Tutorial: Relativity of simultaneity

[James R]

Here is a worked example of the relativity of simultaneity, using the postulates of Special Relativity, for those who wish to understand how that comes about.

The relativity of simultaneity means simply that observers in relative motion do not, in general, agree on whether events occur simultaneously or at different times.

The only assumption in the example given here is that the speed of light is measured to be constant for both observers - namely, the speed of light is c, the same value for each. This is one of the two postulates of Einstein's theory of special relativity.

The example is as follows:

L--------------M---------------R -> +x

Light detectors are located at the left and right ends (L and R) of a rigid rod. A light emitter is located at the middle of the rod (M). The rod moves in the positive x direction at a constant speed v, relative to the Earth.

We make no assumption that an observer on the rod and an observer on Earth will measure the rod to have the same length, or that clocks held by either of these observers will tick at the same rate.

Data are as follows:

Observer on the ground:

The half-length of the rod is measured to be d.
At time t=0, a flash of light is emitted at M, designated as x=0, and two photons head off towards L and R, initially located at x=-d and x=d, respectively.
The positions of the photons as a function of the time t, measured on the ground observer's clock, are:

PL(t) = -ct
PR(t) = ct

The positions of the detectors as a function of time are:

L(t) = -d + vt
R(t) = d +vt

At what times are the two photons detected? The photon is detected by the left detector when:

PL(t) = L(t)
-ct = -d + vt

Solving, we find t=d/(c+v).

The right-hand photon is detected when

PR(t) = R(t).

Solving, we find t=d/(c-v).

Notice that these two values of the time are different, so the photons are NOT detected simultaneously.

Observer on the rod:

The half-length of the rod is measured to be d' (which might be different to d).
At time t'=0, a flash of light is emitted at M, designated as x'=0, and two photons head off towards L and R, initially located at x'=-d' and x'=d', respectively.
The positions of the photons as a function of the time t', measured on the rod observer's clock (hence t' instead of t, since these might be different), are:

PL(t') = -ct'
PR(t') = ct'

The positions of the detectors as a function of time are:

L(t') = -d'
R(t') = d'

Notice, the detector positions don't change with time in this reference frame (the view of an observer standing on the rod).

At what times are the two photons detected? The photon is detected by the left detector when:

PL(t') = L(t')
-ct' = -d'

Solving, we find t'=d'/c.

The right-hand photon is detected when

PR(t') = R(t').

Solving, we find t'=d'/c.

Obviously, these times are the same, so according to the observer on the rod, the detectors both register the photons SIMULTANEOUSLY.

----

We have just shown that events which are simultaneous in one frame of reference (the rod frame) are not simultaneous in another frame of reference (the ground frame), using only the fact that the speed of light is the same regardless of which frame it is measured in.

This is one of the simple results of Einstein's relativity, although it seems counter-intuitive at first.

If you have questions, please post them in this thread.

 

[geistkiesel]

Here is a worked example of the relativity of simultaneity, using the postulates of Special Relativity, for those who wish to understand how that comes about.

The relativity of simultaneity means simply that observers in relative motion do not, in general, agree on whether events occur simultaneously or at different times.

The only assumption in the example given here is that the speed of light is measured to be constant for both observers - namely, the speed of light is c, the same value for each. This is one of the two postulates of Einstein's theory of special relativity.

The example is as follows:

L--------------M---------------R -> +x

Light detectors are located at the left and right ends (L and R) of a rigid rod. A light emitter is located at the middle of the rod (M). The rod moves in the positive x direction at a constant speed v, relative to the Earth.

We make no assumption that an observer on the rod and an observer on Earth will measure the rod to have the same length, or that clocks held by either of these observers will tick at the same rate.

Data are as follows:

Observer on the ground:

The half-length of the rod is measured to be d.
At time t=0, a flash of light is emitted at M, designated as x=0, and two photons head off towards L and R, initially located at x=-d and x=d, respectively.
The positions of the photons as a function of the time t, measured on the ground observer's clock, are:

PL(t) = -ct
PR(t) = ct

The positions of the detectors as a function of time are:

L(t) = -d + vt
R(t) = d +vt

At what times are the two photons detected? The photon is detected by the left detector when:

PL(t) = L(t)
-ct = -d + vt

Solving, we find t=d/(c+v).

The right-hand photon is detected when

PR(t) = R(t).

Solving, we find t=d/(c-v).

Notice that these two values of the time are different, so the photons are NOT detected simultaneously.

You are correct the photons are not detected at L and R simultaneously. However, you have an error in your calculations.

At the instant that L detects a photon, The right photon is exactly ct from the emission point and is located 2vt from R. In order for the right photon to reach R it must cross the 2vt disrance, but the frame has moved a distance vt0' duiring this time. Therefore, the right photon moves a distance ct0' = 2vt + vt0', where t0' = t(2v)/(C - v). Or in terms of t, t = t0'(C - v)/2v. I use t'0 to be the added time it takes to cross the 2vt distance and to distinguish this from the James R t' below.

here is a more complete picture of what is occuring.

And notice the invariant location of the emission point from which all motion can be referenced, which conicidentally is the same for the stationary observer and the moving observer.

Question James R, do you agree or disagree with my calculation? If you do not agree, then would you please indicate your specific reasons for negating my finding?

Observer on the rod:

The half-length of the rod is measured to be d' (which might be different to d).
At time t'=0, a flash of light is emitted at M, designated as x'=0, and two photons head off towards L and R, initially located at x'=-d' and x'=d', respectively.
The positions of the photons as a function of the time t', measured on the rod observer's clock (hence t' instead of t, since these might be different), are:

PL(t') = -ct'
PR(t') = ct'

The positions of the detectors as a function of time are:

L(t') = -d'
R(t') = d'

Notice, the detector positions don't change with time in this reference frame (the view of an observer standing on the rod).

James R, you are overlooking a crucial fact, motion. The observer standing on the rod cannot see the photons arriving at L and R. The observer can only see the reflected light as it return back to the origin (if it does). Or if there are clocks at L and R and then these clocks must necessarily record the arrival time as sequential with L detecting the arrival of the left photon before R detects the right moving photon. If you grant that the clocks spit out the times of arrival on a piece of paper, or should the lights be immediately reflected back to the physical midpoint of the rod with the times of arrival at L and R embedded ion the return signal, then these times will inform the observer that the lights arrived at L and R at different times.

This is so obvious James R. Assume that the gorund observer had clocks placed exacrly where L and R are positioned when the photons arrive. the stationary observer will measure the times of arrival as sequential. The moving observer being at the midpoint of the rod at this time sees nothing until informed by the stationary observer of his measuremnt.

fact: the clocks can only register one arrival time. Your placing all with the moving observer projecting some impossible arrival scenario at L and R has absolutely nothing to do with the speed of light or arrival times. he is just as much a slave to physical measurement and the speed of light as anyone.


Your statement that the moving observer sees the arrival times as measured by the moving observers clock. The only way this observer can see what you claim is when the lights refrlect back to the observer.There is just no way that the observer's clock can measure the arrival times at L and R.

The observer is limted by the speed of light as to what he can measure and standing on the rod measuring with his clock. You are saying that the observer can measure ct' and -ct' standing at the midpoint of the rod, at the same time? Some trick!!

