• The text in italics is copied from that url
• Immediate followed by some comments
In the last paragraph I explain my own opinion.

### Introduction

The article starts with the following sentence.
According to the theory of relativity, time dilation is a difference in the elapsed time measured by two observers, either due to a velocity difference relative to each other, or by being differently situated relative to a gravitational field.
• Time dilation is the physical difference measured by two clocks in relative movement of two events.
• Time dilation is also the physical difference measured of two clocks differently situated as influenced by gravitational masses.
As a result of the nature of spacetime, a clock that is moving relative to an observer will be measured to tick slower than a clock that is at rest in the observer's own frame of reference
Spacetime is the concept which is used to describe the behaviour of clocks, but Time dilation itself has nothing to do with the concept of spacetime.
Ofcourse it is tricky to claim both that the clock of the moving observer A relatif to observer B is running slower and also that the clock of the moving observer B relatif to observer A is running slower.
When this is the case a detailed explanation is required.

In reality considered from one reference frame only one clock will run slower. The reason is the construction of the clock

A clock that is under the influence of a stronger gravitational field than an observer's will also be measured to tick slower than the observer's own clock.
Also here it is important to give the details because this behaviour belongs to the field of gravitational time dilation.
Clocks on the Space Shuttle run slightly slower than reference clocks on Earth, while clocks on GPS and Galileo satellites run slightly faster.
It requires a very accurate description to describe the differences involved. I assume they use the same clocks.

### 1. Velocity time dilation

The laws of nature are such that time itself (i.e. spacetime) will bend due to differences in either gravity or velocity – each of which affects time in different ways.
This sentence is not clear. The laws of nature (any law) is not the cause of physical behaviour.
Time dilation is caused by differences in either gravity or relative velocity. In the case of ISS, time is slower due to the velocity in circular orbit; this effect is slightly reduced by the opposing effect of less gravitational potential.
Again you need a very accurate description to explain what the differences are.

### 1.1 Simple inference of velocity time dilation

Time dilation can be inferred from the observed constancy of the speed of light in all reference frames dictated by the second postulate of special relativity
The concept of time dilation is only an issue for (moving) clocks which inner workings is based on light signals.
This constancy of the speed of light means that, counter to intuition, speeds of material objects and light are not additive.
The issue is that the physical speed of light (the speed with which photons move in all directions) is completely independent of any human observer.
It is not possible to make the speed of light appear greater by moving towards or away from the light source.
This depends.
Monochromatic light has a specific frequency. When you move towards the source the observed frequency will increase. When you move away from the source the observed frequency will decrease. However independent of what the observer does the speed of light stays the same.

### 1.2 Reciprocity

Given a certain frame of reference, and the "stationary" observer described earlier, if a second observer accompanied the "moving" clock, each of the observers would perceive the other's clock as ticking at a slower rate than their own local clock, due to them both perceiving the other to be the one that's in motion relative to their own stationary frame of reference.
Consider 3 Observers situated at the numbers of a clock. One observer O in the center (at rest). One observer A at 12 o'clock moving towards the right and one Observer B at 6 o'clock moving towards the left.
From the point of view of O both clocks of A and B are running equally behind.
From the point of view of A both clocks of O and B are running behind. B has the slowest rate.
The point of view of B is similar as A's point of view except that A has the slowest rate.

When you have 100 observers, one of them should have the fastest moving clock and one the slowest It does not make sense to assume that all run slower than my clock
Common sense would dictate that, if the passage of time has slowed for a moving object, said object would observe the external world's time to be correspondingly sped up.
Common sense has nothing to do with science. Anyway the sentence is not clear, which is a mistake in science.
Counterintuitively, special relativity predicts the opposite.
The issue is how you decide which opinion is correct. IMO the only way is when you compare clock readings directly.
When two observers are in motion relative to each other, each will measure the other's clock slowing down, in concordance with them being moving relative to the observer's frame of reference.
Again it does not matter what each observer thinks about the other clock. You need a more direct (objective) way to decide who is right.
While this seems self-contradictory, a similar oddity occurs in everyday life. If person A sees person B, person B will appear small to them; at the same time, person A will appear small to person B.
This is a complete wrong comparison, because it based on what we see and no physical permanent effect is discussed.
With time dilation this is different as the twin paradox shows. It is a permanent effect.
The reciprocity of the phenomenon also leads to the so-called twins paradox, where one person staying on Earth and one embarking on space travel should apparently expect each other to age the same amount of time since they both see the other as moving at equal speeds from their own frame of reference.
The importance of the twin paradox is not the aging of the twins, but the fact that you are testing the behavior of the clock where the start and finish are the same, but the path is different.
The dilemma posed by the paradox, however, can be explained by the acceleration changes inherent to the traveling person's trip, which not constituting an inertial frame of reference, is not covered by special relativity.
This is the most tricky explanation by claiming that the Twin paradox does not belong to SR
In reality all experiments, which different speeds, always involve accelerations

