When you say “equally at a loss”, what are you comparing aether theory to, if not to relativity?
There is a big difference between a linearly accelerated frame, and a frame which is rotating. In the linearly accelerated frame, during the transit time of the light beam, assuming of course that that light is shone forward or rearward along the direction of travel, the receiver will have moved away from/towards the light source, leading to a corresponding increase/decrease in travel time. This is because the speed of the frame has increased during the light’s transit time.
A change in velocity can be either a change in speed or a change in direction. In the case of the rotating interferometer, its speed is not changing, its direction is changing. As its speed is not changing, the light beam does not have any further to travel in the rotating interferometer than in the stationary interferometer.
This is according to relativity of course.
[QUOTE=HMHW]
Originally Posted by 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’.
[/quote]
If the speed of light is c relative to that frame, and variable relative to the interferometer, then yes - but you have been arguing against that. Your argument is that the speed of light is c relative to each and every observer/receiver, and that includes the interferometer.
[QUOTE=HMHW]
Originally Posted by tomh4040.
“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.
[/QUOTE]
See my answer above to naita which contains this salient point. “A change in velocity can be either a change in speed or a change in direction”. In the case of the rotating interferometer, the (rotational) speed of the receiver is not changing, its direction relative to an arbitrary IFR is changing. As the rotational speed of the interferometer is not changing, the speed of light within it is c. If your sentence above is changed to “The speed of light is invariant for inertial observers; linearly accelerated observers may well see a speed of light different from c.” then we may be in agreement.
You say that “accelerated observers may well see a speed of light different from c”. You did not say “observers will see”. The interferometer is a good example of an accelerated frame in which the speed of light is c, and this comes about because the acceleration is causing a change of direction, not a change of speed.
Reading comprehension it is. I even bolded the part you ignored the first time, but you failed to notice it again so let me rephrase instead: If we acknowledged your misconceived scenario in which you can just say “the acceleration in a rotated frame is perpendicular to the motion so it doesn’t matter and the speed of light is c in the reference frame the whole time”, and I’d like to point out at this point that this is purely hypothetical, hence the initial “if”, it is in fact false in any explanation of the world suggested by actual physicists, but if we, for the sake of argument pretend it is correct, then we have a situation in which the entrenched aether theory and relativity both fail to explain how we get a fringe shift.
Maybe so, but if the sentence were changed that way, then it’d be wrong. The difference you’re trying to concoct here between linear and rotational acceleration simply doesn’t exist.
By Og I hope I get the quotation formatting right here. I meant to reply yesterday, but my broadband went down. On the plus side, I’ve spotted a couple of extra mistakes you’ve made. Lets see how we go…
Here on Earth we are all in different inertial reference frames as far as Einstein’s theory of Special Relativity is concerned. And please note how I bolded Special Relativity there, it will become important later. We all have different velocities, and different accelerations WRT each other.
Oh really? *Bolding below is mine, not Tom’s
Considering your two examples involve coming back to the start. Please understand why I thought you were talking about complete revolutions.
I noticed you referred to an IFR “Inertial Frame Of Reference” in that quote. Forgive me if I was wrong, and that was just a typo.
What he actually realised was that the interferometer was undergoing constant acceleration. Therefore Einstein’s theory of Special Relativity could not consider the various parts of the interferometer as a common frame of reference. Once again pay attention to the bolded parts, I promise they will become important.
Here is why I brought your attention to the bold parts in the above quotes:
Firstly you have… Look above.
Now look below…
This is why I bolded (emboldend? sounds prettier) the parts about Einstein’s theory of Special Relativity (see done it again). SRTspecifically refers to inertial frames of reference. Just in case that slipped you by…
Here’s the Wiki link! If you can’t find the specific mention, it’s in the first frikkin sentence.