Well, if you insist he did do exactly this would you explain how? And assuming you weren't privvy to someone placing clocks at L and R how do you explain the difference in time the clocks at L and R read, which can only be one time from whom ever is observing the events.

I am saying here that the perception of the observers is totally irrelevant to the physics of the photon motion.

Likewise, the moving observer can just as easily assume he is moving can't he not? Then what is his resut? The same as the stationary observers will it not?

I recognize that you made no SRT assumptions this wouod be a result of what you are asserting in your conclusions here.

If he sees the ground rushing past he will see that the embankment clocks are running slower than his, correct? And the stationary observer will asee that the moving clocks are running slower than his, correct? let us say that at some diostance from where the photons were emitted a mechanical switch simultaneously cuts off the clocks on the embankment and the moving frame. We cxan have the motion of the frame run into the switch which shuts both clocks off, so what will each clock read? Will the readings be SRT reciprocal?

I see from your example that there is a gaping contradiction. The first is it appears that the measured time that you are discusssing has absolutely nothing to do with the photons being measured and has everything to do with what the observer appearently thinks he is observing, in other words SRT is all just psychological appearances is it not? At least according to your example?

At what times are the two photons detected? The photon is detected by the left detector when:

PL(t') = L(t')
-ct' = -d'

Solving, we find t'=d'/c.

The right-hand photon is detected when

PR(t') = R(t').

Solving, we find t'=d'/c.


Obviously, these times are the same, so according to the observer on the rod, the detectors both register the photons SIMULTANEOUSLY.

These are the times that the moving observer measures the times on his clock [singular, as you made the point].This what you said earlier.
Can you explain how the observer can measure even one arrival time?

We have just shown that events which are simultaneous in one frame of reference (the rod frame) are not simultaneous in another frame of reference (the ground frame), using only the fact that the speed of light is the same regardless of which frame it is measured in.

This is one of the simple results of Einstein's relativity, although it seems counter-intuitive at first.

Counter intuitive? No James R it is impossible for the moving observer to measure the photons at both ends of the rod, whether he is moving or stationary.

If you have questions, please post them in this thread.

James R,
There are a number of peripheral questions here

• It is apparent that the moving observer has been given extraordinary powers of being in two places at the same times, that is at L and R when he measures the arrival of the photons there and,
• that the observer arrived at the L and R before the photons that were emitted arrives at L and R.

Can you explain how the observer is able to accomplish these extraordinary feats?


Have you ever seen anything like my figure before james R?

The bottom line James R is that the moving observer has made no measurements in your example, not the arrival times of the arrival of the light at L and R, and certainly not the speed of light.


Geistkiesel​

 

[James R]

geistkiesel:

You are correct the photons are not detected at L and R simultaneously. However, you have an error in your calculations.

I don't think so. The origin of the coordinate system I used is at M at time t=t'=0. That origin does not move with the rod in the (x,t) coordinate system. It DOES moves with the rod in the (x',t') system.

And notice the invariant location of the emission point from which all motion can be referenced, which conicidentally is the same for the stationary observer and the moving observer.

My two reference frames both include that, too.

Question James R, do you agree or disagree with my calculation? If you do not agree, then would you please indicate your specific reasons for negating my finding?

I'll take a closer look at it, and get back to you.

James R, you are overlooking a crucial fact, motion. The observer standing on the rod cannot see the photons arriving at L and R. The observer can only see the reflected light as it return back to the origin (if it does). Or if there are clocks at L and R and then these clocks must necessarily record the arrival time as sequential with L detecting the arrival of the left photon before R detects the right moving photon. If you grant that the clocks spit out the times of arrival on a piece of paper, or should the lights be immediately reflected back to the physical midpoint of the rod with the times of arrival at L and R embedded ion the return signal, then these times will inform the observer that the lights arrived at L and R at different times.

You need to be careful here. If we have two clocks which are stationary (in the ground frame) at the points where the the photons meet L and R, then they will measure the (different) times I gave above. If, on the other hand, we use two clocks which are attached to the ends of the moving rod at L and R, then when these clocks spit out their pieces of paper recording the arrival times of the photons, they will both record the same number, since in this frame the photons arrive simultaneously.

In other words, what you say is true in one frame, but not the other.

This is so obvious James R. Assume that the gorund observer had clocks placed exacrly where L and R are positioned when the photons arrive. the stationary observer will measure the times of arrival as sequential. The moving observer being at the midpoint of the rod at this time sees nothing until informed by the stationary observer of his measuremnt.

I agree with you regarding the ground observer. But having the rod clocks record the arrival times gets around the problem of informing the rod observer. He just checks the pieces of paper at his leisure, after the experiment is done.

fact: the clocks can only register one arrival time. Your placing all with the moving observer projecting some impossible arrival scenario at L and R has absolutely nothing to do with the speed of light or arrival times. he is just as much a slave to physical measurement and the speed of light as anyone.

You are right that one clock can only record one arrival time. The question is: do two clocks in different states of motion record the same or different arrival times. As I have shown, they record different times.

Your statement that the moving observer sees the arrival times as measured by the moving observers clock. The only way this observer can see what you claim is when the lights refrlect back to the observer.There is just no way that the observer's clock can measure the arrival times at L and R.

When I talked about the observer's clock (singular), assume I meant the two synchronised clocks located at the positions where the photons are received. In effect, FOUR clocks are used (two in each reference frame, located at L and R), but the clocks in each reference frame are always synchronised with each other. This synchronisation is easy to achieve, and is explained in my thread "What is a reference frame?", linked in the forum FAQ.

The observer is limted by the speed of light as to what he can measure and standing on the rod measuring with his clock. You are saying that the observer can measure ct' and -ct' standing at the midpoint of the rod, at the same time? Some trick!!

Two synchronised clocks solve the problem. Agreed?

Likewise, the moving observer can just as easily assume he is moving can't he not? Then what is his resut? The same as the stationary observers will it not?

We're interested only in what the clocks record, not in what an observer can calculate. If an observer does a calculation assuming a different state of motion, he is not predicting what his clocks will measure, but what clocks in some other frame of reference will measure.

If he sees the ground rushing past he will see that the embankment clocks are running slower than his, correct?

Yes, though I haven't shown that in my description above.

And the stationary observer will see that the moving clocks are running slower than his, correct?

Yes. Again, I haven't shown that that is true above.

let us say that at some diostance from where the photons were emitted a mechanical switch simultaneously cuts off the clocks on the embankment and the moving frame. We cxan have the motion of the frame run into the switch which shuts both clocks off, so what will each clock read? Will the readings be SRT reciprocal?

If you have a switch, it can only shut off all clocks simultaneously from the point of view of ONE of the two reference frames. It is impossible to shut off all four clocks simultaneously in both frames, since events measured to be simultaneous in one frame are not simultaneous in the other frame, as I have shown. One example of such an event is a single event which shuts off all the clocks.

I see from your example that there is a gaping contradiction. The first is it appears that the measured time that you are discusssing has absolutely nothing to do with the photons being measured and has everything to do with what the observer appearently thinks he is observing, in other words SRT is all just psychological appearances is it not? At least according to your example?

