The discussion centers on the concept of simultaneity in the context of special relativity, specifically analyzing a scenario involving a moving frame with mirrors and emitted light pulses. Participants demonstrate that both stationary and moving observers conclude that light pulses emitted from point M arrive simultaneously, despite the movement of the frame. The analysis shows that the perceived simultaneity is consistent across different frames, contradicting the notion of "loss of simultaneity." The conclusion emphasizes that the relative motion of the observer does not affect the simultaneous arrival of light pulses at the observer's location.
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The answers are in terms of positions located by the times you asked for. primes are in the moving frame..
1, M'(t1).
2. M(t1)
3. M'(t3)
4. M(t3)
5. M(t2).
6. M'(t2) , but this is only a copy of Mary's watch who is heading for a detection fo the A photon when the moving frame observer placed permanently at M'(t0) measures the arrival of the photons in the stationary frame at M'(t2).
Cute. Now give me the actual times, not your label for the times.
Here's what I mean: Given that A and B are a distance L from M in the "stationary" frame, and that M' moves past at speed v, tell me the actual times (in terms of L, v, and c) that all clocks read. Let the M' clock read zero just as it passes M. And let the stationary frame simultaneously trigger both A and B to pulse at t = 0. (If you don't like t = 0, then use t = t0, but that just adds extra work for you.)
You are always trying to change what I say. Why do yiou do this? Can't you defeat me on the principals of SR?
It's just that your posts are so sloppy that I have to translate what you say into unambiguous terms. And, you change terminology from thread to thread.
No I am saying that event number two is the arrival of the photons at the midpoint of the A and B emitters as observed by an observer placed at the original point M'(t0). This arival time is t2 in the moving frame.
Event 3 the arrival of the A photon at M'(t3) is just as I stated.
Well then, why did you say this before:
geistkiesel said:
OkIi will tell you again. Did you reaqd the post that had the figure abiove?
Event #2 M'(t1)
Event #3 M'(t2) Note M(t2) is the simultaeoyus arrival of photons at the midpoint M. M'(t3) is the arrival of the A photon in the moving frame. This can be calculated by the moving observer.
OK, so you are renumbering the events that I had labeled. Here they are again:
Event #1: "moving" observer M' detects photons from B
Event #2: "stationary" observer M detects photons from A and B
Event #3: "moving" observer M' detects photons from A
The M' observer, the original observer at M'(t0) has moved on and in her palce a detector that maintains the position in an ever constant watch of the midpoint in the statiionary frame, ok?
Confused?
Of course I'm confused--what are you talking about? If you wish to find out when the moving frame would say that "Event #2" occurred, then just look at the clock in the moving frame that happens to be there. (You must imagine a huge network of clocks everywhere in each frame.) The moving frame can't have a single clock that "maintains the position" where M is. It's MOVING, get it?
I realize thst SR says the clocks won't agree on the time the photons arrived simultaneously at M(t2) and M'(t2), where each t2 is measured in its own frame. However, this isn't important here as the observed simultaneous arrival of the photons by the moving frame is what is being emphacised. Had the frame not been moving the photons would have arrived at the colocated M'(t0) and M(t0). SR and the implied simultaneity implications is an arbitrary and error generating mechanism created by a definition, and arbitrary and unreasonable definition.
Those "arbitrary and unreasonable" aspects of SR just happen to accurately model the real world, as has been shown repeatedly in actual experiment. Get used to it.
To claim now that the photons were not emitted simultaneously in the moving frame would be specious, just because the frame has moved?.
No. In fact, given our knowledge of how the world actually works we are FORCED to conclude that simultaneity is frame-dependent. Get used to it. I won't hurt you.
#33
geistkiesel
538
1
Doc Al said:
Cute. Now give me the actual times, not your label for the times.
Here's what I mean: Given that A and B are a distance L from M in the "stationary" frame, and that M' moves past at speed v, tell me the actual times (in terms of L, v, and c) that all clocks read. Let the M' clock read zero just as it passes M. And let the stationary frame simultaneously trigger both A and B to pulse at t = 0. (If you don't like t = 0, then use t = t0, but that just adds extra work for you.)
Well I see the mean old task master isn't so brutal after all. I can save a step, he says, and use just t= 0, what a guy!.
Originally Posted by geistkiesel
The answers are in terms of positions located by the times you asked for. primes are in the moving frame..
1, M'(t1).
2. M(t1)
3. M'(t3)
4. M(t3)
5. M(t2).
6. M'(t2) , but this is only a copy of Mary's watch who is heading for a detection of the A photon when the moving frame observer placed permanently at M'(t0) measures the arrival of the photons in the stationary frame at M'(t2).
Before any assumptions or actual findings are made we must determine if the observers in the moving frame observes the emitted photons simultaneously in their frame. We will start, therefore with just the moving frame measured values when making this determination. At M'(t0) the moving frame is colocated with the stationary frame at M, the midpoint of photon sources located at A and B, just as the photons are emitted from A and B simultaneously. The moving frame maintains a photon detector at the position M'(t0) = M which is motorized and is capable of maintaining a constant watch at the photon mirrors (detectors) the observer in the stationay frame is scrutinizing. First at M'(t1), the photon from B arrives. Then the photons arrive simultaneously at M in the stationary frame which is witnessed by the motorized observer who gives a sign noting the arrival of the photons simultaneously in the stationary frame.[/color]
Backing up a bit, when the B photon arrives at t'1v, the photon has traveled a distance t'1c and likewise the A photon is located at -t'1v as it too has traveled t'1c. In the time span the moving observer at -M'(t1) moves to M'(t3) or dt = t'3 - t'1, the A photon must travel to M'(t3) or a distance c(t'3 - t'1) which equals the distance 2vt'1 + v(t'3 - t'1). Shortening the algebra the expression for t'3 = t'1(c + v)/(c - v). Picking some realistic numbers and setting c = 1,, t'1= .1 and v = .1, t'3= (.1)(1.1)/.9 = .222. The A photon traveled .222 - .1 = .0222, which should equal 2(.1)(.1) + (.222 - .1)(.10 )= .02 + .00222 = .0222.
The numbers balance. The fact that the moving observers detected the simultaneous emission of the photons in the moving frame no SR implications are warranted.
Let us see if we can justify this. The definition of simultaneity is: "Events that are simultaneous in a stationary frame are not simultaneous in a moving frame." From this we get the working SR process that discards absolute time and substitutes the individual time unique to all moving frames. This follows from natural definition of the sequential measuing of photons first from B then from A as is historically understood. This simplicity is the basis of SR?[/color]
However, the definition is specious and applies literally to any moving entities including EM radiation. But to alter a basic physical structure, or to create one, based on the fact that the observer in the moving frame has moved and being able to observe the photons arriving simultaneously at M and from this we get SR is asking too much.
The instant the photons were emitted in the stationary frame and the moving observer left the location at M'(t0) is not the kind of physical activity one uses in a basic assumption that maintains a high degree of integrity and longevity, especially to the degree and complexity one sees in SR. You look away and the universe is suddenly devoid of absolute time and space and where the measure of the speed of light in all inertial frames is c, which requires the insertion of time dilation and physical shrinking of mass into the structure of physical models that apply only in the direction of motion of the moving frames.
Trains are assumed stationary and stationary platform move at the velocity of trains, why? Because the mathematics of SR allows this even though no such frame exchanging is physically allowed as these kinds of activities are purely in the mental framework of mathematicians. These mathematicians, some of them, are unaware of the limitations on mathematical modlels that exceed the potential of the physical world to imitate.[/color]
With moving observes each can determine that another moving platfrom's clocks are running slower than his own. Dozens of moving observers, millions and billions of observers can assert that each of the other clocks are slower than his own and all these billions of obserevrs will be correct.
SR tlls us in simultaneity conditions that the moving frame photons or events are not simultaneous. As the photons were simultaneously emitted in the stationary frame then the sequential emission of photons requires that one photon be suppressed from being emitted, even though emitted in the stationary frame. Making the stationary frame to moving frame transition under these conditions is a physical impossibility.[/color]
As detailed by Grounded, the error of SR can be resurrected when the observer's velocity is properly added to the mathematical structures (as was done here) defining the physics of inertial platforms.[/color]
"The moving finger writes and having writ moves on; nor with all your piety nor wit shall ye lure it back to cancel even half a line."[/color]
Well I see the mean old task master isn't so brutal after all. I can save a step, he says, and use just t= 0, what a guy!.
And I see that old geistkiesel still keeps dodging my questions. Well, what are those times? In terms of L, v, and c. And then we'll talk about distances.
Before any assumptions or actual findings are made we must determine if the observers in the moving frame observes the emitted photons simultaneously in their frame.
