offer is cherry-picked quotations.
Now have a read of The Other Meaning of Special Relativity to get to the bottom of this. Einstein knew about the wave theory of matter, so I don't know why he didn't come up with this. But such is life.
Perhaps if you could follow the mathematics you would understand why it's wrong.

14. Originally Posted by chinglu
In the T/E experiment, both observers are equidistant to the light sources when they are fired.

Also, Einstein suggested in the train frame, the train observer moves toward the light.

Let's take a look.

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. Observers who take the railway train as their reference-body must therefore come to the conclusion that the lightning flash B took place earlier than the lightning flash A. We thus arrive at the important result:

http://www.bartleby.com/173/9.html

Let's look at the "Now in reality". There is no reality that can be applied from the E to the T as suggested in Einstein's passage above. That would imply an absolute frame. But, carefully, Einstein claims the E frame imposes reality onto the T frame. That is not SR.
No, there is no absolute frame implied at all. What he does insist is that "local events" must be agreed upon between the two frames. IOW, the train frame and the embankment frame have to agree as to the Train observer's position with respect to the embankment when the two light flashes reach him, and as to the embankment observer's position with repect to the train when the flashes reach him.

For things to be otherwise would violate the principle that the laws of physics are constant across frames.

For instance, we can put recording devices on the embankment and train that record exactly what part f the other frame is exactly opposite them whne the light reaches them (train devices record what part of the embankment is opposite them and embankment devices record what part of the train is opposite them).

If you bring these records back together They have to agree. If train device 10 records that is is opposite embankment device 25 when the light reaches them both, then embankment device 25 has to record that it is opposte train device 10 at that same moment. (here we assume that the physical distance between the two devices as they pass is infinitesimally small enough not to effect the outcome)

The fact that the embankment observer is the one that sees the flashes simultaneously was just a arbitrary choice. This does not "impose its reality" onto the train frame any more than the train frame "imposes its reality" unto the embankment frame.

It's like having two people, one facing West and the other East. The first, from his view point, says that Chicago is to the right of Dallas, while the second says that, from his viewpoint, Chicago is to the left of Dallas.

Neither imposes his view on the other, but they can use their viewpoint to determine what the other sees.

This is what happens in the train experiment, the embankment observer uses what he sees, in respect to the embankment frame, to figure out what the train observer sees in his.

15. Yes, as I said, Chinglu's problems arise from failing to properly apply the postulates of SR.

These postulates are:
1. The principle of Relativity - physical laws should be the same in every inertial frame of reference.
2. The constancy of c to all inertial frames.

If you apply these postulates, then everything in SR makes perfect sense. I really don't understand how Chinglu manages to have such a problem with it, but seeing this thread totally explains the problems Chinglu had in his other thread with the Lorentz transformation of the spherical wave-front!

16. Originally Posted by Janus
Originally Posted by chinglu
In the T/E experiment, both observers are equidistant to the light sources when they are fired.

Also, Einstein suggested in the train frame, the train observer moves toward the light.

Let's take a look.

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. Observers who take the railway train as their reference-body must therefore come to the conclusion that the lightning flash B took place earlier than the lightning flash A. We thus arrive at the important result:

http://www.bartleby.com/173/9.html

Let's look at the "Now in reality". There is no reality that can be applied from the E to the T as suggested in Einstein's passage above. That would imply an absolute frame. But, carefully, Einstein claims the E frame imposes reality onto the T frame. That is not SR.
No, there is no absolute frame implied at all. What he does insist is that "local events" must be agreed upon between the two frames. IOW, the train frame and the embankment frame have to agree as to the Train observer's position with respect to the embankment when the two light flashes reach him, and as to the embankment observer's position with repect to the train when the flashes reach him.

For things to be otherwise would violate the principle that the laws of physics are constant across frames.

For instance, we can put recording devices on the embankment and train that record exactly what part f the other frame is exactly opposite them whne the light reaches them (train devices record what part of the embankment is opposite them and embankment devices record what part of the train is opposite them).

If you bring these records back together They have to agree. If train device 10 records that is is opposite embankment device 25 when the light reaches them both, then embankment device 25 has to record that it is opposte train device 10 at that same moment. (here we assume that the physical distance between the two devices as they pass is infinitesimally small enough not to effect the outcome)

The fact that the embankment observer is the one that sees the flashes simultaneously was just a arbitrary choice. This does not "impose its reality" onto the train frame any more than the train frame "imposes its reality" unto the embankment frame.

It's like having two people, one facing West and the other East. The first, from his view point, says that Chicago is to the right of Dallas, while the second says that, from his viewpoint, Chicago is to the left of Dallas.

Neither imposes his view on the other, but they can use their viewpoint to determine what the other sees.

This is what happens in the train experiment, the embankment observer uses what he sees, in respect to the embankment frame, to figure out what the train observer sees in his.
You have a very good understanding of SR.

What do you mean by:
1) What he does insist is that "local events" must be agreed upon between the two frames.
2) For things to be otherwise would violate the principle that the laws of physics are constant across frames. ( Does this mean to you that LT must preserve the that the spherical light wave (SLW) propagatges away from the light emission point in the frame?) In other words, must LT map the SLW such that it propagates away from the primed origin?

17. Here, this makes the problem simpler.

Let r be the distance between M and A.

When M and M' are co-located, A emits light according to the experiment.

A---------r---------M
--------------------M'

Normally, under SR, we have moving coordinates. The T/E experiment though has a moving light source. that is something completely different.

So, the light postulate says an observer will measure c regardless of the motion of the light source which is A.

Now, that is easy for the M frame, it is just r/c.

Does M' move away from the traveling light in its view? No, that is not SR, an observer is stationary in its own view and it makes no difference whether the light source was moving or not, the distance to measure c is the distancxe to the light emission point in the frame to the "stationary" observer.

So, since that distance is r in the E frame, it is length contracted in the T frame.

Hence, the distance from the light emission point to the M' observer is r/γ.

Since by the light postulate, light must measure c, therefore, t' = r/(cγ).

Does anyone dispute this before I move onto B?

18. Originally Posted by chinglu
Here, this makes the problem simpler.

Let r be the distance between M and A.

When M and M' are co-located, A emits light according to the experiment.

A---------r---------M
--------------------M'
See, you're wrong right there. It is not the case that in every frame that A emits light when M and M' are co-located.

19. Originally Posted by PhysBang
Originally Posted by chinglu
Here, this makes the problem simpler.

Let r be the distance between M and A.

When M and M' are co-located, A emits light according to the experiment.

A---------r---------M
--------------------M'
See, you're wrong right there. It is not the case that in every frame that A emits light when M and M' are co-located.
Really?

Let's see, under standard SR, when the origins are the same a light pulse is emitted.

Are you now claiming that one frame claims light has been emitted at some place whereas another frame claims light has not been emitted at that place? Can you prove this under SR?

That also falsifies any possible agreed upon light emission when the origins are the same based on your conclusions. That is not SR, so your view fails.

20. Originally Posted by chinglu
Let's see, under standard SR, when the origins are the same a light pulse is emitted.
This can happen if the light is emitted from the origin, since that is the location of a particular event. As soon as two events are spatially separated, theyare not simultaneous in all frames.

In the train example, events A and B are spatially separated from each other and from any point that might be thought of as the spatial mid-point between their spatial locations, thus these events and events at such midpoints will not always be simultaneous.
Are you now claiming that one frame claims light has been emitted at some place whereas another frame claims light has not been emitted at that place? Can you prove this under SR?
What do you mean by place? Places have different coordinates in different frames. But I was specifically talking of different times. Additionally, we must remember that in the train example, A and B are events, not locations, so we have to be careful when we use them.

21. Originally Posted by chinglu
Originally Posted by PhysBang
Originally Posted by chinglu
Here, this makes the problem simpler.

Let r be the distance between M and A.

When M and M' are co-located, A emits light according to the experiment.

A---------r---------M
--------------------M'
See, you're wrong right there. It is not the case that in every frame that A emits light when M and M' are co-located.
Really?

Let's see, under standard SR, when the origins are the same a light pulse is emitted.

When the origins are the same. M and M' are not the origins.

If we redraw the picture labeling the ends of both lines thusly:

A---------r----------M
A'-------------------M'

Then the light originates when A and A' are next to each other. Now it looks like when A and A' are next to each other, so is M and M'. However, you have not stated yet the frame from which this picture is taken from. Is it the frame in which M and A are at rest or is it the frame in which M' and A' are at rest.

If it is the former, then you must remember that according to A and M, the Distance between A' and M' is a length contracted one.( in the rest frame of A' and M' the distance between A' and M' is longer than that measured by M and A. or put another way, if A-M and A'-M' were to be brought to a rest with respect to each other, A'-M' would be longer than A-M.)

Conversely, A'-M' will see A-M as length contracted, and since the proper length of A'-M' is already greater than the proper length of A-M, the measured length of A-M will be even shorter than A'-M' as measured from the rest frame of A'-M'.

And might look something like this in the frame of A'-M' ( roughly the case if the relative velocity of the frames to each other is 0.866c)

A----r----M
A'-------------------------------------M'

So when A and A' are at the same position, M and M' are nowhere near each other.

Since both M and M' must agree that the light originated form A'A when A and A were adjacent, they cannot agree that M and M' were adjacent when the light originated. It is true for M but not for M'

Of course if the first image was taken in the rest frame of A'-M' rather than A-M, then it will be M' that says the light originated when M and M' were adjacent and M that say they were not.

22. Originally Posted by PhysBang
Originally Posted by chinglu
Let's see, under standard SR, when the origins are the same a light pulse is emitted.
This can happen if the light is emitted from the origin, since that is the location of a particular event. As soon as two events are spatially separated, theyare not simultaneous in all frames.

In the train example, events A and B are spatially separated from each other and from any point that might be thought of as the spatial mid-point between their spatial locations, thus these events and events at such midpoints will not always be simultaneous.
We are just looking at A here.

As soon as two events are spatially separated, theyare not simultaneous in all frames.

Your statement claims it is logically impossible for frames then to agree that LT is correct since and non origin coordinate will not agree on the common light pulse since they are spatially separated. Think that through.

Are you now claiming that one frame claims light has been emitted at some place whereas another frame claims light has not been emitted at that place? Can you prove this under SR?
What do you mean by place? Places have different coordinates in different frames. But I was specifically talking of different times. Additionally, we must remember that in the train example, A and B are events, not locations, so we have to be careful when we use them.

Let me be specific, if light is emitted at A, then there exists a corresponding observer A' at that location when light emits for A. Now, are you claiming A sees the light but A' does not?

