[David Cooper]

Hi GoC,

If I'm understanding you to any extent, you appear to have backtracked on length-contraction by using some kind of "aura" effect on the MMX. Is that actually what you're saying, and if so, do you think there would be length-contraction acting on it if the apparatus was in deep space and moving at high speed?

[David Cooper]

Whether individual paths or a closed course, the ratio of emitted signals to received signals is what's stated with the drawing, and none are 1:2 or 2:1. It seems like you are using instantaneous light speed. The only thing conserved is the exchange of 4 yr and 2 yr.

The 1:2 ratio is for the average, as stated. Each rocket will calculate that the other rocket is emitting one beep every two seconds, and they will measure it as that too if they count the beeps for a length of time while the rockets approach each other and keep counting beeps for the same length of time after they've passed each other.

The accounts are not false, but valid, since any inertial frame can serve as a reference. I.e. there is no need for an absolute rest frame. That is the 'principle of relativity'.

The accounts contradict each other and cannot all be true, so most of them are false. Our inability to pin down which ones are false and which true does not negate the necessity for one to be true and for any that contradict it to be false. The universe has to run on the basis of one frame and not on an infinite number of them at once where they produce contradictions. If you try to run a simulation without tying everything to a universal time, your simulation will produce event-meshing failures, and the same applies to the running of the real universe.

[GoC (OP)]

Hi GoC,If I'm understanding you to any extent, you appear to have backtracked on length-contraction by using some kind of "aura" effect on the MMX. Is that actually what you're saying, and if so, do you think there would be length-contraction acting on it if the apparatus was in deep space and moving at high speed?

 Highly unlikely. The more likely scenario would be a light clock orientation may affect synchronization between mirrors.
Although if the aura of spacetime energy c rotated with a solar system and then the galaxy we might have the same tick rate at a rotation distance from the sun same as sea level of Earth. Making perpendicular possible and distances traveled forward and back canceling the difference with perpendicular. Try to consider the implications of clocks ticking the same at sea level all over the planet regardless of rotation distance traveled. That may or may not be the same in space. The MMX null result may just be an Earth phenomenon. This should suggest gravity and light are comprised of the same thing (energy). Measured in potential energy dilation.   

Both voyagers increasing signal return past the aura was identified as the voyagers slowing down. More likely the space energy reduced dilation threshold past the solar system increasing the tick rate of their clocks not reducing their velocity. It's all about energy even time is just energy c.

[puppypower]

The two atoms emit their own light at their own frequency, and they have to move so that the light from the other atom looks as if it had the same frequency as theirs. If the left atom has moved towards the right one before its light had the time to reach the right one, that right one will move away from the left one after a while, and it will do so also before its light had the time to reach the left one, so that left one will also move forward after a while. We can very well see how contraction would happen between the two, but it is less clear how dilation would happen. What seems clear to me though is that, even if it happened, it wouldn't affect the contraction or the speed of the system, because it wouldn't affect the synchronization between the two atoms.

Consider this scenario. We two hydrogen atoms, one is stationary and the other is moving away. The moving hydrogen gives off an energy quantum, that red shifts on its way to the stationary hydrogen due to the Doppler shift. The second hydrogen atom will not be able to parallel the electron transition of the first hydrogen, because the energy quanta has red shifted, and therefore defines lower energy.

The impact of the lower energy photon on the stationary hydrogen is to cause an apparent distance contraction, since the electron has to stay closer to the nucleus. Also the frequency is less since the red shifted energy is at lower frequency, making it look time dilated.

The point I am making is even if only one hydrogen is moving, that motion which impacts the energy output that reaches   the stationary reference. The stationary hydrogen gets lower energy, due to the red shift, and physically reacts in ways that appear that parallel the relativistic time and space change, to create the no preferred reference illusion. Only one reference has real relativity the other is a physical reflection.

[guest4091]

David Cooper #341

Diverging, A receives 12 months signals from B in 40.4 months.
B receives 7.6 months signals from A in 12 months.
You cannot get a 2:1 or 1:2 ratio from that!
The clock rates are not the same as the doppler rates of perception.
Each account is the perception of the observer, which is correct. Perception is also reality.

[David Cooper]

Diverging, A receives 12 months signals from B in 40.4 months.
B receives 7.6 months signals from A in 12 months.
You cannot get a 2:1 or 1:2 ratio from that!

