Synchronizing moving GPS clocks

This article with drawings is in these links
https://www.academia.edu/40876793/Synchronizing_moving_GPS_clocks Word, PDF https://pengkuanonphysics.blogspot.com/2019/11/synchronizing-moving-gps-clocks.html PDF

1. Light pulse synchronization
Can we synchronize clocks of a moving frame? Let us see Figure 1 where we have stationary frame F1 and moving frame F’2. The 2 clocks in the frame F1 are synchronized with the master clock through a light pulse.

The 2 clocks of the frame F’2 are moving. At the time t0 in the frame F1, the 2 clocks in the moving frame F’2 coincide with that of the frame F1, but due to relativity of simultaneity the time of the moving clocks are different,
dt’=t’2-t’1=/=0. So, the light pulse does not synchronize the clocks in the moving frame.

2. Fixed clocks in orbit
GPS satellites are moving, so they suffer from relativity of simultaneity. Let us imagine a disk centered at the center of the Earth whose rim is the GPS orbit and on the rim are equally spaced clocks, see Figure 2. The disk does not rotate with the Earth and the frame of the disk is the frame F1. These clocks can be synchronized because they are fixed and they are synchronized with the master clock at the North Pole of the Earth through a light pulse. They all show the time t0 when each GPS satellite pass in front of each corresponding fixed clock. But the clocks of the GPS satellites are not synchronized with the master clock because they are moving.

3. Time in GPS satellites
The event the satellite 1 coincides with a fixed clock and the satellite 2 coincides with the next fixed clock are simultaneous in the frame F1. But these 2 events are not simultaneous in the frame F’2, which is the frame moving with the satellite 1 and containing the satellite 2, see Figure 3.

In the frame F’2, the clock of the satellite 1 reads t’1 and the clock of the satellite 2 reads t’2 because of non-simultaneity. So, we have the gap of time dt’=t’2-t’1=/=0.

In the same way, the event the satellite 2 coincides with a fixed clock and the satellite 3 coincides with the next fixed clock are simultaneous in the frame F1, but not in the frame F’3, which is the frame moving with the satellite 2 and containing the satellite 3, see Figure 3. Also for the same reason, the gap of time in the frame F’3 is dt’=t’3-t”2=/=0.

Note that here in the frame F’3, the time of the satellite 2 is denoted by t”2 with double prime to distinguish it from the time of the satellite 2 in the frame F’2.

4. Time of the satellite 2
We have 2 different notations of the time of the satellite 2: t’2 with single prime in the frame F’2 and t”2 with double prime in the frame F’3. The satellite 2 has only one time which is shown by its clock. For the event the satellite 2 coincides with the fixed clock in Figure 3, its value in the frame F’2 is t’2. Now we ask: does the reading of the clock of the satellite 2 change to a different value t”2 when seen in the frame F’3? Logically no, because t’2 is the time of the satellite 2 while t”2 is the time of the same satellite at the same event. So, we must have t’2 = t”2. In this case, we have
t’3 = t”2+dt’=t’2+dt’=t’1+dt’+dt’=t’1+2dt’

5. Time of the satellite n
For the same reason, the double-primed t”3 equals the single-primed t’3 and each double-primed t”i equals each corresponding single-primed t’i. Then, the n+1th satellite has the time: t'(n+1) =t’1+ndt’.

Because ndt’ is not zero, t'(n+1) =/=t’1 while the satellite n+1 being the satellite 1, hence contradiction.

6. If t'(n+1) =t’1?
We can assert that the satellite n+1 being the satellite 1 and having the same time, that is,
t'(n+1) =t’1. In this case, we first suppose that the single-primed t’2 does not equal the double-primed t”2, but instead their different cancels dt’, that is, t”2=t’2-dt’. So, t’3=t”2+dt’=t’2-dt’+dt’=t’1+dt’ .

In this case, we will have for the n+1th satellites t'(n+1) =t’1+dt’ .

However, we still do not have t'(n+1) =t’1. For achieving this equality, we have to suppose the difference of time between single-primed and double-primed time to make t”2 = t’2 -dt'(1+1/n). This way, we will have at the n+1th satellite: t'(n+1) =t’1+dt’-dt’*n/n=t’1

But this assumption is absurd and cannot be true. So, due to relativity of simultaneity, the satellite n+1 and the satellite 1 are the same satellite but cannot have the same time.

