geistkiesel Posted May 21, 2005 Posted May 21, 2005 The proof of the absolute zero velocity inertial frame is focused on the simple statement that the inertial frame should be proved invariant in absolute space and time for some delta t > 0. This proof is not subject to agreement by any number of scientific thinkers as the proof is transcendent to the mental dynamic of the signifcance of any "acceptance" protocol. The use of two widely understood postulates of light iare incorporated in the proof First is the postulate that the motion of light is independent of any motion of the soutrce of the light and second, the speed of light is a constant c measured form any inertal frame of reference,and included in these postulates are the understanding that, an emitted photon will travel in a straight-line trajectory, potentially eternally, or until acted upon by and ouitside force and likewise, light travels equal; distance in equal times. As stated above we need but discover some delta t where a localized point in space is spatially invariant for sonme delta t >0. Two photons are emitted simultaneously in opposite directions from each other. The motion of each photons expanding wave fromt moves frome the emission point P such that P remains the continuously defined invariant point P. In deep space far from perturbations of nny kind, the P remains defined for the duration of the longevity of the invariant straight-line trajectories of equal length of the two photons motion distance. One of the photons, say the one moving to the left, wrt P, is reflected after moving a distance ct 180 degree back along the outgoing trajectory. After the completion of this second ct leg of travel, the L photon has arrived back at the point P and if the Right moving photon R were reflected at the same time as the L photon the photons would arrive at P sinmultaneously. However, either , or both of the photons can be used to localize and find P at any time. Assume the frame is moving, wrt the embankment, and coincidentally moving wrt P. The reflector/ clocks are located equal distances from P on the Left and right ends of the inertial frame. When the photons are emitted simultaneously from the emission point the L photon heads to the in coming L clock and is recorded arriving at L after moving a distance ct, wrt the point P . L is reflected back another distance ct and is located at point P instantaneously and is located a distance 2vt (v unknown at this instant) from the physical midpoint of the photon sources which is heading away from L (P is also 2vt from the physical midpoint at the same instant L has moved 2ct) After the initial distance of ct traveled by the R photon it is located the small distance 2vt from the R clock as the L photon was later after moving a distance 2ct. Here R is heading towrd the on coming physical midpoint of the moving frame a distance 2vt + 2vt' from the physical midpoint (L is 2vt away from the physical midpoint here where both photons have moved a distance 2ct wrt P. containing the reflector and emiission point and finally the photons converge simultaneously at the physical midpoint measuring the total time difference of the round trip of the photons as t' wrt the stationary and moving conditions of the test. It is coincidental that the point P can be referenced in the moving or stationary frame of reference withouit changing the results of the photon motion tests.The small distance the photon must travel crossing the 2vt distance, that added on vt', accounts for the final frame motion after moving the inital distance 2vt when the photons were moving the distances 2ct each. The coordinated motion of frame and photon are classically derived and beautifully coordinated. Rememebr the t' and v expressions are all measured on the moving frame. All motion and times are measured wrt the constant invariant motion of the photons moving in the directions indicated, or when the frame was at rest wrt the embankment. What is the measured velocity of the speed of light wrt to the physically constructed zero velocity emission point P, as maintained and accounted for in space and time without error or deviation? The answer is the speed of light wrt P is c, eternally, at least in the local conditions observed here. What could a moving observer O have to consider to vary the results stated here? Absolutely nothing. O sees events as they unfold, not before. Therefore , any considerations that O might have regarding the state of motion of the moving frame is purespeculation and will have to wait until the significant data is gathered in one spot and analyzed. First, O sees the simultaneous emission of the photons at the physical midpoint of the frame and nothing else, yet. Because he is unable tio distinguish any motion in his train compartment (or space ship) his assumption that he is at rest wrt to the embankment is premature at the instant the photons were emitted. The photon arrives at L before the R photon arrives at R, by a time t'. This data only reaches O after the photons arrive back at the physical