You're averaging the wrong things. You provided the following numbers before:-

The drawings correctly show 3.73 between signals when separating and .27 between signals when closing, for both A and B, when using the relativistic Doppler expressions.

If you hear 3.73 beeps in a length of time t (measured by your clock) while your rocket approaches the other rocket (and reaching it at the end of that time), and you then hear 0.27 beeps in the the length of time t as you move away from the other rocket (having passed it), you have heard 4 beeps in time 2t. That is a 2:1 ratio, although it's the opposite way round from the one I'd expected, which means the wording on my page may need to change (assuming that time t represents one beep of the local clock - I haven't got round to doing these calculations myself). The idea I wanted to get across is simply that each rocket will, if it assumes itself to be stationary, calculate that the other rocket's clock is beeping half as often due to its speed of travel, as would be observed from the "God view" of their frame of reference (which they can calculate). The God view isn't directly available to them, of course, so perhaps they really do receive twice as many beeps on average (instead of half as many) even though the other clock is beeping half as often. I'm beginning to see how that might be the case, because as they approach each other, a lot of the beeps from the moving rocket that are reaching the stationary rocket were sent out long before the timing period began, thereby over-representing the rate of their production considerably, and that can't be cancelled out by the massive reduction in received beeps during the second half, so I think you have identified an error on my page which needs to be corrected.

[David Cooper]

The MMX null result may just be an Earth phenomenon.

And the MM experiment has never been done anywhere other than on the surface of the Earth, so there is actually some sense in your position. Perhaps it could show co-moving, perpendicular light clocks drifting out of sync in space, although relativistic mass and the precise slowing of decay of fast-moving particles suggests that length contraction should happen in all cases.

[David Cooper]

You may not have read yet how I describe the motion between my two atoms, so here it is again:

We have two atoms A and B that are part of the same molecule. The time interval represents the time the information takes between the two atoms at t0. We accelerate A for a while and observe what happens to the system from t0 to t7. The blue arrows represent the blueshifted information that travels from A to B, and the red arrows represent the redshifted information that travels from B to A. The acceleration of A begins at t0 and ends at t4, so because of the time gap, the acceleration of B begins at t1 and ends at t5. After t5, the two atoms travel at the same speed, but we can easily see that the distance between them has contracted, and we can follow its progression during the acceleration. At that moment, the information on the future speed of each atom with regard to aether is situated between them in the form of doppler effect. The main idea is that, without doppler effect, there would be no motion between bonded particles, so there would be no motion either at our scale. I insist on the fact that we have to exert a force to introduce that doppler effect between them, and that this force represents mass. So with the same principle, we explain mass, motion and contraction. Of course motion is a bit more complicated this way, but we can discard the complicated Higgs, and we can study more closely what happens with motion at the micro scale, which could help us to link Relativity theory to Quantum theory.


As you can see, I use only immediately observable things, and you use entropy, which is not immediately observable, at least for an atom.

You say that the red arrows represent redshifted information, but it looks to me as if that would be blueshifted too because particle A is running into it. You should also consider the case where particle B moves off first (moving away from particle A) to see whether particle A also responds appropriately (and whether you again get length-contraction).

[Le Repteux]

You may not have read yet how I describe the motion between my two atoms, so here it is again:

    We have two atoms A and B that are part of the same molecule. The time interval represents the time the information takes between the two atoms at t0. We accelerate A for a while and observe what happens to the system from t0 to t7. The blue arrows represent the blueshifted information that travels from A to B, and the red arrows represent the redshifted information that travels from B to A. The acceleration of A begins at t0 and ends at t4, so because of the time gap, the acceleration of B begins at t1 and ends at t5. After t5, the two atoms travel at the same speed, but we can easily see that the distance between them has contracted, and we can follow its progression during the acceleration. At that moment, the information on the future speed of each atom with regard to aether is situated between them in the form of doppler effect. The main idea is that, without doppler effect, there would be no motion between bonded particles, so there would be no motion either at our scale. I insist on the fact that we have to exert a force to introduce that doppler effect between them, and that this force represents mass. So with the same principle, we explain mass, motion and contraction. Of course motion is a bit more complicated this way, but we can discard the complicated Higgs, and we can study more closely what happens with motion at the micro scale, which could help us to link Relativity theory to Quantum theory.

You say that the red arrows represent redshifted information, but it looks to me as if that would be blueshifted too because particle A is running into it.