If this is what you require for your argument to succeed, then you have failed.[/quote]

23. Originally Posted by Janus
Originally Posted by chinglu
Originally Posted by PhysBang
Originally Posted by chinglu
Here, this makes the problem simpler.

Let r be the distance between M and A.

When M and M' are co-located, A emits light according to the experiment.

A---------r---------M
--------------------M'
See, you're wrong right there. It is not the case that in every frame that A emits light when M and M' are co-located.
Really?

Let's see, under standard SR, when the origins are the same a light pulse is emitted.

When the origins are the same. M and M' are not the origins.

If we redraw the picture labeling the ends of both lines thusly:

A---------r----------M
A'-------------------M'

Then the light originates when A and A' are next to each other. Now it looks like when A and A' are next to each other, so is M and M'. However, you have not stated yet the frame from which this picture is taken from. Is it the frame in which M and A are at rest or is it the frame in which M' and A' are at rest.

If it is the former, then you must remember that according to A and M, the Distance between A' and M' is a length contracted one.( in the rest frame of A' and M' the distance between A' and M' is longer than that measured by M and A. or put another way, if A-M and A'-M' were to be brought to a rest with respect to each other, A'-M' would be longer than A-M.)

Conversely, A'-M' will see A-M as length contracted, and since the proper length of A'-M' is already greater than the proper length of A-M, the measured length of A-M will be even shorter than A'-M' as measured from the rest frame of A'-M'.

And might look something like this in the frame of A'-M' ( roughly the case if the relative velocity of the frames to each other is 0.866c)

A----r----M
A'-------------------------------------M'

So when A and A' are at the same position, M and M' are nowhere near each other.

Since both M and M' must agree that the light originated form A'A when A and A were adjacent, they cannot agree that M and M' were adjacent when the light originated. It is true for M but not for M'

Of course if the first image was taken in the rest frame of A'-M' rather than A-M, then it will be M' that says the light originated when M and M' were adjacent and M that say they were not.
First, this is all confined to having A, M and B stationary i.e. they are in the same frame, since that is the problem.

Next, I have posted twice about the length contraction, so I have already proven this part and I am glad you understand that.

Finally, length contraction does not change any conclusion about the emission of the light pulse.

If r is the distance A-M, then to translate the light pulse to the same place at light emission for both frames when M and M' are co-located, the primed frame views that distance to the light pulse as r/γ. Yet, even though they measure different distances, it still refers to the same light pulse and same place when the light pulse was emitted. That is SR.

Now, let's ask what the 2 frames conclude for this simple experiment.

M Conclusions
M' moves away from the common light emission at A and t' = r/(γ(c-v)).
t = r/c for light reception.

M' Conclusions
t = r/(γc) for light reception.

Note that M' does not conclude the same as M that it is moving away from the light pulse because it is stationary in its own frame. So, it just sees a moving light source A and it measures the distance to the light emission point in its frame to the target M'. It does not measure from the moving light souce A to M' when M' receives the light because that is emission light theory. Yet, that is what Einstein claimed in the T/E experiment.

24. Originally Posted by chinglu
We are just looking at A here.
No, you are also looking at the event where M and M' are adjacent to each other. That's two events.
As soon as two events are spatially separated, theyare not simultaneous in all frames.

Your statement claims it is logically impossible for frames then to agree that LT is correct since and non origin coordinate will not agree on the common light pulse since they are spatially separated. Think that through.
That has nothing to do with what I said. Different frames may disagree on the time that light was emitted from a location. The use of the Lorentz transformation ensures that all frames will agree on what events occur.
What question? Again you strung together English words in a way that didn't have a clear meaning.
Let me be specific, if light is emitted at A, then there exists a corresponding observer A' at that location when light emits for A. Now, are you claiming A sees the light but A' does not?
A is an event, not a location. Perhaps you mean to talk about the spatial location of A in the frame of the embankment and the spatial location of A in the frame of the train. (By "frame of", I mean the frame in which the named object is supposed to be at rest.) The event A happens in all frames. Frames do not depend on observers, they are mathematical frameworks to describe the locations of events.

Again, you seem not to understand how frames are required to describe events and how they are used to describe events. You seem to assume that events that are simultaneous in one frame are simultaneous in all frames.

25. Originally Posted by PhysBang
Originally Posted by chinglu
We are just looking at A here.
No, you are also looking at the event where M and M' are adjacent to each other. That's two events.
There is one light emission in SR and also under LT, two observers are co-located with the light at light reception.

There are 2 events all day long in SR, light emission and some light reception at some coordinate.

You are trying to make magic with 2 events where is is none.

Here is where you are confused. You are confusing events of receiving light at 2 events that will not be simultaneous to 2 frames to light emission and light reception that are judged simultanous events between 2 frames.

If this is false, then LT is false.

In oterh words, if the frames cannot agree light was emitted when the 2 frames orings are co-located, then SR is false. So, there is no provision that light is at some place and when mapped, light is not at that place for another frame. That is a contradiction.

So, LT is only valid if and only if the frames agree on a common light emission.

A is an event, not a location. Perhaps you mean to talk about the spatial location of A in the frame of the embankment and the spatial location of A in the frame of the train. (By "frame of", I mean the frame in which the named object is supposed to be at rest.) The event A happens in all frames. Frames do not depend on observers, they are mathematical frameworks to describe the locations of events.

Again, you seem not to understand how frames are required to describe events and how they are used to describe events. You seem to assume that events that are simultaneous in one frame are simultaneous in all frames.
A is not an event. it is a light emission and that is not normally taken as an event. A event is taken as a light reception.

If it is the case that a light emission cannot be agreed upon by all frames as a common light emission at the origins, then prove how LT applies.

That is where you have an error in your thinking. Light emission is always considered a simultaneous event between all frames as traceable to the origins of the frames when they werw at the same place i.e. t = t' = 0, x' = x = 0.

26. Originally Posted by chinglu
Originally Posted by PhysBang
Originally Posted by chinglu
We are just looking at A here.
No, you are also looking at the event where M and M' are adjacent to each other. That's two events.
There is one light emission in SR and also under LT, two observers are co-located with the light at light reception.

There are 2 events all day long in SR, light emission and some light reception at some coordinate.

You are trying to make magic with 2 events where is is none.
There are two events that you were discussing: the event A where light left a location and a separate event where M and M' were adjacent. These are two separate events regardless of the system of physics that we use. If you cannot understand that, then you can't understand anything.
Here is where you are confused. You are confusing events of receiving light at 2 events that will not be simultaneous to 2 frames to light emission and light reception that are judged simultanous events between 2 frames.
I haven't even begun to consider when light will be received. Did you not read the very experiment that you are referencing? The light in this experiment is released at events A and B regardless of when we consider them to occur and regardless of where we consider M or M' to be at any time.
If this is false, then LT is false.
In oterh words, if the frames cannot agree light was emitted when the 2 frames orings are co-located, then SR is false. So, there is no provision that light is at some place and when mapped, light is not at that place for another frame. That is a contradiction.[/quote]
The things that you say here make no sense. What do you mean by "place"? If you mean coordinate location, then frames do not assign the same coordinate locations to events. If you mean the place where an event happened, this place is given a different coordinate location in different systems of coordinates and these coordinates do not always match at times other than the time when the event occurs. If an event happens in a frame, it happens in every frame. If an event doesn't happen in a well-constructed frame, it doesn't happen in any frame.
A is not an event. it is a light emission and that is not normally taken as an event. A event is taken as a light reception.
Again, you seem not to have actually read the example you wanted to discuss. Esintein writes, "Are two events (e.g. the two strokes of lightning A and B) which are simultaneous with reference to the railway embankment also simultaneous relatively to the train?" A and B are events, events where light is not received but where light is emitted.

If you actually read the example, perhaps you might understand it.
If it is the case that a light emission cannot be agreed upon by all frames as a common light emission at the origins, then prove how LT applies.
If we use coordinates, we can assign any coordinates to an event where light is emitted. We do not have to use the origin.
That is where you have an error in your thinking. Light emission is always considered a simultaneous event between all frames as traceable to the origins of the frames when they werw at the same place i.e. t = t' = 0, x' = x = 0.
That is simply insanity. Read the example you wanted to discuss: if the event of light emission, A, is at the origin, then clearly the event of light emission B cannot also be at the origin because B happens at a distance from A.

27. Originally Posted by PhysBang
There are two events that you were discussing: the event A where light left a location and a separate event where M and M' were adjacent. These are two separate events regardless of the system of physics that we use. If you cannot understand that, then you can't understand anything.
So what? The light emission activity is a simultaneous event between the frames.

I haven't even begun to consider when light will be received. Did you not read the very experiment that you are referencing? The light in this experiment is released at events A and B regardless of when we consider them to occur and regardless of where we consider M or M' to be at any time.
Yes, but you are confused that light emission is a simultaneous event for all frames. Prove otherwise.

This is the whole key for your understanding of the T/E.

You need the frames to disagree that light emission is not a simultaneous event for all frames.

But, if that is so, then it is impossible for the frames to agree the light emits when the frame origins are common for standard SR logic in the standard configuration.

you are in a bind not me.

I simply accept SR. All light emissions can be agreed ipon with all frames as a simultaneous event between the frames i.e. t'=t=0, x'=x=0

Now, either refite this or start your training process and learn something.

28. Originally Posted by chinglu
So what? The light emission activity is a simultaneous event between the frames.
You were calling two different events one single event.

Still, it is worth pointing out to you that when A and B happen is not guaranteed to be simultaneous with each other and these events are not guaranteed to be simultaneous with every event.

Yes, but you are confused that light emission is a simultaneous event for all frames. Prove otherwise.
Hunh? Again, when A and B happen is not guaranteed to be simultaneous with each other and these events are not guaranteed to be simultaneous with every event. I don't know what you mean by, "light emission is a simultaneous event for all frames."
You need the frames to disagree that light emission is not a simultaneous event for all frames.

But, if that is so, then it is impossible for the frames to agree the light emits when the frame origins are common for standard SR logic in the standard configuration.
You might want to actually show this rather than make claims that something studied for 106 years is inconsistent.
I simply accept SR. All light emissions can be agreed ipon with all frames as a simultaneous event between the frames i.e. t'=t=0, x'=x=0
This doesn't make sense in English.
Now, either refite this or start your training process and learn something.
While I have taught SR, I don't think I'm ready to retire yet.

Do you have any explanation for your mistakes about the example you brought up? Any explanation about what you mean by "place"?

29. Originally Posted by PhysBang
There are two events that you were discussing: the event A where light left a location and a separate event where M and M' were adjacent. These are two separate events regardless of the system of physics that we use. If you cannot understand that, then you can't understand anything.
You might try to explain to chinglu that an "event" is nothing more and nothing less than a point in spacetime. So there is nothing magic about an event that is identified with emission or reception of light.

