Tutorial: Relativity of simultaneity

[geistkiesel]

1) Pro,

The signal in the circuit will always be seen to take the same time to propagate by an observer travelling with the circuit.

Sure, everytime the observer moving with the frame will get the same numbers assuming the frame is moving at the same velocity wrt the embankment in each test. Inf fact each leg the light moves does propagate the same distance in the same time. In the case of the left moving photon it arrives at the oncomiong clock after moving a distance ct and the clock moving vt toward the outgpoing phhoton. The other photon has also moved a distance ct wrt the emission point, but because of the fraqme motion this photon has not arrived at the right clock yet. Notice the dimensions of the frame are not used here in any measuremnts, rather only the the distance traveled in equal times is used.

The frame moves vt during the time the photon moves ct. So what?'​

Once again:

http://www.mathpages.com/rr/s2-07/2-07.htm

From the above article:



Absolute velocity is non-existent. Conclusively show the existence of an "ether" or yield to the power of the dark side. Mwahahaha!!!

quoting the article

"Nevertheless, it remains a seminal tenet of anti-scientific crackpotism (for lack of a better word) that the trivial Sagnac effect somehow "disproves relativity". Initially those who adhere to such views assert that the expressions "c+v" and "c-v" are prima facie proof that the speed of light is not c with respect to some inertial coordinate system. When it is pointed out that those quantities do not refer to the speed of light, but rather to the sum and difference of the speed of light and the speed of some other object, both with respect to a single inertial coordinate system, which can be as great as 2c according to special relativity, the anti-scientific crackpots are undaunted, and merely proceed to construct progressively more convoluted and specious "objections".

SL You should throw some light on your "dark side " reference,

In the first place I never stated that the speed of light changes in Sagnac effects. In fact I never stated that the Sagnac Effect disproves relativity theory. I cannot count the numerous times I have stated that the terms c+v and c-v are measurements of the relative motion re light speed with respect to a moving frame of reference. The statements are innocuous, they say only the speed of light is c -v larger than the frame velocity, or that c + v is the relative velocity of oppositely moving photons and inertial frames. The statements do not change the speed of light.

The article you referred to was not directed at anything I have ever said and I take no repsonsibility for what other SRT dissidents have to offer.

Simply stated Sl, If two clocks are attached to a moving frame and if two photons are emitted simultaneoulsy from the midpoint of the two clocks, in the moving and stationary frames, then light moving indepndently of the moving frame will arrive at the clock approaching the outgoing light before the photon moving in the same direction as the other outgoing clock strikes that outgoing clock. Notice no absolute velocity measurements of the SOL are made here.

We aren't measuring the speed of light, we are comparing distances traveled of the frame and photon wrt the emission point of the light as defined as the midpoint of the outgoing light photons. The speed of light is not an issue here, it is the isotropic motion (straight-line) and constant speed of light that is material, only. I am not giving numbers for the time difference here, only that the arrival times at the clocks are sequential.
The Sagnac Linear frame

is simple and takes advantage of the fact that expanding wavefronts have a unique

midpoint location constant in space and time.


Geistkiesel​

 

[Prosoothus]

superluminal,

The signal in the circuit will always be seen to take the same time to propagate by an observer travelling with the circuit.

Almost. But in the example I illustrated, the observer would not be travelling with the circuit because you only want the time dilation, or change in the speed of light, to affect the circuit and not the observer as well.

 

[geistkiesel]

Quoting Geistkiesel:

2) the speed of light is a constant c measured wrt absolute zero velocity. ”



Proven not to be true. SOL is measured to be c no matter what your state of motion.

SL, I need some enlightenment from from your statement. Is not the absolute velocity v = 0 point a valid inertial frame at rest? Let us assume Yes it is for a bit. Now are you negating the concept of absolute velocity zero such that you use the theory to disprove the model, or do you use physics to disprove the model? OK, I grant you SRT will claim that the theoretical statement that there is no absolute velocity = 0 and that this is substantiated by the the postulates of light and governing laws of physics appropriate to the situation.

