How to test length contraction by experiment?

I do not cite any stuff from the internet. My papers are all my own result. I agree that length contraction occurs, but it is a consequence of time dilation and not an independent phenomenon. If we think in mathematical terms, one condition determines one variable. Constancy of the speed of light is one condition, it does not need 2 variables to be conserved, that is, time dilation and length contraction.

However, special relativity is an established theory and people are not willing to change it. But philosophically, it can still evolve and be improved.

 

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Lorentz has Length contraction and time dilation as hypotheses to derive his transform. Einstein used only one condition, constancy of the laws of physics in all inertial reference frames and then length contraction and time dilation as consequences. This shows that length contraction and time dilation are equivalent mathematically with Lorentz transform. So, only one condition is necessary for SR and it is reasonable that one can derive Lorentz transform solely with time dilation, without the need of length contraction.

 

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If you can derive length contraction from time dilation, then you should also be able to derive time dilation from length contraction.

By the way, there is also a third effect called relativity of simultaneity. Do you derive that from time dilation also? Do you derive it at all?

And finally, if Einstein was able to derive all three from one premise, then why are you trying to do it a different way? And what specific changes/improvements to relativity theory are you suggesting?

 

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Lorentz transform is derived from time dilation only. Once Lorentz transform is there , one can derive all three effects.

Because Lorentz transform creates weird things such as the ruler in my paper. I showed that a massless no elastic ruler will have its far end go backward when the near end is accelerated forward. This is not true because both ends should go in the same direction and with the same speed.

What I'm trying to do is to solve this problem.

 

If you want both ends to go in the same direction, and with the same speed & acceleration, then you need to apply the force to both ends simultaneously in the initial rest frame. Of course this requirement precludes any length contraction from being measured in the initial rest frame, because that requirement says that both ends remain a fixed distance apart in the initial rest frame, because both ends' accelerations are required to be identical. That is what happens in Bell's thought experiment, and the string breaks in that one, because in the string's own frame the ends are getting farther away from each other.

If you only put the force at the left end of your ruler, then the left end will start to accelerate before the right end. This is because SR says that nothing can travel at an instantaneous speed, not even information. So the information that there is a force applied to the left end will not be received by the right end until it propagates through the ruler at some speed which is either c or some slower speed such as the speed of sound in the material. You purposefully neglect all of this, and just incorrectly claim that SR predicts that the right end would move to the left.

Please consider the details of the bug & rivet thought experiment. A good solution to that is given here, by Dr. Glyn George:

https://www.engr.mun.ca/~ggeorge/astron/BugRivet.pdf

Note the words on page 2, "Information cannot travel faster than the speed of light. It takes time for knowledge that the rivet head has slammed into the surface to travel down the shaft of the rivet. Until each part of the shaft receives the information that the rivet head has stopped, that part keeps going at speed..."

In other words, you have not successfully shown that the right end of your ruler moves left, because you have not considered these details. So you are trying to change a theory which you perceive as problematic, but you don't even really know if it is problematic yet. You might as well say that the bug & rivet shows a problem with relativity, because the bug both lives and dies. But all you are doing is showing that you purposefully neglect the details that Dr. George went to great lengths to consider.

 

You didn't show that. What you showed is that you're don't know understand how acceleration works in a relativistic scenario.
Not understanding something does not equate with that thing being false.

EDIT: never mind. I see Neddy beat me to it.

Yeah. Even worse, the physical force cannot travel faster than the speed of sound in the material that the rivet is made out of. (Though in a thought experiment, that could approach the speed of light.)

 

Thanks for the Bug and rivet paradox which I was not aware of.

For the Bell's paradox, would you agree that the forces on both spaceships are identical? If yes, then the string does not have strain in it. What is the force that breaks it?

For the speed of transmission of information, there is no need of transmission of information from one end to the other to make both end to move simultaneously, at least in mathematics. You can imagine 2 fleas on both ends who leap at the same time. No information is transmitted, but the frame of the 2 fleas is accelerated. So, the right flea should go backward while jumping forward.

