Has ayone seen the lost simultaneity? [Archive]

geistkiesel

Jun14-04, 08:11 AM

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

Jun14-04, 08:28 AM

geistkiesel

Jun14-04, 10:39 PM

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 suprise 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

Jun15-04, 08:54 AM

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

Jun19-04, 07:10 AM

Doc Al

Jun19-04, 07:22 AM

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

Jun19-04, 09:52 AM

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

Jun19-04, 10:58 AM

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: http://www.physicsforums.com/showpost.php?p=231002&postcount=115)

geistkiesel

Jun21-04, 12:54 AM

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: http://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

Jun21-04, 02:28 PM

geistkiesel

Jun21-04, 08:21 PM

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

Jun22-04, 04:02 AM

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? :eek:

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

Jun22-04, 06:39 AM

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

Jun22-04, 08:00 AM

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

Jun26-04, 04:15 AM

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.

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.

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.

geistkiesel

Jun26-04, 05:10 AM

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

Jun26-04, 05:18 AM

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

Jun26-04, 05:32 AM

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! :rolleyes:
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.

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

Jun26-04, 06:48 AM

I can hardly wait! :rolleyes:

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 wont it Doc Al? You have lost and it shows.

Doc Al

Jun26-04, 07:20 AM

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? :smile: 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 wont 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

Jun26-04, 08:32 AM

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? :smile: 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

Jun26-04, 09:11 AM

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: http://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

Jun26-04, 11:18 PM

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:
http://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 wont 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

Jun26-04, 11:31 PM

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).

Doc Al

Jun27-04, 06:53 AM

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. :rolleyes:

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

Jun27-04, 07:25 AM

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 wont 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. :smile:

geistkiesel

Jun27-04, 10:33 AM

Cute. Now give me the actual times, not your label for the times. :rolleyes:

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 travelled a distance t'1c and likewise the A photon is located at -t'1v as it too has travelled 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 travelled .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 mathmatics 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." :devil:

Doc Al

Jun27-04, 03:57 PM

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 travelled a distance t'1c and likewise the A photon is located at -t'1v as it too has travelled 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 travelled .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 travelled 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 travelled 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 travelled .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." :devil:
:zzz: Translation: "I refuse to face reality and learn from my mistakes." Is that how you want your tombstone to read?

geistkiesel

Jun29-04, 07:58 AM

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 travelled 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 travelled 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 travelled .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."

geistkiesel

Jun29-04, 09:13 AM

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.

M
|

Above the stationary midpoint just as the photons from A and B arrive simulatneously.
/<_________ Stationary
|
\>_________->B' photon moving |

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.

___>\ 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.

_____>\ /_> 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?

ram1024

Jun29-04, 09:27 AM

You are just in a panicked mode and are starting to moan.
What a beautiful sound.

damn geist you're scary

Doc Al

Jun29-04, 01:57 PM

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.

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?? :rofl: Sober up, man!

Doc Al

Jun29-04, 02:16 PM

geistkiesel

Jul2-04, 05:41 AM

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”.
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.

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.

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.

3rd modified moving frame.
M'
#|#
A-->#|#<--B :simultaneously
#|#
#|#
#|#
M' O’--> 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.

Doc Al

Jul2-04, 09:15 AM

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.
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”.
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. :rofl: ) 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.

geistkiesel

Jul2-04, 12:00 PM

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?

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.

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.

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 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?

[quote=Doc Al]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. :rofl: ) 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 wont read my posts anymore because you are bored?

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.

Doc Al

Jul2-04, 12:26 PM

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.
Learn to read, geistkiesel. Here are Einstein's own words: http://www.bartleby.com/173/8.html
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?
Learn to read, geistkiesel. Here's what I wrote:
And if you put a mirror at M, then yes indeed, those photons will arrive at O' simultaneously.
I'll ignore the incoherence of your phrase "enter the moving frame", whatever that might mean.
No you haven't not even once. You've said iot a million times but have never explained it.
Learn to read, geistkiesel. Here's just one example: http://www.physicsforums.com/showpost.php?p=229410&postcount=2
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???? :rofl: 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. :rofl:

geistkiesel

Jul2-04, 01:13 PM

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?

Doc Al

Jul2-04, 01:53 PM

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?

geistkiesel

Jul2-04, 04:58 PM

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.
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 wont 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 occured 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.

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.

geistkiesel

Jul2-04, 05:16 PM

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 :devil: never :devil: see the emission of the photons, :devil: never. :devil: 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.

Doc Al

Jul2-04, 09:28 PM

(1) yes.
(2) yes.
Right! Good for you.
(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 wont 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 occured 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

Jul2-04, 09:30 PM

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".

geistkiesel

Jul3-04, 09:13 AM

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.

geistkiesel

Jul3-04, 09:19 AM

Quote:
(3) No. You must realize that only after the observers can look at the detctors can they tell when the photons arrived. This is after the event.

Wrong. Both frames detect the photons simultaneously on their own clocks. What do you mean the detection is "after the event"? The detection is the event.

The speed of light is constant. If O' is moving wrt the spot in the stationary frame where the light was emitted then it is physically impossible for the A' observer to detect the photon from the source the same instant the B' observer detects the photon., even if their clocks wee running slower thatn the stationary clocks. You can't have it both ways.

geistkiesel

Jul3-04, 09:36 AM

Right! Good for you.

Wrong. Both frames detect the photons simultaneously on their own clocks. What do you mean the detection is "after the event"? The detection is the event.

Right, but not for your reasons.

Wrong. Each observer sees the others clocks as being out of synch. (And yes I remember your silly calculation where you use time measured in the "stationary" frame, but claim you are using times measured in the moving frame. I assume you are just confused, not willfully trying to lie or deceive.)

SR theory sucks, for the following reasons. If you follow the argument in my analysis that you find so silly you will see that the moving observer will detect the photons were emitted simultaneoulsy in his frame as he will receive the information of the emitted photon simultaneously, even though he has moved to the right. The detetctors will not detetct the photons at the same time, after all the observers are moving with respect to the point the photons were emitted, but this has nothing to do with SR.

The bottom line is that both frames will detect the photons arriving simultaneously. And, since the light is emitted at the midpoint between both observers, both O and O' will conclude that the photons were emitted simultaneously in their own frames.

Of course, but the detectors in the moving frame will not detect the photons simultaneously. It is only after the infromation regarding time of emission, detetction times and velocity will the moving frame determine the photons were emitted simultaneously.

But, since we knew that the photons were all emitted at the same place and time according to all frames, this should be obvious.

Wrong. The stationary observer will have no problem, but for the moving observer it is not obvious anad he will have to go through the calculations as I did in order to see that that the photons were emitted simultaneously; or simply use mirros as per my analysis and he can see the photons were emitted simultaneously..

Don't get your hopes up. You are still stumbling over problems I would assign to high school students.
I am not stumbling over problems i am stumbling over people that believe in the silliness of SR as it was pounded into their minds.

Doc Al

Jul3-04, 10:51 AM

Hence = therefore. This is how I used the word, just like AE used the word. What is your real question here?
Right: Hence = therefore. Thus indicating a conclusion not a definition. My real question: Do you understand english?
#3. No, the A' observer detects the photon from M before the B' observer detects the photon from M. Their clocks will show different arrival times.
Incorrect. You are still viewing things from the stationary frame. Break out that box! To observers A' and B', the light flashed on exactly at the midpoint; which is also what A and B see. So clocks A' and B' read the same when the photons arrive; as do A and B.
#4. After comparing their clock times and knowing their velocity wrt the stationary frame the A' and B' observers will conclude the photons were emitted simultaneously.(See my analysis)
Now why in the world would the O' frame use the velocity of the O frame in figuring out what they can observe directly? The reason that all observers agree that the photons were emitted simultaneously is that (a) the photons were emitted from the midpoint between the two detectors and (b) the photons were detected at the same time by both detectors.
#. 5) Do the observers in ("stationary") frame O agree that A' and B' detected the photons simultaneously according to the O frame clocks?

No. The stationary observers know the A' observer is moving to the oncoming photon and they know the B' is chasing the outgoing photon.
You changed your answer from before. Good for you! (But note that this reasoning works exactly the same for the moving frame!)
(6) Do the observers in ("moving") frame O' agree that A and B detected the photons simultaneously according to the O' frame clocks?

