Is Relativity Incorrect? Examining Motion and Perception Evidence

[wespe]

Geistkiesel, we have a problem.. There seems to be a language barrier between us (possibly on my side, I'm not native english speaker). I'm having difficulty understanding you, and I may not be able to express myself clearly. Therefore this discussion will probably be fruitless if we continue...Please see below...

OK maybe SR predicits this but see the fallacy.

"predicts this".. What did you refer to with "this"?

The moving observer sees two photons of light consecutive in time. She can conclude a number of possibilities.

OK, I'm with you up to this point.

That she was at the midpoint of the light sources as she crossed the midpoint measured from the stationry observer, or

How is this is one of the possible conclusions? This was already a given in the experiment setup.

or that one light preceded the other,

Again, it's not clear what you mean here. "One light preceded the other" is the observation. How is this one of the possible conclusions?

http://frontiernet.net/~geistkiesel/index_files/ and which light it was.

I don't understand what you mean by "which light it was". I checked the link once more. That looks like the same experiment I gave the link for, with added stationary observers. I followed you until "the simultaneous arrival of the light pulses at the midpoint at 2t1". Where did that 2t1 come from? I'm already lost. Would you consider creating a clearer version of the page? Could you list the events in time order, like "t0=A emits light", "t1=O receives light from A" etc?

The most basic of common sense does not restrict the moving observer to conclude only the lack of simultaneity. Blue/red analysis of the photons measured indicates the observer is moving wrt the sources of the light if you assume the light photons are identical when emitted,

As far as I was informed, there is no blue/red shift for a single photon. However, if the light source is continuously emiiting photons, there is blue/red shift, but then they are not single events, and you can already tell there is a movement but it is mutual. I think a lightning is supposed to represent a single event.

plus the stipulated fact of her crossing the midpoint when he light were emitted in the stationary frame.. Unless we define the moving observer as a total science gumball she can evaluate a number of possibilities.

I have no clue what you are saying here...

There is no synchronnization of clocks necessary here. The moving observer times the dt between t1 and t2 in her moving frame, period.

I can't make sense out of this, since I failed to understand your version of the experiment. How can you not need synchronized clocks to measure one way speed of light?

I trust you aren't saying the moving observer knows nothing of her crossing the midpoint at t = 0, in her moving frame? Otherwise the problem is ambiguous.

Yes the moving observer knows her crossing the midpoint. That's the experiment setup. There is no issue here.

What is wrong with assuming the speed of light is constant in all moving frames as measured in that moving frame? Whether she is moving or not she will always measure c constant.

I'm surprised you don't see a problem there. Replace the light signals with bullets. Do you not see the speeds should be measured as c+v c-v relative to the moving observer? Moving observer measures the speed relative to herself, not relative to the stationary frame.

The moving observer you call M' arrives with the incoming light the same instant the stationary observer also measures the oncomig light..

I can't make any sense of this... This cannot be the description of the experiment in my link. I'm sorry.

Place 100 light detectors 10 km apart, approximately, and zero the optical distance distance of each leg. over a 90 degree angle. by monitoring any deviations from the optical path equivalence as the Earth rotates and moves around the sun,etc. temperature and other non light speed perturbations can be statistically accounted for. Each measurement is one way. What is wrong with this experiment? carried over 5 years say?

What is being measured here or how is this measuring one way speed of light?

The above is an exception to your postultion that there is only one way to disprove SR. One does not have to disprove SR to disprove it. One has to merely find an exception to the derivation of the simultaneity consequences predicted by the application of the fundmental postulates of SR. The logic applying in reverse will take care of itself. You surrendered to the SRists prematurely. I did not see any contradiciton in your 'Baez' link of the short analysis that derived the t1 = (c - 1)/2 expression (also seen in my link above),except of course, the implication that any variation from predictions of SR are wrong, period.

Geistkiesel, if you want to continue like this, I'm willing to give it a try. But if you are not willing, just ignore this reply and we go our own ways... no problem.

 

[geistkiesel]

Geistkiesel, we have a problem.. There seems to be a language barrier between us (possibly on my side, I'm not native english speaker). I'm having difficulty understanding you, and I may not be able to express myself clearly. Therefore this discussion will probably be fruitless if we continue...Please see below...


"predicts this".. What did you refer to with "this"?

predicts what SR predicts. Nothing moe, nothing less.

OK, I'm with you up to this point.


How is this is one of the possible conclusions? This was already a given in the experiment setup.


Again, it's not clear what you mean here. "One light preceded the other" is the observation. How is this one of the possible conclusions?

I am saying that the detection of the light first from the front, then the rear has a number of conclusions one may infer other than the two measurments being restricted only to the conclusion that the emitted photons were not simultaneous. If the moving platform had no information of its own motion then the moving observer may, never the less, infer more than that the photons were emitted in sequence, or said another way, were not emitted simultaneously. The moving observer thinking himself stationary can consider that he is closer to the source in front of him than from the source to the rear of him. He need not automaticvally determine the photons were not emitted simultaneously. This is the stationary aspect of the 'moving frame question' we discussed earlier.

I don't understand what you mean by "which light it was". I checked the link once more. That looks like the same experiment I gave the link for, with added stationary observers. I followed you until "the simultaneous arrival of the light pulses at the midpoint at 2t1". Where did that 2t1 come from? I'm already lost. Would you consider creating a clearer version of the page? Could you list the events in

I guess that the midpoint should have been stated as the midpoint measured between the position at +t1 and earlier at -t1. like "t0=A emits light", "t1=O receives light from A" etc?

The sequence is: t0 = o the mutual time the moving platform passed the midpoint M of the photon sources and the time of simultabeous emission of the photons from both sources. t1 is t0 + t1 = t1, the time of measuring the first photon from the front. t2 the tiem t0 = t2 = t2 the time of mesuring the 2nd photon from the rear.