I will try to explain this to my Labrador puppy. I bet I have more success than do you.

June 3 update: I have discussed spacetime with Dexter (the puppy). He is not ready to do computations with the Minkowski metric, but he has not said anything that is incredibly obtuse. I think, therefore, that I am ahead.

30. Originally Posted by DrRocket

I will try to explain this to my Labrador puppy. I bet I have more success than do you.

June 3 update: I have discussed spacetime with Dexter (the puppy). He is not ready to do computations with the Minkowski metric, but he has not said anything that is incredibly obtuse. I think, therefore, that I am ahead.
Awww.. you have a Labrador pup? Congratulations! As you know, I have a female black Lab, and as you have said in the past, there are dogs, and then there are Labradors!

31. Originally Posted by PhysBang
Originally Posted by chinglu
So what? The light emission activity is a simultaneous event between the frames.
You were calling two different events one single event.

Still, it is worth pointing out to you that when A and B happen is not guaranteed to be simultaneous with each other and these events are not guaranteed to be simultaneous with every event.
This 1st thing we must get straightened out is that a light emission event is a simultaneous event between the frames or the origins motivation for Einstein's mirror experiment fails.

So, you must at least make it to here. You have been claiming a light emission event is not a simultaneous event which under SR is logically absurd.

However, if you make it to here, the outcome is already logically decidable and SR falls.

You better break it right here.
[/quote]

32. Originally Posted by DrRocket
Originally Posted by PhysBang
There are two events that you were discussing: the event A where light left a location and a separate event where M and M' were adjacent. These are two separate events regardless of the system of physics that we use. If you cannot understand that, then you can't understand anything.
You might try to explain to chinglu that an "event" is nothing more and nothing less than a point in spacetime. So there is nothing magic about an event that is identified with emission or reception of light.

I will try to explain this to my Labrador puppy. I bet I have more success than do you.

June 3 update: I have discussed spacetime with Dexter (the puppy). He is not ready to do computations with the Minkowski metric, but he has not said anything that is incredibly obtuse. I think, therefore, that I am ahead.
Good, you are so very smart.

So, is A a simultaneous event between the frames?

That means x'=x=0 and t'=t=0.

33. Originally Posted by chinglu
Good, you are so very smart.

So, is A a simultaneous event between the frames?

That means x'=x=0 and t'=t=0.

This proves quite clearly that you are irrational. There is no such thing as "a simultaneous event between the frames". A single event cannot be "simultaneous. Two different events can be simultaneous, but "simultaneity" is a relationship among events and is meaningless when applied to a single event.

One can choose the origin of the primed and unprimed frames arbitrarily, so your condition "x'=x=0 and t'=t=0" is as devoid of content as is your questing.

Physbang: With this incredibly stupid question (whoever said "There is no such thing as a stupid question" never met chinglu) it is clear that I win. Dexter has not uttered a single stupid statement. Moreover, while Dexter and chinglu may both be chasing their tail, Dexter occasionally catches his.

34. Originally Posted by chinglu

Very good.

Now, both frames are equidistant to the flashes A and B when they occur, is that correct or not correct?
I want to go back to this. The answer is no. The observers in both frames do NOT agree as to how far away they were from points A and B WHEN the flashes occurred, because they don't agree as to when they occurred. Because the train is in motion, the observer on the train occupies different positions at different times, and therefore observes events to be different distances away if they happen at different times.

The observer on the train thinks event B happened when they were V(train)*D/(C+V(train)) closer to B than the distance observed by the observer on the embankment. They think A happened when they were V(train)*D/(C-V(train)) further away from event A than the distance observed by the observer on the embankment.

Why? Because the position of the observer on the train relative to events A and B changed by those amounts before the light signal could reach them.

35. Chinglu, here is a simple question for you.

Do you SERIOUSLY believe that everyone has been doing SR wrong for the past 100 years? Do you believe yhat you have found a problem with the relativity of simultaneity that everyone else in the last century has somehow overlooked?

If you are interesting in learning how to use SR properly, why are you arguing, instead of trying to learn how to use SR properly?

36. Originally Posted by chinglu
So, is A a simultaneous event between the frames?

That means x'=x=0 and t'=t=0.
Yeah, this particular statement seems to favor the puppy over chinglu.

37. Originally Posted by DrRocket
Originally Posted by chinglu
Good, you are so very smart.

So, is A a simultaneous event between the frames?

That means x'=x=0 and t'=t=0.

This proves quite clearly that you are irrational. There is no such thing as "a simultaneous event between the frames". A single event cannot be "simultaneous. Two different events can be simultaneous, but "simultaneity" is a relationship among events and is meaningless when applied to a single event.

One can choose the origin of the primed and unprimed frames arbitrarily, so your condition "x'=x=0 and t'=t=0" is as devoid of content as is your questing.

Physbang: With this incredibly stupid question (whoever said "There is no such thing as a stupid question" never met chinglu) it is clear that I win. Dexter has not uttered a single stupid statement. Moreover, while Dexter and chinglu may both be chasing their tail, Dexter occasionally catches his.

Let's see, I hereby quote Einstein.

At the time t=t'=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, is the light emission a simultaneous event between the frames, yes or no.

Or, does one frame think, light emitted at a place and another frame claims light is not at that place with co-located observers for the light emission.

You border on the absurd and I am being nice.

38. Originally Posted by SpeedFreek
Chinglu, here is a simple question for you.

Do you SERIOUSLY believe that everyone has been doing SR wrong for the past 100 years? Do you believe yhat you have found a problem with the relativity of simultaneity that everyone else in the last century has somehow overlooked?

If you are interesting in learning how to use SR properly, why are you arguing, instead of trying to learn how to use SR properly?
Yes, here is a very simple proof.

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

Now, I put forth the proof.

You must counter with math proof.

Good luck!

I will only discuss math from this point on.

39. Originally Posted by chinglu
Now, is the light emission a simultaneous event between the frames, yes or no.
What is the difference between a duck ?

40. Originally Posted by DrRocket
Originally Posted by chinglu
Now, is the light emission a simultaneous event between the frames, yes or no.
What is the difference between a duck ?
Let's see, I hereby quote Einstein.

At the time t=t'=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, is the light emission a simultaneous event between the frames, yes or no.

Or, does one frame think, light emitted at a place and another frame claims light is not at that place with co-located observers for the light emission.

You border on the absurd and I am being nice.

41. Originally Posted by DrRocket
Originally Posted by chinglu
Now, is the light emission a simultaneous event between the frames, yes or no.
What is the difference between a duck ?
You claimed to be a phd in math.

I presented a math proof that should not be beyond you, yet, you crap about these things.

I would attack the math but that is the difference between our educations. You don't.

42. Originally Posted by chinglu
Let's see, I hereby quote Einstein.

At the time t=t'=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/
Because one can construct systems of coordinates almost completely arbitrarily, it is possible to construct ones where the same event is given the same coordinates in two different frames. It is pretty stupid to then assume that one kind of event must always be assigned the same coordinate in every example once it is assigned a coordinate in one example.
Now, is the light emission a simultaneous event between the frames, yes or no.
This makes no sense. There is no simultaneity between frames. Simultaneity is something that happens within frames.
Or, does one frame think, light emitted at a place and another frame claims light is not at that place with co-located observers for the light emission.
This doesn't even make sense. What do you mean?
You border on the absurd and I am being nice.
You're being inane.

43. Originally Posted by chinglu
Let's see, I hereby quote Einstein.

At the time t=t'=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, is the light emission a simultaneous event between the frames, yes or no.
How can a single event be "simultaneous between the frames"? Your question makes no sense, but you don't understand why.

44. Originally Posted by chinglu
Originally Posted by DrRocket
Originally Posted by chinglu
Now, is the light emission a simultaneous event between the frames, yes or no.
What is the difference between a duck ?
You claimed to be a phd in math.

I presented a math proof that should not be beyond you, yet, you crap about these things.

I would attack the math but that is the difference between our educations. You don't.
I suspect that even I can see the point he's trying to make. You're asking a question that's logically analogous to DrRocket's duck question and just as nonsensical. If I'm understanding correctly, an event happens in only one frame. Asking if a single event has different properties in two frames is nonsensical. It's logically analagous to asking the difference between a single object.

Am I getting it right DrRocket?

45. Originally Posted by TheBiologista
Originally Posted by chinglu
Originally Posted by DrRocket
Originally Posted by chinglu
Now, is the light emission a simultaneous event between the frames, yes or no.
What is the difference between a duck ?
You claimed to be a phd in math.

I presented a math proof that should not be beyond you, yet, you crap about these things.

I would attack the math but that is the difference between our educations. You don't.
I suspect that even I can see the point he's trying to make. You're asking a question that's logically analogous to DrRocket's duck question and just as nonsensical. If I'm understanding correctly, an event happens in only one frame. Asking if a single event has different properties in two frames is nonsensical. It's logically analagous to asking the difference between a single object.

Am I getting it right DrRocket?
First, the proof forces SR to operate over an interval.

In all cases in the mainstream, SR only operates light beam by light beam or vector by vector.

However, when SR is forced to operate as in the real universe since it is a theory about electrodynamics, it has several failures.

Next, LT mappings and the light emission point are all "simultaneous" events between the frames.

For the light emission point, both frames agree the light is at the common location.

For LT mappings, the frames may disagree on the space coordinates and the timee coordinate, but light, and both space coordinates are co-located if LT maps as such for light like space time intervals.

Hence, a single event is just that, agreed upon by the frames which means the light emission is a single event for the frames. That breaks ROS.

46. Originally Posted by TheBiologista
I suspect that even I can see the point he's trying to make. You're asking a question that's logically analogous to DrRocket's duck question and just as nonsensical. If I'm understanding correctly, an event happens in only one frame. Asking if a single event has different properties in two frames is nonsensical. It's logically analagous to asking the difference between a single object.

Am I getting it right DrRocket?
Not quite. All events happen in all frames. Frames are just ways of assigning coordinates to events. It's possible to assign the same time coordinate to the same event in different frames, but it doesn't make the frames simultaneous.

Simuntaneity is something that happens within a frame. Once we have a frame, we can ask if two different events are simultaneous; that is, we can ask if the two different events have the same time coordinate.

47. Originally Posted by SpeedFreek
Originally Posted by chinglu
Let's see, I hereby quote Einstein.

At the time t=t'=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, is the light emission a simultaneous event between the frames, yes or no.
How can a single event be "simultaneous between the frames"? Your question makes no sense, but you don't understand why.
I mean, if two frames observers are co-located with light, that becomes a single event or a simultaneous event.

All light emissions are a single event or a simultaneous event.