midpoint of the moving frame, simultaneously. The data is time tagged with the arrival times at L and R as well as the arrival time of the photon a time t' greater than that when the test is done in the stationary frame of reference. the t' s derived from ct' = 2vt + vt' the distance the photon has to cover to arrive at he moving clock froma a distance 2vt from the clock. t' = t(2v)/(c - v) If t' = 0 there is no motion, If t' > 0 motion is assured. This shows a measurable delta time for velocities much less than SR could ever hope to measure, Likewise, the classical t' here is much more selective than the SR time dilation formulae in units of velocity. This is seen in the v/(1 - v) factor that increases in the t' expression (assuming the unit speed of light) faster than at lower velocities than the SRT gamma. Velocity ,then is seen as v = ct'/(2t + t') A reminder to every one still with us here right now: the postulate of light that assures us of the independent motion of light, assuming no velocity components of the moving frame of treference and the measurment of the realtive velocity iof the speed of light with respect to the interial frame of reference the zero velocity of the invariant point P. are the attributes of light motion that allow for thias omple classic analysis of light motion.. Now what does any observer have to do with changing these results? Nothing,absolutely nothing. All earth bound meaurements of relative velocity of object and embankment always is preceeded by an intial acceleration of the object, never the embankment, that provides the velocity later termed as the 'relative motion' of object and frame and embankment. For massive objects the embankment refrerence frame is a perfect absolute zero velocity reference frame. Motion measured wrt to any point on the surface of the planet can be used as reference point. Any measurment of frame andf photons wrt the embankment can be cnducted with any degeree of accuracy desired or budgeted for, The motion of all points on the surface of the embankment can be corrected for in any measuremnt. A point moving on the equater for instance is moving at a speed of .464 km/sec. A photon of light that is measured wrt the moving earth frame will have it turning motion 360 degrees /(24x 3600) or at a rotation rate of .004 degrees/sec. Any error due to acceleration effects of the embankment can easily be detected. Rotational motion without acceleration affects do not distract from the inertial characterstics of the rotating frame of reference, as there are no acceleration effects measurable on the embankment surface. and similalry for the orbiting SATs that can and are treated classically as if the frames were purely inetial. Likewise, in accord withe zero velocity frame of reference described her in simplicity all planetary motion significantly affecting any physical result can be corrected for by eliminating the acceleration affects due to earth motion when and if such accelerations are ever found. The plan was to describe a reference frame in no measurable motion v > 0 as defined by the independency and constancy postuilates governing the motion of light. The emitted photons moving in a straight line wrt P at constant and exactly equivalent speeds proves the invariance of the point P wrt anything in the universe. P i s a unique position in space to the exclusion of all other points in the physical universe. That the rest motion is defined by the motion of light itself will incite some to claim the negation of the absolute zero velocity of the point P. But these are the lost sheep we return to their classical fold. Here is a summary below of three gedankens discussed in the literature not in terms of the Sagnac effect, but as a bases for supporting the SRTin terms of simultabneity for the most part. Those assumptioon of frame at rest is the bugaboo of SRT. The ssoner one sees theat SRT is purely a negation of the conceptf physical motoons. How else to meaure the relative sopeed fliught of ftrame and motion as C when the frame us preseumed at rest, i.e. the motion is negated by SRT theory. I am especially fond of the example B whre the moving Observer sees the forward photon before the one arriving from the rear. Like the desciption in the example of the proof the Observer can make no rational conclusion regarding the assumptions of the moving observer's frame of reference.. We assume the moving observer has conducted thousands of tests like the one described here so we aren't going to allow some technically ignorant geek to make the assumption of motion before the datas is acquired. This is a basic protocol of experimenntal physics . Physical conclusions are made form the experimental results of physical data. acquisition. Be aware that the example gedanken below are constructed with a shorthand in miond. The last stage of the photon moption in case A is expanded to indicate the physical nature of vt, vt' and ct and the invariance of the point P. Like some have suggested in this forum, study the classical view and you shall be free.
