# My Problems With Relativity **URL:** \ **Category:** Factual Questions\ **Created:** [January 20, 2007, 7:48pm UTC](https://boards.straightdope.com/t/my-problems-with-relativity/388750 "2007-01-20T19:48:38Z")\ **Posts on this page:** 20\ **Page:** 23
**Author:** ![BunnyTVS](https://avatars.discourse-cdn.com/v4/letter/b/977dab/32.png) [@BunnyTVS](https://boards.straightdope.com/u/BunnyTVS)\ **Post date:** [March 2, 2012, 12:45am UTC](https://boards.straightdope.com/t/my-problems-with-relativity/388750/441 "2012-03-02T00:45:49Z")
> [@tomh4040](#): > > That is correct according to a different frame, but not according to the frame the Tx and Rx are in. No measurement between the Tx and Rx will show any movement. > > Smack your forehead if you finally realise that the two objects have no relative motion. Same answer as above to HMHW. They are only going in different directions relative to another FR. Just to humour you I drew out your circle etc, and rotated it. To my extreme surprise, the Tx and the Rx stayed in their exact locations (WRT the paper) that they were in before the rotation. Thusly the Tx and Rx have the same velocity WHICH THEY DO (which is zero). > Unfortunately your answer just goes to underscore your complete misunderstanding of some of the most basic laws of physics and motion, yet you are trying to argue for SR. > You are also forgetting that The Sagnac Interferometer works even if the axis of rotation is outside the interferometer itself. > > Then please give me credit for being correct about the bowling ball/ pin velocities, and the fact that the Wiki cart example is wrong - as proved by turning it on its side. If it was correct, we would have perpetual motion. I’ll answer the second part (re the diagram on the wiki page first). In that diagram the difference in velocity between the two objects prior to the collision is +5 (5 - 0). After the collision the difference in velocity is -5 (4.999 - 9.999). Therefore prior to the collision they are closing with a velocity of 5, after they are separating with a velocity of 5. This is exactly in accordance with the preservation of momentum _in a perfectly elastic collision_. [Here’s one I prepared earlier.](http://imageshack.us/photo/my-images/710/syghg.png/) \*Yes it’s imageshack, frankly it’s all this converstion deserves. As you can see Tx and Rx have _different velocities and accelerations_ if they are in differing parts of the apparatus. Remember [Newton’s First Law,](http://en.wikipedia.org/wiki/Newtons_laws#Newton.27s_first_law)Law I: Every body persists in its state of being at rest or of **moving uniformly straight forward** , except insofar as it is compelled to change its state by force impressed. And as Newton states, velocity is always in a straight line _unless a force acts upon the body_.This means that, if they were free to move at the point in time shown on the diagram, Tx and Rx would head off in the directions shown by the arrows VTx and RTx. They only reason they don’t is that they are mounted in place. This creates a centripetal force accelerating them towards the centre. This is reflected in the way the combination of velocity and acceleration due to gravity keep the moon in orbit around the Earth. So please show me how two points with velocities at 90deg to each other can have the _same_ velocity. If you do think two objects travelling at right angles to each other have the same velocities, congratulations, you’ve just proved Newton wrong. ---
**Author:** ![nate](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/nate/32/13085_2.png) [@nate](https://boards.straightdope.com/u/nate)\ **Post date:** [March 2, 2012, 3:13pm UTC](https://boards.straightdope.com/t/my-problems-with-relativity/388750/442 "2012-03-02T15:13:26Z")
**tomh4040** , what is your definition of velocity? ---
**Author:** ![tomh4040](https://avatars.discourse-cdn.com/v4/letter/t/ee7513/32.png) [@tomh4040](https://boards.straightdope.com/u/tomh4040)\ **Post date:** [March 2, 2012, 11:44pm UTC](https://boards.straightdope.com/t/my-problems-with-relativity/388750/443 "2012-03-02T23:44:01Z")
> [@Great\_Antibob](#): > > You really aren’t getting the point. Even if the distance between Tx and Rx remains exactly the same, there is still a velocity difference. > > Think of two kids on a merry go round. Even if they maintain exactly the same distance apart, they’re spinning about a common point, which means they experience different velocities (even different accelerations), hence they are not in the same non-inertial frame of reference. > > And, yes, a measurement will show “movement”. It will show the spinning. So, even if the distance between Tx and Rx remains the same, the spinning itself will register differently between the two. > > This is basic high school Newtonian physics, not relativity. They do not “experience” different velocities. I