[nilak]

Sorry Timey , I forgot you also said that a moving clock frequency should also increase which semms to be against the ecperimental results. However I support the idea of true absolute time. However my model shows the moving clocks slow down without forcing things at all. Basically The atom inner frequency (de Broglie) increases or decreases due to Doppler shift but the tick rate being the revolution motion decreases. The increase / decrease in frequency is exactly as to light only when atoms are oriented along the traveling axis. At c it can't revolve anymore. Regarding mass, mass particles have only relativistic mass just like photons, because the revolving motion keeps them moving at c even though center of rotation is at rest.

[timey (OP)]

No - I most certainly did not say that a moving clocks frequency should increase.   What I said was that if one follows the current physics mathematical rules for an increase in kinetic energy for light, that a moving clocks frequency should increase as per those rules.
The difference here is that in implementing those kinetic energy, relativistic mass rules for light, one is not doing so with the viewpoint of lights experience of its own time, but as a mathematical means of explaining the phenomenon of how light travels across space.

When applying relativistic mass to a moving clock, the experimental observation is that the clock has a decreased frequency relative to the stationary clock in the same gravity potential.  The only means (that I can see, please someone point out to me if I have gone astray), of describing this decrease in frequency observed of the moving clock is to state that the greater gravity field caused by the addition of relativistic mass has caused the decrease in frequency of the moving clock.

But then one gets into a tautology problem, as I explained in last post, where adding kinetic energy increases mass, increasing mass slows time, and a constant speed in a slower time will reduce speed, which will reduce kinetic energy, which will reduce mass...
There is no means of avoiding this inherent seesaw action.  This being my point of dissatisfaction in the current physics remit.
(these considerations being based on the concept of people ageing at differing rates in reference frames of dilated or contracted rates of time, as per NIST and NASA seem to purport.  Which is directly contradictory to the notion that a clock only 'appears' dilated or contracted from another reference frame, and that the clock in the other reference frame will be running at the same frequency as the observation points clock is, if you were to be in that other reference frame with the other clock)

[nilak]

Oh, I need to apologize again, Timey, I understand your point now. You trust the evidence but not the explanation.  If added kinetic energy increases mass and decreases time rate, you say the decrease in time should counteract the kinetic energy increase. I will confront this to my model which uses your idea of a variable second ( I use absolute time) to see what I get. At the moment it seesm that my model says that kinetic energy is constant in the absolute frame no matter of how the particle travels so time dilation indeed counters the increase of mass. It only changes shape and becomes more elongated until it mimics a photon. The tick rate is correct. This means you also seem to be correct. But I have to study more aspects of my model.

[nilak]

Yes, Janus, I thought I saw something wrong but I understand now, the Doppler shift seems to be correct.
The point at the right sees the source comming at v and light at c. Drawing this on a spacetime chart shows the speed difference.  I don't see any issues anymore with regard to Doppler Effect. Thanks.

[Jaaanosik]

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All three counters still read the same when they meet.

The left diagram shows the spaceship A not moving and the spaceship B moving.
Where is the aging coming from in the twin paradox then if all counters show the same time when they meet?

[Janus]

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All three counters still read the same when they meet.

The left diagram shows the spaceship A not moving and the spaceship B moving.
Where is the aging coming from in the twin paradox then if all counters show the same time when they meet?

The counters reading the same when they meet up is a consequence of how you set up your experiment.  You started the counters at the same time while they were an equal distance from the midpoint and traveling at equal speeds as measured by the midpoint frame.  It is obvious that they will read the same when they meet in this frame. And if they do in one frame they have to in all frames.  The other frames disagreed as to whether the counters started simultaneously, were an equal distance from the midpoint when they started or traveled at the same speed with respect to the Midpoint.

The twin paradox is a different set up. Here we have two clocks starting at the same point and initially synchronized to each other. Since they are at the same point when synced they both agree that they started synced.  Then the two clocks separate at some speed v, and then one of the clocks changes velocity so that it rejoins the other. We now compare clock when they met up again to find that the one that changed velocities midway through has accumulated less time.

Here are the space-time diagrams for this.