A typically ambiguous statement. If you mean, Does the moving frame agree that the photons from A arrive at M at the same time that the photons from B arrive at M? If so, of course the moving frame agrees with that. (All frames agree with that!) But if you mean, Does the moving observer M' detect the photons from A at the same time that she detects the photons from B? If so, of course she doesn't: she detects the photons sequentially. (And the stationary frame agrees!)
We will start, therefore with just the moving frame measured values when making this determination. At M'(t0) the moving frame is colocated with the stationary frame at M, the midpoint of photon sources located at A and B, just as the photons are emitted from A and B simultaneously. The moving frame maintains a photon detector at the position M'(t0) = M which is motorized and is capable of maintaining a constant watch at the photon mirrors (detectors) the observer in the stationay frame is scrutinizing.
Give me a break! Are you seriously saying that the moving observer keeps a clock in the stationary frame? And uses that clock (which is in the STATIONARY frame) to make her measurements. Is THAT what you mean by times measured in the moving frame?
First at M'(t1), the photon from B arrives. Then the photons arrive simultaneously at M in the stationary frame which is witnessed by the motorized observer who gives a sign noting the arrival of the photons simultaneously in the stationary frame.
Well... Duh! Everyone knows the photons arrive simultaneous at M. When are you going to tell me WHEN the photons arrive at M? I want two answers: one according to M himself (the "stationary" frame) and the other according to what a clock in the M' frame (which just passed M at the exact moment that M detects those photons) would read for that time. I'll wait.
Backing up a bit, when the B photon arrives at t'1v, the photon has traveled a distance t'1c and likewise the A photon is located at -t'1v as it too has traveled t'1c. In the time span the moving observer at -M'(t1) moves to M'(t3) or dt = t'3 - t'1, the A photon must travel to M'(t3) or a distance c(t'3 - t'1) which equals the distance 2vt'1 + v(t'3 - t'1). Shortening the algebra the expression for t'3 = t'1(c + v)/(c - v). Picking some realistic numbers and setting c = 1,, t'1= .1 and v = .1, t'3= (.1)(1.1)/.9 = .222. The A photon traveled .222 - .1 = .0222, which should equal 2(.1)(.1) + (.222 - .1)(.10 )= .02 + .00222 = .0222.
The typical confusing morass of ambiguity, now with a little arithmetic thrown in. Let's see if we can decode it, step by step. What do you mean by t'1? One might think, since you use primed notation, that you mean the time according to the moving observer M' when she detects the photons from B. But you seem to think that "t'1v" represents the position where this event takes place. Ah... so when you say t'1 you really must mean t1: the time that clocks in M must be reading when that event takes place. (Do you understand why I keep pestering you for the EXACT TIMES of these events as observed in each frame? That way we can stop the nonsense!) So you must mean that the location of event #1 is t1v from M as measured by M.
OK, next phrase: "the photon has traveled a distance t'1c". Obviously, you are must mean that the B photon must have traveled a distance t1c according to the M observers. Yes, according to the M frame, the B photon traveled a distance t1c. Where is it at t1? At location t1v from M. So t1v + t1c = L (the distance from B to M).
Next phrase: "the A photon is located at -t'1v as it too has traveled t'1c". According to the M observers, photon A has traveled the same distance, t1c in that time, so it is now a distanc t1c from its starting point A according to the M observers. And yes, it's a distance of -t1v from M at time t = t1.
Next phrase: "the time span the moving observer at -M'(t1) moves to M'(t3) or dt = t'3 - t'1". According to the M observers, M' moves from t1v to t3v, in a time of t3 - t1. (Do you see why you are obviously confused by your own notation? You still think you are talking about times measured by M', but you are really talking about times observed in the M frame.)
Next phrase: "the A photon must travel to M'(t3) or a distance c(t'3 - t'1) which equals the distance 2vt'1 + v(t'3 - t'1)." First off, where does event #3 take place according to the M frame? At location t3v from M. So the photon A must travel from -t1v to t3v (which obviously equals a distance of t3v + t1v) in a time of t3 - t1.
Next phrase: "Shortening the algebra the expression for t'3 = t'1(c + v)/(c - v)." Translation: t3 = t1(c+v)/(c-v). So what?
The grand conclusion: "Picking some realistic numbers and setting c = 1,, t'1= .1 and v = .1, t'3= (.1)(1.1)/.9 = .222. The A photon traveled .222 - .1 = .0222, which should equal 2(.1)(.1) + (.222 - .1)(.10 )= .02 + .00222 = .0222." I'll overlook your arithmetic errors. You seem to be shocked by the obvious: you derive an expression for t3 in terms of t1, you pick t1 and v, use your expression to find t3, then look on in dumbfounded amazement that your expression for distances and times are consistent. Well, yes, distance does equal speed x time. So what?
The numbers balance. The fact that the moving observers detected the simultaneous emission of the photons in the moving frame no SR implications are warranted.
The fact that your "numbers" balance means nothing. Furthermore, ALL of your "numbers" are values measured in the "stationary" M frame.
Let us see if we can justify this. The definition of simultaneity is: "Events that are simultaneous in a stationary frame are not simultaneous in a moving frame."
As usual, you are hopelessly lost. This happens to be a conclusion of SR, not a definition or even an assumption. You are wasting people's time with this nonsense.
"The moving finger writes and having writ moves on; nor with all your piety nor wit shall ye lure it back to cancel even half a line."[/color]
:zzz: Translation: "I refuse to face reality and learn from my mistakes." Is that how you want your tombstone to read?
#35
geistkiesel
538
1
Doc Al said:
And I see that old geistkiesel still keeps dodging my questions. Well, what are those times? In terms of L, v, and c. And then we'll talk about distances.
A typically ambiguous statement. If you mean, Does the moving frame agree that the photons from A arrive at M at the same time that the photons from B arrive at M? If so, of course the moving frame agrees with that. (All frames agree with that!) But if you mean, Does the moving observer M' detect the photons from A at the same time that she detects the photons from B? If so, of course she doesn't: she detects the photons sequentially. (And the stationary frame agrees!)
You should look a little closer. The photons ariving at M simultabeously are deflected into the moving frame and deflected to the front of the moving frame where the A' and B' photons (primes mean in the moving frame now) are running neck and neck with the A photon that was not caught at M. The B photon detected at t'1v allows me to place the A photon in a symmetric position at '-t'1v to the left of M' at t'1. The A photon will arrive at the same time it has always arrived at t'3 in the moving frame. Here the deflected A and B photons are also directed forward in perfect alignment with the A photon. AT this instant this example is outside the scope of the definition of simultaneity as the A and B photons have been detected simultaneously in the moving frame.
Doc Al said:
Give me a break! Are you seriously saying that the moving observer keeps a clock in the stationary frame? And uses that clock (which is in the STATIONARY frame) to make her measurements. Is THAT what you mean by times measured in the moving frame?
No that isn't what I mean, or said, at all. I said the moving frame had a series of detectors that reflected the A and B photons into the moving frame at the same time the stationary observer saw the arrival of the photons. I still don't care about the times on the clocks. Stick around to find out why.
Doc Al said:
Well... Duh! Everyone knows the photons arrive simultaneous at M. When are you going to tell me WHEN the photons arrive at M? I want two answers: one according to M himself (the "stationary" frame) and the other according to what a clock in the M' frame (which just passed M at the exact moment that M detects those photons) would read for that time. I'll wait.
If you impose SR then the stationary clock will read ahead of the moving clock. But I don't have a clock at M' when the photons meet in the stationary frame. I only know from observation that the photons were detected in the moving frame at the same instant, simultaneously, that the photons were detected in the stationary frame.. These photons, now A', B' and A are heading neck and neck to the time the A photon was originally scheduled to arrive . This time, however, all three photons arrived at t'3 simultaneously, thus bringing this experiment outrside the definition of the loss of simultabeity re SR.
There was too much complexity stuck tio the definition of simultaneity which was only defined as such in order to discard the absolute characteristic of time.
Doc Al said:
The typical confusing morass of ambiguity, now with a little arithmetic thrown in. Let's see if we can decode it, step by step. What do you mean by t'1? One might think, since you use primed notation, that you mean the time according to the moving observer M' when she detects the photons from B. But you seem to think that "t'1v" represents the position where this event takes place. Ah... so when you say t'1 you really must mean t1: the time that clocks in M must be reading when that event takes place. (Do you understand why I keep pestering you for the EXACT TIMES of these events as observed in each frame? That way we can stop the nonsense!) So you must mean that the location of event #1 is t1v from M as measured by M.
No again. I meant that t'v is the position of B as measured in the moving frame as is the -t'1v the location of the A photon wrt the moving frame. Got it?
Doc Al said:
OK, next phrase: "the photon has traveled a distance t'1c". Obviously, you are must mean that the B photon must have traveled a distance t1c according to the M observers. Yes, according to the M frame, the B photon traveled a distance t1c. Where is it at t1? At location t1v from M. So t1v + t1c = L (the distance from B to M).