What is psychological about my example, specifically? (Bear in mind the modification to the four clocks with paper readouts).

Counter intuitive? No James R it is impossible for the moving observer to measure the photons at both ends of the rod, whether he is moving or stationary.

Good point. The four clocks solve this problem, though. Agreed?

 

[geistkiesel]

geistkiesel:



I don't think so. The origin of the coordinate system I used is at M at time t=t'=0. That origin does not move with the rod in the (x,t) coordinate system. It DOES moves with the rod in the (x',t') system.

Aren't you assuming in the x't' frame that the frame is stationary?

The photons do not concern themselves with your coordinate system. The distances I had referred to were the distances the photons moved. First the left moving photons arrives at L while the right moving photon has not arrived at R. The observers and the coordinate reference system you use are entirely irrelvant here.

I used the emission point as an invariant point in space and only coincidentally is this point referenced in the stationary frame, and could be maintained in the moviing frame as well.

My two reference frames both include that, too.

The stationary frame and the at rest assumed frame ?
OK, James R, ignore the staionary frame and use just the point iof emission of the photosn as a coordiante frame point of zero. Make all you calculations and theory from this invariant point in space.

I'll take a closer look at it, and get back to you.

James R I have made a fuss about these calculations on many instances, which you nor anyone else has seen fit to respond to, certainly not Superluminal.

However, I will be waiting your reply.

You need to be careful here. If we have two clocks which are stationary (in the ground frame) at the points where the the photons meet L and R, then they will measure the (different) times I gave above. If, on the other hand, we use two clocks which are attached to the ends of the moving rod at L and R, then when these clocks spit out their pieces of paper recording the arrival times of the photons, they will both record the same number, since in this frame the photons arrive simultaneously.

Impossible.

Janmes R, I was very careful. Now it is your turn to be careful. Look at what you have just posted. The ground clocks and the moving clocks at L anad R (ground and moving frame) are co-located in space. There is no possible way in which the clocks in the moving anad stationary frame cannot all measure a different (that is sequential) time of arrival for the photns at L then at R.

here is the position at the time the first photon arrives at L that has not arrived at R.

In other words, what you say is true in one frame, but not the other.

What is true in one frame but not another? How does the L and R detectors know to record a different time? The photons kniow nothing of SRT, coordinate frames or observers. The photons move all indenpendently of all of these aspects of reality. We do not even measure the speed of light in this test.

Be candid JAmes R, look at the fact that photons are generated at moving targets. One tarhget is moving toward an on coming photon, the other photon is moving at a target that is moving away. This is the reality, and sorry to say, is intuitive as well as reasonable, rational and consistent with experiment. Theorize your way out of this. No observer's point of view can fix this for you.

I agree with you regarding the ground observer. But having the rod clocks record the arrival times gets around the problem of informing the rod observer. He just checks the pieces of paper at his leisure, after the experiment is done.

Yes both observers can do this. And like Istated the clocks can embed their arrival times as recorded in the reflected message which is returned to the observer on the moving fraqme. He also can look at the results, paper and embedded signal.

You are right that one clock can only record one arrival time. The question is: do two clocks in different states of motion record the same or different arrival times. As I have shown, they record different times.

When have you ever shown that the clocks will record differently? Certainly not in this thread. Besides you bringing SRT nonsense into the thread as you said in the opening thread you were not going to do, does not help you.
We aren't concerned with the specific time on the clocks or whether the stationary and moving frame clocks tick at the same rate, or even frame contraction, we are concerned with whether thre arrival times are sequential or simultaneous.

When I talked about the observer's clock (singular), assume I meant the two synchronised clocks located at the positions where the photons are received. In effect, FOUR clocks are used (two in each reference frame, located at L and R), but the clocks in each reference frame are always synchronised with each other. This synchronisation is easy to achieve, and is explained in my thread "What is a reference frame?", linked in the forum FAQ.

OK I will let you skate on this, but you still have the oprobklem of whjaqt the moving observ "sees", which is onlhy the output of his clocks. Forget the stationary clocks. have the frame in outer space and the frame performs the exzperiment 10,000 tuimes. the result will alwaus be the same :sequential arrival of photons at L and R.

You must agree that the test made in the stationary frame will result in the photons arriving simultaneously at L and R. Then if we take your frame into space (the observer is asleep) at some velocity, decelerate to a velocity Vf = Ve = 0 (we assume the embankment is stationary) he will repeat the earth results, unambiguously. Then if he is acceleraqted, and knows it, he will get another result, differeing by the length of time the photon take for the roundtriop, or t' =t (2v)/(c - v). How does he explain the difference?

You are saying, and I cannot see anyway out of the conundrum you have p0klaced yourself in, that under no circumstances will the moving frame ever detect sequential arrival of photons, in other words moving objects coordinated (L and R the same distance apart), are suddenly placed in the path of the moving photons that were emitted simultaneously in the moving frame, even if the observer is not on board the frame? This insertion of the detectors has no affect on the outcome? Even though every one in the universe can see that the arrival times are sequential, except the moving observer.

Two synchronised clocks solve the problem. Agreed?

The observer at the moving midpoint still cannot observe any event untll the photons arrive back at the midpoint. In fact the moving observer can see four arrival times 2 from each of the L and R clocks. All four will record that the photons arrived sequentially at L then at R. Also, there are clocks at the invariant and varying midpoints

So what does the moving obsebver do with this data?

We're interested only in what the clocks record, not in what an observer can calculate. If an observer does a calculation assuming a different state of motion, he is not predicting what his clocks will measure, but what clocks in some other frame of reference will measure.

The observer can do both, can't he?

Caslculations have absolutrely nothing to do with what the clocks record, which will be sequsntial arrival times of the photons.

The observer can make two calculations based on the data recveived. He may assume he is moving wrt Ve and he may assume he is at rest wrt Ve. He has these two options.

Yes, though I haven't shown that in my description above.

So then the data will show that clocks, on the stationary and moving frame will record thagt they are moving slower than the other?

Yes. Again, I haven't shown that that is true above.

Same question: Both frame's clock data will show that each frame clock is moving at a rate slower than ther other frame clocks?

If you have a switch, it can only shut off all clocks simultaneously from the point of view of ONE of the two reference frames. It is impossible to shut off all four clocks simultaneously in both frames, since events measured to be simultaneous in one frame are not simultaneous in the other frame, as I have shown. One example of such an event is a single event which shuts off all the clocks.

There is a mechanical switch on both L and R extending into the stationary frame. From previous experiments at this same velocity the stationery clocks are placed within a wavelength of light from the passing moving frame clocks. When the mechanical switches on the moving frame strike the clocks that are effectively colocated with the moving frame clocks all four clocks stop simultaneously as measured by all clocks that are synchronized within each frame.

Therefore, regardless of time dilation and frame contraction, the clocks in both frames will show a sequential arrival of the photons.

Can you find fault with this method of shutting off all clocks simultaneously? If so please indicate your objections dpecifically.

What is psychological about my example, specifically? (Bear in mind the modification to the four clocks with paper readouts).

You mean modification by taking your singular clock and placing the two clocks at L and R? OK they spit out paper and they also embed the arrival times in the reflected signal that returns to the moving midipoint and arrives their simultaneiously as measured by the moving and stationary observer.