### 1.5 Formulation

The formula for determining time dilation in special relativity is:
Delta(t') = gamma*Delta(t) = Delta(t)/sqrt(1-v^2/c^2)
where Delta(t) is the time interval between two co-local events (i.e. happening at the same place) for an observer in some inertial frame (e.g. ticks on his clock), this is known as the proper time, Delta(t') is the time interval between those same events, as measured by another observer, inertially moving with velocity v with respect to the former observer, v is the relative velocity between the observer and the moving clock, c is the speed of light, and the Lorentz factor (conventionally denoted by the Greek letter gamma) is
When you select Reflection 1 you will see that in order to explain this formula no moving observer is involved. Delta(t') is t1 and DElta(t) = t0. The velocity v is measured in the frame at rest. The importance is that all the observers are at rest.
Under all possible proper times between two events, the proper time of the unaccelerated clock is maximal, which is the solution to the twin paradox.
In all physical experiments (with different speeds) accelerations are involved.
When you consider the diagram that demonstrates the twin experiments two specific situations are sketched:
• One twin which travels from a starting position to an end position with a constant speed in a time t. His clock monitors 11 ticks.
• A second twin which also travels between the same starting position and end position, however not in a straight path but via a detour. His speed is the same on the first leg, but much higher on the second leg. His total travel time is t' + t''. The total count on his clock is 4 + 4 = 8 ticks. That means the clock runs slower.
The final conclusion is that the clock which takes the longest between two simultaneous events runs the slowest.
This is a physical permanent effect.

### 2. Gravitational time dilation

Gravitational time dilation is experienced by an observer that, being under the influence of a gravitational field, will measure his own clock to slow down, compared to another that is under a weaker gravitational field.
If you want to demonstrate this then you should compare the behaviour of two clocks under different conditions
Contrarily to velocity time dilation, in which both observers measure the other as aging slower (a reciprocal effect), gravitational time dilation is not reciprocal.
The idea to call that time dilation is reciprocal, is wrong. In a twin type experiment both observers will agree that the clock that travels the longest will run the slowest.
This means that with gravitational time dilation both observers agree that the clock nearer the center of the gravitational field is slower in rate, and they agree on the ratio of the difference.
The same for time dilation.

### 3.1 Experimental testing

• A comparison of muon lifetimes at different speeds is possible. In the laboratory, slow muons are produced, and in the atmosphere very fast moving muons are introduced by cosmic rays.
Taking the muon lifetime at rest as the laboratory value of 2.2197 µs, the lifetime of a cosmic ray produced muon traveling at 98% of the speed of light is about five times longer, in agreement with observations.
Unfortunate this text does not give enough information to validate Special relativity.
1 / Sqr(1 - 0.98*0.98) = 5,025 That is mathematical correct.
The issue is how do you know that the speed of the muon is 98% the speed of light. IMO that is impossible which implies that you can not test (using this method) SR

To read the document "Test of relativistic time dilation with fast optical atomic clocks at different velocities" of November 2007 select the following link: http://www.researchgate.net/publication/228980964_Test_of_relativistic_time_dilation_with_fast_optical_atomic_clocks_at_different_velocities/links/0912f50f3def960458000000.pdf