Yes. The photons from M have the same distance to travel. The moving obsever can see this clealy. Even if the clocks were slower in the moving fame the distance the photons cover is the same, therefore the clocks will determine that the photons had the same time of flight to arrive at A and B. If the clocks do not agree then they are niot operating popely.
No. The moving observer clearly sees that A is moving away from the oncoming light, and B is moving towards it. This is the exact reasoning you used to (finally) get the correct answer to #5. Have you forgotten it so soon?

geistkiesel

Jul4-04, 01:44 AM

Right: Hence = therefore. Thus indicating a conclusion not a definition. My real question: Do you understand english?

Incorrect. You are still viewing things from the stationary frame. Break out that box! To observers A' and B', the light flashed on exactly at the midpoint; which is also what A and B see. So clocks A' and B' read the same when the photons arrive; as do A and B.

Hence means 'therefore'. AE is using the term in the context that the observers on the train see the B pulse first then the A pulse, Hence, they conclude he photons were not emitted simultaneously.And they conclude erroneously. It is easier teaching average 8th graders this stuff than old men whose minds have been corrupted by ignorant graduate advisors. Drink a glass of warm milk Doc and get a good night's rest.

No the speed of light is constant and measuring from different frames does not impose restrictions. A' and B' are moving wrt the pulse source and hence as the pulse source does not move., the A' and B' detectors will record different times. therfore, [HENCE] A' and B' read the photons at different times.

Now why in the world would the O' frame use the velocity of the O frame in figuring out what they can observe directly? The reason that all observers agree that the photons were emitted simultaneously is that (a) the photons were emitted from the midpoint between the two detectors and (b) the photons were detected at the same time by both detectors.

Only the stationary detectors measure simulaneous signals fronm the source.

On;y the moving observer can measure the simultaneous arrival of the pulses and then only if the A' and B' detectors are replaced by mirrors.

Look at the physics of the problem and put your SR crap to sleep. A stationary and moving frame are going to measure different times of emission using absolute time as a reference. And there aren't "both" detectors, there are four detectors, two moving, two stationary. Let uis keep it that way OK?.

You changed your answer from before. Good for you! (But note that this reasoning works exactly the same for the moving frame!)

No you are wrong. I said the moving detectors will not simultaneoulsy detect the emitted photons and the stationary obserbver will also not see simultaneousl detection of the photons by A' and B'. The moviing OBSERVER will detect the photons ariving at the shifted position of O', if the detectors are replaced by mirrors. The reasonong is not the same using the moving frame as a staionary frame, because the moving frame is moving, it is not stationary. Do you undertstand?

No. The moving observer clearly sees that A is moving away from the oncoming light, and B is moving towards it. This is the exact reasoning you used to (finally) get the correct answer to #5. Have you forgotten it so soon?
Doc it is you gedunken stick with it. O is stationary thrpugh out, O' is moving through out.

See, here is where you and SR are full of it. The moving observer O'cannot "see" A moving at all. There is the speed of light restriction. All that the moving observer O' can "see" is the first return of the signals, either at the shifted origin, if the detectors are replaced by mirrors, or after she has looked at the timing of the detectors at A and A' and B and B', which is post experiment time.

Remember Doc Al, this is your gedunken. You said that the O' was the MOVING observer, now you want to play the old "reversee who is moving" game. Stick to what you described and answer the question, if you are able. Are you intentionally doing this (I cannot believe you are other than misguided, which is why I have exercised an infnite degree of patience with you) or is it just a carry over of what you learned in graduate school?

You are putting the moving observer in the stationary frame and doing some kind of logical flim flam dance. Get back in your proper frame Doc Al bwefore you fall off.

I am responding with another post here.

geistkiesel

Jul4-04, 04:04 AM

The Basic System
There are two photon sources at A and B with the midpoint marked at M. Photons are emitted in a pulsed mode, | | | | | | |, where each pulse, |, is coded such that the A and B pulses can be distinguished. This can be accomplished using coded MW signals with pulse width of | ~ λ, the wavelength of a light pulse emitted at A and B.

|A->_____________________M________________________<-B|

Extended System
In the middle of the sources is a mirror system that reflects the A and B photons back to their respective source position as well as pairing each of the reflected photons with a photon from the opposite source. On the first line A and B pulses are deflected into mirrors, on the second line, that reverses the directions of the pulses. On the third and fourth lines the reversed pulses are paired with pulses of the opposite type: B paired with A and A paired with B.

At this point A and B photons are moving in both directions with the same left-right spatial locations.

Organization of Pulses

|_____________________->\/<-___________________________|
A <-____<-/\->____-> B
B <-____<-/
\->____-> A

Final System at Equilibrium
Located at A and B is a system of flat reflecting mirrors that reflect the AB photon pairs back to the opposite set of mirrors. After a complete reflection cycle is completed equilibrium is established.

The vertical lines that stretch from A to B along the bottom can selectively measure AB photon pairs moving in either direction. Therefore, any position along the AB axis will measure AB photon pair pulses simultaneously. If we install an electronic system to enable any one of the detectors randomly or sequentially, AB photon pairs will always be measured simultaneously. If we program the detectors to be triggered sequentially we may simulate motion along the AB axis, or we can insert a detector to actually move along the AB axis.

Now the question: Is a detector moving in the AB axis going to measure the same as a static detector? This question must be answered in the context of the experiment described. Using SR theory to invent an answer will be circuitous logic and therefore useless.

|-----------------------------------<- A--------------------------------|
|-----------------------------------<- B--------------------------------|
|----------------> B ---------------------------------------------------|
|----------------> A ---------------------------------------------------|
|||||||||||||||||||||||||||||||||||||||||||||||||| |||||||||||||||||||||||

Definition of Simultaneity
Pairs of reflected photons are coded as pairs.
AE's definition that :smile: “Events that are simultaneous in stationary frames are not simultaneous in moving frames.” :smile: means must a conflict between definition and experiment exists. Some might consider the 'definition' as a 'conclusion'. The statement is a definition as offered by AE ,and we can debate the point later, however ,it must still be proved, one way or the other, before it is universally adopted. We must be very careful not to implement the definition iin a way that results in circuitous reasoning.

Those implications derived from the definition have resulted in discarding the concept of “absolute time” and as such are fatally flawed. If we use the definition above then we discard absolute time, insert time dilation, shrink matter in the direction of motion and measure the speed of light always as c = 300,000km/sec without regard to the motion of the ftrame from which the measurement is taken. and all of this to conform to the definition, however phrased. [Notice there is no attempt to measure the speed of light with respect to the motion of the frame of reference in practice.].

Definitional flaw in Simultaneity
There is an obvious error in the implementation of the definition of simultaneity which originated with A. Einstein in his widely discussed gedunken experiment using a train as a moving frame, with one observation point on that train passing through the midpoint M of the A and B sources just as photons are emitted from A and B. It is only the first emitted photon pair that are considered in AE's analysis of the definition

As defined any system of moving entities will fall uner the deinitional umbrella of simultaneity.
Before we continue let us look at the condition where A and B are the starting point for NASCAR vehicles leaving A and B at 300km/hr as substitutes for the photons. For the train we substitute a single NASCAR detector moving on a straight track at 100km/hr in the B direction. The AB distance is 800km.The NASCAR detector will detect the ‘B’ NASCAR first, followed by the ‘A’ NASCAR. If we now look at the detector we see the B object detected before the A object. What conclusions may we draw from this data alone? Nothing, other than B was detected before A. May we conclude that the moving detector has proved the definition of simultaneity? No.

Simultaneity is functionally related to information of the time an observer passes through M, the velocity of the train wrt stationary frame, the AB sgtationary distance and the speed of light wrt stationary frames.
:rofl: With only the fact that a train observers in AE's gedunken only see the different arrival times, no conlcusions regarding simultaneity are reasonable. If the observers on the train consider their velocity wrt the stationary frame they are still unable to conclude when the photons were emitted. If they consider that they were at M when the photons were emitted they are still unable to determine simultaneity. [the photons could have been emitted any time before they arrived at A and/or B]. If the observers then add the consideration of the distance between A and B they must first make the calculations to determine if the photons were emitted simultaneously. In a previous post I derived that t2 = t1(C + V)/(C - V) where t1 is the time of the B measurement wrt t = 0 when the observers were at M, and t2 the time of the A measurement wrt t = 0 at M. Using the AE definition, each observer strung along the train must make her own calculations to determine her own state of observing simultaneity. Each will determinme the photons were emitted simultaneously into the moving frame of he train. Imposing time dilations is premature here. The observers who are at the midpoint M when the photons arrive their simultaneously need not make any calculaiopns, they will be able to see the photons arrive at the same place they are located. :rofl:

There can be only one law for all the observers on one train, right?