We have aether in the background, so we can look at the waves as if we were at rest in aether. This way, the light from the right particle is redshifted, and the one from the left particle is bueshifted. Once both lights hit the opposite particles though, they moves to stay on sync with it, so they kind of both gobble the incident doppler effect while both moving in the same direction and at the same speed. Notice that it still takes time before the motion takes place. It takes time during the motion, and it also takes time during the acceleration that precedes the motion, so what the left particle is running into is not the light that is actually emitted by the right one, but the one that is actually producing its motion towards it. My former animation was more explicit about that since we could distinguish the steps from both particles. They are now executed at the same time, but it still takes time before the light from a step reaches the other particle.

You should also consider the case where particle B moves off first (moving away from particle A) to see whether particle A also responds appropriately (and whether you again get length-contraction).

Nice shot! Molecules from a gaz hit each other, but they can't really pull each other. The best they can do is hit another molecule sideways, which could induce a rotation. If they could then kick one of the atoms a bit away from the other atom, the distance between the components of that atom would contract while it is accelerating away from the other, and the distance between the atoms would expand, which is not easy to figure out. Another way to test my hypothesis would be to accelerate my two particles so that the contraction stops. If we take the system at time t7 and accelerate particle B to the left, its motion should produce blueshift on the light it emits towards A, what should subtract to the redshift (red arrows) it was already producing, and the same interference should happen to the light emitted by A towards B, so the distance between them should expand back to the one they had at T0 if the acceleration is the same. It is not easy to imagine, but the math works, so I guess it is right.

[guest4091]

Davod Cooper;

Apparently I used the wrong numbers (from memory) in the last post, which doesn't change the issue.

B leaves A at .866, and sends a signal once a sec. The 1st occurs at Bt=1 ( At=2). Transit time for light is 2(.866)/1 = 1.73. A receives 1st at At= 3.73. The doppler ratio is  3.73:1.
A also sends a signal once a sec. The 1st occurs at At=1 (Bt=.5). Transit time for light is .866/(1-.866) = 6.46. B receives signal at At=7.46.  (Bt=3.73). The doppler ratio is the same. Each perceives the same Doppler effect.
The 2:1 ratio would occur if the relative speed was .6c.
Show how you conclude it’s 2:1 or 1:2.

[David Cooper]

B leaves A at .866, and sends a signal once a sec. The 1st occurs at Bt=1 ( At=2). Transit time for light is 2(.866)/1 = 1.73. A receives 1st at At= 3.73. The doppler ratio is  3.73:1.

That's the same figure you gave for this before, so that's fine.

A also sends a signal once a sec. The 1st occurs at At=1 (Bt=.5). Transit time for light is .866/(1-.866) = 6.46. B receives signal at At=7.46.  (Bt=3.73). The doppler ratio is the same. Each perceives the same Doppler effect.

That's fine too as they should perceive the same rate of beeps from the other.

The 2:1 ratio would occur if the relative speed was .6c.
Show how you conclude it’s 2:1 or 1:2.

The 2:1 ratio comes from averaging the number of beeps while the rockets are approaching each other with the number of beeps while they're moving apart (over the same length of time). As I said before: "If you hear 3.73 beeps in a length of time t (measured by your clock) while your rocket approaches the other rocket (and reaching it at the end of that time), and you then hear 0.27 beeps in the the length of time t as you move away from the other rocket (having passed it), you have heard 4 beeps in time 2t. That is a 2:1 ratio, although it's the opposite way round from the one I'd expected..."

(3.73+0.27) / 2 = 2

[guest4091]

The 2:1 ratio comes from averaging the number of beeps while the rockets are approaching each other with the number of beeps while they're moving apart (over the same length of time). As I said before: "If you hear 3.73 beeps in a length of time t (measured by your clock) while your rocket approaches the other rocket (and reaching it at the end of that time), and you then hear 0.27 beeps in the the length of time t as you move away from the other rocket (having passed it), you have heard 4 beeps in time 2t. That is a 2:1 ratio, although it's the opposite way round from the one I'd expected..."

Now I see what you are saying.
The calculated avg.freq. for B is 4/2 = 2.
The calculated avg.freq. for A is 2/4 = 1/2.
This verifies the conservation of the number of signals.
These are not perceived real time freq. was my point.