Doc Al you are reversing everything I would turn you into a mentor if you wern't one yourself. You can't be as stupid as you are pretending
Doc Al said:
Next phrase: "the A photon is located at -t'1v as it too has traveled t'1c". According to the M observers, photon A has traveled the same distance, t1c in that time, so it is now a distanc t1c from its starting point A according to the M observers. And yes, it's a distance of -t1v from M at time t = t1.
No Doc A' those mesuremnt are wrt the moving frame. You know it don't you?
Why cheat Doc Al? Why cheat?
Doc Al said:
Next phrase: "the time span the moving observer at -M'(t1) moves to M'(t3) or dt = t'3 - t'1". According to the M observers, M' moves from t1v to t3v, in a time of t3 - t1. (Do you see why you are obviously confused by your own notation? You still think you are talking about times measured by M', but you are really talking about times observed in the M frame.)
If I forgot a prime or something then I had a typo m, but I was consistently using moving ftrame calculations. In any event the paper demonstrates the detection of photons arriving simulatneosuly at O' in the moving frame. Hence SR doesn't apply, hence absolute time returned to its rightful position.
No the A photon at -t'1v moves to t'3. I don't care what the M obsever see, calculate or whatever. They are out of the hunt as we are only determining whether the photons are detetcted simultaneously in the movuing frame. It has nothing to do with clocks or differences. You are trying in your stumbling fashion to keep a facade of SR running here, but you are premature in that respect.
Doc Al said:
Next phrase: "the A photon must travel to M'(t3) or a distance c(t'3 - t'1) which equals the distance 2vt'1 + v(t'3 - t'1)." First off, where does event #3 take place according to the M frame? At location t3v from M. So the photon A must travel from -t1v to t3v (which obviously equals a distance of t3v + t1v) in a time of t3 - t1.
Doc Al if you want to calculate using SR go for it. All I am looking at is whether the photons Meet at t'3 simultaneously. It looks like they are .
Doc Al said:
Next phrase: "Shortening the algebra the expression for t'3 = t'1(c + v)/(c - v)." Translation: t3 = t1(c+v)/(c-v). So what?
I calculated the t'3 event using c, v and t'1, is the so what.
Doc Al said:
The grand conclusion: "Picking some realistic numbers and setting c = 1,, t'1= .1 and v = .1, t'3= (.1)(1.1)/.9 = .222. The A photon traveled .222 - .1 = .0222, which should equal 2(.1)(.1) + (.222 - .1)(.10 )= .02 + .00222 = .0222." I'll overlook your arithmetic errors. You seem to be shocked by the obvious: you derive an expression for t3 in terms of t1, you pick t1 and v, use your expression to find t3, then look on in dumbfounded amazement that your expression for distances and times are consistent. Well, yes, distance does equal speed x time. So what?
Doc Al the times were calculated in the moving frame. The distances were as calculated hence the photon arrived at t'3 simultabneously. SR loses.
Doc Al said:
The fact that your "numbers" balance means nothing. Furthermore, ALL of your "numbers" are values measured in the "stationary" M frame.
The numbers were measured in the moving frame. But at this end of the picture show it really doesn't matter as the photons were seen simultaneousl;y arriving at t'3.
Doc Al said:
As usual, you are hopelessly lost. This happens to be a conclusion of SR, not a definition or even an assumption. You are wasting people's time with this nonsense.
You sure went all out to obliterate what I have been saying. Scared Doc Al?
Read the books Doc Al. You might learn something.
You poor man. Read Einstein where he says the discarding of absolute time folllows the natural definition of simultaneity. He uses the phrase that because the B photon was seen beforee the A photon that the passengers on the train "MUST conclude" the photons were emitted sequentially in the moving frame. This is bunk. The passengers know the train is moving and so does the moving observer. She knows that her motion will operate to detect the B photon first, but this does not preclude the observers in the moving frame to detect the photons simultaneousoly in the moving frame. You have used the definition I gave yourself, so forked tongued wonder, what have you to say for yourself? You have given this definition yourself. You are just in a panicked mode and are starting to moan.
What a beautiful sound.
Dc Al said:
:zzz: Translation: "I refuse to face reality and learn from my mistakes." Is that how you want your tombstone to read?
Wow, I should copy that down. You think you are going to be vioctorious in this by your smirking nature?
Maybe so, and the maybe no.
"Some for the glories of this world and some sigh for the prophet's paradise to come, ah, take the cash and let the credit go, nor heed the rumble of the distant drum."
#36
geistkiesel
538
1
Here is how it works Doc Al;
The photons from A and B arrive simulatneously at M in the stationary frame, and are immediately deflected by mirrors into the moving frame, hence the moving frame measures the arrival of the A and B photons simulaneously in the moving frame. Again we will let the photons go as they must until they all meet where the A photon met in the unedited experiment, at t'3. It seems that to those screaming out loud that the photons did not arrive at one spot in the moving frame simultaneously were a bit hasty, in this case at least.
Code:
M
|
Above the stationary midpoint just as the photons from A and B arrive simulatneously.
Above: The B photon reaches the midpoint M in the stationary frame and is deflected down and then to the right toward the moving observer in themovinf frame.
Code:
___>\ Stationary
|
\>__________> A' photon moving |
At the same time the A photon arrives at M, and is also deflected down and formward toward the observer in the moving frame.
Code:
_____>\ /_> A photon moving |
\>________>/
|
M' (t'2)
Which is ovekill as we have an unreflected photon A running head to head with the A' and B' photons, where the primes refer to the moving frame.
The photons are directed into the moving frame just the the photons arrive simultaneously at the moving frame. The deflections are then deflected in the direction of the moving observer.
All photons arrive simultaneously in the moving frame at t'3 and , there is no SR here, hence Einstein made a booboo.
_________________________________________________
Also, in the case where two photons are emitted simultaneously from A and B and arrive at the midpoint of A and B just as the moving observer arrives-we will clear this matter up right here and now!.
The moving observer measures the simulatneous arrival of A and B in the moving frame. Some will scream that no, no the photons were emitted one after the other, with the A photon emitted first.
This is an impossibility as the photons arrived at M simultaneously in both frames. If the A photon was emitted first and the sources were equidistant from the midpoint as they obviously were, then the A photon would have arrived before the B phpoton and the photons would not have arrived simultaneously. You see, the speed of light is the same for both photons in the moving frame. Therefore each must have used the same time of flight in arriving at M and M' simultaneously with the moving observer.
As soon as there are two photons then, from that instant on, there is only one midpoint and each photon thereafter spends equal amounts of time in flight. Draw it out Doc Al before you begin to look silly.
It is easy here as the A and B sources have their midpoint at M and M', get it?
#37
ram1024
301
0
You are just in a panicked mode and are starting to moan.
What a beautiful sound.
I'm no longer going to waste my time giving detailed corrections for all your errors in every inane post of yours. What's the point? (I feel like a cat playing with a retarded mouse. I'm getting bored.)
When you are ready to break out of your delusional slumber and see what's really going on, start by computing the arrival times of the photons according to each clock in terms of L, v, and c. (Still dodging my questions, eh?) Until then, you are just BSing around.
geistkiesel said:
You should look a little closer. The photons ariving at M simultabeously are deflected into the moving frame and deflected to the front of the moving frame where the A' and B' photons (primes mean in the moving frame now) are running neck and neck with the A photon that was not caught at M.
Now what? Are you introducing MIRRORS into this muddle of misunderstanding?? Sober up, man!
The photons from A and B arrive simulatneously at M in the stationary frame, and are immediately deflected by mirrors into the moving frame, hence the moving frame measures the arrival of the A and B photons simulaneously in the moving frame. Again we will let the photons go as they must until they all meet where the A photon met in the unedited experiment, at t'3. It seems that to those screaming out loud that the photons did not arrive at one spot in the moving frame simultaneously were a bit hasty, in this case at least.
Again with the mirrors? When did you get this brilliant idea?
I am amused that after all of your repetitious postings about the "train gedanken" you still haven't a clue. We started off many, many threads ago agreeing that both the stationary and moving frames agree that the moving observer detects the photons at different times. Have you really forgotten that or are you just yanking my chain?
#40
geistkiesel
538
1
Doc Al said:
Again with the mirrors? When did you get this brilliant idea?
Brilliant ideas came from the protracted and heated discussions with Doc Al.
I went looking at the simulaneity question from a pragmatic point of view and I discovered that Einstein's definition of simultaneity basically sucks. And here is why. Read this closely.
Here is your task, should you choose to perform it. At least, among all the stuff you are going to throw at the arguments, even your 'rounded eyes smileys', find the flaw in the examples that show the photons are not emitted simultaneously in the moving frame, using more than the definition, of course.