I made an assumption, (to be tested later) that when the 1st photon arrived at the moving frame at t1, the time of transit between the midpoint M and the arrival of the 1st photon, that the moving platform had in fact passed through the midpoint of the moving sources and photons. This would place the photon arriving later from the rear at a position to the left of the midpoint at -(1)xt1. The second photon was measured at t2 and I assumed for convenience that dt = t2 - t1 = 1. Now then, the time of flight of the second photon to the point determined by t2 is (2t1)x(v) + (dt)x(1). (x is the mathematical symbol for 'times') Since both dt and v = 1, conveniently, we may set the distance between the second photon at the time of arrival of the first photon at 2t1 +1 to the left of the point determined by t2. (remember we have tenmporarily assumed the midpoint where the moving platform zeroed his clock was the midpoint of the emitted photons). The distance the 2nd photon must travel in the time t2 - t1 = 1, is (1)x(c) or the distance 2t1 +1 = 1xc = c and therefore after rearranging the expression we get, t1 = (c - 1)/2 = k. As t1 was a measured quantity, a comparison with k will determine if the photons were emitted simultaneously from sources located equidistant from the midpoint where the moving platform zeroed his clock as we assumed. For a finding that t1 = k, the assumption will prove correct. Otherwise one or the other sources was turned on first. Here the given condition was the stationary M, coincident with the moving M' was the midpoint of the sources of light emitted simultaneously. Hence for t1 = k, this is confirmed.

As far as I was informed, there is no blue/red shift for a single photon. However, if the light source is continuously emiiting photons, there is blue/red shift, but then they are not single events, and you can already tell there is a movement but it is mutual. I think a lightning is supposed to represent a single event.

As an after thought, a series of coded photon pulses can be used to keep a running track of the sources, but this is outside or given conditions.

I agree otherwise, but noting the wavelength of both photons a blue./red shift determination can be accomplished. Consider the hypothetical varying slightly. The light sources are attached at both extreme ends of the moving platform coincident with the sources located in the stationary frame. Then to the moving frame a bleu/red analysis indicates the soures are attached to the moving paltform and both photon wavelength are identical, but the stationary frame will see a non-null blue/red shift determination from sources attached to the moving platform.

I have no clue what you are saying here...

I can't make sense out of this, since I failed to understand your version of the experiment. How can you not need synchronized clocks to measure one way speed of light?

From the moving platform, with dilated clocks, this platform determes the t1 = k value. t1 is tested for equality with k. The moving observer having assumed that M ws the midpoint of the two photons, measured strictly in the moving frame, the t1, measured against the constant k, will detemine simultaneity of the emitted photons with respect to the moving platform. The stationary t1, will be a different number than the measured t1' in the moving frame because the clock speed is assumed to have slowed in the moving frame consistent with SR theory, correct? Under the conditions of what is given, that is the passing of the moving frame through the midpoint when the photons were emitted , the clocks are arbitrarily set at t0 = 0 in all frames. Hence, knowing the velocity of the moving frame as v =1 everything seen strictly from the moving platforms observation, the t1' = k derived above is the t1' of the moving platform. If at anytime the stationary observer and the moving observer comares their respective t1 and t1' both will recognize that moving platform t1' is a number less than the number measured by the stationary observer. Then the clocks can be synchronized, but for what purpose? Simultaneity of the emitted photons has already been established independently by both observers.

I had to sever this post in two parts because of the length.

 

[geistkiesel]

Wespe,this is the continuation of my previous poss in responding to your last post.
First, I am not sure of any need to measure the speed of light as all observers can look up the value in a table of physical constants .The speed of light is assumed to be c in both frames, which is fundamenmtally consistent with SR postultes, is it not? When t1 is determined in the moving frame the observer their uses c = 299xxxkm/sec in calculating k, as does the stationary observer. t1, however is a different number on the output channels of the respective clocks, but the lack of clock synchronization between frames is totally unnecessay for the analysis of the problem..

I endeavor for you to see that the derivation ot t1 = k requires no clock synchronization with the stationary frame. The clock 'synchronization' can be striclty limited to giving the moving platform a reference to the instant of passing through the midpoint at M and only this. The moving platform can either 'zero' the clocks physically (in both frames, neither observer knowing of the other's measurement or even existence) or merely note the current reading in the output channels of the moving clock.

Wespe, I am ready willing and able to discuss this matter until it is resolved. Your use of the English language would shame many who claim English as a first language. I don't feel particlualry shamed, but I am very aware that you are able to understand my corriupted English as well as expressing yourself more than adequately in English. You need not offer any disclaimers or apologies for the form or content of any of your posts. Were I in you shoes, I would even feel a bit smug and pomous of my grasp of a foreign language to the high level that you exhibit, but then our own unique and self imposed limits of personal modesty are not a barrier to understanding each other's posts. It is the subject matter of the thread that provides the difficulty not the mutial use of the English language. This being the case as I see it, there is only the SR problem of simultaeity that we need concern ourselves with. I am not in any particlular hurry to rush the conclusion of this problem to some hasty and unambiguous end.

Please, if any of the above has not been clearly understood by the slightest amount do not cut the questions off with an assumtpion that myself or others, will find your questions as trivial or unimportant. I mean this in every sense of the word. Any, the slightest misunderstanding I beg you to bring to my attention. I will assume the same from yourself.

geistkiesel

 

[geistkiesel]

Wespe,
I am making a guess that you will understand the old adage that applies to the current SR theorists in a way that you might discover some familiarity with.

"Alles kunst ist umsonst wen ein Angel das zündloch prünst"

Anon

 

[wespe]

Wespe,
I am making a guess that you will understand the old adage that applies to the current SR theorists in a way that you might discover some familiarity with.

"Alles kunst ist umsonst wen ein Angel das zündloch prünst"

Anon

Geistkiesel, hi.

Google translation: "All art is in vain whom fishing rod the zuendloch pruenst". The last part is bad so I couldn't even get your joke. :(

Actually, at the moment I was trying to decipher your long reply. I suggest that we first focus on clarifying the experiment. All other issues can wait. My reply is in order, but I'm a bit slow with math and I have some other matters to take care of, so there will be some delay.

To avoid any confusion, I'm planning to label some events and rename some variables, if you don't mind. Then we need the equations for both observers, equations for all measured times and lengths, and then we may even find some nice round numbers to plug into the equations. I will post my reply when it's ready. I hope this is OK with you.

 

[geistkiesel]

Geistkiesel, hi.

