Has ayone seen the lost simultaneity?

[geistkiesel]

Doc Al, Take a look at my last post in "No postulate of light violated in galilean trasnformation". It is in regard to my famous "sneer" t3 = t1(C + V)/ (C - V) expression. You might be interested. I addressed the post to tom_mattson, as that is what I was working on when I came across something in the intenet. take a look. No bull **** on this one.

 

[Doc Al]

Doc Al, Take a look at my last post in "No postulate of light violated in galilean trasnformation". It is in regard to my famous "sneer" t3 = t1(C + V)/ (C - V) expression. You might be interested. I addressed the post to tom_mattson, as that is what I was working on when I came across something in the intenet. take a look. No bull **** on this one.

Take a look at my response, where I explain once again how that trivial expression is derived and what it means. Take notes this time.

 

[Eyesaw]

O' considers her frame as stationary when the train station rushes by. A' and B' photons are emitted simultaneously in her considered stationary frame just as O' is at M in the train station frame rushing by. The A' and B' photons arrive simultaneously at M. O' is considering that the photons B' and A' must reach her simultaneously as she has considered herself stationary. However, phyisics doesn't cooperate. The photons arrive simultaneously at M in the train station considered by O' to be moving as light speed is invariant under motion of the source. O', considering herself stationary will be waiting patiently for the simultaneous arrival of the B' and A' photons. When the B' photons arrives first, she becomes confused, and starts to blabber SR mantras in a vain attempt to correct what she knows to be wrong, physically wrong. The photons have to arrive at her position simultaneously because, dammit, she considered herself stationary, Doc Al told her she could do that. However, as the light speed is still invariant to O' she is still going to collide with the B' photon earlier than she detects the A' photon for the same reasons she did so when the sources were in the stationary frame, the reasons being the laws of physics makes her considerations of her stationary frame silly and I might add, impossible.

When you are moving, you have to keep up the speed or else you auger in if considering yourself stationary. This is what is known in SR theory as the SR stall speed.

Another point to broadcast is that in all these cases the realtive velocity of the observers and photons is crucial to determining and predicting the correct outcome of the events.

Hence, what is simultaneous in the moving frame is simultaneous in the stationary frame, hence a violation of the relativity of simultaneity.

And the real joker here is that what is simultaneous in the moving and stationary frames, the emission of the A' anad B' photons, will be determined by O' not to be simultaneous at her position when she detects the staggered arrival of the B' and A' photons, even if she considerd herself stationary, damn, damn, damn.

So we have a return to absolute speed, zero in this case, also absolute time, I might add.

SR stall speed- that's funny. A very succinct and clear analysis I might add.

 

[geistkiesel]

Take a look at my response, where I explain once again how that trivial expression is derived and what it means. Take notes this time.

Doc will yopu please look at the link I referenced. My expression is used in a very sophisticated simultabneity analysis. Look at it godamnit!.

 

[Doc Al]

Doc will yopu please look at the link I referenced. My expression is used in a very sophisticated simultabneity analysis. Look at it godamnit!.

Do you really think that link which discusses relativity will support your anti-relativity crusade? Yes, that kind of expression does come up. But it's not what you think it is. That article is way too advanced for you.

Why not try reading it from the beginning? You may learn something.

 

[geistkiesel]

Do you really think that link which discusses relativity will support your anti-relativity crusade? Yes, that kind of expression does come up. But it's not what you think it is. That article is way too advanced for you.

Why not try reading it from the beginning? You may learn something.

How in the hell do you know what I think it is? I thought it was interesting dingdong. You are the one avoiding discussing the link. This makes me think I just found out what I have been thinking all along. I put you out of your league, didn't I? I do notice that when I write something decent, at least one line per post, that you avoid the difficult stuff, or you do your frame swapping routuine, or chortle, or start shouting i am using stationary times, or ...anything but physics, or, when you are up against the wall and they are tying the blindfold to cover your eyes, you bring out your most potent weapon, the:

 

[geistkiesel]

Take a look at my response, where I explain once again how that trivial expression is derived and what it means. Take notes this time.