Simultaneity is defined as:
“Events simultaneous with respect to a stationary frame are not simultaneous with respect to a moving frame” (A. Einstein).
From the sequential detection of the photons in the moving frame AE tells us that the moving observers, “ . . .must therefore come to the conclusion . . . ” that the photons emitted simultaneously in the stationary frame are not emitted simultaneously in the moving frame. Experimental arrangement for scrutiny of “simultaneous events”.
Code:
M
A_______________________|________________________ß
O’ t’0 O’#t’1 O’#t’2
M' B A
AE tells us that the mere fact that photons are measured sequentially, that this is enough by itself to discard the concept of simultaneity, even if we were able to detect the photons simultaneously in the moving frame by modification of the detection apparatus.
A simple modification of the arrangement will show the flaws AE has presented to us.
We place mirrors at the midpoint M and deflect the simultaneous arriving photons into the moving frame. The photons are immediately deflected simultaneously into the moving ftame and then forward to the observer O'.
The presence of the human observer O’ at the point the photons are emitted into the moving frame is not necessary for verifying experimental results. Inanimate detection and recording equipment is sufficient for experimental purposes.
Code:
1st alternative stationary and moving frames.
M
/<--B
/ |
/ |
/ |M'
/ \\\--> B'#|moving frame.
A-->\
M|
|
|
|
\\\\--> A'#| moving frame.
M' O'
The stationary frame has mirrors (“/” deflecting the "B" photon, and “\” deflecting the "A” photon.) arranged to deflect the photons as they arrive simultaneously at the midpoint of the A and B sources in the stationary frame. M is along the upper slanted line indicating the simultaneous arrival of the A and B photons (The figure is an attempt at a perspective drawing). The mirrors in the moving frame are strung out along the length of the frame so as not to require any timing protocols between the frames.
There is no way the moving observers can distinguish which source the photons arrived from, certainly not from a sequential arrival.
Other than the use of the definition can you see any direct logical reason, as a factor in the experiment, that will give the observer in the moving platform cause that she "must, therefore" conclude the photons were emitted nonsimultaneously in the moving frame?
You are probably wanting to jam a lot of past experimental data on the scene here in order to wash away the hideous spectre in front of you. I know I would if I were the 'died in the wool' special relativity theorist.
We simplify the previous modified arrangement: one more notch.
Code:
2nd modified moving frame.
M /<-- B stationary frame
|
. |
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M' # B' detector in moving frame.
A-->\ M
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|
|
M' # A' detector in moving frame.
The photons are shown on top of each other, the viewer is asked to see the photons entering simultaneously. Here M’ in the moving frame is aligned with M in the stationary frame as usual. The photons are detected upon arrival in the moving frame. There are a series of detectors ### strung along the moving frame.
The photons diverted from the stationary frame are immediately detected (“#” detectors) in the moving frame simultaneously.
This clearly shows the trivial fix for AE’s insistence that the moving observers “must,” conclude the photons were emitted non-simultaneously in the moving frame. The down arrows are the photons diverted simultaneously into the mirrors in the moving frame and are then immediately detected in the moving frame by #.
We simplify the above arrangement by omitting the mirrors in the moving frame and having the detectors (“Õ”) in the moving frame detect the photons the instant of arrival of the O' observer.
--> motion.
Here the photons A and B are arriving at the midpoint M in the “stationary frame”. The photon detectors, #, are extending out from the moving frame and substitute themselves for the mirrors the stationary observer uses to detect the photons.
The final modification is simply to extend the detectors in the moving frame into the stationary frame and detect the photons when they arrive simultaneously at M and M' in the stationary frame.
This is all pre application of SR theory.
Here we simply extended a photon probe, #|#, from the moving frame into the path of the photons arriving simultaneously in the moving and stationary frame, just like the observer O' could have done when she was passing by. We may do this without regard to any timing technicalities by stringing a series of photon detectors along the full length of the moving frame. We show just one of the detectors in the 3rd modified moving frame arrangement above.
The significance of simultaneity loss.
I have shown some simple adjustments that negate the concept of the loss of simultaneity that AE tells us is "a natural definition"; one that flows easily from the observations we are restricted to considering.
From the arguments AE presented we are told:
To discard the concept of absolute time and to substitute the concept of relative time where all moving frames have their own time.
The conflict between the propagation of light and the concept of “relativity” disappears and each moving frame has its own time. (If you are traveling at .99c, you will still measure the speed of light passing you as c. In other words we are told not to consider the motion of the observer in measuring the relative velocity of the observer and the light photons.)
All frames measure c = 300,000 km/sec in all frames regardless of their speed.
The clocks in moving frames move slower than clocks in stationary frames.
Physical mass shrinks in the direction of motion of the moving frame.
Trains are motionless while train stations assume velocities.
Each frame may consider all other frame clocks moving slower than its own.
That maintaining a rational view of the world around us is contrary to physical law and we are commanded to adjust our thinking to accept the irrationality of special relativity.
The laws of physics are in direct proportion to what we perceive as real, by what we are taught to perceive as real as opposed to what is actually real.
Special relativity has made some truly astounding physical assumptions such as allowing moving trains to assume a velocity of v = 0 and to measure the stationary frame, the land and train station, trees, buildings, lakes and highways, to move instead. Also, between moving frames each may consider the other’s clocks as moving slower than their own and both would be correct (according to special relativity theory). Special relativity mathematics allow these assumptions, it even encourages them. Trains are seen to stop, unload and load passengers and to accelerate to some velocity > 0. Train stations and the surrounding countryside never do this, ever, and it never will.
As I've already stated, I am no longer going to spend time wading through and correcting all your errors. I'll just point out a few of your more egregious boners and move on.
geistkiesel said:
Brilliant ideas came from the protracted and heated discussions with Doc Al.
But not from you, apparently.
Here is your task, should you choose to perform it. At least, among all the stuff you are going to throw at the arguments, even your 'rounded eyes smileys', find the flaw in the examples that show the photons are not emitted simultaneously in the moving frame, using more than the definition, of course.
Read on.
Simultaneity is defined as:
“Events simultaneous with respect to a stationary frame are not simultaneous with respect to a moving frame” (A. Einstein).
If you think that's the definition of simultaneity, you are completely out of your mind. For Einstein, as for the rest of us, simultaneity simply means "happens at the same time". Your so-called "definition" is Einstein's conclusion, not a definition or assumption.
From the sequential detection of the photons in the moving frame AE tells us that the moving observers, “ . . .must therefore come to the conclusion . . . ” that the photons emitted simultaneously in the stationary frame are not emitted simultaneously in the moving frame. Experimental arrangement for scrutiny of “simultaneous events”.
Code:
M
A_______________________|________________________ß
O’ t’0 O’#t’1 O’#t’2
M' B A
AE tells us that the mere fact that photons are measured sequentially, that this is enough by itself to discard the concept of simultaneity, even if we were able to detect the photons simultaneously in the moving frame by modification of the detection apparatus.
In the context of Einstein's original train gedanken, it is true that if the moving observer detects the photons arriving at different times, then she can justifiably conclude that in her frame the photons must have been emitted at different times. All true! (I've explained this many times.)
Note that Einstein is not making the imbecilic general statement "if ever an observer detects photons at different times, they must have been emitted at different times" or "if ever an observer detects photons at the same time, they must have been emitted simultaneously". It is only in the context of Einstein's example that she can so conclude.
A simple modification of the arrangement will show the flaws AE has presented to us.
We place mirrors at the midpoint M and deflect the simultaneous arriving photons into the moving frame. The photons are immediately deflected simultaneously into the moving ftame and then forward to the observer O'.
The moving frame of course agrees that the photons arrive at M simultaneously (all frames do!). And because of that she is forced to conclude that they could not possibly have been emitted simultaneously. After all, from her frame M is moving towards one source and moving away from the other!
And if you put a mirror at M, then yes indeed, those photons will arrive at O' simultaneously. (Of course, this has nothing to do with Einstein's original example. ) You need not go to the trouble of putting a mirror at M: why not just have the moving observer pass by M at exactly the moment that the photons arrive at M. Then both observers will detect the photons simultaneously. So what? The moving observer will still conclude that the emissions were not simultaneous in her frame.
As usual, your "simple modification" shows nothing other than the depth of your misunderstanding of Einstein's simple argument.
#42
geistkiesel
538
1
Doc Al said:
As I've already stated, I am no longer going to spend time wading through and correcting all your errors. I'll just point out a few of your more egregious boners and move on.
But not from you, apparently.
Read on.
If you think that's the definition of simultaneity, you are completely out of your mind. For Einstein, as for the rest of us, simultaneity simply means "happens at the same time". Your so-called "definition" is Einstein's conclusion, not a definition or assumption.