Google translation: "All art is in vain whom fishing rod the zuendloch pruenst". The last part is bad so I couldn't even get your joke. :(

Actually, at the moment I was trying to decipher your long reply. I suggest that we first focus on clarifying the experiment. All other issues can wait. My reply is in order, but I'm a bit slow with math and I have some other matters to take care of, so there will be some delay.

To avoid any confusion, I'm planning to label some events and rename some variables, if you don't mind. Then we need the equations for both observers, equations for all measured times and lengths, and then we may even find some nice round numbers to plug into the equations. I will post my reply when it's ready. I hope this is OK with you.

Sounds good to me. The adage: All skill is in vain when an Angel ______ on your flintlock.
How about one experiment at a time to thoroughly wring it out? If I may suggest the one we were supposedly discussing. This is Einstein's example in huis book "Relativity". The famous moving train. Just as the midpoint M' of the train coincides with the M of the stationary frame and midpoint of two light sources located at A and B (A behind B in front) the sources emit a photon. M sees the light from B first and then from A later. AE claims that because of the different arrival times of the photons the moving observer thinks the pulses were emitted at different times, period.

What ever seems easiest to you, or hardest for that matter, we'll do it.

 

[wespe]

Geistkiesel.. here's my primitive attempt.. it turns out I'm really bad with math. My final result is that photons were not emitted simultaneously in moving frame. I suspect your analysis omits the time dilation effect, but I'm not sure if I'm doing this properly.. let me know..

.....M->v
A...d...O...d...B

t=0, O and M coincide and reset clocks, A emits PhotonA (EventPA) , B emits PhotonB (EventPB)

EventMB: M detects PhotonB
EventOD: O detects PhotonA and PhotonB
EventMA: M detects PhotonA

// For O:

xO = 0
xM = v*t
xPA = -d+c*t
xPB = d-c*t

EventPA: xPA = -d => tPA = 0
EventPB: xPB = d => tPB = 0
EventMB: xM = xPB => tMB = d/(c+v)
EventOD: xPA = xPB = 0 => tOD = d/c
EventMA: xM = xPA => tMA = d/(c-v)

// For M:

gamma = 1/sqrt(1-v^2/c^2)
Length contraction => d' = d/gamma
Time dilation=> tMA'-tMB' = (tMA-tMB)/gamma

xM' = 0
xO' = -v*t'
xPA' = -c*tMA'+c*t'
xPB' = c*tMB'-c*t'

EventPA: xPA' = -d' => tPA' = tMA'-(d'/c)
EventPB: xPB' = d' => tPB' = tMB'-(d'/c)
EventOD: xPA' = xPB => tOD' = (tMA'+tMB')/2
xPA' = xO' => tOD' = c*tMA'/(c+v)
xPB' = xO' => tOD' = c*tMB'/(c-v)
=> tMA' = (c+v)/(c-v)*tMB'


//say c=3, v=1.5, d=9

//For O:
tPA = tPB = 0
tMB = d/(c+v) = 2
tOD = d/c = 3
tMA = d/(c-v) = 6

Event order:
tPA=tPB= 0 < tMB=2 < tOD=3 < tMA=6

//For M:
gamma=1.16
Length contraction => d' = d/gamma = 7.76
Time dilation=> tMA'-tMB'= tMA-tMB/gamma = 3.45
tMA'=(c+v)/(c-v)*tMB'=3.tMB'
tMB'=1.725
tMA'=5.175
tOD'=(tMA'+tMB')/2=3.45
d'/c=2.59
tPA'=tMA'-2.59=2.585
tPB'=tMB'-2.59=-0.865

Event order:
tPB'=-0.865 < tMB'=1.725 < tPA'=2.585 < tOD'=3.45 < tMA'=5.175

 

[geistkiesel]

Motion with respect to coordinates of emitted photon sources.

Geistkiesel.. here's my primitive attempt.. it turns out I'm really bad with math. My final result is that photons were not emitted simultaneously in moving frame. I suspect your analysis omits the time dilation effect, but I'm not sure if I'm doing this properly.. let me know..

.....M->v
A...d...O...d...B

t=0, O and M coincide and reset clocks, A emits PhotonA (EventPA) , B emits PhotonB (EventPB)

EventMB: M detects PhotonB
EventOD: O detects PhotonA and PhotonB
EventMA: M detects PhotonA

// For O:

xO = 0
xM = v*t
xPA = -d+c*t
xPB = d-c*t

EventPA: xPA = -d => tPA = 0
EventPB: xPB = d => tPB = 0
EventMB: xM = xPB => tMB = d/(c+v)
EventOD: xPA = xPB = 0 => tOD = d/c
EventMA: xM = xPA => tMA = d/(c-v)

// For M:

gamma = 1/sqrt(1-v^2/c^2)
Length contraction => d' = d/gamma
Time dilation=> tMA'-tMB' = (tMA-tMB)/gamma

xM' = 0
xO' = -v*t'
xPA' = -c*tMA'+c*t'
xPB' = c*tMB'-c*t'

EventPA: xPA' = -d' => tPA' = tMA'-(d'/c)
EventPB: xPB' = d' => tPB' = tMB'-(d'/c)
EventOD: xPA' = xPB => tOD' = (tMA'+tMB')/2
xPA' = xO' => tOD' = c*tMA'/(c+v)
xPB' = xO' => tOD' = c*tMB'/(c-v)
=> tMA' = (c+v)/(c-v)*tMB'


//say c=3, v=1.5, d=9

//For O:
tPA = tPB = 0
tMB = d/(c+v) = 2
tOD = d/c = 3
tMA = d/(c-v) = 6

Event order:
tPA=tPB= 0 < tMB=2 < tOD=3 < tMA=6

//For M:
gamma=1.16
Length contraction => d' = d/gamma = 7.76
Time dilation=> tMA'-tMB'= tMA-tMB/gamma = 3.45
tMA'=(c+v)/(c-v)*tMB'=3.tMB'
tMB'=1.725
tMA'=5.175
tOD'=(tMA'+tMB')/2=3.45
d'/c=2.59
tPA'=tMA'-2.59=2.585
tPB'=tMB'-2.59=-0.865

Event order:
tPB'=-0.865 < tMB'=1.725 < tPA'=2.585 < tOD'=3.45 < tMA'=5.175

I am sure you are doing it properly but there may be an inherent flaw due to your use of time dilation which you are assuming. Here is the argument that explains the flaw, if any flaw exists.