I know what it measn Doc.

 

[Doc Al]

I do notice that when I write something decent...

When did this happen?

 

[geistkiesel]

When did this happen?

We put the emitters in the moving frame and just as the sources A' and B' arrive at A and B in the stationary frame, the A' and B' sources emit photons. The O', who was at M in the stationary frame when the photons were emitted simultaneously in the moving frame still heads to the B' photon on a collision course and detects the B' photon first, then later the A' photon.


Therefore, the passengers on the railway train must, therefore, conlcude the photons emitted simultaneously in the moving frame, were not emitted simultaneously in the moving frame. Hence the passengers all come to the conclusion, simultaneously, that special relativity theory sucks.
[/size]

 

[Doc Al]

lights on the train? No problem!

We put the emitters in the moving frame and just as the sources A' and B' arrive at A and B in the stationary frame, the A' and B' sources emit photons. The O', who was at M in the stationary frame when the photons were emitted simultaneously in the moving frame still heads to the B' photon on a collision course and detects the B' photon first, then later the A' photon.

I'm not sure what the set up is in this scenario, but the following is true. It doesn't matter if the emitters are on the train or on the embankment. If the embankment observes them flashing simultaneously, then the train will not. And vice versa.

 

[ram1024]

again, Doc Al is playing with the word "simultaneously".

we don't care what they OBSERVE simultaneously. we care what they CALCULATE as EMITTED simultaneously.

you know that's what we want and you dodge behind your relativity facade every chance you get :D

 

[Doc Al]

again, Doc Al is playing with the word "simultaneously".

No I'm not.

we don't care what they OBSERVE simultaneously. we care what they CALCULATE as EMITTED simultaneously.

That's what I mean by "observe". This is fairly common usage in discussing relativity, but I agree I could be clearer. By "observe" I mean: calculate, measure, deduce, etc. When I say that one frame will "observe them flashing" I mean deduce the time of emission based upon measurements made in that that frame. Of course, if that frame happened to have an observer at the right place and time, they could directly "observe" the emission. But that's hardly necessary.

you know that's what we want and you dodge behind your relativity facade every chance you get :D

Nice try. Now go back and reread every post I've made with the proper understanding of "observe".

If you were really paying attention, the meaning should be clear from context. That is if you had a clue what SR says. But I see your point.

 

[ram1024]

just trying to ruffle your feathers really <chuckle>

 

[geistkiesel]

I'm not sure what the set up is in this scenario, but the following is true. It doesn't matter if the emitters are on the train or on the embankment. If the embankment observes them flashing simultaneously, then the train will not. And vice versa.

This Einstein gedunken is different from Einstein's original experiment only that the photons are emitted from the moving frame at A' = A and B' = B when the moving obser O' is located at O' = M, where M is the midpoint of the A and B photon detectors (when used) and from data from 1000 previous experiments identical to this experimetn, the exact midpoint of A' and B' on the moving frame is located in the stationaray frame M''.
Here's the set up:

1. I and others have claimed that special elativity is a perception of the observer game - it isn't a physical dynamic being imposed - it is purely in the minds of beholding observers, So, question #1. If the photons were emitted simultaneously in the moving frame and the moving frame observers detect the B' and A' phoptons sequentially with the same moving frame clock indications as wehen the photons had been emitted in the stationaray frame,, why do the mpoving frame observers still conclude the photons were not emitted simultaneously in the moving frame?
2. The stationary frame has photon detectors at A = A' and B = B' and record the time the photons were emitted in the moving frame, say distrinuted in a density of 1000/photon-wavelength. Relative simultaneity demands that the stationary frame, even with its high number of photon detectors will not record the photons emitted from the moving platform as being simultaneously emitted into the stationary frame. Question#2: In the context of Einstein's gedunken, what measuring scheme will the stationary observer use to support the conclusion of relative simultaneity? Will they record the photons in a mirror image reflection of the case where the photons were emitted simultaneously in the stationary frame and the moving frame observer detected the photons sequentially,first at B then A?
3. The A' and B' photons were emitted spatially at locations mirroring AE's original gedunken where the photons were emitted in the stationary frame. The observer on the moving frame will see the photons sequentially with the clocks synchronized wrt the moving frame.
   • The moving frame observer detects the B' and A' photons in the following order:
   • The photns at A' and B' simultaneously by a' and b' on the moving frame located at A' and B'.
   • By O' the B' photon arriving on a collision course with O'.
   • At the original midpoint of the moving frame when O' = M.
   • by O', A' arriving from behind.
   • By this time all passengers located between A' and B' when the photons were emitted have observed and clocked the A' and B' photon's arrival times at their locations.
   Question #3: Will the observers/passengers on the moving frame conclude the photons were emitted simultaneously in the moving frame?
4. The photons in the moving frame are detected in the exact sequence the photons were detected when emitted simultaneously in the stationary frame. From the clocks of the a' and b' observers located at A' and B' in the moving frame the times of emission are exactly the same, identical, simultaneous.Question #4: Will the observers on the moving frame conclude the photons were emitted simultaneoulsy in the moving frame?
5. Question #5:Under what conditions, if any, will observers on the moving frame be able to conclude photons are emitted simultaneously in the moving frame[/B]?
6. The photons are detected in the same order they were detected when emitted in the stationary frame. Question #6:Explain the application of relative simultaneity in the context of this gedunken.
7. This post says the moving observer's conclusions are dependent on what the stationary frame observers detect. Describe the the stationary observers photon detection process, in terms of a time history in this experiments if the stationary observers conclude:
   • the photons were not emitted in the stationary frame simultaneoulsy or,
   • the photons were emitted simultaneously.

There is no observable stationary frame. In isolated space the moving observer has no stationary observer to communicate with in order to know what the stationary observers concluded. The moving observer detectced the A' and B' photns (emitted in the moving frame) in the same sequence as when the photons were emitted by the stationary observer, when there was a stationary observer. Therefore, I conclude the response is incomplete as there being no frame/frame comparison possibility. Will the moving frame here conclude the photons were emitted simultaneously in the moving frame?

 

[Doc Al]

This Einstein gedunken is different from Einstein's original experiment only that the photons are emitted from the moving frame at A' = A and B' = B when the moving obser O' is located at O' = M, where M is the midpoint of the A and B photon detectors (when used) and from data from 1000 previous experiments identical to this experimetn, the exact midpoint of A' and B' on the moving frame is located in the stationaray frame M''.

Before we follow the yellow brick road to the land of Oz, let's define our terms. I assume you mean by A' and B' certain locations on the train. I further assume that these locations are the positions in the train that were directly opposite A and B when the lights flashed at A and B. Is that what you mean by A' and B'?

If so, then what is the set up? You now have lights at A' and B'? (These points happen to be equidistant from M'.) And they flash simultaneously according to who?

Note that if A' flashes when A' passes A, and B' flashes when B' passes B, then they obviously do not flash simultaneously in the train frame.

 

[quantumdude]

Ram's little outburst can be found in the Feedback Forum at this link:

https://www.physicsforums.com/showthread.php?p=254873#post254873

For future reference, feedback doesn't belong in the Theory Development Forum.

 

[ram1024]

this is the argument that shut you down, Tom.

although it will just get deleted again anyways...

Speed is defined as:

[u]Distance[/u]
  Time

now i want you to specifically take note that you say the speed of light is constant to the observer.

that specifically means no matter HOW the observer moves he will always measure the speed of light to be the same RELATIVE to him as calculated if he were stationary.

For Example, an observer moving 40,000 km/sec towards a light source would still measure light coming towards him at 300,000 km per second. and the stationary observer would too.

the argument was as follows. speed consists of two components Distance, and Time.