 

That's great for math. You can have infinite speeds and 26 dimensions and infinitely rigid objects.
Relativity is a real world model.

Still no. Keep studying.

 

If there were no string connecting them, then the forces on both spaceships would be identical. In that simpler case, SR says that each spaceship (in its own rest frame) finds the other spaceship to be getting farther away, as they separate to greater and greater distance between them. Obviously then, a string tied between them which is as weak as a spider web would have to stretch and eventually break.

But if you connect them with a stronger string, then the forces on both spaceships could not be identical, because the string would be resisting the separation to greater and greater distance between them. If you do not understand these things in SR, then please first try studying the details first, before deciding it is wrong and trying to replace it yourself.

That is the same as applying the force to both ends of your ruler simultaneously, which is what happens in the Bell scenario. If you only apply a force to one end, and not the other, then the force does not instantaneously appear at the other end, (as previous explained, but automatically disputed by you anyway). I am seeing a pattern here.

I probably won't be replying to this thread anymore, as it seems no progress is being made. Best of luck to you in the future.

 

Kuan Peng;


The left graphic shows ref. frame A with a rod of length d.
Frame B is passing A at speed v with an identical rod of length d.
Without lc (length contraction), A measures dB as d.
B measures dA as d/gg, or lc of 2 orders of gamma, and measures d in his own frame as gd using light, contradicting his internal measurement of d.
Add to that, the observations are not reciprocal, as required by SR.
Each passes the other at relative speed v, each should observer the same physical phenomenon.
The right graphic shows the B rod lc to d/g, and just by inspection of similar geometry where the right ends cross, we see reciprocal lc.
This shows the need for both effects to satisfy SR requirements.

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This is the kind of smoking gun diagram that I think is perfect to show the discrepancy.
Unfortunately, I can't reconcile the diagram with your instructions on how to read it.
Any way you could spell it out more verbosely?

 

Sorry for lack of detail.
A and B have a mirror M on the end of a rod of length d.
In the A frame, the length of the rod is ct=d.
The red hyperbola, aka calibration curve, denotes the same A-time where it intersects the time line of a moving observer.
Left:
Bt=2gt at the return signal, event D. B assuming a pseudo rest frame (because there is no absolute rest frame) perceives his signal to M as equal out and back, giving d=gct, greater than d.
Right:
With length contraction, B measures the B rod as ct=d.
A measures the length of the B rod as d/g.
B measures the length of the A rod (on the Bx axis) as d/g.
The results are reciprocal only if the B frame is affected by lc.

 

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

 

Why did you go from 2 satellites to n? Why add unnecessary complexity of n satellites arranged in some pattern?
This obfuscates the problem and the solution.
Rewrite the article using only two satellites.

 

Analysis of Einstein's derivation of the Lorentz Transformation
Einstein's derivation of the Lorentz Transformation is purely theoretical. This study shows how it is related to the physical phenomenon of time dilation and length contraction. The Lorentz Transformation was first derived using the conditions of time dilation and length contraction. Later, Albert Einstein has given a different derivation of the Lorentz Transformation by using constancy of the speed of light only, making time dilation and length contraction subsequent to the Lorentz Transformation which then acquired the status of fundamental law.
Einstein exposed his derivation in the Appendix 1 of his 1920 book « Relativity: The Special and General Theory ». But the physical significance of this derivation is blurry. I will analyze Einstein's theoretical derivation to find out how it is related to the physical conditions of time dilation and length contraction.
In his derivation Einstein only used the fact that the speed of light is c in K and K'. However, another claim was used without being properly declared, which is: (x,t) the co-ordinates of an event in K and (x',t') the co-ordinates of the same event in K' satisfy equations (3) and (6) in general. As these equations are in direct proportional form, we give this claim the name "Proportionality assumption". This claim is an assumption because it is not proven.
I analyze this assumption in the article below.
https://www.academia.edu/41712834/Analysis_of_Einsteins_derivation_of_the_Lorentz_Transformation
https://pengkuanonphysics.blogspot.com/2020/01/analysis-of-einsteins-derivation-of.html

 

I know this is an old thread, but the discussion (at first at least) seemed to focus on how to do step 4, which actually seems pretty easy to me.
The hard part (according to the OP) is getting a measurable object up to a speed where contraction is measurable.
The solution to that is just to consider an object already moving at near light speed, like say Earth itself.

