Wang / Sagnac Devices
in [Relativity]
|
Prev: Relativity ring problem - what shape is this?
Next: BUY CHEAP TEXTBOOKS | College Textbooks | Used Textbooks |
From: tominlaguna on 23 Oct 2009 08:59 On Thu, 22 Oct 2009 16:44:03 +0100, "Androcles" <Headmaster(a)Hogwarts.physics_p> wrote: > ><tominlaguna(a)yahoo.com> wrote in message >news:bqs0e5lqmuqtjqft1lvurh8ui21i974qp0(a)4ax.com... >> On Thu, 22 Oct 2009 22:49:58 +1100, "Inertial" <relatively(a)rest.com> >> wrote: >> >>> >>><tominlaguna(a)yahoo.com> wrote in message >>>news:0qc0e5dhbaf4vu8qhib1s6omos1cv3e8m6(a)4ax.com... >>> >>>As I understand, ballistic theory means light behaves light a ballistic >>>particle .. so it's velocity in some frame is the sum of the velocity of >>>source and that of the light wrt the source. >> >> That is correct to this extent: If the observer has relative motion >> with respect to the source, he will experience that the light at c +/- >> v, where v is the value of the relative motion. >> >>>Similarly light would reflect >>>from a mirror at the same speed as it hits the mirror .. so if it hits a >>>mirror at c+v, it will leave the mirror at the same speed. >> >> Almost correct. For example, in the situation where a mirror is >> moving normally toward a source at velocity v, the mirror will >> experience the light as arriving at c + v. Upon reflection, the light >> will be traveling at c + 2v with respect to the source; and, as you >> state, at c + v with respect to the mirror. >> > > >With source and reflector relatively at rest, the ray reflects at >(MINUS) -c, just as a bouncing ball would with a perfectly elastic >collision. >With the bat/mirror swinging at -v toward the source, the >approach velocity is c+v, the return velocity relative to the >pitcher is -c-v. > >That is, the ball leaves the pitcher at c and leaves the bat at -c-v. >One can imagine the ball leaving the bat at -c and the bat moving >at -v, so the ball is caught by the pitcher at -c-v. There is no 2v, >you cannot add v twice. >Just imagine the bat is moving backwards at v = c, so that c-v = 0. >You would not add v twice for that, would you? The ball >would have to go through the bat if you did. > Wrong. This is the same situation you have with Doppler radar.(#) A car is speeding toward the Doppler radar gun parked aside the road. The formula used by the gun's software to calculate the speed of the car is: f = fo(1 + 2v/c*COS(theta)) where: v*COS(theta) is the component of the car's velocity toward the gun. f is the frequency measured by the gun fo is the frequency emitted by the gun Guess where the "2" came from? __________________________________________ (#) C. L. Andrews, "Optics of the Electromagnetic Spectrum", (1960) page 30.
From: bz on 23 Oct 2009 08:20 tominlaguna(a)yahoo.com wrote in news:8qm2e594ha17fniorfc2fjktli1n1f03b3(a)4ax.com: > On Fri, 23 Oct 2009 02:32:10 +0000 (UTC), bz > <bz+mspep(a)ch100-5.chem.lsu.edu> wrote: > >>tominlaguna(a)yahoo.com wrote in news:bqs0e5lqmuqtjqft1lvurh8ui21i974qp0@ >>4ax.com: >> >>> Almost correct. For example, in the situation where a mirror is >>> moving normally toward a source at velocity v, the mirror will >>> experience the light as arriving at c + v. Upon reflection, the light >>> will be traveling at c + 2v with respect to the source; and, as you >>> state, at c + v with respect to the mirror. >> >>Easily tested by experiment: >>a) Two parallel mirrors, moving toward and away from each other (one >>attached to the voice coil of a loud speaker, or plated onto a surface >>of a quartz crystal). >>b) laser beam bouncing back and forth between the mirrors many times. >>If the bounce is n times, then the final velocity of the light exiting >>from the pair of mirrors should >>be c+n*v and c-n*v > > That is a very interesting concept. I will try to model it and see if > it can be done easily. I am thinking the result might be c+/-2n*v. Correct, if both mirrors are moving, toward or away from each other. Then they would have a peak 'closing speed' of +/- 2v. I was only thinking of moving one of them but my wording was ambiguous. -- bz please pardon my infinite ignorance, the set-of-things-I-do-not-know is an infinite set.