put experience in quotes because a while ago on this forum, I was derided for using that phrase, even though it is convenient shorthand. Only when looked at from a different FR will they show different velocities. Any measurement **internal** to that merry go round will not show that it is spinning. Look at it from the fairground and you will see that it is spinning. Look from the merry go round to another part of the fairground and you will see it is spinning - or more precisely, you will see the rest of the fairground spinning. Look at where you like on the merry go round, but do not look outside it, and you cannot tell it is spinning. This is what led Ptolemy to believe that the Earth was the centre of the known universe, as everything was spinning round it. According to your line of logic, here on Earth we are all in different FRs. [QUOTE : BunnyTVS] I’ll answer the second part (re the diagram on the wiki page first). In that diagram the difference in velocity between the two objects prior to the collision is +5 (5 - 0). After the collision the difference in velocity is -5 (4.999 - 9.999). Therefore prior to the collision they are closing with a velocity of 5, after they are separating with a velocity of 5. This is exactly in accordance with the preservation of momentum in a perfectly elastic collision. Here’s one I prepared earlier. \*Yes it’s imageshack, frankly it’s all this converstion deserves. As you can see Tx and Rx have different velocities and accelerations if they are in differing parts of the apparatus. Remember Newton’s First Law,Law I: Every body persists in its state of being at rest or of moving uniformly straight forward, except insofar as it is compelled to change its state by force impressed. And as Newton states, velocity is always in a straight line unless a force acts upon the body.This means that, if they were free to move at the point in time shown on the diagram, Tx and Rx would head off in the directions shown by the arrows VTx and RTx. They only reason they don’t is that they are mounted in place. This creates a centripetal force accelerating them towards the centre. This is reflected in the way the combination of velocity and acceleration due to gravity keep the moon in orbit around the Earth. So please show me how two points with velocities at 90deg to each other can have the same velocity. If you do think two objects travelling at right angles to each other have the same velocities, congratulations, you’ve just proved Newton wrong. [/QUOTE] There are two answers to this question. As is so often the case, the argument is flawed. The two objects in question which are travelling at right angles to each other are going round in a circle. Because they come back to their starting point their average velocity is zero (see below). Also this is a matter of using the correct frame of reference. I am standing on the surface of the Earth. Some distance away stands somebody else. There is no relative motion between us. Even if we are standing on the equator and are 90 degrees apart, there is still no relative motion between us. Look at us from the moon, and yes, we will have different velocities. That is what relativity is all about. I have no problem with some aspects of relativity, that being one of them. > [@](#): > > : nate. tomh4040, what is your definition of velocity? Velocity is speed in a given direction, usually given as a vector. A car racing round a track 3 miles long comes back the start point after one minute. Its average speed was 180 mph, but its average velocity was zero. Does that answer your question? ---
**Author:** ![BunnyTVS](https://avatars.discourse-cdn.com/v4/letter/b/977dab/32.png) [@BunnyTVS](https://boards.straightdope.com/u/BunnyTVS)\ **Post date:** [March 3, 2012, 2:11am UTC](https://boards.straightdope.com/t/my-problems-with-relativity/388750/444 "2012-03-03T02:11:00Z")
> [@tomh4040](#): > > They do not “experience” different velocities. I put experience in quotes because a while ago on this forum, I was derided for using that phrase, even though it is convenient shorthand. Only when looked at from a different FR will they show different velocities. Any measurement **internal** to that merry go round will not show that it is spinning. > [@tomh4040](#): > > Look at it from the fairground and you will see that it is spinning. Look from the merry go round to another part of the fairground and you will see it is spinning - or more precisely, you will see the rest of the fairground spinning. Look at where you like on the merry go round, but do not look outside it, and you cannot tell it is spinning. This is what led Ptolemy to believe that the Earth was the centre of the known universe, as everything was spinning round it. According to your line of logic, here on Earth we are all in different FRs. Totally incorrect. Any object rotating about a