How to test length contraction by experiment?

No. There is one.

No. They are the same thing.

Create 10,000 spaceships, and stack them. end-to-end from your observation point to the Earth's surface.
Q: Which one of your "two types of contraction" applies?

A: Flawed question. They are the same thing.

 

Log in or Sign up to hide all adverts.

Perhaps it would be better for somebody with an understanding to write such articles.

The ship's proper length is L2. O1 and O2 seem to refer to respective objects or observers(?), each stationary in their respective frames. Frames themselves don't have locations like that. An observer is not a frame. You should call the frames frame 1 and frame 2.

Distance contraction is shown in both figures, depicted as a shorter magenta ship as opposed to the blue full proper length on top. Figure 2 seems to show 'after time t in their respective frames' and figure 1 shows the state of each system at the event when the two origins are co-located.

This is wrong. It (observer 1 watching the ship go by) moves in frame 2 a distance of L1, or L2/gamma, as pictured. You need to fix that in the text.

This makes no sense. You've been talking about frame 2, in which the backend of the ship is stationary and is never 'at' a different location. You're treating O1 as a location in frame 2, which it isn't.

If O1 is an observer, then after time t, O1 is located at -L1 down the length of the ship because O1's velocity is -L1/t in that frame, per your text.

This is what I mean by this:

You're deliberately striving to be wrong.

 

Log in or Sign up to hide all adverts.

Borrowing the hyperphysics muon experiment
( http://hyperphysics.phy-astr.gsu.edu/hbase/Relativ/muon.html )

Vertical distance between count stations in the Earth frame is 10E4 metres
Velocity of the muon in the Earth frame is 0.98c
Time in the Earth frame for muon to move between count stations T_Earth is 34e-6 seconds
Distance between count stations in the Earth frame (D_Earth) = 0.98c*T_Earth=10E4 metres

Time (T_muon) in the muon frame is (T_Earth/5)

The muon observer sees each count station flash past the observation port at 0.98c so concludes the distance between stations (D_muon) is speed * T_muon
so
D_muon = 0.98c*(T_Earth/5)=2E4 metres
where (clearly) D_muon = D_Earth/5

 

Log in or Sign up to hide all adverts.

Time relativity transformation of coordinates

Without length contraction, time relativity transformation solves paradoxes and explains incongruent relativistic experiments, which allows us to build a transformation of coordinates without length contraction. For abscissa transformation, Figure 1 shows a spaceship in the frame of O1, its backend is at O1 and frontend at A1. At time zero the spaceship is stationary, from time zero to time t1, it is accelerated. The trajectories of its backend and frontend are the parallel curves from O1 to O2 and from A1 to A2. At time t1 the spaceship moves at the velocity v, its backend is at O2 and frontend at A2. So, the frame of O2 moves with the spaceship at the velocity v in the frame of O1.
Because the trajectories of the backend and frontend are parallel, the distance O2A2 constantly equals the distance O1A1 which is the proper length of the spaceship. O2A2 is the length of the moving spaceship in the stationary frame O1. So, the length of the moving spaceship constantly equals its proper length.
…
For Time transformation, The transformation of time from the stationary frame to the mobile frame is like that in special relativity. In Figure 3, at time zero a light signal is sent in the mobile frame from O2 to the mirror M which reflects it back to O2. The time the light signal takes to complete the journey is tO2=2L2/c, with c being the speed of light and L2 the distance from O2 to M. tO2 is also the time of O2. In the stationary frame the light signal is sent when O2 coincides with O1 and goes from O1 to the moving M which reflects it to O2. The time of this journey is tO1=2L1/c, with L1 being the distance from O1 to M. tO1 is also the time of O1. Because L2= L1√(1-v^2/c^2 ), tO1 is related to tO2 by equation (7).
Read the article below.
https://www.academia.edu/42129223/Time_relativity_transformation_of_coordinates
https://pengkuanonphysics.blogspot.com/2020/03/time-relativity-transformation-of.html

 

Your example describes Newtonian physics: a ship accelerating without length contraction. If you put this to the test (two ships accelerating identically), a string between them would stretch and break as its contracted length no longer spans the distance between the two ships.
One ship accelerating like that will accelerate harder at the rear than at the front, as required by the equivalence principle.
So your article is wrong from the very beginning. It describes physics which had been falsified 115 years ago.

 

For discussion convenience, I put the link of my paper here: « Discrepancy of length contraction»

Figures 1 and 2 are for different purposes. Object contraction is different from distance contraction because the length of an object is constant. The front of an object stays at the same distance from the back.


Distance between 2 objects like spaceship and star is varying. The spaceship moves, so the distance from the spaceship to the star diminishes. Varying distance does not contract the same way than constant length.