So, given the light emission at A is a single event between the frames, it cannot be the case that a primed observer can claim light is at its location with light and a co-located unprimed observer claims light is not at that location.

therefore, in the T/E experiment light must travel a distance r for the E frame and r/\gamma for the T frame and light emission is a single event.

But, then when that same logic is applied to B, the same conditions hold. Henc e, both frames with claims the strikes are simultaneous to it and not to the other.

48. Originally Posted by PhysBang
Originally Posted by TheBiologista
I suspect that even I can see the point he's trying to make. You're asking a question that's logically analogous to DrRocket's duck question and just as nonsensical. If I'm understanding correctly, an event happens in only one frame. Asking if a single event has different properties in two frames is nonsensical. It's logically analagous to asking the difference between a single object.

Am I getting it right DrRocket?
Not quite. All events happen in all frames. Frames are just ways of assigning coordinates to events. It's possible to assign the same time coordinate to the same event in different frames, but it doesn't make the frames simultaneous.

Simuntaneity is something that happens within a frame. Once we have a frame, we can ask if two different events are simultaneous; that is, we can ask if the two different events have the same time coordinate.
I am not saying time is the same for the frames.

I am saying light emission is a single event for the frames. Under SR, that assigment is as follows:

x'=x=0, t'=t=0.

You think I believe simultaneous means they have the same time on their respective clocks. I know how to do that also, but that is not this case.

So, the word simultaneous means a single event in your terms.

49. Originally Posted by PhysBang
Originally Posted by TheBiologista
I suspect that even I can see the point he's trying to make. You're asking a question that's logically analogous to DrRocket's duck question and just as nonsensical. If I'm understanding correctly, an event happens in only one frame. Asking if a single event has different properties in two frames is nonsensical. It's logically analagous to asking the difference between a single object.

Am I getting it right DrRocket?
Not quite. All events happen in all frames. Frames are just ways of assigning coordinates to events. It's possible to assign the same time coordinate to the same event in different frames, but it doesn't make the frames simultaneous.

Simuntaneity is something that happens within a frame. Once we have a frame, we can ask if two different events are simultaneous; that is, we can ask if the two different events have the same time coordinate.
Well that's somewhat clearer- though I grasped DrRocket's broad point even without the subtle understanding of the subject matter. For some reason Chinglu assumed it was nonsense.

50. Originally Posted by kojax
Originally Posted by chinglu

Very good.

Now, both frames are equidistant to the flashes A and B when they occur, is that correct or not correct?
I want to go back to this. The answer is no. The observers in both frames do NOT agree as to how far away they were from points A and B WHEN the flashes occurred, because they don't agree as to when they occurred. Because the train is in motion, the observer on the train occupies different positions at different times, and therefore observes events to be different distances away if they happen at different times.

The observer on the train thinks event B happened when they were V(train)*D/(C+V(train)) closer to B than the distance observed by the observer on the embankment. They think A happened when they were V(train)*D/(C-V(train)) further away from event A than the distance observed by the observer on the embankment.

Why? Because the position of the observer on the train relative to events A and B changed by those amounts before the light signal could reach them.
Here is the logic you violate under SR. The light postulate clearly states the frame receiver will measure c regardless of the motion of the light source, this case the "light sources" are A and B. That means if the light emission point in the frame is d away, then t = d/c. Now, if A happens to be moving away fron the observer after the light emission, the case of the T observer, that does not matter. The distance for ligt to travel is still d and it is still the case t=d/c.

That is the light postulate.

Keep in mind, A and B are simply moving light sources in the view of T and that makes absolutely no difference to T in its calculations. The answer would have been the same if A and B were stationary to T except for length contraction.

So, you cannot apply a (c+v) and (c-v) argument to any stationary observer regardless of the motions of the light sources.

When stationary, it is always c, period, under SR.

51. Originally Posted by TheBiologista
Originally Posted by PhysBang
Originally Posted by TheBiologista
I suspect that even I can see the point he's trying to make. You're asking a question that's logically analogous to DrRocket's duck question and just as nonsensical. If I'm understanding correctly, an event happens in only one frame. Asking if a single event has different properties in two frames is nonsensical. It's logically analagous to asking the difference between a single object.

Am I getting it right DrRocket?
Not quite. All events happen in all frames. Frames are just ways of assigning coordinates to events. It's possible to assign the same time coordinate to the same event in different frames, but it doesn't make the frames simultaneous.

Simuntaneity is something that happens within a frame. Once we have a frame, we can ask if two different events are simultaneous; that is, we can ask if the two different events have the same time coordinate.
Well that's somewhat clearer- though I grasped DrRocket's broad point even without the subtle understanding of the subject matter. For some reason Chinglu assumed it was nonsense.
Nope, I did not assume it is nonsense, I know it is.

Further, He is supposed to be a PHD in math. He should be able to follow my proof.

If he does follow the proof, he will see SR contradicts its own light postulate.

But, for ROS, light emission is one event for all frames. That means if light is emitted at some location, then all frame observers at that same location will agree light is there. This is a trivial conclusion.

Now, note I am not talking about two different events when discussing light emission.

52. Originally Posted by PhysBang
Originally Posted by TheBiologista
I suspect that even I can see the point he's trying to make. You're asking a question that's logically analogous to DrRocket's duck question and just as nonsensical. If I'm understanding correctly, an event happens in only one frame. Asking if a single event has different properties in two frames is nonsensical. It's logically analagous to asking the difference between a single object.

Am I getting it right DrRocket?
Not quite. All events happen in all frames. Frames are just ways of assigning coordinates to events. It's possible to assign the same time coordinate to the same event in different frames, but it doesn't make the frames simultaneous.

Simuntaneity is something that happens within a frame. Once we have a frame, we can ask if two different events are simultaneous; that is, we can ask if the two different events have the same time coordinate.
I am able to create simultaneity between frames. This is not it though and I am not discussing that.

there eixts a line of simultaneity between all frames such that t'=t since the origins were common and the light was emitted.

53. Originally Posted by chinglu
I am not saying time is the same for the frames.

I am saying light emission is a single event for the frames. Under SR, that assigment is as follows:

x'=x=0, t'=t=0.
It could be, but it doesn't have to be. So what?

It's a serious problem that you do not understand how to represent events in frames and how to translate between them.
You think I believe simultaneous means they have the same time on their respective clocks. I know how to do that also, but that is not this case.

So, the word simultaneous means a single event in your terms.
No. As I said before, two events are simultaneous in a frame if they have the same time coordinate assigned to them in that frame.

54. Originally Posted by chinglu
I am able to create simultaneity between frames. This is not it though and I am not discussing that.
It's like you're saying that you can make green red and red green.
there eixts a line of simultaneity between all frames such that t'=t since the origins were common and the light was emitted.
You're free to do that in your own, special theory, but not in special relativity. The conclusion to the very thought experiment that you brought up is: " Events which are simultaneous with reference to the embankment are not simultaneous with respect to the train, and vice versa (relativity of simultaneity). Every reference-body (co-ordinate system) has its own particular time; unless we are told the reference-body to which the statement of time refers, there is no meaning in a statement of the time of an event."

You really need to read these things that you reference.

55. Originally Posted by PhysBang
Originally Posted by chinglu
I am not saying time is the same for the frames.

I am saying light emission is a single event for the frames. Under SR, that assigment is as follows:

x'=x=0, t'=t=0.
It could be, but it doesn't have to be. So what?

It's a serious problem that you do not understand how to represent events in frames and how to translate between them.
You think I believe simultaneous means they have the same time on their respective clocks. I know how to do that also, but that is not this case.

So, the word simultaneous means a single event in your terms.
No. As I said before, two events are simultaneous in a frame if they have the same time coordinate assigned to them in that frame.
Are you able to discuss the light emission point since that is the topic? Why does that confuse you?

56. Originally Posted by PhysBang
Originally Posted by chinglu
I am able to create simultaneity between frames. This is not it though and I am not discussing that.
It's like you're saying that you can make green red and red green.
there eixts a line of simultaneity between all frames such that t'=t since the origins were common and the light was emitted.
You're free to do that in your own, special theory, but not in special relativity. The conclusion to the very thought experiment that you brought up is: " Events which are simultaneous with reference to the embankment are not simultaneous with respect to the train, and vice versa (relativity of simultaneity). Every reference-body (co-ordinate system) has its own particular time; unless we are told the reference-body to which the statement of time refers, there is no meaning in a statement of the time of an event."

You really need to read these things that you reference.
I will show you that some other time.

Let's stay on task of light emission.

I changed my mind, here it is.

Set
x = vtγ /( 1 + γ ) and z = 0.

LT
t' = ( t - vx/c² )γ
t' = ( t - v((vtγ/(1+γ) /c² )γ

t' = ( t - tv²γ/(c²(1+γ)) )γ
t' = t( γ - v²γ²/(c²(1+γ)) )
t' = t( (γ(c²(1+γ))/(c²(1+γ) - v²γ²/(c²(1+γ)) )
t' = t(( (c²γ²- v²γ²+c²γ))/(c²(1+γ)))
t' = t(( (c²+c²γ))/(c²(1+γ)))
t' = t(( c²(1+γ))/(c²(1+γ)))
t' = t

57. Originally Posted by DrRocket
Originally Posted by chinglu
Good, you are so very smart.

So, is A a simultaneous event between the frames?

That means x'=x=0 and t'=t=0.

This proves quite clearly that you are irrational. There is no such thing as "a simultaneous event between the frames". A single event cannot be "simultaneous. Two different events can be simultaneous, but "simultaneity" is a relationship among events and is meaningless when applied to a single event.

One can choose the origin of the primed and unprimed frames arbitrarily, so your condition "x'=x=0 and t'=t=0" is as devoid of content as is your questing.

Physbang: With this incredibly stupid question (whoever said "There is no such thing as a stupid question" never met chinglu) it is clear that I win. Dexter has not uttered a single stupid statement. Moreover, while Dexter and chinglu may both be chasing their tail, Dexter occasionally catches his.
OK, I saw in the math threads, you do have the concept of a topological space down.

So, you do have a higher level of math education.

Yet, you could not stop my missing SR vector theorem.

And, you will not be able to stop this one either with dt'/dt<0.

Why don't you confess since that is what math folks do.

58. Originally Posted by chinglu
Let's stay on task of light emission.
We can't even begin, since you keep saying things that are nonsense. You began with the claim that an event is simultaneous accross two frames. This makes no sense.

Until you can make sense, you can't discuss anything. If you want to assign coordinates to a particular event, go ahead. Try to assign coordinates to all the relevant events.

59. Originally Posted by PhysBang
Originally Posted by chinglu
Let's stay on task of light emission.
We can't even begin, since you keep saying things that are nonsense. You began with the claim that an event is simultaneous accross two frames. This makes no sense.