point in space will “experience” an accelerating force. For a real world example, have you ever been on a merry-go-round? I have, and I guarantee you will “experience” a force. [QUOTE=BunnyTVS] I’ll answer the second part (re the diagram on the wiki page first). In that diagram the difference in velocity between the two objects prior to the collision is +5 (5 - 0). After the collision the difference in velocity is -5 (4.999 - 9.999). Therefore prior to the collision they are closing with a velocity of 5, after they are separating with a velocity of 5. This is exactly in accordance with the preservation of momentum in a perfectly elastic collision. Here’s one I prepared earlier. \*Yes it’s imageshack, frankly it’s all this converstion deserves. As you can see Tx and Rx have different velocities and accelerations if they are in differing parts of the apparatus. Remember Newton’s First Law,Law I: Every body persists in its state of being at rest or of moving uniformly straight forward, except insofar as it is compelled to change its state by force impressed. And as Newton states, velocity is always in a straight line unless a force acts upon the body.This means that, if they were free to move at the point in time shown on the diagram, Tx and Rx would head off in the directions shown by the arrows VTx and RTx. They only reason they don’t is that they are mounted in place. This creates a centripetal force accelerating them towards the centre. This is reflected in the way the combination of velocity and acceleration due to gravity keep the moon in orbit around the Earth. So please show me how two points with velocities at 90deg to each other can have the same velocity. If you do think two objects travelling at right angles to each other have the same velocities, congratulations, you’ve just proved Newton wrong. [/QUOTE] > [@tomh4040](#): > > There are two answers to this question. As is so often the case, the argument is flawed. The two objects in question which are travelling at right angles to each other are going round in a circle. Because they come back to their starting point their average velocity is zero (see below). There are more than two flaws to this answer: Firstly you are assuming the two points make a complete multiple of revolutions. No-one said that. Secondly this is only the case if, when the object reaches the starting point of measurement, it comes to rest. If it retains any velocity at all… _then it’s velocity is greater than 0_ Think about it, the only way an objects ‘average velocity’ can be 0 is if it returns to its point of origin _and stays there_. If it continues to move it cannot have a velocity of 0. Thirdly averages are, yet again, irelevant. What is important is the behaviour of Tx, Rx and the light beams _at any moment in time_ You’ve still failed to explain the facts that Tx and Rx are experiencing different velocities at each point in time (making them unsuitable as a common frame of reference), and are also experiencing a measurable acceleration (meaning they are not an inertial frame.) Please show some proof that at any point in time Tx and Rx have a relative velocity of 0. ---
**Author:** ![Exapno\_Mapcase](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/exapno_mapcase/32/1051_2.png) [@Exapno\_Mapcase](https://boards.straightdope.com/u/Exapno_Mapcase)\ **Post date:** [March 3, 2012, 4:05am UTC](https://boards.straightdope.com/t/my-problems-with-relativity/388750/445 "2012-03-03T04:05:05Z")
I can’t take it any more. It’s [quote=name] not [quote _name_] or [quote : _name_] or any other variation. And each > [@](#): > > must be paired with [noparse] [/noparse] Use Preview! ---
**Author:** ![Andy\_L](https://avatars.discourse-cdn.com/v4/letter/a/c67d28/32.png) [@Andy\_L](https://boards.straightdope.com/u/Andy_L)\ **Post date:** [March 3, 2012, 5:58pm UTC](https://boards.straightdope.com/t/my-problems-with-relativity/388750/446 "2012-03-03T17:58:36Z")
> [@BunnyTVS](#): > > Totally incorrect. Any object rotating about a point in space will “experience” an accelerating force. For a real world example, have you ever been on a merry-go-round? I have, and I guarantee you will “experience” a force. Yes, but in a small enough volume for a short enough time, the force would be indistinguishable from a gravitational field (thus returning this discussion to the equivalence principle). ---
**Author:** ![BunnyTVS](https://avatars.discourse-cdn.com/v4/letter/b/977dab/32.png) [@BunnyTVS](https://boards.straightdope.com/u/BunnyTVS)\ **Post date:** [March 4, 2012, 1:34am UTC](https://boards.straightdope.com/t/my-problems-with-relativity/388750/447 "2012-03-04T01:34:47Z")
> [@Andy\_L](#): > > Yes, but in a small enough volume for a short enough time, the force would be indistinguishable from a gravitational field (thus returning this discussion to the equivalence principle). True, however **Tomh4040** has never mentioned General Relativity or the equivalence principle wrt the sygnac experiment. My point was that the presence of this force, means that the points cannot be considered to be in an inertial frame. Therefore the argument that **Tomh4040** presents, that SR states that the two beams of light should reach the receiver in phase is incorrect. ---
