The discussion centers on the concept of simultaneity in the context of special relativity, specifically analyzing a scenario involving a moving frame and light pulses emitted from a midpoint. Observers in both stationary and moving frames conclude that light pulses meet simultaneously at the midpoint, despite the movement of the frame. The analysis suggests that the perceived simultaneity of events is maintained regardless of the observer's frame of reference, challenging the notion of "loss of simultaneity." It is emphasized that disagreements about simultaneity arise only when events occur at different spatial locations. Ultimately, the discussion reinforces that both moving and stationary observers agree on the simultaneous arrival of light pulses at the midpoint.
#1
geistkiesel
538
1
The figure is of a moving frame moving to the right at velocity = 1/3c. On each end of the frame is a mirror (L and R) located at 4(1/3 + 1/3 + 1/3 + 1/3) x (1 second) from the midpoint of the frame. A pulse of light emitted at M (on either the moving or stationary frame) . We define the stationary point M when the pulse is emitted.
Code:
L | |____|____|____M____|____|____| |R
1. In the first second the photon headed left will arrive at 1c to the left of M while the left mirror has moved to a point where the photon is located during the first second. The right photon is located 1c to the right of M at the end of 1 second..
1 sec. L |<____________M___________> |R
2. After 2 seconds the photon on the left has reflected back to M while the right photon has just reached the right mirror:
2 sec. ____________>M ____________>|
3. After 3 seconds the left photon is now located at the shifted M' (3)(1/3) units to the right of M. The right photon reflecting off the R mirror is also located (3)(1/3) units to the right of M.
_____________>M'<____________
3 sec.
As the frame has moved to the right (3)(1/3) units the photons meet at the position of the moving observer who, quite naturally, concludes the photons were emitted simiultaneously.
We conducted the above analysis from the stationary frame. From symmetry considerations conducting the analysis in the moving frame gives the same result: The moving observer concludes the photons were emitted simultaneously in the moving frame.
This you can easily demonstrate for yourself. Where is the loss of simultaneity? If any?
neither your observer/emitter or reflectors are moving in relation to each other.
net result = same as stationary frame
#3
geistkiesel
538
1
ram2048 said:
neither your observer/emitter or reflectors are moving in relation to each other.
net result = same as stationary frame
The point I was making (among others) is that the emitter can be moving or stationary.
Right: the reflectors are attached to the moving frame. The Left reflector moves to the oncoming photon heading Left. The right photon begins to chase the moving reflector to the Right. Watching from the stationary platform we see the photons finally meeting where the obseserver on the moving platofrm is then located.
The moving observer will see the lights simultaneously meeting her, after the reflections, whatever the time/space dilations may be, or not be. Pick your frame, you always get simultaneous emissions of the photons.
The moving frame contains L reflectror, the Observer midpoint to the reflectors and the Right reflectors. The photons are onblivious to the motion of the frame and only reflect when a mirror is present.
#4
ram2048
220
0
yep yep, according to Einstein's relativity all inertial frames will act the same way
and function under the same physics as something "supposedly" stationary, since he didn't define an absolute rest frame
#5
ram2048
220
0
this is sweet i learn something in another thread and get to come here and talk like i know what's going on kekeke
#6
wisp
179
0
geistkiesel said:
This you can easily demonstrate for yourself. Where is the loss of simultaneity? If any?
You have demonstrated that the observer will always measure light pulses meeting at the middle, regardless of whether or not the observer frame is moving.
So if light is moving at c+v and c-v the results are similar to a frame in which light travels at c, as the losses and gains in the speed of light cancel out.
This is effectively a two-way light speed test, and it is very difficult to record the fact that the speed of light could be different from c.
#7
geistkiesel
538
1
wisp said:
You have demonstrated that the observer will always measure light pulses meeting at the middle, regardless of whether or not the observer frame is moving.
So if light is moving at c+v and c-v the results are similar to a frame in which light travels at c, as the losses and gains in the speed of light cancel out.
This is effectively a two-way light speed test, and it is very difficult to record the fact that the speed of light could be different from c.
You might recognize this as the cerulean etc example of another thread that 'proved' the moving observer would conclude the photons were not emitted simultaneously. If you look closely, you are able to use either frame as a basis of calculations. Where is the loss of simultaneity? Howver, the moving observer sees only the final arrival of both photons after some time dilation. The stationary observer in the stationary case sees the emitted to return time as 2 + 2x(1/3) = 8/3 sec. The stationary observer sees the return time in the moving frame as 9/3 sec.due to the movement of the point M in 3 sec time span.
Inany event, whatever time is determined by the moving observer for the round trip of the reflected lights, thephotons always arrive simultaneously at the moving observer.
If I am not mistaken, the Cerulean model had the moving observer concluding the photons were not emitted simultaneously because the photon first hit the left reflector then the right reflector. The moving observer cannot make this claim as he cannot see when the photons struck the left and right mirrors. He has to wait, just like the stationary case, until the photons arrive at the midpoint M.
I am not sure what you mean by the losses and gains in the speed of light. Each photon simply travels the same distance in the same of amount of time. Again, the cerulen example concluded othewise.
Wespe quoting Paul Newman in movie, 'The Color of Money', "I'm back."
yes of course both the observer moving relative to what you have defined as the 'stationey frame' and the observer in the stationery frame agree that both photons were emitted simulataneously.
this comes as no great surprise as simultaneity only fails at distance.