I am generalizing so as not to bias any physical construct you may use here.
SL. please see the following as physically "true" as defined by the laws of physics we both are familiar with. Some things just aren't open to opinion or point of view.

Two photons emitted in opposite directions move along the same straight-line. Thus the motion is isotropic. The photons move with a constant velocity, whatever that is, which we will call c. As both photons cover the same distance traveled since they were emitted, the emission point is a mark of the midpoint of the ever moving and expanding photons. Lets us stay reasonably local here OK?

Any problem with statement number 1?

The emitted photons are physically untouched by the presence of the inertial frame containing the L and R clocks (it is indenpendence day for photons, so "photons", "ulight!!!"). The distance traveled by the right photon, with respect to the point P, is greater than than the distance of the MP to the photon after the right photon moves a distance ct.

Any problems with statement #1a?


The continued motion of the photons is potentially eternal (until acted on by an external force).

Any problem with statement # 2.

If one of the photons is reflected 180 degress after moving a distance ct, the photon will arrive back at the emission point after traveling another distance ct.

Any problem with statement #3?
The return to the emission point is guaranteed by the motion of the light, governed by the postulates that, the motion is isotropic (straight-line) , is constant velocity and is moving independent of the inertial frame containing the source of the photons.

Any problems with statement #4"
The observer is on the moving frame so he does not see the rearrival of the left photon at point P as the observer has moved wrt P. If the frame is moving at some velocitry v wrt the point P, P is a legitnmately defined point for the velocity reference purposes of this particlular frame. (Conicidentally, the point P is also referenced to the stationary frame, and the entire observuing unuiverse, the embankment). But not yet. For we know not what the velocity v is just yet, nor do we know t. If the left photon moves a distance ct then in that same time t the frame has moved a distance vt, where v and t are unknown.

If the photon could send a reverse signal back to the source (a form of side lobe motion) during its outbound travels, the right photon would measure a shorter time for the reflection of the secondary radiation from the physical MP than the reflection from a mirror located at P.

Any problems with statement # 5?

If the photon that moved a distance ct is immediately reflected 180 degrees the photon will move back to the initial emission point. The physical midpoint and physical source iof the photons stared at by the moving observer, has moved a distance 2vt from the point of emission of the photons. This is where the left photon is after moving 2ct, exactly 2vt from the physical midpoint of the frame that is moving away. For the photon catch the midpoint in time t' the photon moves ct' = 2vt + vt', where vt' is the distance the frame moves when the photon catches up from a distance 2vt away.

Any probklem with statement # 6?

After the right photon has moved a distance ct to the right we know that ct is 2vt short of the R clock. This is the same distance the left photon faces later after moving a total distance 2ct.The observer on the moving frame has no data as yet.The Observer knows nothing.

Problems with 7?


After both photons move a distance 2ct the left photon is 2vt from the MP, while the right photon is 2vt + 2vt' from the oncoming MP. The left and right photons converge after moving the fianl distance of 2vt + vt' at the MP, this final arrival all being simultaneous.

8?

All distances are measured wrt the emission point. There are no direct velocity measurements in the arangement. From the expression ct' = 2vt + vt' the velocity is extracted as v = ct'/(2t + t'). The L and R clocks have the time of photon arrival embedded in the photon signals that arrive simuiltaneously at the physical midpoint of the inertial frame. t' is the measured time difference L(t2) - R(t1).

9?
All this based on the known and agreed postulates of light governing the isotropic and independent constant motion of light.

10?


Geistkiesel​

 

[geistkiesel]

Well, so say you guys. I will bet however that c is measured to be c no matter how the source is moving. If it's not then much of physics, that has already been proven correct, would have to be wrong.

