Johnny5

This will be useful to me. I have never actually seen a michelson interferometer' date=' only the very loose discussion involving (c+v), (c-v). I've seen the derivation before, but there is an error in it (the derivation). At any rate, about umm 5 years ago, I read a book which I have, which went into detail about the device used by Fizeau to measure the speed of light, and somewhere else I read that the Michelson interferometer was quite similiar to the device used by Fizeau. You say that the "sagnac effect" is equivalent to the "michelson interferometer." That isn't exactly what you mean, so what do you mean? PS: I can find the book again, if necessary. As I recall... 1) Fizeau wanted to measure the speed of light experimentally. (Really the speed of light in an earth frame... since speed is a frame dependent quantity) He had access to something that could spin very fast, at a known rpm. I think the way the rpm was known was an ingenious one, but I don't recall the details right now. There were cogs or something. And mirrors were involved. Now, the way things worked, light was aimed at a distant stationary mirror. So it went back and forth a known distance. And obviously that takes some amount of time. And measuring that amount of time was the hard part. As I recall, Fizeau made his measurement in the 18th century, i think 1751 or something. I will google it in a moment. Then around the same time, someone else used his technique to also measure the speed of light, and came up with a similiar measurement. Now, Michelson did his experiments in the late nineteenth century, I think 1889, I could be wrong about that, it might have been 1898, I'll check. But the point is, the experiments were done roughly one century apart. So exactly how did Michelson modify the original device used by Fizeau? if you know, and how do you relate a Michelson interferometer to the Sagnac Device, which appeared in 1913, approximately a decade or so after Michelson's work? Regards PS: Here is one source dating Fizeau's experiment: Fizeau's measurment of speed of light And the relevant quote: Yep, a wheel with teeth. And the date was 1849. It was Foucalt who did it slightly later 1862, and Albert Michelson slightly later at 1880. According to this source. Here are two adjacent diagrams, comparing Fizeau's technique to Foucalt's: Measuring light speed Oddly, the link above has the date of Fizeau's measurement as 1851, disagreeing with the 1862 value given at the other site. Here is a site devoted to a timeline of electromagnetism/optics: Timeline of Electromagnetism and Optics So all three experimenters performed their experiments in the nineteenth century. Here is a quote from the site above: Fizeau had control over the rotation rate, and adjusted that until... ... the time it took the wheel to move the width of one tooth, was equal to the time it took light from a candle to travel a distance of about 10 miles, 5 to a mirror and 5 back. It's hard to imagine this experiment actually being done, one would think the intensity of the light too small. It's not like he had a flashlight. Here is a link which shows how Fizeau incorporated lenses into the experiment, as well as commenting on the fact that one mirror was partially transmitting, partially reflecting, and another was "fully silvered." I've read that before. The article above, says that the wheel had 720 teeth. 360+360=720 Hence fizeau wanted his mathematical analysis to be simple. He thought ahead. Reason being, 360 has a lot of integer factors. Right now I am reading about Michelson's 1927 Mt. Wilson experiment. I saw a picture of a giant iron interferometer too.

A hall probe measures magnetic field strength.

Suppose that in some inertial reference frame, some object is at rest. Thus, its speed is zero in the inertial reference frame. By Galileo's law of inertia, the center of mass of that object will remain at rest in the frame, for so long as no external force acts upon it. now, suppose that you want it to jump in some direction instantly, say one mile whatever. Is such a feat possible? Seems to me that's what you guys mean when you say "teleportation of matter" Well I can say this much: Suppose something external does act upon it at moment in time t, to accelerate the center of mass of the object in frame S. Then there will be two consecutive moments in time, one at which the CM of the object was at one position, and the very next moment in time, the CM would not be there anymore, it would have jumed some distance. So in a sense, what you are calling teleportation not only is possible, it is necessary. It has to happen in order for anything to move. What is not possible is for the distances jumped to be arbitrary. How far the CM jumps in the frame will be a function of several variables. The inertial mass of the object for one. And also, the magnitude of the applied external force for two. And I can make the answer depend upon other variables as well. Regards