First we have things according to the frame of our "Stay at Home" twin.  Again, the yellow lines represent clock tick info being exchanged between twins.
His twin travels away at some relative speed, stops, reverses direction and returns at the same speed. during this whole time his twin's clock runs slow compared to his. What he sees via light information is his twin aging slow for a long time and then aging fast for a short time.  Both lead to the conclusion that his twin's net age will be less than his own when they meet up.

For the traveling twin we have to consider two different diagrams, the first is for his outbound leg. and second for his return leg. First, we we examine what he sees happening to his brothers clock. For the Outbound trip he sees it as running slow by a factor of ~0.577, and on the return he sees it run fast by a factor of ~1.732. Both legs take the same time, ~3.5, so the total time he sees pass on the stay at home twins clock is 3.5 x 0.577+3.5 x 1.732 = ~8 for the 7 he measures by his own clock.

What actually is happening to the stay at home clock according to our traveler during both legs is that it is running slow.   But how can it run slow for both legs and yet show more time has past when they meet up.  The answer is in what happens when our traveler transitions from outbound to inbound legs.  Looking at the middle image (out bound leg, when he reaches the turnaround point, he is just getting the image of the stay at home clock reading 2, And at that moment the stay at home clock is just about reading three ,while his own reads ~3.5.  Now we look at the third image, which just after he has made his velocity change. He is still seeing the same image of the stay at home clock, but at that moment it is actually reading ~5.  His transition from going from traveling in one direction to the other causes the stay at home clock to jump forward by 2.

Now what physical change could his velocity change have on the operation of the Stay at home clock? None.  But that is not a problem, because, as I've said before, Relativity is not about outside physical influences effecting how clocks tick, but about the nature of time and space and its measurement.

Think of space-time as a map that you use to plot events with respect to you in space and time. (for instance, I had lunch two hrs ago and 20 feet from where I am now).

But here is the strange thing about this map. Unlike a typical map, which mark out defined directions that everyone agrees upon (everyone agrees on the North-South direction for example.)  Your map is marked out left-right, forward-backward. If you turn, your map turns with you.  For example, if you start with an object 3 feet to your left and 4 feet to the front of you, and you turn, it is no longer 3 feet to your left and 4 feet to the front of you.

Now replace one of the spatial dimension with time.  If an event took place 3 sec ago and 4 meters from me by my measurement, and I change my velocity, it will no longer have happened 3 sec ago and 4 meters from me by my measurement.  By changing my velocity, I have changed how I measure the time and spatial separation between my here and now and the event.

Relativity forces us to rethink our notions of time and space.

Either you are willing and capable of this, or you will forever flounder in trying to understand Relativity.

[Jaaanosik]

Hi Janus,
your last space diagram does not show up properly. I cannot see it.
Is it possible you can fix it?
Thanks.

[Jaaanosik]

Here is a food for thought why the straight linear acceleration is not equivalent to the rotational acceleration.
Rotational includes gravity of a curved spacetime.


Figure 1: The pendulum, a simple harmonic motion (a clock), in the straight line accelerated frame of reference.


Figure 2: The pendulum, a simple harmonic motion (a clock), in the rotationally accelerated frame of reference.

The pendulum in the rotational frame of reference is being accelerated by additional Coriolis’ acceleration that comes from dR therefore we can see that even though point B is being accelerated by the acceleration of the same magnitude 'a' in both scenarios the magnitude of point A acceleration is different for the straight line acceleration and the rotational acceleration.

This is a very simple but a major reason why the Equivalence principle cannot hold.

The accelerated rocket has Figure 1 'pendulum-clock-harmonic motion' at the bottom and at the top of the rocket.
The gravity field has Figure 2 'pendulum-clock-harmonic motion' at the bottom and at the top with different R and dR between the bottom and the top.

[Jaaanosik]

Thanks Janus, I can see the diagram now.

I wonder is there a scenario with a different planet with more mass that the twin traveling away at 0.6c and coming back would return older than the twin that stayed on the planet?

[Jaaanosik]

...
Since there is no absolute frame by which to measure motion, The question is pretty meaningless.

Which ship is "moving" or "not moving" simply depends on the frame of reference it's being it being measured from. 
...