The only way for the satellite n+1 and the satellite 1 to have the same time is that relativity of simultaneity is zero, that is, dt’=0.

Testing relativity of simultaneity using GPS satellites

In Special Relativity relativity of simultaneity is the fact that 2 simultaneous events occurring in a stationary frame does not appear simultaneous in a moving frame. For example, in Einstein’s train thought experiment 2 simultaneous flashes of light on the platform do not appear simultaneous for the observer in the train. But relativity of simultaneity has never been tested with real simultaneous events.

For testing relativity of simultaneity we need 2 synchronized clocks moving at high speed and we will read them in a stationary frame. Fortunately, we have at hand many GPS satellites which carry precision clocks and broadcast their time, with which we can check relativity of simultaneity.

Suppose that 2 satellites are separated by the distance L in the same orbit. Their clocks are synchronized with one clock on Earth, that is, the event “time of the satellite 1 is t0” and the event “time of the satellite 2 is t0” occur simultaneously on Earth at the time te. In the frame of these 2 satellites, due to relativity of simultaneity, these same events occur at time t1 on the satellite 1 and t2 on the satellite 2 and the difference of time is dt= t2- t1.

Suppose that we have n satellites equally spaced in the same orbit which is circular. The nth satellite is the last satellite and the (n+1)th satellite is the first satellite, which complete the circle of the orbit.

Due to relativity of simultaneity, the difference of time is always dt from one satellite to the next and the difference of time between the ith satellite and the first satellite equals (i-1)*dt. Then, the difference of time between the (n+1)th satellite and the first satellite equals n*dt. So, The time of the (n+1)th satellite is t1+n*dt.

We notice that the (n+1)th satellite is the first satellite but its time, t1+n*dt, is different from t1 the time of the first satellite. How can the time of a satellite is not the time of itself?

I explain this phenomenon in the article below.
PDF: https://pengkuanonphysics.blogspot.com/2019/10/testing-relativity-of-simultaneity.html
Word: https://www.academia.edu/40736335/Testing_relativity_of_simultaneity_using_GPS_satellites

From Michelson–Morley experiment to length contraction

Length contraction is used to explain Michelson–Morley experiment. A Michelson interferometer is formed by 2 perpendicular arms. The journey of light along the vertical arm gives the time dilation law. The journey of light along the horizontal arm gives the length contraction law. The derivation scheme of the length contraction law is:
1. The time of the journey along the horizontal arm is computed using a method that does not give directly the correct time.
2. Length contraction was invented to adjust the computed time.
3. The Michelson–Morley experiment dictates the ratio of length contraction.

But the variation of distance due to time dilation is more appropriate to explain this experiment. We expose a method that gives directly the correct time called Time Dilation Approach which does not need the length contraction assumption and the experimental result. So, these 2 conditions are not necessary for explaining the Michelson–Morley experiment.

Special Relativity cannot work with acceleration because length contraction creates contradiction. But in real world objects accelerate all time, which call for a theory to explain the acceleration of high speed objects.

Please read the article at
PDF: From Michelson–Morley experiment to length contraction https://pengkuanonphysics.blogspot.com/2019/08/from-michelsonmorley-experiment-to.html
or
Word: https://www.academia.edu/40208137/From_Michelson-Morley_experiment_to_length_contraction

Astrophysical jet and length contraction

Astrophysical jets are flows of matter that moves at relativistic speed. They are opportunity to see length contraction in action. An astrophysical jet is analyzed to explain the length contraction effect.

Astrophysical jets are ejected from compact objects such as black holes. https://en.wikipedia.org/wiki/Astrophysical_jet#/media/File:M87_jet.jpg is a photograph of the jet ejected by the supermassive black hole at the center of the galaxy M87 which stretches over 5 000 light-years. Matter in this jet moves at almost the speed of light and undergoes relativistic effects.

Almost all predictions of Special Relativity are proven by experiment except one: length contraction. Indeed, it is impossible to accelerate chunk of matter to relativistic speed to directly measure length contraction on Earth. Fortunately, length contraction should occur in astrophysical jets where we could finally see this effect for real.

