Today Special Relativity dies

[geistkiesel]

I'd say more like frustrating. At some point you will wear everyone out. What you guys are doing isn't anything like physics. You are just rejecting SR because it seems to defy common sense, which we all understand (no we aren't crazy). But you can't ignore experiment results which are in accordance with SR, not the intuitive Gallilean relativity (I'm not sure Geistkiesel understands even that). Other people aren't stupid, not to mention 100 years of brain power combined. You can see I'm frustrated with you Ram and Geistkiesel too.

No I don't reject SR just because it defies common sense, even though it surely does that, at least you do see that. If you have been following my threads on the subject I use a hell of a lot more than "violations of common sense" if you are paying attention like you claim you are. You might not agree with me and my techniqoues and you might be frustrated, so what? Do you see that ram1024, myself, grounded and eyesaw, to name a few, even yourself once, remember wespe, are all approaching from a different point of view, but guess what there is a focus and we aren't as far apart as we were last month last week or yesterday, right? I suspect each has a secet desire fror recognition at the very least. There is a purpose, personal I mean, in ewhat anybody does, right? And for physics forums, theory development it sure isn't the money, is it? Are you getting paid? I wopuld be soely poissed if Wspe was getting some quid pro quo and I was only getting my regular monthly DSL bills. :surprise: I can assure you that my pockets are as bare as they were when I started this.

You were a lot more interesting and energetic, to yourself as I observed, even when you realized how far off the nornmal path you actually were. But wasn't it energetic, stimulating, got the old gray cells moving in ways you didn't know they could move?
:surprise:
Wespe no one is saying experimental results aren't what they are.

A stupid trout example. Let us say you have a watch that runs fast. The watch is expensive, the best in town. The only thing you use the watch for is its designed purpose: measuring the speed that trout swim in cold water. Let us say the watch has trout moving 1.2 times what was measured using ordinary watches. Since your watch was so expensive and calibrated, wrongly it seems, it was hard to reject such expensive technology. Trout around the world are now believed to swim in cold water 1.2 faster than heretofore thought since All trout swimmer measurers got a wespe watch that has the same built in error. So when a trout is found not to swim 1.2 faster than wespe's watch expects, it is probably due to using a watch other than wespe's so someone comes up with the excuse that wespe's watch which is used under water etc. is the only accurate measure, therefore the trout found out of sink, (pun intended) get explained away by some inane and inexplicable nonsnese, like dark matter, or ptolemy's circles within circles, restricted fin action, you know hre drill. But every time wespe's watch is used the trout swim at 1.2v as expected. Therefore, only use wespe's watch and all the associated techniques of using the watch will get you back on the proper physical tack if you want to measure trout swimming speed according to physical law: go wespe's watch!. After a while truth takes on a different turn that is coaxed along with each finding of a conflict. Dissidents are exiled to Theory Development forums, oh the horror, the horror... :yuck:

I said the example was stupid, but the metaphorcal inanity is in synch with built in "always observed in experiental results" rhetoric.

One thing you have never done is publically or privately, made a decent and honest effort, your own standards, in taking soemone's appraoch and see if there is anything reasonable or appropriate. Check out some of Doc Al responses to my posts. Sometimes he is on, other times he is just screaming in my face and he claims it is physics I am getting. You know all about that don't ya?

I consider tom_mattson a very bright guy, but he doesn't always approach physics as I feel he should, he would agree at least partially on this (not openly, after all he is a mentor and like America's greatest thepian of all times, yes I speak of John Wayne, was fond of saying, "Don't apologize it is a sign of weakness." He might have used the word "pilgrim' in this context here I am not sure.) sometimes just like a normal person he goes screaming at someone repeating SR mantras, do you dig it? If it weren't for dissidents of fundamental concepts of science and our beloved physics all worhtwhile activity of human kind would stagnate, wouldn't it, hasn't it?

I can tell from this post it is way past my bedtime.

 

[russ_watters]

Wespe no one is saying experimental results aren't what they are.

What you and the others don't understand or simply won't accept is that what the experimental results say directly contradicts you. When this is pointed out and we ask that you give an example of one experiment that contradicts Relativity, you guys turn conspiracy theorist, claiming the results are faked, experimenters lie, and there is a vast conspiracy among scientists to hide the flaws in Relativity. This is why I (and finally, just about everyone else) have stopped playing this little game of obfuscate-the-thought-experiment with you guys. Are you dishonest? Are you blind? At this point, I don't really care.

 

[Doc Al]

No SR is the key result of simultaneity, read the history sir.

Gibberish.

BS. NASCARS are running invariant in speed in my example and you ignored that example.

Unless your NASCARS are running at the speed of light, their speeds are not invariant.

Eintein does not use any SR postulate in arriving at his conclusion.