### Reflection 1 - Lorentz Contraction

In this experiment the train travels in the x direction with a speed v. In the train are two "mirrors". One at the bottom of the train and one at the ceiling. The two mirrors are used as a clock. The light signal is supposed to go in the y direction.
 ``` B1------B-------B2------B3 /|\ / \ / | \ / t1/ | \ / / | \ / D L D / / |t0 \ / v--> / t2 | \ / -.A-.-.-.-E-.-.-.-C.-.-.-.-.- Figure 3 ``` The time for a light signal to travel (v=0) from E to B and back is 2*t0 In this case t0 = BE / c = L /c The time for a light signal to travel from A to B and back to C is 2 * t1 The time for a "train" to travel with speed v, from A to E to C is 2 * t2 If both arrive at the same time than: t1 = t2 = t AB = D = c*t1 = c*t and AE = EC = v*t2 = v*t Using some arithmatic we get: AB^2= AE^2 + BE^2 = c^2*t^2 = v^2*t^2+L^2 c^2*t^2 - v^2*t^2 = L^2 : t^2 = L^2/(c^2-v^2) : t^2 = L^2/c^2/(c^2-v^2)/c^2 Now we get t = L/c / SQR(1-v^2/c^2) or t1 = t0/ SQR(1-v^2/c^2)) = gamma * t0 The factor SQR(1-v^2/c^2)) with v > 0 is smaller than 1.
That means t = t1 = t0/SQR(1-v^2/c^2)) that t1 is larger than t0.
That means as measured in the rest frame that it takes longer for a time signal to go from A to B to C than from A to B1 to A or from E to B to E or from C to B2 to C.
The importance is that if the signal A,B,C,B3 etc is used as a clock (this is than a moving clock) this clock ticks slower as a clock in the rest frame.
When the moving clock stops and returns this delay compared with the clock in the rest frame will contuinue to increase.
No length contraction is involved Also no moving frame is involved. All values are measured in the rest frame.

### Reflection 2 - Moving Observer

 ``` B3 B1------B2------.-------B2 | / \ . . / \ | / \ . . / \ | t1/ .\t1 t1/. \ t0| / . \ / . \ L D . v D / . \ | / . --> \ / . \ |/. t2 t2 \ / .\ -.A-.-.-.-.-.-.-.-C.-.-.-.-.-.-.-.C1 Figure 4 ``` ``` B2 | B1 B1 | / \ / \ |t1 t0/ \t0 t0/ \t0 | / \ / \ D L L L L | / \ / <-- v1 \ | / t3 t3 \ / t3 t3 \| .-.-.-.-.-.-.-.-.-.-.-.-.-.AC Figure 5 ``` ``` B3 /\ B2 / \ | / \ B1 | / \ / \ |t1 / \ / \t0 | / \ \ D / L | / v2--> \ | / t3 \|/ -.-.-.-.AC-.-.-.-.-.-.-.- Figure 6 ```
Figure 4 is basically the same as Figure 3 but with some minor modifications.
In Figure 4 we have two obervers A and C0 and C1 at rest.
Observer A will issue a light signal at time zero.
• Observer A will see a reflection a time 2*t0 later.
• Observer C will see a reflection a time 2*t1 later and C1 4*t1 later
For a moving observer on a train to see both signals (events) at A and C the speed of the train has to be v. v = distance between C and A divided by 2*t1. In that case t1 = t0/ SQR(1-v^2/c^2)).
It is important to remark that this exeperiment is symetrical.
It does not matter if the train moves with a speed v to the right towards observer C,C1 or to the left towards C-1,C-2 with a speed -v.

