Is Relativity Really Wrong? Exploring the Evidence on Motion and Perception

[Doc Al]

same old, same old

The experiment
Clocks are positioned at A, M and B in a stationary inertial frame and at A’, M’ and B’ in an inertial frame moving along the AB line. M is the midpoint of photon sources located at A and B. When A’ = A, M’ = M and B’ = B, the clocks are set to zero and photons are emitted from A and B.

As usual, you merely assume that there is a single instant when "A’ = A, M’ = M and B’ = B".

Setting the clocks to zero just makes them laughably unsynchronized.

The experimental results
The subject photons emitted simultaneously in all frames.

Consider how silly this statement is, given the conditions of your "experiment". It's like saying to everyone in the world "When you wake up this morning, set your clocks to 6:00 am", and then concluding "Look everybody, I've just proven that everyone wakes up at exactly the same time!".

geistkiesel, don't you find it ironic that even wespe, who started this thread trying to expose a flaw in relativity, has ripped apart your circular arguments?

(When I say "even wespe", I don't mean that as an insult to wespe. I'm just pointing out the irony. wespe is doing a excellent job!)

 

[wespe]

M will receive the A and B photons simultaneously. I don't care when M' receives he photons. The question is whether the photons were emitted simultaneously or not, in the moving frame. The photons were emitted in the moving frame simultaneously. If M' makes a calculation that results in his determination that the photons were not emitted simultaneously the calculation is flawed. Why flawed you may ask? Because the photons were emitted in the moving frame simultaneously. Read the expeiment. Of course M' won't receive the photons at the same time, he is moving into and away from the oncoming photons. The question can apply to moving ants or moving photons.

The question is not when M' receives the photons, the question is not whether the photons are perceived as being emitted simultaneously, in the moving frame. The questin is whether the photons were actually emitted in the moving frame at the same time and whether this fact can betdetermined by M'. M' can look at the expeimental results..

If calculations indicate the photons were emitted at diffeent times in the moving frames, then the calculations corrupted the physical reality as indicated by the experimental results.

Do not ever say to me that, "SR says that . . . ", you will be wasting your breath. I won't be listening to you. Do you understand?

Question Wespe: When were the photons emitted in the moving frame? Assume for the sake iof argument that the moving frame is the only frame in the universe.

Specifically, what time were the clock readings at the instant the photons were emitted?


The A' clock was reading what time when the photon from A was emitted?

The B' clock was reading what time when the photon from B was emitted?

Answer these questions, period.

You started this thread, do you want me to finish it for you or can you finish what you started yourself?

Well, the answers depend on what theory you want to use. But only a real experiment can decide what is true. You don't want to hear SR predictions so what theory should I use?

OK, since you now think speed of light is not constant, let's say it is c+v and c-v relative to M'. Then, of course, we are saying that the stationary frame is stationary relative to aether. Now, should we consider length contraction? Because, when length of the moving train is shortened, A' and B' will not meet A and B at the same time. We could use a longer train to compansate, but what if length contraction is mutual? Calculations should agree for both frames. What will we assume about time dilation?

Without these, I can't make any calculations, and I won't just assume things like you do. You should really decide on your version of a theory and get your own answers. As far as I know, lorentz ether theory (LET) is most compatible with c+v c-v. But its predictions are generally the same as SR, so you won't want to hear that either. I'm sorry.

 

[wespe]

After having been accelerated @ one g for one minute along the line AB photons are emitted from A and B. Will the observer at M, the midpoint between A and B, receive he photons at the same time? What does SR say about this?

I don't claim to be an expert, but SR says (lol) that (I think), the scenario is symmetric so the results will be symmetric. Neither M nor M' will receive A/B and A'/B' photons at the same time (or they will both do, it may depend on distance and velocity, we need to calculate). Someone correct me if I'm wrong.

 

[Hurkyl]

After having been accelerated @ one g for one minute along the line AB photons are emitted from A and B. Will the observer at M, the midpoint between A and B, receive he photons at the same time? What does SR say about this?

This is poorly specified; one couldn't even answer this experiment... using classical mechanics. Furthermore there are all sorts of relativity issues that make it even worse in SR.

 

[geistkiesel]

Well, the answers depend on what theory you want to use. But only a real experiment can decide what is true. You don't want to hear SR predictions so what theory should I use?

OK, since you now think speed of light is not constant, let's say it is c+v and c-v relative to M'. Then, of course, we are saying that the stationary frame is stationary relative to aether. Now, should we consider length contraction? Because, when length of the moving train is shortened, A' and B' will not meet A and B at the same time. We could use a longer train to compansate, but what if length contraction is mutual? Calculations should agree for both frames. What will we assume about time dilation?

Without these, I can't make any calculations, and I won't just assume things like you do. You should really decide on your version of a theory and get your own answers. As far as I know, lorentz ether theory (LET) is most compatible with c+v c-v. But its predictions are generally the same as SR, so you won't want to hear that either. I'm sorry.

Wespe:, get a hold of yourself. Do you see in the post that theere were clocks at A=A', B=B' and M=M' that were set to 0 when photons were emitted? Do you see this above in the post yoyu are answering??
Now you answered , or were trying to snawer another question.