Until you can make sense, you can't discuss anything. If you want to assign coordinates to a particular event, go ahead. Try to assign coordinates to all the relevant events.
Here it is very simple.

Is the light emission one simultaneous event between the frames? Or, does one frame conclude light emitted with a co-located observer of another frame at the same location claiming light was not emitted. That is the issue.

60. Originally Posted by chinglu
Here it is very simple.

Is the light emission one simultaneous event between the frames?
Events aren't "between" frames. They are in every frame. Events can only be considered simultaneous to other events that have coordinate locations within a given frame. One cannot ever compare simultaneity between frames. It's like asking if Peter Pan met Dracula.
Or, does one frame conclude light emitted with a co-located observer of another frame at the same location claiming light was not emitted. That is the issue.
What is that supposed to mean?
I changed my mind, here it is.

Set
x = vtγ /( 1 + γ ) and z = 0.

LT
t' = ( t - vx/c² )γ
t' = ( t - v((vtγ/(1+γ) /c² )γ

t' = ( t - tv²γ/(c²(1+γ)) )γ
t' = t( γ - v²γ²/(c²(1+γ)) )
t' = t( (γ(c²(1+γ))/(c²(1+γ) - v²γ²/(c²(1+γ)) )
t' = t(( (c²γ²- v²γ²+c²γ))/(c²(1+γ)))
t' = t(( (c²+c²γ))/(c²(1+γ)))
t' = t(( c²(1+γ))/(c²(1+γ)))
t' = t
None of that makes you any less crazy for saying things like, "Is the light emission one simultaneous event between the frames?"

If one restricts consideration of the events in a frame to x = vtγ /( 1 + γ ) and z = 0 as you are doing, then one is speaking only of a very restricted space in a given frame. Assuming that you've done the algebra right, you have found some locations in a frame where the time coordinate assigned to those events happen to be the same coordinate in another frame. This does not mean that this will be the case for any locations other than those in the particular region of the frame that you have identified. It may be that the only region in space where your conditions hold is x=y=z=t=0.

Regardless, your algebra does not make your nonsense have any more sense.

61. Originally Posted by PhysBang
None of that makes you any less crazy for saying things like, "Is the light emission one simultaneous event between the frames?"

...

Regardless, your algebra does not make your nonsense have any more sense.
Dexter has still not said anything stupid.

62. Originally Posted by chinglu
Yes, here is a very simple proof.

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

Now, I put forth the proof.

You must counter with math proof.

Good luck!

I will only discuss math from this point on.
Your "proof" is bullshit and what you claim to prove is false, as (almost) any idiot can see:

For reference frames in standard configuration, as you propose

So,

QED

63. Originally Posted by DrRocket
Originally Posted by chinglu
Yes, here is a very simple proof.

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

Now, I put forth the proof.

You must counter with math proof.

Good luck!

I will only discuss math from this point on.
Your "proof" is bullshit and what you claim to prove is false, as (almost) any idiot can see:

For reference frames in standard configuration, as you propose

So,

QED

You propose x is not functional in t when considering the spherical light wave and its behavior. This is a common error for a beginner.

x is the coordinate for the expanding spherical light wave at some time t (SLW).

You can't take the partial derivative holding x fixed while t increases since x increases as t increases with the expanding SLW.

Therefore, as I showed with the simple easy math,

you must first translate x into an equation functional in t.

I set z = 0 and set y=k fixed to some value.

So, whereas normally x² + y² + z² = c² t²*, hence

x = ± √( c² t² - y² - z² ), we then have with the fixed values of z = 0 and y=k

x = ± √( c² t² - k² ).

I ignore the negative for this exercise so,

x = √( c² t² - k² ).

Now I have x functional in t, hence,

t' = ( t - vx/c² )γ

Substitute the x above,

t' = ( t - v(√( c² t² - k² ))/c² )γ

Now, I can legally take the partial derivative without making a simple minded error.

64. Originally Posted by PhysBang
Originally Posted by chinglu
Here it is very simple.

Is the light emission one simultaneous event between the frames?
Events aren't "between" frames. They are in every frame. Events can only be considered simultaneous to other events that have coordinate locations within a given frame. One cannot ever compare simultaneity between frames. It's like asking if Peter Pan met Dracula.
Or, does one frame conclude light emitted with a co-located observer of another frame at the same location claiming light was not emitted. That is the issue.
What is that supposed to mean?
I changed my mind, here it is.

Set
x = vtγ /( 1 + γ ) and z = 0.

LT
t' = ( t - vx/c² )γ
t' = ( t - v((vtγ/(1+γ) /c² )γ

t' = ( t - tv²γ/(c²(1+γ)) )γ
t' = t( γ - v²γ²/(c²(1+γ)) )
t' = t( (γ(c²(1+γ))/(c²(1+γ) - v²γ²/(c²(1+γ)) )
t' = t(( (c²γ²- v²γ²+c²γ))/(c²(1+γ)))
t' = t(( (c²+c²γ))/(c²(1+γ)))
t' = t(( c²(1+γ))/(c²(1+γ)))
t' = t
None of that makes you any less crazy for saying things like, "Is the light emission one simultaneous event between the frames?"

If one restricts consideration of the events in a frame to x = vtγ /( 1 + γ ) and z = 0 as you are doing, then one is speaking only of a very restricted space in a given frame. Assuming that you've done the algebra right, you have found some locations in a frame where the time coordinate assigned to those events happen to be the same coordinate in another frame. This does not mean that this will be the case for any locations other than those in the particular region of the frame that you have identified. It may be that the only region in space where your conditions hold is x=y=z=t=0.

Regardless, your algebra does not make your nonsense have any more sense.
You are not getting what I am saying. My math above shows light reception simultaneity between the frames.

I am trying to show you that light emission is one event.

Are you claiming one light emission is more than one event between frames?

No, you can't. Therefore, both frames will conclude the strikes are simultaneous and wil conclude the other frame will not see the strikes as simultaneous.

That is simple SR. Simultaneity is relative. This that struggle with SR never realize each frames claims at any instant simultaneity. IE each frame sees a light sphere.

But, each then claims the other does not view the same light sphere i.e. as simultaneous.

But, this requires 2 different light spheres.

65. Originally Posted by chinglu

My logic is just fine, thank you very much.

66. Originally Posted by DrRocket
Originally Posted by chinglu

My logic is just fine, thank you very much.

That's odd, you used personal attackes rather than refuting the correct math I presented.

I think I may know your mental error. You do not understand how to pass from the SLW in the unprimed frame, using LT, to the primed SLW. You will then see, x is functional in t based on the unprimed SLW and LT.

Assume x² + y² + z² = c² t² is true.

*x²/γ² + y²/γ² + z²/γ² = c² t²/γ²
*x²( 1 - v²/c² ) + y²/γ² + z²/γ² = c² t² ( 1 - v²/c² )
*x² - x²v²/c² + y²/γ² + z²/γ² = c² t² - t²v²
*x² + t²v² + y²/γ² + z²/γ² = c² t² + x²v²/c²
*x² + t²v² - 2xvt + y²/γ² + z²/γ² = c² t² + x²v²/c² - 2xvt
*(x² - vt)² + y²/γ² + z²/γ² = c²( t² + x²v²/c^4 - 2xvt/c² )
*( x² - vt)² + y²/γ² + z²/γ² = c²( t² - vx/c² )
*( x² - vt)²γ² + y² + z² = c²( t - vx/c² )²γ²*LT

x'² + y² + z² = c² t'²

standard configuration.

x'² + y'² + z'² = c² t'²

67. Originally Posted by chinglu
You are not getting what I am saying. My math above shows light reception simultaneity between the frames.
No, you're not getting what you're saying. First prove that there exists a point where x = vtγ /( 1 + γ ) and z = 0. Next prove that there is a point other than x=y=z=t=0 that matches this restriction.
I am trying to show you that light emission is one event.
We already know that one event is one event. We do not need your convolutions of algebra to know this.
Are you claiming one light emission is more than one event between frames?
No, I'm claiming that we can't claim that any events are simultaneous between frames.
No, you can't. Therefore, both frames will conclude the strikes are simultaneous and wil conclude the other frame will not see the strikes as simultaneous.
Read the chapter you referenced. You will see clearly that one frame sees the strikes as simultaneous and one does not. There are two different light emission events, A and B. These are two events, not one.

68. Originally Posted by PhysBang
Originally Posted by chinglu
You are not getting what I am saying. My math above shows light reception simultaneity between the frames.
No, you're not getting what you're saying. First prove that there exists a point where x = vtγ /( 1 + γ ) and z = 0. Next prove that there is a point other than x=y=z=t=0 that matches this restriction.
I proved this above.

I am trying to show you that light emission is one event.
We already know that one event is one event. We do not need your convolutions of algebra to know this.
Are you claiming one light emission is more than one event between frames?
No, I'm claiming that we can't claim that any events are simultaneous between frames.
What is it eactly that you think LT tells you? I don't have the patience for this.

No, you can't. Therefore, both frames will conclude the strikes are simultaneous and wil conclude the other frame will not see the strikes as simultaneous.
Read the chapter you referenced. You will see clearly that one frame sees the strikes as simultaneous and one does not. There are two different light emission events, A and B. These are two events, not one.

I never said A and B are the same event.

I am talking about the light emission.

Are you able to understand this?

69. Originally Posted by chinglu
No, you're not getting what you're saying. First prove that there exists a point where x = vtγ /( 1 + γ ) and z = 0. Next prove that there is a point other than x=y=z=t=0 that matches this restriction.
I proved this above.[/quote]
No, you assumed x = vtγ /( 1 + γ ) and z = 0 and then did some algebra. You never proved that there is any location where this assumption is true. Try to pick some actual numbers and see if you can make it work.

I never said A and B are the same event.

I am talking about the light emission.

Are you able to understand this?
Well, this proves (yet again) that you can read something and not understand it.

Einstein: "When we say that the lightning strokes A and B are simultaneous with respect to the embankment, we mean: the rays of light emitted at the places A and B, where the lightning occurs, meet each other at the mid-point M of the length A —> B of the embankment.... 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."

The only light emission in this example happens at A and B. You don't know what you are talking about.

70. Let me try another angle, to see if it helps you with the concept you currently cannot seem to understand...

Originally Posted by chinglu
Is the light emission one simultaneous event between the frames?
Simultaneous according to whom?

How can you tell if a different observer (in motion relative to you) will calculate something happened at the same time that you calculate it to have happened?

In order to ascertain this, you will need to set up a simultaneity convention with the other observer, and when you do that you will find that it is only a convention, rather than anything absolute, as it is based on the relative motion between you and the other observer.