I got the information from Einstein's book "Relativity". He says specifically that the "natural definition" of simultaneity works to discard the concept of absolute time. "Events which are simultaneous wrt embankment are not simultaneous wrt the train and vice vesa. And he later says in context with absolute time that "this assumption is incompatible with the most natural defintiion of simultaneity". So AE did consider the 'simultaneity definition' just that.
So you are saying the photons will not enter the moving frame at the same time when deflected there by the mirrors I set up?
Doc Al said:
In the context of Einstein's original train gedanken, it is true that if the moving observer detects the photons arriving at different times, then she can justifiably conclude that in her frame the photons must have been emitted at different times. All true! (I've explained this many times.)
No you haven't not even once. You've said iot a million times but have never explained it.
Doc Al said:
Note that Einstein is not making the imbecilic general statement "if ever an observer detects photons at different times, they must have been emitted at different times" or "if ever an observer detects photons at the same time, they must have been emitted simultaneously". It is only in the context of Einstein's example that she can so conclude.
The simultaneous event we are discussing is the emission of photons on the stationary frame being tested as being simultaneously emitted in the moving frame. The photons did not enter the moving frame when detected by the moving observer, the photons entered the moving frame when they were emitted from the A and B source, and likewise emitted friom the stationary platform through the mirror assembly. The same photons, emitted simultaneously in the stationary frame, again, are emitting into the moving frame, simultaneously, again.
Doc Al said:
The moving frame of course agrees that the photons arrive at M simultaneously (all frames do!). And because of that she is forced to conclude that they could not possibly have been emitted simultaneously. After all, from her frame M is moving towards one source and moving away from the other!
because the photons arrived simultaneously at the midpoint she is forced to conlcude the photons could not have been emitted sinultaneousl? And after all etc?
Yes but the photons arrive at the detector for an instant. You are saying that just because she is moving to the B source and away from the A source the photons could not possibly have been emitted simultaneously, in which frame Doc Al, which frame could they not have been emitted simultaneously? You left this out. On purpose i believe.
The mirrors emit photons into her frame simultaneously as the mirrors are placed side by side to each other to insure integrity of the midpoint. She will not be able to tell which photon is from which source as they arrive in her frame simultaneously.
Doc Al]After all, from her frame M is moving towards one source and moving away from the other.
No this is a physical impossibility. Perhaps some mathematical interpretation allows her tio adjust he sight, she sees M moving to one source and away from the other.
Doc Al said:
Doc Al trains move, stationary platforms do not move. She knew she acccelerated, she has never seen a moving stationary frame, neiither have you or I. and we never will. This silly statement is nothing more than mathematical BS, a contrivance without a scintilla of realism truth or physical law to it, amd you know it. But thios is the law of SR isn't?
Doc Al said:
And if you put a mirror at M, then yes indeed, those photons will arrive at O' simultaneously. (Of course, this has nothing to do with Einstein's original example. ) You need not go to the trouble of putting a mirror at M: why not just have the moving observer pass by M at exactly the moment that the photons arrive at M. Then both observers will detect the photons simultaneously. So what? The moving observer will still conclude that the emissions were not simultaneous in her frame.
Based on what law of physics will the moving observers conclude the photons were not emitted simultaneously in the moving frame?
Listen to yourself :rolf: One side of the mouth says, "those photons arrive at O' simultaneously" and then the last line: "The moving observer will still conclude that the emissions were not simultaneous in her frame."
Isn't an "emission" of the photons the same as the photons were emitted? You say, "Then both observers will detect the photons simultaneously. So what?" And you ask "So what?" She concludes from what Doc Al? The deifinition that is not a definition, but a conclusion? What does she use to base her conclusuion? I can't see it, except for the definition.
AE saying the loss of simultaneity is the basis for discarding absolute time and you say "so what"? and you threaten that you won't read my posts anymore because you are bored?
Doc Al said:
As usual, your "simple modification" shows nothing other than the depth of your misunderstanding of Einstein's simple argument.
OK what is AE's simple argument? It sure sounds like the defined simultaneity of events as seen in stationary and moving frames, doesn't it? You can understand why I think like this can't you?
If you look at the problem properly you will see that the discussion preceeds any SR implications as that is what we are talking about. If I were you Doc Al I would be one of the first to get on the band wagon, not one of the last.
You tell me, that simple argument, when I already think I know the answer. Explain it to me so I can understand it. You are vey close to pulling me over the line, but yiou cannot apply BS this time. I need the above answers.
I got the information from Einstein's book "Relativity". He says specifically that the "natural definition" of simultaneity works to discard the concept of absolute time. "Events which are simultaneous wrt embankment are not simultaneous wrt the train and vice vesa. And he later says in context with absolute time that "this assumption is incompatible with the most natural defintiion of simultaneity". So AE did consider the 'simultaneity definition' just that.
The simultaneous event we are discussing is the emission of photons on the stationary frame being tested as being simultaneously emitted in the moving frame. The photons did not enter the moving frame when detected by the moving observer, the photons entered the moving frame when they were emitted from the A and B source, and likewise emitted friom the stationary platform through the mirror assembly. The same photons, emitted simultaneously in the stationary frame, again, are emitting into the moving frame, simultaneously, again.
Don't tell me you are calling the reflection at the mirror the emission of the photons? Damn, you are a goofball. As I've said over and over and over: Everyone agrees that the photons arrive at M simultaneously. So what? The question is WHEN DID THEY LEAVE THE SOURCES (what we used to call A and B).
because the photons arrived simultaneously at the midpoint she is forced to conlcude the photons could not have been emitted sinultaneousl? And after all etc?
If all you are saying is "the moving observer sees the photons reflected by the mirror at M simultaneously", then of course. So what does this have to do with Einstein's argument?
If you had even a slight clue, you would know that there is no issue with events happening simultaneously if they happen at the same place.
Yes but the photons arrive at the detector for an instant. You are saying that just because she is moving to the B source and away from the A source the photons could not possibly have been emitted simultaneously, in which frame Doc Al, which frame could they not have been emitted simultaneously? You left this out. On purpose i believe.
I, like Einstein, have always been talking about whether the emissions from A and B are simultaneous, not their reflection from some mirror at M.
You are quite the comedian, geistkiesel.
#44
geistkiesel
538
1
Doc Al said:
Don't tell me you are calling the reflection at the mirror the emission of the photons? Damn, you are a goofball. As I've said over and over and over: Everyone agrees that the photons arrive at M simultaneously. So what? The question is WHEN DID THEY LEAVE THE SOURCES (what we used to call A and B).
What is the difference? Call it two experiments if you must. The photons left the source mirrors and immediately entered the moving frame immediately and did so simultaneously. And were detected simultaneously, immediately. Cannot you see this? Your petty language is getting old can't you conduct a debate w/o smirking? but then it wouldn't be Doc Al would it? You sir are the King of Denial.
Everybody agrees that the photons entered the moving frame simultaneously also, right?
What is the difference? Call it two experiments if you must. The photons left the source mirrors and immediately entered the moving frame immediately and did so simultaneously. And were detected simultaneously, immediately. Cannot you see this? Your petty language is getting old can't you conduct a debate w/o smirking? but then it wouldn't be Doc Al would it? You sir are the King of Denial.
Everybody agrees that the photons entered the moving frame simultaneously also, right?
But your new experiment is trivial and has nothing whatsoever to do with the relativity of simultaneity. Of course the photons hit the M mirror at the same time. So what? Of course photon A and the reflected photon B will arrive at O' at the same time. So what? It doesn't change the fact that the unreflected photon B arrives at O' before photon A. And it doesn't change the conclusion: According to the O' clocks, photons from A and B were not emitted simultaneously. Case closed.
Simultaneity is only interesting when the events are separated by a distance.
Here's a similar experiment for you to ponder: Consider a "stationary" frame O with A, B, and midpoint M. But instead of A and B being the source of the light, let's have the light source be at M. Let A and B be detectors.
And let's have a similar "moving" frame O' with A', M', and B': A' and B' are light detectors and M' is the midpoint.
Let the experiment be such that just as M' passes M, the light at M flashes. Questions:
(1) Do the observers A and B detect the photons from M simultaneously according to their clocks?
(2) Do the observers A and B conclude that the photons were emitted simultaneously in their frame?
(3) Do the observers A' and B' detect the photons from M simultaneously according to their clocks?
(4) Do the observers A' and B' conclude that the photons were emitted simultaneously in their frame?
Extra credit:
(5) Do the observers in ("stationary") frame O agree that A' and B' detected the photons simultaneously according to the O frame clocks?
(6) Do the observers in ("moving") frame O' agree that A and B detected the photons simultaneously according to the O' frame clocks?