Figure 5 and figure 6 are drawn from the perspective that the whole platform (rails etc) moves to the right with a speed v. The central line in both is the line AC-B2 which is supposed to be at rest.
The difference between the two is:
• In Figure 5 the train is supposed to have a speed v to the left. That means compared to Figure 4 the train is at rest,
• In Figure 6 the train is supposed to have a speed v to the right. That means compared to Figure 4 the speed of the train is larger than v. In Figure 4 this is the line A-B3.
It is important to remark that this experiment is a-symetrical
 ``` B3 B1------B2------.-------B2 | / \ . . / \ | / \ . . / | t1/ .\t1 t1/. t0| / . \ / . L D . v D / | / . --> \ / |/. t2 t2 \ / -.A-.-.-.-.-.-.-.-A2.-.-.-.- Figure 4 ``` ``` A2 . | . . | . . | . .B3 t| . . | . . | . /. A1 | .B2 | . | ./ | ./ | . / | / .|. / | / .|B1/ | / . | / |/. |/ A------B---------------- Distance Figure 7 ``` ``` | | A1 t| | X2 / | |. . / | / . . / | X1 . | B2 | ./ , . | ,/ | . / ., | , / | . / . , | , / | . / . , | , / A1 / . ,B1 / | . /. , | / |A2/. , | / | / . , | / |/ . , |/ A---------X---------B-------- Distance Figure 8 ```
• Figure 7 is the same as Figure 4 except that the horizontal axis is the dsitance and the vertical axis is the time.
The line A-B1-A1 in Figure 7 represents the line A - B1 - A in Figure 4. This is the line of an observer at rest which issues a light signal and which is reflected at a distance by B.
The line A-B2-A2 in Figure 7 and in Figure 4 are identical. This is the line of an observer at rest which issues a light signal and which is reflected at a larger distance by B2. For a moving observer the speed has to be v.
The line A-B3-A3 in Figure 7 and in Figure 4 are identical. This is the line of an observer at rest which issues a light signal and which is reflected at a even larger distance by B3. The speed of the moving observer has to be larger than v.
• Figure 8 shows the typical arangement when the mirrors are not parallel in the direction of the movement of the train but perpendicular. The observer, at the center of the train, issues two light signals. In this case both the train travels in the +x direction and the light signals in +x and -x direction.
• The path X - A1 - X1 and X - B1 - X1 show the path when the train is at rest.
• The path X - A2 - X2 and X - B2 - X2 show the path when the train is moving.
When the distance AX is L, then the time t0 = 2*L/c
When the train moves with the speed v, then to calculate A2 we get v*t2 + c*t2 = L or t2 = L/(c+v)
To calculate B2 we get: c*t3 = L + v*t3 or t3 = L/(c-v)
In total to calculate X2 we get t1 = t2+t3 = L/(C+v) + L/(c-v) = 2Lc/(c^2-v^2) = t0*c^2/(C^2-v^2) = t0/(1-v^2/c^2) = gamma^2 * t0 The importance is that also here the moving clock ticks slower.
• What is also important that you should not perform neither setup with an airplane which circles around the earth.
The reason is the construction of the clock which can only be guaranteed to work properly when you fly in a straight path.

### Reflection 3 - LC simple

An even simpler demonstration of the mathematical relations in Figure 3 (See Reflection 1 ) goes as follows:
c^2*t1^2 = c^2*t0^2 + v^2*t2^2
With t1 = t2 = t we get:
c^2*t^2 = c^2*t0^2 + v^2*t^2 or (c^2-v^2)*t^2 = c^2*t0^2 or (1-v^2/c^2)*t^2 = t0^2
now we get: t = t0 / sqr (1-v^2/c^2) = t1
No moving observers are involved.
What this law demonstrates is that when you have a clock, which way of operation can be described by this law, than there are two possibilities when you move one clock from A to B and back to A with a speed v:
1. That the moving clock actual lacks behind when it returns back to A compared with the stay at home clock. In that case than no length contraction is involved.
2. That the moving clock returns and both times are the same, than you can explain this by assuming that length contraction is involved.
However in that second case (contradicted by actual experiments) you cannot be absolute sure if the above mentioned mathematical relations apply. In that same case also an even simpler explanation could be used i.e. that neither time dilation nor length contraction are involved.

### Reflection 4 - LC Application from two frames

Also in this case the train goes in the x direction and the light signal in the y direction.
 ``` | | | .t3 | |. | | . | | . | | . | | . | | | . | | | .| | t1. .t2 | . | .| | . | . | | . | . | .t0 | . | | . | . | | . | . | | . |. | L A L Figure 9 Y ``` ``` | / | ,t5 | ,/ t4. , / | . , / | . . / | . / | . . / | . xt3 t1. / | / | / | / | / | / | / |/ A---------C Figure 9 X ``` Figure 9 y shows the moving light flash in the y direction. t1 is the duration of one count of the stay at home clock. i.e at A. t1 = 2 * L / c t3 is the duration of one count of the moving clock from A to C t3 = t1 / sqr(1 - v^2/c^2) Figure 9 x shows the moving light flash in the y direction. t4 is the time when the observer at rest receives the first count of the moving clock t4 = AC/c + t3 : v*t3/c + t3 = (v+c) * t3 /c AC = v*t3 = (t4-t3)*c : t4*c = (v+c) * t3 t5 is the time when the moving observer receives the first count of the clock at rest t5 = L1/(c-v) + t1 : L1 = v* t1: t5 = v * t1/(c-v) + t1 = (v*t1 + (c-v)*t1)/(c-v) = t1 * c / (c-v) Using this information the following two fractions can be calculated: f1 = t4/t1: This is the fraction of the clock at rest. f2 = t5/t3: This is the fraction of the moving clock.
 v/c t1 t2 t3 t4 t5 f1 f2 f1/f2 f2/f1 0.666 2 0 2.683 4.472 6 2,236 2,236 1 1 0.5 2 0 2.309 3.464 4 1,732 1.732 1 1 0.333 2 0 2.121 2.828 3 1.414 1.414 1 1 0.166 2 0 2.028 2.366 2.4 1.183 1.1183 1 1
What the table shows is that there is a no difference between the frame that is at rest and which is not. because both f1/f2 and f2/f1 are the same using this method.