#9
geistkiesel
538
1
jcsd said:
yes of course both the observer moving relative to what you have defined as the 'stationey frame' and the observer in the stationery frame agree that both photons were emitted simulataneously.
this comes as no great surprise as simultaneity only fails at distance.
I don't understand " . . .simultaneity only fails at distance". The example here was argued extensively where the moving observer was deemed to conclude the photons were not emitted simultaneously, by some, not me.. However, in the example discussed the lack of simultaneity was based, as far as I was able to determine, on the fact that the left mirror met the photon before the right mirror received a photon. This was the "physical state" that indictated the moving observer would conclude the photons were not emitted simultaneously. The analysis did not extend to to the expanded reflection scenario that I described. I could never understand how the moving observer could observethe photons before they reached her, do you know what I mean?
So again, what is the distance condition you refer to?
I don't understand " . . .simultaneity only fails at distance". The example here was argued extensively where the moving observer was deemed to conclude the photons were not emitted simultaneously, by some, not me.. However, in the example discussed the lack of simultaneity was based, as far as I was able to determine, on the fact that the left mirror met the photon before the right mirror received a photon. This was the "physical state" that indictated the moving observer would conclude the photons were not emitted simultaneously. The analysis did not extend to to the expanded reflection scenario that I described. I could never understand how the moving observer could observethe photons before they reached her, do you know what I mean?
So again, what is the distance condition you refer to?
Two 'events' (really they are the same event) having the same spatial and temporal location will appear to be simultaneous to all observers.
#11
geistkiesel
538
1
jcsd said:
Two 'events' (really they are the same event) having the same spatial and temporal location will appear to be simultaneous to all observers.
Could you please elaborate on the meaning o "distance". Thanx
Two 'events' (really they are the same event) having the same spatial and temporal location will appear to be simultaneous to all observers.
geistkiesel said:
Could you please elaborate on the meaning o "distance".
Disagreements about the simultaneity of two events only happen when those events take place at different places (that is, when they are separated by a distance).
For example: Let's say you and I bang our heads together. Since that "event" takes place at the same place, every frame will agree that we hit our heads together "at the same time".
But if instead, you and I are miles (light years?) apart and we decide to slap ourselves in the head at the same time (how? we check our synchronized atomic watches!). In this case, observers moving with respect to us will disagree that we slapped ourselves at the same time. Some will say you slapped yourself before I did; others will say the opposite.
#13
geistkiesel
538
1
Doc Al said:
Disagreements about the simultaneity of two events only happen when those events take place at different places (that is, when they are separated by a distance).
For example: Let's say you and I bang our heads together. Since that "event" takes place at the same place, every frame will agree that we hit our heads together "at the same time".
But if instead, you and I are miles (light years?) apart and we decide to slap ourselves in the head at the same time (how? we check our synchronized atomic watches!). In this case, observers moving with respect to us will disagree that we slapped ourselves at the same time. Some will say you slapped yourself before I did; others will say the opposite.
Thanx for clearing that up. I make the semi-educated guess that the events at our A and B locations are a "distance" problem. However, what of equating the emission of the photons at A in the stationary frame at the same time the photons were emitted at A' in the moving frame, and simularly for the photons emitted at B and B', and M and M'. The photons are measured simultaneously in the moving frame at the thee positions indicated.
The question here: Are the photons emitted from A' and B' simultaneously in the moving frame from symmety considerations.
I make the semi-educated guess that the events at our A and B locations are a "distance" problem. However, what of equating the emission of the photons at A in the stationary frame at the same time the photons were emitted at A' in the moving frame, and simularly for the photons emitted at B and B', and M and M'. The photons are measured simultaneously in the moving frame at the thee positions indicated.
I'm not sure what you are talking about. You must describe the precise situation you have in mind. (Are you mixing this thread with another?)
If A and B are the locations where photons are emitted simultaneously in the stationary frame (t = 0), and A' and B' are the observed locations of those emissions in the moving frame, then no, A' and B' are not simultaneous: the moving frame will measure those events (the photon emissions) to be at different times. (For details, see: https://www.physicsforums.com/showpost.php?p=231002&postcount=115)
#15
geistkiesel
538
1
Doc Al said:
I'm not sure what you are talking about. You must describe the precise situation you have in mind. (Are you mixing this thread with another?)
If A and B are the locations where photons are emitted simultaneously in the stationary frame (t = 0), and A' and B' are the observed locations of those emissions in the moving frame, then no, A' and B' are not simultaneous: the moving frame will measure those events (the photon emissions) to be at different times. (For details, see: https://www.physicsforums.com/showpost.php?p=231002&postcount=115)
Doc Al, I have agreed that SR will predict whatever it predicts. The following does not change any of the parameters we have been using in this thread.
All moving frame values are non-primed with the exception of M’, the consistent location of the observer O in the moving frame.
At no time is there an inference that M’ was at the midpoint of the A and B photons emitted in the stationary frame.
To demonstrate the following:
Einstein’s moving train calculation indicating when the oncoming B photon is detected at t1 the A photon was located at a position consistent with –t1. Said in other words, as t1 is determined from t0 which locates M’ at t0, the A and B were equidistant to M’(t0) when t = t1.