You will finally come to the conclusion that simultaneity cannot be absolute, due to the constancy of c to all inertial frames, and once you have this realisation, you will see how your above question makes no sense.

71. Originally Posted by chinglu
x'² + y² + z² = c² t'²

standard configuration.

x'² + y'² + z'² = c² t'²
This does not represent a general rrelationship, but only a statement about light propagation.

x,y,z,t and x',y',z',t' are quite arbitrary and simply coordinatize spacetime.

The Lorentz transformations relate arbitrary spacetime coordinates in one inertial reference frame to coordinates of the same spacetime point in another inertial reference frame.

My derivation above is quite valid. And yours is utter nonsense.

BTW your equations do imply that light waves from a point source produce a spherical wavefront in all inertial reference frames, a fact which you dispute.

72. Originally Posted by DrRocket
Originally Posted by chinglu
x'² + y² + z² = c² t'²

standard configuration.

x'² + y'² + z'² = c² t'²
This does not represent a general rrelationship, but only a statement about light propagation.

x,y,z,t and x',y',z',t' are quite arbitrary and simply coordinatize spacetime.

The Lorentz transformations relate arbitrary spacetime coordinates in one inertial reference frame to coordinates of the same spacetime point in another inertial reference frame.

My derivation above is quite valid. And yours is utter nonsense.

BTW your equations do imply that light waves from a point source produce a spherical wavefront in all inertial reference frames, a fact which you dispute.

1) My proofs do not show a spherical wave front. They show each light beam measure c. that does not prove a spherical wave front when mapped. Further, since I proved a special case in which dt'/dt < 0, which in nature is logically absurd, then the price that is paid for each light beam measuring c when mapped is logical absurdity. But, that is the whole point of the proof you cannot refute.

2) You do not understand the relationship between the light postulates in the frames.

I will let Einstein explain it to you since you don't undertstand it. Now, hurry and upgrade.

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. If (x, y, z) be a point just attained by this wave, then
x² + y² + z² = c² t²*

Transforming this equation with the aid of our equations of transformation we obtain after a simple calculation
ξ² + η² + ς² = c² τ²*The wave under consideration is therefore no less a spherical wave with velocity of propagation c when viewed in the moving system. This shows that our two fundamental principles are compatible.5

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

73. Originally Posted by PhysBang
Originally Posted by chinglu
No, you're not getting what you're saying. First prove that there exists a point where x = vtγ /( 1 + γ ) and z = 0. Next prove that there is a point other than x=y=z=t=0 that matches this restriction.
I proved this above.
No, you assumed x = vtγ /( 1 + γ ) and z = 0 and then did some algebra. You never proved that there is any location where this assumption is true. Try to pick some actual numbers and see if you can make it work.

I never said A and B are the same event.

I am talking about the light emission.

Are you able to understand this?
Well, this proves (yet again) that you can read something and not understand it.

Einstein: "When we say that the lightning strokes A and B are simultaneous with respect to the embankment, we mean: the rays of light emitted at the places A and B, where the lightning occurs, meet each other at the mid-point M of the length A —> B of the embankment.... 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."

The only light emission in this example happens at A and B. You don't know what you are talking about.[/quote]

1) x = vtγ /( 1 + γ )

I have been through all this with the math and it is valid. You can pick numbers so you can learn something. For each t based on the expanding light sphere, there exists a clock in the primed frame and a clock in the unpirmed frame that read exactly the same time and are co-located when the SLW reaches them. In other words, the measurement is absolute between the frames.

2) Well, this proves (yet again) that you can read something and not understand it.

The questions you continue to fear and run from, when M and M' are co-locate is the strike at A a single event for the frame yes or no.

Now, don't forget for the Einstein mirror experiment to develop LT, he assumed light emission is a single common event between the frames.

Once you get this figured out, then you will finally learn something.

74. Originally Posted by SpeedFreek
Let me try another angle, to see if it helps you with the concept you currently cannot seem to understand...

Originally Posted by chinglu
Is the light emission one simultaneous event between the frames?
Simultaneous according to whom?

How can you tell if a different observer (in motion relative to you) will calculate something happened at the same time that you calculate it to have happened?

In order to ascertain this, you will need to set up a simultaneity convention with the other observer, and when you do that you will find that it is only a convention, rather than anything absolute, as it is based on the relative motion between you and the other observer.

You will finally come to the conclusion that simultaneity cannot be absolute, due to the constancy of c to all inertial frames, and once you have this realisation, you will see how your above question makes no sense.
We are not talking about time with this idea of simultaneous.

You are caught up in terms in an absolute sense.

When light emits, is that a single agreed upon event between the frames yes or no.

75. Originally Posted by chinglu
1) x = vtγ /( 1 + γ )

I have been through all this with the math and it is valid. You can pick numbers so you can learn something. For each t based on the expanding light sphere, there exists a clock in the primed frame and a clock in the unpirmed frame that read exactly the same time and are co-located when the SLW reaches them. In other words, the measurement is absolute between the frames.
To me, this looks like you can't work through a problem.
The questions you continue to fear and run from, when M and M' are co-locate is the strike at A a single event for the frame yes or no.
What you say makes no sense. We already know that A is one event. Not only did I say this before, I pointed it out to you. Aditionally, When M and M' are co-located is one event. This is the case for every frame.
Now, don't forget for the Einstein mirror experiment to develop LT, he assumed light emission is a single common event between the frames.
Like I told you before, all events are in all well-formed frames.

76. Originally Posted by PhysBang
Originally Posted by chinglu
1) x = vtγ /( 1 + γ )

I have been through all this with the math and it is valid. You can pick numbers so you can learn something. For each t based on the expanding light sphere, there exists a clock in the primed frame and a clock in the unpirmed frame that read exactly the same time and are co-located when the SLW reaches them. In other words, the measurement is absolute between the frames.
To me, this looks like you can't work through a problem.
Yea?

Looks like I am working through it pretty well. Do you understand math proof?

Set
x = vtγ /( 1 + γ ) and z = 0.

LT
t' = ( t - vx/c² )γ
t' = ( t - v((vtγ/(1+γ) /c² )γ

t' = ( t - tv²γ/(c²(1+γ)) )γ
t' = t( γ - v²γ²/(c²(1+γ)) )
t' = t( (γ(c²(1+γ))/(c²(1+γ) - v²γ²/(c²(1+γ)) )
t' = t(( (c²γ²- v²γ²+c²γ))/(c²(1+γ)))
t' = t(( (c²+c²γ))/(c²(1+γ)))
t' = t(( c²(1+γ))/(c²(1+γ)))
t' = t

The questions you continue to fear and run from, when M and M' are co-locate is the strike at A a single event for the frame yes or no.
What you say makes no sense. We already know that A is one event. Not only did I say this before, I pointed it out to you. Aditionally, When M and M' are co-located is one event. This is the case for every frame. [/quote]

Good, now A is one event for the frames.

At light emission, if r is the distance to M, then r/γ is the distance to M'.

Is this true for you with length contraction?

Now, that means tA = r/c andf t'A = r/(cγ).

Do you understand this?

77. Originally Posted by chinglu
Originally Posted by DrRocket
Originally Posted by chinglu
x'² + y² + z² = c² t'²

standard configuration.

x'² + y'² + z'² = c² t'²
This does not represent a general rrelationship, but only a statement about light propagation.

x,y,z,t and x',y',z',t' are quite arbitrary and simply coordinatize spacetime.

The Lorentz transformations relate arbitrary spacetime coordinates in one inertial reference frame to coordinates of the same spacetime point in another inertial reference frame.

My derivation above is quite valid. And yours is utter nonsense.

BTW your equations do imply that light waves from a point source produce a spherical wavefront in all inertial reference frames, a fact which you dispute.

1) My proofs do not show a spherical wave front. They show each light beam measure c. that does not prove a spherical wave front when mapped. Further, since I proved a special case in which dt'/dt < 0, which in nature is logically absurd, then the price that is paid for each light beam measuring c when mapped is logical absurdity. But, that is the whole point of the proof you cannot refute.

2) You do not understand the relationship between the light postulates in the frames.

I will let Einstein explain it to you since you don't undertstand it. Now, hurry and upgrade.

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. If (x, y, z) be a point just attained by this wave, then
x² + y² + z² = c² t²*

Transforming this equation with the aid of our equations of transformation we obtain after a simple calculation
ξ² + η² + ς² = c² τ²*The wave under consideration is therefore no less a spherical wave with velocity of propagation c when viewed in the moving system. This shows that our two fundamental principles are compatible.5

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

78. Originally Posted by chinglu
Originally Posted by PhysBang
To me, this looks like you can't work through a problem.
Yea?

Looks like I am working through it pretty well. Do you understand math proof?

Set
x = vtγ /( 1 + γ ) and z = 0.

LT
t' = ( t - vx/c² )γ
t' = ( t - v((vtγ/(1+γ) /c² )γ

t' = ( t - tv²γ/(c²(1+γ)) )γ
t' = t( γ - v²γ²/(c²(1+γ)) )
t' = t( (γ(c²(1+γ))/(c²(1+γ) - v²γ²/(c²(1+γ)) )
t' = t(( (c²γ²- v²γ²+c²γ))/(c²(1+γ)))
t' = t(( (c²+c²γ))/(c²(1+γ)))
t' = t(( c²(1+γ))/(c²(1+γ)))
t' = t
Yet again you show that you don't understand your own problem. You haven't worked through this problem. You can set x = vtγ /( 1 + γ ) and z = 0, but if you can't show that any points match these criteria then you might as well set x = x+1 or x = orange.
Good, now A is one event for the frames.
I am amazed that you accept this.
At light emission, if r is the distance to M, then r/γ is the distance to M'.
What is "at light emission"? Are you assigning a time? In what frame are you talking about? How are you determining the distance to M' and in what frame is that distance? What is the time you are assigning to A and in what frame? What is the spacial coordinates you are assigning to A and in what frame?

79. Originally Posted by DrRocket
Originally Posted by chinglu
Originally Posted by DrRocket
Originally Posted by chinglu
x'² + y² + z² = c² t'²

standard configuration.

x'² + y'² + z'² = c² t'²
This does not represent a general rrelationship, but only a statement about light propagation.

x,y,z,t and x',y',z',t' are quite arbitrary and simply coordinatize spacetime.

The Lorentz transformations relate arbitrary spacetime coordinates in one inertial reference frame to coordinates of the same spacetime point in another inertial reference frame.

My derivation above is quite valid. And yours is utter nonsense.

BTW your equations do imply that light waves from a point source produce a spherical wavefront in all inertial reference frames, a fact which you dispute.