If they have the same time when they meet and the clock postulate is true then both spaceships have to have the same speed in clock postulate frame of reference. The  "momentarily comoving inertial frame" (MCIF) of both spaceships have to move at the same speed in the CP frame. There is only one type of CP frame where this can be true.

If we move the CP frame to the spaceship A then SA MCIF is not moving and SB MCIF is moving fast.
If the CP is true then SBC counter should be less then SAC when they meet.
If SAC=SBC then there are two options how this can happen.
Either the CP is false or the CP frame cannot be moved to SA.
Either we deny the experiments that support CP or we say there is only one proper CP frame.

[guest4091]

No - I most certainly did not say that a moving clocks frequency should increase.   What I said was that if one follows the current physics mathematical rules for an increase in kinetic energy for light, that a moving clocks frequency should increase as per those rules.

Remember,light speed is independent of its source.
If light does not acquire the speed of the source, then it does not acquire additional ke.

[timey (OP)]

Granted, if you look at a light source as being velocity red shifted or velocity blue shifted instead of the light being shifted via the gravitational shift equation (a notion which in itself raises more questions than it answers), then your comment has logic...
BUT... is it actually physically possible to separate the notion of light having an association with the electron?

http://physics.bu.edu/~duffy/PY106/Potential.html

[Jaaanosik]

No - I most certainly did not say that a moving clocks frequency should increase.   What I said was that if one follows the current physics mathematical rules for an increase in kinetic energy for light, that a moving clocks frequency should increase as per those rules.

Remember,light speed is independent of its source.
If light does not acquire the speed of the source, then it does not acquire additional ke.

Light can acquire energy. When it is being blue-shifted it is increasing frequency/energy.
Light can not travel faster then c but it can increase the frequency and that's how it gains energy.
Just check the second image of the Janus' post #136.
That's how the kinetic energy of the source is added to the light - increased frequency.

[Janus]

Here is a food for thought why the straight linear acceleration is not equivalent to the rotational acceleration.
Rotational includes gravity of a curved spacetime.


Figure 1: The pendulum, a simple harmonic motion (a clock), in the straight line accelerated frame of reference.


Figure 2: The pendulum, a simple harmonic motion (a clock), in the rotationally accelerated frame of reference.

The pendulum in the rotational frame of reference is being accelerated by additional Coriolis’ acceleration that comes from dR therefore we can see that even though point B is being accelerated by the acceleration of the same magnitude 'a' in both scenarios the magnitude of point A acceleration is different for the straight line acceleration and the rotational acceleration.

This is a very simple but a major reason why the Equivalence principle cannot hold.

The accelerated rocket has Figure 1 'pendulum-clock-harmonic motion' at the bottom and at the top of the rocket.
The gravity field has Figure 2 'pendulum-clock-harmonic motion' at the bottom and at the top with different R and dR between the bottom and the top.

Coriolis effect is a second order effect that diminishes as R increases relative to the pendulum length.  Or put another way, If you continue to shrink our pendulums in both the linearly and rotational accelerated frames, they converge towards each other in how they behave and the pendulum in the rotating frame acts more and more like the one in under linear acceleration. 
The equivalence principle applies locally, or over a "sufficiently small" region.   Your objection to it is based on applying it in a way it is not meant to be applied.   The fact that a linearly accelerated object does not behave exactly like a rotating accelerated object in all respects does not invalidate the equivalence principle. ( I mean really, if it was that easy to invalidate the Equivalence principle, it would not have made it past the first peer review let alone till now.)

[Jaaanosik]

Coriolis effect is a second order effect that diminishes as R increases relative to the pendulum length.  Or put another way, If you continue to shrink our pendulums in both the linearly and rotational accelerated frames, they converge towards each other in how they behave and the pendulum in the rotating frame acts more and more like the one in under linear acceleration. 
The equivalence principle applies locally, or over a "sufficiently small" region.   Your objection to it is based on applying it in a way it is not meant to be applied.   The fact that a linearly accelerated object does not behave exactly like a rotating accelerated object in all respects does not invalidate the equivalence principle. ( I mean really, if it was that easy to invalidate the Equivalence principle, it would not have made it past the first peer review let alone till now.)

Here is a chapter from the same Feynman book, this one precedes the rocket and the clocks chapter.


I'll repeat the last paragraph.