If the omission of the photons were in fact simultaneous, with emphasis on if, then SR is in error. Therefore, in testing the truth of the matter we must start at some point that does not infer SR and at the same time does not deny SR, yet the analysis must still be physically correct.

I have just posted the following in response to doctordick who rejected my 'link analysis', get this, because it is "simple minded". This is the sum total of his analysis. It may be simple minded, which I gleefully confess, but see if the following makes sense to you. I only suggest that you are of a frame of mind that is purely objective before inserting 'objections' from preconceived ideas regarding SR. The objections are satisfied. I have the same mental aberrations as anyone though I can still maintain an 'open mind' when reading and analyzing theories and ideas that conflict even grossly with my own. I retain the objections rising from the subconscious as background noise until finishing absorbing the ideas in front of me. Someimes I surrender, some times I take no prisoners, which is the case here as there are too many prisoners to take.

As peviously submitted
You have misread the sequence of events. I shall be very careful here in describing what is occurring.

The numbers in the figure can apply to moving or stationary fames with ‘dilation corrections’ appropriately inferred.

Midpoint coordinates of emitted photons are invariant as photon speed is not source or frame dependent. Midpoint coordinates of expanding light spheres are invariant from the instance of emission of the photons. Therefore, any statements, derivations or conclusions drawn from experimental results, and/or theoretical constructs or postulates, denying the reality or concept of absolute space, meaning invariance in the spatial coordinates of events, must be discarded.

[Wespe I add here: The motion of the observer can be determined with respect to the invariant coordinates of the emitted photon source. The mirrors are moving, not with respect to the photons such that we add and subtract c and v, rather the mirrors move with respect to the coordinates of the photon source]

Photons and mirrors converge to predictable coordinates. In the moving frame the left photon and mirror converge before the right photon and mirror converge. Photons and observer converge simultaneously in all frames. The Δt of emission-to-convergence is greater in moving frames due to the greater photon distance traveled.

The observers’ perceptions cannot be determined until the convergence of the photons, hence all determine that the reflections were not simultaneous in the moving frame, but the emissions were simultaneous. The dilations of moving clocks can be explained, partially at least, due to the extended distance traveled during emission and convergence times.

Michelson-Morley experimental results were not null. An aether drift induced wave shift of 1/20 of that predicted was subsequently confirmed by Dayton Miller.

 

[geistkiesel]

Wespe - please accept m apologies. The xample in my reply to yours was different than our agreed example which you properly analyzed. However, as the subject matter is consistent whatever value thee is in my reply is still relevant to your analysis as the postulates of SR imposing the time dilation contraints must be modified on a fundamental level.

Let me epeat the basic argument based on a physical assumption that the speed of light is constant from whatever frame measured [which may have some mdification due also].

If two photons are emitted in opposite directions or a short burst of photons expand as a sphere we see that the midpoint is invriant as:

<- P ->
<- P ->
<- P ->
with t moving down and distance along the horizontal. The only way the midpoint P can vary is if the speed of one of the photons varies, or if the relative speed of both photons vary. If hee were a photon speed variation then the instantaneous location of P would shift. Of course thee could not be in reverse time implications rather there would result a dri8ft of the measured midpoint. In an experiment with spatial dimensions in the order we have been discussing only the measured postions of the photons can be compared as a test of theory, which is what my derivation of t1 in our chosen experiment was or t1 = (c - 1)/2 = k, for a velocity of the moving platform v =1 and the distance traveled between dt = t2 - t1 = 1 where v and dt are convenient and arbitrary.

In my analysis, and correct me if I am in error, I assumed the invariance of the midpoint M of the moving photons throughout the analysis, hence the times and distances were measured from an absolute position coincidental with the midpoint of the sources at the instant the photons were emitted. The irony of the situation here is that the physical location of M measured does drift with respect to the stationary platform as the platform moves (the Earth's rotationsal and solor orbit motion) with respect to the absolute position of the midpoint defined by the emitted photons.

 

[ahrkron]

Michelson-Morley experimental results were not null. An aether drift induced wave shift of 1/20 of that predicted was subsequently confirmed by Dayton Miller.

In the middle of http://www.weburbia.demon.co.uk/physics/experiments.html (section II, "Repetitions of the MMX") there is a table with experimental results of Michelson-Morley type experiments. The last one in the table (by Joos, performed in 1930!) gave a result of 1/350 of the expected shift.

The missing piece of information here is the experimental error. A shift of 1/10 would be enough to declare SR correct (or rather, to declare wrong the alternative way of computing the expected shift) if the expected error is of the same size. Basically, you have two theories, one predicting 350, one predicting 0, and the result turns out to be 1+-1. I think that settles things in favor of SR.

 

[wespe]

I am sure you are doing it properly but there may be an inherent flaw due to your use of time dilation which you are assuming. Here is the argument that explains the flaw, if any flaw exists.

If the omission of the photons were in fact simultaneous, with emphasis on if, then SR is in error. Therefore, in testing the truth of the matter we must start at some point that does not infer SR and at the same time does not deny SR, yet the analysis must still be physically correct.

I have just posted the following in response to doctordick who rejected my 'link analysis', get this, because it is "simple minded". This is the sum total of his analysis. It may be simple minded, which I gleefully confess, but see if the following makes sense to you. I only suggest that you are of a frame of mind that is purely objective before inserting 'objections' from preconceived ideas regarding SR. The objections are satisfied. I have the same mental aberrations as anyone though I can still maintain an 'open mind' when reading and analyzing theories and ideas that conflict even grossly with my own. I retain the objections rising from the subconscious as background noise until finishing absorbing the ideas in front of me. Someimes I surrender, some times I take no prisoners, which is the case here as there are too many prisoners to take.

As peviously submitted
You have misread the sequence of events. I shall be very careful here in describing what is occurring.