I happen to have one of these, but geistkeisel also seems to know that there is a problem with SR, perhaps via some other argument. Possibly their argument is based upon the Sagnac effect, which is exactly what I am now investigating. I should have the answer in less than a week. Although it appears to me as though their argument, however it runs, is based upon linearizing the Sagnac effect, I plan to do it both ways. Why would you need such an argument though? Regards

Let logbx = m, and let logba = n. Therefore: x=bm and a = bn So now, take the mth root of both sides of x=bm to obtain: x1/m = (bm)1/m =bm/m = b So we now know that: b = x1/m and a = bn, so by substitution it follows that: a = (x1/m)n And multiplying the exponents it follows that: a = xn/m Now, raising both sides to the power m, it follows that: am = xn But go back one step to this result here: a = xn/m If we raise both sides to the nth power we obtain: an = xnn/m Therefore, the only way that: an = x is if nn/m=1 From which it follows that m cannot be equal to zero. let it be the case that not (m=0). Now, multiply both sides by m to obtain: nn=m From which it follows that: n2=m From which it must follow that n is equal to the square root of m. That is: n = m1/2 So the line you circled in red is not unconditionally true, it is conditionally true. If logbx = m, and logba = n, and an = x then n = m1/2 if logbx = m, and logba = n, and not(n = m1/2) then not(an = x). Regards

Geistkiesel, You seem to know alot about the Sagnac effect, you have made me aware of it, which in and of itself is an achievement, so I thank you for that. I am going to keep working on it, I like viewing physics from a historical persepective, something I didn't get a chance to do when studying it to pass an exam. It's refreshing to go about it this way. I don't have any intelligent questions yet, so I am not going to make up one just to ask you something, but eventually I will have one, which I will need help answering so stick around, and be nice to Dr Swanson. Also.. if you don't mind me asking, are you male? If that is too personal don't answer it. Kind regards

Can we please not fight? I want to hear what geistkeissel has to say about the sagnac effect, I do not want to see them kicked off the forum by having their IP address permanently blocked. I think an apology is in order, they know better than to act like that, since their level of competency is so high. This person can be asked to stop nicely and will. I know things about the future, I am good like that. So no more attitude please. Swansont answered your objections quite professsionally. I am simply interested in any knowledge that you or he might possess. And he has been published in the IOP so your insulting him certainly isn't furthering your career. Please don't let this childish thing go on, I would like to understand the Sagnac effect, and see it as you do. And I want to read his responses as well. Antagonist/protagonist is good, but not to extremes. Regards to all

I don't think you are going to get a general concensus on the meaning of "Planck length." I myself am not sure how to correctly interpret it yet, and I've been working on it. Your question is centered around something existing and being smaller than the Planck length, but there is another possibility you might consider. Changes in the position of the center of inertia must occur in jumps, over consecutive moments in time, and the Planck length is the smallest amount through which the center of mass of something can jump, when that something's motion is being viewed in an inertial reference frame. Thus, Planck length would tie into the meaning of inertial reference frame, and possibly have nothing to do with actual size of objects. But as I said, I haven't decided what it means yet. Regards

Classical electrodynamics is an example of a field theory, and gravity can also be formulated as a field theory. Gauss' law is a consequence of the mathematics of a field theory, in general it stems from the mathematical nature of an inverse square law, as Meir pointed out. So the answer is yes. Regards