#46
geistkiesel
538
1
Read the Einstein version of the moving trrain as stated in "relativity" he is quite explicit that the definition is as I stated it. Also, you are correct my analysis had nothing to dio with SrRas SR was negated by the showing of the contradiction of experimental results and defintion, or theory.
Doc Al said:
Simultaneity is only interesting when the events are separated by a distance.
Here's a similar experiment for you to ponder: Consider a "stationary" frame O with A, B, and midpoint M. But instead of A and B being the source of the light, let's have the light source be at M. Let A and B be detectors.
And let's have a similar "moving" frame O' with A', M', and B': A' and B' are light detectors and M' is the midpoint.
Let the experiment be such that just as M' passes M, the light at M flashes. Questions:
(1) Do the observers A and B detect the photons from M simultaneously according to their clocks?
(2) Do the observers A and B conclude that the photons were emitted simultaneously in their frame?
(3) Do the observers A' and B' detect the photons from M simultaneously according to their clocks?
(4) Do the observers A' and B' conclude that the photons were emitted simultaneously in their frame?
Extra credit:
(5) Do the observers in ("stationary") frame O agree that A' and B' detected the photons simultaneously according to the O frame clocks?
(6) Do the observers in ("moving") frame O' agree that A and B detected the photons simultaneously according to the O' frame clocks?
yes.
yes.
No. You must realize that only after the observers can look at the detctors can they tell when the photons arrived. This is after the event.
Yes see below.
Yes, though if you are using SR the clocks won't tell the same time but the clocks will show the time the photons arrive simultaneously at M, or M'. In other words assuming the stationary observer is located just at the spot where the O', observer and the arriving signals meet he will detect simultabeousl arrival of the signald and O'.
Yes, though if SR dilation occurred then the clock readouts would be different, but the moving observer would determine the photons arrived simultaneously at A and B. This is the same problem where you gave your cynical "so what" to my finding t3 in terms of t1, c, and v. t3 = t1(c + v)/(c - v), remember?
Or said another way: When the light arrives at A and B the detectors will record a time. If SR is operating and O' knows he is moving he knows the photons will arrive a A and B simultaneously, and will arrive at A' before B'. but knowing his velocity he can back calculate and detemine the simultaneity of the arrival at A and B.
He later checks the A and B clocks which will indicate the same time. When O' gets to B his clock will read somethingwhich he ntes but he hasn't arrived at B' yet. but he will only infer the photons arrived at A and B simultaneously as he knew they were stationary wrt the emitted photons.
Number 4.
Code:
A'|_|_|_|_M''_|_|_|_|B' t = 0
|<_|_| M'|_|_>|_|_| t = 1 The the left moving photon arrived at A' and M' moved a distance to the right as shown
|>|_|_M'_|_|_|_ | >| t=2 The The left moving photon is now(signal wise) at the original M' t = 0, The right photon has just caught up with B' six equal marks away from M' when t = 0.
|_|_|_'| |>M'<|_|_|_|_| t = 3.
The scale might be off but the O' observer is shifted three units to the right.
Here both photon signals arrive at M' three units shifted to the right. I do not mean to change your question but this is the fastest way the O' observer (both observers) can receive information. The clocks will confirm simultaneous arrival of the signal at the M' position shfted three units to the right. If all you wanted was the arrival of the photons at the detectors, then no, the signals will not arrive simultaneously at the detectors A' and B'.
However, the O' will conclude the photons were emitted simultaneously in his frame as the signal will arrive simultaneously to the shifted M', position. just as he arrives. This is the first observation he, O', can receive.
I will dispute your assumption that like the original Einstein experiment we have been discussing, the fact that the moving observer will detect the photons simultaneously negates the assumption, or conclusion, that the photons were not emitted simultaneously just because he detected he B photon before he detected the A and B photons simultaneously, later. Think about it. It follows the defintion, the only problem for SR is that it is left bleeding at the side of he road.
These kinds of problems geneally do not need clock measurements in order to determine simultaneity. One only needs an instantaneous inference that two entities did something together.
Doc Al, in spite of your persistent sneering, I stopped long ago, you still have the opportunity to get a good seat in the stands as the late arriving SR theorists will be parading slowly in their professional funeral march with sadness written all over their faces. Then you can really sneer.
Last edited by a moderator:
#47
geistkiesel
538
1
Doc Al, I reread AE and his definition of simultaneity, yes it is a definition. He says: "Hence the observer will see the light emitted earlier from B than that emitted from A". The observers never see the emission of the photons, never. The observer sees the photons long after the photons have been emitted.
AE is presumptive in this definition and he is very unscientific in doing so. AE knows what he is saying and doing and what he is doing is pulling the scam of scams on a trusting audience.
(3) No. You must realize that only after the observers can look at the detctors can they tell when the photons arrived. This is after the event.
Wrong. Both frames detect the photons simultaneously on their own clocks. What do you mean the detection is "after the event"? The detection is the event.
(4) Yes see below.
Right, but not for your reasons.
(5) Yes, though if you are using SR the clocks won't tell the same time but the clocks will show the time the photons arrive simultaneously at M, or M'. In other words assuming the stationary observer is located just at the spot where the O', observer and the arriving signals meet he will detect simultabeousl arrival of the signald and O'.
(6) Yes, though if SR dilation occurred then the clock readouts would be different, but the moving observer would determine the photons arrived simultaneously at A and B. This is the same problem where you gave your cynical "so what" to my finding t3 in terms of t1, c, and v. t3 = t1(c + v)/(c - v), remember?
Wrong. Each observer sees the others clocks as being out of synch. (And yes I remember your silly calculation where you use time measured in the "stationary" frame, but claim you are using times measured in the moving frame. I assume you are just confused, not willfully trying to lie or deceive.)
The bottom line is that both frames will detect the photons arriving simultaneously. And, since the light is emitted at the midpoint between both observers, both O and O' will conclude that the photons were emitted simultaneously in their own frames.
But, since we knew that the photons were all emitted at the same place and time according to all frames, this should be obvious.
Doc Al, in spite of your persistent sneering, I stopped long ago, you still have the opportunity to get a good seat in the stands as the late arriving SR theorists will be parading slowly in their professional funeral march with sadness written all over their faces. Then you can really sneer.
Don't get your hopes up. You are still stumbling over problems I would assign to high school students.
Doc Al, I reread AE and his definition of simultaneity, yes it is a definition. He says: "Hence the observer will see the light emitted earlier from B than that emitted from A".
Please pull out your dictionary and look up the word "Hence".
#50
geistkiesel
538
1
Doc Al said:
Please pull out your dictionary and look up the word "Hence".
Hence = therefore. This is how I used the word, just like AE used the word. What is your real question here?
#3. No, the A' observer detects the photon from M before the B' observer detects the photon from M. Their clocks will show different arrival times.
#4. After comparing their clock times and knowing their velocity wrt the stationary frame the A' and B' observers will conclude the photons were emitted simultaneously.(See my analysis)
#. 5) Do the observers in ("stationary") frame O agree that A' and B' detected the photons simultaneously according to the O frame clocks?
No. The stationary observers know the A' observer is moving to the oncoming poton and they know the B' is chasing the outgoing photon.
(6) Do the observers in ("moving") frame O' agree that A and B detected the photons simultaneously according to the O' frame clocks?
Yes. The photons from M have the same distance to travel. The moving obsever can see this clealy. Even if the clocks were slower in the moving fame the distance the photons cover is the same, therefore the clocks will determine that the photons had the same time of flight to arrive at A and B. If the clocks do not agree then they are niot operating popely.
#51
geistkiesel
538
1
Quote:
(3) No. You must realize that only after the observers can look at the detctors can they tell when the photons arrived. This is after the event.
Doc Al said:
Wrong. Both frames detect the photons simultaneously on their own clocks. What do you mean the detection is "after the event"? The detection is the event.
The speed of light is constant. If O' is moving wrt the spot in the stationary frame where the light was emitted then it is physically impossible for the A' observer to detect the photon from the source the same instant the B' observer detects the photon., even if their clocks wee running slower thatn the stationary clocks. You can't have it both ways.
#52
geistkiesel
538
1
Doc Al said:
Right! Good for you.
Wrong. Both frames detect the photons simultaneously on their own clocks. What do you mean the detection is "after the event"? The detection is the event.
Right, but not for your reasons.
Wrong. Each observer sees the others clocks as being out of synch. (And yes I remember your silly calculation where you use time measured in the "stationary" frame, but claim you are using times measured in the moving frame. I assume you are just confused, not willfully trying to lie or deceive.)
SR theory sucks, for the following reasons. If you follow the argument in my analysis that you find so silly you will see that the moving observer will detect the photons were emitted simultaneoulsy in his frame as he will receive the information of the emitted photon simultaneously, even though he has moved to the right. The detetctors will not detetct the photons at the same time, after all the observers are moving with respect to the point the photons were emitted, but this has nothing to do with SR.
