# A light speed gedanken

Discussion in 'Physics & Math' started by CANGAS, Apr 28, 2006.

1. ### Neddy BateValued Senior Member

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I suppose I should have explained more clearly what I was trying to do. Since Geistkiesel accidentally posted two different equations (probably a typographical error), I was not sure which one was right, so I tried to derive it myself. I think my expression #10 is the one that Geistkiesel meant to post.

Anyway, I got an idea from this post of GK's:
I thought, perhaps there could be a time dilation formula, based on Geistkiesel's ideas, but one which includes both the forward and the rearward "delta t" that you mentioned. More specifically, I suppose this would be clock dilation, and not time dilation. But on a local scale it might be construed as time dilation by an unwitting experimenter on the train.

I suppose this is no different from Aether theory, but I had never heard much about "clock dilation" within that theory, so I thought it would be interesting to see what I could derive.

I realize that there is no length contraction in this theory, and that c is only isotropic in the absolute rest frame. It is obviously different from relativity (I even compare it to relativity in the last section). However, it still does predict a time dilation effect (clock dilation) for the train, and I thought that this does support some of Geistkiesel's earlier posts to that effect.

3. ### kevinalmRegistered Senior Member

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993
Yes, I thought it looked like some kind of time dilation. Essentially what Geistkiesel does in his derivation is t' = (t2-t1) - (t3-t2) where t1, t2, t3 are emission, reflection, and reception times respectively along the forward parallel arm of an M/M type apparatus in classical universal time. He expresses this in terms of the total trip time t = (t2-t1) + (t3-t2) He derived it in another thread a while back, iirc for some reason he wanted to eliminate the length of the arm, don't recall why. Of course the L is implicitly embedded in t.

>> edit Anyway, what he seems to think is more along the lines of clock skew, and that somehow a century of experiments have missed what he considers to be a measurable and correctable clock setting error, or something along those lines.

Last edited: Jul 27, 2006

5. ### geistkieselValued Senior Member

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This cannot be as you describe. Regardless of the claims of "others", measuring the same time for light to strike the two detectors, both in motion with the same speed as the "physical light emitter", cannot be.

This is equivalent to saying distances the light travels in the two cases is identical. Try this one for consideration: From the embankment determine the exact point that the light strikes the two different moving detectors. Now, remove the two detectors and install them on the embankment in the two positions measured.

The light pulses emitted on the moving frame or on the embankment, it makes no difference, will now strike the two detectors exactly as when the detectors were actually moving, sequentially.

You keep forgetting that the light motion knows nothing of the moving frame and that the pulses are indepenmdent of any motion of the source of the light.

Using the xpression "ct" the distance the light travels before striking the rearward and oncoming detector is less than the light must travel in order to strike the detector bei8ntg chased by the pulse.

Your problem here is resolving the contraction and time dilation that is consisistent with both detectors being struck by light pulses that are moving in complete disregard to the motion of the source of the pulses.

Finally, try this for consideration. When the light striking the oncoming detector/mirror has traverled a distacnce ct and is reflected a distance ct, the pulse is now loicated at a point in space exactly identical with the original location of the light source. The light source has moved a distance 2vt in the interim and this light pulse in now located 2vt from the original emission point. To arrive back at the physical device that emitted the pulses the light must travel 2vt plus a small amount the device has moved in the interim, or vt'. This distance ct' = 2vt + vt'. Solving for c, c = (2v)(t + t'/2)/t'.

Thoi allows a straight forward measurement of the speed of light from a moving platform.

Geistkiesel​

7. ### kevinalmRegistered Senior Member

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993
None the less, that is what experiment shows. Argueing that it cannot be is pointless. If you choose to ignore the results of experiment you cease to be doing science and enter the realm of philosophy or religion.

Most of the experiments that have been done over the past century have been atempting to measure this time difference your keep insisting on, either directly or indirectly. They all fail. More properly, they all put an unbelievably low upper limit on the Earth's velocity wrt the light propagation medium. The best modern experiments put that limit somewhere around 30 m/s. Whereas one would expect that at some point in its orbit the Earth should be moving on the order of 30 km/s.

8. ### Neddy BateValued Senior Member

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1,588
That is far from straight forward, in my opinion. If you want a straight forward calculation of the speed of light as measured on the moving platform, why don't you just use these expressions from my derivation on the previous page?