**Author:** ![Andy\_L](https://avatars.discourse-cdn.com/v4/letter/a/c67d28/32.png) [@Andy\_L](https://boards.straightdope.com/u/Andy_L)\ **Post date:** [March 4, 2012, 1:41am UTC](https://boards.straightdope.com/t/my-problems-with-relativity/388750/448 "2012-03-04T01:41:47Z")
> [@BunnyTVS](#): > > True, however **Tomh4040** has never mentioned General Relativity or the equivalence principle wrt the sygnac experiment. My point was that the presence of this force, means that the points cannot be considered to be in an inertial frame. Therefore the argument that **Tomh4040** presents, that SR states that the two beams of light should reach the receiver in phase is incorrect. I know - I was just amused by the return of an issue that the equivalence principle is relevant too. ---
**Author:** ![tomh4040](https://avatars.discourse-cdn.com/v4/letter/t/ee7513/32.png) [@tomh4040](https://boards.straightdope.com/u/tomh4040)\ **Post date:** [March 5, 2012, 9:22pm UTC](https://boards.straightdope.com/t/my-problems-with-relativity/388750/449 "2012-03-05T21:22:00Z")
> [@BunnyTVS](#): > > Totally incorrect. Any object rotating about a point in space will “experience” an accelerating force. For a real world example, have you ever been on a merry-go-round? I have, and I guarantee you will “experience” a force. > > There are more than two flaws to this answer: > > Firstly you are assuming the two points make a complete multiple of revolutions. No-one said that. > > Secondly this is only the case if, when the object reaches the starting point of measurement, it comes to rest. If it retains any velocity at all… _then it’s velocity is greater than 0_ Think about it, the only way an objects ‘average velocity’ can be 0 is if it returns to its point of origin _and stays there_. If it continues to move it cannot have a velocity of 0. > > Thirdly averages are, yet again, irelevant. What is important is the behaviour of Tx, Rx and the light beams _at any moment in time_ > > You’ve still failed to explain the facts that Tx and Rx are experiencing different velocities at each point in time (making them unsuitable as a common frame of reference), and are also experiencing a measurable acceleration (meaning they are not an inertial frame.) > > Please show some proof that at any point in time Tx and Rx have a relative velocity of 0. Your arguments are irrelevant. I am - we are all - on a merry go round at this very moment while tapping away at our keyboards. In general, I am assuming nothing about complete revolutions, multiple or not. You asked me what is the difference between speed and velocity, and I answered you with a specific example. You seem to want a 20 page thesis instead of a simple answer. I have shown proof that Tx and Rx have a relative velocity of zero. London has a relative velocity of zero relative to Bath or any other point in the world you care to mention. There is your proof. If I travel from London to Bath, I can measure the acceleration of Bath can I? Or can I measure the accleration of Bath from London? Tx and Rx are only experiencing different velocities when viewed from another frame. That is called relativity. All you are (all) doing is trying to deflect attention away from the fact that relativity cannot explain the Sagnac effect. The Sagnac effect has been known for 100 years, and in that time many great minds have been unsuccessful in trying to explain it using relativity . All “explanations” have viewed the interferometer from a separate reference frame (which may or may not be within the body of the interferometer). If the speed of light is c WRT the observer/receiver, a separate FR is not required, and the effect cannot be explained. ---
**Author:** ![naita](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/naita/32/5862_2.png) [@naita](https://boards.straightdope.com/u/naita)\ **Post date:** [March 5, 2012, 9:58pm UTC](https://boards.straightdope.com/t/my-problems-with-relativity/388750/450 "2012-03-05T21:58:28Z")
> [@tomh4040](#): > > All you are (all) doing is trying to deflect attention away from the fact that relativity cannot explain the Sagnac effect. > The Sagnac effect has been known for 100 years, and in that time many great minds have been unsuccessful in trying to explain it using relativity . All “explanations” have viewed the interferometer from a separate reference frame (which may or may not be within the body of the interferometer). If the speed of light is c WRT the observer/receiver, a