So far you have shown flaws in your understanding of:
Momentum
Elasticity
Velocity
Acceleration
Special Relativity
Your previous arguments
Also you claimed earlier to have shown “proof” of your claims. No you haven’t. Proof is not, yet another, example of your perceptions of the world. We have all taken time to break down the fundamentals of physics, as much as possible, to show that you are working from a flawed premise. Now by stating that you weren’t even talking abount inertial frames in the first place, you’re basically re-defining the whole argument.
Admission here: A week ago I flagged this thread to the mods. I felt that this had moved from a GQ into a debate. I also felt that you were witnessing. In the last week your change of arguments, combined with your denial of previous positions, has done nothing but bolster my opinion.
Well, what I meant is just that the analysis carried out in the laboratory frame is valid in classical as well as relativistic contexts – you can get a Sagnac effect also with sound waves propagating in a circular, rotating cavity (or any other kind of wave: here’s an article about an experiment with electron waves propagating at non-relativistic speeds). It’s thus just that special relativity needs to be consistent with it.
“The acceleration in a rotated frame is perpendicular to the motion so it doesn’t matter and the speed of light is c in the reference frame the whole time” is not a quote of mine, and I have never said this or anything like it – rephrased or not..
You still have not answered what you are comparing aether theory to. “Aether theory is equally at a loss in your misconceived scenario” is what you said. What does “equally” refer to ?
The fringe shift is easily explained using aether theory. Here it is again :-
We will use an arbitrary IFR where the speed of light is c, and use that as reference to the interferometer, where the speed of light is not WRT the interferometer, but is WRT the IFR. The speed of light in the interferometer is of no consequence. As the interferometer rotates, the light going with the rotation has further to go than when it is stationary, due to the fact that the receiver has moved away from the leading edge of the light during the transit time of the light. The light going against the rotation has less distance to travel because the receiver has moved towards the leading edge of the light during the transit time of the light. This causes the fringe shift. The arbitrary IFR is the aether.
“Does the Coriolis effect exist in your universe?” is an example of a question designed to belittle. Of course it does. In the interferometer it is a very small effect, and does not affect the results. In all attempted explanations by relativity of the Sagnac effect for the past 100 years, it has not been mentioned. That is a recognition of that fact.
From http://www.anti-relativity.com/sagnac.htm
The Aether based explanation of this effect and the Relativity explanation are suspiciously similar:
Relativity: Though we know there is no universal frame of reference and therefore never any need for more than two frames of reference for an experiment, for the explanation of this experiment, we’re going to add a third frame of reference outside of the emitter and receiver and call it “proper time”. In regard to this third frame of reference the light beam is traveling a different distance so that is why it is out of sync.
Aether: Because the light traveled further in one direction than the other in regard to an absolute reference frame(the medium through which it traveled), the two signals are out of sync.
Here is a link to a project apparently in New Zealand to more precisely gauge the Earth’s rotation using the Sagnac Effect. The funny part is that their indoctrination into relativity is so deep that they talk about proofs of Aether without even realizing that they are doing so. They do also mention that the Sagnac Effect does cause a Doppler shift in frequency which it should not if relativity were true. Overview
I’ll give you a Link to a page with a very lengthy explanation of why I’m wrong and relativity is right. I’ll let him have 20 Pages to my 2 and let you decide what is logical. Pay close attention to the automatic hostility towards anyone not on the writer’s side. The opposing side is painted with the broad brush of “Anti-Scientific”, “Crack-Pots”. Whereas a historian can tell you that there were Nobel Prize winning scientists of that day and current day he just painted with that brush. Also notice his erroneous proclamation of no Doppler in Sagnac it’s a very important aspect.
My contention with his explanation is not that it’s incorrect exactly. I’m saying that the method he’s using excludes relativity from the equation. I have no doubt that his math is impeccable and that it reflects test results and reality. However the problem is not with the math but with the logic. Any argument of relativity that includes a third frame of reference other than the emitter and the detector is inferring a universal frame of reference. Logically he’s contradicting relativity in his explanation of how it fits with relativity. He does a great job of explaining what is really going on because the truth is that it really does travel through more space (more Aether)
This guy is actually explaining that the light is having to travel further in regard to an outside reference frame. So, in other words, he’s saying that if three equidistant planets are all traveling in one direction with light emitted from the middle one, the light traveling to the lead planet is having to go through more space than the light going to the trailing planet when we add an outside reference point. This is implying a universal frame of reference. This is implying an Aether. If there is no Aether then there is no “Proper Time” and the motion of the three planet system should not affect how long it takes for light to get to the two outer planets according to relativity.