The defnition is ambiguous regarding conditions of information.
If we know ahead of time that the detector passed through M just as each A and B NASCAR left A and B may we then approve of the definition of simultaneity? No, obviously not. If two objects are headed toward each other and a third object is behind the middle object and is trying to catch up (under the conditions established here) the objects moving in a collision course will always meet before the trailing object catches up. Think about it, any objects will suffice :earth worms, olympic sprinters, even photons confined to the situation described above. Are we, then going to assign AE’s definition to the NASCAR/detector system in order to bring the definition of simultaneity into a universally operating reality, for all objects? No.

Simultaneity not universally defined.
Simultaneity, as discussed in the context of the definition refers specifically to electromagnetic motion, or radiation. In this sense, the definition is used to justify the time dilation, mass shrinking and the implication from these concepts that the measure of the speed of light as constant with respect to all inertial frames. In other words, once one accepts the definition the concepts mentioned must be implmented in order to maintain consistency with the definition.

:yuck: Circuitous analysis substututed for physical reality. :yuck:
One should look very closely at what is being trashed in order to bring about a compromise that justifies the mental limitations of the scientific industry.

:cry: We must be very careful that definitions do not bootstrap coincidental observations imposed by the defintion. To infer preconceived ideas about concepts of natural phenomena and surrepticously constructing defintions is always fatally hazardous. :cry:

The Speed of Light is C in vacuo :smile:
So light moves at a constant velocity of c in vacuo. Is this any reason, by itself, to conclude that we must measure the relative velocity of light as c with respect to any frame of reference? Not including the velocity of the referenced frame when measuring c is equated to voluntarily placing the observer firmly on the horn of a dilemma. :yuck: This omission will always result in an erroneous conclusion that time dilates and mass shrinks., which results in, voila! the speed of light measured as c = 300,000km/sec [with dialted time and mass, of course].

Are we thinkers ot robots? :approve:
What about the implications of Galilean vs. Lorentzian transformations and the implications of Maxwell’s equations? Just crank out some maths? What about the relative motion of 8km/sec between light and the aether found by Dayton Miller [who confirmed the Michelson-Morely's results]? Those confining themselves to the rote meanings of concepts as they were taught, without an in depth and critical analysis, simply get back on their horn and proceed along in the pain of collective and singular confusion :yuck: .

Doc Al

Jul4-04, 08:37 AM

Doc it is you gedunken stick with it. O is stationary thrpugh out, O' is moving through out.

You are still stuck thinking that "stationary" somehow means something absolute. It's just an arbitrary label. I never said what O was stationary with respect to. It just so happens that O is stationary with respect to a spaceship that is traveling alongside it. And, what do you know, O' is moving with respect to that ship... but it's stationary with respect to the Earth! (Which is perhaps a billion miles away.)

Who knew? Time to change all your goofy arguments around, geistkiesel. I guess O' was really stationary all the time! :rofl:

Doc Al

Jul4-04, 08:58 AM

geistkiesel

Jul4-04, 09:17 AM

You are still stuck thinking that "stationary" somehow means something absolute. It's just an arbitrary label. I never said what O was stationary with respect to. It just so happens that O is stationary with respect to a spaceship that is traveling alongside it. And, what do you know, O' is moving with respect to that ship... but it's stationary with respect to the Earth! (Which is perhaps a billion miles away.)

Who knew? Time to change all your goofy arguments around, geistkiesel. I guess O' was really stationary all the time! :rofl:

Oh I get it. If you are an SR theorist, you can make it up as you go along. So a stationary train depot is not staionary anymore? Oh I get it arbitrary labels allowed in SR theory, but rational analyiss is verboten? You have just introduced the silliest things you have ever posted in your career. talk about out in the clouds.

SR error that is fatal, every time: Assuming stationary and moving platforms then swapping them in the middlde of analysis to sound like you know what you are talking about. :rofl:
Not only do you make it up arbitrarily as you go along, you are in lock step wuith all the other SR robots: you make up impossibke situations and conditions. I admit my latest post has some engineering problems to ovecome, but theoretically the gedunken is "rational" Doc Al, a word you should look up in the dictionary.

geistkiesel

Jul4-04, 09:31 AM

As I have explained at length in an earlier post, introducing a mirror at the midpoint changes the original gedanken and makes it trivial:
- Everyone (O and O') agrees that the photons reach the mirror at the same time
- Everyone (O and O') agrees that the "paired photons" left the midpoint at the same time
So what?

Once again I request that you go back to the original Einstein gedanken and answer my questions. There is no escape, geistkiesel! :smile:

Excuse me I got the order of your posts reversed. You are correct the gedunken is trivial, as trivial as Einstein's is in error. I have answered your questions many times over. It is your attitude that needs answering.

- Everyone (O and O') agrees that the photons reach the mirror at the same time
- Everyone (O and O') agrees that the "paired photons" left the midpoint at the same time[/INDENT]
So what?

So what? You didn't complete the logically next third sentence. Let me assist you in this trivial matter:

Geistkiesl is finishing Doc Al's post for him. Apparently Doc fell asleep at the wheel before the simultaneity wall appeared in front of ---. :rofl:

Everyone (O and O') agrees that the "paired photons" were emitted simultaneously (and detected as such) in the moving frame at the same time.

You are very welcome Doc. Can I call you Doc? :smile:
There, Doc Al, let me be the first to welcome you back to a rational world. :smile:

Janus

Jul4-04, 10:02 AM

Oh I get it. If you are an SR theorist, you can make it up as you go along. So a stationary train depot is not staionary anymore? Oh I get it arbitrary labels allowed in SR theory, but rational analyiss is verboten? You have just introduced the silliest things you have ever posted in your career. talk about out in the clouds.

The idea that there is a state of absolute rest died over 350 yrs ago. Since then, "stationary" has always been an arbitrary label; That which is considered "stationary" is considered so for convenience only. This is not a idea new to SR.

Until you come to grips with this you are no better off than the members of the Flat Earth Society, who refuse to come to grips with the concept of a spherical world.

geistkiesel

Jul4-04, 10:51 AM

The idea that there is a state of absolute rest died over 350 yrs ago. Since then, "stationary" has always been an arbitrary label; That which is considered "stationary" is considered so for convenience only. This is not a idea new to SR.

Until you come to grips with this you are no better off than the members of the Flat Earth Society, who refuse to come to grips with the concept of a spherical world.
The "idea" of absolute rest dying ove 350 years ago is a non sequitor, even if universally true, which it isn't. That idea held by some may be true in their own abstract world, but only there can the idea be scrutinized.

So stationary is a "convenience only". Absolute rest is merely a mathematical convenience, I think you mean, don't you?

Railway stations are not stationary, they are what, conveniently at rest?

This is not an idea new to SR? So what, and who said it was new to SR? And what is the difference of the newness of the idea? Ptolemy theorists held the concept of an earth centered universe for almost 2000 years, but that old idea died didn't it Janus? You think that just because you are the modern holder of a silly theory that you couldn't be possibly in error, could you?

Have you ever seen a railway station accelerate wrt a moving train? Or even a stationary train? When a relative velocity is measured between train and train station, which people felt the acceleration that produced the motion? The ones on the stationary platform or the people on the moving train? What you term is "for convenience only" you offer as a peversion of physics by assuming your mathematical garbage that trains are stationary and that it is the train station that moves. But then you wanted to be a physicist and no way are you going to make that grade even thinking about challenging your graduate advicor on SR theory, right?

This frame swapping you SRists are so perverted fond of are physical impossibilities and you know it. But then if you really knew it, you wouldn't be able to honestly use the convenience as a substitute for physical law and reality.

Make some earth frame mesurements and detect roundness. Sure add a ton of energy, get in orbit, take a photograph, decelerate and show me your photograph and say, "See, the earth is round." I then show you a zillion earth frame based (low energy) experimental results using '"flat laser beams" that are unable to show roundness. And everybody points their finger at you and laughs, or they just point their finger at you.

Bottom line then from the frame of Janus: The concept of SR simultaneity implications are based on arbitrary and unreal physical concepts, which are substituted for by abstract and mathematically silly coveniences.

geistkiesel

Jul4-04, 10:59 AM

:
#. 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 an on coming photon and they know the B' is chasing the outgoing photon. They know it when they get the information.