Figures 1 shows only Object contraction. The spaceship is the object and stationary in frame 2, its proper length is L2. Frame 2 moves in frame 1, so an observer in frame 1 sees the spaceship contracts. Its length in Frame 1 is contracted and is L2/gamma. The length of the spaceship is constant in time.

Figure 2 shows distance contraction in comparison with object contraction. The observer at O1 in frame 1 sees O2 moves the length of the spaceship. So, the observer sees the back of the spaceship at O1, that is, the frontend is at x=L2/gamma.


So, O2 has moved x=L2/gamma during time t in frame 1.

By time dilation, at this event, the time in frame 2 is t/gamma. So, O1 has moved during the time t/gamma in frame 2.


As the velocity of O1 is -v in frame 2, the distance from O2 to O1 is the distance done by the travel of O1, that is, x2=-vt/gamma.


In frame 1, the spaceship has moved vt= L2/gamma. So x2=-vt/gamma = -L2/gamma^2.

So, O1 is at the position x2= -L2/gamma^2 in frame 2.

In frame 1, O1 is the backend of the spaceship. So, the back of the spaceship is at x2= -L2/gamma^2 in frame 2. Until here, Do you agree?


In frame 2, the back of the spaceship is at its proper length, xs=-L2.

So, x2=xs, that is -L2/gamma^2 =-L2

 

Objects contract in length if they are moving.

Not if it's moving.

Or maybe increases, if the spaceship is moving away from the star.

Your paper linked in post 144 says otherwise, so your papers contradict each other.

These statements contradict each other. If its length is a function of gamma, then it isn't constant if gamma (a function of its speed) changes.

I've pointed out some errors in the earlier paper in prior posts. You don't seem interested in critique, so unclear why you're posting any of this here.

 

Your link is good because we agree on the parameters.

Hyperphysics muon experiment http://hyperphysics.phy-astr.gsu.edu/hbase/Relativ/muon.html

I agree that:

T_muon = T_Earth/5

We can see Earth in the frame of the muon as well. In this case, the ground moves at -0.98c. According to time dilation, Time of the ground is Time in the muon’s frame /5:

T’_Earth = T_muon/5.


Because time of the muon is 6.8 micro second, then Time of the ground is 1.36 micro second.

But the measured time on the ground is 34 micro second. What is wrong?

 

You misunderstand me. The front of an object stays at the same distance from the back in time; That is, in the observer's frame, the front of the spaceship is at the distance L from the back, at time t1, t2, t3.....

 

34 usec is time to traverse 10 km at .98c, measured by a stationary clock.
1.36 usec is time to traverse 2 km at .98c, measured by a clock moving at .98c.

2 km is the length-contracted thickness of the moving atmosphere.

Only if the observer is stationary WRT the object.

 

time of the muon measured by a clock moving at .98c is 6.8 micro second.

Please read carefully before replying.

 

if the object moves at constant velocity, its length is constant in the observer's frame.

 

Please specify exactly what you want answered then. Your wording (specifically 'time of the muon') is quite ambiguous without specification of 1) the frame in which the time is to be meaningful, 2) the clock doing the measuring, and 3) the interval being measured (which pair of events).
You asked specifically about the 34 usec, so I answered what that represents: 1) Earth frame, 2) Earth clock, and 3) the events of the creation and detection of the muon.
While I'm being specific, the 1.36 usec represents 1) muon frame, 2) Earth clock, and 3) The location of the detector when the muon is created and the detector reaching the muon.

 

Halc #136;

[No one on earth observes any contraction of the atmosphere. What is observed is the slowing process of muon decay, i.e. time dilation for the muons. This is equivalent to observing a muon clock. The observer cannot observe any effects of his motion, he can assume a pseudo rest frame.]


PengKuan#140;

[Length contraction (lc) results in elements of composite objects becoming more closely spaced. There is no difference if the elements are particles or other composite objects.

Put an anaut in a ship moving at high speed v, parallel to an earth-moon alignment with a separation of d. The anaut and ships clock would indicate less time than d/v. The anaut would conclude, based on his local time that the distance d has contracted. An observer on earth or moon would describe the ship as lc. The first is altered perception, the second altered structure. Even though the constant, independent speed of light is responsible for both td and lc, there is a significant difference. Time is cumulative with varying rates and losses, but always increasing. Lc is temporary.]
____________________________________________________________________

[Einstein developed his SR theory to account for the known anomalies in physics. He could have extended it as a theory of perception, but did not. This aspect is lost in the mathematical expositions, only considering mass, light, and motion. If reality is defined as the behavior of the universe without human observation, then perception is reality confined to the mind. Perception is also a biological process.
There would be cases where reality and perception are not distinguishable, such as two anauts in remote space with a relative velocity.
The observers motion does not and cannot alter distant processes, but it does alter measurement and perception.
When the particle accelerators are wound up, astronomers don't see the universe contract. Only someone moving with the particles would have that experience.

