How to test length contraction by experiment?

I could be wrong but I think the problem is much fundamental than that - it isn't the sychronisation - it's the time dilation itself causing the problem.

 

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There is no problem.

From my post #185:
IN THE MUON REFERENCE FRAME:
Event: Muon passes HIGHER earth clock:
Higher Clock Time: 0.0 micro seconds
(Lower Clock Time: 32.6 micro seconds)
Muon Clock Time: 0.0 micro seconds
Event: Muon passes LOWER earth clock:
(Higher Clock Time: 1.4 micro seconds)
Lower Clock Time: 34.0 micro seconds
Muon Clock Time: 6.8 micro seconds

Notice that the higher earth clock goes from 0.0 to 1.4 in the time the muon's clock goes from 0.0 to 6.8. That is the earth clock's rate being about 1/5 of the rate of the muon's clock. That is time dilation.

Notice also that the lower earth clock goes from 32.6 to 34.0 in the time the muon's clock goes from 0.0 to 6.8. That is also the earth clock's rate being about 1/5 of the rate of the muon's clock. That is also time dilation.

Yet there is no disagreement between the muon and the earth regarding the fact that when the muon passed the higher earth clock, it displayed 0.0. And there is no disagreement between the muon and the earth regarding the fact that when the muon passed the lower earth clock, it displayed 34.0. So the muon would understand (if it could) that the earth must have measured the life of the muon to be 34.0 microseconds, which is 5 times more than the muon's own lifetime of 6.8 microseconds.

Why, did you really think was a problem with SR (like PengKuan does)? Or do you suppose it might have just been your possible misunderstanding of it?

 

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Usually one think of the earth and the spaceship in the twin paradox, which are 2 objects. But in fact, there are 3 objects in action, the earth and the spaceship and the star. If we see only the earth and the spaceship, we cannot say which one is moving, which one is stationary, because motion is relative.

Once you take the star into account, you can say that the spaceship is moving because it goes to the star but the earth does not. So, the earthling will say that I'm stationary with respect to the star and the time on the spaceship is slow. Here, the spaceship encounters the third object, the star, the traveler agrees that his time runs slow.

In the case where the traveler considers the spaceship as immobile and the earth mobile, we cannot use the star to be the third object because the earth is fixed with respect to it. Rather, the traveler has to find a sister spaceship which is at constant distance from him. Than, because the spaceship and sister spaceship are in relative motion with respect to the earth, the sister spaceship will meet the earth. In this case, at the encounter the time on the earth is slower than the time in the spaceship. Here the third object is the sister spaceship.

So, in the first case, the spaceship meets the star which is the third object and its time runs slow.
In the second case, the earth meets the sister spaceship which is the third object and its time runs slow.

The third object tells who is the mobile one between the earth and the spaceship.

 

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Thank you for having read all my papers and understood that my purpose is to discredit relativity.

Indeed, it has fatal flaws and has to be replaced.

 

I just want to show you that in the formula “-c²t'²=v²/t²-c²t²”
v²/t² should be v²*t² . You have put "/ " at the place of "*"

 

By the below

I just show that for whatever velocity of the spaceship, the distance between it front and back is not contracted in the frame of an observer. That is, there is no length contraction.

As I said above, because the trajectories of the front and back are parallel, the distance between them is identically the same when the spaceship moves at 0.0001c or 0.5c. Length contraction is not correct. The astronaut does not see the moon length contracted, nor the mooner see the spaceship length contracted.

 

Humble apology.
-C2.

 

Lorentz transformation transforms the number of clock ticks from muons's frame to earth's frame. That is, 6.8 ticks in muons's frame gives 1.4 ticks in earth's frame. That corresponds to the time dilation that you said. The paradox is that in earth's frame, there are 34 clock ticks in reality, not 1.4.

Lorentz transformation is not used the way you described: muon sees the clock at high station, then sees the clock at the ground station. This is not the right way.

 

There is no paradox. The muon is correct that both of the earth clocks are ticking slow compared to the muon's own clock, according to the muon's own rest frame, just as I showed. Your mistake was in your post #181 where you assumed that the two earth clocks which are synchronised in the rest frame of the earth, would also be synchronised in the rest frame of the muon. They are not. That is called relativity of simultaneity.

The Lower earth clocks is already displaying 32.6 microseconds at the time the muon is formed right next to the Higher earth clock which displays 0.0. So the Lower earth clock only has to elapse 1.4 microseconds to reach 34.0. Your so-called "paradox" would have us wondering how the Lower clock could display both 34.0 and 1.4 at the same time. It doesn't, it only displays 34.0, but it ELAPSED 1.4 microseconds since the muon was formed.

Once you understand that SR requires relativity of simultaneity, time dilation, and length contraction, (ALL THREE!), then you will start to understand. Until then, you will be making mistake after mistake, as you have been doing.

 

Clearest post yet. Nice one. No I wasn't seeing that.

 

Your method can solve the twin paradox this way:
A traveler goes for a round travel to a star. The earthling sees the travel lasts 4 years, and set a clock to prove it.
The traveler sees his travel lasts 2 years back on earth. He computes the earthling would be 2 times younger than him because the earth is moving.

At the reunion, the earthling is 4 years older and the traveler 2 years older. The traveler is surprised because he expected the earthling to be one year older. So, he thinks that this is paradoxical. But the the earthling shows him the clock and the traveler sees that the clock shows effectively 4 years. Then he agrees: 'I'm wrong, there is no paradox.'