Johnny5 Posted May 21, 2005 Posted May 21, 2005 I'm reading through this right now. You might want to know that you only have six hours to edit your post. Right now I have read up to here: If t' - 0 there is no motion, If t' > 0 motion is assured. I think you mean: If t' = 0 there is no motion, If t' > 0 motion is assured. You might want to check for any other tiny errors, since you only have six hours to edit your post. I'm still reading. Regards PS: Ok I've read the whole thing through once. Now, I am going back carefully. Consider where you write the following. The use of two widely understood postulates of light iare incorporated in the proofFirst is the postulate that the motion of light is independent of any motion of the soutrce of the light and second' date=' the speed of light is a constant c measured form any inertal frame of reference,and included in these postulates are the understanding that, an emitted photon will travel in a straight-line trajectory, potentially eternally, or until acted upon by and ouitside force and likewise, light travels equal; distance in equal times.[/quote'] Am I to understand that this is a proof by contradiction? I just need to make sure I follow you. You want me to start off with the standard set of assumptions of SRT, and then reason, and then arrive at a contradiction? Yes or no? Assumption 1: Motion of light is independent of any motion of the emitter Assumption 2: Speed of light must be measured as c=299792458 m/s in any inertial frame of reference. Assumption 3: Unless influenced by an external force, a photon/wavefront will travel in a straight line at a constant speed... equivalently... equal distances traveled in equal times. Assumption 3 is built into the concept of inertial frame of reference, and really isn't an additional assumption, but rather something incorporated into the definition of an IRF. You might want to add to the list of assumptions/Postulates of SRT, that the laws of physics are the same in any inertial frames, but then again that is tacitly assumed anyway. I'd leave it out, since people will just assume it without being aware that they assumed it. Ok, the next thing you say is this: As stated above we need but discover some delta t where a localized point in space is spatially invariant for sonme delta t >0. To make sure I follow, here you are talking about there being a point, a place in space, which is "spatially invariant" for some "change in the universe." In other words a place which doesn't move, is that what you mean by "spatially invariant" or do you mean something else? I presume you are saying this is connected to the idea of "absolute rest," which of course is the subject of this thread in the first place. You are saying, i think, that you need to show that there is a place in space that doesn't move, yet other things did, because dt>0. Correct me if I'm wrong, or just add comments. Just trying to follow you here. Next you say this: Two photons are emitted simultaneously in opposite directions from each other. The motion of each photons expanding wave fromt moves frome the emission point P such that P remains the continuously defined invariant point P. In deep space far from perturbations of nny kind, the P remains defined for the duration of the longevity of the invariant straight-line trajectories mof equal length of the two photons motion distance. You are now describing the linear Sagnac experiment, and the paragraph above is related to the attached diagram. To paraphrase you: There is an emitter at rest in space, in an inertial frame of reference, and it is about to emit two photons simultaneously in this inertial frame, one to the left, the other to the right (see diagram). Actually, I'm not sure if your emitter is at rest or not. It now seems like you intend the point P of space to be at rest, quite independently of the emitter, and that what happens is that there comes a point in time where two photons are emitted simultaneously, and that happened at location P of space, and you have location P at absolute rest relative to the center of the universe. Am i right here? From the looks of your diagram, the experimental setup is moving from left to right, relative to point P, which is at absolute rest. I understand now. Next you say: One of the photons, say the one moving to the left, wrt P, is reflected after moving a distance ct 180 degree back along the outgoing trajectory. After the completion of this second ct leg of travel, the L photon has arrived back at the point P and if the Right moving photon R were reflected at the same time as the L photon the photons would arrive at P sinmultaneously. However, either , or both of the photons can be used to localize and find P at any time. Ok so... there is a photon moving to the left wrt P, and after time t has passed in the appropriate frame, it strikes a mirror and is reflected 180 degrees, so that it is now moving back towards P. Now, the frame in which t is measured is stipulated to be an inertial reference frame. Therefore, by one of the postulates of SRT, it must be the case that the speed of this photon will be measured to be c, in the IRF. Using the postulate of SR that the formulas of physics are the same in all inertial frames, it must be the case that the definition of speed formula is true, hence: c = D/t Thus, the distance traveled by the photon in time t is ct, as you say. Now, during the time it took the leftgoing photon to go from point P to the mirror, the rightmoving photon traveled identical distance, in identical time, since