separate FR is not required, and the effect cannot be explained. No, **you** believe the SR explanation of the Sagnac effect is wrong because the available explanations use the simplest available inertial frame. In your attempts to convince us of the correctness of this belief you again and again give an appearance of not understanding that the Earth, or some other rotating object, is an accelerated frame, and that doing calculations in it requires you to take the acceleration into account. Early 20th century physicists accepted that the Sagnac effect was consistent with both SR and an aether completely dragged by the rotation Earth. Do you really think you’ve got a superior understanding of the issue? If you travelled from London to Bath you could measure the acceleration of the Earth, given a sufficiently accurate mode of transportation and sensitive instruments. It’s called the Coriolis effect and together with the fundamental difference between accelerated and inertial reference frames is basic physics, not Einsteinian relativity. Now, which **inertial** reference frame do you think it’s correct to do the calculations for the Sagnac interferometer in? ---
**Author:** ![BunnyTVS](https://avatars.discourse-cdn.com/v4/letter/b/977dab/32.png) [@BunnyTVS](https://boards.straightdope.com/u/BunnyTVS)\ **Post date:** [March 5, 2012, 10:37pm UTC](https://boards.straightdope.com/t/my-problems-with-relativity/388750/451 "2012-03-05T22:37:52Z")
> [@naita](#): > > No, **you** believe the SR explanation of the Sagnac effect is wrong because the available explanations use the simplest available inertial frame. In your attempts to convince us of the correctness of this belief you again and again give an appearance of not understanding that the Earth, or some other rotating object, is an accelerated frame, and that doing calculations in it requires you to take the acceleration into account. > > Early 20th century physicists accepted that the Sagnac effect was consistent with both SR and an aether completely dragged by the rotation Earth. Do you really think you’ve got a superior understanding of the issue? > > If you travelled from London to Bath you could measure the acceleration of the Earth, given a sufficiently accurate mode of transportation and sensitive instruments. It’s called the Coriolis effect and together with the fundamental difference between accelerated and inertial reference frames is basic physics, not Einsteinian relativity. > > Now, which **inertial** reference frame do you think it’s correct to do the calculations for the Sagnac interferometer in? Sensitive instruments! Bah, I could do it with a lump of plasticine, a bootlace and a pair of bathroom scales :D\* \*I make no assertion regarding the accuracy however. ---
**Author:** ![Trinopus](https://avatars.discourse-cdn.com/v4/letter/t/2bfe46/32.png) [@Trinopus](https://boards.straightdope.com/u/Trinopus)\ **Post date:** [March 5, 2012, 11:30pm UTC](https://boards.straightdope.com/t/my-problems-with-relativity/388750/452 "2012-03-05T23:30:57Z")
> [@tomh4040](#): > > . . . London has a relative velocity of zero relative to Bath or any other point in the world you care to mention. There is your proof. If I travel from London to Bath, I can measure the acceleration of Bath can I? . . . Since Bath is a bit north of London, it has less overall distance to travel in its daily trip around the earth’s axis. So, Bath moves a bit more slowly than London. The speeds aren’t the same, although the direction of travel is pretty much thesame: eastward. Points on the equator move approximately 1,000 mph eastward. At the North Pole, one doesn’t move at all. (Although one rotates!) However: someone in London is moving at right angles to, say, someone in India or someone in the heartland of the U.S. Velocity is (in modern physics) a vector, having both a quantity and a direction. ---
**Author:** ![The\_Hamster\_King](https://avatars.discourse-cdn.com/v4/letter/t/8edcca/32.png) [@The\_Hamster\_King](https://boards.straightdope.com/u/The_Hamster_King)\ **Post date:** [March 5, 2012, 11:33pm UTC](https://boards.straightdope.com/t/my-problems-with-relativity/388750/453 "2012-03-05T23:33:21Z")
> [@tomh4040](#): > > If I travel from London to Bath, I can measure the acceleration of Bath can I? Or can I measure the accleration of Bath from London? Well, if you carry an accelerometer from London to Bath you’ll notice that their acceleration vectors deviate by about 1.5 degrees. The only reason London and Bath maintain a constant distance from each other is that the difference in acceleration is offset by a difference in velocity. ---