The numbers in the figure can apply to moving or stationary fames with ‘dilation corrections’ appropriately inferred.

Midpoint coordinates of emitted photons are invariant as photon speed is not source or frame dependent. Midpoint coordinates of expanding light spheres are invariant from the instance of emission of the photons. Therefore, any statements, derivations or conclusions drawn from experimental results, and/or theoretical constructs or postulates, denying the reality or concept of absolute space, meaning invariance in the spatial coordinates of events, must be discarded.

[Wespe I add here: The motion of the observer can be determined with respect to the invariant coordinates of the emitted photon source. The mirrors are moving, not with respect to the photons such that we add and subtract c and v, rather the mirrors move with respect to the coordinates of the photon source]

Photons and mirrors converge to predictable coordinates. In the moving frame the left photon and mirror converge before the right photon and mirror converge. Photons and observer converge simultaneously in all frames. The Δt of emission-to-convergence is greater in moving frames due to the greater photon distance traveled.

The observers’ perceptions cannot be determined until the convergence of the photons, hence all determine that the reflections were not simultaneous in the moving frame, but the emissions were simultaneous. The dilations of moving clocks can be explained, partially at least, due to the extended distance traveled during emission and convergence times.

Michelson-Morley experimental results were not null. An aether drift induced wave shift of 1/20 of that predicted was subsequently confirmed by Dayton Miller.

Geistkiesel,

Well, I am not sure if I did the analysis properly. I may have also made a mathematical mistake somewhere. But I did not assume anything except the given setup conditions, plus the SR effects. Then, the general equations of motion forced the outcome as I calculated.

I think you are suggesting that we remove time dilation and length contraction effects from the analysis. OK, I will try that and see. But, my guess is, the main problem is "constancy of speed of light in all frames". The photons cannot be simultaneously emitted in all frames, mathematically, if speed of light is to be constant in all frames. Your reply mentions about MMX result being not null. So are you suggesting that we remove constant light speed assumtion from the analysis too? Anyway, if you remember my post about one way light speed, you know that I don't dismiss the (c+v)(c-v) possibility. So I'll try that, too. (there is a discussion going on currently about one way light speed (OWLS) on sci.physics.relativity newsgroup, SR experts don't deny OWLS hasn't been measured and it is an unproven postulate).

Well I have to go now. I'm having trouble finding spare time for this discussion, so I may not always be able post a reply promptly.

 

[geistkiesel]

How to find simultaneity.

Geistkiesel,

Well, I am not sure if I did the analysis properly. I may have also made a mathematical mistake somewhere. But I did not assume anything except the given setup conditions, plus the SR effects. Then, the general equations of motion forced the outcome as I calculated.

I think you are suggesting that we remove time dilation and length contraction effects from the analysis. OK, I will try that and see. But, my guess is, the main problem is "constancy of speed of light in all frames". The photons cannot be simultaneously emitted in all frames, mathematically, if speed of light is to be constant in all frames. Your reply mentions about MMX result being not null. So are you suggesting that we remove constant light speed assumtion from the analysis too? Anyway, if you remember my post about one way light speed, you know that I don't dismiss the (c+v)(c-v) possibility. So I'll try that, too. (there is a discussion going on currently about one way light speed (OWLS) on sci.physics.relativity newsgroup, SR experts don't deny OWLS hasn't been measured and it is an unproven postulate).

Well I have to go now. I'm having trouble finding spare time for this discussion, so I may not always be able post a reply promptly.

Draw a picture of the events as they happen without any reference to moving or statioanry frames. We have the two sources. We have M we have he point to the right where the B photon reaches moving platform at B' and the point where the photon fom A reaches the moving platform at A'. We ask, what are we trying to determine? I asked, Does he scenario just described prove the moving observers determine the photons left A and B simultaneously or otherwise?

First I notice that a platform coming from the opposite direction as a mirrored representation is going to get the opposote "relativity" based answer as her twin. A third platform moving perpendicular and equal distant from M as the twins must be at some point along the line joining A and B when a photon is met also is aimed at the midpoint..

If we carefull point a million observers to reach the midpoint M from a million different directions (and say a million experimental arrangements) those crossing M with the arriival of the photons will, if relativity is correct, all get different answers for the calculated photon emissions, except for the observers heading to the midpoint a a plane perpendiclular to, and centered at, M. These all calculate the same emission times. Remember, in all cases we are using relativity theory to make any calculation, therefore our experimental arrangement is not designed to prove anything regarding raltivity or the consequential loss of simultaneoty. Therefore using relativity theory proves nothing other than a claimed discovery of a scenaio consistent with relativity theory. A fix crap game up to here.

OK let us see. The marked coordiantes of A and B, M and the points B' and A' where B and A photons are detected. Laying along this path we mark locations on the stationay platform at each place the train observer detected photons and the midpoint. You can even insert two clocks at all points and label them S and M, where the M clocks are clicking at a relativity rate of 1/2 that of the S clocks. We were clever enough to have our stationary detectors located at the same spatial locations throughout. There is no need to make any calculations to determine whether the emitted photons are physically sequential in the moving frame. We know they are physically simultaneous in the stationary frame. We have a moving clocks coincidental spatially with the stationary clocks at A and B. Each singular event at A and B therefore is, by definition simultaneous, such as the instant the photons were emitted. SO if one B' clock concidental with B stationary clock reads 9, then the moving clock A' at A also reads 9, and the fact that the stationary clocks both read 10 means what regarding simultaneity?

If you calculated that the photons were emitted at different times in the moving frame, then how do you explain the fact that both of your clocks read he same value, yet your calculations concluded otherwise. Where did you assume it was necessary to calculate the time the photons were emitted? Likewise, if you are on the moving platform where the laws of physics are the same as the stationary frame how do you explain that the laws of physics in one frame gives a grossly different conclusion as the moving frame? This is not within the definition of the physical equivalence of inertial frames, is it within your defintion? Still, even with the measured clocks having the same reading on the moving platform, under what theoretical imperative must you adjust the already dilated clocks to give the nonsimultaneous emission of the photons? iF the platforms are unique within themselves what is the necessity, if any, of recognizing the very existence of other frames in order to determine if the photons are simultaneously emitted? If you were on the moving paltform would you need any nonmoving platform data in order to make a calculation on your platform? What if every one on the moving platform were the twntieth generation of the oiginal crew without a scrap of paper even suggesting relativity theory and all physics problems were void of relativity content but f = ma in the moving frame differing in calculatied results from the stationary calculation by a time fiunction that is lorentz biased?