What is psychological?

Your model requires a conscious observer on the moving frame, for one psychological factor.

Assume the recorded time at L is "100" and at R "101" The observer that subsequently reads the data must necessarily read both as "100, or "101" in order to maintain consistency with SRT.. This is also what I mean by psychological.

Assume that the clocks on the moving frame L and R radiate the time of arrival to the statioary receivers adjacent to the clocks at a distance no greater than a few wavelengths of the signal wavelength. Will the stationary observer who has measured sequential arrival times read sequential arrival times as radiated by the moving clocks?

Good point. The four clocks solve this problem, though. Agreed?

It solvbes the probklem of negating the claim that the moving frame will record simultaneous arrival time when the moving frame will ecord sequential arrival times, but you still have the problem of showing the loss of simultyabneity. If all six clocks, two each for the L and R detector/reflectors and the stationary and moving midpoints are involved they will all be consistent tio the unambiguous fact that the photons will arrive sequentially at L and R.

Geistkiesel​

 

[James R]

geistkiesel:

Aren't you assuming in the x't' frame that the frame is stationary?

A correct statement is that the rod does not move in the x',t' coordinate system.

The photons do not concern themselves with your coordinate system. The distances I had referred to were the distances the photons moved. First the left moving photons arrives at L while the right moving photon has not arrived at R. The observers and the coordinate reference system you use are entirely irrelvant here.

The coordinate systems are simply ways of quantifying the observations of the observers. And I have shown mathematically that in one frame, the photons arrive simultaneously, while in the other frame they do not.

If you can see an error in my mathematics, please point it out.

I used the emission point as an invariant point in space and only coincidentally is this point referenced in the stationary frame, and could be maintained in the moviing frame as well.

Yes, it could. For example, the person on the rod could drop a flag off the rod onto the ground at the instant the flash is triggered at M.

OK, James R, ignore the staionary frame and use just the point iof emission of the photosn as a coordiante frame point of zero. Make all you calculations and theory from this invariant point in space.

I did. That is the (x,t) frame. See above.

Janmes R, I was very careful. Now it is your turn to be careful. Look at what you have just posted. The ground clocks and the moving clocks at L anad R (ground and moving frame) are co-located in space.

This is true at the times of the photons hitting L and R, but not at any other time.

There is no possible way in which the clocks in the moving anad stationary frame cannot all measure a different (that is sequential) time of arrival for the photns at L then at R.

I have shown mathematically that this is indeed possible, and is in fact what occurs (unless, of course, you want to dispute the constancy of the speed of light for all observers).

What is true in one frame but not another? How does the L and R detectors know to record a different time? The photons kniow nothing of SRT, coordinate frames or observers. The photons move all indenpendently of all of these aspects of reality.

It seems that clocks are dependent on the motion, doesn't it? See my analysis above.

Be candid JAmes R, look at the fact that photons are generated at moving targets. One tarhget is moving toward an on coming photon, the other photon is moving at a target that is moving away.

Only in the Earth frame. In the rod frame, neither target is moving at all.

When have you ever shown that the clocks will record differently? Certainly not in this thread.

Can't you follow my mathematics? I think it is quite clear.

Besides you bringing SRT nonsense into the thread as you said in the opening thread you were not going to do, does not help you.

Of course I'm bringing "SRT nonsense" into this thread - that is the whole point of it. The relativity of simultaneity is a relativistic phenomenon. It would not occur if our universe was Newtonian. I stated quite clearly what I assumed, in my very first post of the thread.

We aren't concerned with the specific time on the clocks or whether the stationary and moving frame clocks tick at the same rate, or even frame contraction, we are concerned with whether thre arrival times are sequential or simultaneous.

Right. I have shown that different frames give different answers. Hence, simultaneity is relative.

You must agree that the test made in the stationary frame will result in the photons arriving simultaneously at L and R.

Yes. Put v=0 in my mathematical equations above, and that becomes very obvious.

Then if we take your frame into space (the observer is asleep) at some velocity, decelerate to a velocity Vf = Ve = 0 (we assume the embankment is stationary) he will repeat the earth results, unambiguously.

Yes.

Then if he is acceleraqted, and knows it, he will get another result, differeing by the length of time the photon take for the roundtriop, or t' =t (2v)/(c - v). How does he explain the difference?

He explains the difference by realising that he is now in a different reference frame.

Caslculations have absolutrely nothing to do with what the clocks record, which will be sequsntial arrival times of the photons.

If you think my calculations above are incorrect, please point out where I went wrong.

So then the data will show that clocks, on the stationary and moving frame will record thagt they are moving slower than the other?

What do you mean by "moving slower"? Do you mean that each clock would observe the other as running slower? If so, then yes, if we collected that kind of data (which is not part of the original post above) that is what would be recorded.

There is a mechanical switch on both L and R extending into the stationary frame. From previous experiments at this same velocity the stationery clocks are placed within a wavelength of light from the passing moving frame clocks. When the mechanical switches on the moving frame strike the clocks that are effectively colocated with the moving frame clocks all four clocks stop simultaneously as measured by all clocks that are synchronized within each frame.

The clocks won't stop simultaneously in both frames. Your switch is analogous to the photons travelling, and doesn't add anything fundamentally new to the situation.

 

[Xgen]

Observer on the rod:

The half-length of the rod is measured to be d' (which might be different to d).
At time t'=0, a flash of light is emitted at M, designated as x'=0, and two photons head off towards L and R, initially located at x'=-d' and x'=d', respectively.
The positions of the photons as a function of the time t', measured on the rod observer's clock (hence t' instead of t, since these might be different), are:

PL(t') = -ct'
PR(t') = ct'

The positions of the detectors as a function of time are:

L(t') = -d'
R(t') = d'

Notice, the detector positions don't change with time in this reference frame (the view of an observer standing on the rod).

At what times are the two photons detected? The photon is detected by the left detector when:

PL(t') = L(t')
-ct' = -d'

Solving, we find t'=d'/c.

The right-hand photon is detected when

PR(t') = R(t').

Solving, we find t'=d'/c.

Obviously, these times are the same, so according to the observer on the rod, the detectors both register the photons SIMULTANEOUSLY.

James, isn't there a problem here? According to you both photons will pass the same distance - d' to the left and right. This can happen only if photons 'move with' the frame. Then to the other frame - the rest frame (I prefer the Absolute Space) it would appear that photons move with c+v and c-v which is violation to light invariance principle. The photons do not move together with frames. You can consider them existing only in the totally inmovable absolute space.

Why not try to make a computer program demonstrating this phenomenon? You will see that 'photons moves with the frame' is the only way L and R photons to reach simultaneously both detectors. Well, there is and another option but I dont consider it seriously - we can speculate that the space is 'smaller' in the direction of movement. i.e that relativity is anisotropic. In this case the R photon will pass smaller distance then the L photon. It is legally then his velocity to be bigger then c and we can say acually that photons are moving with the frame and the detectors will register the photons SIMULTANEOUSLY. Hmm, I had to think about this.