**Author:** ![Andy\_L](https://avatars.discourse-cdn.com/v4/letter/a/c67d28/32.png) [@Andy\_L](https://boards.straightdope.com/u/Andy_L)\ **Post date:** [March 6, 2012, 1:09am UTC](https://boards.straightdope.com/t/my-problems-with-relativity/388750/454 "2012-03-06T01:09:46Z")
> [@The\_Hamster\_King](#): > > Well, if you carry an accelerometer from London to Bath you’ll notice that their acceleration vectors deviate by about 1.5 degrees. The only reason London and Bath maintain a constant distance from each other is that the difference in acceleration is offset by a difference in velocity. And if you fire a cannon from London towards Bath, you’d better account for the relative acceleration if you want an accurate shot. ---
**Author:** ![tomh4040](https://avatars.discourse-cdn.com/v4/letter/t/ee7513/32.png) [@tomh4040](https://boards.straightdope.com/u/tomh4040)\ **Post date:** [March 7, 2012, 8:08pm UTC](https://boards.straightdope.com/t/my-problems-with-relativity/388750/455 "2012-03-07T20:08:59Z")
> [@naita](#): > > No, **you** believe the SR explanation of the Sagnac effect is wrong because the available explanations use the simplest available inertial frame. In your attempts to convince us of the correctness of this belief you again and again give an appearance of not understanding that the Earth, or some other rotating object, is an accelerated frame, and that doing calculations in it requires you to take the acceleration into account. > Early 20th century physicists accepted that the Sagnac effect was consistent with both SR and an aether completely dragged by the rotation Earth. Do you really think you’ve got a superior understanding of the issue? > Now, which **inertial** reference frame do you think it’s correct to do the calculations for the Sagnac interferometer in? I believe the SR explanation of the Sagnac effect is wrong not because the available explanations use the simplest available inertial frame, but because the speed of light is assumed to be c in **that** IFR, and this then allows the (incorrect) argument to be used about the light having to “catch up with” the receiver because the receiver has moved during the transit time of the light pulse. As the Sagnac effect is consistent with an aether completely dragged by the rotation of the Earth, I do not and never have considered myself to have a superior understanding of that issue. Relativity theory tries to explain the phenomenon by introducing another FR, assuming the speed of light is constant with that one, and making arguments such as HMHW did when trying to prove me wrong in his postings of about 26 Feb. He then realised I was correct, and changed his argument to say that SRT does not apply here. I have already answered the question about which FR to use. As the speed of light according to SRT is always and without exception WRT the observer/receiver, then there is no need of another FR. Light has no “catching up” of the receiver to do, so it takes the same amount of time to travel clockwise and anticlockwise round the interferometer. No fringe shift seen, no calculations to do. The data available about the existence or not of a fringe shift can be transmitted out of the interferometer in the form of a video link into your IFR, and would show exactly the same result as someone looking at it directly from within the interferometer. But as we all know, there is a fringe shift which SRT is at a loss to explain. ---
**Author:** ![naita](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/naita/32/5862_2.png) [@naita](https://boards.straightdope.com/u/naita)\ **Post date:** [March 8, 2012, 8:45am UTC](https://boards.straightdope.com/t/my-problems-with-relativity/388750/456 "2012-03-08T08:45:58Z")
> [@tomh4040](#): > > I have already answered the question about which FR to use. As the speed of light according to SRT is always and without exception WRT the observer/receiver, then there is no need of another FR. Light has no “catching up” of the receiver to do, so it takes the same amount of time to travel clockwise and anticlockwise round the interferometer. No fringe shift seen, no calculations to do. The data available about the existence or not of a fringe shift can be transmitted out of the interferometer in the form of a video link into your IFR, and would show exactly the same result as someone looking at it directly from within the interferometer. But as we all know, there is a fringe shift which SRT is at a loss to explain. Aether theory is equally at a loss in your misconceived scenario. A rotating reference frame is not inertial and you can’t treat it as one. As there actually is a fringe shift, you must agree that there is some flaw in your description of what happens. A rotation is an objective movement in space and time which can’t be disregarded. In **any and all** inertial frames you will get the result that one path is shorter. ---
