Has ayone seen the lost simultaneity?

[geistkiesel]

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.

 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?

 

[ram2048]

neither your observer/emitter or reflectors are moving in relation to each other.

net result = same as stationary frame

 

[geistkiesel]

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.

 

[ram2048]

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

 

[ram2048]

this is sweet i learn something in another thread and get to come here and talk like i know what's going on kekeke

 

[wisp]

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.

 

[geistkiesel]

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."

 

[jcsd]

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.

 

[geistkiesel]

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?

 

[jcsd]

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.

 

[geistkiesel]

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

 

[Doc Al]

the relativity of simultaneity

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".

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.

 

[geistkiesel]

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.

 

[Doc Al]

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)

 

[geistkiesel]

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.

 

[Doc Al]

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.

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!

 

[geistkiesel]

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]

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!

 

[geistkiesel]

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

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.

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.

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?

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'.

 

[Doc Al]

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.

 

[geistkiesel]

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.

Einstein’s stationary platform moving train experiment.

1. 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.
2. A photon from B is detected at M’(t1).
3. Photons from A and B arrive simultaneously at M(t2) = M’(t2).
4. A photon from A arrives at M’(t3).
5. Primed notation refers to moving frame measurements.

  M(t0)                   M(t2)                         M(t0)
  A                         M(t0)                         B 
___|_______________________|__________________________|__
___________________________|_____________|______|__________  
                         M’(t0)        M'(t1)   M'(t3)
                         M'(t2)

Defining photons.

1. Red ants.
2. Chevrolet Camarros.
3. 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.

 

[geistkiesel]

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?

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.

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.

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..

 

[ram1024]

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.

 

[Doc Al]

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.

I can hardly wait!

Einstein’s stationary platform moving train experiment.

1. 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.
2. A photon from B is detected at M’(t1).
3. Photons from A and B arrive simultaneously at M(t2) = M’(t2).
4. A photon from A arrives at M’(t3).
5. Primed notation refers to moving frame measurements.

  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".

 

[geistkiesel]

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).

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.

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.

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.

 

[Doc Al]

a cold day for geistkiesel

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. :rofl:

 

[geistkiesel]

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.

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?

Einstein’s stationary platform moving train experiment.

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.

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?

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?

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?

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. :rofl:
[/QUOTE]

They have all been answered Doc Al. If you are missing some let me know.

 

[Doc Al]

Einstein’s stationary platform moving train experiment.

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! :rofl:

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? :rofl:

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?

Been there, done that: https://www.physicsforums.com/showpost.php?p=229410&postcount=2

I don't assume simultaneity is absolute, I say it as a fact of physical law.

Don't just say it, prove it.

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. :rofl:

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? :rofl:

 

[geistkiesel]

Sure it's confusing. What are all those times?

I think you are confusing yourself with your own notation! :rofl:

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? :rofl:

Been there, done that:
https://www.physicsforums.com/showpost.php?p=229410&postcount=2

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?.

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? :rofl:

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.

 

[geistkiesel]

o make this discussion productive, why not start off by telling us:

1. - what time will Mary's watch read when she detects photon B?
2. - what time will Stu say that his watch reads when Mary detects photon B?
3. - what time will Mary's watch read when she detects photon A?
4. - what time will Stu say that his watch reads when Mary detects photon A?
5. - what time will Stu's watch read when he detects photons from A and B?
6. - 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).

 

[Doc Al]

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.)

 

[Doc Al]

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:

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.

 

[geistkiesel]

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.

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?

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.

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.

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.

"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."

 

[Doc Al]

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? :rofl: :rofl:

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. :rofl:

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."

:zzz: Translation: "I refuse to face reality and learn from my mistakes." Is that how you want your tombstone to read?

 

[geistkiesel]

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.

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? :rofl: :rofl:

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.

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.

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?

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

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?

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.

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 .

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.

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.

The fact that your "numbers" balance means nothing. Furthermore, ALL of your "numbers" are values measured in the "stationary" M frame. :rofl:

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.

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.

: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."