# Is a length contraction just a visual thing?

Discussion in 'Pseudoscience' started by absolute-space, Feb 22, 2016.

1. ### river

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Disagree

Measured to outside perspective or within perspective ?

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3. ### paddoboyValued Senior Member

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Do you have trouble in comprehension? Or just trolling?
As I said, no one questions your right to disagree: But don't presume anyone is taking you seriously on cosmological matters.

Let me again explain...Every frame of reference is as valid as any other frame of reference. Time dilation and length contraction are real results caused by the finite speed of light.
That's been well known now for a 100 years...You disagree?

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That's OK, but you are wrong.

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5. ### rpennerFully WiredValued Senior Member

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This question was answered in post #363 as well as numerous times in other posts.
Empirical high-precision evidence of phenomena at high relative speed has always supported $K= \frac{1}{c^2}$ over $K=0$ when the precision was high enough to distinguish the two.
Thus the math used in post #62 is directly based on the best available model of the behavior of the universe. Thus there is no obvious mistake in relying on said mathematics and the burden is on you to make a coherent, evidence-based argument that supports a positive position that explains all these experimental results and your unusual claim.

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7. ### James RJust this guy, you know?Staff Member

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Moderator note: absolute-space has been permanently banned from sciforums, after admitting that he was using a sock puppet.

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8. ### paddoboyValued Senior Member

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While that is admirable and the right course of action to take, should not this thread and its ridiculous unsupported claims that have inevitably been shown to be wrong, be also moved to the appropriate section?

9. ### James RJust this guy, you know?Staff Member

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The thread question is a legitimate one, and it's an interesting matter to discuss. Unfortunately, this particular thread went down a cranky path fairly early on. In the end, I've decided to move it and to leave a redirect in Physics.

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Hooray.

11. ### exchemistValued Senior Member

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Agreed. It is a pity it went off track.

12. ### Confused2Registered Senior Member

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A while back there was a question about the muon experiment "What if it were the other way round?" From the context this was clearly - assume the Earth is moving and muons are stationary. In reality the only movement is relative and "the other way round" would be for the muons to have 'Earth detectors' spaced (say) 10km apart - then the events would be 'detector 1 hits Earth' and 'detector 2 hits Earth'. The distance in the muon frame would be 10km and zero in the Earth frame. The invariant spacetime interval would be the same in both cases.

13. ### paddoboyValued Senior Member

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http://www.atmosp.physics.utoronto.ca/people/strong/phy140/lecture32_01.pdf

Muons - An Example of Time Dilation and Length Contraction Muons are unstable particles created when cosmic rays interact with the upper atmosphere. They move at very high velocities (β ~ 0.9999) and have very short lifetimes, τ = 2 × 10-6 s, as measured in the lab. Do muons reach the ground, given an atmospheric "thickness" of about 10 km? "Classical" answer: distance = velocity × time = 0.9999c × 2 × 10-6 s ≅ 0.6 km ∴ conclude that muons will never reach the ground. However, they do! What is wrong? Because muons move so quickly, relativistic effects are important. The classical answer mixes up reference frames. • τ - refers to lifetime in the muon reference frame (lab where τ is measured) • atmospheric thickness - refers to length in the Earth’s reference frame (1) Time dilation approach – from Earth frame → treat the muons as a clock in A' t' = 0 muons are created t' = τ muons are destroyed ∆t' = τ The creation and destruction of muons occurs at the same place in A'. What is the time interval as measured in A? t t' 1.4 10 s −4 ∆ = γ∆ = γτ = × i.e., it takes 70 times longer ∴ distance = 0.9999c × 1.4 × 10-4 s ≅ 42 km Since 42 km > 10 km, the muons will reach the ground. (2) Length contraction approach – from muon frame In reference frame A (now with the muons) lifetime = τ = 2 × 10-6 s velocity of ground = 0.9999c So the distance that the ground travels before the muon decays is 0.6 km. But what is the thickness of the atmosphere that the muon sees? proper length of atmosphere = 10 km length of atmosphere in muon frame is x x' 1 v c 10 1 0.9999 0.14 km 2 2 2 ∆ = ∆ − = − = i.e., the atmosphere that the muon sees is 70 times thinner 0.6 km > 0.14 km and so the ground will reach the muon. Thus, length contraction and time dilation are real!

14. ### paddoboyValued Senior Member

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or this...................

