# A light speed gedanken

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

1. ### PeteIt's not rocket surgeryRegistered Senior Member

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If V1 and V2 are both positive, and V1 is less than V2, then train A has faster clocks in the implied reference frame.

3. ### MontecRegistered Senior Member

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248
Hi Pete
Hehe I did not say what "some velocity" was in reference to. Hence your " If V1 and V2 are both positive" preface.

If the reference made "some velocity" equal to zero or negative then the answer is ambiguous. There is always references that will make "some velocity" equal to zero, positive, or negative.

5. ### geistkieselValued Senior Member

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So you are a proponent of the postulate that the motion of light, including speed, is dependent on the motion of the source of light. I hadn't realized that the independence postulate of light motion had been scuttled. In this light , apparently the laws of physics are not the same in all inertial frames.

Geistkliesel

7. ### geistkieselValued Senior Member

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Einsteins use of the expression w = c - v is very misleading. The conditions that were established leading to the expression are due to measurements taken from the embankment as inertial reference frame. The discussion here is the attempt by Einstein to uise the expression as if the measurements were taken from the trains as reference frame, which is not the case. This w = c - v is not an expression developed from the train as reference frame, therefore the discussion following is totally meaningless.
Geistkiesel

8. ### kevinalmRegistered Senior Member

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It is not misleading at all. He is simply using the classical prediction of ether theory, w=c-v. In Chapter VII he mearly points out that if this is true, both observers cannot measure the progress of the light pulse as travelling at c, each measuring with respect to their own reference frame.

9. ### przyksquishyValued Senior Member

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You're making it sound like there are multiple competing versions of STR floating around. There aren't. Also, in the example I gave, the speed of the photon was the same in both frames.
STR is based on the same postulates it was 100 years ago: The laws of physics and the speed of light are the same in all inertial frames.
The laws are invariant; the measurements transform from one inertial frame to another according to Lorentz.

10. ### Neddy BateValued Senior Member

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In your opinion, what equation should be used to calculate the speed of light relative to the train? Or, what equation do you think Einstein would like us to use to calculate the speed of light relative to the train?

I agree with kevinalm, and this is exactly why relativity would be in trouble if light were measured to be w=c-v or w=c+v (in some inertial frame) instead of just w=c.

Below is a link to an article in which Associate Professor Cahill claims that light does propagate at different speeds in different directions. I don't understand why the effect was so small in the MM experiment -- he claims that the calibration of the apparatus was wrong! He also fails to address the great experimental successes of relativity, and focuses only on the problems (dark matter, etc). Here is a blockquote, and then the link:

<blockquote>"In 2002, we showed that the calibration of the experimental apparatus had been very wrong, and the small effect seen by Michelson and Morley and others actually corresponded to the solar system moving through space at more than 400 kilometres per second, which is more than 1,000th of the speed of light," Associate Professor Cahill said.</blockquote>

http://www.flinders.edu.au/?news=77

Last edited: Jul 15, 2006
11. ### kevinalmRegistered Senior Member

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Interesting Neddy. Have you read any of Cahill's papers downloadable at his website? Haven't got a link atm but I'll add one to this post if anyone is interested. If I'm reading correctly he accepts matter contracts, etc. ala Lorentz Ether Theory. Which makes absolute motion undetectable if the optical path is a vacuum. Only when the index of refraction in the path is greater than 1 is there a measurable residual effect. The preference over the years to using vacuum paths in experiments has hindered detection of absolute motion, according to Cahill.

He also claims that reanalyzing the data from a number of experiments that used optical media instead of a vacuum path, that a consistant absolute velocity can be determined. (he gives an exact speed and direction but I don't recall them right now, speed was something like 350 kps)

Last edited: Jul 15, 2006
12. ### PeteIt's not rocket surgeryRegistered Senior Member

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That's right. Hence the "in the implied reference frame" restriction.

13. ### Neddy BateValued Senior Member

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Excellent find, kevinalm. The only thing I had read at the time of my earlier posting was the article to which I linked. Now that I have read some more (from your link), I see that he <i>does</i> address the experimental successes of relativity in his papers. If I understand correctly, much of the successful predictions of relativity are explained as simply having been caused by absolute motion rather than relative motion. There have been many similar claims on this forum, and usually their arguments are disproven using the classical interpretation of M&M. Cahill has shown that it requires a relativistic interpretation, if I am reading him correctly.