Proof:
A moving observer located at M’ on a moving frame passes through the midpoint M of photon sources located at A and B in the stationary frame just as A and B emit photons. M’ is moving along a line connecting A and B, toward B.
At this instant the moving source t = t0. Later the moving observer detects the photon from B at t1, and later the photon from A at t2. The observer has measured her velocity wrt the stationary frame as v. Determine the position of the A photon at tx in terms of t0, t1, t2, and v when the B photon was detected at t1.
The photon from A must reach the position of M’ when t = t2. Therefore, the distance traveled by the A photon during Δt = t2 – t1, is Δtc. This is equal to the distance cΔt = vΔt + vt1 + vtx . Now we rearrange somewhat to arrive at, vtx = vΔt – cΔt + –vt1. Now as vΔt - cΔt is just -vtx - vt1
vtx = -vtx - vt1 – vt1
2tx = -2t1
tx = -t1
Therefore, in the moving frame the photon from A and the photon from B were equidistant from M’(t0) at t1.
Sorry geistkiesel, but I think these discussions are pretty much a waste of time. I don't think you understand enough SR to have a meaningful discussion about it, never mind provide an intelligible critique.
My offer remains: When you are ready to discuss Einstein's simple argument showing the relativity of simultaneity (no calculations to confuse you!) and you are prepared to point out the flaw in it, let me know.
geistkiesel said:
The photon from A must reach the position of M’ when t = t2. Therefore, the distance traveled by the A photon during Δt = t2 – t1, is Δtc.
Right. As measured by the moving frame, the A photon travels a distance equal to (t2 - t1)c between photon detections by O'.
This is equal to the distance cΔt = vΔt + vt1 + vtx .
No it's not. Where did you get this quaint relationship? Looks like someone's been swapping frames again!
#17
geistkiesel
538
1
Doc Al said:
Sorry geistkiesel, but I think these discussions are pretty much a waste of time. I don't think you understand enough SR to have a meaningful discussion about it, never mind provide an intelligible critique.
My offer remains: When you are ready to discuss Einstein's simple argument showing the relativity of simultaneity (no calculations to confuse you!) and you are prepared to point out the flaw in it, let me know.
Right. As measured by the moving frame, the A photon travels a distance equal to (t2 - t1)c between photon detections by O'.
No it's not. Where did you get this quaint relationship? Looks like someone's been swapping frames again!
Doc Al, c(t2 -t1) is the distance A miust travel during t2 - t1 in the moving frame to arrive at M'(t2) when it does. v(t2 -t1) to get the distance the frame moves in t2 -t1. vt1 is the distance moved from the M'(t0) to M'(t1) and vTX is the distance A is from M'(t0), everything carefully restrained to the moving platform.
Come on Doc Al, no one is swapping frames, that's my charge against you; I would appreciate it if you would get your own metaphors. If you thought someone was "swapping frames" , you wouldn't hesitate to point directly to the source, would you? Doc Al you are beginning to relapse into your old ways. Get a hold of yourself man. The world hasn't come to an end, yet. We may be close, but there is time left for plenty to do.
Doc Al, c(t2 -t1) is the distance A miust travel during t2 - t1 in the moving frame to arrive at M'(t2) when it does.
That distance c(t2 - t1) is measured in the moving frame.
v(t2 -t1) to get the distance the frame moves in t2 -t1.
But now you claim it's also the distance measured from the stationary frame?
vt1 is the distance moved from the M'(t0) to M'(t1) and vTX is the distance A is from M'(t0), everything carefully restrained to the moving platform.
Gee... I hope you're not using times measured in the moving frame to do calculations in the stationary frame?
Come on Doc Al, no one is swapping frames,
Telegram to geistkiesel: Get a clue!
Oh, and be sure to see my post in the other thread. I'll walk you through these calculations step by step. Don't worry... baby steps!
#19
geistkiesel
538
1
Quote: Originally Posted by geistkiesel
Doc Al, c(t2 -t1) is the distance A miust travel during t2 - t1 in the moving frame to arrive at M'(t2) when it does
Doc Al said:
That distance c(t2 - t1) is measured in the moving frame.
Right, that's what I said.
Quote:geistkiesel
v(t2 -t1) to get the distance the frame moves in t2 -t1.
Doc Al said:
But now you claim it's also the distance measured from the stationary frame?
This is the distance calculated by the moving observer. And yes t0 has a stationary connection. She knows she is moving wrt stationary frame and this is what she calculates, using moving frame data. All she has to do is look out the window.
Quote geistkiesel
vt1 is the distance moved from the M'(t0) to M'(t1) and vTX is the distance A is from M'(t0), everything carefully restrained to the moving platform.
Doc Al said:
Gee... I hope you're not using times measured in the moving frame to do calculations in the stationary frame?
Well Doc the A photon is in the stationary frame, anmd the moving frame as well. IS thee some jurisdictional problem we are javing her? All the calculaions two lines uop are all moving frame calculations. You got yourself confused when you tried to change the units and notation to confuse anyone following the thread, remember?
What is this "I hope " stuff? You can see what I am doing. The calculations are in the moving frame. It just so happens the t0 is referenced to a stationary landmark. I can't help it if the gedunken was designed as it was. Talk to A if you have a complainl. If you want to claim that the moving values are going to be dilated then do it. Don't forget to squeeze time down to an appropriate value.