**Author:** ![tomh4040](https://avatars.discourse-cdn.com/v4/letter/t/ee7513/32.png) [@tomh4040](https://boards.straightdope.com/u/tomh4040)\ **Post date:** [March 8, 2012, 7:04pm UTC](https://boards.straightdope.com/t/my-problems-with-relativity/388750/457 "2012-03-08T19:04:06Z")
> [@naita](#): > > Aether theory is equally at a loss in your misconceived scenario. A rotating reference frame is not inertial and you can’t treat it as one. As there actually is a fringe shift, you must agree that there is some flaw in your description of what happens. > > A rotation is an objective movement in space and time which can’t be disregarded. In **any and all** inertial frames you will get the result that one path is shorter. I like the word equally that you used there. So you acknowledge that relativity cannot explain the fringe shift. Go back on the postings and read mine again. The fringe shift is very easily explained by an entrained aether theory. I have not called the interferometer an IFR, I have called it an FR. In any and all IFRs, when you assume the speed of light to be c relative to that frame, yes you will get the result that one path in the interferometer is shorter than the other. But as I have said repeatedly, in SRT, the speed of light is c with respect to the receiver/observer - no exceptions. It is therefore WRT the interferometer as a whole, and there is no “catching up” to do. The receiver/observer in the interferometer is just that - **in the interferometer**. ---
**Author:** ![naita](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/naita/32/5862_2.png) [@naita](https://boards.straightdope.com/u/naita)\ **Post date:** [March 8, 2012, 7:22pm UTC](https://boards.straightdope.com/t/my-problems-with-relativity/388750/458 "2012-03-08T19:22:45Z")
> [@tomh4040](#): > > I like the word equally that you used there. So you acknowledge that relativity cannot explain the fringe shift. > Go back on the postings and read mine again. The fringe shift is very easily explained by an entrained aether theory. I have not called the interferometer an IFR, I have called it an FR. > In any and all IFRs, when you assume the speed of light to be c relative to that frame, yes you will get the result that one path in the interferometer is shorter than the other. But as I have said repeatedly, in SRT, the speed of light is c with respect to the receiver/observer - no exceptions. It is therefore WRT the interferometer as a whole, and there is no “catching up” to do. The receiver/observer in the interferometer is just that - **in the interferometer**. What I wrote was “Aether theory is equally at a loss in **your misconceived scenario**.”. Calling that an acknowledgement from me is either blatant misrepresentation or a lack of reading comprehension. And you can’t just “call it an FR” and say the speed of light is then constant in it. It’s an accelerated frame, there’s an objective difference between it and one rotating faster, or one rotating slower. That difference results in an objectively shorter path in one direction. I’d ask you to give us this easy explanation that aether theory gives for the fringe shift, but I’m going to give up on you again instead. ---
**Author:** ![Frylock](https://avatars.discourse-cdn.com/v4/letter/f/ce7236/32.png) [@Frylock](https://boards.straightdope.com/u/Frylock)\ **Post date:** [March 8, 2012, 7:24pm UTC](https://boards.straightdope.com/t/my-problems-with-relativity/388750/459 "2012-03-08T19:24:10Z")
What is the fringe shift you guys are referring to? ---
**Author:** ![Half\_Man\_Half\_Wit](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/half_man_half_wit/32/21766_2.png) [@Half\_Man\_Half\_Wit](https://boards.straightdope.com/u/Half_Man_Half_Wit)\ **Post date:** [March 8, 2012, 8:04pm UTC](https://boards.straightdope.com/t/my-problems-with-relativity/388750/460 "2012-03-08T20:04:57Z")
> [@tomh4040](#): > > In any and all IFRs, when you assume the speed of light to be c relative to that frame, yes you will get the result that one path in the interferometer is shorter than the other. That sentence says ‘special relativity explains the Sagnac effect’. > [@](#): > > But as I have said repeatedly, in SRT, the speed of light is c with respect to the receiver/observer - no exceptions. That sentence is simply wrong. The speed of light is invariant only for inertial observers; accelerated observers may well see a speed of light different from c. > [@Frylock](#): > > What is the fringe shift you guys are referring to? The shift of the interference pattern in the [Sagnac effect](http://en.wikipedia.org/wiki/Sagnac_effect). Basically, light propagating around a rotating interferometer assembly will travel different distances based on whether it moves in the direction of rotation or against it, leading to different travel times, which causes a difference in the interference pattern. [Previous page](https://boards.straightdope.com/t/my-problems-with-relativity/388750.md?page=22) [Next page](https://boards.straightdope.com/t/my-problems-with-relativity/388750.md?page=24)