15. ### arfa branecall me arfValued Senior Member

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This video starts with a short clip of the cloud chamber in operation, which shows vapor trails appearing suddenly and fading, much like vapor trails in the sky. The tracks are made by muons (what else?).

That is, knowing what is known about cosmic rays, that's the explanation for the observations.
And, yes, you can build a cosmic-ray muon detector, all you need is a way to contain alcohol vapor in a dense enough state. And a dark room plus a torch.

Last edited: Mar 1, 2016
16. ### arfa branecall me arfValued Senior Member

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Here's another DIY alcohol vapor particle detector (quite a bit more professional). A small piece of pitchblende is the object in the centre, and a source of alpha particles. The trails appear where the alcohol vapor is condensed, i.e. near the cool surface at the floor of the chamber (pardon, chambre).

In the 45 minutes or so of nice classical music and particle trails, you can see trails that can't have been made by particles from the sample.

Last edited: Mar 1, 2016
17. ### rpennerFully WiredValued Senior Member

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Howdy. The muon example is so nice because it only works if special relativity has a special* type of consistency.

Let's start with the coordinates where the Earth is stationary in 1+1 Special Relativity. Top of atmosphere: $x= 0$, Bottom of Atmosphere: $x=L_0$, Birth of Muon: $t=0$, Detection of Muon: $t = \frac{L_0}{u}$. We have four points:
$\left( x_O, t_O \right) = \left( 0, 0 \right), \; \left( x_Q, t_Q \right) = \left( 0, \frac{L_0}{u} \right), \; \left( x_P, t_P \right) = \left( L_0, \frac{L_0}{u} \right), \; \left( x_R, t_R \right) = \left( L_0, 0 \right)$
This gives a space-time volume OQPR of $\frac{L_0^2}{u}$ in which decays either happen or don't happen. For consistency, that space-time volume needs to be the same in any other frame.

We transform to the coordinates where the muon is stationary by applying:
$x' = \frac{x - u t}{\sqrt{1 - \frac{u^2}{c^2}}}, \; t' = \frac{t - \frac{u x}{c^2}}{\sqrt{1 - \frac{u^2}{c^2}}}$
or alternately by substituting in
$x = \frac{x' + u t'}{\sqrt{1 - \frac{u^2}{c^2}}}, \; t = \frac{t' + \frac{u x'}{c^2}}{\sqrt{1 - \frac{u^2}{c^2}}}$

Interlude. World-lines.
Let I be the worldline for the top of the atmosphere, J the worldline for the bottom of the atmosphere, and K the worldline of the muon. Thus I must pass through O and Q, J through P and R and K through O and P. We can also add a worldline H which zips from R to Q at the same speed as the muon.
In the Earth-stationary coordinates, the equations of those lines are:
$x_I + 0 t_I = 0 \\ x_J + 0 t_J = L_0 \\ x_K - u t_K = 0 \\ x_H + u t_H = L_0$
Crudely substituting expressions for the Earth-stationary coordinates in terms of the muon-stationary coordinates we have:
$\frac{x'_I + u t'_I}{\sqrt{1 - \frac{u^2}{c^2}}} = 0 \\ \frac{x'_J + u t'_J}{\sqrt{1 - \frac{u^2}{c^2}}} = L_0 \\ \frac{x'_K + u t'_K}{\sqrt{1 - \frac{u^2}{c^2}}} - u \frac{t'_K + \frac{u x'_K}{c^2}}{\sqrt{1 - \frac{u^2}{c^2}}} = 0 \\ \frac{x'_H + u t'_H}{\sqrt{1 - \frac{u^2}{c^2}}} + u \frac{t'_H + \frac{u x'_H}{c^2}}{\sqrt{1 - \frac{u^2}{c^2}}} = L_0$
Multiplying left and right sides by $\sqrt{1 - \frac{u^2}{c^2}}$ we have:
$x'_I + u t'_I = 0 \\ x'_J + u t'_J = L_0 \sqrt{1 - \frac{u^2}{c^2}} \\ x'_K + u t'_K - u t'_K - \frac{u^2 }{c^2}x'_K = 0 \\ x'_H + u t'_H+ u t'_H + \frac{u^2 }{c^2} x'_H = L_0 \sqrt{1 - \frac{u^2}{c^2}}$
The last two expressions can be simplified:
$x'_K = 0 \\ x'_H + \frac{2 u}{1 + \frac{u^2 }{c^2} } t'_H = \frac{\sqrt{1 - \frac{u^2}{c^2}}}{1 + \frac{u^2 }{c^2} } L_0$
Doing geometry, we can figure out where O, Q, P, and R are in muon-stationary coordinates by intersecting these lines.
$\left( x'_O, t'_O \right) = I \cap K = \left( 0, 0 \right), \; \left( x'_Q, t'_Q \right) = H \cap I = \left( -\frac{L_0}{\sqrt{1 - \frac{u^2}{c^2}}}, \frac{L_0}{u \sqrt{1 - \frac{u^2}{c^2}}} \right), \; \left( x'_P, t'_P \right) = J \cap K = \left( 0, \frac{L_0 \sqrt{1 - \frac{u^2}{c^2}} }{u} \right), \; \left( x'_R, t'_R \right) = H \cap J = \left( \frac{L_0}{\sqrt{1 - \frac{u^2}{c^2}}}, - \frac{u L_0}{c^2 \sqrt{1 - \frac{u^2}{c^2}}} \right)$
Finding these points by intersection is perhaps more work than simply transforming the original coordinates for the points, but gives you a feel for the geometry.