Anyway there is following options for this L and R experiment:

1) photons are registered SIMULTANEOUSLY, SRT is not anisotropic, but invariance principle is invalid

2) photons are not registered SIMULTANEOUSLY, SRT is not anisotropic, invariance principle is valid

3) photons are registered SIMULTANEOUSLY, SRT is anisotropic, invariance principle is valid

I had never seen the notion that SRT is anisotropic. This phenomenon may be harder to track then it seems. But if the both detectors register the photons SIMULTANEOUSLY this is the only option (and it is by the way not in the violation of the existence of the Absolute Space, as not is SRT and GRT)

What do you think James?

 

[geistkiesel]

Geistkiesel:



A correct statement is that the rod does not move in the x',t' coordinate system.

And it would be incorrect to say the rods aren't moving, period, however you construct your model.

Down below in the body of this text you admit the frame is moving wrt the ground. You cannot have it both ways James R.

The coordinate systems are simply ways of quantifying the observations of the observers. And I have shown mathematically that in one frame, the photons arrive simultaneously, while in the other frame they do not.

Quantifying observations. This is measuremnt isn't it? You read sequential times of arrival on the L and R clocks. Your quantification is erroneous and you use no physics to come to your calculated conclusion. You ignore data, you ignore the taking of data, you iognore the motion of the frame in you "quantification". You can only quantify the result from the data acquired during experiment.

Your quantification is erroneous as you negate the motion of the moving frame.

Why did you not negate the "at rest" condition of the stationary frame when you "quantified" those results? Why did you not have the frame moving when you quantified that aspect of the test?

If you can see an error in my mathematics, please point it out.

1 + 1 = 2. You seem have followed the rules of algebra above. You didn't follow the rules of physics.

Yes, it could. For example, the person on the rod could drop a flag off the rod onto the ground at the instant the flash is triggered at M.

Then you have determined the velocity of the frame. You drop the flag to the ground; the flag strikes the ground and bounces in the direction of its acquired momentum. If, for instance if the moving observer was at rest, a physical impossibility when it is moving, the ground would seem to be moving to the left. If he ground were moving and the frame at absolute rest the flag would bounce in the direction of the acquired momentum from the ground.

But this cannot happen as the ground cannot move while the frame remains at rest.

You have just determined the motion of the frame to the right.

I did. That is the (x,t) frame. See above.

No, you think you did. I am discussing the independent absolute location of the emission point that is invariant and only coincidentally measurable wrt the stationary frame and the moving frame as well.
This is an absolute reference point in space.

This is true at the times of the photons hitting L and R, but not at any other time.

The times the photons hit L and R are the only significant times.

The four clocks are colocated and read the arrival times. L and R on the moving frame are indistinguishable from their counterparts located on the ground. The detectors read the arrival times at L and R simultaneously with their ground counterparts, L and the ground detector measure the same arrival instant, as does R and the counterpart to R colocated with R.

I have shown mathematically that this is indeed possible, and is in fact what occurs (unless, of course, you want to dispute the constancy of the speed of light for all observers).

No the speed of light is constant for all observers. The only thing is that the relative velocity of frame and photon are described as C - V and C + V depending on the realtive directions of frame and photon. These two simple expression do not lower and/or increase the claimed speed of light. The measurement is for the relative velocity of light, not the velocity of light. The simplest and most direct way to measure this is to use a common reference point say the invariant point of emission of the photons.

You can use your flag to measure the speed of the frame, and you can use the value 3 x 10^5 km/sec for light and voila, you have the relative speed of frame and photon. This is basic physics 101.

It seems that clocks are dependent on the motion, doesn't it? See my analysis above.

The clocks read arrival times. The clocks are synchronized on the moving frame. The clocks are located symmetrically wrt the emission point loated at the midpoint of L and R. No the clocks are not dependent on the motion of the frame. The measured arrival times are the only measurement of significance here in determining whether the photons arrive simultaneously at L and R. They do not.

The L and R detectors are moving wrt the emission point, the stationary observer is non-existent. All we have is photons and detectors and the photons must arrive at different times at L and R. This should be, and is, embarrassingly obvious.

The arrival times difference between L and R is

t' = t(2v)/(c - v)

or in terms of velocity

v = t'c/(2t + t').

You promised to reply to the mathmatics here. I would like to see your analysis. I am sure the readers might also be interested in this matter.

Only in the Earth frame. In the rod frame, neither target is moving at all.

You have proved the incorrectness of your statement here when you dropped the flag to the ground. Clearly the midpoint is moving, as the moving observer saw himslef move, and therefore the targets are moving.

Can't you follow my mathematics? I think it is quite clear.

Do you think that your condescending tone and attitude are going to get you points with readers? To quote 2inquisitive in another thread, " I am not a fucking moron". If you are a "mentor", then act like one.

Your mathematics are grossly flawed as you have inserted physical errors in your assumptions. The mathematics is grade school simple and describe clearly your errors.

Of course I'm bringing "SRT nonsense" into this thread - that is the whole point of it. The relativity of simultaneity is a relativistic phenomenon. It would not occur if our universe was Newtonian. I stated quite clearly what I assumed, in my very first post of the thread.

Well then you have just declared the Newontian Nature of the universe with your faulty assumptions regarding the"mathematics" you hold so dear to your heart.

Right. I have shown that different frames give different answers. Hence, simultaneity is relative.

No you have shown the opposite. The sequerntial arrival times of the photons at L and R prove motion The simultaneous arrival of the photons back at the physical midpoint prove motion when making calulations.

Yes. Put v=0 in my mathematical equations above, and that becomes very obvious.

Put V = 0 in my equations and you get the obvious and correct version.

Yes.

Good

He explains the difference by realising that he is now in a different reference frame.

Good again. In other words he realizes he is moving. Now realising as he does his own motion the conumdrum of SRT is solved.

If you think my calculations above are incorrect, please point out where I went wrong.

I have done this see above.

[quiote=James R]What do you mean by "moving slower"? Do you mean that each clock would observe the other as running slower? If so, then yes, if we collected that kind of data (which is not part of the original post above) that is what would be recorded.[/quote]
Explain how each clock can be measured slower than the other.

The clocks won't stop simultaneously in both frames. Your switch is analogous to the photons travelling, and doesn't add anything fundamentally new to the situation.

The correct terms is switches, plural ,not switch. Didn't you read the statement clearly?

No, the mechanical switch is on the frame at L and R and each mechanical switch triggers the stopping of the L and R detectors on the frame and the stationary ground. Coincidentally the photons and switch arrive at the ground located detectors at the same time, first at L then at R.

Geistkiesell​

 

[geistkiesel]

James, isn't there a problem here? According to you both photons will pass the same distance - d' to the left and right. This can happen only if photons 'move with' the frame. Then to the other frame - the rest frame (I prefer the Absolute Space) it would appear that photons move with c+v and c-v which is violation to light invariance principle. The photons do not move together with frames. You can consider them existing only in the totally inmovable absolute space.

I agree Xgen, 100%

Why not try to make a computer program demonstrating this phenomenon? You will see that 'photons moves with the frame' is the only way L and R photons to reach simultaneously both detectors. Well, there is and another option but I dont consider it seriously - we can speculate that the space is 'smaller' in the direction of movement. i.e that relativity is anisotropic. In this case the R photon will pass smaller distance then the L photon.