Quote geistkiesel
Come on Doc Al, no one is swapping frames,
I told you all the parameters were in the moving frame and that is what I used. Are you angry at me for something? You seem rather piqued. Was it something I said?
Doc Al said:
Telegram to geistkiesel: Get a clue!
Oh, and be sure to see my post in the other thread. I'll walk you through these calculations step by step. Don't worry... baby steps!
After that I'd be scared to read your post, . . . oh the horror, the horror . . .. Doc Al the 'slice and dice prof'', to baby step Geistkiesel through the throes of a dying theory. See the Final Chapters of, "Throw SR from the Frame".
But you did that in the last post didn't you. You even stole the term "frame jumping " from me. Oh Doc Al how you do stress me out. I was going to coin the phrase about "frame jumping SR theorists" , but you stole my thunder Doc il what a dreadful day I am having.
What clue? Tell me here and now. Embarrass me in front of my friends. Why threaten a dismal future. You might not have me to kick aorund anymore Doc Al, I may be turning professional; I caught a glimpse of thread where they were discussing something about, 'giving me the business'.
This is the distance calculated by the moving observer. And yes t0 has a stationary connection. She knows she is moving wrt stationary frame and this is what she calculates, using moving frame data. All she has to do is look out the window.
v(t2 -t1) is the distance that the stationary frame is observed to move according to the moving frame. It's got to be a measurement made by the moving frame, since it uses times measured by the moving frame! It is not the distance that the moving frame moves with respect to the stationary frame.
Well Doc the A photon is in the stationary frame, anmd the moving frame as well. IS thee some jurisdictional problem we are javing her? All the calculaions two lines uop are all moving frame calculations. You got yourself confused when you tried to change the units and notation to confuse anyone following the thread, remember?
You are mixing measurements and frames left and right. Maybe this will help to clear your head:
There are two events of interest in this problem. Event #1 is the detection of a B photon by moving observer O' (who is located at x' = 0). Event #2 is the detection of an A photon by moving observer O' (still located at x' = 0). You do realize that the two frames disagree as to when those events took place? You can't just mix and match measurements made in the different frames.
#21
geistkiesel
538
1
The f0llowing is a response to Doc Al in this thread. This is a geneal response. I will answer Doc Al word for word in his last post.
Physical Implications of Inherent Errors in Special Relativity
Two events simultaneous in a stationary platform are not simultaneous events in a moving inertial frame. M is the midpoint of two photon sources A and B. Photons emitted simultaneously from A and B arrives simultaneously at M. This is the definition of simultaneity assumed by special relativity, either expressly or impliedly, and is the source of a fundamental contradiction to physical law. This fault is not related to any theoretical postulates of SR regarding the measurement of the speed of light.
Einstein’s explanation of the loss of simultaneity is best described in his popular stationary platform and moving train gedunken. Be prepared to discover the contradiction in physical law before any issues of SR arise.
Two photons are emitted at A and B at t0 when M’(t0) moving with velocity v is located at M(t0) the midpoint of A and B in the stationary platform.
A photon from B is detected at M’(t1).
Photons from A and B arrive simultaneously at M(t2) = M’(t2).
A photon from A arrives at M’(t3).
Primed notation refers to moving frame measurements.
Code:
M(t0) M(t2) M(t0)
A M(t0) B
___|_______________________|__________________________|__
___________________________|_____________|______|__________
M’(t0) M'(t1) M'(t3)
M'(t2)
Defining photons.
Red ants.
Chevrolet Camarros.
Light – electromagnetic radiation.
Definitional Limitations of Simultaneity Exceeded
Any of the three definitions will be sufficient in describing the significant events as long as v < v(photons) and are offered only to show that all three qualify as photons as used in SR theory for the purposes of defining simultaneity. Of course the photons from A and B will arrive simultaneously at points identified as M’(t0) and M’(t2), these points are the original location of the observer in the moving frame at M’. This physical point has moved since the photons were emitted, so naturally the photons from and B will not follow the moving frame, rather the photons will arrive simultaneously M’(t2) = M(t2), the midpoint of the emitted photons.
The original point M’(t0) has moved and it would be impossible for any photons to arrive simultaneously at M’(t1) or M’(t3) without adjusting their speed and time then they were emitted. Therefore, to use the definition of simultaneity with respect to SR, SR must fail having exceeded physical limitations of mathematical definitions. Certainly, the measurement of the constancy of the speed of light is not significant here as all photons are moving at a constant velocity. For SR to survive there must exist independent causes that account for the claims of SR without the use of the definition of simultaneity as previous used by SR.
Physical Law Exceeded by Special Relativity Postulates
It is recognized that SR will conclude that the photons were not emitted simultaneously in the moving inertial fame. This follows from the sequential detection of photons at M’(t1) and M’(t3). This means, of course using SR theory, that photon B must have been emitted before photon A was emitted. This requires some ∆t > 0 that only one photon has been emitted, which is in direct contradiction with the experimental results where the photons arrived at M’(t2) = M(t2) simultaneously. Even if the clocks were moving slower in the moving frame they will be able to determine the time when M’(t0) = M(t0). From the known velocity, v, the original position of M’ with respect to the stationary platform can be determined and hence the simultaneous arrival photons can be determined.