Are we calculating or observing? If both, then why do measurements, even with time slowing, differ from your calculated version? What does, "perceive the events to be nonsimultaneous" mean? Is this difeent from emasuring the events? Does it mean simply that the clock difference of the two frames define nonsimultaneity?

Remember, there was no direct observtion that the photons were not emitted simultaneously. WE must compare clocks to infer the fact, but using the moving clocks, is unambiguous that we infer the photons were emitted ast the same time in the moving frame.

Wespe, unless you see the unconditional necessity of using lorentz mathematics to determine the perception, I say the problem went away. I read Einstein. He uses the trivial fact of the different times that the A and B photons arrived as sufficient to discard concepts of simultaneity. But again AE didn't refer to the mere possibility that both emission times can be observed by all frames, hence the contradiciton.. Nor did AE bring up the possibility that the different times of photon arrival does not imply a conclusion of nonsimultaneity as the observers could reason that they were moving and having a picture of the arrangements consider the possibility the photons were emitted simultaneously.

 

[wespe]

I don't want to continue this discussion for it will be a waste of time. I spent many hours on that analysis. Simply put: if you are midpoint of two light sources, and if they emit light at the same time according to your clock, and if light speed is constant, you will detect the lights at the same time. If you don't detect them at the same time, that means either 1- the distances were not the same, 2- light speed was not constant, 3- they didn't emit the light at the same time. I'm not even sure which one of these is your claim. I see no other possibility if x=vt. (And if you are moving towards one of the light sources, you cannot possibly measure light speed the same in both directions, that would correspond to #2 above. And if you measure them the same, either #1 or #3 must be true. AE says it's #3)

Take care.

 

[geistkiesel]

Devious Experimenters; Not reccomended as a regula Experimental Diet.

Geistkiesel,

Well, I am not sure if I did the analysis properly. I may have also made a mathematical mistake somewhere. But I did not assume anything except the given setup conditions, plus the SR effects. Then, the general equations of motion forced the outcome as I calculated.

Wespe,
An Intwoah Tivleep pounced upon me just I published the previous post. I fought and struggled, but was powerless against the overwhelming force, and in the end, my defenses spent and shattered I surrendered to the Intwoah Tivleep. With a shaky moral hand, I did, with a reasonbly proper plethora of regret and angst, of course, force myself to post you the following:

Let us say two conspiring and devious physicists Wes Devfizz and Geis Devfizz conduct an experiment giving the relativity industry the opportunity to reproduce the Einstein gedunken:

The two scoundrels set the stage by first calculating the same difference in time for the emission of two photons and actually emit the photons such that the times are such that the moving observers will perceive the photons were emitted simultaneously. Now the times of arrival are such that the moving frame should calculate the photons were emitted simultaneously, right? And the genius of SR will discover this situation, right? Are the moving frame relativitists going to get to perceive a correct answer from their perspective? We ask them how their theory can provide such a horendously and grossly incorrect answer, a deviation from physical reality, such that one event is made into two events? The strongest and most committed of the RTs will say that as to their perception the photons were emitted at the same time and that's just how it is down at the relativity factory. Then, we show them the secret clocks we had hidden that gave the actual moving and stationary clock readings equating the A' and B' readings in the moving frame and the A and B readings in the stationary frames. A - A' = B - B' > 0 for A' = B'.

We should be standing at some safe distance removed when we ask the RTs "the question": howd ja splainal dis? (Translated: How do you explain all this, coming from AE, "you know, the name yas got tooed on yers theeree der"?)

The problem Wespe ain't a simultaneity problem. It is a problem with relativity theorists who demand the right to perceive the universe in any way they choose. They are teaching this stuff to our children in public and private schools you know?

 

[wespe]

OK from me to you too. I don't care. What the hell, I'm anonymous here.

OK. So what you are saying is: "after we show the secret clocks to the moving observer, he should be convinced that the lights were emitted at the same time". But according to those secret clocks, moving observer should have measured light speed as c+v and c-v too, right? So he should be convinced about that too, or should he trust his own measurements? (I already told you that his measurement of light speed was not one-way, but in fact two-way, so it is possible that c+v and c-v was real and canceled each other.) Just tell what is your claim about the light speed measurements.

 

[geistkiesel]

I don't want to continue this discussion for it will be a waste of time. I spent many hours on that analysis. Simply put: if you are midpoint of two light sources, and if they emit light at the same time according to your clock, and if light speed is constant, you will detect the lights at the same time. If you don't detect them at the same time, that means either 1- the distances were not the same, 2- light speed was not constant, 3- they didn't emit the light at the same time. I'm not even sure which one of these is your claim. I see no other possibility if x=vt. (And if you are moving towards one of the light sources, you cannot possibly measure light speed the same in both directions, that would correspond to #2 above. And if you measure them the same, either #1 or #3 must be true. AE says it's #3)
Take care.

I won't let you quit yet. Now we have it all settled.

A. You are at the midpoint when the lights are emitted. You detect the light from B before it reaches M and then you detect the light from A after it has passed M. You can't possibly detect the lights from the sources if you are moving and located at the the midpoint when the lights turn on. You see the lights in sequence first B, you are closer because you are both heading to each other, then A, as A has to catch up from behind.

B. You forgot one. Let us say your train is really long and you have sycnchronized clocks A' and B' that secretly measure the emitted photons from A and B simultaneously and the stationary frame also measures the emitted photonas simultaneously and the coordinates of the A and A' pair, and the B and B' pair of clocks are < 10^-12m of each other, and even if A' = B' < A = B and these A' and B' clock outputs are consistent with SR time slowing theory, what, if any corrections to your post do you make now that you've seen the moving and stationary platform data?

 

[wespe]

I won't let you quit yet. Now we have it all settled.