Doc Al said:
The bottom line is that both frames will detect the photons arriving simultaneously. And, since the light is emitted at the midpoint between both observers, both O and O' will conclude that the photons were emitted simultaneously in their own frames.
Of course, but the detectors in the moving frame will not detect the photons simultaneously. It is only after the infromation regarding time of emission, detetction times and velocity will the moving frame determine the photons were emitted simultaneously.
Doc Al said:
But, since we knew that the photons were all emitted at the same place and time according to all frames, this should be obvious.
Wrong. The stationary observer will have no problem, but for the moving observer it is not obvious anad he will have to go through the calculations as I did in order to see that that the photons were emitted simultaneously; or simply use mirros as per my analysis and he can see the photons were emitted simultaneously..
Doc Al said:
Don't get your hopes up. You are still stumbling over problems I would assign to high school students.
I am not stumbling over problems i am stumbling over people that believe in the silliness of SR as it was pounded into their minds.
Hence = therefore. This is how I used the word, just like AE used the word. What is your real question here?
Right: Hence = therefore. Thus indicating a conclusion not a definition. My real question: Do you understand english?
#3. No, the A' observer detects the photon from M before the B' observer detects the photon from M. Their clocks will show different arrival times.
Incorrect. You are still viewing things from the stationary frame. Break out that box! To observers A' and B', the light flashed on exactly at the midpoint; which is also what A and B see. So clocks A' and B' read the same when the photons arrive; as do A and B.
#4. After comparing their clock times and knowing their velocity wrt the stationary frame the A' and B' observers will conclude the photons were emitted simultaneously.(See my analysis)
Now why in the world would the O' frame use the velocity of the O frame in figuring out what they can observe directly? The reason that all observers agree that the photons were emitted simultaneously is that (a) the photons were emitted from the midpoint between the two detectors and (b) the photons were detected at the same time by both detectors.
#. 5) Do the observers in ("stationary") frame O agree that A' and B' detected the photons simultaneously according to the O frame clocks?
No. The stationary observers know the A' observer is moving to the oncoming photon and they know the B' is chasing the outgoing photon.
You changed your answer from before. Good for you! (But note that this reasoning works exactly the same for the moving frame!)
(6) Do the observers in ("moving") frame O' agree that A and B detected the photons simultaneously according to the O' frame clocks?
Yes. The photons from M have the same distance to travel. The moving obsever can see this clealy. Even if the clocks were slower in the moving fame the distance the photons cover is the same, therefore the clocks will determine that the photons had the same time of flight to arrive at A and B. If the clocks do not agree then they are niot operating popely.
No. The moving observer clearly sees that A is moving away from the oncoming light, and B is moving towards it. This is the exact reasoning you used to (finally) get the correct answer to #5. Have you forgotten it so soon?
#54
geistkiesel
538
1
Doc Al said:
Right: Hence = therefore. Thus indicating a conclusion not a definition. My real question: Do you understand english?
Incorrect. You are still viewing things from the stationary frame. Break out that box! To observers A' and B', the light flashed on exactly at the midpoint; which is also what A and B see. So clocks A' and B' read the same when the photons arrive; as do A and B.
Hence means 'therefore'. AE is using the term in the context that the observers on the train see the B pulse first then the A pulse, Hence, they conclude he photons were not emitted simultaneously.And they conclude erroneously. It is easier teaching average 8th graders this stuff than old men whose minds have been corrupted by ignorant graduate advisors. Drink a glass of warm milk Doc and get a good night's rest.
No the speed of light is constant and measuring from different frames does not impose restrictions. A' and B' are moving wrt the pulse source and hence as the pulse source does not move., the A' and B' detectors will record different times. therfore, [HENCE] A' and B' read the photons at different times.
Now why in the world would the O' frame use the velocity of the O frame in figuring out what they can observe directly? The reason that all observers agree that the photons were emitted simultaneously is that (a) the photons were emitted from the midpoint between the two detectors and (b) the photons were detected at the same time by both detectors.
Only the stationary detectors measure simulaneous signals fronm the source.
On;y the moving observer can measure the simultaneous arrival of the pulses and then only if the A' and B' detectors are replaced by mirrors.
Look at the physics of the problem and put your SR crap to sleep. A stationary and moving frame are going to measure different times of emission using absolute time as a reference. And there aren't "both" detectors, there are four detectors, two moving, two stationary. Let uis keep it that way OK?.
You changed your answer from before. Good for you! (But note that this reasoning works exactly the same for the moving frame!)
No you are wrong. I said the moving detectors will not simultaneoulsy detect the emitted photons and the stationary obserbver will also not see simultaneousl detection of the photons by A' and B'. The moviing OBSERVER will detect the photons ariving at the shifted position of O', if the detectors are replaced by mirrors. The reasonong is not the same using the moving frame as a staionary frame, because the moving frame is moving, it is not stationary. Do you undertstand?
No. The moving observer clearly sees that A is moving away from the oncoming light, and B is moving towards it. This is the exact reasoning you used to (finally) get the correct answer to #5. Have you forgotten it so soon?
Doc it is you gedunken stick with it. O is stationary thrpugh out, O' is moving through out.
See, here is where you and SR are full of it. The moving observer O'cannot "see" A moving at all. There is the speed of light restriction. All that the moving observer O' can "see" is the first return of the signals, either at the shifted origin, if the detectors are replaced by mirrors, or after she has looked at the timing of the detectors at A and A' and B and B', which is post experiment time.
Remember Doc Al, this is your gedunken. You said that the O' was the MOVING observer, now you want to play the old "reversee who is moving" game. Stick to what you described and answer the question, if you are able. Are you intentionally doing this (I cannot believe you are other than misguided, which is why I have exercised an infnite degree of patience with you) or is it just a carry over of what you learned in graduate school?
You are putting the moving observer in the stationary frame and doing some kind of logical flim flam dance. Get back in your proper frame Doc Al bwefore you fall off.
I am responding with another post here.
#55
geistkiesel
538
1
The Basic System
There are two photon sources at A and B with the midpoint marked at M. Photons are emitted in a pulsed mode, | | | | | | |, where each pulse, |, is coded such that the A and B pulses can be distinguished. This can be accomplished using coded MW signals with pulse width of | ~ λ, the wavelength of a light pulse emitted at A and B.
Extended System
In the middle of the sources is a mirror system that reflects the A and B photons back to their respective source position as well as pairing each of the reflected photons with a photon from the opposite source. On the first line A and B pulses are deflected into mirrors, on the second line, that reverses the directions of the pulses. On the third and fourth lines the reversed pulses are paired with pulses of the opposite type: B paired with A and A paired with B.
At this point A and B photons are moving in both directions with the same left-right spatial locations.
Organization of Pulses
Code:
|_____________________->\/<-___________________________|
A <-____<-/\->____-> B
B <-____<-/
\->____-> A
Final System at Equilibrium
Located at A and B is a system of flat reflecting mirrors that reflect the AB photon pairs back to the opposite set of mirrors. After a complete reflection cycle is completed equilibrium is established.
The vertical lines that stretch from A to B along the bottom can selectively measure AB photon pairs moving in either direction. Therefore, any position along the AB axis will measure AB photon pair pulses simultaneously. If we install an electronic system to enable anyone of the detectors randomly or sequentially, AB photon pairs will always be measured simultaneously. If we program the detectors to be triggered sequentially we may simulate motion along the AB axis, or we can insert a detector to actually move along the AB axis.
Now the question: Is a detector moving in the AB axis going to measure the same as a static detector? This question must be answered in the context of the experiment described. Using SR theory to invent an answer will be circuitous logic and therefore useless.
Code:
|-----------------------------------<- A--------------------------------|
|-----------------------------------<- B--------------------------------|
|----------------> B ---------------------------------------------------|
|----------------> A ---------------------------------------------------|
|||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
Definition of Simultaneity
Pairs of reflected photons are coded as pairs.
AE's definition that “Events that are simultaneous in stationary frames are not simultaneous in moving frames.” means must a conflict between definition and experiment exists. Some might consider the 'definition' as a 'conclusion'. The statement is a definition as offered by AE ,and we can debate the point later, however ,it must still be proved, one way or the other, before it is universally adopted. We must be very careful not to implement the definition iin a way that results in circuitous reasoning.
Those implications derived from the definition have resulted in discarding the concept of “absolute time” and as such are fatally flawed. If we use the definition above then we discard absolute time, insert time dilation, shrink matter in the direction of motion and measure the speed of light always as c = 300,000km/sec without regard to the motion of the ftrame from which the measurement is taken. and all of this to conform to the definition, however phrased. [Notice there is no attempt to measure the speed of light with respect to the motion of the frame of reference in practice.].