What time do the clocks indicate that the photons were emitted?
To get the correct answer you must supply the correct number, just one number.

Where was A', B' and M' when the photons were emitted?
Hint:The experimental given was that A'=A, B'=B and M'=M at the very instant the photons were emitted. These locations are exact. Further hint: If your answer is other than A'=A, B'=B M'=M then your answer is wrong and you fail the course.

Why do you want to say A' was not at A and that B' was not at B?

You probably want to say that M' was not at M also, don't you?

You recognize, don't you that it is given that these locations were identical , colocated, at the same place, together, on top of each other the same when the photons were emitted?

Do you recognize this?

Why do you want to change the data? Looking for a theory you say? Why not try to use the laws of physics?

 

[geistkiesel]

As usual, you merely assume that there is a single instant when "A’ = A, M’ = M and B’ = B".

Setting the clocks to zero just makes them laughably unsynchronized.


Consider how silly this statement is, given the conditions of your "experiment". It's like saying to everyone in the world "When you wake up this morning, set your clocks to 6:00 am", and then concluding "Look everybody, I've just proven that everyone wakes up at exactly the same time!".

geistkiesel, don't you find it ironic that even wespe, who started this thread trying to expose a flaw in relativity, has ripped apart your circular arguments?

(When I say "even wespe", I don't mean that as an insult to wespe. I'm just pointing out the irony. wespe is doing a excellent job!)

Come Mac Al , look at the given. The experiment is set up like it is.
The values are given. [
The train is long enough that A'=A and B'=B and M' = M at the instant the photons are emitted @ t = 0 for all the clocks, six in all.This is the given time for the emission of the photons..

This is a given do you understand? I have modified the starting conditions of the gedunken, do you understand? So whatever the frame contractions or clock slowing parameters you want to invoke to time events after the photons are emitted go right on with your 'physics'.

I never saw a response to the post where I told you that either your memory was pathologically impaired or you were dishonest. Do you have any comment?

 

[geistkiesel]

I don't claim to be an expert, but SR says (lol) that (I think), the scenario is symmetric so the results will be symmetric. Neither M nor M' will receive A/B and A'/B' photons at the same time (or they will both do, it may depend on distance and velocity, we need to calculate). Someone correct me if I'm wrong.

This is the same problem we have been discussing/

Just as a check let's us see if we are working on the same problem.

SR says that events that are simultaneous in one frame are not simultaneous in a movung frame.(this OK?)

At least under the conditions of our current gedunken, which was Einstein's train gedunke?
.
BTW Mac Al is rooting for you to win something..

 

[Hurkyl]

The train is long enough that A'=A and B'=B and M' = M at the instant the photons are emitted @ t = 0 for all the clocks

In which frame are you asserting this?

A and A' can certainly agree the photon from A was emitted at t = 0.
Similarly, B and B' can agree the photon from B was emitted at t = 0.

Every other combination cannot simply be observed; it has to be determined via some means. e.g. how does M' know what time the photon was emitted from A, since his clock wasn't there when it happened?

 

[geistkiesel]

This is poorly specified; one couldn't even answer this experiment... using classical mechanics. Furthermore there are all sorts of relativity issues that make it even worse in SR.

Hurkyl; I merely wanted to give a platform some velocity and have photons emitted at both ends of the moving platform, which is the experimental arrangement under discussion. This could very well be the moving platform used in the current gedunken. The thread (the current state of the thread) is a relativity question specifically intrinsic to SR theory.

This is the latest experimental arrangement posted earlier and is the arrangement to which I am responding (below). Notice Doc Al has a problem wih the zeoring of the clocks being a phony scynchronization of clocks. This is not intended or implied. The zero state only marks when the emision of the photons occurred when all six clocks were located: A'=A, B'=B and M'=M. Its a long train and what ever shrunken state the train is in, is not my concern as the locations of the clocks are simultaneusly colocated when the photons left A and B, which is the question to be determined, which is:

are events simultaneous in a stationry frame simultaneous in moving frames?.

What is being skewed is the fact that the crucial events, the time the photons were emitted, is identical in both frames. This is an experimental given!

What we are talking about.

The experiment
Clocks are positioned at A, M and B in a stationary inertial frame and at A’, M’ and B’ in an inertial frame moving along the AB line. M is the midpoint of photon sources located at A and B. When A’ = A, M’ = M and B’ = B, the clocks are set to zero and photons are emitted from A and B.

The experimental results
The subject photons emitted simultaneously in all frames.

Conclusion
Special relativity is based on the postulates that the laws of physics and the measured speed of light are inertial frame independent. The concept of simultaneity variance is derived from the postulates of special relativity. The laws of physics are universally consistent, therefore the measured speed of light is frame dependent.

Geistkiesel
June 2004[/QUOTE

 

[wespe]

Geistkiesel,

According to SR, length contraction is mutual. Therefore, according to SR, your setup is impossible. A cannot meet A' at the same time as B meets B'. You can't make things possible by saying they are experimental given.