We know the proper height of say mount Washington (~1900m), so we want to see if that altitude contracts when moved vertically at near light speed. This has been demonstrated with the muon experiment. In the frame of the muon, an approaching object (sea level) needs to be within about 475 meters of the stationary muon if the object is expected to get there within its half life. The proper altitude of the mountain's is 4 times that, yet the majority of the time the sea level gets to the muon before the decay, confirming a length contraction of the mountain's altitude by a factor of about 8.8 (Frisch/Smith, 1963). With sea level coming at the muon at near light speed from only about 215 meters away we get the matching muon flux difference between those two locations.

 

We ran the muon experiment past him. He doesn't get it. Start at post 10 and read his responses.

 

Right. Didn't even read the entire first page before posting that.
He doesn't want to learn it seems.
The method is very accurate for measuring length contraction. You have an object of precise known length (a linear accelerator say) and a stationary clock (muons) which measures the time taken for the particle accelerator to pass by. That gets its length down to a couple digits of accuracy at least, plenty to verify the effect. If that's rejected, then rejection is the goal.

 

Hi, sorry for answering you so late. I was preparing an article about length contraction, which perfectly replies your concern.

Object contraction is shown in figure 1 of https://pengkuanonphysics.blogspot.com/2020/02/drawing-relativity.html

Frame O2 is the Spaceship P moving at v with respect to frame O1. Its length is L2. In frame O1 P is seen as P’, P is length contracted and its length is

L1=L2/gamma, shorter than L2.

Time is t=0

Distance contraction is shown in figure 2

Frame O2 moves L1 in frame O1. Time is t=L1/v. The backend of the spaceship is at O1. In Frame O2, O1 moves L1/gamma, shorter than L1. So, O1 in Frame O2 is at the position

O’1=-L1/gamma=-L2/gamma^2

But because the backend of the spaceship is at O1, O1 should coincide with the backend of the spaceship in Frame O2. That is, at O’’1=-L2

The question is, what is the position of O1 in Frame O2?

Is it -L2? Or -L2/gamma^2?


See: https://pengkuanonphysics.blogspot.com/2020/02/drawing-relativity.html
and
https://www.academia.edu/41972922/Length_distance_and_Michelson-Morley_experiment
https://pengkuanonphysics.blogspot.com/2020/02/length-distance-and-michelsonmorley.html

 

Length, distance and Michelson–Morley experiment

There are 2 types of length contraction, the physical meaning of each is explained below with the help of the example shown in Figure 1. The Earth and the star are stationary and the spaceship mobile. The 2 types of length contraction are:

1) Object contraction: In the frame of the Earth the length of the moving spaceship appears shorter than its proper length.

2) Distance contraction: In the frame of the spaceship the distance from the star to the Earth appears shorter than in the frame of the Earth.


a) Measurement of distance by travel

Let us focus on Distance contraction and see why distance is shorter in the moving frame. Suppose that the spaceship coincides with the Earth at time zero and coincides with the star at the end of its travel. The spaceship moves at the velocity v in the frame of the Earth and the star moves at the velocity –v in the frame of the spaceship.


In the frame of the Earth time is tE at the end of the travel and the spaceship has traveled the distance v•tE which is the distance from the Earth to the star measured by the travel of the spaceship. This distance is denoted by DE and expressed by equation (1).


In the frame of the spaceship time is tS at the end of the travel and the star has traveled the distance v•tS which is the distance from the Earth to the star measured by the travel of the spaceship. This distance is denoted by DS and expressed by equation (2)…

Read the article below.

https://www.academia.edu/41972922/Length_distance_and_Michelson-Morley_experiment

https://pengkuanonphysics.blogspot.com/2020/02/length-distance-and-michelsonmorley.html