by postulate speed of all photons in any IRF is c. There is only one frame here, so that amount of time is t. So let me carefully track the events here, in order. Let t0 denote the moment in time at which the photons were emitted, and let t1 denote the moment in time at which the leftgoing photon strikes the mirror. So the following event took time t, in the appropriate inertial frame: [t0,t1] During that event, the leftgoing photon traveled a distance ct relative to P, and also, the rightgoing photon traveleed a distance ct relative to P, as in your diagram 1. The initial state was labeled 0. Now, it appears to me that the experimental device is traveling to the right, relative to point P, with uniform constant speed v, throughout the experiment. So at the moment in time at which the leftgoing photon strikes mirror L, namely moment in time t1,the rightgoing photon does not strike mirror R simultaneously, because both mirrors have advanced some nonzero distance vt, to the right relative to P. You say: After the completion of this second ct leg of travel, the L photon has arrived back at the point P So there is a third moment in time at which the photon reflected from mirror L returns to the point P in space. Denote this instant by t2. So, during the event [t1,t2], the photon which struck mirror L must travel an identical return distance, in an identical time. This is important to see. Moment in time t2 corresponds to your diagram labeled 2. That is the instant at which the photon which struck mirror L coincides with point P. Certainly, the speed of the photon must be c, since the rest frame of P is inertial, and we have assumed einstein postulate 1. Thus, the time of event [t1,t2] is also t, like during the first half of its trip away from P, and its speed on the return journey is again c, therefore the distance traveled by that photon on the return trip satisfies: ct=D All as you say. The very next thing you say is this: and if the Right moving photon R were reflected at the same time as the L photon the photons would arrive at P sinmultaneously. However, either , or both of the photons can be used to localize and find P at any time. Correct... AND IF the initially rightmoving photon hit mirror R simultaneously to the moment in time the other photon hit mirror L then it would simultaneously arrive at P with the other photon, yet the photons did not hit the mirrors simultaneously in the P frame, because the experimental device was moving to the right with nonzero speed v in that frame. The photon that initially was moving right strikes mirror R after the other photon struck mirror L, in the inertial rest frame P. Next, you say this: Assume the frame is moving' date=' wrt the embankment, and coincidentally moving wrt P. The reflector/ clocks are located equal distances from P on the Left and right ends of the inertial frame. [/quote'] Now I have to think about this for a second. There are two frames here, the rest frame of P, and the rest frame of the moving mirror device thing. Since you say refer to this frame as moving with respect to P, i presume you mean the rest frame of the Sagnac device. You go on to say: When the photons are emitted simultaneously from the emission point the L photon heads to the in coming L clock and is recorded arriving at L after moving a distance ct, wrt the point P . L is reflected back another distance ct and is located at point P instantaneously and is located a distance 2vt (v unknown at this instant) from the physical midpoint of the photon sources which is heading away from L (P is also 2vt from the physical midpoint at the same instant L has moved 2ct) Right. At the moment in time that photon L has returned to P, photon L is now a distance 2vt from the physical midpoint of the Sagnac device. That's simple to see from your diagrams. Its also conceptually easy to see. The time it took that photon to hit mirror L was amount of time t, t measured in rest frame P. The time it took that photon to travel back to P was amount of time t, as measured in rest frame P. Total time of travel of Sagnac device t+t=2t. The speed of the Sagnac device was constantly v, during event [t0,t2]. Hence, during this event, the total distance traveled by the Sagnac device in rest frame P is given by: v = D/2t Hence D=2vt. Right. The next thing you say is this: After the initial distance of ct traveled by the R photon it is located the small distance 2vt from the R clock as the L photon was later after moving a distance 2ct. Here R is heading towrd the on coming physical midpoint of the moving frame a distance 2vt + 2vt' from the physical midpoint (L is 2vt away from the physical midpoint here where both photons have moved a distance 2ct wrt P. containing the reflector and emiission point and I think you made an error here. I believe you meant to say "located small distance vt from the R clock." Correct me if I'm wrong. After photon R has traveled a distance ct away from P, it is at that moment in time, a distance vt away from mirror/clock R. You say 2vt, but i think you meant vt. I could be wrong, I'm not sure what you said here just yet. Ok, its been about ten minutes and I still don't get the previous paragraph. So right here is where I got lost. I can see from your diagrams, exactly what's going on though. They were drawn perfectly. Let me just try to keep going. At the moment in time photon L strikes mirror L, photon R is a distance ct away from point P, a distance vt away from mirror R, and a distance ct-vt away from the physical midpoint of the Sagnac device.