Well I cannot edit my old post, so I will pick it up here. I want to at least derive their formula for t`. I said that if you're willing to define speed in the spinning frame then you can obtain their result. The speed of the photon in the spinning frame will exceed the speed of the photon in frame F. So the question now is, what is the speed of the photon, in the spinning frame, using variables already introduced? Certainly, distance divided by speed has units of time. And the photon is to be moving faster in the spinning frame, than it is in frame F, and in frame F, its speed was denoted by c, where c = 2pR/Dt0 So now the question is how much faster. The answer obviously depends on the tangential speed vt. Suppose that the tangential speed was also c. In that case, the emitter would have traveled through a semicircle in frame F, at which moment in time, the counterclockwise moving photon would have also traveled in a semicircle in F, and they would have met in half the time it takes the counterclockwise moving photon to go full circle, which would thus have be: Dt0/2 The way this works, is as follows: There are two other frames here, frame G, and frame H. Frame G is the rest frame of the photon. Frame H is the rest frame of the detector. A little more information has to be added because this is a particularly complicated problem, in that both frames G,H are in uniform circular motion in frame F, and frame F is the inertial frame. Let the z axes of frames G,H be perpendicular to the orbital plane. Let the origin of frame G be permanently located where the counterclockwise moving photon is, and let the center of the record player be permanently situated at a distance R away from the origin, on the positive axis. Similarly, let the origin of frame H be permanently located where the detector is, and let the center of the record player be permanently situated at a distane R away from the origin of frame H, on the positive x axis of frame H. Frames G,H are both non-inertial frames. Question: What is the relative speed of the photon, to the detector? Whatever the answer is, the answer is constant in time, because of the way the problem has been set up. Now, there will come a moment in time at which the distance between the photon and the detector is equal to 2R. Initally it is zero, then after some unknown amount of time the distance between them is 2R, and then at moment in time t2, the distance again is zero, and this is the amount of time we are interested in. The detector and the photon begin to separate at moment in time t1, and then the meet up at moment in time t2, and A.G. Kelly used t` to denote that amount of time. But at the moment in time at which they are a distance 2R apart, the velocity vectors point in opposite directions. ---------------------------------------------------------------------- ---------------------------------------------------------------------- I made a mistake up above, and it carried on down below. In the above work, when I say photon, I meant the counterclockwise moving photon. And so, at the moment in time at which the counterclocwise moving photon, and the detector are a distance 2R apart their velocity vectors point in the same direction. The point is there are two photons in the setup, so by my just saying "the photon" you cannot follow the work properly. I was thinking of the counterclockwise moving photon first. ---------------------------------------------------------------------- ---------------------------------------------------------------------- In frame F, the speed at that moment in time of the photon is c, and the speed of the detector is vt. I will fix the work below, by changing "photon" to "photon A". Photon A is the counterclockwise moving photon, and photon B is the clockwise moving photon. From a simple diagram, we can infer that at the precise moment in time at which the distance between photon A and the detector is 2R, the relative speed between them is equal to c+v. ___________________________________________ And now because I have fixed the error, the statement above is false. At the moment in time at which the photon and the detector are a distance 2R apart, their velocity vectors have the same direction, and so if they were in linear motion, instead of circular motion, the speeds would do the opposite. in other words, if at this moment in time, the motion were converted to linear instead of circular, the two things would depart slower than c, namely c-v. I think I'm correct this time. Now it is plain to see though, that as the photon swings back around, when it finally passes through the detector, its velocity vector at that moment in time will point exactly opposite to the velocity vector of the detector, and so at that moment in time the relative speed will be c+v. I think this has totally fixed the problem. It would pay to have a more systematic approach to this, but for now this will have to suffice. At any rate, A.G. Kelly was not as clear as they should have been. Since I know that the motion, both of the photon and the detector is uniform circular motion in frame F, it follows that I can express the velocity vectors of both objects in reference frame F, using sines and cosines quite nicely. Then there will be a way to transform into the rest frames of photon A, and the dector, and define the relative speed properly, for all moments in time. The main thing, is that the relative speed (photon to detector) is not constant in time, as the counterclockwise moving photon moves around, being greatest just when it leaves the detector, and the moment before it is about to pass through after having made its first orbit. It decreases on its way to the 2R point, and then increases through the 2nd half of its journey. The best way to discuss this, would be mathematically.

Yeah neither could I.

Revision? You mean review. It's going. I have been gathering some books on other topics, and reading. There are some other things I am working on first. But obviously I have to regain the linear algebra information, so that has to happen. But I want it done right. I have gone through the Gauss elimination method many times. That's not a problem, and honestly I never forgot how to do it. And I remember how to multiply matrices together too. The thing I wanted to work on next, was knowing what the answers mean, after you get the coefficient matrix into row echelon form(i think thats what its called zeros everywhere ones down the main diagonal). Like, suppose you are dealing with equations for planes. You are in three dimensional space. And you get something inconsistent, like 0=2, after you did all that work. What exactly does that mean? That the planes are parallel? And also, if underdetermined... I forget how you write the answer. Something like (3,2,4) + r(0,0,1) you get these arbitrary scalars in your answer. I forget that. I was hoping to find my original textbook, but I still haven't located it. Oh and I was reading up on eigenvectors and eigenvalues of a matrix just yesterday. It will come back, but i've been reading up on some things in real analysis lately. Convergence of series, that kind of thing.