You changed your answer from before. Good for you! (But note that this reasoning works exactly the same for the moving frame!)
No I didn't change my answer, I said earlier that the moving observer would see the photons arriving at his shifted position simultaneously. Yopu are perverting your position as a reviewer and mentor. There you go swapping frames again.

geistkiesel

Jul4-04, 11:10 AM

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.
In the first thread you and I discussed this matter, where the moving observer and the emitted photons from A and B arrived at the midpoint M simultaneously, you argued, and to this day I presume, that the photons were not emitted simultaneously in the moving frame. Some passengers on the train see the photons such that they "must, therefore" conclude the photons were not emitted simultaneously in the moving frame, all except those sitting adjacent to the photons arriving simultaneously at M and their position, right Doc Al? Well ,I guess that simultaneity is functionally dependent on where one is located in the moving frame, right?
The old peception of the :rofl: eye :rofl: in the beholder, right?

geistkiesel

Jul4-04, 11:21 AM

Who knew? Time to change all your goofy arguments around, geistkiesel. I guess O' was really stationary all the time.

Your physics is that you "guess" the moving observer O' was really stationary all the time?

I thought this was a physics forum, then why are so many of you acting out some silly nonsense "guessing charades?" Fie on you Doc Al, for shame.

geistkiesel

Jul4-04, 11:27 AM

Doc Al

Jul4-04, 04:09 PM

Why don't you just admit it? SR is a mathematical contrivance allowing SRists to pervert physcial reality by implementing the perversion of frame swapping which is physically impossible to do, right?
Discussing SR with you is rather pointless, since you haven't even caught on to Galilean relativity yet. Let us know when you reach the 20th century.

geistkiesel

Jul4-04, 04:54 PM

Discussing SR with you is rather pointless, since you haven't even caught on to Galilean relativity yet. Let us know when you reach the 20th century.
Doc are you saying that you have run out of specifics to attack? Why not just focus on the defintion (ok conclusion) of simultaneity. Changing subject matter is lequivalent to frame swapping, isn't it?

Let us take a non SR example. Two NASCARS with velocity of 300km/hr pass are emitted from A and B seaprated by 800meters just as the observer in 1979 Ford Futura passes the midpoint of A and B The B NASCAR is detected first 100 meters to the right of M and later the A NASCAR catches the Furtura at 200 meters past M.

Everybody in the track can see the NASCARS emitted into the moving frame as the NASCARS leave A and B, including the observer in the Futura. Am I going too fast for you? Just kidding. :smile:

Now Einstein didn't bring uniqeness of photons into the discussion as a parameter of his defintion (conclusion) that events simultaneous in stationary frames are not simultaneous in moving frames.So how do you rationalize the NASCAR equivalence of Einstein's model? :smile:

Or, on the passenger train, there are four seats that are adjacent to the Midpoint M just as the A and B photons arrive and are immediately detected in the moving frames by the exceptions to simulaneity that are sitting in those four exceptional seats. These four are the only ones on the train that the photons are emitted simultaneously to aren't they? What a perception, Ahh es, quite a perception of those four exceptional observers, right? :smile:

You sound like an angy man Doc Al, chill man, all is not lost, even if you are a loser in this instance. :smile:

How about :uhh: a truce? :smile:

How does one create attachments that are posted as URLs?

Doc Al

Jul4-04, 06:30 PM

Doc are you saying that you have run out of specifics to attack?
Exactly! When you come up with something new, I'll be happy to destroy it as I have all your previous "arguments". It is getting tedious to keep repeating myself.
Why not just focus on the defintion (ok conclusion) of simultaneity. Changing subject matter is lequivalent to frame swapping, isn't it?
There are two meanings to "frame swapping": one good and necessary, the other bad and silly. The good kind is to realize that physics should work the same from any frame. So any frame is as good as any other to describe the world.

The bad kind is mixing up measurements made in one frame and confusing them with measurement made in a different frame. Now... which kind of "frame swapping" do you keep doing?
Let us take a non SR example. Two NASCARS with velocity of 300km/hr pass are emitted from A and B seaprated by 800meters just as the observer in 1979 Ford Futura passes the midpoint of A and B The B NASCAR is detected first 100 meters to the right of M and later the A NASCAR catches the Furtura at 200 meters past M.
Silly rabbit! You think SR doesn't apply to NASCARS? There is no escape!
You sound like an angy man Doc Al, chill man, all is not lost, even if you are a loser in this instance. :smile:

How about :uhh: a truce? :smile:
Me angry? I'm as calm as can be. :smile: A truce? Have you run out of material already?
How does one create attachments that are posted as URLs?
I don't understand the question. You've posted URLs before.

geistkiesel

Jul5-04, 07:54 AM

Exactly! When you come up with something new, I'll be happy to destroy it as I have all your previous "arguments". It is getting tedious to keep repeating myself.

Silly rabbit! You think SR doesn't apply to NASCARS? There is no escape!

I don't understand the question. You've posted URLs before.

I am under thje belief that SR is a photon phenomnon. If we can see the NASCARS entering the moving frame and see them entering simultaneously then what law of physics says we aren't allowed to add velocities, or subtract for that matter with regard to NASCARS?..

I understand simultaneity, I just happen not to agree with it.

Regarding attachments it is what I am asking about. How do I make a figure and present the figure as a physicsforum link? or copy some figure from a website and present as an attachment?
Thanx

Doc Al

Jul5-04, 08:06 AM

I am under thje belief that SR is a photon phenomnon.
Add that to your pile of mistaken beliefs.
If we can see the NASCARS entering the moving frame and see them entering simultaneously then what law of physics says we aren't allowed to add velocities, or subtract for that matter with regard to NASCARS?..
Not sure what you are talking about. For low speeds you can certainly use Galilean addition of velocities. But realize it's only an approximation. (An incredibly good one, for NASCAR speeds!)

Also: You often talk about things "entering the moving frame". Bad habit. Things just are. They can be viewed from many different frames at once. They don't belong to one frame or another, but to all frames.
I understand simultaneity, I just happen not to agree with it.
I know you don't agree with it, but I'm still awaiting proof that you understand it.
Regarding attachments it is what I am asking about. How do I make a figure and present the figure as a physicsforum link? or copy some figure from a website and present as an attachment?
When you make a post or reply to one you'll see a tool called "attach files". Just put your diagram in one of the acceptable file types and go for it. Let me know if it doesn't work for you.

geistkiesel

Jul7-04, 04:35 PM

Add that to your pile of mistaken beliefs.

Not sure what you are talking about. For low speeds you can certainly use Galilean addition of velocities. But realize it's only an approximation. (An incredibly good one, for NASCAR speeds!)

[quote=Doc Al]Also: You often talk about things "entering the moving frame". Bad habit. Things just are. They can be viewed from many different frames at once. They don't belong to one frame or another, but to all frames.

We have been talking about photons emitted simukltabeously in the stationary frame being simultaneously emitted in the moving frame. What are you talking about. I think you have some bad habit neediong correciton here not me.

I know you don't agree with it, but I'm still awaiting proof that you understand it.

What do you have that makes you feel that I owe you some proof that i undertsand SR? I undertstand it and you know I do,. iIjust believe it is garbage, that's all.

When you make a post or reply to one you'll see a tool called "attach files". Just put your diagram in one of the acceptable file types and go for it. Let me know if it doesn't work for you.

Thanx for the "attachment" info.

Doc Al you and I both know that the gedunken we have been working on here does not use any SR postulates requiring time dilation, mass shrinking or any of the other SR inferences. Einstein's gedunken and the example given here is all he used. Einstein will have us believe because the oncoming photon was measured before the one approaching from the rear that this is suffciient to discard the simultaneity of events and to discard absolte time. read the reference you gave me, This is the same book I have been quoting from which you ridicule with your school yard jimmer jammer. Don't be dishonest DOc Al, There is only one life you have to live, don't let it be a lie that is as grossly uttered as SR.

geistkiesel

Jul8-04, 12:52 PM

Einstein gedunken experiment has been much misinterpreted in this thread. This discussion is intended to clear up any misconceptions.

For those following this thread you can see the intensity of those defending the concept of simultaneity, for if simultaneity goes, so goes SR, out the window. Doc Al has made a valiant but fruitless effort to insert time dilation and mass shrinking in to this discussion of AE’s gedunken, primarily by Doc Al.

Read and be your own judge.

_-> M <-_.
A|__________________|____________________|B.
_a__________________O’___________________b__.

This is where we start. The moving observer O’ is at M the midpoint of A and B photon sources when photons are emitted. Passengers designated a and b detect these photons when first emitted. The train extends beyond the two A and B source locations.

As the train moves to the right, second line, there are four significant events to consider.

1. The photons are emitted simultaneously in the embankment observed by passengers a and b sitting adjacent to the A and B sources when the photons were emitted- call this time1
2 The observer O’ detects the on coming B photon - time 2.
3 The photons A and B arrive simultaneously at M in the stationary frame and are observed simultaneously by passengers sitting in those adjacent seats in the moving frame. We call these passengers a|b - time 3. These are not the same a and b passengers observing the original omission of the photons.
4 O’ observes the photon A from the rear – time 4.