However, the photons were emitted simultaneously in the stationary platform and therefore there was no time that two photons did not exist simultaneously, at least in the stationary platform. Once existing there is no possible physical activity that can dissolve one photon, or mask its existence, in order to agree with SR. As SR does not postulate the sequential independent emission of photons from ghost sources at A’ and B’, or that the moving frame measurements of the emission process could possible affect the simultaneous measurement of the photons at A and B in the stationary frame, SR is patently interpreted falsely here.
Simultaneity Definition Constraints are Satisfied in Experiment
The definition of simultaneous events had been adequately followed in the experiment in both stationary and moving frames. The photons are emitted simultaneously in both inertial frames.
SR theory is Constructed with an Inherent Error in Measurement
The posts by Grounded have shown unambiguously that the failure to include the observer’s velocity in measuring the speed of light will always result in an erroneous finding that the speed of light will always be measured as c in all inertial frame. The speed of light is constant in all inertial frames; however, all inertial frames are able to determine their relative velocity with respect to the velocity of the source of the photons and therefore observers on the moving inertial frames are able to determine their relative speed with respect to the speed of light.
#22
geistkiesel
538
1
Quote:
Originally Posted by geistkiesel
This is the distance calculated by the moving observer. And yes t0 has a stationary connection. She knows she is moving wrt stationary frame and this is what she calculates, using moving frame data. All she has to do is look out the window. So she uses SR clocks, so what?
Doc Al said:
v(t2 -t1) is the distance that the stationary frame is observed to move according to the moving frame.
Here you make the mathematicians classic error. Your mathematics may allow you to say that the stationary frame is moving and the train is stationary. This is a physical impossibility and you know it. The train has stopped and started hundreds of times, each with a starting and stopping acceleration. The platform has never accelerated. The moving platform is the train, the stationary platform is the stationary platform.
This is the way it is.
It is the fault of equating your mathematical freedoms indiscriminately as applying to the physical world in all cases. You can discuss SR without the phony switch of frames.
Doc Al said:
It's got to be a measurement made by the moving frame, since it uses times measured by the moving frame! It is not the distance that the moving frame moves with respect to the stationary frame.
You are confused. I specifically stated that v(t2 - t1) is the distance moved as determined in the moving frame during the time interval noted. The moving observer knows the t0 when she was at the midpoint of the A and B in the stationary frame. Using her own measurements of v and t she is able to determine exctly where she was located at t0 - remember she is using her moving frame calculations. She never measures the speed of light.
Doc Al said:
You are mixing measurements and frames left and right. Maybe this will help to clear your head:
There are two events of interest in this problem. Event #1 is the detection of a B photon by moving observer O' (who is located at x' = 0)[originally m'(t0) geistkiesel]. Event #2 is the detection of an A photon by moving observer O' (still located at x' = 0)[originally M'(t2). geistkiesel] You do realize that the two frames disagree as to when those events took place? You can't just mix and match measurements made in the different frames.
I didn't mix and match. But t0 is t0 in the moving and stationary frames. These times are identical.I don't care what the two frames agree on or not. I never said I did. Your changing the notation is forgiven. When I say M'(t0) this locates the observer in the moving frame when the photons are emitted in the stationary frame. M'(t1) is the location of the moving observer when the photon from B arrives, and M'(t3) the observer's location when the photon from A arrives (see my general answer. I have included the simultaneous time of arrival of the photons at the original M'(t0) = M(t0) to be M'(t2) = M(t2). All of these M' times are moving frame times. In other words the t0 in M'(t0) is not the t0 in M(t0). The former a moving platform time, the latter determined in the stationary frame, however, the t0 are determined to be identical here, wherever determined..
#23
ram1024
301
0
However, the photons were emitted simultaneously in the stationary platform and therefore there was no time that two photons did not exist simultaneously, at least in the stationary platform. Once existing there is no possible physical activity that can dissolve one photon, or mask its existence, in order to agree with SR. As SR does not postulate the sequential independent emission of photons from ghost sources at A’ and B’, or that the moving frame measurements of the emission process could possible affect the simultaneous measurement of the photons at A and B in the stationary frame, SR is patently interpreted falsely here.
but SR gives everyone their own reality and timeline. a simple thing like existing or not existing at a time is irrelevant to SR. in FACT, nothing is relevant(relative) to SR that's why it's exceedingly difficult to disprove.
Two photons are emitted at A and B at t0 when M’(t0) moving with velocity v is located at M(t0) the midpoint of A and B in the stationary platform.
A photon from B is detected at M’(t1).
Photons from A and B arrive simultaneously at M(t2) = M’(t2).
A photon from A arrives at M’(t3).
Primed notation refers to moving frame measurements.
Code:
M(t0) M(t2) M(t0)
A M(t0) B
___|_______________________|__________________________|__
___________________________|_____________|______|__________
M’(t0) M'(t1) M'(t3)
M'(t2)
Before we go off on another wild goose chase, please define your notation and the set up unambiguously. Until we do that, there's no point in discussing this set up. Each event (the emission or arrival of a photon) should have two descriptions: it's position and time as measured in each frame. For example, I have no idea whose clocks are taking those time measurements. I take it that you use M now to mean one frame and M' to mean the other? (So M is no longer the midpoint?) M(t0) means what?