Of course he does: the invariant speed of light.

He uses only the the staggrered arrival of theh B photon and A photon as detected by the O' observer. You have said on many occasions that this is what all agree.

That "staggered arrival" by itself means nothing. But when coupled with Einstein's postulate of the invariant speed of light, then you can conclude that simultaneity is relative.

quote Doc Al "The fact (agreed by all) the light from B hits O' before the light from A leads the O' observer to conclude the lights could not flash simultameously."

Right--assuming that the speed of light is frame invariant.

How do you know this? They must check the times they are on opposite ends of the train, but each immediately relay their findings to O' who receives the data along with the simultaneous observation of the A and B photons arriving at M simultaneously.

I know it because I know relativity. Realize that your sleight of hand of "relaying their findings" doesn't change their findings.

You are invoking SR time dilation and mass shrinking right?

(What the hell is mass shrinking?)

What else could you be invoking to make your statement that the observers at A and B will not record the same time of the emitted photons.

You are babbling again: Observers at A, B, a, and b all record the time that the photons were emitted. A and B record the same times; a and b record different times.

The photons know nothing of SR theory or even that there are two observers dutifully recording the times the photons are emitted.

The photons don't have to know SR; but if you wish to make an accurate prediction of how they behave, you'd better know SR. You don't.

the a and b observers record the same time.

On what basis do you make that wild and incorrect claim?

Remember, what you said back in the Lost simultaneoity thread. let me remind you:
quote by Doc Al
"If you understood the Eisntein simple argument you would know that nowhere does "time dilation" or " mass shrinking" enter into it"

Right! Let's be perfectly clear:

Does Einstein need to invoke "time dilation" or "length contraction" to prove that simultaneity is frame dependent in his Train Gedanken: No! All he needs to invoke is the invariant speed of light.

Does that mean that "time dilation" and "length contraction" are irrelevant to a complete analysis of the Train Gedanken? No! To go beyond Einstein's simple argument, and completely describe all events from each frame you must include the effects of time dilation, length contraction, and the relativity of simultaneity. For example: Where is observer a on the train? observer b? What times do their clocks read when they detect those photons being emitted from A and B? Where is the train when these photons are detected? All these questions are from the train frame: to get the answers, you need to understand SR.

 

[geistkiesel]

Einstein tells us that the mere staggered arrival time of the photons and that c is constant as everybody knows, is all that the passenger have that they "'must, therefore come to the conclusion" the photons were not emitted simultaneously. The same would hold for the NASCAR hypothetical.

That "staggered arrival" by itself means nothing. But when coupled with Einstein's postulate of the invariant speed of light, then you can conclude that simultaneity is relative.

AE said the staggered arrival times was everything. So you alter AE's conclusions do you? To win an argument, or to keep it going from your naive prerspective and to hell with physics. AE did not inlcude the invariant speed og light as you are trying to stuff it in some crack here in the conversation where it doesn't belong and where only a classic DOc Al smog bank is created. Your light invariant argument doesn't get off the grouind. It was redlined Doc. That kite doesn't fly.


Einstein says: "Hence, the observer will see the the beam of light emitted from B earlier than he wills see that emitted from A. Observers who take the railway train as their reference-body must therefore come to the conclusion that the lightning flash at B took place earlier than the lightning flash A. We thus arrive at the important result: Events which are simultaneous with reference to the embankment are not simultaneous wrt the train and vice versa."

Also if O' weren't moving O' would see the lights emitted simultaneously.
I read, motion, staggered arrival and the speed of light is c.

So what I infer from this is that the observers that see the lights arrive simultaneously at M', here frame wise no different than where O' had just measured the arrival of B, there would be no loss of simultaneity as the photons would have arrived simultaneously. All your postulates and arguments are included. I also conclude that to the O' observer the photons were not emitted simultaneously in the moving frame, but only by definition, and that for the a|b observers at the midpoint M when the photons A and B arrive simultaneously, the emitted photons were simulaneously emitted in the stationary and moving frame.

Therefore the Sr theory predicts that on the same frame there are a multitude of observers seeing the simultaneous arrival of the photons as well as a multitude seeing no simultaneity.

The observers at A and B, a and b, are also recorded the times the photons were emitted and must also conlcude the photons were emitted simultaneously. Assuming the observes a and b know they are measuring photons that are simultaneously emitted in the stationary frame, they must conclude they are measuring the same as their partner on the olther end of the frame. The a and b clocks are moving frame synchronized, therefore each clock always reads the same. Therefore the a and b clocks a must read the same as the clocks used by the observers at M viewing the arrival of the A and B photons arriving simultaneously there. and of course, O'.

Another problem you have. Assume the stationary frame emits ionly one photon instead of two. Will the photon anticipate which photons it was that had photons been emitted simultaneously from both sources?