Doc Al will scream that I haven’t said which frame we are counting time. I am using the same time frame Einstein used in his gedunken.

By this time all passengers sitting to the rear of O’ have detected the A and B photon, including the passengers sitting adjacent from M when the photons arrived at M in he stationary frame and observed simultaneoulsy in the moving frame by a|b pasengers simultaneously. simultaneously. simultaneously.

a. In describing this experiment Einstein did not invoke any aspects of special relativity certainly not time dilation or mass shrinking in the set-up of the experiment, or the conduct of the experiment.
b. AE drew all of his conclusions from this experiment based on the sequential arrival of the A and B photons at O’.
c. AE states that “Now in reality (considered with reference to the railway embankment) he [O’] is hastening toward the beam of light from coming from B, whilst he is riding on ahead of the light emitted from A. Hence the observer [O’] will see the beam from of light emitted from B earlier than he will that emitted from A.
d. When AE stated that the, ”Observers who take the railway train as their reference-body must, therefore come to the conclusion that the lightning flash B took place before lightning flash A. We thus arrive at the important result: Events, which are simultaneous with reference to the embankment, are not simultaneous with respect to the train, and vice versa (relativity of simultaneity). Every reference body (coordinate system) has its own particular time; unless we are told the reference body to which the statement of time refers, there is no meaning in a statement if the time of an event.” All of this in quotes follows from the sequential arrival and detection of the photons by O’ on the train, or so claims AE.
e. AE imposes the conclusions in B to all passengers on the train including those passengers that were located at the midpoint of the embankment when the photons arrived simultaneously there. Apparently AE has some special conditions associated with the mysterious O’ observer that do not attach to the other passengers on the train. Again, AE makes his conclusion based purely in the sequential detection of the A and B photons by O’. From this AE makes the conclusions stated in D above. This is essential and for those truly interested in following the discussion you should clarify the points to you satisfaction. There should not be any remaining ambiguity on the correct interpretation of the gendunken.

Now ask yourself: if there is one exception to a definition then does the definition fail?
In the world of physics and mathematics the answer is yes, the definition fails, or the conclusion is erroneous.

What Einstein actually found was that in the emission of photons into a moving frame at least one location on that moving frame will locate the simultaneous arrival of the emitted photons. What is simultaneous in a stationary platform is simultaneous in the moving platform.

What if the moving platform is only a grain of sand and is smaller than a fame extending to include the midpoint of the emitted photons in the stationary frame? One can calculate the spatial midpoint positions. It is not necessary to have a moving frame of the size and velocity that will always include the sources and midpoint s of the photons emitted.From time 4, or t4, we can calculate the correctness of whether the photons were emitted simultaneously in the moving frame as

T4 = T2(C + v)/(C – v)

where T4 is the time the A photon was detected and T2, the time the B photon was detected. As T2 and T4 are both measured, or detected values, the accuracy of the expression can be tested – if the calculation agrees with observation, the photons were emitted simultaneously. This would be a very simple experiment to conduct.

Further, the original observers of the emitted photons a and b, may compare their respective clocks later and determine the photons were emitted simultaneously in the moving frame and so on down the line of passengers from a ->, to the right as the A photons go past, and from <-b to the left as the B photons go past to the left.

Doc Al

Jul8-04, 01:43 PM

Einstein gedunken experiment has been much misinterpreted in this thread.
By you.
This discussion is intended to clear up any misconceptions.
I assume you meant to say: "In this post I intend to repeat my misconceptions yet again."
For those following this thread you can see the intensity of those defending the concept of simultaneity, for if simultaneity goes, so goes SR, out the window. Doc Al has made a valiant but fruitless effort to insert time dilation and mass shrinking in to this discussion of AE’s gedunken, primarily by Doc Al.
If you understood Einstein's simple argument with the train gedanken, you would know that nowhere does "time dilation" or "mass shrinking" :yuck: enter into it.
Read and be your own judge.

_-> M <-_.
A|__________________|____________________|B.
_a__________________O’___________________b__.

This is where we start. The moving observer O’ is at M the midpoint of A and B photon sources when photons are emitted. Passengers designated a and b detect these photons when first emitted. The train extends beyond the two A and B source locations.
Your diagram implies that there are observers on the train (you call them a and b) who see those flashes happen at the same time. :rofl:
As the train moves to the right, second line, there are four significant events to consider.

1. The photons are emitted simultaneously in the embankment observed by passengers a and b sitting adjacent to the A and B sources when the photons were emitted- call this time1
By claiming that train occupants a and b observe the flashes at a single time, you merely assume what you intend to prove. You may as well stop right here.

geistkiesel

Jul8-04, 04:14 PM

I have not read your replies. I just came up with another proof that the moving observer must conclude the photons were emitted in the moving frame simultaneously with the emitted photons in the stationary frame. All clocks on the moving frame are synchronized within the moving frame. The clocks in the vicinity of A and the clocks in the vicinity of B are showing the same time throughout the train at all times. When the photons are emitted at A and B they are then detected by a and b, which are clocks nearest to A and B in the moving frame. These clocks times are immediately relayed to O'. As these times are now electromagnetic radiation they will travel exactly as fast as the photons they just timed. Therefore the B photon arrival at O' is joined by the time the b clock in the moving frame detected the photon emission into the moving frame. Later the photon from A is also joined with the time stamp of the a clock, giving the time the A photon was emitted at A. O' then compares the a and b clocks and gets the same number in the moving frame clock system. O' must, therefore, come to the conlcusion the photons were emitted simultaneously in the moving frame. The timing information is also joined by the times the a|b observers detected the A and B photon arriving simultaneously at M.

That tap-tap-tap you hear in the background is just another SR Theory coffin nail being set, it's no big deal.

Alkatran

Jul8-04, 04:17 PM

If passengers A and B are on the train, the photons were emitted simultaneously in that frame and will meet at O' (who won't be in the center at the time).

In the stationary frame the left photon will be emitted first.

geistkiesel

Jul8-04, 05:20 PM

Originally Posted by geistkiesel
Einstein gedunken experiment has been much misinterpreted in this thread.
By you.Quote:
This discussion is intended to clear up any misconceptions.

quote; Doc Al. I assume you meant to say: "In this post I intend to repeat my misconceptions yet again.

Quote geistkiesel:
For those following this thread you can see the intensity of those defending the concept of simultaneity, for if simultaneity goes, so goes SR, out the window. Doc Al has made a valiant but fruitless effort to insert time dilation and mass shrinking in to this discussion of AE’s gedunken, [insidiously inserted by Doc Al.]

If you understood Einstein's simple argument with the train gedanken, you would know that nowhere does "time dilation" or "mass shrinking" enter into it

Why did you lie about this Doc?. You have been perverting my efforts from the get go. You haven't a clue what "mentor" means you creep.

Quote:Geistkiesel
Read and be your own judge.
Code:
_-> M <-_.
A|__________________|____________________|B.
_a__________________O’___________________b__.

Geistkiesel says:
This is where we start. The moving observer O’ is at M the midpoint of A and B photon sources when photons are emitted. Passengers designated a and b detect these photons when first emitted. The train extends beyond the two A and B source locations.

The a and b passenges seen at A and b detect the photons when they were emitted. a knows nothing of b and vice versa, except that they are cousins. They only determine the train clock time when the photons were emitted and these a and b passengers detected the photons.

Your diagram implies that there are observers on the train (you call them a and b) who see those flashes happen at the same time.

Not necessarily, because they are on opposite ends of the train. (See my proof in the following post). I merely say they record the time the photon was emitted at their respective stations. The gedunken tells us that the photons were emitted simultaneously in the stationary frame. Whether the a and b passenges are aware of this information is not given here. But if the photons were emitted simultanously in the stationary frame what conclusion do you come to when you see one of the photons emitted and you know your partner on the other end of the train is probably measuring the same thing you are measuring right? Right? Seems reasonable doesn't it?

In any event any errors are going to be swamped by the delta times between the B and A photon detetction by O'.

Quote:Geistkiesel
As the train moves to the right, second line, there are four significant events to consider.

1. The photons are emitted simultaneously in the embankment observed by passengers a and b sitting adjacent to the A and B sources when the photons were emitted- call this time1

By claiming that train occupants a and b observe the flashes at a single time, you merely assume what you intend to prove. You may as well stop right here.