Definitional flaw in Simultaneity
There is an obvious error in the implementation of the definition of simultaneity which originated with A. Einstein in his widely discussed gedunken experiment using a train as a moving frame, with one observation point on that train passing through the midpoint M of the A and B sources just as photons are emitted from A and B. It is only the first emitted photon pair that are considered in AE's analysis of the definition
As defined any system of moving entities will fall uner the deinitional umbrella of simultaneity.
Before we continue let us look at the condition where A and B are the starting point for NASCAR vehicles leaving A and B at 300km/hr as substitutes for the photons. For the train we substitute a single NASCAR detector moving on a straight track at 100km/hr in the B direction. The AB distance is 800km.The NASCAR detector will detect the ‘B’ NASCAR first, followed by the ‘A’ NASCAR. If we now look at the detector we see the B object detected before the A object. What conclusions may we draw from this data alone? Nothing, other than B was detected before A. May we conclude that the moving detector has proved the definition of simultaneity? No.
Simultaneity is functionally related to information of the time an observer passes through M, the velocity of the train wrt stationary frame, the AB sgtationary distance and the speed of light wrt stationary frames. With only the fact that a train observers in AE's gedunken only see the different arrival times, no conlcusions regarding simultaneity are reasonable. If the observers on the train consider their velocity wrt the stationary frame they are still unable to conclude when the photons were emitted. If they consider that they were at M when the photons were emitted they are still unable to determine simultaneity. [the photons could have been emitted any time before they arrived at A and/or B]. If the observers then add the consideration of the distance between A and B they must first make the calculations to determine if the photons were emitted simultaneously. In a previous post I derived that t2 = t1(C + V)/(C - V) where t1 is the time of the B measurement wrt t = 0 when the observers were at M, and t2 the time of the A measurement wrt t = 0 at M. Using the AE definition, each observer strung along the train must make her own calculations to determine her own state of observing simultaneity. Each will determinme the photons were emitted simultaneously into the moving frame of he train. Imposing time dilations is premature here. The observers who are at the midpoint M when the photons arrive their simultaneously need not make any calculaiopns, they will be able to see the photons arrive at the same place they are located.
There can be only one law for all the observers on one train, right?[/size]
The defnition is ambiguous regarding conditions of information.
If we know ahead of time that the detector passed through M just as each A and B NASCAR left A and B may we then approve of the definition of simultaneity? No, obviously not. If two objects are headed toward each other and a third object is behind the middle object and is trying to catch up (under the conditions established here) the objects moving in a collision course will always meet before the trailing object catches up. Think about it, any objects will suffice :earth worms, olympic sprinters, even photons confined to the situation described above. Are we, then going to assign AE’s definition to the NASCAR/detector system in order to bring the definition of simultaneity into a universally operating reality, for all objects? No.
Simultaneity not universally defined.
Simultaneity, as discussed in the context of the definition refers specifically to electromagnetic motion, or radiation. In this sense, the definition is used to justify the time dilation, mass shrinking and the implication from these concepts that the measure of the speed of light as constant with respect to all inertial frames. In other words, once one accepts the definition the concepts mentioned must be implmented in order to maintain consistency with the definition.
Circuitous analysis substututed for physical reality.
One should look very closely at what is being trashed in order to bring about a compromise that justifies the mental limitations of the scientific industry.
We must be very careful that definitions do not bootstrap coincidental observations imposed by the defintion. To infer preconceived ideas about concepts of natural phenomena and surrepticously constructing defintions is always fatally hazardous.
The Speed of Light is C in vacuo
So light moves at a constant velocity of c in vacuo. Is this any reason, by itself, to conclude that we must measure the relative velocity of light as c with respect to any frame of reference? Not including the velocity of the referenced frame when measuring c is equated to voluntarily placing the observer firmly on the horn of a dilemma. This omission will always result in an erroneous conclusion that time dilates and mass shrinks., which results in, voila! the speed of light measured as c = 300,000km/sec [with dialted time and mass, of course].
Are we thinkers ot robots?
What about the implications of Galilean vs. Lorentzian transformations and the implications of Maxwell’s equations? Just crank out some maths? What about the relative motion of 8km/sec between light and the aether found by Dayton Miller [who confirmed the Michelson-Morely's results]? Those confining themselves to the rote meanings of concepts as they were taught, without an in depth and critical analysis, simply get back on their horn and proceed along in the pain of collective and singular confusion .
Doc it is you gedunken stick with it. O is stationary thrpugh out, O' is moving through out.
You are still stuck thinking that "stationary" somehow means something absolute. It's just an arbitrary label. I never said what O was stationary with respect to. It just so happens that O is stationary with respect to a spaceship that is traveling alongside it. And, what do you know, O' is moving with respect to that ship... but it's stationary with respect to the Earth! (Which is perhaps a billion miles away.)
Who knew? Time to change all your goofy arguments around, geistkiesel. I guess O' was really stationary all the time!
The Basic System
There are two photon sources at A and B with the midpoint marked at M. Photons are emitted in a pulsed mode, | | | | | | |, where each pulse, |, is coded such that the A and B pulses can be distinguished. This can be accomplished using coded MW signals with pulse width of | ~ λ, the wavelength of a light pulse emitted at A and B.
Extended System
In the middle of the sources is a mirror system that reflects the A and B photons back to their respective source position as well as pairing each of the reflected photons with a photon from the opposite source.
As I have explained at length in an earlier post, introducing a mirror at the midpoint changes the original gedanken and makes it trivial:
- Everyone (O and O') agrees that the photons reach the mirror at the same time
- Everyone (O and O') agrees that the "paired photons" left the midpoint at the same time
So what?
Once again I request that you go back to the original Einstein gedanken and answer my questions. There is no escape, geistkiesel!
#58
geistkiesel
538
1
Doc Al said:
You are still stuck thinking that "stationary" somehow means something absolute. It's just an arbitrary label. I never said what O was stationary with respect to. It just so happens that O is stationary with respect to a spaceship that is traveling alongside it. And, what do you know, O' is moving with respect to that ship... but it's stationary with respect to the Earth! (Which is perhaps a billion miles away.)
Who knew? Time to change all your goofy arguments around, geistkiesel. I guess O' was really stationary all the time!
Oh I get it. If you are an SR theorist, you can make it up as you go along. So a stationary train depot is not staionary anymore? Oh I get it arbitrary labels allowed in SR theory, but rational analyiss is verboten? You have just introduced the silliest things you have ever posted in your career. talk about out in the clouds.
SR error that is fatal, every time: Assuming stationary and moving platforms then swapping them in the middlde of analysis to sound like you know what you are talking about.
Not only do you make it up arbitrarily as you go along, you are in lock step wuith all the other SR robots: you make up impossibke situations and conditions. I admit my latest post has some engineering problems to ovecome, but theoretically the gedunken is "rational" Doc Al, a word you should look up in the dictionary.
#59
geistkiesel
538
1
Doc Al said:
As I have explained at length in an earlier post, introducing a mirror at the midpoint changes the original gedanken and makes it trivial:
- Everyone (O and O') agrees that the photons reach the mirror at the same time
- Everyone (O and O') agrees that the "paired photons" left the midpoint at the same time
So what?
Once again I request that you go back to the original Einstein gedanken and answer my questions. There is no escape, geistkiesel!
Excuse me I got the order of your posts reversed. You are correct the gedunken is trivial, as trivial as Einstein's is in error. I have answered your questions many times over. It is your attitude that needs answering.
Doc Al above said:
- Everyone (O and O') agrees that the photons reach the mirror at the same time
- Everyone (O and O') agrees that the "paired photons" left the midpoint at the same time[/INDENT]
So what?
So what? You didn't complete the logically next third sentence. Let me assist you in this trivial matter:
Geistkiesl is finishing Doc Al's post for him. Apparently Doc fell asleep at the wheel before the simultaneity wall appeared in front of ---.
Everyone (O and O') agrees that the "paired photons" were emitted simultaneously (and detected as such) in the moving frame at the same time.
You are very welcome Doc. Can I call you Doc?
There, Doc Al, let me be the first to welcome you back to a rational world.
Oh I get it. If you are an SR theorist, you can make it up as you go along. So a stationary train depot is not staionary anymore? Oh I get it arbitrary labels allowed in SR theory, but rational analyiss is verboten? You have just introduced the silliest things you have ever posted in your career. talk about out in the clouds.
The idea that there is a state of absolute rest died over 350 yrs ago. Since then, "stationary" has always been an arbitrary label; That which is considered "stationary" is considered so for convenience only. This is not a idea new to SR.
Until you come to grips with this you are no better off than the members of the Flat Earth Society, who refuse to come to grips with the concept of a spherical world.
We must be very careful that definitions do not bootstrap coincidental observations imposed by the defintion. To infer preconceived ideas about concepts of natural phenomena and surrepticously constructing defintions is always fatally hazardous. 