Look at Einstein’s book, chapter XX, last paragraph. He has the train accelerating towards a gravitational source.
This is absolutely not correct. The light in the Sagnac interferometer is traveling at its fixed speed c relative to the receiver. That is the fundamental tenet of relativity. Because the Tx and Rx are both immobile in the interferometer (in the same FR), as the Rx moves towards or away from the light, so does the Tx, and the light still has the same distance to travel from Tx to Rx. The distance does not change during the flight time of the light pulse.
If I send a pulse of light to a car, and the car is moving away from me, yes the distance changes, so the pulse will take longer to get there. The two cars are not in the same FR.
If I am in a car doing 50 mph and send a pulse of light at a car doing 50 mph in front of me, the distance does not change, so the light will take the same time as if we are both stationary (which according to SRT we both are- we are both in the same FR).
The comparison between the cars and the interferometer is not valid.
I’m honestly not sure what you fail to comprehend here. The interferometer rotates, such that the co-propagating light pulse has to ‘catch up’ with it, while the counterpropagating pulse is intercepted. If you don’t want to believe me, take it from wiki:
Or from here:
From here:
Or from any other place the effect is discussed.
Yes, the interferometer rotates. As in my example with the cars both doing 50 mph, the light pulse has no catching up to do. Both cars are at the same speed, so there is no relative motion. If the cars are 300m apart, the light pulse takes one micro-second to travel the distance between them - regardless of their motion relative to another frame. It also takes one micro-second to travel in the opposite direction. That is basic SRT.
In the interferometer, both Tx and Rx are at the same speed, so there is no relative motion. If the interferometer path is 300m long, light takes one micro-second to travel between Tx and Rx - regardless of their motion relative to another frame. It also takes one micro-second to travel in the other direction. That is basic SRT. According to relativity, the two pulses should arrive at the Rx together. They do not.
In my previous posting I said “The comparison between the cars and the interferometer is not valid.”
I should have said “Your comparison between the cars and the interferometer is not valid.”
It’s actually got nothing to do with SRT at this point – the simple fact is that in the one microsecond, the interferometer rotates a certain angular distance, such that the light source is no longer at the same point it was when it emitted the light. To the point of origin, both pulses would take the same time; but since the source is no longer at that point, one pulse has a longer path than the other.
In any case, I’ve provided you with three independent cites – oh heck, I’ll just throw in another two:
from here, and:
from here. This is about all I’m going to say about the issue.
If you want to continue deluding yourself that, despite having been shown wrong about elementary notions again and again, you somehow have a better grasp of relativity, or physics in general, than the experts do – well, be my guest.
“Jerk” is the third derivative of position/first derivative of acceleration. Does it play any role in the SRT?
Those links tell me nothing that you haven’t already said. They refer to an arbitrary FR, and measure the speed of light from that.
You have been arguing that the speed of light is the same for all observers all through this discussion, and that is as you should, as it is the basic tenet of SRT. Now you abandon that belief when it suits you and refuse to discuss the matter any further. This statement “…the simple fact is that in the one microsecond, the interferometer rotates a certain angular distance, such that the light source is no longer at the same point it was when it emitted the light. To the point of origin, both pulses would take the same time; but since the source is no longer at that point, one pulse has a longer path than the other…” is only true as seen from the arbitrary reference frame, so if you are making the speed of light relative to that frame you are absolutely correct, and that FR is in effect the aether. You cannot, however make that claim as a relativist, because SRT says that c is WRT the observer/receiver. As seen from within the interferometer, the light source does not move, and neither does the receiver. You claim that it has nothing to do with SRT at this point, and that is simply because the statements I am making about SRT are correct, so you try to remove SRT from the argument.
This question is open to anybody. What is wrong with the following statement?
If I am stationary and send a pulse of light to a car which is moving away from me, the distance increases during the transit time of the pulse, so the pulse will take longer to get there than if the car was stationary. The two cars are not in the same FR.
If I am in a car doing 50 mph and send a pulse of light to a car doing 50 mph in front of me and traveling in the same direction, the distance does not change, so the light will take the same time as if we are both stationary (which according to SRT we both are - we are both in the same FR).
‘Arbitrary reference frame’ means you’re free to choose any frame of your liking, and thus, means the same as ‘for all observers’ (since every (inertial) observer defines a reference frame). If there were a reference to an ether, these things wouldn’t hold in arbitrary frames, but would depend on the choice of frame – namely, things would be different in the special frame singled out by the ether. That a statement holds in arbitrary frames means there is no special frame.
Anyway, I’ve realized that I’m no longer in this because I believe, or even genuinely hope, that you’re going to learn anything from this discussion, and haven’t been for a while. So it’s best for me to leave things here. I sincerely hope that eventually, you’ll find the humility to consider that maybe it’s not everybody else that’s wrong, and go and learn some serious physics; but if not, well, it’s certainly possible to live one’s life without an understanding of relativity, after all, people did so for thousands of years.