Good catch. I meant only that 'a' detected the A photon when omitted. 'b' detetcted the B photon when emitted. By my adding the photons were emitted simultaneoulsy in the stationary frame merely repeats the experimental given. No conclusions are made at this time, but I sure would like to bet a lot of money on the outocme of any experiment.

By you.

I assume you meant to say: "In this post I intend to repeat my misconceptions yet again."

If you understood Einstein's simple argument with the train gedanken, you would know that nowhere does "time dilation" or "mass shrinking" :yuck: enter into it.

No Doc Al I meant in my post that you imposed these constrictions by making all your silly time is different to the moving and stationary observers. demands. You have been doing this consistently and you know it. You have been conniving, not too successfully, to interject confusion and smog into a seriously offered point of view, whether you agree with it or not.

So whatever you say,

Frankly, my Doc Al, I don't give a damn.

You just made a calculated dishonest move Doc Al. Tsk, tsk. Where are the scientific rules of integrity? Have you ever heard of them? i bet the rules are strictly relative, aren't they?

Your diagram implies that there are observers on the train (you call them a and b) who see those flashes happen at the same time. :rofl:

I say and after correcting myself above, that the a and b passengers detetcted the photons when emitted at A and B,. In other words their clocks recorded the time and then I add here, relayed the information immediately to O'.

:rofl: I also said that two different passengers a|b were adjacent to the A and B photons when the A and B photons arrived simultaneously at M the midpoint of the A and B photon sources. You would object if I claimed the a|b passengers (the train is 8 seats wide and the train is packed) detected the photons simultaneously with their arrival at M? :rofl: OK whatever you call the simultaneous arival of the A and B photon at the midpoint M when the a|b passengers, 8 of them, time stamped their observed arrival of the photons at M, the midpoint of A and B in the stationary frame. :rofl:

What did you think of the form and structure of the post Doc Al? Ease of reading? Information content? Persuasivenss of arguments?

Doc Al

Jul8-04, 06:23 PM

I just came up with another proof that the moving observer must conclude the photons were emitted in the moving frame simultaneously with the emitted photons in the stationary frame.
Yeah, "another" one.

Let's cut through the nonsense, once again. Let's say a and b are the observers on the train located right next to the flashing lights at A and B when they flash. When the see the lights flash, they check the time. No need for any "relaying" of clock times anywhere. Assuming, like you did, that all clocks on the train are synchronized then--like it or not--observers a and b will record different times for the two photon emissions.

Doc Al

Jul8-04, 06:31 PM

What did you think of the form and structure of the post Doc Al? Ease of reading? Information content? Persuasivenss of arguments?
Just the usual crap. You've said the same thing--including the obnoxious accusations of my "lying"--many, many times.

geistkiesel

Jul8-04, 08:01 PM

Just the usual crap. You've said the same thing--including the obnoxious accusations of my "lying"--many, many times.

OK maybe one, or both of us had a mental lapse. I went out of my way to avoid using SR imperatives when makingh my calculations. I do remember you often criticzing me for using stationary times with moving frame times (which I denied doing).You use of the :rofl: was proficient.

If I err then I publically apologize. If it was a misinterpretation of what you were saying I was doing vs what I perceived you doing I will leave it at that. At least, at this time there is no ambiguity is there? I am now stating unequivacally that I reject the necessity of the use of any SR aspects or imperatives in determining the conclusions of AE's gedunken. If this isn't good enough to salve your obvious anger at the accusations then so be it. I can do no more.

Ok what usual crap. Spell it out. Everybody wants to know especially myself.

Did I make any improper claims of simultaneiity of events? I did make corrections, or clairifications, are you aware of these?

Is it not feasible that I can conclude that a moving frame has at least one point associated with a midpoint of photons emitted simultaneously from a stationary frame? Even if the midpoint falls off the frame, or cannot be determined from the information available.

You said the NASCAR example fell under the umbrella of SR. The low velocity makes it difficult to measure, but we are working theory here. What distinguihes NASCAR and the AE gedunken?

Do you agree that the difference in time that the O' observer measuresd the arrival of the B and A photon is the basis of AE concluding that the passengers "must , therefore conclude" the photons were not emitted simultaneously in the moving frame?

Did not the determination of the a|b passengers adjacent to the the midpoint when the A anad B photons arrived at M simultaneously fullfill the definition of the sources being separated equally at M and that the a|b passengers marking the time the photons arrived is a detection of simultaneously emitted photons in the stationary frame, at the very least? If so then what is it that prevents you from agreeing that the a|b passengers seeing the arrival of the A and B photons at M simultabeously is not a measurement of the photons simultaneously in the in the moving frame in the same sense that the O' observer could not so detect the photons simultaneously?

Do you agree that any SR constraints, if any, of the problem is swamped by the difference in time of measuring the B and A photons by the O' observer?

Fact assumed: the clocks of the a and b passengers located at A and B when the photons were emitted were synchronized wrt the moving frame. If the clocks later show the same time for the emission of the photons, does not this constitutes a detection of the the photons simultaneously such that the event of the simultaneously emitted photons in the stationary frame also constitutes simultaneously emitted photons in the moving frame as well? If no why not?

Under the given circumstances, could any of the passengers have ever detected or determined the emission of the A and B photons simultaneously into the moving frame? Why?

You seem to think I have not described a situation that is either proved, or provable, where is the error, so I don't crowd the forum with this pitifull junk? Shiow me I will stop this as sudeenly as I started it. Just prove it. Those watching can determine my "honesty" in this regard.

What usual crap are you referring? Be brief, or lengthy, just don't be ambiguous, ok?

I have concluded that we are at at an impasse. Like Robin Hood and Friar Tuck meeting in the middle of the stream joisting for the only dry passage over the stream to the other side. Where is the crap? Point to it directly, I implore you. In fact you owe it to yourself to be as clear with a physical description as you are able at least to the level of certaqinty you show by the level of your sarcastic quips, your use of :rofl:, and your anger and cycnicism.

Priove it Doc Al, or get another profession, bcause sooner or later someone just might ask you to do just that, prove it, I mean, like your boss.Or is she the one that directed you to stuff this line of reasoning where it would never see the light of day?

I know the rhetorical dangers of being painted into a corner and painting another into a corner [Huis Clos], but once the paint has been spread onto the floor, the dye is cast, n'cest pas? (pun intended)

Doc Al

Jul8-04, 08:41 PM

Is it not feasible that I can conclude that a moving frame has at least one point associated with a midpoint of photons emitted simultaneously from a stationary frame? Even if the midpoint falls off the frame, or cannot be determined from the information available.
I have no idea what you are talking about. Here's a guess: Is there a point on the train that coincides with M at the moment that the light from A and B reaches M? Of course there is. So what?

You said the NASCAR example fell under the umbrella of SR. The low velocity makes it difficult to measure, but we are working theory here. What distinguihes NASCAR and the AE gedunken?
What are you talking about now? SR applies to EVERYTHING!
Do you agree that the difference in time that the O' observer measuresd the arrival of the B and A photon is the basis of AE concluding that the passengers "must , therefore conclude" the photons were not emitted simultaneously in the moving frame?
The fact (agreed to by ALL observers) that light from B hits O' before the light from A leads the O' frame to conclude that the lights did not flash simultaneously.

Did not the determination of the a|b passengers adjacent to the the midpoint when the A anad B photons arrived at M simultaneously fullfill the definition of the sources being separated equally at M and that the a|b passengers marking the time the photons arrived is a detection of simultaneously emitted photons in the stationary frame, at the very least? If so then what is it that prevents you from agreeing that the a|b passengers seeing the arrival of the A and B photons at M simultabeously is not a measurement of the photons simultaneously in the in the moving frame in the same sense that the O' observer could not so detect the photons simultaneously?
I have no idea what you're saying here. Get this straight: Observers a and b DO NOT detect the flashes simultaneously.

Do you agree that any SR constraints, if any, of the problem is swamped by the difference in time of measuring the B and A photons by the O' observer?
What's that supposed to mean?
Fact assumed: the clocks of the a and b passengers located at A and B when the photons were emitted were synchronized wrt the moving frame. If the clocks later show the same time for the emission of the photons, does not this constitutes a detection of the the photons simultaneously such that the event of the simultaneously emitted photons in the stationary frame also constitutes simultaneously emitted photons in the moving frame as well? If no why not?
If observers in O' measure two events to occur at the same time according to their clocks, then of course those events are considered simultaneous in the O' frame. That's what is meant by simultaneous.

Under the given circumstances, could any of the passengers have ever detected or determined the emission of the A and B photons simultaneously into the moving frame? Why?
No. See Einstein's Train Gedanken for a simple proof that events simultaneous in the O frame cannot be simultaneous in the O' frame. We've discussed this many times. Of course, you keep dodging that one. (Why do you insist on adding stuff to Einstein's simple argument?)