The relative speed of photon & train for light pulses travelling toward the front of the train
w<sub>front</sub> = c - v ......................... 1

The relative speed of photon & train for light pulses travelling toward the rear of the train
w<sub>rear</sub> = c + v ...........................2

Then, using a coordinate system attached to the train, you can determine the distance traversed by any given wave-front of light (at any given time after emission) using the following formula:

d<sub>front</sub> = t * (c - v) ....................1a

d<sub>rear</sub> = t * (c + v) ....................2a

Of course reflections will have to be treated as two seperate rays: one "frontward" and one "rearward" using the appropriate formula for each.​

(Note: At this point, GK and I are disscussing an Aether-type of model rather than the usual Relativity model.)

Last edited: Aug 2, 2006
9. ### geistkieselValued Senior Member

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If I have answered this already then please accept this as a mere 'emphasis'. AE made no such assertion as you state above.

The gedanken was set up with all observations made from the embankment as frame of reference. You are in error when you state that observers on the train and embankment cannot measure the same speed of light. The error comes form your misunderstanding (or ignoring) the difference in measuring the speed of light and the measure of the relative speed of frame and photon.
AE did not merely “point out” anything of the nature you suggest. He switched frames and attempts top convince us that the measurements were taken from the train as reference frame, which they surely were not.

Here is what AE said:

“The velocity of propagation of a ray of light relative to the carriage thus comes out smaller than c.”

AE was referring to the expression, w = c – v. W is the relative motion of frame and photon, c the speed of light measured form the embankment and v, the speed of the train. Remember, he got this by substituting the light motion for the motion of the man walking on the train. The speed of the man on the train, seen from the embankment, is the sum of the train motion and the man walking relative to the train. To make this substitution assumes that the speed of the light is dependent dent on the speed of the train, as seen from the embankment. Surely you do not believe this do you? Chapter VII is a travesty of justice and recognition of AE’s misuse of the laws of physics. This isn’t a mere mistake, as AE had recognized the independence postulate of the motion of light being independent from the motion of the source of the light earlier in chapter VII. Read it and weep.. Then he follows with:

“ But this result comes into conflict with the principle of relativity set forth in Section V. For, like every other general law of nature, the law of the transmission of light in vacuo must, according to the principle of relativity, be the same for the railway carriage as reference-body as when the rails are the body of reference.”

There are two errors here. First, AE switches reference frames and attempts to guile the reader into believing the measurement was from the train as reference frame, which it was not.

Second, AE misstates the law of relativity by implying that the laws of transmission of light in vacua and with the train as reference frame means that the “SPEED” must be measured the same. The laws of relativity mean only what is says, that the “laws of transmission of light must be the same” as they must for all material objects being measured in all reference frames. This is an AE distortion – a major “lie” if you will.

Why not claim that the measure of the speed of ducks be measured the same in all reference frames? Remember, all measurements were from the embankment and to state here that the measurement from the train as reference frame would be the same is simply not supported by his arguments and the manner in which the problem was constructed.

Geistkiesel ​

10. ### geistkieselValued Senior Member

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“ Originally Posted by geistkiesel
When the light striking the oncoming detector/mirror has traveled a distance ct and is reflected a distance ct, the pulse is now located at a point in space exactly identical with the original location of the light source. The light source has moved a distance 2vt in the interim and this light pulse in now located 2vt from the original emission point. To arrive back at the physical device that emitted the pulses the light must travel 2vt plus a small amount the device has moved in the interim, or vt'. This distance ct' = 2vt + vt'. Solving for c, c = (2v)(t + t'/2)/t'.

This allows a straight forward measurement of the speed of light from a moving platform.

Geistkiesel ”
You ask why I don’t use these expressions from the previous page. The answer is simple” The expressions are expressions of the relative motion of frame and photon and not the speed of light expressions as are the one you complain of in my post.

You say it it far from “straight forward”. Is this because you view any variation form SRT models as other than “straight forward”?
No I am not discussing an “ether type model”. I am discussing Einstein’s discussion in Chapter VII of his book “Relativity” published ten years after his 1905 publication of SRT.

Neddy Bate, do you agree with the following:

When the rearward light pulse has moved a distance ct and struck the oncoming clock/mirror and reflected another distance ct (now a total of 2ct) that the light has returned to the exact point in space it originally was emitted (from)?

Further, that the physical device from which the light was emitted is now located a distance 2vt from the point in space in which the light was emitted (from)?