Transforming the coordinates for the four points we have:

$\left( x'_O, t'_O \right) = \left( 0, 0 \right), \; \left( x'_Q, t'_Q \right) = \left( \frac{ - L_0}{\sqrt{1 - \frac{u^2}{c^2}}}, \frac{L_0}{u \sqrt{1 - \frac{u^2}{c^2}}} \right), \; \left( x'_P, t'_P \right) = \left( 0, \frac{ L_0 \sqrt{1 - \frac{u^2}{c^2}}}{u} \right), \; \left( x'_R, t'_R \right) = \left( \frac{L_0}{\sqrt{1 - \frac{u^2}{c^2}}}, - \frac{ u L_0}{c^2 \sqrt{1 - \frac{u^2}{c^2}}} \right)$
That was rather simpler than transforming equations for world-lines and finding intersections. But the exact same results are found. That's one type of consistency related to the fact that the Lorentz transforms are linear in coordinates.

To find the space-time volume, we need the cross product between two sides of the figure OQPR which is a parallelogram in muon-stationary coordinates.

$(R-O) \times (Q-O) = (x'_R-x'_O)(t'_Q - t'_O) - (x'_Q - x'_O)(t'_R-t'_O) = \frac{L_0}{\sqrt{1 - \frac{u^2}{c^2}}} \frac{L_0}{u \sqrt{1 - \frac{u^2}{c^2}}} - \frac{ L_0}{\sqrt{1 - \frac{u^2}{c^2}}} \frac{ u L_0}{c^2 \sqrt{1 - \frac{u^2}{c^2}}} = \frac{L_0^2}{u}$

That's the same as in Earth-stationary coordinates. This is a second type of consistency -- the Lorentz transform is a linear transform with determinant of one, so it preserves areas.

Thus while the coordinates are distorted, we are still addressing the same patch of space-time, with the same "area". The muon either decays in that region of space-time or doesn't and our physics is self-consistent.

* A linear operator or matrix with determinant of 1 is called "special" and so, that's a bit of double entendre. You are welcome

Last edited: Mar 1, 2016
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18. ### SchmelzerValued Senior Member

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I think banning absolute space from a scientific forum is not a good idea, given that it is part of many important scientific theories like Newtonian mechanics.

(Yes, I have understood that this is about a user with this name, but SCNR)

19. ### exchemistValued Senior Member

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Yes. The muon experiment is one of my favourite ways to visualise the relation between length contraction and time dilation.

20. ### exchemistValued Senior Member

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Nobody has banned the concept of absolute space. A user who pretends to ask an innocent question, when in fact it is a pretext for introducing a crank agenda, has been banned from part of the forum. That is posting in bad faith. It wastes people's time on explaining science, when the user in fact has no interest in the answers.

Personally, I find such behaviour really offensive, as it abuses the goodwill of the people who take the trouble to respond.

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21. ### SchmelzerValued Senior Member

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Hm. Do you read a post before answering? In case not, I repeat: Yes, I have understood that this is about a user with this name, but SCNR

I apologize for having not added 20 smilies and a "warning: satirical content" spoiler.

22. ### exchemistValued Senior Member

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Yes I do read them and I read yours. Somehow I missed the "satire".

P.S. What does SCNR stand for? Not the French Railway system, obviously (that's a joke.)

23. ### SchmelzerValued Senior Member

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SCNR means "sorry, could not resist".