As I understand, you have no grasp of SR, neither of scientific methods. Therefore I will not discuss further with you. I don't want you to post anymore to this thread, if you will respect my request. Create a new thread for yourself please. To admins: I would like this thread locked please, if possible.

 

[Hurkyl]

Notice Doc Al has a problem wih the zeoring of the clocks being a phony scynchronization of clocks.

That is true; either A and B will not be synchronized in the "stationary" frame, or A' and B' will not be synchronized in the "moving" frame, or both.


Anyways, let me bring up some of my points again:

A and A' happen to be there when the photon from A is emitted, thus they can set their clocks to zero when that happens.
B and B' happen to be there when the photon from B is emitted, thus they can set their clocks to zero when that happens.

Question: How do M and M' know to set their clocks to zero?

Question: Suppose we did not assume a priori that the photons were emitted "simultaneously"; how can we tell from the results of the experiment whether or not the emission was "simultaneous"?

Question: How would one go about determining the relationship between clocks A and B? What about clocks A' and B'?

 

[geistkiesel]

Geistkiesel, just a friendly advice, your attitude is immature and you won't get answers like that. Not much longer even from me.


Clocks at A,A',B,B' indicate 0, since you reset them. While A/A' and B/B' are synchonized pairs at that instant, As and Bs are not synchronized with each other, due to length contraction. And the pairs won't remain synchronized due to time dilation. Clocks at M and M' looks uncertain. You can't make them know and reset instantly when the photons were emitted. They can reset when they meet, but that time is not evident yet.

Remember that we don't know how to send information instantly, so you must consider finite speed of light while sending information in a realistic experiment.

What information are you trying to send and to whom? and why? Please be specific.

All six clocks were synchronized in a stationary frame. The clocks at A', B' and M' remain synchronized during a gentle. very slow acceleration . When the moving frame reaches our experimental velocity the clocks are still synchronized with respect to each other on the moving frame. When one clock say '10' all the clocks say '10'. Let the clocks speed up, slow down whatever, as lomg as the moving clocks all keeping the same time in the moving frame.

OK, the moving frame shrunk. However, we have adjusted the position of the clocks such that when M'=M, then A'=A and B'=B. This is true even if we leave the clocks bolted into place, stop the moving train, and discover, lo and behold, says SR, the distance A'B'> AB, OK?

A was at A' at a time. And B was at B' a time. That doesn't mean A was at A' at the same time when B was at B'. What I just said doesn't mean they were not the same, it meant you cannot automatically assume they are the same. You also can't make things possible by saying "it is given in the experiment".

You are wrong, we designed the train such that A'B' = AB when the train velocty was v = 1. This is more of an engineering problem. We are using the best physicists in the universe, "Texas A&M Aggies".

Sure I can. I make some SR calculations and determine the moving frame will shrink when reaching the velocity used in the experiment, say v = 1. Also, we can very easily rig the experiment such that when M'=M then our caefully placed clocks are at B'=B and A'=A and the moving train triggers the photons leaving A and B at the same time. The A'B' distance is physically equivalent to AB. The starting times ar equivalent as we set them sio. We could very well have simply noted the times of all the clocks when M'=M, A'=A and B'=B and sorted out an agreed starting time, but setting the clocks to zero is easier and removes ambiguities.

That means you are not really seeking an answer and my time is wasted.

You can quit anytime you choose. If you are unable to grasp what is just written above, then maybe you had better quit, because there isn't anything conceptually complex. I keep seeing that you are making every effort to invoke SR, which is OK, but you aen't getting the starting conditions proper.

To summarize: A'B' = AB when v(A'B') = 1. t0(A'=A) = t0(M'=M) = t0(B'=B).
All calculations are perormed after the B and A photons are detected on the moving frame. So whatever the clocks speeds are, we know that all the clocks read '0' when the photons are emitted. If you still insist at this time, right now, that t0(A'=A) is not equal to t0(B'=B), you will never see it, and you can "do lunch" with Doc Al.

I am still not sure what you are trying to calculate?

Will you be as specific as you are able?

Ae you rying to determine an event time or what?

If you are trying to change the starting times of the photons forget it. If you want to change an experimental fact and assume that t0(A'=A) is not equal to t0(B'=B), or if you are trying to calculate what the t0(A'=A) and t0(B'=B) are, then you are wasting both our times. We designed the experiemtn to assure ourselves that the photn starting times were identical at both ends of the train. Technically difficult and expensive as hell, but reasonable achievable.

t0(A'=A) = t0(B'=B) expresses an instant in time, and this is all that these expressions indicate.

 

[Hurkyl]

The clocks at A', B' and M' remain synchronized during a gentle. When the moving frame reaches our experimental velocity the clocks are still synchronized with respect to each other on the moving frame.

Why would one think that?

 

[ram2048]

my take on this is you're trying to work the theory backwards-to-forwards, defining an outcome then working your way towards the "set-up"

the problem with your set-up is set-up thus, there is no WAY to arrive at the outcome. and starting from the outcome there is no WAY you can produce a logical backtract of processes to get to the beginning.

saying "that's just the way it is" doesn't make it right.