This suggestion as you recognized has some problems. Is not he most glaring problem a violation of the symmetry of length of the rods on both sides of the midpoint?

It is legally then his velocity to be bigger then c and we can say acually that photons are moving with the frame and the detectors will register the photons SIMULTANEOUSLY. Hmm, I had to think about this.

Symmetry rears it correcting head here also, does it not?

Anyway there is following options for this L and R experiment:

1) photons are registered SIMULTANEOUSLY, SRT is not anisotropic, but invariance principle is invalid

2) photons are not registered SIMULTANEOUSLY, SRT is not anisotropic, invariance principle is valid

3) photons are registered SIMULTANEOUSLY, SRT is anisotropic, invariance principle is valid

I had never seen the notion that SRT is anisotropic. This phenomenon may be harder to track then it seems. But if the both detectors register the photons SIMULTANEOUSLY this is the only option (and it is by the way not in the violation of the existence of the Absolute Space, as not is SRT and GRT)

What do you think James?

I think you have outlined the possibilities very nicely. An experiment designed around this moving frame condition should be conducted, but who has the balls to conduct the experiement?

Geistkiesel​

 

[James R]

Xgen:

James, isn't there a problem here? According to you both photons will pass the same distance - d' to the left and right. This can happen only if photons 'move with' the frame. Then to the other frame - the rest frame (I prefer the Absolute Space) it would appear that photons move with c+v and c-v which is violation to light invariance principle.

Note that I assumed the invariance of the speed of light right at the start of my explanation. geistkiesel, for example, doesn't believe in the invariance of the speed of light, which is why he disagrees with my presentation (even if he doesn't realise that is why he disagrees).

It is not really a case of the photons "moving with the frame". It is simply that the speed of photons is determined to be the same in all frames, despite the relative motion of the frames. That is the counterintuitive assumption of special relativity.

The photons do not move together with frames. You can consider them existing only in the totally inmovable absolute space.

Relativity has no concept of absolute space.

Well, there is and another option but I dont consider it seriously - we can speculate that the space is 'smaller' in the direction of movement. i.e that relativity is anisotropic. In this case the R photon will pass smaller distance then the L photon.

That is not necessary. In fact, d' is a smaller distance than d in my example (though I haven't proven that). That result is relativistic length contraction, which follows from the assumption of the invariance of the speed of light. Length contract is anisotropic. It occurs only in the direction of motion, and not perpendicular to that direction. However, in this example it does not act differently to the left and right of M.

 

[MacM]

That is not necessary. In fact, d' is a smaller distance than d in my example (though I haven't proven that). That result is relativistic length contraction, which follows from the assumption of the invariance of the speed of light. Length contract is anisotropic. It occurs only in the direction of motion, and not perpendicular to that direction. However, in this example it does not act differently to the left and right of M.

In which case length contraction in no way alters the outcome of timing since the distance is foreshortened equally between both the L & R reflectors.

 

[James R]

geistkiesel:

And it would be incorrect to say the rods aren't moving, period, however you construct your model.

Down below in the body of this text you admit the frame is moving wrt the ground. You cannot have it both ways James R.

The key word is "relative".

The rod moves relative to the ground. It does not move relative to the (x',t') frame. Setting up the (x',t') frame is equivalent to laying out a ruler on top of the rod, such that the ruler moves with the rod. At various times, we note the positions of the left and right ends of the rod, as measured with the moving ruler. What do we find? The positions are always the same! Surprise surprise. That's because the rod is stationary relative to the ruler (i.e. stationary in the (x',t') frame).

The (x,t) frame, on the other hand, involves laying our ruler on the ground, so that it does not move with the rod.

See?

Quantifying observations. This is measuremnt isn't it? You read sequential times of arrival on the L and R clocks. Your quantification is erroneous and you use no physics to come to your calculated conclusion.

My physics is transparently laid out for you in my first post. Didn't you read it?

You ignore data, you ignore the taking of data, you iognore the motion of the frame in you "quantification".

I have explicitly not ignored the motion. Do you see the term "v" in my mathematics? That's the motion.

Your quantification is erroneous as you negate the motion of the moving frame.

I haven't negated the motion of the moving frame, relative to the ground. Of course, in its own frame, the rod doesn't move, as I have explained.

Are you getting a feel for frames of reference yet?

Why did you not negate the "at rest" condition of the stationary frame when you "quantified" those results?

I never said the ground was at rest. For example, the ground usually moves through space at about 200 km/second relative to the centre of the galaxy. But who cares? It doesn't affect the analysis in this problem, does it?

1 + 1 = 2. You seem have followed the rules of algebra above. You didn't follow the rules of physics.

Now now. You shouldn't make statements like that unless you can back them up. Which rules of physics did I get wrong?

But this cannot happen as the ground cannot move while the frame remains at rest.

Can you see the rod move, if you're standing on the ground? Can you still see the rod move even if you have a ruler laid out on the ground? I think so.

The four clocks are colocated and read the arrival times. L and R on the moving frame are indistinguishable from their counterparts located on the ground.

Indistinguishable in a physical sense, yes, but not in the times they record, as I have shown.

The detectors read the arrival times at L and R simultaneously with their ground counterparts, L and the ground detector measure the same arrival instant, as does R and the counterpart to R colocated with R.

No. See my mathematical description above, with proves the exact opposite. There is no error in my mathematics.

No the speed of light is constant for all observers. The only thing is that the relative velocity of frame and photon are described as C - V and C + V depending on the realtive directions of frame and photon. These two simple expression do not lower and/or increase the claimed speed of light. The measurement is for the relative velocity of light, not the velocity of light. The simplest and most direct way to measure this is to use a common reference point say the invariant point of emission of the photons.

I don't think you understand what is meant when we say "the speed of light is the same for all observers".

Suppose you throw a ball to the right at speed v. Suppose I am moving to the left at speed w. Then, I would see the ball moving to the right at speed v + w. Do you agree?

Suppose you throw a photon to the right at speed c. Suppose I am moving to the left at speed v. Then, I would see the photon moving to the right at speed v + c. Do you agree?

Well, relativity doesn't agree with the second example. It says that I would see the photon moving to the right at speed c, and NOT at speed v + c. Rant and rave all you want to about how it "doesn't make sense". The fact is, this is a conclusion amply supported by experimental evidence.

The arrival times difference between L and R is

t' = t(2v)/(c - v)

I don't know how you calculated that. Do your t' and t refer to the times measured in the different reference frames, as mine do, or is this some kind of different notation for two different times in the one reference frame?

I calculate the time difference in the ground frame between the arrival of the photons at L and R to be 2dv/(c^2 - v^2). Do you agree? The time diffence in the moving frame is, as I have shown above, zero.

Do you think that your condescending tone and attitude are going to get you points with readers? To quote 2inquisitive in another thread, " I am not a fucking moron". If you are a "mentor", then act like one.

I tend to respond in kind, geistkiesel. If you claim, without justification, that I don't know what I'm talking about, that I use no physics, that my mathematics is incorrect, then I think a bit of condescension is quite justified, don't you?

Why don't you stop whining, and start explaining where I went wrong?