14. ### kevinalmRegistered Senior Member

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Yes, what he suggests is that in a "two way" path device, a vacuum path has zero sensitivity. Also that 2 way transparent solid (glass path) devices don't work well, athough I don't understand why, as 1 way glass paths are ok, according to Cahill. That's if _I_ am reading him right. For a 2 way, gas is best. The fringe shift of a gas MM interferometer is a combination of a small "ether" drift shift and a smaller fresnel drag effect.

Anyway, according to his analysis the early experiments were much less sensitive than expected and more modern vacuum apparatus are virtually useless.

Honestly, I don't know what to think of Cahill's work. In his favor in my mind is that he is making some definate testable predictions. Be interesting to see if anything developes.

>>edit
Reread some of his paper's. I had the optical solid (glass) part wrong. It _never_ works, according to Cahill.

Last edited: Jul 17, 2006
15. ### geistkieselValued Senior Member

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I would use the expression t' = t(2c)/(c - v)which is the additional time a light pulse will strike a detector moving away from the pulse as compared to a pulse heading toward an oncoming detector, where the detectors are located equal distances from the point of the pulse emissisons. It takes longer for the light to reach the parallel moving detector because there is more distance to travel, than the pulse aimed at the oncoming detector.
You (Kevinalm) are mistaken in what Einstein was saying when he developed the expression w = c - v. This is an accurate statement of the relative velocity of frame and photon measured from the embankment. Kevinalm's error is his failure to distinbguish between the "velocity of light" and the "relative velocity of frame and photon". The measure of the speed of light is the same in all frames, without the necessity of imposing special relativity theory, however, the relative velocirty of frame and photon is frame dependent. Einstein did not get to the point that Kevinalm is discussing until later and after he had already made the construction of special relativity theory. Einstein based his formulatiin in chapter VII based on the statement that the "laws of motion of light should be the same in all inertial frames". (paraphrasing)His error is assuming the laws of motion of light require that the speed of light be measured as c, and therefore neglected the motion of the inertial frame moving wrt the embankment. In other words, Einstein separated, without discussion, light motion from the motion of all other material objects, which wqas an arbitrary act and not discussed in chapter VII, at least he did not justify his separating light and other material onjects as requiring two separte physical law structures.
Einstein stated that w = c - v could not be correct as they had lready stated that the speed of light should be measured the same. Therefore, he was assuming that w = c - v applied to the measurement of light when the train was the inertial frame, when the expression was developed with the embankment being the inertial frame of reference. It is a sham.
NO there is no problem. I have shown dozens of times that all of Einstein's train gedanken can ve resolved without the necessity of invoking SRT.

I believe also that you owe me a reponse to the "twin paradox" resolution where Feynman et al, resolved the paradox in a straight forward manner. The accelerating twin was the one who aged the least in his round trip. Ergo, the equivalence of inertial frames was effectively discarded to avoid a very embarrassing situation, of an answering why both twins, earth bound and space ship traveller could not both see the other as moving and themselves as stationary.

I will get back to you on this after reviewing the
Geistkiesel ​

16. ### kevinalmRegistered Senior Member

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993
Geistkiesel,
Ok, let me pose the question like this. The station master uses a meter stick to lay out distance marks along the track. He has his own set of clocks, and has determined that to the best of his ability to determine, they keep good time and are syncronized. The conductor also uses a meter stick to lay out distances along the floor of the train cars. Likewise, he has his own set of syncronized, accurate clocks. The train is moving at speed v wrt the station as measured by the station master.

According to classical, pre Einstein physics, if the station master measures the progress of a light pulse at speed c through his distance marks using his clocks, what does the conductor measure as the speed of progress of the same pulse, using the train distance marks and clocks?

17. ### Neddy BateValued Senior Member

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1,675
The standard SR response is that it is not acceleration itself that determines time dilation, but relative velocity and/or differences in gravitational potential. Yet acceleration by itself seems to provide a very intuitive resolution to the twin paradox, as you have noted. Acceleration encompasses both gravitational potential (by the equivalence principle), and also all changes in relative velocity. So it is a convenient, although technically incorrect substitution.