geistkiesel Posted May 21, 2005 Author Posted May 21, 2005 Johnny5, Thank you for the comments. I wanted this to be a two, three paragraph proof but it got away from me. I had intended a contraqdiction proof, by assuming that no localized point in space could be determined then show the converse. Let's do that. There is no unique point in space that is possibly invariant as defined by the location of physical processes. Two photon are emitted in opposite directions from each other at t = 0. As the photons are moviong at the same speed they move the same distance in equal times. Likewsie, as the distance to P from both photons is the same the point P is defined continuously as the midpoint of the photon wave fronf expanding as independent photons. As the photons move in a straight-line until acted on by an external force, and no external force being found here the trajectory of the two photons will be maintained along the same straight-line. The distances traveled also being identical establishes the invariance of the point P. If after moving a distance ct the left photon say is reflected back to the zero point and the initial photon vector, while p has moved from the original point of emission. If the photon reflects back on the same trajectory , which by definition is a straight line, then the continued straight line must be invariant as any deviation from the straight line would mean the constancy of direction is not maintained and the straight line woul d be broken. The return to the starting point of the initial velocity vector is a return to the emission point defined as P. But P was assumed to have moved, which, by the laws of light motion the straight-line trajectory is preserved the invariance of the point P along any line drawn through the midpoint except on the photon trajectory, If motion of p was along either of the two photon;s uniique trajectory space, then at least one of the reflected photon will overrun the moved pont P before arriving back at at the start of the initial photon vector. However, the simultaneously reflected photons each equal distance from the point P must necessarily arrive back at reflected photon emission point. An invriant straight line plus the equality of equal motion in equal time insists the point P is determined for all time remaining in the universe. Even perturbations of the photon trajectory will only obscure the location of the point, bot nothing wll move the point. hence the assumption that the point P is variant is diosproved and the invarianc of the emission point of the photons is maintained. QED[/indent]
swansont Posted May 22, 2005 Posted May 22, 2005 Your definitions are far too confusing and awkward to wade through. You've assumed c is constant. Good. L__________P__________R LP and PR are equal distance. If the system is at rest, photons emitted from P will reflect off L and R and arrive back at P at the same time. Is it your contention that this is not true if the system is thought to be moving?
geistkiesel Posted May 22, 2005 Author Posted May 22, 2005 Your definitions are far too confusing and awkward to wade through. You've assumed c is constant. Good. L__________P__________R LP and PR are equal distance. If the system is at rest' date=' photons emitted from P will reflect off L and R and arrive back at P at the same time. Is it your contention that this is not true if the system is thought to be moving?[/quote'] Are you suggesting I have ever considered c other than constant c?- Basically if moving aor at ret, the photons wioll always wrrive simultaneousl bact at the phusical MP after the MP has moved a distance 2vt + vt' Absolutely not to your question. Remember the title of the thread:proof of absolite zero velocity. This means b defeinion that any point P that is at absolute rest, absolute zero velocity will reain invariant .That is a point P is at absolure rest if it is not moving. A physical frame of reference such as you have described has not coem close t the definition, probably because as you say the arguments are covoluted. So allow we to straightened them a bit, Two photons are emitted simultaneously from the emitter, The absolute xero poingt absolute zero velocity is defined and located by gthe movng pjotns, AS the phootns move in straight-line trajectories, unless acted on by external forces. AS the photons are emitted opposite to teach other the photons define an invariant line in space, The line is not moving because the photons motion is isotropis and constant in space and time. the continued motin of the photonwave trains provide a contonuous definioton and location of the midpoint of the two photons' motion. When we focus on the point P in space which from the instant the photons were emtted the physical frame becomes usless and insignificant of having any affect on the invariant point P. In general, any two moving wave trains or photons define an absolutely zero velocity oiut inb space. If after a time t one of the emitted photons is reflected back in the direction of the outgoing photon, the reflected photon will arrive at the point P after having traveled another distance ct. -- will arrive exactly at point P. Were the point P, not invariant and if the point moved then the back reflected photon would not arrive at the point P, which is not found to be tbe case. And Mr. Swansont, please make note of the claim that here, that the invatriant point P defjned by the midpoint of the moving photon wave trains is not a physical massive thing, The point P is just that an abstravcrtion defined by the constant motion of the photons. Now, to your question and your figure. The question you ask will be answered purely from the objective hand of the laws of physics and specifically various postulates of light, We will determine of your frame is at rest with respect to absolute zero velocity or if the frame is in motion. First, the photons are emitted at the physical; midpoint of the frame, the halfway point halving the LR