To be strictly correct, that is true only for one point inside the ship. The gravitational field of the earth is not precisely uniform, so a freely falling ball has a slightly different acceleration at different places—the direction changes and the magnitude changes. But if we imagine a strictly uniform gravitational field, it is completely imitated in every respect by a system with a constant acceleration. That is the basis of the principle of equivalence.

The Equivalence principle stands on imaginary, unrealistic uniform gravitational field.

I am just a messenger here, please don't shoot.

How is this possible? Unbelievable!!!
So yes, it passed the peer review.

I guess that's why Einstein never got the Nobel prize for the Relativity.

[guest4091]

Jaaanosik

Light can acquire energy. When it is being blue-shifted it is increasing frequency/energy.
Light can not travel faster then c but it can increase the frequency and that's how it gains energy.
Just check the second image of the Janus' post #136.
That's how the kinetic energy of the source is added to the light - increased frequency.

Post 136 concerns doppler shift. This is a changed perception of the base/original frequency by the receiver due to relative motion between emitter and receiver.

[guest4091]

Coriolis effect is a second order effect that diminishes as R increases relative to the pendulum length.  Or put another way, If you continue to shrink our pendulums in both the linearly and rotational accelerated frames, they converge towards each other in how they behave and the pendulum in the rotating frame acts more and more like the one in under linear acceleration. 
The equivalence principle applies locally, or over a "sufficiently small" region.   Your objection to it is based on applying it in a way it is not meant to be applied.   The fact that a linearly accelerated object does not behave exactly like a rotating accelerated object in all respects does not invalidate the equivalence principle. ( I mean really, if it was that easy to invalidate the Equivalence principle, it would not have made it past the first peer review let alone till now.)

Here is a chapter from the same Feynman book, this one precedes the rocket and the clocks chapter.


I'll repeat the last paragraph.

To be strictly correct, that is true only for one point inside the ship. The gravitational field of the earth is not precisely uniform, so a freely falling ball has a slightly different acceleration at different places—the direction changes and the magnitude changes. But if we imagine a strictly uniform gravitational field, it is completely imitated in every respect by a system with a constant acceleration. That is the basis of the principle of equivalence.

The Equivalence principle stands on imaginary, unrealistic uniform gravitational field.

I am just a messenger here, please don't shoot.

How is this possible? Unbelievable!!!
So yes, it passed the peer review.

I guess that's why Einstein never got the Nobel prize for the Relativity.

All theories rest on postulates that are accepted on faith until an exception is discovered.
At least the article cited includes the restriction regarding applicability. For nonuniform g-fields that converge, such as a spherical body, other effects (tidal forces) show up.
For practical applications there are no significant differences.

[Jaaanosik]

...
Since there is no absolute frame by which to measure motion, The question is pretty meaningless.
...

There is an island on the equator with a 122.5m hill, the open ocean to the east and to the west. The observer shoots projectiles horizontally to the east and to the west at the same time with a velocity v=465m/s, we will ignore the air resistance. What is the time of the falling till the projectiles hit the ocean surface?
The Earth equatorial radius: 6,378.1km
The Earth equatorial rotation velocity: 1,674.4km/h →1674400/3600=465m/s

The g = 9.814m/s² for non-rotating Earth. The centrifugal acceleration for v=465m/s is a_cf = v²/r =4652/6378100=0.0339m/s².

The rotating Earth gravitational acceleration is g =9.814−0.0339=9.781m/s². The projectile that was shot to the west is not moving horizontally in the gravitational field therefore it’s going to be falling down with the gravitational acceleration approximately g =9.814m/s². The projectile that was shot to the east is moving horizontally in the gravitational field with the velocity v=930m/s. The centrifugal acceleration for v = 930m/s is a_cf = v²/r = 9302/6378100 = 0.1356m/s² therefore it’s going to be falling down with the gravitational acceleration approximately g=9.814−0.1356=9.678m/s².

What are the times of falling? The westward projectile falls 5s so h=0.5gt²=0.5*9.814*25=122.675m The eastward projectile falls from the same height in t=sqrt(2h/g)=sqrt(2*122.675/9.678)=5.035s.
The conclusion is that the eastward projectile will be falling 0.035s longer than the westward projectile due to the difference in g.