[David Cooper]

The calculated avg.freq. for B is 4/2 = 2.
The calculated avg.freq. for A is 2/4 = 1/2.
This verifies the conservation of the number of signals.

There must be an error in that or you could tell which rocket's moving - the perceived frequency rate must be the same for both rockets (taking into account the clock of one rocket running half as fast as the other).

These are not perceived real time freq. was my point.

I only claimed that the average was a 2:1 ratio, although it looks as if I got it the wrong way round because I was thinking about the calculated ratio once the real rate of beeps is calculated from the number received (to cancel out the Doppler shift).

[GoC (OP)]

There must be an error in that or you could tell which rocket's moving - the perceived frequency rate must be the same for both rockets (taking into account the clock of one rocket running half as fast as the other).

There is a way to know which ship is moving faster than the other. You each send a signal to the other at one signal per your second. Then you send a radar type signal to bounce back to you. In this way you can subtract one signal from the other to get the speed of the other vs. your speed relative when you compare distance between returned signals vs. your second. This would only work when you are on the same line. Each of you would measure a difference between the sent signal and the returned signal frequency between beeps. Then you compare your red and blue shifts in your signals. It should be the same as the gradient red shift in gravity GR measured in momentum. But of course it would not be momentum of light changing. Light does not change momentum down a gravity well either. Its just the dilation of energy change.

[David Cooper]

There is a way to know which ship is moving faster than the other. You each send a signal to the other at one signal per your second. Then you send a radar type signal to bounce back to you. In this way you can subtract one signal from the other to get the speed of the other vs. your speed relative when you compare distance between returned signals vs. your second. This would only work when you are on the same line. Each of you would measure a difference between the sent signal and the returned signal frequency between beeps. Then you compare your red and blue shifts in your signals...

You obviously haven't checked out your claim by doing the maths. If you had, you'd know that it won't work.

[Le Repteux]

Hi David,

I went on a french scientific forum this afternoon, and I asked why they didn't use simulations like yours to convince people. While answering questions, I realized that you did not put enough importance on the reason why it is always the twin that turns around that ages less. When it is him that is at rest and the earth traveling, you say that he will have to travel faster to catch up with the earth at the end, thus you change reference frames without saying why. I suspect it is because we can't accelerate without knowing that we did, so the twin on the earth cannot say he did, but the people at the forum will probably tell me that acceleration is not part of SR, so what am I going to tell them? That they were wrong?

[GoC (OP)]

There is a way to know which ship is moving faster than the other. You each send a signal to the other at one signal per your second. Then you send a radar type signal to bounce back to you. In this way you can subtract one signal from the other to get the speed of the other vs. your speed relative when you compare distance between returned signals vs. your second. This would only work when you are on the same line. Each of you would measure a difference between the sent signal and the returned signal frequency between beeps. Then you compare your red and blue shifts in your signals...

You obviously haven't checked out your claim by doing the maths. If you had, you'd know that it won't work.

You are correct. I did not mean to send that thought after I wrote it. When I closed the screen it sent. The energy left by c is counteracted by kinetic energy (speed) in the signal auto correcting duration between ticks. You never know which ship is causing the duration of the received signal. This is also why measured speed of light is the same in every frame.

Why do the clocks tick at the same rate (electron cycle) at sea level? How do they know to stay in sync? Some time in the future science will understand there is a control mechanism. I come from an analytical chemistry background and realize there is always a physical reason for a physical condition. It can be no different for physics. The clock's electron cycle is controlled by an outside force (energy from space) or it is a coincidence that happens on all planets. Chose your side carefully if you want the best understanding. Coincidence is a choice but it is a coincidence at every level the sea would occupy on any planet. Energy is of space and not mass. E=c and in the presence of mass E= c * c moving electrons. Electrons do not move themselves. Unless you believe in magic, you have to come to the same conclusion. Mass has entropy. Energy may have entropy also but it is relative to chemical reactions. We could not perceive energy slowing down since we are part of the measurement system. If energy was stored in atoms they would not change in concert with other atoms electrons during frame changes. The standard model of space being a void is not logical to clocks changing in concert.

[David Cooper]

Hi Raymond,

I went on a french scientific forum this afternoon,...

That might be exactly what I need to improve my French scientific vocabulary. Ou est-ce que je peux le trouver?