I have been reading the article on the circular sagnac effect, and I think the following formula in the article is wrong: The Sagnac Effect and Uniform Motion t’ = 2pr/(c + v) (equation 3 in the article) I am only just beginning to think about it now. Multiplying both sides by (c+v) gives: t’ (c+v) = 2pr Distributing amount of time t`gives t’c+vt’ = 2pr The problem I have is with the RHS. The photon hasn't really travelled in a full circle of radius R. And the reason is certainly clear, the disc is spinning. To help me understand this better, I am thinking of a tiny marble, on a frictionless disc of ice, with no friction whatsoever. Then the disc starts spinning, as if someone turned a record player on. Because there is no friction, the marble remains at rest in an outside frame. Yet to an observer anchored to the spinnin disc, the marble "appears" to have travelled in a circle. But, a frame fixed to the ice disc, is a non-inertial frame. In particular, if an observer fixed to the ice disc completely disregards external objects (which too appear to be orbiting) The marble appears to have a centripetal acceleration. But the net force on the marble is zero. So something strange is going on in the physics of this. You can see it as you switch from the record frame, to the inertial frame, and back again. I would be the first to argue that you can do physics in any frame, inertial or otherwise, but you need to have transformation formulas. At any rate, right now I am only concerned with their equation (3). Here it is again: t’ = 2pr/(c + v) (equation 3 in the article) The writer also says that t' is the time on the spinning disc, as if time is different there, as opposed to measuring time using a clock, at rest in an inertial frame, which clock is necessarily external to the disc. t’c+vt’ = 2pr I think the simplest way to clear up the problem is to use the precise definition of "relative speed" which ironically is an absolute in all frames. You know, there is the same problem with measurements made on earth, as is going on on the record player. The stars don't stay fixed in the sky. I can do this correctly, but I just need to think for a few moments carefully. Let me refer to the external frame, as frame F. When the motor is off, the disc isn't spinning in frame F. Now, suppose that the following experiment is performed in frame F. Somehow, a photon is fired, and made to go around the circumference of the "record player" in time Dt0 as measured by a clock at rest in frame F. Maybe frame F is inertial maybe not, but that is the measurement by the clock. Now, using a "meter string" the circumference of the record player is measured to be: 2p R So we have two solid measurements in the frame, and we define the ratio of them to be the tangential speed of the photon in frame F. That is: vt= 2p R/Dt0 Now, we can relate the tangential speed, to dq/dt as follows: vt= Rw w = dq/dt Ok firstly... The circumference of the record player is stipulated to be an absolute in all frames of reference. Now, if we turn on the record player, and the marble remains at rest while the record player turns, after some time T has elapsed (as measured in F), the marble will again coincide with the starting point (which is just a mark dug into the record). WE could also give the marble a "push," and if there is a lip around the record players perimeter, the marble would travel in a circle in frame F (instead of remaining at rest in frame F), in some amount of time which could be measured by a clock at rest in F. So these are the kinds of measurements we can make. The case of the Sagnac device is interesting for the following reason: Not only is the photon traveling in a circle in frame F, but the disc is spinning as well. Ah HAH! Ok so now, in symbolic imitation of the article, let us do this: c = 2p R/Dt0 Now, that isn't necessarily 299792458 meters per second, it is whatever quantity is obtained upon taking the ratio of the measurements (as already described). I chose the letter c, because the author of that article used the letter c for that ratio. At least this way, I can check what they are doing. So when the device isn't spinning you get whatever you get for c, as measured by devices at rest in frame F. Now let me glance at the article. Their very next step is to simply write the following formula: t’ = 2pr/(c + v) (equation 3 in the article) I would much prefer to write things in terms of dq/dt, and attempt to derive that formula, rather than simply pull it out of thin air. For what it is worth, that is the first time that symbol v is brought into their argument, and the author calls it "speed of a point on the periphery of the disc." If you