Not so SL. The end as you projected were the matters as stated by "the guys" to be true is not quite as tragic as you may think. Your just uttered, "I will bet that c is measured c no matter how the source is moving" I will make the same bet and we will make a lot of money.

You are aslo no doubt thinking about all measured velocities of the SOL result in a measurement o c. This, I take it to mean that the relative velocity of frame and photon will always be measured as c, as if the frame were at rest wrt the value of the speed of light c. I have my reservations about this but we assume the truth of the assumption. The measured velocity of light wrt the emission point of the two photons will always be measured as c. Likewise the moving frame of reference moving wrt the emission point may also measure its velocity with repect to the point P. No harm no foul.

One could measure the distance the photon covered in an arbitrary, but useful time difference, to define the "second", the "meter" the "yard", the "foot", the micron.

Anyway, SL, the moving photon heading to the left uin a collision course of with the approaching clock moving at velocity v, wrt P, unknown at this instant.. The photon will strike the left clock before the right moving photon strikes the right moving clock.Always.The clock moving away is a farther distance wrt to the distance the photon must travel to catch the clock,. than is the distance the left photon must travel to reach the left clock.

Can you see the universal;ity of this?

The photon moving at a constant velocity will always travel a greater distance than the instantaneous distance of the photon and a moving physical object in order to catch that moving object.


WE'll advertise that one.

The simple fact is that the photon can determine its relative motion wrt moving physical objects just as one would think. The photon doesn't know it though, photons being dumber than cows, but fast cows of course.
Here is yet another clear, one would think, description of the elusive v = 0


How does the observer explain the difference in times of the arrival of photons at L and R, if indeed this were what the data handed to the observer showed?
Geistrkiesel​

 

[superluminal]

Geist,

I freely admit that I cannot completely follow your reasoning. I gather you are stating that the emission point constitutes an absolute reference. Keeping it simple, if your entire apparatus is moving to the right, how is the emission point absolute?

I also admit that I cannot even conceive of an absolute reference in the universe as we know it. This would imply a "center" to the universe and preferred directions. We see no evidence of this in the universe at large. All MM type tests that seek an ether are ambiguous at worst and completely negative in the eyes of almost all scientists.

 

[geistkiesel]

superluminal,
“ The signal in the circuit will always be seen to take the same time to propagate by an observer travelling with the circuit. ”





Almost. But in the example I illustrated, the observer would not be travelling with the circuit because you only want the time dilation, or change in the speed of light, to affect the circuit and not the observer as well.

To Prosoothus and Superluminal

I trust this question to both is pertinent. Can someone describe the meaning of the phrase "will always be seen to take" for me please? I ask in the context, of just exactly what does the observer "see"? If as photon is emitted from his very presence does this give any information regarding motion? Is the obsverver as mentioned in SL's statement really seeing things within the limits of the speed of light? Is the observer given powers to see the results before they occur? Especially crucial are the cases where the observer has abolutely no clue as to whether he is moving or some other inertial frame is moving. Besides some theory, is there any physical possibility that the moving observer is observing events outside of the measurement limits of the speed of light, and, no pun intended of course, but superluminally?
Geistkiesel​

 

[superluminal]

I think this is just standard SR. Two circuits (or whatever) will have a proper time. Observers with each circuit will see the signal take its proper time to propagate. They will each report the others circuit signal to take longer to propagate based on gamma.

 

[geistkiesel]

Geist,

I freely admit that I cannot completely follow your reasoning. I gather you are stating that the emission point constitutes an absolute reference. Keeping it simple, if your entire apparatus is moving to the right, how is the emission point absolute?

I also admit that I cannot even conceive of an absolute reference in the universe as we know it. This would imply a "center" to the universe and preferred directions. We see no evidence of this in the universe at large. All MM type tests that seek an ether are ambiguous at worst and completely negative in the eyes of almost all scientists.