The inertial reference frame at the top of the hill is an approximation only, it is not accurate for the above calculation. The reference frame that is ’linked’ to the gravitational field background is preferred one for more accurate calculation. Here is a grid of the gravitational field.



The horizontal lines are the gravity equipotential lines and the vertical lines are the gravity field lines. There are two reference frames. The first reference frame, the gravity field frame, has [0,0] at the sea level, straight below the observer that is on the hill at the time T0=0s when the observer shoots the projectiles. This frame is linked to the non-rotating gravitational field. The x-axis values are labelled in blue. The second reference frame, the Earth frame, has [0,0] at the sea level, straight below the observer that is on the hill at the time T0=0s when the observer shoots the projectiles but this frame rotates with the Earth at v=465m/s. The x-axis values are labelled in red.

The position of the projectiles at T1 = 1s where fallen height for the blue projectile falling straight down in the gravity field reference frame is h_b=0.5gt²=0.5*9.814*1²=4.907m and for the red projectile flying eastward and falling down is h_r=0.5gt²=0.5*9.678*1² =4.839m.



The position of the projectiles at T2 = 2s where fallen height for the blue projectile falling straight down in the gravity field reference frame is h_b=0.5gt²=0.5*9.814*2²=19.628m and for the red projectile flying eastward and falling down is h_r=0.5gt²=0.5*9.678*2² =19.356m.



The position of the projectiles at T3 = 3s where fallen height for the blue projectile falling straight down in the gravity field reference frame is h_b=0.5gt²=0.5*9.814*3²=44.163m and for the red projectile flying eastward and falling down is h_r=0.5gt²=0.5*9.678*3² =43.551m.

The position of the projectiles at T4 = 4s where fallen height for the blue projectile falling straight down in the gravity field reference frame is h_b=0.5gt²=0.5*9.814*4²=78.512m and for the red projectile flying eastward and falling down is h_r=0.5gt²=0.5*9.678*4² =77.424m.

The position of the projectiles at T5 = 5s where fallen height for the blue projectile falling straight down in the gravity field reference frame is h_b=0.5gt²=0.5*9.814*5²=122.675m and for the red projectile flying eastward and falling down is h_r=0.5gt²=0.5*9.678*5² =120.975m.




There is the International Space Station flying with the velocity v=7.67km/s at about h=400km above the Earth and g=8.682m/s² at that height. Let us shoot a projectile in the direction of the ISS velocity vector with v=7.67km/s and in the opposite direction of the ISS velocity vector with v=7.67km/s. Now based on what we learned above we know that the projectile in the direction of the ISS velocity vector will fly away with 2v=15.34km/s velocity, a_cf=v²/r=153402/6778100=34.717m/s², the proper centrifugal acceleration needs to be adjusted by subtracting the gravitational acceleration a_cf−adj=34.717−8.682=26.035m/s². The projectile in the opposite direction of the ISS velocity vector will be falling straight down because its velocity is v=0m/s in the gravity field reference frame. The ISS 'inertial reference frame' comes short in predicting what happens to the projectiles but we are being taught by the General Relativity that there is no force acting on a free falling body. We are left with a conclusion that a free falling reference frame is an approximation only at some velocities.

We are left with a conclusion that a free falling reference frame is an approximation only at some velocities.

It appears the gravitational field reference frame is a preferred one. It provides better prediction of the projectile fall within the gravity field. The question might arise why not to use the reference frame at the center of the gravity field, the Earth center?
The issue is that the locality of the gravitational field determines the gravitational acceleration g. We ignored the influence of the Moon in our calculations, but if we want to be accurate we would need to account for the gravitational field coming from the Moon as well, and the Sun, ...


Now before saying anything about the tidal effect and the size/length of the experiment let us think about a hollow ring of 1m radius on the ISS. It is in a horizontal position, the z axis going straight through the ring points to the center of the Earth.
There are 2 (4, 6, ... ) evenly distributed through the ring friction less balls with accelerometers in the balls moving at 7.67km/s inside of the ring.

Are the accelerometers going to report different instantaneous accelerations in z direction when flying in ISS direction (2v) and the opposite one (0v)?