duistance. The clocks we have installed at poihts L and R measure the arrival time of the photons at L and R. The clocks at L, P and R have been synchronized with mechanical switches set in the stationary frame of reference that start the three clocks simultaneously counting from zero. What can the observer tell us?From his position observing the photons emitted, we know absolutely nothing regading the frame motion. When the photons arrive at L and R the arrival sequence is crucial in determining the state of motion of the frame wrt absolute zero. which is the point in space,from which the photons were emitted, which if the frame is in motion, that point is definitely not the midpoint, MP, on the pysical frame , If the platform is moving from L to R direction the L clock will record the L photon arriving at L first, then the R phioton arrives at R, later. The motions of the photons then ar all directed in the direction of the physical midpoint, The Observer is restricted in reviewing the test data. When the photons arrive, simultaneously at the physical miidpoint MP (as I have renamed your point P to avoid confusion) the moving observer may now review the data and determine any states of motion of the physical inertial frame with wrt the point P. t' derived I have derived the term for t' the extra time required for the round trip of the photons when the physical frame is moving with respect to P (and, only coincidentally, moving wrt the stationary platform). We construct he t' term using he zero velocity point in space. After the L phioton has reflected back fromL, after being duly recorded there, tyeh L photon has arrived at P and the is distance 2vt from the physical midpoint MP, which is oving away. there fore the L photon must cross the distance 2vt to arrive at the MP and in this time the physical frame is moving so the photon has an extra small distance to cross vt;, to arrive at MP. simply said ct' = 2vt + vt'from which t' is determined t'= t(2v) /(c - v). Or, since t' is a measured quantity, the cvlocity of the fdrame wrt P is, v = ct'/(2t + t'). Did Swansont's emitted photons arrive simultaneously at the physical midpoint of his inertial frame moving wrt v(P) = 0. Now you answer your iown question: If t' = 0, v = o, otherwise if t' > 0 the physical frame is movinmg at velocity v wrt the invariant point P. If you trace the distances the two photon moved you will see that the photons moved the same distance in the same amount of time and ergo the photons are duty bouind to arrive simultaneously at the MP always, under the same test,conditions, though any velocity v will suffice.. If, for some unknown reason the physical frame were moving wrt P and the photons arrived simultaneously at L and R then the photons could not arrive simultaneosly at MP with anyvelocity.. This is easiest seen looking up from the point P (invariant in spatial location) and use the stationary frame as a measure. If the frame is at rest wrt v(P) = 0, then and only then will the emitted photons arrive simultaneously at L and R and if arriving simutaneousley at Land R the photons can return to the physical MP simultaneoulsy if and only if, the physical frame is at rest wrt v(P) = 0. If the frame velocity v(f) is > 0 wrt v(P) = 0 and the photons have arrived simultaneously at L and R then there is no way the photons will arrive simultaneoulsy at MP. The only way the photons arrive simultaneoulsy at L and R when the v(f) > 0 is if the photons were either not emitted simultaneously or were divereted from the stragtht-line trrajectory during their time of flight. Tricksters would be doing this Sawnsont, I suspect you have a word or two to discuss the moving oberver's perspective here and perhaps want to suggest that he "sees" the lights ariving simultaneously at the L and R clocks. Not so, I mean 'not so' he "sees" anything except the emission of the photons. The L and R clocks are sufficiently far away at any distance from from MP, such that the observer must wait until the data that includes the arrival times at L and R arruves at MP.. Any assumption that his frame of reference is at rest wrt v(P) =0 before the arrival of the data is pure speculation. Also, Swansont, and you can appreciate this more than most in this forum, if the speed fo light is deterimined wrt to v(P) = 0, absolutely, then the measured speed of light will be measured classically as c. SRT effectively negates the motion of inertial frames when making the measurement of the speed of light wrt the moving frame [the postulate remember]. SRT effectively guarantees the measured v = c for light by assuming the physical frame at rest. Now what is the result when making the measurement when the reference is really at rest? Another ball game that's what. Whether your physical frame is moving or not under the conditions you described the photons will always arrive simultaneously back at the physical frame MP. Notice from the expression for the velocity, that the MP to L and Mp to R distance is not significant in determining the velocity of the frame. parameter in the experiments or tests. Let us see what happens when the moving observer assumes a state of rest wrt the embankment. Hencanot see the photons arriving at aL and R and I have seen no analyses of this particlul;ar condition where the moving observer actually measures the photon arrival at L and R. However, the photons will arrive siultaneously back at the MP while the Observers assumptions are totally insignificant tothe coutcome. Th e moving observer will measure thge round trip times of the L anad R photons he will idscover thediscrepenacy inround trip arrival times compared to the test done in the stationaary frame of referfence . The discrempancy is in the measuresd tr' time whoich SRT has tried to purloin as its own when SRT claim the clocks run slower on the moving frame and that accounts for the t' > 0, when this clearly is not the case.
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