If you are REALLY interested, go back and read all the many, many posts in which I have painstakingly explained every inch of the "Einstein Train Gedanken" problem. Even better, pick up a relativity book.

geistkiesel

Jul9-04, 01:56 AM

Quote:
Originally Posted by geistkiesel
I just came up with another proof that the moving observer must conclude the photons were emitted in the moving frame simultaneously with the emitted photons in the stationary frame.
Yeah, "another" one.

Let's cut through the nonsense, once again. Let's say a and b are the observers on the train located right next to the flashing lights at A and B when they flash. When the see the lights flash, they check the time. No need for any "relaying" of clock times anywhere. Assuming, like you did, that all clocks on the train are synchronized then--like it or not--observers a and b will record different times for the two photon emissions.

Based on what Doc? You are so emphatic. Based on what?

You have a problem here Doc Al. Let us assume that there is only one emitter say A, then can the a passenger located at A when A emits the photon accurately determine the time the A photon was emitted? Sure 'a' can do thisa slam dunk easy as pie task..

Now A and B are inanimate photon sources that emit photon simultaneously in the stationary frame right?, each "knowing absolutely nothing of the other source."?
Similarly for the a and b detectors that do not have to be people. Now if one detector can accurately determine the emission of a photon, how in hell can two photons be deteced sequentially? Ram1024 suggested god intervened. Is this what you SRists have done, recruited god to your side?If so then this isn't playing fair.

And how do the sources determine which photon gets to be emitted first. Do they check with someone before acting? Maybe its nonlocal activity? hmmmm?

In other words, how do the A and B photon sources know which is at which end of the train and, hence, which photon must be emitted before the other, in the moving frame that is? Is this one of those "that's just the way it is?"

I have no idea what you are talking about. Here's a guess: Is there a point on the train that coincides with M at the moment that the light from A and B reaches M? Of course there is. So what?

If the a|b passengers, those located at M when the photons arrived simultaneously, wouldn't this satisfy the definition of simultananeity when they see and record the simultaneous arrival of the photons? We are still in the Eisntein train gedunken and haven't resorted to SR imperatives as Einstein did not so resort in his discussion of the gedunken.

What are you talking about now? SR applies to EVERYTHING!

OK if SR applies to everything how does it apply to the measurement of the B and A photon sequentially? And how is this measurement related to SR, in light of the other simultaneous measurements I have discussed here, [that you disagree with of course]?.

The fact (agreed to by ALL observers) that light from B hits O' before the light from A leads the O' frame to conclude that the lights did not flash simultaneously.

Is the difference in the time the B and A photon were detected such as to swamp any SR effects that you alluded to where "SR applies to EVERYTHING!"? In other words can SR effects be calculated or measured in the Einstein gedunken?

Get this straight: Observers a and b DO NOT detect the flashes simultaneously.

I discussed this elsewhere, but are you saying that if there were only one photon emitter, say A, that the 'a' passenger could properly detect the time the A pohoton was emitted?

If observers in O' measure two events to occur at the same time according to their clocks, then of course those events are considered simultaneous in the O' frame. That's what is meant by simultaneous.

No. See Einstein's Train Gedanken for a simple proof that events simultaneous in the O frame cannot be simultaneous in the O' frame. We've discussed this many times. Of course, you keep dodging that one. (Why do you insist on adding stuff to Einstein's simple argument?)

BECAUSE EINSTEINS ARGUMENT IS SPECIOUS, INCOMPLETE A CONTRIVANCE, INANE NOT IN ACCORD WITH OPHYSICAL LAW AND EXPERIMENTAL RESULTS.

]
If you are REALLY interested, go back and read all the many, many posts in which I have painstakingly explained every inch of the "Einstein Train Gedanken" problem. Even better, pick up a relativity book.

You haven't explained 1/16 of an inch to me. I have read a lot of posts like he one above that says: quote DOc Al "Get this straight: Observers a and b DO NOT detect the flashes simultaneously."
This statement proves nothing. It is just another SR mantra.

I read your link regarding Einstein's "Relativity" a treasured book I have been quoting from the get go. You even sneered at this once.

geistkiesel

Jul9-04, 04:59 AM

Another subject. If I vow not to post on any but Theory Development can my privileges regarding other forum activities be reatored? Access to my profile, private messages etc. One violation 86 me, OK?

Alkatran

Jul9-04, 09:53 AM

Quote:
You haven't explained 1/16 of an inch to me. I have read a lot of posts like he one above that says: quote DOc Al "Get this straight: Observers a and b DO NOT detect the flashes simultaneously."
This statement proves nothing. It is just another SR mantra.

Awww, too bad you repeating something doesn't make it true either. I also like how you said SR goes against experimental results. BS at its finest.

Doc Al

Jul9-04, 02:05 PM

Even though I keep telling myself to stop wasting time, I just can't resist. You are too funny, geistkiesel!

You have a problem here Doc Al. Let us assume that there is only one emitter say A, then can the a passenger located at A when A emits the photon accurately determine the time the A photon was emitted? Sure 'a' can do thisa slam dunk easy as pie task..
Let's be clear. Light A flashes. At that very instant, observer "a" on the train is directly opposite point A. Observer "a" detects the flash and records the time. Where's the problem?
Now A and B are inanimate photon sources that emit photon simultaneously in the stationary frame right?, each "knowing absolutely nothing of the other source."?
So far, so good.
Similarly for the a and b detectors that do not have to be people. Now if one detector can accurately determine the emission of a photon, how in hell can two photons be deteced sequentially?
Uh... because one was detected before the other? :rofl:
And how do the sources determine which photon gets to be emitted first. Do they check with someone before acting? Maybe its nonlocal activity? hmmmm?
Their time of emission is determined by clocks in the O frame. No mysterious nonlocal forces between photons.
In other words, how do the A and B photon sources know which is at which end of the train and, hence, which photon must be emitted before the other, in the moving frame that is? Is this one of those "that's just the way it is?"
The photons know nothing. They just flash when triggered to do so. It's kind of trivial.
If the a|b passengers, those located at M when the photons arrived simultaneously, wouldn't this satisfy the definition of simultananeity when they see and record the simultaneous arrival of the photons?
All measurements made in the O' frame agree with the fact that the flashes were not simultaneous in the O' frame.
We are still in the Eisntein train gedunken and haven't resorted to SR imperatives as Einstein did not so resort in his discussion of the gedunken.
Einstein invokes the invariant speed of light: and from that deduces all of SR.
OK if SR applies to everything how does it apply to the measurement of the B and A photon sequentially? And how is this measurement related to SR, in light of the other simultaneous measurements I have discussed here, [that you disagree with of course]?.
Your "other simultaneous measurements" are just things you made up. But, looking at things from the O frame it's trivial to show that M' detects a photon from B before detecting a photon from A. After all, M' is moving towards B and the light (which travels at speed c with respect to O) is moving towards M'. This staggered arrival of photons at M' is a real physical effect that everyone will agree upon. The O' frame can use this information to deduce (from the invariant speed of light as viewed from the O' frame) that the lights could not have flashed simultaneously at the moment M' passed M according to the O' clocks. (Damn, I must have explained this trivia about 1000 times by now.)
Is the difference in the time the B and A photon were detected such as to swamp any SR effects that you alluded to where "SR applies to EVERYTHING!"? In other words can SR effects be calculated or measured in the Einstein gedunken?
Don't know what you are talking about. What "swamps" SR effects? The entire discussion of the train gedanken is an "SR effect"!

Can the train gedanken be used to illustrate all the SR effects? Of course! But for that you'd have to know some relativity. (For example: Tell me how far is "a" from M'? How far is "a" from M at the moment that A flashes? What time does the "a" clock read when A flashes? etc, etc.)
I discussed this elsewhere, but are you saying that if there were only one photon emitter, say A, that the 'a' passenger could properly detect the time the A pohoton was emitted?
The "a" observer could detect the time that the A light flashed according to the O'-synchronized clock that he uses. Why is this so difficult for you?
BECAUSE EINSTEINS ARGUMENT IS SPECIOUS, INCOMPLETE A CONTRIVANCE, INANE NOT IN ACCORD WITH OPHYSICAL LAW AND EXPERIMENTAL RESULTS.
A laughable comment. (1) SR is in complete--and overwhelming--accord with experiment, (2) you know nothing of physical law, and (3) why do you keep quoting Einstein if you don't agree with his simple (HS level) argument? I'm still waiting for you to point out the flaw in it. (And without your usual smoke screen of mirrors and extra observers.)
I read your link regarding Einstein's "Relativity" a treasured book I have been quoting from the get go. You even sneered at this once.
This "treasured book" remains a mystery to you. And I wasn't sneering at the book. :smile:

geistkiesel

Jul9-04, 05:51 PM

Even though I keep telling myself to stop wasting time, I just can't resist. You are too funny, geistkiesel!
Quote:
Originally Posted by geistkiesel
You have a problem here Doc Al. Let us assume that there is only one emitter say A, then can the a passenger located at A when A emits the photon accurately determine the time the A photon was emitted? Sure 'a' can do thisa slam dunk easy as pie task..
Let's be clear. Light A flashes. At that very instant, observer "a" on the train is directly opposite point A. Observer "a" detects the flash and records the time. Where's the problem?