From: bz on 23 Oct 2009 08:46 tominlaguna(a)yahoo.com wrote in news:hbi2e51p85ia4o310h1gi8oph2l5rbbb25@ 4ax.com: > On Fri, 16 Oct 2009 16:08:07 +0100, tominlaguna(a)yahoo.com wrote: > >> > [snip] > > Tom Roberts has prompted a refinement to my description of the Sagnac > experiment which I have incorporated below: > > Now let's refer to the actual Sagnac diagram which is located at: > http://commons.wikimedia.org/wiki/File:Sagnac-Interferometer.png > > Now we'll follow the clockwise beam from start to finish and numerate > each step for later identification: > > 1. Light is emitted from source O at c with respect to that source. ..... > 16. After refraction, Beam T passes through mirror j and after a > second refraction proceeds toward telescope L at speed c. > 17. It is at mirror j that Beam T and Beam R are mixed to produce > interference fringes. Why should there be fringe_s_ when the 'line-of-sight' distances traveled are identical for beams traveling in both directions? There should be a fixed phase relationship between the two beams in the ballistic theory. > 18. Beam T arrives at telescope L at c since there is no > "line-of-sight" relative motion between mirror j and telescope L. > 19. Beam T proceeds down the telescope and arrives at the photographic > plate PP' at speed c since there is no "line-of-sight" relative motion > between telescope L and the film at PP'. > 20. The diagram properly shows that counter-clockwise beam R arrives > before the clockwise beam T since it has traversed a shorter optical > path length; made shorter due to rotation. You just jumped from the rotating frame of reference "along the 'line-of- sight'" to the fixed frame of reference, which is the frame in which the distances traveled are different. Naughty, naughty! The path lengths do NOT change as measured along the 'line-of-sight' distances traveled in the rotating frame of reference, so unless you have some magic Wilsonian effect to insert, you seem to be out of luck. In the rotating frame of reference, in the ballistic theory, the _only_ place that you can look for any difference between rotating and non rotating saganac apparatus is the coriolis effect, and if I remember correctly, that is insufficient in magnitude to account for the observed saganac effects. > 21. The reader can reconstruct the path steps in a similar manner for > the counter-clockwise beam. > > Summarizing: > > A. The Ballistic Theory of Light has 2 Postulates: (1) Light is > emitted at c with respect to its source and (2) light is reflected a c > with respect to the mirror image of the source. > B. With the Ballistic Theory of Light, the beams of light traverse > the optical circuit at speed c in each direction since there is no > "line-of-sight" relative motion element-to-element. AND no change in distance traveled along that line of sight. > C. The angle of incidence equals the angle of reflection at all > points of reflection. > D. The counter-clockwise beam arrives sooner than the opposing beam > because it has a shorter optical path length to traverse. bzzzt. wrong! > E. The Ballistic Theory of Light is compatible with the Sagnac > experiment. Sorry but you seem to be wrong. -- bz please pardon my infinite ignorance, the set-of-things-I-do-not-know is an infinite set.
From: Dono. on 23 Oct 2009 10:02 On Oct 22, 11:11 pm, tominlag...(a)yahoo.com wrote: > > >Bottom line: such an emission theory is refuted by Sagnac's observations, as I said. > > Wrong. > Math says that you are an idiot.
From: doug on 23 Oct 2009 11:31
Jonah Thomas wrote: > doug <xx(a)xx.com> wrote: > >>Jonah Thomas wrote: > > >>>When I started to pay attention to problems in Einstein's original >>>paper various people pointed me to a later Einstein paper which they >>>said was better written and which cleared things up. If we've had a >>>hundred years to learn better how to derive SR and how to teach it, >>>and we are no better than Einstein in 1920, then something is very >>>wrong. >>> >>>In the last hundred years explanations have been found for the >>>various self-contradictions and failures of SR, so the problems have >>>either been fixed or covered over. >> >>There are no contradictions or failures of SR in its domain of >>applicability. There was nothing to fix. >> >> There's no need to pretent there were never any >> >>>problems if the problems have been resolved. >> >>Point out a problem is you think there was one? > > > To you? Whatever for? You would deny it and argue that it never existed, > you would misunderstand and confuse the issue. I have better things to > do than argue with fanatics. > So you have no problems to point out. I just wanted you to be clear on that. > And anyway it's of purely historical interest. Einstein's mistakes are > no more important than Darwin's mistakes or Newton's mistakes provided > they've been fixed. What mistakes did Einstein make that needed to be fixed? |