[QUOTE=HMHW]
Quote:
Originally Posted by tomh4040
If your sentence above is changed to “The speed of light is invariant for inertial observers; linearly accelerated observers may well see a speed of light different from c.” then we may be in agreement.
Maybe so, but if the sentence were changed that way, then it’d be wrong. The difference you’re trying to concoct here between linear and rotational acceleration simply doesn’t exist.
[/QUOTE]
As I have pointed out, there is a great deal of difference between a linear and a rotational accelerated frame, or a linearly accelerated frame with the acceleration applied at an angle to the line of motion.
The following is true according to relativity.
In an accelerating space ship, the acceleration is linear. The acceleration increases the speed of the spaceship, so during the transit time of a beam of light shone forward the spaceship has moved away from the light beam, so the light has further to travel, and takes longer to get to the nose.
Acceleration can also change the direction of the spaceship, if applied sideways to the direction of travel. Now the spaceship does not change its speed, but it changes its direction.
To make the maths simple, assume the spaceship is 300,000 Km long. Light shone from tail to nose takes one second to reach the nose. The spaceship is stationary or traveling with a constant velocity (v) as I switch on the light. I then accelerate the spaceship to v + 30,000 Km/s during the transit time of the light. It takes the light 0.1s to travel the extra distance. The light arrives at the nose after 1.1s.
Now I am going to apply the acceleration 90 degrees sideways to the line of travel instead of forward. The spaceship is assumed to be 30,000 Km wide. The acceleration is applied during the transit time of the light. Because the receiver has moved sideways during the transit time, the light has further to travel. The base line (fore and aft) is 300,000 Km long, the line along which the receiver has traveled is 30,000 Km long and at right angles to the base line, making a right angle triangle. The light now travels along the hypotenuse, which is slightly longer than the base line, and takes 1.005s to reach the receiver.
In the case of acceleration increasing the speed of the spaceship, the light takes 1.1 seconds to reach the nose of the spaceship. In the case of the acceleration at right angles to the line of flight, causing a change of direction (but no change of speed) of the spaceship, the light takes 1.005 s to reach the nose. There is a very discernable difference.
Now I am going to introduce the acceleration from a point which is not at 90 degrees to the line of flight, but at around 85 degrees approximately from the nose. With the acceleration in this orientation, the light takes once again takes one second to reach the nose. This is because the line along which the receiver has traveled is no longer at right angles, but is at 85 degrees aprx to the base line, causing the receiver to move 1500 Km closer to the advancing light beam. The spaceship can be accelerated, and yet the apparent speed of light is still c.
I used approximations in the above discussion. I used 85 degrees as an example, but that is not an exact figure. The fact remains that in relativity, the speed of light can be c in an accelerated frame.
Remember HMHW’s statement. “Maybe so, but if the sentence were changed that way, then it’d be wrong. The difference you’re trying to concoct here between linear and rotational acceleration simply doesn’t exist.”
The aether is not an arbitrary IFR in any aether theory. In some theories it’s an IFR, but my perception so far have been that you champion the entrenched aether theory where there is a local aether co-rotating with the Earth. No other aether theory is consistent with light speed measurements taken in varying directions at varying locations on Earth. So they are all equally irrelevant. But the entrenched aether, co-rotating with the Earth, is not an inertial reference frame, it’s the same accelerated Earth locked reference frame you claim relativity can’t explain the Sagnac effect in.
Thus, if there’s an explanation for the Sagnac effect it applies equally to a Sagnac interferometer rotating with the Earth in a universe where an entrenched aether exists, and any rotating Sagnac interferometer in a universe where relativity applies.