But why Einstein himself acknowledged that there was really a paradox and proposed that it was the acceleration that caused it?

 

He didn't.

 

PengKuan#158;

d1=ct, d2=vt+L

if d1=d2 then

t=L/(1-v)

The part (1-v) is classical/Newtonian, popularly referred to as 'closing speed'.
A spatial gap shrinks or expands at a rate equal to the difference in speed of the objects defining the gap. This is the common stumbling block for LET advocates, who insist 1-v is a modified light speed, when in fact it is a relation between two object speeds, as shown above. There is no physical object with a speed of 1-v. Also, light is considered a particle in SR, ref, Compton scattering, photoelectric effect.
All this is true for the return signal at L/(1+v).
SR is already present since L is actually L/gamma, but the lab dudes can't detect it with their shortened rulers.

 

This was on file, and posted in the belief that graphics help understanding.

In this case t= x/c = 10 km/3*(10)8 = 33 µsec.

Please Register or Log in to view the hidden image!

 

Sorry, what is LET advocates?

For t=L/(1-v), let us see the line below, where O is the source of Light, L is the light and M the mirror.

O……………….L…………………M

For O, L moves at the speed c. M moves at the velocity v. So, the point where L meets M is L/(1-v/c) from O. This is computed by using c as the velocity of a classical ball.

But this is not true for M because from the standpoint of M, L moves also at c. What is the distance from O to M when L meets M? M cannot compute this distance in the frame of O because O is going away for M.

The correct method is that I used in « Length, distance and Michelson–Morley experiment » section 3. “Why Object contraction” in which I have explained in detail how to compute this distance which is found to be not contracted. The title is clickable.

 

This is the result of Lorentz transformation. I'm talking of a different transformation which gives different result.

 

For that scenario, the earth and the star measure the time of the round-trip journey to be 4.00 years, but the traveler measures it to be 2.00 years using his own clock. We can put a clock on the star, and synchronise it to the clock on earth, but they are only synchronised in the earth/star reference frame. The traveler who is moving at a speed of 0.866c relative to that reference frame will say that those clocks display different times simultaneously, and therefore are not synchronised in the traveler's reference frame.

When the traveler is moving at a speed of 0.866c toward the star, during the trip from the earth to the star, the traveler's clock goes from 0.00 years to 1.00 years (so it elapses 1.00 year). But in the traveler's own rest frame, the clock on earth goes from 0.00 years to 0.50 years (so it elapses 0.50 year), and the clock on the star goes from 1.50 years to 2.00 years (so it elapses 0.50 year). When the traveler gets to the star he can stop momentarily, and then he will say that the earth clock and the star clock are syncrhonised, because for that moment he is not moving relative to them. So both the star clock and the earth clock say 2.00 years at that point, and the traveler's own clock says 1.00 year.

Then, when the traveler is moving at a speed of 0.866c back toward the earth, during the trip from the star to the earth, the traveler's clock goes from 1.00 years to 2.00 years (so it elapses 1.00 year). But in the traveler's own rest frame, the clock on the star goes from 2.00 years to 2.50 year (so it elapses 0.50 year), and the clock on earth goes from 3.50 years to 4.00 years (so it elapses 0.50 year). When the traveler gets to the earth he can stop, and then he will say that the earth clock and the star clock are syncrhonised, because then he is not moving relative to them. So both the star clock and the earth clock say 4.00 years at that point, and the traveler's own clock says 2.00 years.

 

I'm glad you see it now. Relativity of simultaneity (ROS) is just as important as time dilation and length contraction. All three are usually needed to fully understand these SR thought experiments. Many times people get confused because they know about length contraction and time dilation, but not ROS.

 

What if there is no star? The traveler goes for 1 year, finds nothing, and goes back to earth for another year. He counted 2 years and expected that it would have passed 1 year on earth, but he finds finally that it has passed 4 year on earth.

 

If there is no star, it is still the same, as far as the earth clock in concerned. When the traveler is moving at a speed of 0.866c away from the earth, the traveler's clock goes from 0.00 years to 1.00 years (so it elapses 1.00 year). But in the traveler's own rest frame, the clock on earth only goes from 0.00 years to 0.50 years (so it elapses 0.50 year). When the traveler gets to where the star was supposed to be, he can stop momentarily, and then he will say that the earth clock is showing 2.00 years at that point, and the traveler's own clock says 1.00 year. Notice that the earth clock went from 0.50 to 2.00 during the time the traveler decelerated and came to a stop.

Then, when the traveler is moving at a speed of 0.866c back toward the earth, during the trip from the star to the earth, the traveler's clock goes from 1.00 years to 2.00 years (so it elapses 1.00 year). But in the traveler's own rest frame, the clock on earth goes from 3.50 years to 4.00 years (so it elapses 0.50 year). Notice how the earth clock went from 2.00 to 3.50 during the time the traveler accelerated toward the distant earth. Whenever there is acceleration or deceleration, distant clocks change the time they are displaying "now" because the traveler's own simultaneity changes. (Relativity of simultaneity).

The resolution to the so-called twin paradox (which is not really a paradox) is that the stay-home twin never accelerates, but the traveler has to decelerate & accelerate when the stay-home twin is far away from the traveler. This causes the traveler's simultaneity to change, and the distant clock (or age of the stay-home twin) changes as explained above.