Further, the light must travel a distance 2vt, plus a small amount vt’ the frame moves in the interim, in order to reach the physical device from which the light was emitted? Do you agree? What is the problem when making all of these measurements from the moving frame?

Using the clocks on the moving frame, and even assuming some “time dilation” the fact remains that the rearward clock will be struck before the forward clock is struck, assuming the clocks are placed equal distances from the point of the emission. How does time dilation and frame contraction occur such that both pulses reach the clocks at the same instant, from wherever measured?

Geistkiesel ​

11. ### geistkieselValued Senior Member

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The 100 experiments proving SRT is a specious argument, unless you can provide the details.

For instance, with the earth orbiting the sun, the earth spinning on its axis and the sun moving through space, what experiment has been conducted that separate all these various motions from each other? Have you read Dayton Miller’s paper on the 100,000, plus, experiments redoing the Michleson-Morely experiments?

How about the famous “1919 eclipse experiments" that thrust AE into the spotlight? Did you know that the resolution of the telescopes used in those experiments were at least an order of magnitude greater than the requires “hundredths of a millimeter” required, as admitted by Einstein himself? Also, that the data used by Einstein in “Relativity” was a selected list of data points, not all of the points. Some of the unused data points were on opposite sides of the photographs used, others with differences in the order of centimeters. Explain the first experiment before you use some generalization of others that you claim prove your point(s).

You aren’t writing to me here are you.

You are attempting to communicate with the casual reader that comes along, as it appears to me.
Einstein’s use of the eclipse data is a violation of basic principles of scientific ethics, don’t you agree? Or is this just a common and recognized scientific practice of "discretion of the experimentor"? Throw out the bad data and keep the good data - is this your concept of scientific standards that you practice?

Your reference to the religious content of my posts is an insult, which I am sure you have no apology. I have carefully made references to sources used, and arguments using the laws of physics. I do not know, nor care, of your belief system, but this is what your post here is all about, as I read it.

Why do you bother to diuscuss the matters if your input degenerates into such lapses of trviality and useless rhetoric? It seems you are merely repeating what your graduate advisor schooled you on and to which you have swallowed most eagerly and wholeheartedly. After all you did want to pass the exams, correct? Careers and peer acceptance are much more important than (embarrassing) critical analysis - have you heard of "critical analysis"?
Geistkiesel ​

12. ### geistkieselValued Senior Member

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WRONG. L is not implicitly or explicitly embedded in t. You expression bears no relation to the one I obtained. My expression t’ = t(2v)/(c – v) is grossly different form your attempt to rewrite what I developed. Why do you do this?

Wrong again. Get it right Ka and please stop misstating what I am doing. Neddy Bate knows how to discuss without perjuring himself or misstating what I am doing, I suggest you take a lesson from him. Just because you believe un SRT and find a comrade in arms in Neddy Bate does not give you justification for misrepresentations.

Didn’t your mommy and/or daddy teach you or expose you to any concepts of decency, fair play; to always search for the truth; to be honorable among all else and to have your deeds always reflect good taste and manners? Am I too harsh here? If so, then please accept my most humble and seriously intended apology.

Geistkiesel ​

13. ### geistkieselValued Senior Member

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The explanation here is that t’ = t(2v)/(c – v) is a determination of motion. If zero the frame is at rest.

In any event, when reading the times recorded by the two clocks, there will be a clear showing that the rearward clock is struck before the forward clock, regardless of any clock or time f dilation. The fact that the clocks measure a sequential arrival of the light disproves SRT and the so called, simultaneity loss.

Likewise, t’ the time difference foe the round trip of the light where the embankment round trip is t, the moving frame round trip is t + t’.

Geistkiesel ​

14. ### Neddy BateValued Senior Member

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I wasn't complaining of your expressions, I just want to understand them. Theory of relativity aside, and assuming that the clocks on the train are truly synchronized, I would think that the experimenter on the train in chapter VII would expect to measure the speed of light to be either c + v or c - v depending on whether it was frontward or rearward. If you do not consider that to be the "speed of light" on the train, then perhaps you are looking for something like this?

c = w<sub>front</sub> + v ......................... 1b

and

c = w<sub>rear</sub> - v ......................... 2b

For frontward and rearward rays of light, respectively, such that the measurements of light speed are adjusted according to the known velocity of the train, and then the speed of light can always be considered the constant c.​

No, the reason I say that your approach is is not straight forward is simply because I find it difficult to follow. I will try to work it through here, and hopefully you will see where I am getting lost, and then perhaps you can help me to better understand.​

I agree with this so far. Let time t=0 be the time the light is emitted, and then after one time increment &Delta;t=+1 the light is a distance d=ct toward the rear (where it meets the reflector) and then after one more time increment &Delta;t=+1 the light is d=ct in the opposite direction, so yes, it is now back where it started from.