Let me repeat it for clarity: If you champion an aether theory with an inertial aether, you’re really lost to reason. All of those have been thoroughly falsified. If you champion an entrenched aether you have to deal with a rotating, and thus non-inertial reference frame.
And just so I’m not accidentally promoting entrenched aether theories here, they have also been thoroughly falsified, but lets stick to the errors closest at hand.
No, it’s an example of a question designed to illuminate. You cannot treat an accelerated frame the same as an inertial one. When you asked the question “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?” earlier in the tread you showed you don’t have this very basic physics knowledge internalized and since you haven’t acknowledge this blunder and keep repeating ideas about the Sagnac effect that requires not understanding the important difference between inertial and non-inertial frames, it bears repeating.
For simplicity’s sake, let’s consider a rigid disk one meter in radius rotating at one radian per second (makes the math really easy).
From an IFR stationary with respect to the center of the disk, a point at the edge of the disk has a velocity of sin(t+d), and thus an acceleration of cos(t+d) (where d depends on which point on the edge of the disk we’ve picked), while the center of the disk doesn’t have any acceleration at all.
From any IFR, the center of the disk and the edge of the disk will have different accelerations.
Per Newton, a pseudoforce (centrifugal) would exist at every point on the disk, of magnitude k meters per second squared (where k is the distance from the center), always directed outward, caused by the disk’s inertia as its tensile strength forces its parts to not follow their “natural” motion (constant velocity in a constant direction). A rotating frame of reference, per Newton, is not an inertial frame, for this very reason, because the rotation induces this pseudoforce.
Per Einstein, even a rotating frame of reference can be treated as inertial - but adjustments must be made. That is, calculations of physics on the spinning disk can consider the disk to be non-rotating, but only if those calculations also include a new force that locally looks just like gravity is also included in the calculations. Per Einstein, any experiment on the disk will follow the known laws of physics for a small enough area of the disk over a small enough time, if that local gravity-like force is included in the calculations (and the results of experiments that cover larger areas of the disk can be determined by incrementally accounting for the variations from place to place).
So for a spinning object of any kind, you can treat it as spinning, in which case, its parts are certainly accelerating relative to each other, or you can treat it as non-rotating, and add the new force that that viewpoint requires. Is there any disagreement with this?
Why on earth do you insist on creating hypotheticals with such bizarre differences between states? Is that the only way you can get your figures to agree with your assumptions?
For example, why do you have the initial setup with transmitter and receiver 300,000Km, and in the latter one only 30,000?
Why say the object “accelerated to x” instead of actually providing us with the value of the acceleration?
Why have the linear acceleration occur “during the transit time of the beam”, yet the perpendicular acceleration occur instantly. And before you argue that point, yes you have assumed instantaneous acceleration. If the perpendicular acceleration had been constantly applied during the transit, the path the light takes would be a parabola.
Why are you approximating in the final example? Why not give us the exact figures you obtained.
Why have you abandoned your insistance that two points of constant distance, co-rotating about a common centre share the same velocity and are not accelerating?
Why have you abandoned your claim that a beach ball leaves a baseball bat at a greater speed if the bat is in motion prior to the collision, than if they start off in contact.
Why have you abandoned your claim that a beach ball leaves the bat at a greater velocity than the bat was originally travelling because the bat keeps moving and the balloon squashes, then springs back quickly.
Why have you abandoned your claim that the “squashiness” of the balloon means it would leave the bat faster than a rigid body such as a billiard ball.
Do you still believe a “heavy” object experiences greater acceleration due to gravity than a “light” one?
There was a mistake in the following paragraph - I omitted to say that the acceleration in the fore and aft direction is still present during the introduction of the sideways acceleration.
Strange how this was not picked up when you can certainly find errors which may or may not exist.