The of the speed of light is COMPLETELY dependant on fudging with those two components to maintain "Constancy" to all observers.

If i take a value C made up of two values A / B (A divided by B) can you not see that i can make C any value i want by fudging the values A and B to suit my whims?

Distance is Immutable
Time is Immutable

try your hand at calculating the speed of light using real measurements not fake elastic ones and the constancy of the speed of light goes in the crapper.

 

[quantumdude]

this is the argument that shut you down, Tom.

No, it isn't. First of all, I didn't delete this argument. I deleted the argument about monkeys and bamboo (that was a real stroke of genius). Second, I already answered this argument. And third, this argument is so idiotic, there is no way it could shut anybody down.

I've already said it several times, but here I go again: All you are doing here is the following:

1. Assume that Galilean relativity is correct.
2. Assume that measurements taken will agree with Galilean relativity.
3. Assume that any measurements that don't agree with it are wrong.
4. Conclude that Galilean relativity is correct based on 1-3.

You are the very definition of irrationality, as you make no reference to the real world. The data that you tell me to look at will back up SR, which is what you won't accept. Too bad for you.

although it will just get deleted again anyways...

I'll leave it here, but be assured that the next time you post this same drivel, it will be deleted. Do it once, and it's an honest error. Do it twice, and it's annoying. Do it three times, and it's spam.

edit: By the way, your monkey and bamboo argument has been reposted in the thread in the Feedback Forum, so everyone can see how brilliant you are.

 

[russ_watters]

Morbid curiosity here...

Distance is Immutable
Time is Immutable

On what, specifically, do you base those two statements?

 

[krab]

I just looked in on this thread and, whew! I am surprised and impressed by the amount of work people have put into responding to Ram. It is positively heroic. But Ram is irrational. In a math thread, he stated that "zero point nine nine nine repeating is not equal to one, and nothing anyone can say will convince me otherwise" (or words to that effect). That's his attitude in Math. So if he denies solid mathematical proofs (many were given), teaching him Physics is even more hopeless.

My advice (if anyone cares) is: do not respond to an irrational person. It's entirely pointless. Such a person is not accepting the rules of logic that you are using. He laughs, you waste your time, and the argument will continue to the point where your blood pressure rises, and even if you are the most rational person, you finally become irrational. (Yes, it has happened to me.)

 

[geistkiesel]

Simultaneity? Yes or no?

I'm not sure what the set up is in this scenario, but the following is true. It doesn't matter if the emitters are on the train or on the embankment. If the embankment observes them flashing simultaneously, then the train will not. And vice versa.

This is the qustion: On the moving platform the lights were emitted such that the observers at A' and B', that is a and b, mesurd the time of emission. The lights emitted just as O' was located at M. The O' observer detected the A' and B' photons in the same sequence as she detected the photon emitted from the moving frame.
Will the moving frame come to the same conclusion she came to under the standard Einstein version where the photons were emitted simultaneously in the stationry fame? Remember, the only difference here is "the photons were emitted n the moving frame as O' was at M."

Your response to the first version of thois indicated that simultaneity could be determined by what each observer did. If the stationary observer concluded the photons were emitted simultaneously in the stationary fram,e then the photons were not emitted simulltaneoulsy in the stationary frame simultaneity. Can you please discuss the four cases below? Thanx.

1. This sounds like O's reality is determined by what the stationary observer detects say: true or False?
2. the stationary observer did not have his detectors working at the instant the photons were emitted in the moving frame:What eaffect on determining simultaneity does have? and
3. the stationary observer did have his detectors working.
4. The A' and B' photons were detected arriving simultaneously at M. What affect, if any?

 

[geistkiesel]

Before we follow the yellow brick road to the land of Oz, let's define our terms. I assume you mean by A' and B' certain locations on the train. I further assume that these locations are the positions in the train that were directly opposite A and B when the lights flashed at A and B. Is that what you mean by A' and B'?