But even if we knew all of the accelerations that occurred in any particular twin paradox, that is not really enough. We must also know the relative velocity for all of the inertial portions of the voyage as well, if we are to be able to predict to what extent one twin is younger then the other. This is a more specific task than simply identifying which twin is younger. So acceleration is not really the key factor in fully understanding time dilation.​

18. ### geistkieselValued Senior Member

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2,471
Neddy Bate,
I did not intend to imply that it was the acceleration itself that created the condition of time dilation, but that the acceleration created the motion of one of the twins, while leaving the other twin in a state of zero relative velocity. AS Feynman stated, 'it is the accelerated twin that is actually moving' (paraphrasing)and therefore this moving twin is the one whose aging rate is slower than the earthbound twin. I believe we have agreed on this, at least as the discussion applies to SRT.

Oncve we agree that the accelerated twin, be he/her on a space ship or a train moving wrt the embankment, there is still the problem of maintaining the "equivalence of inertial frames" postulate. Once the twin paradox is resolved by the motion of one twin, we are no longer at the liberty of stating that each twin can rationally claim to be at rest and assert that the other twin is really moving. Especially when we add the observed condition that the twin on the space ship knows he is on a space ship, is cognizant of the acceleration of the ship, knows to a certainty that the earth is not in a state of acceleration, ever, and all this known by the earthbound twin as well. The experiment can just as well be conducted with the earthbound twin removed to a gravity free position in space at a zero relative velocity wrt the earth.

Trains or space ships, it makes no difference, does it? The train observer is in an identical condition as the space ship twin. The only difference is the possible state that the train observer does not know he is on a train, is completely ignorant of the reality that the scene whizzing by is actually the embankment, he must therefore guess as to which frame is actually moving and which is at rest. However, the laws of physics are not determined by the psychological state of the observer, and having negated the implications of the equivalence of inertial frames as understood by SRT theorists, the theory itself vaporizes.
Geistkiesel ​

19. ### geistkieselValued Senior Member

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2,471
Both observers wilkl nmeasure the same velocity for the speed of light. The train observer, however, will mneasure a relative velocity of frame and photon less than the speed of light.

One doesn't need meter sticks and clocks as you have described.

On the moving frame, when at rest, place detectors equal distances from a fixed point on the frame. Emit a series of pulses to what will be the "forward" detector". Then repeat the the trests with the "rearward detector". AS the frame is stationary and the distances are the same, the time of arrival for both pulses will be identical.

Now, accelerate the frame in the "forward" direction. Emit a series of pulses in the rearward direction. When the pulse has moved a distance ct the light strikes the 'oncoming' detector. Twenty thousand such tests confirm the time of detection is the same for all emitted pulses. Now, repeat the tests directing the pulses to the forward detector. When the pulse has moved a distance ct, the detector has moved distance vt, where v is the velocity of the moving (once accelerated) frame realtive to the embankment, which is verified by the observer on the moving frame. The detector is now 2vt from the light pusle trying to catch up with the moving detector. In order to catch the detector the light must move an additional distance 2vt, plus the distance the frame moves in the interim time t', or vt'. Therefore, after the forward pulse has moved a distance ct, it will strike the detector after moving an additional distance ct' = 2vt + vt'. The additional time for the light to strike the moving away detector is therefore, t' = t(2v)/(c -v). This time is certainly different than the time calculated using SRT, but the analogy is clear.

However, Kevinalm, look at the condition from a purely classical perspective for just an instant. What is the difference in the two conditions where in one instance the frame is at rest wrt the embankment and the other where the frame is moving wrt the embankment?

Place mirrors at the detector positions and measure the roundtrip times for the pusles in the moving and at rest conditions. We see immediately a condition where the reflected light pulses lose their state of symmetry. Throughout the trajectory of the pulses the light beams, in the stationary test, are moving in opposite directions at all times . They reflect at the same time on each mirror. The forward/rearward reflection tests need not be conducted at the same time. The clocks will determine the time differences. Even if you assume that the clocks dilate in time, they do so equivalently. In other words, we are not looking at absolute time difference, but merely that there are some measureable time differences.

In the moving case, when the rearward pulse has struck the rearward detector (after moving a distance ct) and is reflected back toward the source, the forward pulse is still moving forward and is a distance 2vt from the detector. For a small instant of time, t' actually, both pulses are moving the same direction which accounts for the time difference of the round trip of the light motion in the two conditions of the test frame. This is the loss of light pulse symmetry of motion. One cannot deny that the embankment observer will observe what was just described

Simultaneity is not an issue here.