Yes, it is as simple as you suggest, but completely unreklated to the concerns expressed. The moving expanding photons have a natural midpoint, Within local limits, the midpoint is not moving in space and time. The point is at absolute rest. How do we know this?

When the lights move apart they both cover the same distance in the same time. The midpoint of the photons is the emission point of the photons, even though the physical device emitting the photons has moved with respect to the midpoint P of of the photons.

How do we know the point is simply not moving in space and time?
The postulates of light tell us.

The photons emitted from the moving frame move independently of the sources of the photons, the physical source. The photons igniore the moving frame.

Second the photons move isotropically, meaning moving in the same straight-line.until acted on by an external force.

Finally the speed of the photons wrt the emission point is constant and identical for both photons.

The photons measure out equal distances in equal times along and invariant straight-line.

If a [photon has traveled a distance ct and is reflected 180 degrees it will arrive at point P aftre traveling another ct, just further verification of P and its invarinat attributes.

Using these postulates in their most simple and straight forward meaning, what can possibly be moving P, the emission point, if in deep space away from all electrctromagnetic and massive disturbances? Even if the photon trajectories are perturbed this merely moves the midpoint of the same photons, but the old midpoint is still 'there', if unused and perhaps unusable.

Can P possibly move, even by some agreement? We can write coordinate transforms, but these rely on the invariance of P, and ergo any transform is nothing but a rotated clone of the invariant coordinate system.

To this day Sl, as many times as I have used the words absolute space and absolute velocity, they mean absolutely nothing intuitively to me. I can comprehend the postulates of light discussed above and see the point not moving by what ever definition one comes up with on what moving means. How about dx/dt = constant for starters.

So the fucking universe is moving all of its trllions of massive objects around our scrawny invariant emission point P, so what, and who cares anyway? The point still isn't moving. Check it out: how many electromagnetic expanding spheres are there are in the universe? Infinitely more numerous than the massive stellar objects. All matter can be indexed to the absolute motionless frame, defined by as many midpoints of expanding photon wave fronts as one needs to do the job.

Geistkiesel​

 

[geistkiesel]

I think this is just standard SR. Two circuits (or whatever) will have a proper time. Observers with each circuit will see the signal take its proper time to propagate. They will each report the others circuit signal to take longer to propagate based on gamma.

I can see that it is off the shelf SRT SL, but can you go to the point of the observer who assumes he is at rest wrt the "moving inertial ftrame". I think this is different than your answer. What assumptions are the moving observers making wrt the current speed in the circuits?

Geistkiesel​

 

[superluminal]

Geist,

I must think about your photon sphere midpoint idea.

 

[geistkiesel]

Geist,

I must think about your photon sphere midpoint idea.

That's cool. I am only saying that the speed of light is constant even measured from the physical point from which it was emitted. Whether it be a single photon moving in a straightline, or a burst of photons expanding as a sphere, the expansion process is uniformly that of a sphere, where all radial lines are increasing their instantaneous distances from the common emission point exactly the same in the same time. Measure just one of the radials and you have the entire sphere.
Geistkiesel ​

 

[Pete]

Check it out: how many electromagnetic expanding spheres are there are in the universe?

Hi geistkiesel,
If the speed of light is constant in all frames, then all observers of an expanding EM sphere will say that the sphere's centre is motionless in their frame, yes?

So how does this suggest an absolute motionless frame?

 

[geistkiesel]

Hi geistkiesel,
If the speed of light is constant in all frames, then all observers of an expanding EM sphere will say that the sphere's centre is motionless in their frame, yes?

So how does this suggest an absolute motionless frame?

If everybody looking at the midpoint of the expanding wavefront see the same thing, then the midpoint cannot be moving can it? The midpoint is the motionless zeropoint of the absolute frame. This is what we mean by absolute motionless point is it not-not moving to all observers.​

 

[Pete]

If everybody looking at the midpoint of the expanding wavefront see the same thing, then the midpoint cannot be moving can it?​

They don't see the same thing. They each see a midpoint that is stationary in their frame... ie they each see a midpoint that is moving at the same speed as themselves.​

 

[geistkiesel]

They don't see the same thing. They each see a midpoint that is stationary in their frame... ie they each see a midpoint that is moving at the same speed as themselves.