Your mathematics are grossly flawed as you have inserted physical errors in your assumptions. The mathematics is grade school simple and describe clearly your errors.

So show me! Don't talk about how easy it would be to show me the error in my ways. Do it!

Good again. In other words he realizes he is moving. Now realising as he does his own motion the conumdrum of SRT is solved.

The rod observer realises the ground is moving, relative to him.

 

[kevinalm]

James R

t' = t(2v)/(c - v) is an absolute space/absolute time "etheric" equation Geist derived. For what it is it is actually correct. t is the time for a wavefront to travel the parallel arm in the direction counter to the motion of the apparatus. t+t' is the time in the same direction as the apparatus is moving and t' is the difference. 2t + t' is the total time for the parallel arm of the M/M experiment. He uses an unusual method that is a little hard to follow. What he doesn't know is that since t=l/(c+v), 2t+t'=(2l/c)*gamma^2 in agreement with the standard pre SRT analysis of the M/M.

Geist doesn't accept the idea that c is constant for _all_ inertial frames. He seems to think that the composite nature (t para vs. t perp the total para instead of it's forward and reverse parts) of the M/M is hiding a measurable t' . I don't believe he realizes how small t' actually is or that it is almost certainly beyond current technology to measure. Those of us that accept SRT as correct are of course of the opinion that if you could directly measure t' it would =0 for any inertial frame.

In fairness to Geist I would say if you could measure t' and found it to not equal 0 always it would be very bad for SRT/GR.

 

[Quantum Quack]

Maybe someone can clarify for me whether simultaneousness NOW's are applicable [see animated diagram below]

Each "Row" of seconds represents relative time duration but in all circumstances the NOWs are simultaneous for both Frames.

Is this a correct assessment?

 

[Xgen]

Note that I assumed the invariance of the speed of light right at the start of my explanation. geistkiesel, for example, doesn't believe in the invariance of the speed of light, which is why he disagrees with my presentation (even if he doesn't realise that is why he disagrees).

James, invariance of the speed of light means that photons do not cares and do not knows that there are moving frames, they exist in their own frame which I call abs. space and which is a concept which is nessecary for SRT equations.

Please tell me, if invariance of the speed of light is valid for any frame, then how is possible both photons to move with c in both frame - the rest and this moving and to arrive at the same times in second and at different in first? If the second frame moves wrt the absolute frame then photons , remaining in their own frame, will arrive at different times and will pass different paths in all frames except the frame of the absolute space. It is just the opposite of what you had given, in absolute space photons arrive simultaneously and in ALL other frames they will arrive at different times (if time is homogeneous for entire frame).

It is not really a case of the photons "moving with the frame". It is simply that the speed of photons is determined to be the same in all frames, despite the relative motion of the frames. That is the counterintuitive assumption of special relativity.

Yes. But photons can be considered in many frames at the same time. Then their speed should be c in all frames. That is why photons are not "carried away" from the frame. They do not care that the frame is moving. It is not possible they to pass the same distance d' to the left and right in the second frame and their speed to remain c in the first.

What if second frame moves with v = c?

1 <-------|------->2
L-----------------------R ----> c

Then 2 will never reach R while 1 has no problem reaching L. The time for 2 to travel to R is ininity while for 1 to travel to L is finit.

That is not necessary. In fact, d' is a smaller distance than d in my example (though I haven't proven that). That result is relativistic length contraction, which follows from the assumption of the invariance of the speed of light. Length contract is anisotropic. It occurs only in the direction of motion, and not perpendicular to that direction. However, in this example it does not act differently to the left and right of M.

What about the opposite to the direction of motion? In relativity equations v the relative velocity participate only as absolute quantity. Maybe it shoudn't be so. I dont know how had been proved that lenght contraction in the direction of motion is the same as the opposite direction:

S2<----------|---------->S1
--------------|-------------> v

I know that it is not considered in current SRT but if you insist that both photons arrive at the same time for every frame in motion you will see that this is the only option (it should be called non-homogeneos space contraction, sorry)
- S1 is different then S2. Time also will be unhomogeneous.

I will show what follows if T1 = T2.
This can happens only if light is "carried away" with the moving frame. Then

T1 = T2 = T
V1 = c - V
V2 = c+ V
S2/T = V2 = c+V
S1/T = V1 = c-V

then S2 = S / c - v
S1 = S / c + v

Then v' - the speed of photons in second frame may appears c even if it is bigger then c because space is contracted differently in both directions. We will observe the same with a totally different theory about relativity!

 

[James R]

Xgen:

James, invariance of the speed of light means that photons do not cares and do not knows that there are moving frames, they exist in their own frame which I call abs. space and which is a concept which is nessecary for SRT equations.

SR does not require an "absolute space" frame. In fact, it says there is no such thing.

Please tell me, if invariance of the speed of light is valid for any frame, then how is possible both photons to move with c in both frame - the rest and this moving and to arrive at the same times in second and at different in first?

See my first post in this thread. Do you follow the math?

It is just the opposite of what you had given, in absolute space photons arrive simultaneously and in ALL other frames they will arrive at different times (if time is homogeneous for entire frame).

So, would you say my moving rod (from the first post) is "absolute space", while the Earth is not absolute space? The photons arrive simultaneously in the rod frame, but not in the Earth frame, as I have shown.

Yes. But photons can be considered in many frames at the same time. Then their speed should be c in all frames.

That's right. That's one of the postulates of SR, which I used in my example. The "c" factor in the equations for both frames is the same.

It is not possible they to pass the same distance d' to the left and right in the second frame and their speed to remain c in the first.

It is possible. But (and it's a big "but"), it requires us to throw away our intuitive concept of absolute universal time, and absolute space, which is exactly what SR does.

What if second frame moves with v = c?

Nobody knows. SR doesn't deal with frames moving at v = c. It can only handle frames moving at less than the speed of light. This is reasonable, since no material object can ever move at the speed of light.

What about the opposite to the direction of motion? In relativity equations v the relative velocity participate only as absolute quantity. Maybe it shoudn't be so. I dont know how had been proved that lenght contraction in the direction of motion is the same as the opposite direction:

Maybe I'll post something about that soon.

I know that it is not considered in current SRT but if you insist that both photons arrive at the same time for every frame in motion you will see that this is the only option

If you insist that both photons arrive at the same time in all frames, then you have universal time. But then you throw away the invariance of the speed of light in all frames. You can't have both. The problems with universal, absolute time, are part of what led to the development of special relativity in the first place. These days, we know it doesn't exist. It is inconsistent with measurable aspects of reality.

 

[geistkiesel]

James R

t' = t(2v)/(c - v) is an absolute space/absolute time "etheric" equation Geist derived. For what it is it is actually correct. t is the time for a wavefront to travel the parallel arm in the direction counter to the motion of the apparatus. t+t' is the time in the same direction as the apparatus is moving and t' is the difference. 2t + t' is the total time for the parallel arm of the M/M experiment. He uses an unusual method that is a little hard to follow. What he doesn't know is that since t=l/(c+v), 2t+t'=(2l/c)*gamma^2 in agreement with the standard pre SRT analysis of the M/M.