 

[geistkiesel]

As usual, you merely assume that there is a single instant when "A’ = A, M’ = M and B’ = B".

Setting the clocks to zero just makes them laughably unsynchronized.

Who said this synchronized the clocks? You did not me. Is this what is so laughable?

Consider how silly this statement is, given the conditions of your "experiment". It's like saying to everyone in the world "When you wake up this morning, set your clocks to 6:00 am", and then concluding "Look everybody, I've just proven that everyone wakes up at exactly the same time!".

If everybody did in fact wake up at the same instant, the your statement would be true. Here we merely set the clocks to zero when A'=A, B'=B and M'=M, which is the instant the photons were emitted. Why do you make such inane arguments as you have just posted? We are triggering photo emission times, not creating a universal 'wake up call'.

geistkiesel, don't you find it ironic that even wespe, who started this thread trying to expose a flaw in relativity, has ripped apart your circular arguments?

(When I say "even wespe", I don't mean that as an insult to wespe. I'm just pointing out the irony. wespe is doing a excellent job!)

If you are like Rush Limbaugh and have 'half your brain tied behind your back' I suggest you release that other half of your thinking power, you need both brain cells for this one.

Sure, you didn't mean an insult with your "even wespe" statement, this was just characteristically instinctive, wasn't it?

 

[geistkiesel]

Considering with simultataneous events...

Suppose that cell phone technology gets real good over the next 10 years so that one can generate 100GHz microwave transmitions from small compact transmitters no more then the size of a small cell phone. Now assume that one sets up an array of say 10,000 microspopic microwave transmitters' antennas around the circumference of a metalic disc with a 1 ft diameter with a reciever at the center of the disc . The transmitters transmit microwaves at a frequency of 100 GHz through a disc to the receiver at the center of the disc. Now the wave length of the microwave tranmission is 0.00984 ft. 1 ft divided by 0.00984ft is how many wave cycles fit in the diameter of the 1ft disc, which happens to be 101.6260162601626 wave cycles. Now if we set each transmitter to transmit at plus 360degrees/10,000 of a degree out of phase with the transmitter to its right, and minus 360degrees/10,000 of a degree out of phase with the transmitter to its left; then a spiral pattern will be formed by the sum of 10,000 transmitters waveforms. The spiral will have the illusion of spinning one 360 degree revolution every cycle. Now, since the transmitters are all transmitting at 100GHz, the spiral will appear to spin at a rate of 1 revolution 100,000,000,000th of a second. Since microwaves are traveling across the diameter of the disc at the speed of light, and the circumference of the disc is (pi)(diameter), and since the number of cycles that reside within the disc's diameter is 101.6260162601626, thus the velocity of a point on the spiral at the outermost circumference of the disc, would appear to be (pi)(101.6260162601626)c, that is approximately 319.1056910569106 times the speed of light! The question is, does time dilation play a role in this particular case? That is, would the spiral be spiralling back in time?

Inquisitively,

Edwin G. Schasteen

Assuming the truth of Special Relativity, time dilation would not be tied to your huge factor multiplying c, necessarily.. If I read your experiment correctly you are describing 'phase velocity" which is not 'signal carrying' and hence time dilation is not functionally related to these waves.

 

[geistkiesel]

Okay. Tell me, how far apart are your two lights? According to the Devfizz's clocks, when was each light lit? Pick a moving observer and describe his motion relative to the lamps. How does the moving observer calculate the photons were emitted simultaneously?

I could give a scenario, but the Devfizz's will have lit the lights at different times, according to their clocks.




If two lights are being activated, it's two events (unless they happened to be at the same place at the same time). Even if they occurred simultaneously, and even if time is absolute, why would you think it should be described as a single event?

I was merely suggesting, probably poorly, that we emit the light in the stationary frame such they reach an observer moving towards one and away from the other such that the photons reach the moving observer at the same time. We emit the lights nonsimultaneously.

 

[geistkiesel]

So how do you explain moving observer measures light speed from behind and ahead the same, if the light is trying to catch up from behind

I would locate a SR Theorist, yourself, say, and you, uisng the posulated fact that the speed of light is constant in all frames would measure the postulated prediction.

Do you understand that: When A' B' clocks are synchronized with a light signal from M', they will not look synchronized in the stationary frame. Or if you snychronize them with a light signal from M, they won't look synchronized in the moving frame. In fact that's the whole point. It's like reverse of this experiment. Time slowing is not sufficient to explain this experiment because both clocks would slow down equally, but the synchronization causes a shift in clocks.

I am not synchronizing clocks as you suggest. Long before the experiment gets going the moving frame places the clocks at A' and B' and M' and synchronizes these clocksonly to that frame.. A'B' > AB when both frames are statioanary, because we anticipate shrinking when we get going.. We slowly accelerate the moving frame to v(A'B' = AB) = 1 from the calculaions of our scienitfic team. In any event once the moviing frame is moving at the proper velocity when A'=A, B' = B and M' = M, the six clocks are set to zero by a mechanical switch. We have never made any attempt to synchronize the stationary and moving clocks with respect toi each other. Everybody knows he clocks slow down. Everyone is a SR Theory expert. We have stationary teams and moving frame teams. Thank you, you are correct, the moving frame clocks slow down equally.