are rigidly attached to the disc, then all points on the disc have no speed relative to you, so that v which they introduced, has to be the speed of a point fixed to the edge of the disc, and that speed is defined in frame F, which is external to the disc. And that speed is appropriately the tangential speed of the disc, in frame F. So forget about the photon, and focus on the disc. The center of inertia of the disc is at rest in frame F. But a fixed point on the periphery of the disc completes one revolution in T seconds. Thus, the tangential speed of a periphery point in frame F is given by: vt= 2p R/T = Rw where w = dq/dt Now, what is really going on is this... At the moment in time that the photon has completed one circular revolution in frame F, a fixed point on the perimeter has advanced through some total angle Q And the amount of time this happened in, is the amount of time it took the photon to make one revolution, in frame F, which is the period of the photon. The period of the photon in frame F was denoted by: Dt0 This is different than the period of the spinning disc, which was denoted by T. Now, the moment in time at which the event began was denoted by t1. And, the moment in time at which the photon completed one revolution in frame F was denoted by t3. And so the amount of time of event [t1,t3] is Dt0 Now, the tangential speed of the disc was denoted by vt vt= 2p R/T = Rw And w = dq/dt Therefore: vt= 2p R/T = R dq/dt So, let us stipulate that the tangential speed of a fixed point on the record player is constant. (this has nothing whatsoever to do with the photon) So we have this: vtdt= 2p R/T dt= R dq Now, we can integrate from moment in time t1=0, to moment in time t3=t1+ Dt0 The total change in angle, the integral of dtheta, has been denoted by: Q So, we have: (2p R/T) Dt0 = R Q Now, I need to connect this to what they call t`. Here is their quote: So t` is the amount of time it takes the counterclockwise moving photon to travel from the detector around and back to the detector. And this has to be less than the amount of time Dt0, you can just figure that out from their diagram. So my question now, is how do I figure it out. In other words, I now need the time of the following event [t1,t2] using the variables already introduced. I see very clearly their solution, but I am not the type to use something I cannot either outright understand, or derive myself. What bothers me is that they have the distance traveled as being 2piR, but that isnt the case. There is something significant here about spinning frames, but I can'tquite put my finger on it. If you permit yourself to switch between the two frames, then you can make sense out of things, but you have to be very careful about using the definition of speed. I'msure if you've already solved the problem it's a piece of cake, and I've solved this one before, probably over 15 years ago. Its been so long I have to redo it.I think it was in a book on classical mechanics by Kleppner/ Kolenkow, a book used at MIT. They solved this or something similiar. It probably didn'tmake sense the first time, which would explain why it didn't keep. If you are willing to define the speed of the photon, in a frame fixed to the record player, then the photon will have a greater speed in that frame, then in frame F. And the reason is obvious. You will use 2piR for the distance traveled, and the same amount of time as measured in frame F for event [t1,t2] but in frame F, the distance traveled by the photon is less than 2 pi R, in the same time, as passed in the spinning frame. And hence the speed of the photon in frame F, will be less than the speed of the same photon in the spinning frame. But of course, this has turned distance traveled into something meaningless. Special relativity is the least of my concerns right now. Well let me see if I can at least get their formula for t`.

Is anyone here familiar with George Boole's work entitled "Treatise On The Difference Calculus" I was wondering if it was any good.

What is moving at 8 kilometers per second, and in what frame? Regards

I can't find the derivation. Can you send me a link in which the off axis solution is derived?

Ok will do. Thanks

It was Max Born's idea. Schrodinger thought that psi was connected to energy density, but Max Born suggested that psi times the complex conjugate of psi represented the probability that the electron would be in some region of space about the proton of a hydrogen atom. Regards Here is a link to Born's probability wave

Not quite. As in if X is a moment in time, and Y is a different moment in time, and X before Y, then in all reference frames, at all moments in time, X before Y.