If the B photon is exclusively emitted will it be measured the same instant of emittance as in the case the A photon were emitted also? How do the photons know they are a pair and that one must precede the other? Nonlocal entanglement?

How do you come to the conclusions you do? If the a and b observers and the a|b observers all relay their times to O' the signals will arrive simultaneously with the A photon. Hence O' will have two unambiguous data sources indicating simultaneously emitted photons. But you know different, why?

I know it because I know relativity. Realize that your sleight of hand of "relaying their findings" doesn't change their findings.

I never said it did change their findings. I was merely giving you some alternative to observations of passengers on Einstein's train that conclude the photons were emitted simultaneously in the moving frame, obviously.

This hypo of AE is no different tha the conclusion spectators at the NASCAR race I described will also conclude the NASCARS emitted form A and B emitted sinmultaneously as determined by the moving Ford Futura, the analog of O', as well as 100,000 spectators viewing the event.

(What the hell is mass shrinking?)
You are babbling again: Observers at A, B, a, and b all record the time that the photons were emitted. A and B record the same times; a and b record different times.
The photons don't have to know SR; but if you wish to make an accurate prediction of how they behave, you'd better know SR. You don't.

Prove the a and b and the a|b passengers will not measure the same as the A and B observers. "I know it because I know relativity." This is your physics, "I know", is your physics?. The forum is there for you to put your knowledge on the line and to prove what you know.

What do we do in the alternative, just start calling each other names and making the wise cracks? By the way, if you were looking for the job of smart ass, that job is already filled, and by a real professional. Need I say more?

On what basis do you make that wild and incorrect claim?

Wild guess, maybe or finally, but it turned out OK, like some of the other wild guesses that turn out ok. Incorrect though, you must show those with more than "I know".

From the reason and logic learned in the study of physics. Einstein didn't use a gamma in his gedanken, nor did he infer the existence of a gamma regarding the conclusions derivable form the gedanken, but Doc Al can hardly wait. However,I anticipate no such physics will be forthcoming from Doc Al.

Right! Let's be perfectly clear:

Does Einstein need to invoke "time dilation" or "length contraction" to prove that simultaneity is frame dependent in his Train Gedanken: No! All he needs to invoke is the invariant speed of light.

That is what I thought, there is no definitive SR postulate or imperative that proves the a and b observers did not observe the photons emitted simultaneously, and the same ogres for th a|b [passengers agt the midpoint when A and B pohoton s arrive there simultaneously.

Does that mean that "time dilation" and "length contraction" are irrelevant to a complete analysis of the Train Gedanken? No! To go beyond Einstein's simple argument, and completely describe all events from each frame you must include the effects of time dilation, length contraction, and the relativity of simultaneity.

I knew it, you flim flammed us, or are trying to.

Try looking at this again:
"Hence, the observer will see the the beam of light emitted from B earlier than he wills see that emitted from A. Observers who take the railway train as their reference-body must therefore come to the conclusion that the lightning flash at B took place earlier than the lightning flash A. We thus arrive at the important result: Events which are simultaneous with reference to the embankment are not simultaneous wrt the train and vice versa."

You mean that if we just use different observers than O', then you have to use time dilation and length contraction, AKA mass shrinking.? Sounds like a fast swimming trout story to me.
Bogus Doc, you can do much better. You seem to think nobody remembers what you just posted above, in this post. AE didn't need it but Doc Al needs it in order to smog up the discussion and keep some semblance of SR alive. Chill DOc. ram1024 has given everybody another couple of days, Relax, prepare yourself for the inevitable. Don't embarrass yourself.

For example: Where is observer a on the train? observer b?

a and b are at A and B when the photons were emitted into the moving frames, which is recorded by their clocks. a nad b obseved the phtons being emitted. a and b recorded the time of the emitted photons when emitted simulataneoslh at A and B.

What times do their clocks read when they detect those photons being emitted from A and B?

t' = 0. The same as the O' observer who zeroed her clock when she arrived at the midpoint just as the photons were emitted simultaneously from A and B. Long before the O' detected the staggered arrival of the B and A photons. Come on we have done this scene many times before.

Where is the train when these photons are detected?

The train? Come on Doc, please. Doc the passengers are riding on the train. a is at A, b is at B and O' is at M, when the clocks were set to zero, remember?

All these questions are from the train frame: to get the answers, you need to understand SR.

To get the answers? I do know SR. I even understand it. It's just isn't true that's all. The a and b observer record the A and B photon simultaneously when emitted at A and B, as do the a|b observers when the A and B photons arrive at M simultaneously.

You think you are going to blind side the maths ignorant geistkiesel do you? Try me Doc. I dare you to try me.

Ese el mentor!: la li'nea habe dibujado en el desierto con lamis espada. Aqui! ahora!