There are other subtle factors that question this form of reciprocity.
If the anaut can accelerate to high speed in a short time interval, how do the distant (1000 ly) masses do the same? This would require fictitious forces, as in the earth vs sun planetary system. The heliocentric system represents reality as defined above. The geocentric system has to reproduce the perceptions of the observer, with its retrograde motions. Not sure if this relates to what PengKuan is thinking.]

 

PengKuan;


The graphic reduces the MMX to the bare essentials. (g is gamma)
left:
The direction of motion is x, any direction perpendicular to the x direction is p.
The distance to each mirror is 1 unit.
The light transit time to reflect from the p mirror (Rp) is 2g.
The light transit time to reflect from the x mirror (Rx) is 2gg.
Obviously without any revisions, the two signals do not return simultaneously.
If each mirror was designated as a light clock, in a rest frame, the local time would be 2 for each clock.
In a moving frame calculations predict each clock is affected differently via time dilation.
Thus the issue is not time dilation, but distance.
right:
Length contraction of the x arm provides the resolution.

An oddity:
If the p-clock is designated as the standard (with a long second), then its value t' times c equals gtc which equals a distance of g. I.e. seemingly providing the correct answer without length contraction!

Please Register or Log in to view the hidden image!

 

What known anomalies are your talking about? The only anomaly of which I am aware is the frame-independent measurement of light speed, something not predicted by the current theories of the day.

 

Your "The anaut would conclude, based on his local time that the distance d has contracted." is my distance contraction, see below. Your lc is my object contraction. lc is a constant length, because the ship is constant length. But d from the ship to the moon is varying, so, it is not the same as lc.

In post 146, I explained the difference between object contraction and distance contraction. For discussion convenience, I put the link of my paper here: « Discrepancy of length contraction»

Figures 1 and 2 are for different purposes. Object contraction is different from distance contraction because the length of an object is constant. The front of an object stays at the same distance from the back. Distance between 2 objects like spaceship and star is varying. The spaceship moves, so the distance from the spaceship to the star diminishes. Varying distance does not contract the same way than constant length.

 

You have already done this graphic. But the derivation of Lorentz transformation from Michelson–Morley experiment is flawed.

In short, they have computed the length of the arm by determining where the light meets the mirror. And this point is found to be L/(1-v/c). This is wrong because the mirror is a moving object at velocity v, but light travels at c and should not be taken as moving object with classic law.

If you have a ball traveling at vb, then the point where it meets the mirror is L/(1-v/vb), this is correct because v and vb are classical velocity. But light’s speed is c and should not be computed like the ball. So, the light does not meet the mirror at L/(1-v/c). This flaw makes the point of meeting further away and thus, one has to make a mathematical contraction. But in reality, the arm does not contract.

This flaw is described in the section 3 “Why Object contraction?” of « Length, distance and Michelson–Morley experiment ».

 

Try measuring the distance to the back of the ship (of length L) using a pulse of light reflected off a mirror at the back of the spaceship. This is a well recognised way of measuring distance. In the spaceship frame there is Event A when the pulse leaves front of ship and Event B when the pulse arrives back at the front of the ship. In the ship frame the time t between events is 2L/c. If the ship is travelling at velocity v in the Earth frame what is the time interval in the Earth frame between the two events? Can we not compute the length of the ship in the Earth frame from the time it takes light to travel to the back and back to the front? We just need to 2t'/c. What is the relationship between t and t'? Does this give us the length of the ship in the Earth frame?

 

Sorry I'm not used to this form - edits (if possible) in progress.

What is 'wrong' is what you are not seeing. [late edit]

Invariant spacetime interval (s) .. see geometry of spacetime..
In one dimension
s²=x²-c²t²

The important point is that in the Earth frame the count stations are 10km (x) apart. The muons actually pass through the count stations (after time t') so for them x'=0.

Muon
s²=0-c²t'²
Earth
s²=x²-c²t²

s²=s²
so
-c²t'²=x²-c²t²
v=x/t (see Newton)
-c²t'²=v²/t²-c²t²
Take it from there.

Edits look OK.
Absolutely vital .. the muon only moves through time between count stations - in the Earth frame they are (say) 10km apart in distance as well as time. That makes all the difference.

The invariant spacetime interval s is precisely equivalent to the invariant speed of light. Another day if you doubt this.
s²=x²-c²t²