... and I asked why they didn't use simulations like yours to convince people. While answering questions, I realized that you did not put enough importance on the reason why it is always the twin that turns around that ages less. When it is him that is at rest and the earth traveling, you say that he will have to travel faster to catch up with the earth at the end, thus you change reference frames without saying why. I suspect it is because we can't accelerate without knowing that we did, so the twin on the earth cannot say he did, but the people at the forum will probably tell me that acceleration is not part of SR, so what am I going to tell them? That they were wrong?

The first thing you need to point out is that I don't change frame when the rocket accelerates. (1) If the planet's stationary throughout, the rocket accelerates, moves away from it, decelerates to halt, accelerates in the opposite direction to the original acceleration, moves back to the planet, then decelerates to a halt. At no point in that have we changed the frame we're using for our analysis of the events. (2) If the planet's moving throughout, the rocket decelerates to a halt while the planet continues to move through the frame of reference in which the rocket is now stationary, then the rocket accelerates for twice as long as its original deceleration to set off in pursuit of the planet, closes in on it (moving at high speed through our chosen frame of reference), then decelerates so that it can land on the planet. Again, at no point in this have we changed the frame used for the analysis of events.

There are a lot of people out there who misunderstand how acceleration relates to SR, but the simple way to illustrate its role is to introduce more space ships which don't accelerate at any point during the analysis. In case (2), for example, when the rocket decelerates at the start, it could decelerate to sit alongside a space ship which happens to be there already (sitting stationary in the frame used for the analysis), and when the rocket then accelerates hard to chase after the planet, another spaceship can happen to come along such that it moves alongside the rocket throughout that part of the journey. The clock in the rocket must tick at the same rate as the clock in any space ship that's co-moving with it at the time. This shows us that the mechanism used to control the clocks and determine how quickly they tick relative to each other is set purely by the speed at which they move through space - there is no fancy voodoo done by the accelerations other than to change the speed at which a clock is moving through space.

Also, if anyone gets obsessed with the length of time taken by the accelerations and imagines that this somehow invalidates the thought experiment, that can be dealt with as follows. Accelerations take time to happen, so there will be short stretches of time when the rocket's clock is not ticking at any of the fixed rates used in our calculations, but if we make the accelerations quicker and quicker or make the distances travelled longer and longer, we can make these complications increasingly insignificant - we can see how they get closer and closer to being instantaneous accelerations, and this gives us full justification for using instantaneous accelerations in our thought experiments - while they may be impossible in reality, we can see that the time the rocket spends accelerating can be made shorter and shorter and that the deviation this would give us from the numbers we get out of our thought experiments tends towards zero.

Be aware that it's possible (in thought experiments) to accelerate a rocket in such a way that the acceleration is not felt by the rocket - you can hold a planet near a stationary rocket to accelerate it and move the planet along ahead of it at faster and faster speed as the rocket speeds up, then whip the planet away and leave the rocket moving at high speed through space with its clock running slow and the people inside it having no idea that they're no longer stationary (unless they were watching what happened).

[David Cooper]

Why do the clocks tick at the same rate (electron cycle) at sea level? How do they know to stay in sync?

The maths of relativity is full of amazing coincidences, so this is just one more of them. In a way though, none of them are coincidences as there is simply no alternative way for things to be without breaking mathematics. Thanks again for pointing this one out to me though, because I'd never heard of it before.

[Le Repteux]

(2) If the planet's moving throughout, the rocket decelerates to a halt while the planet continues to move through the frame of reference in which the rocket is now stationary, then the rocket accelerates for twice as long as its original deceleration to set off in pursuit of the planet, closes in on it (moving at high speed through our chosen frame of reference), then decelerates so that it can land on the planet. Again, at no point in this have we changed the frame used for the analysis of events.

To me, taking the ship as a reference frame would mean that it is the planet that would make the whole roundtrip: it would thus first accelerate away from the ship, then it would decelerate, accelerate back to the ship, and decelerate to land it. It is easier to figure it out with two ships though, and this way, I don't see how we could get out of it without knowing which one has accelerated, and my diagram about the distance between two atoms being contracted during acceleration shows how it would work. Of course, it doesn't fit with relativity, so I'm probably going to be moderated on the forum if I use that kind of argument, but I think it fits with LET. If it doesn't then we may need a new theory.

Here is the link to the topic I opened on the french forum:
forums.futura-sciences.com/physique/794268-outil-comprendre-relativite.html

I'm french, so I often use this translator for the words I'm not used to:
wordreference.com/enfr/