### Reflection 5 - Worldline in Special Relativity

Special Relativity uses both the concept of Worldline and the concept of proper time. The proper time is the time of the moving clock.
The length of the world line tau is equal to:
In formula: tau = sqrt(T^2 - R^2)
With T the time of the clock at rest and R the total distance travelled. The purpose of this reflection is to compare this formula with Reflection 1 - Lorentz Contraction
Reflection 1 starts with the following text:
In this experiment the train travels in the x direction with a speed v. In the train are two "mirrors". One at the bottom of the train and one at the ceiling. The two mirrors are used as a clock. The light signal is supposed to go in the y direction.
 ``` B1------B-------B2 /|\ / | \ t1/ | \ / | \ D L D / |t0 \ v--> / t2 | \ / -.A-.-.-.-E-.-.-.-C.-. R Figure 3 ``` The time for a light signal to travel (v=0) from E to B and back is 2*t0 In this case t0 = BE / c = L /c The time for a light signal to travel from A to B and back to C is 2 * t1 The time for a "train" to travel with speed v, from A to E to C is 2 * t2 If both arrive at the same time than: t1 = t2 = t AB = D = c*t1 = c*t and AE = EC = v*t2 = v*t Using some arithmatic we get: AB^2= AE^2 + BE^2 = c^2*t^2 = v^2*t^2+L^2 c^2*t^2 - v^2*t^2 = L^2 : t^2 = L^2/(c^2-v^2) : t^2 = t0^2*c^2/(c^2-v^2) Now we get t^2 = t0^2/(c^2-v^2)/c^2 or t1 = t0/ SQR(1-v^2/c^2)). The factor SQR(1-v^2/c^2)) with v > 0 is smaller than 1. You can also rewrite the last two lines and then you get: c^2*t^2 - v^2*t^2 = L^2 or c^2*t^2 - R^2 = L^2 or L = sqrt (c^2*t^2 - R^2) with R = AE
The concept of worldline is discussed in the book GRAVITATION by MTW at the pages 21 and page 315. The equation that describes the situation is the same (tau = sqrt(T^2 - R^2) but the pictures are different. It should be mentioned that the equation is confusing. IMO this should be (tau = sqrt(C^2*T^2 - R^2)
Figure 5A shows the picture of page 21. Figure 5B shows the picture of page 315.
 ``` Z | t+x .Z |\ | \ | \ | \ | \ t |-----.B | /. | / . | / . | / . |/ . t-x .P . | . | . A----------> x Figure 5A (p21) ``` ``` ^ t | T B |\ | \ | \ | \ | \ 0.5T|-t2--.P | /. | / . | /t1. | / . |/ . 0 A--t2------->Z 0.5R Figure 5B (p315) ``` Figure 5A page 21 shows the worldline of a particle at rest. B is an event outside this worldline. This event creates a light ray which strike the worldline at Z. P is an earlier event that creates a lightray which coincides with B. The angle PBZ is 90 degrees. Figure 5B page 315 demonstrates a non straight worldline. In this case a particle moves with uniform velocity v from A to P and back to B. In reality the particle moves along along the Z axis The angle APB is larger than 90 degrees because v

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