Geistkiesel ​

20. ### kevinalmRegistered Senior Member

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993
Except that experimentally, the observer _doesn't_ measure any time difference. This holds across many, many different types of experiments, not just Michaelson interferometers. We seem to detect isotropic propagation regardless of the velocity of the apparatus. (inertial motion of course)

edit>> Just for clarity, our experiments show that an observer moving with the apparatus always measures the "relative velocity of frame and photon" as you term it, to be isotropically c.

Last edited: Jul 25, 2006
21. ### dav57Extraordinary Thinker ThingyRegistered Senior Member

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2inquisitive,

I think you're wrong about mass attracting the ether though. Try thinking about it the other way; mass / ether repulsion. This fits in much better, I find.

22. ### Neddy BateValued Senior Member

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1,675
You have shown two different expressions in two different posts. I assume one of them contains a typographical error. I would be interested in seeing how you derived your equations, because I came up with something different.

Here is my derivation for "Geistkieselian Time Dilation":

d = train's length
v = trains velocity

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

The time required for a light beam to propagate the train's length (based on a light beam travelling toward the front of the train)
t<sub>front</sub> = d / w<sub>front</sub> ......................3

The time required for a light beam to propagate the train's length (based on a light beam travelling toward the rear of the train)
t<sub>rear</sub> = d / w<sub>rear</sub> ........................4

Substituting expression 1 into expression 3
t<sub>front</sub> = d / (c - v) .....................5

Substituting expression 2 into expression 4
t<sub>rear</sub> = d / (c + v) ......................6

With the train at absolute rest in the embankment
v = 0
t<sub>embank</sub> = d / c .........................7

The ratio of expresssion 5 to expression 7
t<sub>front</sub> / t<sub>embank</sub> = (d / (c - v)) / (d / c)
t<sub>front</sub> / t<sub>embank</sub> = (1 / (c - v)) / (1 / c)
t<sub>front</sub> / t<sub>embank</sub> = c / (c - v) ..............8

The ratio of expresssion 6 to expression 7
t<sub>rear</sub> / t<sub>embank</sub> = (d / (c + v)) / (d / c)
t<sub>rear</sub> / t<sub>embank</sub> = (1 / (c + v)) / (1 / c)
t<sub>rear</sub> / t<sub>embank</sub> = c / (c + v) ..............9

Soving expression 8 for t<sub>front</sub>
t<sub>front</sub> = t<sub>embank</sub> * c / (c - v) ..............10

Solving expression 9 for t<sub>rear</sub>
t<sub>rear</sub> = t<sub>embank</sub> * c / (c + v) ..............11

Averaging the two different "time dilations" in expressions 10 and 11
t' = ( (t * c / (c - v)) + (t * c / (c + v)) ) / 2
t' = t/2 * ((c / (c - v)) + (c / (c + v))) ..............12

Checking expression 12 against the absolute rest frame in which v = 0
t' = t/2 * ((c / (c - 0)) + (c / (c + 0)))
t' = t/2 * ((c / c) + (c / c))
t' = t/2 * (1 + 1)
t' = t/2 * 2
t' = t

Testing expression 12 with v = 0.866c to compare with relativity theory
t' = t/2 * ((c / (c - 0.866c)) + (c / (c + 0.866c)))
t' = t/2 * ((c / (0.134c)) + (c / (1.866c)))
t' = t/2 * (7.462 + .536)
t' = t/2 * 7.998
t' = 3.999t
(Different from relativity's predictions of t'=2t, but interesting none-the-less)

Last edited: Jul 27, 2006
23. ### kevinalmRegistered Senior Member

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993
Neddy,
Geist's t' equations aren't transforms. They are a time difference calculation based on classical ether theory, for the difference in the forward and reverse transit times. Most would use delta t for this, in fairness I don't know how to insert a greek character either. For what it is, it is likely correct. (assumes ether, universal time, no lorentz contraction, etc.)

From what I can gather he doesn't believe the experimental evidence for isotropic propagation in all inertial frames. At least that's my impression, I don't like to speak for others, and possibly he's trying to express something else.