“Now I am going to introduce the acceleration from a point which is not at 90 degrees to the line of flight, but at around 85 degrees approximately from the nose. With the acceleration in this orientation, the light takes once again takes one second to reach the nose. This is because the line along which the receiver has traveled is no longer at right angles, but is at 85 degrees aprx to the base line, causing the receiver to move 1500 Km closer to the advancing light beam. The spaceship can be accelerated, and yet the apparent speed of light is still c.”
I have been very busy of late and have not had time to read or respond since my last posting. I am sure you are waiting impatiently.
Before we move on I would like to go back a few posts to clarify something. Have I got this correct? According to relativity theory, the rotating interferometer is a non inertial frame of reference – for the sake of argument it can be called a rotating frame of reference or an RFR. This means that the speed of light in it is not c, but is c +/- v, where v is the rotational speed of the RFR. This gives rise to the fringe shift because the light traveling in opposite directions is traveling at different speeds.
A rotational acceleration is the sum of two linear accelerations, so this is simply a non-starter.
It certainly can, but in general, it won’t be.
Only for an observer co-rotating with the interferometer, and standing on the axis of rotation. This is one way of seeing that the frame used in the explanation of the Sagnac effect does not constitute an ether – since the effect is independent of the choice of frame, so you can use any frame, while the ether only describes a single, special frame of reference, in which the effect would exist.
Standing on the axis of rotation is not what I remember from previous postings.
[QUOTE= Half Man Half Wit ]
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.
[/QUOTE]
You have not specified where the observer is in the above explanation.
What would the observer co-rotating with the interferometer, but not on the axis of rotation see or measure? For instance he might be at the receiver, looking at the light fringes.
I’m not interested in calculations either. What do you mean by nothing out of the ordinary? That is remarkably vague. What I am interested in is if the observer has to be in a specific location to see the fringe shift. The fringe shift shows up where the beams are brought together again after their trip round the interferometer. This point is not at the axis of rotation, but the fringe shift shows up there. From our previous discussions that statement must be correct.
The thing is, without calculations, it’s impossible to say anything definitely. But the observer will see the Sagnac effect – as I said, it’s frame-independent. He’ll not see the light moving at c +/- rw, though.
Also, it seems that you’re still not quite getting the idea of reference frames and observers. Both terms can be used interchangeably – a reference frame characterizes how things look for a given observer; it’s the frame ‘attached’ to this observer. So an observer at the axis of rotation of the Sagnac interferometer and one at some point on the disk will be in different frames of reference.
So any observer will see the Sagnac effect. In your your previous postings, you said that from any IFR the Sagnac effect would be seen. When you posted this then :-
you were referring to the observer seeing c +/- rw, not the Sagnac effect.
The Earth is a rotating frame of reference (RFR), so any observer standing on its surface will see the Sagnac effect
When these two signals are brought together, there would be a fringe shift corresponding to the Earth’s rotational speed -that is implicit in the above paragraphs. According to relativity, during the transit time of the light from W to E around the equator, the Rx has moved away from the light beam by 0.0614 Km, so changing the apparent speed of the light from 300,000 Km/s to 300,000.12 Km/s (ignoring the index of refraction of the atmosphere). During the transit time of the light from E to W, the Rx has moved towards the light beam by 0.0614 Km, changing the apparent speed of the light to 299,999.54 Km/s. This change of speed is WRT any observer on the Earth who measures the distance involved, which is fixed at 40,075 Km, and the difference in arrival times, which is 0.000000328 seconds. The observer would therefore come to the conclusion that on Earth, c is not constant.
When the MMX was performed, it proved that the speed of light was the same in all directions because no fringe shift showed up, no matter what the orientation of the apparatus. There should have been one. When orientated with a north-south leg and an east-west leg, the east-west leg mirror has moved during the transit time of the light, the north-south leg mirror has not, so causing a fringe shift.
Why was no fringe shift seen?
The effect is much too small to be observable for an interferometer on the scale of the Michelson-Morley apparatus; besides, it is only observable in geometries enclosing a certain area. However, when they made it bigger, the effect was indeed observed.