Ovserbers on each frame see themselves as staionary?

Is this because there are no other entities ariound that would allow them to see if one is stationary wrt an embankment for instance? These guys are in deep space with only each other to reference and it is for this reason that each sees the other as moving , themselves at rest?
G

 

[Pete]

It doesn't matter whether the observer considers themselves stationary or not, in this instance.
Every observer will see a midpoint that is moving at the same speed as they consider themselves to be moving.

I should add that I'm not using PP to guide my responses in this thread, but rather the postulate that the speed of light is the same for all observers.

 

[geistkiesel]

Hi geistkiesel,
If the speed of light is constant in all frames, then all observers of an expanding EM sphere will say that the sphere's centre is motionless in their frame, yes?

So how does this suggest an absolute motionless frame?

AS you read the response be aware that I am gong to remove the word "absolute" in the description of frames that are not moving.

Let's try this.

When I say "motionless frame" I mean that the frame is not moving. There is no spatrial translation of the frame which in this case is a point. All pointing finger tips of ovserbers wll converge to the collective colocated point when they stretch their fingers out to touch the point as they pass by in their individually and uniquely uniform moving frames. All ovserbers will measure the value C for the speed of light, but will all measure, in general, different relative motions of frame and photon. In other words, constant motion is constant .

There is a common instantaneous "now' fpr all ovservers in all inertial frames, universaly, though each may record, in general, different observation times for the for any of the events. The tiniest fraction of events in the uiniverse are observed and a tiny fraction of these are ovserbed simultaneously.any single or multi-point event, though each may record different point or multi point simultaneous event. n general tghe closest ti the evet is a distace
The measuremnt of the relative velocity of the anty frame

All these ovserbors are knowledgeable about motion, especially the postulate that assures us that the the current state of perceived uiform motion in no way implies the state of motion is one of absoluterest, wrtgabsolute zero.
The inability to measure motion does not imply a sstate of rest.

I should apologize for the apology (John Wayne said that apologies are a sign of weakness), but the term Absolute Motionless Frame is redundant in the use of the term, a"bsolute motionless" and unqualifiedly ambiguous. Th eword motionless is a complete and unambiguous word with an unconditional meaning.Tom saynow "absolute" of tegh motionless frame os to apply a condition to a n unconditional meaning.. Therefore i am goingt to use the word "motionless" as it means standing alone, not moving. I differnentiate this from not moving with respect to X, as this is conditional. I dop not intend here to place any special meaning to the wprd "motionless" it means what it says.

So let me ask you to be the first to cast tocast .
Whaty do you mean by, "absolute motionless frame" .

 

[Johnny5]

Are you the same Geistkiesel I was talking to at another forum, who asked me to read chapter 34, article 7, from Feynman's lectures on physics? If so, i finally got the book, and am now curious as to what you wanted me to read it for. The article is about w,k, and four vectors. It's a short article.

Regards

 

[geistkiesel]

Are you the same Geistkiesel I was talking to at another forum, who asked me to read chapter 34, article 7, from Feynman's lectures on physics? If so, i finally got the book, and am now curious as to what you wanted me to read it for. The article is about w,k, and four vectors. It's a short article.

Regards

If i told anyone to read Feynman;'s Lectures ion Physics, it would be only ch 5 vol III. The chapter you refer to is not of great interest to me for the simple reason I have not read it in detail. I am a Feynman hunter. I look for mistakes the man made that are of such a level of signifcance to science and physics that I am motivated enough to comment.
If I did urge you to read 34-7 I cannot remember the context of why I would reccomen 34-7.
I tell you what though, johnny5, I will read the article and comment.