I asked the question using A and a. Now ask the same question using B andb and I get the same answer, via a symmetry argument OK? You cannot say otherwise.
If a and b, separately can record the time of the photons where and when emitted in the moving frame separately, when emitted in the stationary frame, why cannot they do so when the a and b observers are both observing the A and B flashes that flash simultaneously?

geistkiesel

Jul9-04, 05:55 PM

Awww, too bad you repeating something doesn't make it true either. I also like how you said SR goes against experimental results. BS at its finest.
For now we settle for a gedunken experiment. observer 'a' in the moving frame is adjacent to A the source of A photons in the stationary frame just as A emits the photon. Does 'a' in the moving frame detect the photon when emitted? assuming the 'a' detetcor is one wavelength from the emitting source A?

geistkiesel

Jul9-04, 06:10 PM

If O' detects the A and B photons simultaneously in her frame she therefore concludes the photons were emitted simultaneously in the moving frame. Now she must conclude the photons were not emitted simultaneously in the stationary frame, right?

geistkiesel

Jul9-04, 06:27 PM

O' considers herself stationary and watches the station move past her to the left. Just as M in the moving station is adjacent to O' photons are emitted simultaneously in the moving station frame. These photons when emitted were equidistant from M and hence equidistant from O' when emitted. Hence, O', knowing the speed of light is constant in her stationary frame will absolutely be guaranteed that the photons from the A and B emitters have the exact distance to travel form A and B in order to reach her. Hence the photons will spend the same time of flight in reaching her stationary position, hence the photons will arrive at her position simultaneously. Light is constant for the A and B photon when measured from O' frame, no problem correct?

geistkiesel

Jul9-04, 06:47 PM

The moving frame with O' is at M when photons are emitted simultaneously from A' and B' sources where O' is the midpoint of the A' and B' sources. A' and B' will arrive simultaneously at M in the stationary frame as the speed of light is invariant in the stationary frame.. However, as the light is still invariant to O' she is still going to collide with the B' photon earlier than she detects the A' photon for the same reasons she did so when the sources were in the stationary frame.

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

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

geistkiesel

Jul9-04, 07:14 PM

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

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

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

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

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

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

geistkiesel

Jul9-04, 11:35 PM

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

Doc Al

Jul10-04, 05:57 AM

Eyesaw

Jul10-04, 09:15 AM

geistkiesel

Jul10-04, 10:37 AM

Take a look at my response, where I explain once again how that trivial expression is derived and what it means. Take notes this time.
Doc will yopu please look at the link I referenced. My expression is used in a very sophisticated simultabneity analysis. Look at it godamnit!.

Doc Al

Jul10-04, 10:51 AM

Doc will yopu please look at the link I referenced. My expression is used in a very sophisticated simultabneity analysis. Look at it godamnit!.
Do you really think that link which discusses relativity will support your anti-relativity crusade? :rofl: Yes, that kind of expression does come up. But it's not what you think it is. That article is way too advanced for you.

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

geistkiesel

Jul10-04, 11:08 AM

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

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

geistkiesel

Jul10-04, 11:12 AM

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

Doc Al

Jul10-04, 11:18 AM

I do notice that when I write something decent...
When did this happen?

geistkiesel

Jul10-04, 12:38 PM

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

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

Doc Al

Jul11-04, 07:30 AM

We put the emitters in the moving frame and just as the sources A' and B' arrive at A and B in the stationary frame, the A' and B' sources emit photons. The O', who was at M in the stationary frame when the photons were emitted simultaneously in the moving frame still heads to the B' photon on a collision course and detects the B' photon first, then later the A' photon.
I'm not sure what the set up is in this scenario, but the following is true. It doesn't matter if the emitters are on the train or on the embankment. If the embankment observes them flashing simultaneously, then the train will not. And vice versa.

ram1024

Jul11-04, 06:59 PM

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

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

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

Doc Al

Jul11-04, 07:24 PM

again, Doc Al is playing with the word "simultaneously".
No I'm not.
we don't care what they OBSERVE simultaneously. we care what they CALCULATE as EMITTED simultaneously.
That's what I mean by "observe". This is fairly common usage in discussing relativity, but I agree I could be clearer. By "observe" I mean: calculate, measure, deduce, etc. When I say that one frame will "observe them flashing" I mean deduce the time of emission based upon measurements made in that that frame. Of course, if that frame happened to have an observer at the right place and time, they could directly "observe" the emission. But that's hardly necessary.
you know that's what we want and you dodge behind your relativity facade every chance you get :D
Nice try. Now go back and reread every post I've made with the proper understanding of "observe". :rofl:

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

ram1024

Jul11-04, 08:11 PM

just trying to ruffle your feathers really <chuckle>

geistkiesel

Jul12-04, 01:17 AM

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

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

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

Doc Al

Jul12-04, 09:09 AM

This Einstein gedunken is different from Einstein's original experiment only that the photons are emitted from the moving frame at A' = A and B' = B when the moving obser O' is located at O' = M, where M is the midpoint of the A and B photon detectors (when used) and from data from 1000 previous experiments identical to this experimetn, the exact midpoint of A' and B' on the moving frame is located in the stationaray frame M''.
Before we follow the yellow brick road to the land of Oz, let's define our terms. I assume you mean by A' and B' certain locations on the train. I further assume that these locations are the positions in the train that were directly opposite A and B when the lights flashed at A and B. Is that what you mean by A' and B'?

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

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

Tom Mattson

Jul12-04, 12:21 PM

ram1024

Jul12-04, 12:41 PM

this is the argument that shut you down, Tom.

although it will just get deleted again anyways...

Speed is defined as:

Distance
Time

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

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

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

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

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

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

Distance is Immutable
Time is Immutable

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

Tom Mattson

Jul12-04, 12:48 PM

this is the argument that shut you down, Tom.

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

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

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

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

although it will just get deleted again anyways...

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

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

russ_watters

Jul12-04, 01:14 PM

Morbid curiosity here...
Distance is Immutable
Time is Immutable On what, specifically, do you base those two statements?

krab

Jul12-04, 01:14 PM

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

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

geistkiesel

Jul12-04, 03:56 PM

I'm not sure what the set up is in this scenario, but the following is true. It doesn't matter if the emitters are on the train or on the embankment. If the embankment observes them flashing simultaneously, then the train will not. And vice versa.
This is the qustion: On the moving platform the lights were emitted such that the observers at A' and B', that is a and b, mesurd the time of emission. The lights emitted just as O' was located at M. The O' observer detected the A' and B' photons in the same sequence as she detected the photon emitted from the moving frame.
Will the moving frame come to the same conclusion she came to under the standard Einstein version where the photons were emitted simultaneously in the stationry fame? Remember, the only difference here is "the photons were emitted n the moving frame as O' was at M."

Your response to the first version of thois indicated that simultaneity could be determined by what each observer did. If the stationary observer concluded the photons were emitted simultaneously in the stationary fram,e then the photons were not emitted simulltaneoulsy in the stationary frame simultaneity. Can you please discuss the four cases below? Thanx.This sounds like O's reality is determined by what the stationary observer detects say: true or False? the stationary observer did not have his detectors working at the instant the photons were emitted in the moving frame:What eaffect on determining simultaneity does have? and the stationary observer did have his detectors working. The A' and B' photons were detected arriving simultaneously at M. What affect, if any?

geistkiesel

Jul12-04, 04:17 PM

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

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

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

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

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

geistkiesel

Jul12-04, 04:33 PM

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

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

For future reference, feedback doesn't belong in the Theory Development Forum.
Speaking of outbursts it appears to me that your anger got the best of you when you closed my thread. My post describing the zero velocity absolute inertial frame you cionssidered as circular? why becfause no one caved on to your demjands that we succumb to SR doma?

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

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

Here is another ove your head.

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

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

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