But notice that you have the time increments equal(&Delta;t=+1 in both cases). This means that the speed of light is the same in both directions. I thought you wanted to analyze the train frame in which the speeds are different, the times are different, and the math is expressed by

w<sub>front</sub> = c - v ......................... 1

w<sub>rear</sub> = c + v ...........................2

d<sub>front</sub> = t * (c - v) ....................1a

d<sub>rear</sub> = t * (c + v) ....................2a

Yes, but I am not sure why we need to know where the strobe-lamp is located at this time. Using my expressions we can locate the light rays relative to the train without worrying about that. The strobe lamp does not move in the train frame (assuming it is mounted to the train).​

I agree, but now you have introduced t' which is completely unknown at this point. We will need further derivation.​

There is nothing wrong with using the moving frame, but you seem to be trying to do all of this from the embankment frame. My expressions do use the moving frame, and they are far more straight forward.​

Now I am lost. So I am getting lost somewhere in the above quote. Plus, recently you have introduced this expression​

the derivation of which was not shown. However, when substituting these expressions together, (I don't have the energy to do it right now, but I might try it later), it seems to me that it must simplify to either

c<sub>train</sub> = c<sub>embank</sub> - v

or

c<sub>train</sub> = c<sub>embank</sub> + v

(These are essentially the same as my expressions 1 and 2 from my original derivation post.)

or

c<sub>train</sub> = c<sub>embank</sub>
(This is essentially the same as my expressions 1b and 2b in this post.)

I don't see how it can be otherwise in the non-relativistic world being considered in chapter VII. But then, think of what life would be like if light rays propagated at different speeds in different directions. It is perhaps even stranger than the relativistic world.

I provided a reference to Cahill's work earlier in the thread, and he seems to reaching conclusions alongf those lines (for certain non-vacuum media). Of course he does not use my simple late-19th century derivation, but takes a much more sophisticated approach that still requires relativity theory.​

15. ### CANGASRegistered Senior Member

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If Special Relativity is not legitimate, then the observation from the embankment will not agree with the observation from inside the Pullman car. In such an incredible circumstance, the two observations must be understood on different terms.

16. ### geistkieselValued Senior Member

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Neddy Bate
Registered User (268 posts)
Not quite Neddy Bate but let me explain.

As I stated earlier the two expressions above are expressions of relative motion of frame and photon. If the train is moving in the same direction as the light motion the relative velocity of frame and photon is clearly, Cr = Ce – Ve, where the subscripts refer to “relative” and “embankment” respectively. In other words, Cr is how much faster the light beam speeds is over the train speed. Here Cr, the relative velocity of frame and photon is the difference of the speed of light wrt the embankment (“the vacua” in Einsteins’s terms) and the speed of the train wrt the embankment. If the train and light directions are reversed then Cr = Ce + Ve Here the relative speed of frame and photon is greater than Ce. From the embankment the speed of light emitted on the embankment and the speed of light emitted from the train is identical.
There are a number of ways of deriving the expression. Consider the frame, velocity v, and light, velocity c, directions originally opposite. During the time t, the light moves a distance ct, where (and when) the pulse strikes the oncoming mirror/clock. We assume that the clocks at the emission point and oncoming mirror are synchronized.

During the time t, the light pulse has moved a distance ct, the frame has moved a distance vt, correct? Now, after the reflected light has moved an additional distance ct (now totaling 2ct) the frame has moved a total distance 2vt, from the original emission point. It may be helpful to consider this point relative to some indicator on the embankment, or “vacua”, to use AE’s terminology.

At this point the light beam is bearing down on the physical emission device on the frame and will arrive at this point on the frame after crossing the distance 2vt, plus the distance the frame moves in the interim, vt’, or ct’ = 2vt + vt’. We run this experiment a few thousand times getting the same result every time, using the frame clocks only.