Only when A' = A , B' - B , M' = M are the clocks set to zero by mechanical devices appropriately located at the three locations. This is not intended for any thing more than a convenience, such that the start times are known in both frames when the photons are emitted.

AS I stated in another post, this informaion ios not provided tot he moving observers until after they have deternined whether or not the photons were emitted simultaeously in both frames.

edit: Look what you have done to me. I am defending relativity in my own "why relativity is wrong" thread. I'm going offline now.

No I didn't do this to you, it was all arranged by a cycnical conspiracy of SR theorists. I am here to help you sir, trust me. You are doing a remarkably tenacious job.

 

[geistkiesel]

That is true; either A and B will not be synchronized in the "stationary" frame, or A' and B' will not be synchronized in the "moving" frame, or both.


Anyways, let me bring up some of my points again:

A and A' happen to be there when the photon from A is emitted, thus they can set their clocks to zero when that happens.
B and B' happen to be there when the photon from B is emitted, thus they can set their clocks to zero when that happens.

Question: How do M and M' know to set their clocks to zero?

All the clocks can be triggered by a mechanical switch when A' =A, B' = B and M' = M. Human switches ae elativiely unreliable.

Question: Suppose we did not assume a priori that the photons were emitted "simultaneously"; how can we tell from the results of the experiment whether or not the emission was "simultaneous"?

My infamous and http://frontiernet.net/~geistkiesel/index_files/

Question: How would one go about determining the relationship between clocks A and B? What about clocks A' and B'?[/QUOTE]

All this is done is the stationary frame before he experiment. Locate the midpoints M and M' and synchronize the clocks within their respective frames. Set A'B' > AB based on the calculated pediction of shrinking such that v(A'B' = AB) = 1 when the moving frame is moving.

 

[wespe]

G, will you continue to do this to yourself? Why do you care what happens if some train goes very fast? Are you seeking attention? Everybody kind of does, no shame in that. Do you want a nobel prize? Sure, but this is not the way. You know, I have been doing what you are doing now. Powered by my ignorance, I would come up with faulty arguments and expect people to devote their time to prove me wrong. I now feel guilty for what I did and I now understand why I was treated badly or ignored. Maybe we shoud quit doing this and start asking questions and reading and learning. I will probably change my nickname to start with a clean slate now, so this is my last advice to you. Forget these experiments, first shape up your discussing abilities. I found good advice here:
http://www.ephilosopher.com/Sections-article36-page1.html

Take care.

 

[geistkiesel]

G, will you continue to do this to yourself? Why do you care what happens if some train goes very fast? Are you seeking attention? Everybody kind of does, no shame in that. Do you want a nobel prize? Sure, but this is not the way. You know, I have been doing what you are doing now. Powered by my ignorance, I would come up with faulty arguments and expect people to devote their time to prove me wrong. I now feel guilty for what I did and I now understand why I was treated badly or ignored. Maybe we shoud quit doing this and start asking questions and reading and learning. I will probably change my nickname to start with a clean slate now, so this is my last advice to you. Forget these experiments, first shape up your discussing abilities. I found good advice here:
http://www.ephilosopher.com/Sections-article36-page1.html

Take care.

Feeling guilty because you are wrong? get a grip wespe, who isn't? Treated badly? people who hold their beliefs strongly will more often than not treat you badly when you try to undermine their belief systems, even though you might not define what you are doing as undermining.
Start a clean slate and change your name? Come on wespe, you haven't got anything to hide. Hey, Doc Al is on your side, cheer up!. You aren't thinking of going into engineering are you? If so, you had better change your name. Talk about being the town laughing stock.

 

[Hurkyl]

All the clocks can be triggered by a mechanical switch when A' =A, B' = B and M' = M. Human switches ae elativiely unreliable.

Ok. So, as the experimenter, how do we check that the switch at A was triggered at the "same time" as the switch at M and as the switch at B?

My infamous and undiscussed link has short description of the process.

I don't see an undiscussed link; we've told you multiple times you have a problem at the very start of your analysis, which you've ignored. Recall your infamous quote:

"However, before we even begin to consider relativity constraints we make note of the unambiguous physical reality of the simultaneous measurements."

where you make clear that you deny relativity before you even consider it.


In order to simplify the situation, let's consider a single moving observer who passes by A, receives photons, then passes by B.


The moving observer records this sequence of events, not necessarily in this order:

A passed by me at time w. A had a velocity of v.
B passed by me at time x. B had a velocity of v.
I received a photon from A at time y.
I received a photon from B at time z.

Give me a formula that tells me whether or not A and B emitted photons simultaneously.

 

[geistkiesel]

Ok. So, as the experimenter, how do we check that the switch at A was triggered at the "same time" as the switch at M and as the switch at B?

The time of emission is sent to M when when the photons wee emitted. M is the midpoint of A and B therefore the time of flight of the message and photon is identical as the speed of light is identical for both photons and signals. To keep a running track of the clocks you can have the clock signals sent to M on a continuous basis checking every cycle to assure that t(A) = t(B), Another way is to run he experiment a few millions time until a confidence level based on repeatability of the paameters can be assured.