A. You are at the midpoint when the lights are emitted. You detect the light from B before it reaches M and then you detect the light from A after it has passed M. You can't possibly detect the lights from the sources if you are moving and located at the the midpoint when the lights turn on. You see the lights in sequence first B, you are closer because you are both heading to each other, then A, as A has to catch up from behind.

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

B. You forgot one. Let us say your train is really long and you have sycnchronized clocks A' and B' that secretly measure the emitted photons from A and B simultaneously and the stationary frame also measures the emitted photonas simultaneously and the coordinates of the A and A' pair, and the B and B' pair of clocks are < 10^-12m of each other, and even if A' = B' < A = B and these A' and B' clock outputs are consistent with SR time slowing theory, what, if any corrections to your post do you make now that you've seen the moving and stationary platform data?

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.

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.

 

[Hurkyl]

The two scoundrels set the stage by first calculating the same difference in time for the emission of two photons and actually emit the photons such that the times are such that the moving observers will perceive the photons were emitted simultaneously.

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.

one event is made into two events?

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?

 

[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

The moving observers and the staionary are making use of the physical fact that the speed of light in their respective frame is constant at 300,000km/sec. There is no measurement of the soeed of light. Everybody knows what the speed of light is. The observers are recording when the light is recorded in their frame.

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.

yes.

Or if you snychronize them with a light signal from M, they won't look synchronized in the moving frame

Yes.

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.

SO what? And why sycnchronize clocks, especially the way you are trying to do? See below.

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.

This is soon to change. Read very closely and do not look for ways to jump out of the box I am stuffing you into, not just yet. See below, OK?

Wespe, you are so close to seeing it. It does not matter if the clocks in both the frames are synchronized to each other. The question is whether the moving observer will predict, or calculate or perceive that the emitted photons were were simultaneously emitted. Did the A' precede the B' emission, or did B' precede A'. And this is all. Can you see this? You asked some questions above. What is the pourpose of those questions? Why is the answer necessary?

Why are you so obsessed with synchronizing the clocks? or measuring the speed of light? Who needs iit and for what purpose?

Even without any reference to any stationary or other frame of reference the moving observer is asked a question: Do you perceive the photons were emitted at the same instant in the moving frame?, or if you prefer, "emitted simultaneously in your frame"? The moving platform had carefully synchronized its clocks in the stationary frame then slowly accelerated to the constant speed it was moviing when the midpoint of the moving platform passed the midpoint of the sources at M on the stationary frame just as the sources emitted a photon simultaneously at A and B.So what if the moving clocks do not give the same output as the stationary clocks? Why is this importatnt to you? And what does this have to do with determining whether the photons A and B were emitted at the same time as perceived by the moving platform?

There is absolutely no reason for the moving platform to receive some synchrionizing signal friom the stationary platform, especially as M' passes therough M. This isn't an issue relevant to detemining simultaneity.

If you want to include the question that even if the photons were turned on simultaneously in the moving frame does the moving frame observer still have the imperative to determine that the time of emission of the photons be identical to that seen in the stationary frame? If so then you might be suggeing that the stationary clocks say "photons leave A and B at 12:00 noon", while the moving platform clocks say "photons leave A and B at 11:00 A.M." . If this is he case what do you say if the moving platform had clocks in the extreme forward and exreme rear position of the train such that at exactly the iinstant the photons left A and B the moving clocks recorded the same event at that same instant.

What I mean is that assume there is an observer at the front and rear of the train reading the clocks as well as observers at the A and B locations on the stationary platform. Also, let us add that each observer pair at each end of the train and stationary platforms saw directly that the other made the same measurement at the same instant, which is a different meaning than saying "the clocks read the same". No this is not what "at the same instant" means. Each observer can determine the photon left the source because the moving observer saw the photon leave at the same instant the stationary observer saw the photon leave.They aren't asking each other to synchronize clocks to tell anyone "what time it was that the simultaneious event occured" They could do this but why?
.
Ok we will add that the observers at each end swaps his ckock reading with the observer in the oher frame who is right there. The stationary observers sends clock readings of 10, at both ends, the moving observers send clock readings of 9 at both ends. Later, when the forward and rearward observers compare data, they see that their moving platform clocks read the same time. In other words the photons were measured to have been emitted at the same time in the moving frame.

These observers also note that the stationary observers sent the same clock readings to both moving observers at each end of the train. The moving observers both received a "12:00 PM" and the stationaray observes both received an 11:00 AM" at both ends. These observers say to each other "It sure looks like the photons were emitted simultaneously doesn't it.?" , "Yes" each replied.

These two obsercvers then apprioach the O' observer who made calculations based on whatever he uses to determine if the photons are to be perceived as having been emitted simultaneously in the moving frame. Naturally, he finds that the photons were not emitted simultaneously. He uses an algorithm wespe might use which is striclty obedient to SR dictates. He finds no simultaneity. Then the clock observers show their measurements to the Observer who made the calculations? What is he going to say when the measuremnets from his frame shows a different result than that he had just calculated using SR theory?

As the train observer in the middle of the train at M' when M' and M shared identical coordinates at the same instant, then the photons were emitted. The observer at M' did not know of the measuremnts made at both ends of his rather long train. SR tells us that the O' observer should determine that the B light went on before the A light. And so the O' observer so dutifyully determines. y=using hisSR algorithyms.

 

[wespe]

The moving observers and the staionary are making use of the physical fact that the speed of light in

their respective frame is constant at 300,000km/sec. There is no measurement of the soeed of light. Everybody knows what the

speed of light is. The observers are recording when the light is recorded in their frame.

"Everybody knows this, no need to measure". How very unscientific of you. You should in fact be objecting to the constancy of

speed of light. Everything else in SR has derived from this assumption. Clock slowing, length shortening, relative

simultaneity, all made up just to satisfy the assumption! Yet, surprisingly, you want to keep the assumption, but you object

to the results. Do you care to check what happens if you write simple x=vt equations and solve them? You get the SR results,

all this just because of that assumption. [/QUOTE]

SO what? And why sycnchronize clocks, especially the way you are trying to do? See below.