1. Question Doc Al: How do the a and b observers know there are two photons emitted from the stationary frame?
   
2. or said another way, how do the photons know they must change their emittance protocol because the photons are about to be emitted into the moving frame?
3. How do the photons know which one must emit first,
4. and at what what time? [*] I think you are digging yourself a perception of the observers hole instead of building a phyisics arguement.

.

 

[Doc Al]

Einstein tells us that the mere staggered arrival time of the photons and that c is constant as everybody knows, is all that the passenger have that they "'must, therefore come to the conclusion" the photons were not emitted simultaneously.

Yes, and your point is what? "c is constant" means "the speed of light is invariant".

AE said the staggered arrival times was everything. So you alter AE's conclusions do you? To win an argument, or to keep it going from your naive prerspective and to hell with physics. AE did not inlcude the invariant speed og light as you are trying to stuff it in some crack here in the conversation where it doesn't belong and where only a classic DOc Al smog bank is created. Your light invariant argument doesn't get off the grouind. It was redlined Doc. That kite doesn't fly.

It's time to up the dosage on your medication, geistkiesel. What do you think "constant speed of light" means?

The observers at A and B, a and b, are also recorded the times the photons were emitted and must also conlcude the photons were emitted simultaneously.

Wrong.

Assuming the observes a and b know they are measuring photons that are simultaneously emitted in the stationary frame, they must conclude they are measuring the same as their partner on the olther end of the frame.

Wrong. All a and b do is observe the times of emission and record it. They observe different times.

The a and b clocks are moving frame synchronized, therefore each clock always reads the same. Therefore the a and b clocks a must read the same as the clocks used by the observers at M viewing the arrival of the A and B photons arriving simultaneously there. and of course, O'.

After all these posts you still haven't the foggiest notion of what's going on. Clocks a and b are syncronized in the O' frame; so of course they read "the same time" in the O' frame. But emissions at A and B do not happen at the same time in the O' frame.

Another problem you have. Assume the stationary frame emits ionly one photon instead of two. Will the photon anticipate which photons it was that had photons been emitted simultaneously from both sources?

Huh? Totally incoherent. What are you talking about?

If the B photon is exclusively emitted will it be measured the same instant of emittance as in the case the A photon were emitted also? How do the photons know they are a pair and that one must precede the other? Nonlocal entanglement?

Huh? Totally incoherent. What are you talking about?

How do you come to the conclusions you do? If the a and b observers and the a|b observers all relay their times to O' the signals will arrive simultaneously with the A photon. Hence O' will have two unambiguous data sources indicating simultaneously emitted photons. But you know different, why?

You keep claiming that a and b measure the same time? Where do you get this? Read Einstein's train gedanken one more time.

Try looking at this again:
"Hence, the observer will see the the beam of light emitted from B earlier than he wills see that emitted from A. Observers who take the railway train as their reference-body must therefore come to the conclusion that the lightning flash at B took place earlier than the lightning flash A. We thus arrive at the important result: Events which are simultaneous with reference to the embankment are not simultaneous wrt the train and vice versa."

I think you are the one who should look at this again. It obviously hasn't sunk in yet.

a and b are at A and B when the photons were emitted into the moving frames, which is recorded by their clocks. a nad b obseved the phtons being emitted. a and b recorded the time of the emitted photons when emitted simulataneoslh at A and B.

a and b record the times that A and B emitted the photons. OK, so what time is that, geistkiesel? I want the actual time seen on the O' clocks in terms of L, v, and c. (Hint: look back over the many posts I've made: I spell it all out for you.) A and B only flash simultaneously in the O frame. (Or have you already forgotten the Einstein quote you just entertained us with?)

t' = 0. The same as the O' observer who zeroed her clock when she arrived at the midpoint just as the photons were emitted simultaneously from A and B. Long before the O' detected the staggered arrival of the B and A photons. Come on we have done this scene many times before.

Yes we have. And yet you still get it wrong!

Clocks at A and B time the flashes to occur at t = 0. All of the clocks in O' (including a and b) are synchronized in the O' frame. We agree that when M' passes M their two clocks both read zero. Everything else must be deduced.

1. Question Doc Al: How do the a and b observers know there are two photons emitted from the stationary frame?
   
2. or said another way, how do the photons know they must change their emittance protocol because the photons are about to be emitted into the moving frame?
3. How do the photons know which one must emit first,
4. and at what what time? [*] I think you are digging yourself a perception of the observers hole instead of building a phyisics arguement.

What are you babbling about now? What's this nonsense about photons changing their "emittance protocol"? And photons conspiring together? :rofl:

Einstein's train gedanken is as simple a set up as you are going to get (but obviously not simple enough!). Two lights flashing. No conspiracy. No "emittance protocol". Just the invariant speed of light.