If so, then what is the set up? You now have lights at A' and B'? (These points happen to be equidistant from M'.) And they flash simultaneously according to who?

Note that if A' flashes when A' passes A, and B' flashes when B' passes B, then they obviously do not flash simultaneously in the train frame.

No, the A' and B' photons are emitted simultaneously in the moving frame as determined by the watches on observes at A' and B' when O' was located at M in the stationary frame. The O' observer will naturally detect the photons in the same sequence she detected the photons emitted simultaneously in the stationary frame. It was this sequene of detection that Einstein referred to when he said thyat the passengers n the train 'must, therefore come to the conclusion that the photons were not emitted simultaneously in the moving frame.

Once the photons have been emitted in the moving frame the moving observer may not rationally consider herself at rest and the train station moving, this is a physical impossibility.She cannot do this under anycircumstances the consuideration being a physical impossibility.

Do any of your answers include the use of gamma in xdetermining whether the moving observer determines the photons were, or were not emitted simultaneously in the moving frame/

 

[geistkiesel]

Ram's little outburst can be found in the Feedback Forum at this link:

https://www.physicsforums.com/showthread.php?p=254873#post254873

For future reference, feedback doesn't belong in the Theory Development Forum.

Speaking of outbursts it appears to me that your anger got the best of you when you closed my thread. My post describing the zero velocity absolute inertial frame you cionssidered as circular? why becfause no one caved on to your demjands that we succumb to SR doma?

Tom,. the post was anything but circular. It was my impression that you weren; up to the task of refuting my model,. so you simply exercised what you thought was your absolute right to close a thread for any reasson what so ever.

It was a good paper, too bad it was over your head.

Here is another ove your head.

Just as O' in the moving frame arrives at M the midpoint of to stakes in the ground that once held photon emitters At and B. At this instant phoitons are emitted in the moving frame, As the photons speed opf light is not fdependent on the velocity of the sources A' and B' in the moving frame, the O' observer detects the photons first from A' coming at her from the front, then the A' and B' photons are detected simultaneously arriving at M in the stationary frame then the A' photon is detetcted by O' in the moving frame. Effectively the photons were detected in the same order by O' as when the photons were emitted simultaneoulsy in the stationary frame.

Does the moving ovserver conclude the photons were not emitted simultaneoulsy in the moving frame? Remember the photons had to be detected sequentially as when emitted in the stationary frame from the laws of the propagation of photons a la di Sitter. In other words, O' may not consider herself at rest when the photons were emitted and hence detect the photons arrived at the midpoint of A' and B' in the moving frame.

SR does not allow physical impossibilities in its mathematical modeling, does it?

 

[quantumdude]

Speaking of outbursts it appears to me that your anger got the best of you when you closed my thread.

No anger whatsoever. Just doing my job.

My post describing the zero velocity absolute inertial frame you cionssidered as circular? why becfause no one caved on to your demjands that we succumb to SR doma?

If a "zero velocity absolute inertial frame" is one that can be considered at absolute rest, then it does not exist in nature. Period.

Tom,. the post was anything but circular.

Yes, the thread was going around in circles. You post the same crap over and over, and no matter how many times it's refuted, you keep coming back with the same lame arguments.

It was my impression that you weren; up to the task of refuting my model,. so you simply exercised what you thought was your absolute right to close a thread for any reasson what so ever.

With the possible exception of ram1024, you are the last person here who is qualified to make that kind of assessment.

It was a good paper, too bad it was over your head.

Ditto for these assessments.

Here is another ove your head.

Oh, please. Your silly thought experiments are like a special ed version of a high school textbook.

In other words, O' may not consider herself at rest when the photons were emitted and hence detect the photons arrived at the midpoint of A' and B' in the moving frame.

If O' is inertial, then O' may consider herself at rest. It's as simple as that.

 

[terrabyte]

if O' considers himself at rest, the rest of the universe moves opposite O's velocity (same speed, opposite direction)

it's simple

what seems to be the problem?