To simplify this some, let us turn our attention to a light pulse originally moving in the direction of the frame motion. When this pulse moves a distance ct the pulse is a distance 2vt from the detector moving away from the light pulse. It might simplify things to draw a sketch and verify the distances. Like the previous test the pulse will catch the up to the detector moving away after traveling 2vt plus the distance the frame moves in the interim, or vt’, then ct’ = 2vt + vt’ from which we again obtain the time t’ = t(2v)/(c – v). Instinctively one might think the distance from the pulse and detector is vt, but trust me it is 2vt. Like the previous example we run this test a few thousand times.

What is this t’ you ask?
We agree and we are thinking in parallel so far.
If I left you with this belief, then it is my poor powers of communication that have left you confused. We needn’t collect data from the two detectors at the same time. We only need to keep v constant and, of course, the distance from the point of emission to each detector the same in each series of tests.
We don’t need to do this but it helps to understand what is occurring. What is important at this point is to realize that the clocks on the moving frame will measure two different arrival times of the light pulse at each detector. The rearward moving pulse arrives at the detector in time t, as we agreed. However, as the forward detector is moving away from the pulse the time of arrival of the forward moving pulse at the forward detector is t + t’.
OK. Let us assume T is the measured time the light pulse travels needs to arrive at the forward detector. Hence, T = t + t’, or T – t = t’.
We needn’t fuss about whose expression is more straight forward, do we?
It is confusing, I agree.
Let us solve for c in the expression t’ = t(2v)/(c – v).

t’(c – v) = t(2v)
t’c = t(2v) + t’v

c = v(t2 + t’)/t’
This is an expression to determine the measured speed of light. T is measured, t’ determined from the measured T, and of course, v is measured wrt the embankment (or vacua).
I don't see how it can be otherwise in the non-relativistic world being considered in chapter VII. But then, think of what life would be like if light rays propagated at different speeds in different directions. It is perhaps even stranger than the relativistic world.

I provided a reference to Cahill's work earlier in the thread, and he seems to reaching conclusions along those lines (for certain non-vacuum media). Of course he does not use my simple late-19th century derivation, but takes a much more sophisticated approach that still requires relativity theory. [/quote]

ctrain = cembank - v

or

ctrain = cembank + v

You have indicated that the speed of light is different depending on the direction of measurement, correct? What these two expression indicate ion the relative velocity of frame and photon in the two cases and are correct as you have expressed them.

The final expression ctrain = cembankment is also correct as the speed of light is the same in both frames, not the relative speed of frame and photon.. This last expression is different from the first two, as it does not reflect the relative speed of frame and photons; it is the absolute speed of the light in both frames.

You have complained of the problems if the speed of light were different in different directions, a problem that SRT (you here) have indicated by the two relative velocity expressions, which are accurate and pose no problems.

ctrain = cembank is a true statement as we all know. Neddy Bate, there should be no problem once you recognize the fundamental difference in relative and absolute motion which you have so simply and clearly described here (above).
Below is a copy of a discourse above.
“ Originally Posted by geistkiesel
Further, the light must travel a distance 2vt, plus a small amount vt’ the frame moves in the interim, in order to reach the physical device from which the light was emitted? Do you agree? ”

I agree, but now you have introduced t' which is completely unknown at this point. We will need further derivation.

When you have “agreed” here you have conceded the error in SRT. AS I understand it, in the gedanken of the two directions of the pulses, SRT claims that the pulses will arrive at both detectors simultaneously, as the pulses do when the frame is at rest wrt the embankment.

Are you sure you want to agree her? With anybody else Neddy bate, I would not be so generous as to allow a revision of the “agreement”, but you deserve the courtesy, so think about it.

Our maths do not disagree. My expression shows how to measure the speed of light wrt the train. Your expressions described the relative speeds of the two measured directions and make the fundamental statement that the speed of light is the same in all inertial frames, while the relative speeds differ according to the speed of the frame relative to the vacua, or embankment.
Geistkiesel ​

17. ### geistkieselValued Senior Member

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By the way all, and I write obvious mirrored image:

CANGAS|SAGNAC.
Geistkiesel

18. ### geistkieselValued Senior Member

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Neddy Bate,

After reflecting on my last post I realize I have made a grievous miscalculation. Using your two expressions for the relative speed of light as measured in the two directions. I think you can readily see what I am driving at.
Here is what I mean: The relative velocity of the light measured in both directions cannot possible be correct! While you added the terms and arrived at the expression that

ctrain = cembank

and to which I agreed, we were horribly in error. Do this, simply equate the two expressions to arrive at,

cembank - v = cembank + v, which results in

2v = 0, which is true only if v = 0. However, we both assumed the frame was moving. It is essential that the two expression for ctrain be distinguished between the two directions for which the expressions apply, or,

ctrain- = cembank - v

or

ctrain+ = cembank + v

where the added ‘-‘ and ‘+’ are used to distinguish the two relative motions. Now when you add the terms (as you did in your post), the results is,

ctrain- + ctrain+ = 2cembank .