1) My proofs do not show a spherical wave front. They show each light beam measure c. that does not prove a spherical wave front when mapped. Further, since I proved a special case in which dt'/dt < 0, which in nature is logically absurd, then the price that is paid for each light beam measuring c when mapped is logical absurdity. But, that is the whole point of the proof you cannot refute.

2) You do not understand the relationship between the light postulates in the frames.

I will let Einstein explain it to you since you don't undertstand it. Now, hurry and upgrade.

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. If (x, y, z) be a point just attained by this wave, then
x² + y² + z² = c² t²*

Transforming this equation with the aid of our equations of transformation we obtain after a simple calculation
ξ² + η² + ς² = c² τ²*The wave under consideration is therefore no less a spherical wave with velocity of propagation c when viewed in the moving system. This shows that our two fundamental principles are compatible.5

http://www.fourmilab.ch/etexts/einstein/specrel/www/
Here I will try to show you how all this works.

That means x² + y² + z² = c² t²*

You then go through my math proof and eventually apply LT as Einsein suggested.

LP = light postulate equation

So, LT(LP) = LP'.

Now, this is a requirement of SR.
However, with the math proof I showed, we find this to be false.

If you are going to prance around here as the resident math expert, I suggest you attempt to take out my math or concede.

So far, your feeble attempt to call x fixed in the equation t' = ( t - vx/c² )γ when x represents the x location of the expanding light sphere as a function of t was a complete failure.

That is where we are.

80. Originally Posted by PhysBang
Originally Posted by chinglu
Originally Posted by PhysBang
To me, this looks like you can't work through a problem.
Yea?

Looks like I am working through it pretty well. Do you understand math proof?

Set
x = vtγ /( 1 + γ ) and z = 0.

LT
t' = ( t - vx/c² )γ
t' = ( t - v((vtγ/(1+γ) /c² )γ

t' = ( t - tv²γ/(c²(1+γ)) )γ
t' = t( γ - v²γ²/(c²(1+γ)) )
t' = t( (γ(c²(1+γ))/(c²(1+γ) - v²γ²/(c²(1+γ)) )
t' = t(( (c²γ²- v²γ²+c²γ))/(c²(1+γ)))
t' = t(( (c²+c²γ))/(c²(1+γ)))
t' = t(( c²(1+γ))/(c²(1+γ)))
t' = t
Yet again you show that you don't understand your own problem. You haven't worked through this problem. You can set x = vtγ /( 1 + γ ) and z = 0, but if you can't show that any points match these criteria then you might as well set x = x+1 or x = orange.
OMG, there is no intelligent life here.

Given x = vtγ /( 1 + γ ) and z = 0, then

c²t² = x² + y²

Substitute for x.

c²t² = [vtγ /( 1 + γ )]² + y²

Solve for y.

y = ±√(c²t² - [vtγ /( 1 + γ )]² )

There, you have the light postulate satisfied in the rest fame.

To finalize the proof we must show c²t² - [vtγ /( 1 + γ )]² > 0 because of the square root function.

Well,

( 1 + γ ) > γ

Since c > v, then

c ( 1 + γ ) > vγ

c > vγ/( 1 + γ )

ct > vγt/( 1 + γ )

c²t² > [vγt/( 1 + γ )]²

Hence,

c²t² - [vγt/( 1 + γ )]² > 0

QED

81. Originally Posted by chinglu
OMG, there is no intelligent life here.
Given x = vtγ /( 1 + γ ) and z = 0, then
Prove that any points actually fit this restriction.

82. Originally Posted by chinglu
OMG, there is no intelligent life here.
Is that why we all understand and agree with the meaning of the train and embankment experiment, as understood by mainstream physics, whereas you do not?

83. Originally Posted by PhysBang
Originally Posted by chinglu
OMG, there is no intelligent life here.
Given x = vtγ /( 1 + γ ) and z = 0, then
Prove that any points actually fit this restriction.
Just look at the proof I posted.

I verfied mathematically there exists points of the nature because they are in the range of the light sphere at any time t.

Again, look at the proof I posted.

84. Originally Posted by SpeedFreek
Originally Posted by chinglu
OMG, there is no intelligent life here.
Is that why we all understand and agree with the meaning of the train and embankment experiment, as understood by mainstream physics, whereas you do not?
I agree you all are wrong.

Again, I have attempted to get you to look at the light emission at A and understand it.

When M and M' are co-located, the A is a distance r from M and A is a distance r/γ from M'.

Both frames must agree the light is at A.

Now, the same logic applies to B when M and M' are co-located.

Hence, under SR, both frames will claim the strikes are simultaneous to its frame and not to the other.

Can you refute this logic without resorting to the logical fallacies argumentum ad populum and argumentum ad verecundiam?

85. Originally Posted by chinglu
Originally Posted by SpeedFreek
Originally Posted by chinglu
OMG, there is no intelligent life here.
Is that why we all understand and agree with the meaning of the train and embankment experiment, as understood by mainstream physics, whereas you do not?
I agree you all are wrong.

Can you refute this logic without resorting to the logical fallacies argumentum ad populum and argumentum ad verecundiam?
You have been shown repeatedly, with rigorous mathematics and rigorous physics, by professionals, precisely why you are wrong, and what is in fact correct.

This was done largely for the benefit of innocent lurkers with an interest in real science. Logical and mathematically rigorous arguments are wasted on you.

Nobody gives a damn what you agree to. Your opinion is irrelevant. You are nuts.

86. Originally Posted by chinglu
Just look at the proof I posted.

I verfied mathematically there exists points of the nature because they are in the range of the light sphere at any time t.
No, your proof assumes x = vtγ /( 1 + γ ) and z = 0. That's why it begins, "Given x = vtγ /( 1 + γ ) and z = 0, then..." You need to show just what points in a frame meet the criteria that x = vtγ /( 1 + γ ) and z = 0.

When someone writes, "Given x = vtγ /( 1 + γ ) and z = 0, then...," they are assuming something for the purposes of a proof. That proof does not prove that the assumption actually holds, though. You need to prove that any points match your assumption and show where those points are.

I could prove the same thing that you do by assuming that x = x+1. However, since no point matches that criteria, my proof wouldn't be very useful. You need to show that your proof is worth something.
When M and M' are co-located, the A is a distance r from M and A is a distance r/γ from M'.
This doesn't make any sense if you do not specify frames.
Both frames must agree the light is at A.
As we all told you before, A is an event at which light is emitted and it is in every frame and every frame might assign it a different time coordinate and different spacial coordinates. If one does not take care, talking about A can lead to the same confusion that you have.
Now, the same logic applies to B when M and M' are co-located.
What logic? You have failed to specify a frame for your distances and your times, so you have not yet got anything we can consider properly.
Hence, under SR, both frames will claim the strikes are simultaneous to its frame and not to the other.
You need to actually read through the article that you brought up. You haven't read it.

87. Originally Posted by DrRocket
Originally Posted by chinglu
Originally Posted by SpeedFreek
Originally Posted by chinglu
OMG, there is no intelligent life here.
Is that why we all understand and agree with the meaning of the train and embankment experiment, as understood by mainstream physics, whereas you do not?
I agree you all are wrong.

Can you refute this logic without resorting to the logical fallacies argumentum ad populum and argumentum ad verecundiam?
You have been shown repeatedly, with rigorous mathematics and rigorous physics, by professionals, precisely why you are wrong, and what is in fact correct.

This was done largely for the benefit of innocent lurkers with an interest in real science. Logical and mathematically rigorous arguments are wasted on you.

Nobody gives a damn what you agree to. Your opinion is irrelevant. You are nuts.

Really, what post refuted my proofs.

You just talk and that is an indication of a failure.

The only attempt made at my proofs was your worthless failed attempt at a partial derivative where you claimed x was fixed where x represented the x coordinate of the expanding light sphere which was functional in t. I taught you about this failure of yours.

Anyway, at this point, all my proofs stand and cannot be refuted. All you can do is sit by and nod your head up and down in agreement.

Otherwise, do something with math other than talk.

88. Originally Posted by PhysBang
Originally Posted by chinglu
Just look at the proof I posted.

I verfied mathematically there exists points of the nature because they are in the range of the light sphere at any time t.
No, your proof assumes x = vtγ /( 1 + γ ) and z = 0. That's why it begins, "Given x = vtγ /( 1 + γ ) and z = 0, then..." You need to show just what points in a frame meet the criteria that x = vtγ /( 1 + γ ) and z = 0.
I already showed this in the proof above. Take any v and time t give z = 0 and the y above. You will find t = t'.

Now, in order to refute math, an opinion does not work like you are doing. I presented the clear math proof that any child in China could understand. So, you will need to refute it with an example or find a flaw in the math.

You have 2 ways you can do that, noway and nohow.

When someone writes, "Given x = vtγ /( 1 + γ ) and z = 0, then...," they are assuming something for the purposes of a proof. That proof does not prove that the assumption actually holds, though. You need to prove that any points match your assumption and show where those points are.

I could prove the same thing that you do by assuming that x = x+1. However, since no point matches that criteria, my proof wouldn't be very useful. You need to show that your proof is worth something.
I proved a valid y and x that meets the conditions of x² + y² + z² = c² t²*given x = vtγ /( 1 + γ ) and z = 0.

that is all I have to prove.

You do not understand math proof like Rocket, isn't that correct.

When M and M' are co-located, the A is a distance r from M and A is a distance r/γ from M'.
This doesn't make any sense if you do not specify frames.

I specified the frames, M and M'. What else do you need?

Both frames must agree the light is at A.
As we all told you before, A is an event at which light is emitted and it is in every frame and every frame might assign it a different time coordinate and different spacial coordinates. If one does not take care, talking about A can lead to the same confusion that you have.
However you want to put it, it is one event between the frames. that is all I need to refute ROS.

Now, the same logic applies to B when M and M' are co-located.
What logic? You have failed to specify a frame for your distances and your times, so you have not yet got anything we can consider properly.
Hence, under SR, both frames will claim the strikes are simultaneous to its frame and not to the other.
You need to actually read through the article that you brought up. You haven't read it.
Oh, I understand the article.

That is why you cannot refute anything I have said about the T/E experiment.

Are you interested in learning anything or do you just follow the herd?