Physical Law Exceeded by Special Relativity Postulates
It is recognized that SR will conclude that the photons were not emitted simultaneously in the moving inertial fame. This follows from the sequential detection of photons at M’(t1) and M’(t3). This means, of course using SR theory, that photon B must have been emitted before photon A was emitted. This requires some ∆t > 0 that only one photon has been emitted, which is in direct contradiction with the experimental results where the photons arrived at M’(t2) = M(t2) simultaneously.
You still confuse "photons arriving at the same time and place" with simultaneity. But no point in getting into any detailed discussion (for the nth time!) until you clean up that notation and properly define the setup.
Why not define the train experiment like so: Two light sources A and B are a distance L from a midpoint M (all measured from a frame in which A, B, and M are at rest). Call that frame O. The lights flash simultaneously at t = 0 (measured in the O frame). There is an observer sitting at M. Let's call him Stationary Stu.
Now introduce a moving frame O'. In that frame there is a second observer, Moving Mary. (Mary and Stu move past each other at a speed v.) Mary passes by Stu at the exact moment that Stu's watch reads t = 0. Mary sets her watch to match, thus she passes Stu at t' = 0.
To make this discussion productive, why not start off by telling us:
- what time will Mary's watch read when she detects photon B?
- what time will Stu say that his watch reads when Mary detects photon B?
- what time will Mary's watch read when she detects photon A?
- what time will Stu say that his watch reads when Mary detects photon A?
- what time will Stu's watch read when he detects photons from A and B?
- what time will Mary say that her watch reads when Stu detects those photons?
Once we agree on that, then we can talk about "simultaneity".
Last edited:
#25
geistkiesel
538
1
Doc Al said:
I can hardly wait!
Before we go off on another wild goose chase, please define your notation and the set up unambiguously. Until we do that, there's no point in discussing this set up. Each event (the emission or arrival of a photon) should have two descriptions: it's position and time as measured in each frame. For example, I have no idea whose clocks are taking those time measurements. I take it that you use M now to mean one frame and M' to mean the other? (So M is no longer the midpoint?) M(t0) means what?
M(t0) is the location of the midppoint of A and B in the stationary frame. M'(t0) the locaion of the observer when the photons emitted from A and B and she was colocated at M(t0) . All primes are moving frame values.
M'(t1) and M'(t3) is he observer when detecting the photons from B and A respecticvely. M'(t2) the time the photons from A and B arrive simultaneously at M(t2).
Doc Al said:
You still confuse "photons arriving at the same time and place" with simultaneity. But no point in getting into any detailed discussion (for the nth time!) until you clean up that notation and properly define the setup.
No simultaneity is defined at the time t he photons were emitted. However, a definition of simultaeity is that if these photons emitted at A and B arrive at M(t0) simultaneously, then they were emitted simultaneously. The notation is clean, you just want to pervert the notation to your own satisfaction and for whatever you motives happen to be in this regard.
doc Al said:
Why not define the train experiment like so: Two light sources A and B are a distance L from a midpoint M (all measured from a frame in which A, B, and M are at rest). Call that frame O. The lights flash simultaneously at t = 0 (measured in the O frame). There is an observer sitting at M. Let's call him Stationary Stu.
Now introduce a moving frame O'. In that frame there is a second observer, Moving Mary. (Mary and Stu move past each other at a speed v.) Mary passes by Stu at the exact moment that Stu's watch reads t = 0. Mary sets her watch to match, thus she passes Stu at t' = 0.
To make this discussion productive, why not start off by telling us:
- what time will Mary's watch read when she detects photon B?
- what time will Stu say that his watch reads when Mary detects photon B?
- what time will Mary's watch read when she detects photon A?
- what time will Stu say that his watch reads when Mary detects photon A?
- what time will Stu's watch read when he detects photons from A and B?
- what time will Mary say that her watch reads when Stu detects those photons?
Why do you need to compare clocks in order for Mary to determine the photons were not emitted simultaneously in the moving frame? Mar doesn'y need Stu to determine simultabeity using SR does She?
You ask the trivial questions and ignore the difficult questions. Let's ask a preliminary question first: If the photons were emitted sinultaneously in the stationary frame and the moving obvserver determines the photons were not emitted simultaneously in her frame then where was the photon that followed the first emitted photon when ithe first emitted was all by itself in the universe? You must take into account the physics of the simultaneously emitted photons in the stationary frame.
You must also be able to distinguish between mathematics that is accurately describing physical activity from mathematics that is merely convenient to prove anything you want to prove and to hell with reality.
You can use your SR theory to make the calculations. I am not going to bother as Grounded's posts, as well as my own, have already negated any "speed of light" postulates used in SR theory.
Doc Al said:
Once we agree on that, then we can talk about "simultaneity".
When you want to discuss where the lost A photon was when the other B photon was emitted "before", will be a cold day in hell won't it Doc Al? You have lost and it shows.
M(t0) is the location of the midppoint of A and B in the stationary frame.
Bad terminology. Just call it M. (Why is there a time associated with the midpoint?)
M'(t0) the locaion of the observer when the photons emitted from A and B and she was colocated at M(t0).
More bad terminology. Call the moving observer M'. M and M' were collocated at t = t0; t' = t0. (Why not call that time t = 0?)