I have no idea who actually said what in the above quotes, please learn to love the quote function .
But the glaring mistake is that, in the case of the interferometer, Tx and Rx are not in the same reference frame. If they are both rotating about a common point, then at no point in time are their velocities or the acceleration they are experiencing identical. Therefore they must be assumed to be in different reference frames.
The only way they could possibly have identical velocities and acceleration is if they occupied the same point in space. Even if that were the case, if you picked any other point on the paths the light beams travelled, those points again have different velocities and acceleration.
Well, ideally, the ‘transmitter’ and ‘receiver’ (or what I think tomh4040 means by those designations) are indeed at a common point (the light is detected at the same point with respect to the interferometer geometry that it is emitted from). The picture on wiki is very helpful: both the blue and red arrows, representing pulses of light, start from the same point, the location of the light source when the light is emitted; they then travel around the interferometer, which itself rotates, such that their point of origin moves during their travel time; eventually, they both reach that point again, the red one before the blue one. This difference in propagation time causes a phase shift, which in turn leads to a shifting interference pattern. That’s about all there is to the effect.
Hmm, your point about no two points on the ring being in the same frame made me think of something. Basically, the thing is that the ‘co-rotating frame’ of the interferometer is not an inertial frame, so the light going around the ring in either direction goes at different velocities – c + rw and c - rw, if w is the angular velocity of the interferometer’s rotation. Perhaps that’s where tom sees a conflict with relativity – after all, according to relativity, light should always move at c! But that’s of course only true in inertial frames, i.e. those corresponding to linear, constant motion. A rotation is non-inertial (as you can tell by the existence of pseudoforces), so the speed of light differing from its vacuum value isn’t a problem. Another way to put this is to observe that unlike translational motion, rotation is absolutely detectable, i.e. it’s not ‘relative’; the Sagnac effect is precisely one way to perform this detection.
I was trying to keep My explanation based in “real world” situations with a physical transmitter and screen to show the interference pattern. Yes the rotation means we aren’t in an inertial frame, the entire apparatus is undergoing constant acceleration. Don’t know how that effects c in this specific situation as I’m at work atm. My understanding has always been that the speed of the 2 beams can be considered identical, but the paths taken differ. Hence the discrepancy (or basically what you said :))
I was surprised by this, but sat down and worked through the numbers for the case of a mass of 10kg, moving 5m/s bashing into a mass of 1kg, not moving at all. The difference of the speeds after collision was still 5m/s.
Next, I solved it in general terms, as above, and, yep, that’s what falls out.
I had had it in my mind that the bowling pin flew away from the bowling ball much faster than the bowling ball was moving, but, apparently not so!
This is why I’m sticking with this thread; I’m learning stuff!
The Tx and Rx are both in the same reference frame because they are solidly connected together. By that I mean that a rigid measuring rod can be bolted to both. Their distance apart will not vary over time or with velocity or acceleration. They are demonstrably in the same FR. Light generated in the interferomer travels at c not only WRT the Rx, but WRT the whole of the interferometer. When the interferometer is rotating, nothing changes internally, so the light is still measured as c WRT it.
I thought you had learned something when I saw that first sentence. In that sentence you are agreeing with me that that Tx and Rx are indeed in the same FR (they could be at a common point, but don’t have to be), therefore there is not and cannot be any relative motion between the two. Unfortunately you put in the word “ideally” - the word, if any is needed, is “actually”. Then you go back to your old way of thinking and say that the point of origin moves during the transit time (of the light). The point of origin does not move relative to the point of recepion during the travel time because the two points are fixed together. Look back at the example of the two cars which can be stationary or moving at the same velocity. They are in the same reference frame. You appeared to have grasped that concept.
Anyway, I’ve realized that I’m no longer in this because I believe, or even genuinely hope, that you’re going to learn anything from this discussion, and haven’t been for a while. You were nearly there, but you are so engrained with SRT, that you overlook an obvious truth and go back to dogma.
I sincerely hope that eventually, you’ll find the humility to consider that maybe I have been correct in a lot (even most, but I admit not all) of what I have posted. You are again confusing not agreeing with relativity with not understanding it. You don’t agree with aether theory therefore you don’t understand it.
Well done! Will you now look at other things I have posted with your newly opened eyes?
Especially look at the cart collision on Wiki, where if you turn it through 90 degrees anti-clockwise, each time the small mass bounces it bounces higher than the previous bounce, which I sincerely hope I am not alone in saying is impossible.