I don't see an undiscussed link; we've told you multiple times you have a problem at the very start of your analysis, which you've ignored. Recall your infamous quote:

"However, before we even begin to consider relativity constraints we make note of the unambiguous physical reality of the simultaneous measurements."

where you make clear that you deny relativity before you even consider it.

http://frontiernet.net/~geistkiesel/index_files/
Hurkyl, you have been so thorough up to to now. The measurements were of the arrival of the A and B photons on the moving frame that were simultaneously recorded in the moving and stationary platforms. This is not an assumption of an ivariance of simultabneity that contradicts SR. Right?

In order to simplify the situation, let's consider a single moving observer who passes by A, receives photons, then passes by B.


The moving observer records this sequence of events, not necessarily in this order:

A passed by me at time w. A had a velocity of v.
B passed by me at time x. B had a velocity of v.
I received a photon from A at time y.
I received a photon from B at time z.

Give me a formula that tells me whether or not A and B emitted photons simultaneously.

You seemed to have switched frames. Was this ifor a reason? Not to worry, there is no problem either way.

The distance between A and B is vw - vy, which is the extra distance the farthest photon must travel in reaching the observer. So, when the B and A photon arrive at I, we assign t the value y - z. The distance the farthest photon travels is simply tc.
We start by assuming the photons were emitted simultaneously.

The extra distance for the A photon to travel to I is vw - vx.. Therefore we equate (vw - vx) = tc, where t = y -z or tc = (vw - vx). Just let v = 1 for convenience and we have the expression t = (w - x)/c. If we have measured w and x correctly then t should equate as the expression says. If t is less than the calculated value the the photon left A first, If t is greater than the expression the photon left A after B.

If the observer is moving and we assert SR postulates of time dilation we simply use the moving clock, as the speed of light is constant in the moving frame and we are only interested in determining differences anyway. If the observer is stationary he uses the stationary clock and makes straight forward calculations. Even if there were a time and space shrinking this mthod of calculation can deternmine whether the photons were emitted simultaneously.
I know you only asked if simultaneity was preserved, but you get the bonus of knowing which photon was emitted first if they were not emitted simultaneously.

 

[geistkiesel]

Summary of intentions re thread

In which frame are you asserting this?

A and A' can certainly agree the photon from A was emitted at t = 0.
Similarly, B and B' can agree the photon from B was emitted at t = 0.

Every other combination cannot simply be observed; it has to be determined via some means. e.g. how does M' know what time the photon was emitted from A, since his clock wasn't there when it happened?

I want to insert a summary here of where I am going with this.

Everything I have added to the experiment is to determine whether we can find a contradiction between experiment and SR theory such that any contradiction will be unambiguously clear.

Therefore I put clocks at A,B,M,A',B',M' and then perfectly made the experimental condition reasonably and physically performable. This is why I had 10000 runs on placing the clocks such that the condition, A=A', B=B' and M=M' is repeatable with a high degree of confidence and an effective experimental error of 0.

Our hired team of SR theorists calculated the time dilation and physical shrinking, for an expected velocity of v = 1, that could be repeatably achieved such that the experimental error in reaching v = 1 is measurably 0.

The acceleration parameters calculated by our mechanical physics group were tested until we were able to get one velocity expectation that was effectiively invariant -- repeatably invariant. We called this v = 1.

The test and instrument engineering firm developed a switch such that as the moving platform passed (A=A', B=B' and M=M') three events occured. The two clocks at each position immediately were set to zero and left to their own timing frequency. This was only for convenience. We could have simply read the current time from each clock and stored it somewhere.

The Physics calibration group synchronized the clocks in each frame such that the expected time of each clock could be determined with equal resolution for an error effectively 0; by reading any of the clocks. Or t(A) = t(M) = T(B) since we triggger each clock to zero by the switch set by the moving platform. Each clock's function does not depend on any other clock for any reason. Of course t(A') = t(M') = T(B') even after slow accleration to v = 1, which was verified. By the same mechanism each clock on the moving platform is dependent on no other clock.

The third event triggered at the same time is the emission of the photons at A and B.

As we ran and reran the experiment we were able to repeat without fail the experiment where, when A = A' and B = B' and M = M', then t(A = A') = 0, t(B = B') = 0, and t(M = M') =0. We refer to this primary test condition as: all clocks are set to zero at the same instant. With the newer version of clocks we are able to maintain an uninterrrupted ticking of an unperturbed twin clock.at each station. The clocks are intra-frame synchronized (calibrated) to any significant figure required, This means we have to edit our primary test condition by adding the current times are recorded along with the zero instant.

Why all of this elaboration? When the photon from B arrives at the moving platform at t(Pb) and the photon arrives from A at t(Pa), does not SR theory predict the events of the photon emissions from A and B were not simultaneous in the moving frame? If this is true, then the experimental results are unambiguously contradictory. The photons were not only measured to be emitted simultaneously there was the slam dunk guaranteed and pedictable experimental result that the photons from A and B were emitted simultaneously in the moving frame.