All primes are moving frame values.
M'(t1) and M'(t3) is he observer when detecting the photons from B and A respecticvely. M'(t2) the time the photons from A and B arrive simultaneously at M(t2).
Even more bad terminology. Who is measuring those times? Your notation is obscure; think of these events:
Event #1: M' detects a photon from B
Event #2: M' detects a photon from A
Event #3: M detects photons from A and B
Each of these events happens at a different time in each frame. Tell me those times.
No simultaneity is defined at the time t he photons were emitted. However, a definition of simultaeity is that if these photons emitted at A and B arrive at M(t0) simultaneously, then they were emitted simultaneously. The notation is clean, you just want to pervert the notation to your own satisfaction and for whatever you motives happen to be in this regard.
Again, you just assume that simultaneity is the same for all observers. So far, all we can agree upon is that according to frame M, those lights were flashed at t = t0. When were they flashed according to M'? Your notation builds in a confusion about whose clocks are being used.
Why do you need to compare clocks in order for Mary to determine the photons were not emitted simultaneously in the moving frame? Mar doesn'y need Stu to determine simultabeity using SR does She?
You are the one obsessed with all those times. As I stated many many times, if you stick to Einstein's simple argument you can prove the relativity of simultaneity with no calculations whatsoever.
You ask the trivial questions and ignore the difficult questions. Let's ask a preliminary question first: If the photons were emitted sinultaneously in the stationary frame and the moving obvserver determines the photons were not emitted simultaneously in her frame then where was the photon that followed the first emitted photon when ithe first emitted was all by itself in the universe? You must take into account the physics of the simultaneously emitted photons in the stationary frame.
If those questions are so trivial, where are your answers? Your question is silly: any frame will simply say that the photon obviously didn't exist before it was emitted. It's only you who have a problem with that, because you assume that simultaneity is absolute. Reality disagrees.
When you want to discuss where the lost A photon was when the other B photon was emitted "before", will be a cold day in hell won't it Doc Al?
I have answered your question. So, when are you going to answer mine? I asked you a series of questions in this thread and in the Einstein Train Paradox thread. Still waiting for those answers.
#27
geistkiesel
538
1
Doc Al said:
Bad terminology. Just call it M. (Why is there a time associated with the midpoint?)
The moving observer is keeping track of the events in her frame.
DocAl said:
More bad terminology. Call the moving observer M'. M and M' were collocated at t = t0; t' = t0. (Why not call that time t = 0?)
This is confusing and you are a trained Ph D physicist, I m,ean a mathematician?
Two photons are emitted at A and B at t0 when M’(t0) moving with velocity v is located at M(t0) the midpoint of A and B in the stationary platform.
A photon from B is detected at M’(t1).
Photons from A and B arrive simultaneously at M(t2) = M’(t2).
A photon from A arrives at M’(t3).
Primed notation refers to moving frame measurements.
Code:
M(t0) M(t2) M(t0)
A M(t0) B
___|_______________________|______________________ ____|__
___________________________|_____________|______|_ _________
M’(t0) M'(t1) M'(t3)
M'(t2)
Here is the original. Is this confusing to you?
I want to distinguish between platofrm and moving frme times, and where the moving oibserver is at all times.
Doc Al said:
Even more bad terminology. Who is measuring those times? Your notation is obscure; think of these events:
Event #1: M' detects a photon from B
Event #2: M' detects a photon from A
Event #3: M detects photons from A and B
Each of these events happens at a different time in each frame. Tell me those times.
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.
Your saying the events are happening at different times doesn't make it so. Let us look at the moving obvserver keeping a running track of the original location of M'(t0) when the photons were emitted in the stationary frame. By keeping a runningf watch on this location she will view the arrival of the photons the same instant the stationary observer detects these photons. So even if time is dilating and the moving frame's time for the arrival of the photons at M'(t2) = M(t2), both frames can witness the simultanaeous arrival of the photons at the original midpoint so briefly visited by the moving frame. This is a trivial technological act. Having so observed the simultaneous arrival of the photons at the opriginal midpoint what is the moving observer going to do with all her SR textbooks?
Doc Al said:
Again, you just assume that simultaneity is the same for all observers. So far, all we can agree upon is that according to frame M, those lights were flashed at t = t0. When were they flashed according to M'? Your notation builds in a confusion about whose clocks are being used.
Doc Al you are some work of art. The photons were emitted at the instant the moving observer was at M(t0). or M'(t0) = M(t0). Do you see where it says all prime measurmkents are in th emoving frame?
Doc Al said:
You are the one obsessed with all those times. As I stated many many times, if you stick to Einstein's simple argument you can prove the relativity of simultaneity with no calculations whatsoever.
So make the proof. You haven't ever done this for me. Is it from the fact that the B photon was detected before the A photon in the moving frame? What else could it be?
Doc Al said:
If those questions are so trivial, where are your answers? Your question is silly: any frame will simply say that the photon obviously didn't exist before it was emitted. It's only you who have a problem with that, because you assume that simultaneity is absolute. Reality disagrees.
I don't assume simultaneity is absolute, I say it as a fact of physical law. So just saying it didn't exist answers the puzzle of the lost photon that has already been emitted in the stationary frame? Brilliant.