Geist doesn't accept the idea that c is constant for _all_ inertial frames. He seems to think that the composite nature (t para vs. t perp the total para instead of it's forward and reverse parts) of the M/M is hiding a measurable t' . I don't believe he realizes how small t' actually is or that it is almost certainly beyond current technology to measure. Those of us that accept SRT as correct are of course of the opinion that if you could directly measure t' it would =0 for any inertial frame.

In fairness to Geist I would say if you could measure t' and found it to not equal 0 always it would be very bad for SRT/GR.

kevinalm, Show me an instyance where I have not recognized the constancy of the speed of light. What i have been saying is that the measure of the SOL and the measure of the relative speed of light wrt ftrame and photon are two sifferent physical measutrements that cannot be erased bu theoretuical fiat.

I have always recognized the constancy of the speed of light, James R is full of shit, and he knows it. My model deriving t' uses the moving frame clocks that know nothing of the photons, which know nothing of the frame or clocks.

Look at the point of emission of the photons as if the frame were in deep space away from any stationary frame. The point of emission of the light is an invariant point in absolute space. I realize how small t' is. The larger the experimental conditions, dimensions, the larger the t'. For a velocity equal to .1c and ct = 10 km what do you calculate?

t' = t (2v)/(c - v) = t (.2c)/.9c = t/4.5.
t =10/3x10^5
t' = (3.33 10^-5)/4.5 = 7.4 x10^-6 sec.

This is on the order of a microsecond for a frame moving at .1c. The velocity of the frame is the limiting factor, i.e. 2v/(c-v).

This experiment could be conducted on the earth surface.

Geistkiesel [/indent]

 

[geistkiesel]

James R

t' = t(2v)/(c - v) is an absolute space/absolute time "etheric" equation Geist derived. For what it is it is actually correct. t is the time for a wavefront to travel the parallel arm in the direction counter to the motion of the apparatus. t+t' is the time in the same direction as the apparatus is moving and t' is the difference. 2t + t' is the total time for the parallel arm of the M/M experiment. He uses an unusual method that is a little hard to follow. What he doesn't know is that since t=l/(c+v), 2t+t'=(2l/c)*gamma^2 in agreement with the standard pre SRT analysis of the M/M.

Geist doesn't accept the idea that c is constant for _all_ inertial frames. He seems to think that the composite nature (t para vs. t perp the total para instead of it's forward and reverse parts) of the M/M is hiding a measurable t' . I don't believe he realizes how small t' actually is or that it is almost certainly beyond current technology to measure. Those of us that accept SRT as correct are of course of the opinion that if you could directly measure t' it would =0 for any inertial frame.

In fairness to Geist I would say if you could measure t' and found it to not equal 0 always it would be very bad for SRT/GR.

kevinalm. I do accept the constancy of the speed of light, I also accept that measuring the relative velocity of a frame and photon moving parallel to each other is C -v, which is a measure of relative velocity and is not a measure of C at a value less than 3x10^5 km/sec. It is a measure of a relative velocity. The SRT effectively zeroes out the frame motion to always have the value c in the realtive velocity expression. This is corrupt physics, ignoring a critical motion in a critical physical measurement.

What next must I demonstrate to you?-That two inertial frames in open space can detect their absolute velocity wrt each other. I mean the velocities that sum to their mutually measured relative velocity.

Here is a figure where the light motion like you analyzed from a previous post is referenced ftrrom the point of emission of the light. Notice the physical midpoint of the LR frame moving wrt the invariant emission point. What is violated here wrt SRT?

L and R clocks that spit out the time of arrival of the photons on pieces of paper (and/or returns the times in the reflected signal reflected back to the physical midpoint). Clearly the L clock records an arrival before the R clock. What observation point can any observer in the universe have that can negate the actuality of the conclusions implied in the figure? Observers aren't necessary here, and what is simultaneous in one frame is simultaneous in another.

Now what do I not accept about the speed of light? Some theory or another perhaps?

Thanks for the post kevinalm,
Geistkiesel.​

 

[kevinalm]

Ok, now consider that those of us who accept SRT see the M/M experiment (and many many other experiments) to utterly fail to measure v. In any direction (orientation). Also, that the conceptual analysis that failed is the exact sort of absolute frame/abs time light propagating at c only in that frame that you are working with. We aren't sure why it failed. Lorentz had some ideas, but made limited progress.

So Einstein broke the deadlock by saying, "what if in any/all intertial frames you always measure the speed of light as a constant c?" That is what is meant by the invariance of c. In my system of coordinates I measure c, isotropic in all directions. In your system of coords which is moving at constant v wrt me, you measure c as constant isotropically as well.

Yep, it's screwy. The only thing screwier is the fact that a lot of the predictions SRT makes are verified experimentally. Pretty much everything we have been able to test.

I certainly understand your objections to SRT. And I for one welcome experiments testing all aspects of SRT. You've seen SL's experiment thread, right. Iirc, you have posted as well.

Anyway, the way I see it is that if an absolute frame exists then the "laws" of physics must make it impossible (or nearly so) to detect. And if that is true then SRT must give very acurate "answers", at least in terms of what can be measured.

Regards,
Kevin

 

[Quantum Quack]

As a photon itself is essentially unobservable would it not be more correct to say that "the effect of light is invariant to all observers"?

There is an important distinction in the use of the word "effect" rather than "Photon"
The thinking:
The effect of light is all that we are measuring, not the photon itself but it's effect.

 

[MacM]

Ok, now consider that those of us who accept SRT see the M/M experiment (and many many other experiments) to utterly fail to measure v. In any direction (orientation). Also, that the conceptual analysis that failed is the exact sort of absolute frame/abs time light propagating at c only in that frame that you are working with. We aren't sure why it failed. Lorentz had some ideas, but made limited progress.

So Einstein broke the deadlock by saying, "what if in any/all intertial frames you always measure the speed of light as a constant c?" That is what is meant by the invariance of c. In my system of coordinates I measure c, isotropic in all directions. In your system of coords which is moving at constant v wrt me, you measure c as constant isotropically as well.

Yep, it's screwy. The only thing screwier is the fact that a lot of the predictions SRT makes are verified experimentally. Pretty much everything we have been able to test.

I certainly understand your objections to SRT. And I for one welcome experiments testing all aspects of SRT. You've seen SL's experiment thread, right. Iirc, you have posted as well.

Anyway, the way I see it is that if an absolute frame exists then the "laws" of physics must make it impossible (or nearly so) to detect. And if that is true then SRT must give very acurate "answers", at least in terms of what can be measured.

Regards,
Kevin

Unfortunately your statements are only partially true.

1 - All tests including M/M have not been null. They merely measured a dinural cycle which was far less than predicted or supported by a stationary ether concept. Numerous tests after M/M, Miller (200,000 data samples over many years) and other more recent tests have found the same unexplained dinural cycles. that should not continue to be viewed as null.

2 - Being undetected does not in of itself pose any basis to reject its reality. That is simply unscientific and short sighted.

3 - The accurate answers you mention are not the whole of the theory. The only data ever collected ONLY supports a one way gamma calculation. SRT is a collection of assumptions including reciprocity which has not once ever been recorded or observed.

I am very much in favor of SL's experiment as long as it is properly thought out. I have every confidence that such a test will result in the falsification of all the troubling aspects of SRT.