This being the case, then it is unavoidable that we recognize that SR theory is not describing a physical event with any correlated or uncorrelated consistency. The fact that experimental results unambiguoulsy prove physical simultaneity does not necessarily disprove SR theory, unless, SR theory is confined exclusively predicting physical conditions and/or events.

SR Prediciton: First one photon was emitted, then another photon was emitted from the sources at A and B in the stationary frame. The event described in the previous sentence is a physical event. This same event was observed in the moving frame though directly contradictory. If the assertion of SR theory is physically and intrinsically exclusive, then SR theory must be revised, if possible, or discarded otherwise. If my understanding that SR theory is exclusive, and confined to the prediction of physical events is correct, then SR theory must be discarded.

It is my position that the burden is on SR theory to accurately predict physical events. The postulates of SR theory that ultimately demand the loss of simultaneity are either singularly or collectively false. To maintain SR postulates invariantly is to maintain the falsehood of simultaneity loss.

 

[Hurkyl]

The distance between A and B is [vw - vx], which is the extra distance the farthest photon must travel in reaching the observer.

I take it you're only considering the case when both photons came from the same direction, then? (I.E. you are not considering the case when the actual sequence of events goes something like "Pass A, see photon A, see photon B, pass B")

(P.S. I'm not faulting you for doing this; it is a fairly messy task to account properly for all the possibilities)


So the criterion for simultaneous emission, according to the moving observer, is that:

c * (y - z) = v (x - w)

Note that I think you have one of the sides reversed; if the moving observer passed A before B, then he should expect to get B's photon first if emission was simultaneous, making both sides positive. (And similarly if B was passed before A)


As an aside, this is a nifty equation, because it renders calculation of time dilation irrelevant; any sort of dilation of times cancels out!


So let's remember this criterion and go back into the stationary frame. We have an experimental setup consisting of two stationary lamps, A and B. We watch a moving observer pass by both lamps with velocity v, and then, by some means, we trigger A and B simultaneously.

Notice that, in particular, this corresponds to the previous setup; the moving observer will pass A, then pass B, then get B's photon, then get A's photon. Thus, let us compute the times these occur.

For simplicity, the stationary observer sets his clock to 0 when the moving observer passes A. Let's suppose A was at position 0, and B was at position p.

Then, the moving observer passes B at time p / v.

Let's trigger the lamps at time 2p/v. The moving observer is at position 2p when this happens...

Thus, it takes time p / (c - v) for the photon from A to reach him, and time 2p / (c - v) for the photon from A to reach him.

So we have the following: In our frame,
He passes A at time 0.
He passes B at time p/v.
He receives the photon from B at time 2p/v + p/(c-v)
He receives the photon from A at time 2p/v + 2p/(c-v)


Remembering from our aside that dilation is irrelevant for the criterion that the moving observer computes the emission to be simultaneous. Let's substitute the times into the equation:

c * (y - z) = v (x - w)

yielding

c * ((2p/v + 2p/(c-v)) - (2p/v + p/(c-v))) = v ((p/v) - 0)
simplifying...
c * p / (c-v) = p
cancel out p...
c / (c - v) = 1
et cetera...
c = c - v
v = 0

Since the observer is moving, this equation clearly cannot hold, thus we conclude the moving observer does not compute that A and B were activated simultaneously.

 

[Hurkyl]

Let's run some actual numbers.

Suppose your setup is 10m long. In the lab frame, all 6 (identical) clocks are stationary and set to 0.

A' is at the left end, M' is in the middle, and B' is at the right end.


Suppose that A', M', and B' are all accelerated in exactly the same way to a velocity of 1 m/s. In this moving frame, the clock readings are no longer synchronous; B' > M' > A'. In particular, according to the moving frame, when A' reads 0, M' reads 5.56 * 10^-17 seconds, and B' reads 1.11 * 10^-16 seconds.

Note that the descrepancy is less than a millionth of a nanosecond. The differnce between B' and A' is less than a millionth of the frequency of a Cs-133 atom. Hardly a surprise it should go unnoticed. Heck, NIST's clock is only accurate to about one part in 10^15, still not enough to detect the difference.

 

[ram2048]

uhhh

if they're all accelerated the same way (i'm assuming you mean same direction and rate of accel) why are they not synchro anymore?

 

[Hurkyl]

If they gave the same readings at the same time in the lab frame...
And they all underwent the exact same acceleration in the lab frame...
Then they still give the same readings at the same time in the lab frame.

But that is no reason to think they give the same readings at the same time in the moving frame, unless you've already assumed simultaneity is absolute.

 

[ram2048]

but why should you think that it isn't the same in the first place.

if all variables for the clocks are the same then how can the result be different

the only answer is there's some external effect happening that we haven't thought of

 

[Hurkyl]

if all variables for the clocks are the same then how can the result be different

They can't.

But, according to SR, there are very few frames where all of the variables are the same. The lab frame is one of them, (and thus the readings in the lab frame would remain synchronized)


In most other inertial frames, the readings were not synchronized before the acceleration began, and they remain unsynchronized when the acceleration finished.