The photon didn't exist? DId the stationary platform exist? Was the moving observer at the midpoint of the A and B photons at the very instant they were both emitted into the universe? YES YES YES. I see SrR takes reality, of the stationary platform, and lo and behold with a couple of postualtes SR says the photon didn't exist. So you ignore the reality of the stationary platform is if the observer on the moving platform were some kind of dunce, isolated from reality?
SR a theory bullt on the foundation of arbitrary convenience.
[quotw=Doc Al]I have answered your question. So, when are you going to answer mine? I asked you a series of questions in this thread and in the Einstein Train Paradox thread. Still waiting for those answers.
[/QUOTE]
They have all been answered Doc Al. If you are missing some let me know.
Two photons are emitted at A and B at t0 when M’(t0) moving with velocity v is located at M(t0) the midpoint of A and B in the stationary platform.
A photon from B is detected at M’(t1).
Photons from A and B arrive simultaneously at M(t2) = M’(t2).
A photon from A arrives at M’(t3).
Primed notation refers to moving frame measurements.
Code:
M(t0) M(t2) M(t0)
A M(t0) B
___|_______________________|______________________ ____|__
___________________________|_____________|______|_ _________
M’(t0) M'(t1) M'(t3)
M'(t2)
Here is the original. Is this confusing to you?
Sure it's confusing. What are all those times?
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.
I think you are confusing yourself with your own notation!
Allow me to translate:
You are claiming that Event #1 happens at t' = t1.
You are claiming that Event #2 happens at t' = t3.
Note that those times are according to the moving frame clock that M' uses, right? OK, so what are those times exactly?
You are claiming that Event #3 happens at t = t2.
I assume that now you are using measurements made with the "stationary" clock that M uses, right? (Careful with those frames!) Well, what time is that exactly?
Your saying the events are happening at different times doesn't make it so. Let us look at the moving obvserver keeping a running track of the original location of M'(t0) when the photons were emitted in the stationary frame. By keeping a runningf watch on this location she will view the arrival of the photons the same instant the stationary observer detects these photons. So even if time is dilating and the moving frame's time for the arrival of the photons at M'(t2) = M(t2), both frames can witness the simultanaeous arrival of the photons at the original midpoint so briefly visited by the moving frame. This is a trivial technological act. Having so observed the simultaneous arrival of the photons at the opriginal midpoint what is the moving observer going to do with all her SR textbooks?
Again you stumble over the same ground. You confuse the simple fact that EVERYONE observes that observer M detects the photons from A and B as arriving together as saying something about WHEN they arrive according to the moving M' clock. When do you think that the M' frame would say the photons arrived at M? You aren't saying that the M' and M clocks would BOTH say that Event #3 happened at t2?... or are you?
So make the proof. You haven't ever done this for me. Is it from the fact that the B photon was detected before the A photon in the moving frame? What else could it be?
I don't assume simultaneity is absolute, I say it as a fact of physical law.
Don't just say it, prove it.
Doc Al said:
I have answered your question. So, when are you going to answer mine? I asked you a series of questions in this thread and in the Einstein Train Paradox thread. Still waiting for those answers.
geistkiesel said:
They have all been answered Doc Al. If you are missing some let me know.
You haven't answered ANY of my questions, geistkiesel. I have asked you to tell me specific times and locations, both in this thread and the other. (In the other thread I made a list of 8 questions.) Where are your answers?
#29
geistkiesel
538
1
Doc Al said:
Sure it's confusing. What are all those times?
I think you are confusing yourself with your own notation!
Allow me to translate:
You are claiming that Event #1 happens at t' = t1.
You are claiming that Event #2 happens at t' = t3.
Note that those times are according to the moving frame clock that M' uses, right? OK, so what are those times exactly?
You are claiming that Event #3 happens at t = t2.
I assume that now you are using measurements made with the "stationary" clock that M uses, right? (Careful with those frames!) Well, what time is that exactly?
Again you stumble over the same ground. You confuse the simple fact that EVERYONE observes that observer M detects the photons from A and B as arriving together as saying something about WHEN they arrive according to the moving M' clock. When do you think that the M' frame would say the photons arrived at M? You aren't saying that the M' and M clocks would BOTH say that Event #3 happened at t2?... or are 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?
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.
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?
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.
To claim now that the photons were not emitted simultaneously in the moving frame would be specious, just because the frame has moved?.
Doc Al said:
You haven't answered ANY of my questions, geistkiesel. I have asked you to tell me specific times and locations, both in this thread and the other. (In the other thread I made a list of 8 questions.) Where are your answers?
If you haven't got the answer by now what have you been loking at. You keep confusing yourself by altering my notation and diagrams, jsu chill out Doc Al, you might get some benfit from all this.
#30
geistkiesel
538
1
Doc Al said:
o make this discussion productive, why not start off by telling us:
- what time will Mary's watch read when she detects photon B?
- what time will Stu say that his watch reads when Mary detects photon B?
- what time will Mary's watch read when she detects photon A?
- what time will Stu say that his watch reads when Mary detects photon A?
- what time will Stu's watch read when he detects photons from A and B?
- what time will Mary say that her watch reads when Stu detects those photons?
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).
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" occured, 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
|
. |
|
M' # B' detector in moving frame.
A-->\ M
|
|
|
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.
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#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.