The last sentence in 34-7 says
"The angular momentum is never "completely along the z-direction.".
Did Feynman prove this in 34-7? Or more generally is the statement true and true for what fe3hynmansaid. Remember he equated quantum mechanics with classical models, which stinks as far as I am concerned.

Geistkiesel​

 

[geistkiesel]

It doesn't matter whether the observer considers themselves stationary or not, in this instance.
Every observer will see a midpoint that is moving at the same speed as they consider themselves to be moving.

I should add that I'm not using PP to guide my responses in this thread, but rather the postulate that the speed of light is the same for all observers.

OK when the photons are movuing oppositely of each other the moving observer does not accuratelhy descrive physical reality . Now Pete, you brough it up, by saying :"

" Every observer will see a midpoint that is moving at the same speed as they consider themselves to be moving."
You afre confusing two points. There kis a physical midpoint between the clocks A and B. When the lights are emitted the frame is already moving to the right say as the photons scurry down their trjaectories left and right;

If we adopted your position theere wouild be a seriojus problem. I can see the observer at the physical midpoint thinking that the physical midpoint is where the photons were emitted from is in space That would be a mistake for a couple of reaons,. The observer would think that the light was expanding aorund him (the physical midpoint of the enmitted light.2. The concept of motion would be negated.

Consider the reality that the moving photons define the midpouint for all time. At t = 0 the emission point was left of the observer who is moving. Now if the postulate is as you believe it to be the truth then by all means stick with it, but then you avoid the physics of the motion of light is another victim. First you insist that the photons arrive at both moving clocks simultaneously, which means that the position of the clocks in space are completely insignificant. You are sdaying that the moving observer wioll have Mother Nature cooperate with the observer's perspective and adjyust clocks to provide one signal with two outputs.

The cklocks are nothing to the photons other than obstructions in their respective motions. The poiition of the left clock with respect to the emission point limits the extent the left phjoton can move before being measured. Liekwise the right moving phioton extends the limit (compared to the left clock) the photons can miove. There is no way one can rationally postulate the light into arriving simultanoeously at the left and right clocks.
Think about it no moving frame can ever be movuing friom the observer ion the frames's point iof view-hey this is just SRT isn't it?

Where does seeing the point P as invariant violate any laws of physics? say the midpoint theorem is a reality, an experimental reality. How do you erase the concept that the point P is defined by the motionof the light itself -"Einstein said that ,. . . ." and Pete believes it. I know Pete claim ignorance, yiou are an amateur, that way you wont have to explain to anyone.

Pete, beside the postulate, what physical constructs can yiou point tio that prove what you believe? If the moving observer sees the emission point along with the motion of th eobserver, the perspective iof the observer is illusory, OK it is also the postuklate of physics that says tghe moving observer will always measure the speed of light aas c. Fine, measuring the speed of light with respect to the light emission point you get s\the SOL = c, wrt P. the emission point. Now you say that isn't gopod enough you ahve to undergo frame contraction, anmd time dilation ansd assume the emission point is moving with the observer in order to measure the SOL equal to C? For christ sake, why do all this buillshit? Why not measure the SOL with respect to the emission point? Answer: because the SRT wouild crumble.And the goal is to save SRT at all costs PP has an inherent bias and is completely useless.

I asked you in the PP thread how did the moving observer measuire anmything Pete. How did the moving observer measure anything?
Maybe all these observers use and need in fact, is a speed of light postulatometer.

The physical midpoint is assumed the midpoint of the expanding wave front, but how do you get the light to cooperate in this ? How does the moving observer measure the motion olf the physical midpoint, when he isn't moving?
What does the speed of light is independent of the motion of the source of the light mean? Your version would have it mean that the point in space whenre the light was emitted is trashed proportioanl to the perspective iof observers.

Again how does the obserber in PP measure anything?????? With uinique perspective?
geistkiesel​