Now the result is,

(1/2)(ctrain- + ctrain+) = cembank

which says the speed of light on the embankment is the average of the two relative speeds as indicated in your expressions.

Geistkiesel ​

19. ### CANGASRegistered Senior Member

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Why is Madonna Lisa DelGiaconda smiling so mischieviously?

20. ### CANGASRegistered Senior Member

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You are striking at the exact heart of the problem and its resolution.

Continue.

21. ### geistkieselValued Senior Member

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A personal opinion follows, which is all I can offer.

I suspect (as have others) that the Madonna Lisa DelGiaconda, is a self portrait, self likeness, a self reflection of the artist's soul. This most cleverly disguised work of Leonardo manifests a sense of beauty and a treasure trove of humor that transcends and simplistic model of 'practical joke'.

The world has viewed the painting as a perpetual question of the source of the generated smile adorning the lips of the Madonna. Unlike analysis of most paintings in terms of value, color scheme, light and composition etc, the viewer is attracted to the work by the smile of the eyes. The subtle, alluring, mischevious and deceptive motion of the lips radiates throughout the face. Though the viewer is drawn to the pose by the eyes which is what attracts viewers in virtually all serious portrait paintings. In this sense the structure of the lips is secondary and supporting, though certainly an essential and crucial element of the painting. The Madonna's pose is simple with the overall effect of drawing the viewer into the work without any distraction to the intended effect.

The Madonna is a perfect example of the strength of simplicity. Unlike modern trends that attempt to emphacize pain, love, anger, confusion, of some one of known human emotional traits, Leonardo throws all that out as fluff, perhaps even, he characterized [in the Madonna] as such attempts as silly.

Leonardo has an interest in soul, a human quality long forgotten and neglected in more modern works. Soul, modernly is equated with personal 'ego' and is reflected, erroneously, in emotional and psychological characteristics with no serious understanding indicated in the modern artist's sense of modt fundamental humaness, of soul.

In terms of 'quanmtum mechanical' renderings, what the oberservers 'sees' is the physical world. The stuff, the stars, trees, his and her own hands, mirror reflections, rocks, oceans, all things 'out there', these are the things of the observable, the measured universe.

When the observer then questions her own part, she catches herself with the inescabale understanding that the observer is other than the observed. The physical body from which all observations are made is suddenly understood as merely 'belonging' to the the person making the observations. It is always, 'my hands', 'I have a head ache'. 'I love you', 'the setting sun is beautifuyl [to me].

The human body is not merely just a complex machine, for without the body there is no observer. Do not we all look out from a distatched perspective, where our 'essence' is quietly in the background, observing the ego at work. Don't we know that 'I' fall in love, I have desires, I bet on the Yankees to win the world series. One's death may be mistaken as a negation of all the physical world, which is a measure of the limits an ego can assume. Yet do not we sense that the death of once living things is merely the absence of 'soul'; that the dead body no longer sustains as a platform of observability, the departed?

Do not we all secrertly adhere to the understanding that 'cowards die a thousand times before their death; the valiant tastes of death only once; It should seem to you most strange that death, a necessary end, will come when it will come'.

Apparrently DaVinci carried the painting throughout ten years of his travels in Europe, adding little bits here and there. He must have known he had something so very special and something that he recognizewd as a treasure. Of course Leonardo was a professional artist and made a very good living at what he did, but the Madonna is something that he never held out to the world as reflecting his ego in demanding recognition for his artistry.

He found a treasure in his work and the fact that he painted the Madonna was inconsequential to the totality of the eternal beauty of soul, all person's souls.

Geistkiesel ​

22. ### CANGASRegistered Senior Member

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I am sure that my comment was understood as a compliment to the beautiful smile on the face of a certain icon. And since many people have wondered what mystery the Mona Lisa smile represented it was fitting that I would respond to your unveiling of what might be the mystery of my handle.