89. Originally Posted by chinglu
I already showed this in the proof above. Take any v and time t give z = 0 and the y above. You will find t = t'.
You will find this if you assume that x = vtγ /( 1 + γ ) and z = 0. Is there anywhere that x = vtγ /( 1 + γ ) and z = 0?
Now, in order to refute math, an opinion does not work like you are doing. I presented the clear math proof that any child in China could understand. So, you will need to refute it with an example or find a flaw in the math.
I'm not talking about your math. I'm talking about your assumptions before you even start any math. You assumed that x = vtγ /( 1 + γ ) and z = 0
I proved a valid y and x that meets the conditions of x² + y² + z² = c² t²*given x = vtγ /( 1 + γ ) and z = 0.

that is all I have to prove.
If you just want to continue to be seen as a crazy person, then yes. You have assumed that x = vtγ /( 1 + γ ) and z = 0 but you haven't shown that any points actually meet these criteria. We know where z=0 is, but you haven't shown where x = vtγ /( 1 + γ ) is. You don't seem to know the scope of your own proof.
When M and M' are co-located, the A is a distance r from M and A is a distance r/γ from M'.
This doesn't make any sense if you do not specify frames.

I specified the frames, M and M'. What else do you need?[/quote]
M and M' are not frames, they are places that can be assigned different coordinates in different frames. They are loosely defined within certain frames in the relevant Chapter, but you haven't said where " distance r" or "distance r/γ" are given.
However you want to put it, it is one event between the frames. that is all I need to refute ROS.
Well, since relativity of simultaneity only holds between two or more events, you would have to be crazy to claim that one can consider only one event to deny the relativity of simultaneity.
Oh, I understand the article.
How do you know this without reading the article? Claims like that make you look crazy.
That is why you cannot refute anything I have said about the T/E experiment.
One cannot refute crazy. So far you haven't said anything that makes sense. You refuse to clearly identify the frames for your supposed distances. This may be because you are just sane enough to know that you haven't got a case.

So, either keep up the same craziness that we've seen here (and that got you banned from bautforum) or try to actually clearly identify the frames you are using (for a start).

90. Originally Posted by PhysBang
Originally Posted by chinglu
I already showed this in the proof above. Take any v and time t give z = 0 and the y above. You will find t = t'.
You will find this if you assume that x = vtγ /( 1 + γ ) and z = 0. Is there anywhere that x = vtγ /( 1 + γ ) and z = 0?
How many times do I need to prove this?

Given x = vtγ /( 1 + γ ) and z = 0, then

c²t² = x² + y²

Substitute for x.

c²t² = [vtγ /( 1 + γ )]² + y²

Solve for y.

y = ±√(c²t² - [vtγ /( 1 + γ )]² )

There, you have the light postulate satisfied in the rest fame.

To finalize the proof we must show c²t² - [vtγ /( 1 + γ )]² > 0 because of the square root function.

Well,

( 1 + γ ) > γ

Since c > v, then

c ( 1 + γ ) > vγ

c > vγ/( 1 + γ )

ct > vγt/( 1 + γ )

c²t² > [vγt/( 1 + γ )]²

Hence,

c²t² - [vγt/( 1 + γ )]² > 0

QED

If you just want to continue to be seen as a crazy person, then yes. You have assumed that x = vtγ /( 1 + γ ) and z = 0 but you haven't shown that any points actually meet these criteria. We know where z=0 is, but you haven't shown where x = vtγ /( 1 + γ ) is. You don't seem to know the scope of your own proof.
You don't get it. That x is a point acquired by the light sphere at time t. It is not an assumption, it is a fact as I proved. Let the SLW acquite a radius r. Then t = r/c.

Then set z=0, x = vtγ /( 1 + γ ) and y = √(c²t² - [vtγ /( 1 + γ )]² )

then, (x,y,z) is on the light sphere. It is that simple.

Well, since relativity of simultaneity only holds between two or more events, you would have to be crazy to claim that one can consider only one event to deny the relativity of simultaneity.

ROS only applies to 2 or more light reception events. We have a light emission and a light reception.

Anyway, are you going to refute both frames see the strikes as simultaneous?

If so, show the math.

91. I want to go back to this statement.

Originally Posted by chinglu
Originally Posted by DrRocket
Originally Posted by chinglu
Good, you are so very smart.

So, is A a simultaneous event between the frames?

That means x'=x=0 and t'=t=0.

This proves quite clearly that you are irrational. There is no such thing as "a simultaneous event between the frames". A single event cannot be "simultaneous. Two different events can be simultaneous, but "simultaneity" is a relationship among events and is meaningless when applied to a single event.

One can choose the origin of the primed and unprimed frames arbitrarily, so your condition "x'=x=0 and t'=t=0" is as devoid of content as is your questing.

Physbang: With this incredibly stupid question (whoever said "There is no such thing as a stupid question" never met chinglu) it is clear that I win. Dexter has not uttered a single stupid statement. Moreover, while Dexter and chinglu may both be chasing their tail, Dexter occasionally catches his.

Let's see, I hereby quote Einstein.

At the time t=t'=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, is the light emission a simultaneous event between the frames, yes or no.

Or, does one frame think, light emitted at a place and another frame claims light is not at that place with co-located observers for the light emission.

You border on the absurd and I am being nice.
I find that nearly all confusions a person can have in mathematics rests with bad assumptions.

While it is certainly possible to choose a coordinate from which, if a spherical light wave were emitted, both the moving and stationary observer would perceive it to have occurred at the same moment in time,.... the train example you are working from is not one such case. In the train example, the events A and B are not simultaneous to both observers.

92. Originally Posted by kojax
I find that nearly all confusions a person can have in mathematics rests with bad assumptions.

.

93. Originally Posted by chinglu
How many times do I need to prove this?

Given x = vtγ /( 1 + γ ) and z = 0, then

c²t² = x² + y²

Substitute for x.

c²t² = [vtγ /( 1 + γ )]² + y²

Solve for y.

y = ±√(c²t² - [vtγ /( 1 + γ )]² )

There, you have the light postulate satisfied in the rest fame.
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. This comes before your proof. You proof only works when x = vtγ /( 1 + γ ) and z = 0 and not anywhere else. So I want to know where all the points are that x = vtγ /( 1 + γ ) and z = 0.
You don't get it. That x is a point acquired by the light sphere at time t. It is not an assumption, it is a fact as I proved.
So far, you've proven nothing of the sort. But at least we're getting somewhere.
Let the SLW acquite a radius r. Then t = r/c.
What is SLW? What does it mean for "the SLW acquite a radius r"?
Then set z=0, x = vtγ /( 1 + γ ) and y = √(c²t² - [vtγ /( 1 + γ )]² )

then, (x,y,z) is on the light sphere. It is that simple.
Why should we set z=0, x = vtγ /( 1 + γ )? What are these points?

ROS only applies to 2 or more light reception events.
So you admit that you need more than one point. That's good. But the relativity of simultaneity holds for any spatial separated points whether or not there is any light.
Anyway, are you going to refute both frames see the strikes as simultaneous?

If so, show the math.
I'll just stick with the refutation that makes up the entirety of the chapter that you initially referenced, since it actually explains in detail why the two frames do not find that the strikes are simultaneous. Despite claiming that you understand the chapter, you clearly have not read it. Read it before you post here again.

94. Originally Posted by kojax
I want to go back to this statement.

Originally Posted by chinglu
Originally Posted by DrRocket
Originally Posted by chinglu
Good, you are so very smart.

So, is A a simultaneous event between the frames?

That means x'=x=0 and t'=t=0.

This proves quite clearly that you are irrational. There is no such thing as "a simultaneous event between the frames". A single event cannot be "simultaneous. Two different events can be simultaneous, but "simultaneity" is a relationship among events and is meaningless when applied to a single event.

One can choose the origin of the primed and unprimed frames arbitrarily, so your condition "x'=x=0 and t'=t=0" is as devoid of content as is your questing.

Physbang: With this incredibly stupid question (whoever said "There is no such thing as a stupid question" never met chinglu) it is clear that I win. Dexter has not uttered a single stupid statement. Moreover, while Dexter and chinglu may both be chasing their tail, Dexter occasionally catches his.

Let's see, I hereby quote Einstein.

At the time t=t'=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, is the light emission a simultaneous event between the frames, yes or no.

Or, does one frame think, light emitted at a place and another frame claims light is not at that place with co-located observers for the light emission.

You border on the absurd and I am being nice.
I find that nearly all confusions a person can have in mathematics rests with bad assumptions.

While it is certainly possible to choose a coordinate from which, if a spherical light wave were emitted, both the moving and stationary observer would perceive it to have occurred at the same moment in time,.... the train example you are working from is not one such case. In the train example, the events A and B are not simultaneous to both observers.
It is one thing to claim without foundation that math assumptions are wrong, it is entirely another subject to prove they are wrong. Therefore, all you need to do is prove the math is wrong. Rocket has struggle unsuccessfully to prove any of the math is wrong.

I think he has given up in frustration and simply resorted to name calling.

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.

95. Originally Posted by DrRocket
Originally Posted by kojax
I find that nearly all confusions a person can have in mathematics rests with bad assumptions.

.
So, you are still left with a sour taste in your mouth after I corrected you on your lack of understanding with how to use the derivative on the LT equations when operating with the SLW spherical light wave.

96. Will you please address the request put to you instead of continuing to evade it for another four pages? Please demonstrate which points meet the criteria of x = vtγ /( 1 + γ ) and z = 0.

97. Originally Posted by PhysBang
Originally Posted by chinglu
How many times do I need to prove this?

Given x = vtγ /( 1 + γ ) and z = 0, then

c²t² = x² + y²

Substitute for x.

c²t² = [vtγ /( 1 + γ )]² + y²

Solve for y.

y = ±√(c²t² - [vtγ /( 1 + γ )]² )

There, you have the light postulate satisfied in the rest fame.
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. This comes before your proof. You proof only works when x = vtγ /( 1 + γ ) and z = 0 and not anywhere else. So I want to know where all the points are that x = vtγ /( 1 + γ ) and z = 0.

You do not understand any of this.

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.

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.

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?

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

(SLW ) = spherical light wave

ROS only applies to 2 or more light reception events.
So you admit that you need more than one point. That's good. But the relativity of simultaneity holds for any spatial separated points whether or not there is any light.
See, we have a problem with your reasoning and mine is correct. Events are light receptions and not light emissions.

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.

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.

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.

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.

98. Originally Posted by inow
Will you please address the request put to you instead of continuing to evade it for another four pages? Please demonstrate which points meet the criteria of x = vtγ /( 1 + γ ) and z = 0.
I am not evading anything. There are an infinite number.

But, I gave an example above in which t'=t=1.

Since the math is valid, I don't feel the need to give examples. The math speaks for itself. But, that is not true for everyone, so if needed I will give more examples.

99. I believe inow asked you demonstrate these were not valid, not simply state these were valid examples. There is a difference. One of the differences is whether or not I bother to continue reading your posts.

100. 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.

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.

101. Originally Posted by Ophiolite
I believe inow asked you demonstrate these were not valid, not simply state these were valid examples. There is a difference. One of the differences is whether or not I bother to continue reading your posts.
I am not understanding your post.