And if you actually want to consider the accelerated frames of A', M', and B', it turns out that the fact that A' is accelerating towards B', but B' is accelerating away from A' is important; it's the temporal analog of centrifugal force.

 

[ram2048]

And if you actually want to consider the accelerated frames of A', M', and B', it turns out that the fact that A' is accelerating towards B', but B' is accelerating away from A' is important; it's the temporal analog of centrifugal force.

i don't think that makes a difference. they're both accelerating away from their starting positions at the same rate, it wouldn't matter what direction they were going in.

 

[Hurkyl]

Well, the general result is that in accelerated frames, observations of clocks are affected by more than just time dilation. In particular, if you're accelerating towards a clock, you perceive it running faster, and if you're accelerating away from a clock, you perceive it running slower (/stopped/backwards).

So in A''s accelerated frame, B' will run fast. However, in B''s accelerated frame, A' will run slow.

 

[ram2048]

sounds like doppler effect, not time dilation

you're intercepting the photons that say what time it is faster than a person who is stationary

you
1:00--------59----------58----------57-----------56----------55---------54-------53(o) <-- clock
guy

guy stays still and waits for the light from clock to hit him, thus registers time as standard.

you move towards the clock and intercept incoming light from clock sooner than him just by virtue of being closer

 

[geistkiesel]

I take it you're only considering the case when both photons came from the same direction, then? (I.E. you are not considering the case when the actual sequence of events goes something like "Pass A, see photon A, see photon B, pass B")

(P.S. I'm not faulting you for doing this; it is a fairly messy task to account properly for all the possibilities)


So the criterion for simultaneous emission, according to the moving observer, is that:

c * (y - z) = v (x - w)

Note that I think you have one of the sides reversed; if the moving observer passed A before B, then he should expect to get B's photon first if emission was simultaneous, making both sides positive. (And similarly if B was passed before A)


As an aside, this is a nifty equation, because it renders calculation of time dilation irrelevant; any sort of dilation of times cancels out!


So let's remember this criterion and go back into the stationary frame. We have an experimental setup consisting of two stationary lamps, A and B. We watch a moving observer pass by both lamps with velocity v, and then, by some means, we trigger A and B simultaneously.

Notice that, in particular, this corresponds to the previous setup; the moving observer will pass A, then pass B, then get B's photon, then get A's photon. Thus, let us compute the times these occur.

For simplicity, the stationary observer sets his clock to 0 when the moving observer passes A. Let's suppose A was at position 0, and B was at position p.

Then, the moving observer passes B at time p / v.

Let's trigger the lamps at time 2p/v. The moving observer is at position 2p when this happens...

Thus, it takes time p / (c - v) for the photon from A to reach him, and time 2p / (c - v) for the photon from A to reach him.

So we have the following: In our frame,
He passes A at time 0.
He passes B at time p/v.
He receives the photon from B at time 2p/v + p/(c-v)
He receives the photon from A at time 2p/v + 2p/(c-v)


Remembering from our aside that dilation is irrelevant for the criterion that the moving observer computes the emission to be simultaneous. Let's substitute the times into the equation:

c * (y - z) = v (x - w)

yielding

c * ((2p/v + 2p/(c-v)) - (2p/v + p/(c-v))) = v ((p/v) - 0)
simplifying...
c * p / (c-v) = p
cancel out p...
c / (c - v) = 1
et cetera...
c = c - v
v = 0

Since the observer is moving, this equation clearly cannot hold, thus we conclude the moving observer does not compute that A and B were activated simultaneously.

Maybe so. The simple model I used had the moving sorces passing by the stationary observer. He was able to make the determination that one or the other photons were emitted before the other. Do all the calulations you desire. The experiment I deswcribed in my summary still holds. If one measures the emission of the photons simulaneously then any theoretical construct that contradicts the expemenatl observation is flawed.

 

[geistkiesel]

Let's run some actual numbers.

Suppose your setup is 10m long. In the lab frame, all 6 (identical) clocks are stationary and set to 0.

A' is at the left end, M' is in the middle, and B' is at the right end.


Suppose that A', M', and B' are all accelerated in exactly the same way to a velocity of 1 m/s. In this moving frame, the clock readings are no longer synchronous; B' > M' > A'. In particular, according to the moving frame, when A' reads 0, M' reads 5.56 * 10^-17 seconds, and B' reads 1.11 * 10^-16 seconds.

Note that the descrepancy is less than a millionth of a nanosecond. The differnce between B' and A' is less than a millionth of the frequency of a Cs-133 atom. Hardly a surprise it should go unnoticed. Heck, NIST's clock is only accurate to about one part in 10^15, still not enough to detect the difference.

Your analysis is totally bogus. Come on Hurkyl. The A', M' and B' clocks are on the same rigid frame and are all accelerated together, at the same time. How in th e heck do you get the inequaliies you do? Do you make this up as you go along? It is the same train man. The clocks are just set at diffeent points in the train. Are you trying to be purposfully obscuring the hypothetical experiment in an attempt to confuse me or any observer to this thread?