Why? Because, to determine simultaneity of two events, you need either: 1-A midpoint observer, or 2-synchronized clocks. For

#1, you must also assume that light speed is constant in both directions, because the observer will use light to see the

events. For #2, you don't need the assumption, but you already used the assumption to synchronize clocks in the first place!

Say, something happened here at 5 o'clock, and something else happened there at 5 o'clock. Did they really happen at the same

time, or were the clocks in different time zones or something? You must make sure the clocks were synchronized, and that

again involves light signals.

Same is true for measuring speed. Say, I will throw a ball and I want to measure its speed. I can either: 1-read my clock,

throw the ball, ball bounces off a wall, I read my clock again, I divide the distance by the difference of readings. Or, 2-

synchronize two distant clocks, throw the ball, divide the distance by the difference in clock readings. Again, for #1, I

have to assume the constant speed of the ball in both directions. For #2, I need to make sure the clocks were synchronized,

which involves light signals. Please tell me you understand...

This is soon to change. Read very closely and do not look for ways to jump out of the box I am stuffing

you into, not just yet. See below, OK?

OK

Wespe, you are so close to seeing it. It does not matter if the clocks in both the frames are synchronized

to each other. The question is whether the moving observer will predict, or calculate or perceive that the emitted photons

were were simultaneously emitted. Did the A' precede the B' emission, or did B' precede A'. And this is all. Can you see

this? You asked some questions above. What is the pourpose of those questions? Why is the answer necessary?

You cannot perceive simultaneity if the events occur at different places. You can only see lights coming from those events.

And then you can make calculations, and then you conclude. To make those calculations, you need to know: 1-Your clock

readings of seeing the events, 2-The speed of light in both directions, 3- The distance to the events. Then, using x=vt, you

can know exactly where the light signals were, as a function your time. Then, you calculate what your clock was showing when

the events occured. Are they equal? Events were simultaneous. Are they not equal, events were not simultaneous. Simple as

that. Why you not trying to check the simple x=vt calculations? Am I wasting my time here?

Why are you so obsessed with synchronizing the clocks? or measuring the speed of light? Who needs iit and

for what purpose?

Please see above.

Even without any reference to any stationary or other frame of reference the moving observer is asked a

question: Do you perceive the photons were emitted at the same instant in the moving frame?, or if you prefer, "emitted

simultaneously in your frame"?

As I said above, you don't perceive the events themselves or their simultaneity. You only see the light coming form them, and

when you see them depends on distance to the events, and the speed of light in both directions.

The moving platform had carefully synchronized its clocks in the stationary frame then slowly accelerated to the constant speed.it was moviing when the midpoint of the moving platform passed the midpoint of the sources at M on the stationary frame jus as the sources emitted a photon simultaneously at A and B.So what if the moving clocks do not give the same output as the stationary clocks? Why is this importatnt to you? And what does this have to do with determining whether the photons A and B were emitted at the same time as perceived by the moving platform?

You can't avoid using light speed. This is a kind of slow transport. But the clocks won't remain synhcronized once you accelerate. How do you assume they remain synchronized? Suppose you push the train from its back to accelarate it. Your push will propagate at the speed of light, causing a delay, which in turn casues the clocks to go out of synch (due to time dilation). Suppose you push the train from the back and pull from the front at the same time. To do this at the same time, you must first have synchronized clocks, with a light signal, so you already assumed speed of light was constant in both directions.

For the rest of your post, I will reply when you consider my points above.

Geistkiesel,if:
you don't see the importance of synchronizing clocks
you don't understand why you can't sychronize clocks without light speed
you don't see a reason to measure speed
you don't see 300.000km/s is two-way-light-speed, not one-way-light-speed
you won't make any calculations as simple as x=vt

you must be making some wrong assumptions or a logical error somewhere which I can't pinpoint because I don't understand what you're saying and I don't want to waste any more time trying

 

[geistkiesel]

The Revision of Special relativity

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

 

[wespe]

Geistkiesel. Chill out.

What you are saying is what any normal person would think. Yes, they gave each other high fives. No problem. Events occurring at the same place is not an issue, nobody disagrees with their simultaneity. Still, that doesn't connect the back and front ends of the train, according to SR. Get this: When seen from the embankement, the back end of the train is living in the past, and the front end of the train is living in the future. That's how SR claims the events did not occur at the same time. Bull****, right? We will just make corrections. Problem is, you can't. According to SR, when seen from the train, the opposite is also true. It's mutual ! Which calculation will you consider correct? Is it embankement's, because it is "stationary"? SR says there is no way to determine this, both are real and true. But, if one-way-light-speed is not the same in both directions as SR postulates, we can determine which calculation was false. In fact, the calculations would change and become consistent. (though time dilation and length contraction remain intact). Aether and absolute time would come back. That's what I want, if the reality is so. But without such proof, you can't accomplish much by discussions.

Whatever, man. Take care.

 

[geistkiesel]

Geistkiesel. Chill out.

What you are saying is what any normal person would think. Yes, they gave each other high fives. No problem. Events occurring at the same place is not an issue, nobody disagrees with their simultaneity. Still, that doesn't connect the back and front ends of the train, according to SR. Get this: When seen from the embankement, the back end of the train is living in the past, and the front end of the train is living in the future. That's how SR claims the events did not occur at the same time. Bull****, right? We will just make corrections. Problem is, you can't. According to SR, when seen from the train, the opposite is also true. It's mutual ! Which calculation will you consider correct? Is it embankement's, because it is "stationary"? SR says there is no way to determine this, both are real and true. But, if one-way-light-speed is not the same in both directions as SR postulates, we can determine which calculation was false. In fact, the calculations would change and become consistent. (though time dilation and length contraction remain intact). Aether and absolute time would come back. That's what I want, if the reality is so. But without such proof, you can't accomplish much by discussions.

Whatever, man. Take care.

I edited the last post. Check out the proper vesion. My suggestion is you learn to remove robotic thinking tendencies.

Geistkiesel

 

[geistkiesel]

The physics of simultaneity is restored to moving inertial frames..

Do you think M will receive photons from A and B at the same time? And will M' receive photons from A' and B' at the same time? Or will M detect A first then B, and M' detect B' first then A'? Which one?

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?

 

[geistkiesel]

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?

 

[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.