An astrophysical jet is a moving cloud of particles whose velocity varies from almost the speed of light in the ejection region to very slow thousand light-years away. We name the jets in these 2 regions fast jet and slow jet. The fast jet should be strongly length-contracted while the slow jet should not. Let us compare a moving cloud of particles with the air inside a ball. The air inside a moving ball moves with the ball and would seem denser than the air inside the same ball at rest because the flat moving ball has a contracted surface. In the same way, fast jet should seem denser than slow jet. Since the particles of an astrophysical jet emit photons, the fast jet should be brighter than the slow jet. As the slowdown of the jet is gradual, we expect that the brightness of a jet decreases gradually from the fast jet to the slow jet. However, we do not see such gradual decrease of brightness in the photograph. Why?

Please read the article at
PDF: Astrophysical jet and length contraction https://pengkuanonphysics.blogspot.com/2019/08/astrophysical-jet-and-length-contraction.html
or
Word: https://www.academia.edu/40066246/Astrophysical_jet_and_length_contraction

How to test length contraction by experiment?

Relativistic length contraction is theoretically predicted but not directly tested, which lead to incorrect interpretation of the theory illustrated by Bell’s spaceship paradox and Ehrenfest paradox. But these paradoxes can help us designing experiments to test length contraction.

Ideal direct experimental proof should contain the following steps:
1. Measure the tested object’s length at rest, the value l0.
2. Put this object in motion.
3. Measure the object’s speed, the value v.
4. Measure the object’s length in motion, the value l.
5. Check if these 3 values verify length contraction law.

For doing this experiment, the difference of length l0  l should be in measurable range. If the object is a chunk of matter, l0  l is not measurable. For example, matter objects with the highest speed we can make are satellites, whose speed is generally 7.8 km/s. If a satellite is made of a string of 100 km long, the value of l0  l would be 0.03 mm, which is absolutely not measurable from the ground. This is why contraction of length has never been measured.

Below I propose two experiments inspired from Bell’s spaceship paradox and Ehrenfest paradox.

Please read the article at
PDF: How to test length contraction by experiment? https://pengkuanonphysics.blogspot.com/2019/06/how-to-test-length-contraction-by.html
or
Word: https://www.academia.edu/39584663/How_to_test_length_contraction_by_experiment

Twin paradox when Earth is the moving frame

We analyze the mathematical mechanism that slows the time of the traveler in the twin paradox and explain what distinguishes the traveler’s frame from the Earth’s frame

Please read the article at
PDF: Twin paradox when Earth is the moving frame https://pengkuanonphysics.blogspot.com/2019/05/twin-paradox-when-earth-is-moving-frame.html
or
Word: https://www.academia.edu/39216040/Twin_paradox_when_Earth_is_the_moving_frame

Graphic of set counting and infinite number

When counting a set, we can plot a graphic that represents the members of the set on the plane (x, y) to observe visually the counting. Also, graphic of counting of infinite set helps us to understand infinite natural number.

PDF Graphic of set counting and infinite number https://pengkuanonmaths.blogspot.com/2018/11/graphic-of-set-counting-and-infinite.html

or Word https://www.academia.edu/37766761/Graphic_of_set_counting_and_infinite_number

 

Analysis of the proof of Cantor’s theorem

Cantor’s theorem states that the power set of ℕ is uncountable. This article carefully analyzes this proof to clarify its logical reasoning

Please read the article at

PDF Analysis of the proof of Cantor’s theorem http://pengkuanonmaths.blogspot.com/2018/09/analysis-of-proof-of-cantors-theorem.html
or Word https://www.academia.edu/37356452/Analysis_of_the_proof_of_Cantors_theorem

Longitudinal magnetic force and high field magnet

Theoretical explanation of longitudinal magnetic force and its practical application in high field magnet. Although longitudinal force is not explained in classical theory, its action has been demonstrated by several experiments long time ago. For example Nasilowski effect. But why is it not recognized in theory? The reason is that it shows no significant effect on practical devices, so no physicist is interested in exploring these experiments. But I have found a huge effect of longitudinal force in high field resistive magnets which could improve their performance.

Please read the article at
Longitudinal magnetic force and high field magnet
PDF http://pengkuanem.blogspot.com/2018/06/longitudinal-magnetic-force-and-high.html
or
Word https://www.academia.edu/36787024/Longitudinal_magnetic_force_and_high_field_magnet