By the way, the bowling pin appears to move much faster than the ball for the following reason. The pin is for the sake of this explanation 400mm tall but only 40mm wide. When the ball hits the pin, it contacts the pin at a distance from the pin’s centre of mass of 20mm. The pin then rotates rapidly as it flies towards the back wall, and contacts the back wall when it is horizontal (either top or base first) not vertical. The centre of mass has been moving at the same speed as the ball, but with the pin horizontal when it hits the wall, the centre of mass is still 200mm away, so if the wall was 800mm away from the standing pin, the pin, on contact with the wall, has travelled only 600mm, not 800mm. This gives the illusion of the pin moving faster than the ball.
I notice that nobody contradicted my example - if the cars are both stationary or both doing 50 mph they are in the same FR. This is so self evident it defies contradiction.
Why then, is it so difficult for some of you to accept that in the interferometer, the Tx and Rx are also in the same FR, and there could be no fringe shift if SRT was correct. The excuse of this example being outside the remit of SRT is a cop-out, and well you all know it. It only became “outside the remit of SRT” after it was discovered that SRT when applied correctly predicts no fringe shift. HMHW was vociferous in his insistence that SRT could explain it, until the penny dropped, and he realised it could not, then he used the fall back excuse, and now refuses to take part in any further discussion.
The distance in this case is irrelevant. They are moving at different velocities AT ALL TIMES. This means by definition that they are NOT the same reference frame. If you disagree, please show me a citation that two points with different velocities can be considered to occupy the same frame of reference in Special Relativity.
Even if you were using a theoretical build, where Tx and Rx were at exactly the same point in space, as I mentioned above the two beams of light must pass through points in space which do have differing velocities. And as HMHW pointed out, because the entire structure is undergoing rotational acceleration, the whole shebang cannot be considered an inertial frame in any case.
The Tx and Rx cannot, and do not, move at different velocities with respect to each other at any time and no matter what velocity or acceleration they are undergoing collectively relative to another frame. They are rigidly fixed together. They are in the same FR precisely because they are rigidly fixed together, so their relative velocity is zero. They may have differing velocities relative to another frame, for instance if that frame is rotating, or if the interferometer is rotating. If you are using that frame as a reference, then in effect that frame becomes the aether.
Please explain to me how they are moving relative to each other. Please also explain how the beam of light “must pass through points in space which do have differing velocities” when that light is generated in the interferometer and never goes outside it. It is, according to SRT, traveling WRT the Rx, and the Rx is fixed rigidly to the interferometer, so the light is also WRT the interferometer.
If they’re rigidly fixed to one another, then, say, the receiver can still rotate around the transmitter. The transmitter’s velocity would be 0, while the receiver’s wouldn’t be.
I knew this would be your answer. Unfortunately it just goes to underscore your complete misunderstanding of some of the most basic laws of physics and motion, yet you are trying to argue against SR.
Ok back to basic physics. Firstly Velocity is a vector quantity. This means it has both a magnitude and a direction. Now get yourself a pencil, pair of compasses, a ruler and a piece of paper. Are you sitting comfortably, then we’ll begin…
1: Plot a point on the paper. Call it Tx
2: Draw a circle whose circumference passes through Tx.
3: Draw a tangent to the circle through point Tx. This shows you the direction of travel of Tx.
4: Pick any other point on the circle, for the sake of obviousness make it 90deg around the circle from Tx. Call it Rx.
5: Again draw a tangent to the circle through Rx. This shows you the Direction of travel of Rx.
6: Smack your forehead if you finally realise that they are going in different directions and hence have different velocities.
The only reason they don’t go flying off is the fact they are fixed to the body of the interferometer. This body continuously imparts an acceleration at 90deg to the direction of travel of both Tx and Rx. Thusly if Tx and Rx have different velocities WHICH THEY DO, then the acceleration is also different as it is ALSO a vector quantity.
As to how the light and Tx are moving relatively to each other; Tx is constantly accelerating, therefore once a photon has left Tx it is no longer subject to this acceleration. In a physical interferometer if you dont use mirrors, fibre-optic cable or some other means to change the lights direction it will never reach Rx.
In fact the Body of the device actually imparts two different kinds on rotation on Tx and Rx. They not only rotate about the cetral axis of the interferometer, they are also forced to revolve around their central point of fixture so that the maintain the same attitude wrt to the centre of the device. It’s somewhat analogous to the moon being tidal locked to the Earth, but for completely different reasons.
If Tx and Rx were free to rotate about their own axis, then they would actually always point on the same direction, regardless of the angle they have rotated around the device. In such a situation an observer on Tx who started out with Rx in his direct line of sight, would see Rx move across his field of vision in the same direction as the interferometer is spinning. This would be visual comfirmation that Tx and Rx are NOT in the same reference frame.
I have never closed my eyes to anything you have said. I have worked as hard as possible to give you every credit.
Half Man Half Wit backed his point with equations, which I was able to solve for specific cases, and this was what persuaded me that I had been in error in one of my assumptions.
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.
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.