# Lorentz Contraction

I don't recall seeing this derivation anywhere but its probably somewhere in the relativity literature since most things are.

A poster at sci.physics.relativity posted what he thought was proof that the Lorentz contraction doesn't exist. See the Fig. 1 in Lorentz Contraction - Version 2 -- www.geocities.com/physics_world/sr/lorentz_contraction_2.htm

Here is the poster's claim
Consider (in this rest frame) an EM Emitter, and a detector placed 10 feet away from said emitter. A mirror, placed 10 feet above the emitter, is angled @ 45 degrees so it will strike the detector . The emitter is on and is sending a continuous beam. The detector has a light that shows it is detecting the EM Wave.
Now consider a relatively moving observer wrt the above frame, he also sees the detector light on that shows the detector is receiving the beam from the emitter continuously.

If length contraction was actually physical, the beam would not strike the detector in any frame. For all frames are said to view the same events, only at different times according to SR. If one frame sees the detector light on, all frames will too eventually. If one frame does not see the detector light on, no frames will ever see it on.

We all know, in this scenario, that all frames will view the detector light on. So length contraction doesn't occur period.

(note: He got the angle of the mirror wrong)

I requested a proof of his claim. In fact I repeated the request many times but he refused to post a proof.

It was kind of fun to write this up so I posted it on the web. It turned out to be another way to derive the Lorentz contraction relation.

In my derivation all that is assumed is the Lorentz transformation.

So the person thought he found proof that there is no Lorentz contraction when in fact all he did was to find another derivation and he didn't even know it! :-D

Ya got to love that! :-)

Pete

Gold Member
Yes I've seen it before, though not in a textbook, it's actually quite a good way of demonstrating length contraction.

quartodeciman
Obviously one needs to derive the velocity transformation rules first.
Why does that poster say the following?

If length contraction was actually physical, the beam would not strike the detector in any frame.

Was the poster assuming the X-Y orientation of the mirror was the same in S and S'? So there was no contraction of the X component of the mirror width?

Obviously one needs to derive the velocity transformation rules first.
Yes. If you take a look right above Eq. (1) I provide a link to that derivation.

Why does that poster say the following?

If length contraction was actually physical, the beam would not strike the detector in any frame.
I'm not sure. I guess he thinks that the layout is exactly the same in the frame in which the layout is moving and that the only thing different is the length of the rod and thus the emitter-detector disatance. I kept asking him to back that claim up but he refused to. In fact I think I asked him to provide a proof almost every single day. All I got was insults - as usual! :-)

Was the poster assuming the X-Y orientation of the mirror was the same in S and S'? So there was no contraction of the X component of the mirror width?

I don't recall. The thread is called "Length contraction doesn't occur period." - check it out if you have the time.

Pete

Originally posted by jcsd
Yes I've seen it before, though not in a textbook, it's actually quite a good way of demonstrating length contraction.

Excellant!! If you recall where you've seen that can you let me know?

Thanks

Pete

Gold Member
I'm pretty sure that a simlair example is in Brian Greene's elegant universe.

Originally posted by jcsd
I'm pretty sure that a simlair example is in Brian Greene's elegant universe.

Thank you.

Pete

quartodeciman
Pete,

Are you going to send that poster a link to your new second Lorentz contraction derivation after you publish it?
Or just leave the growler lie?

quart

Originally posted by quartodeciman
Pete,

Are you going to send that poster a link to your new second Lorentz contraction derivation after you publish it?
Or just leave the growler lie?

quart

But you know how the internet is. They love to divert attention away from the physics. One jerkoff notice that I made an error in my original Lorentz contraction page at

http://www.geocities.com/physics_world/sr/lorentz_contraction.htm

At the bottom I showed that lengths perpendicular to motion do not contract. But I made an error in three places an instead of writing "
perpendicular" I wrote "Parallel" . However it was very clear what I was writing about since I gave a diagram and clearly wrong

However if the rod is aligned perpendicular to the velocity then we can take another, rather easy, approach.

However someone was unable to prove me wrong in another thread and they made some very dumb comments. So to hide their embarassment they wrote
Would *you* put your kids in the hands of a teacher who makes errors like that?
I responded with the most obvious answer
So if you're asking me if I'd put my kids in the hands of a person who makes mistakes I'd say of course. I have no choice since people who don't make mistakes don't exist.

Oy! :-)

In fact that thread was interesting. I should post that one too. It's on relativistic mass as an invariant. See thread I'm starting on this for details

Pete

Originally posted by jcsd
Yes I've seen it before, though not in a textbook, it's actually quite a good way of demonstrating length contraction.

FYI - the standard derivation is here

http://www.geocities.com/physics_world/sr/lorentz_contraction.htm

Do you recall that nut that was tossed out for flaming tom? He was the one who couldn't understand that a scalar was a tensor of ranks zero? Well that same nutcase now claims that anyone who posts such a derivation as in the link above is plagerizing him.

So many nuts on the inernet. Where do they all come from?

Pete

Well this was rather humerous. Here is the response from that person
(poster) Wrote: 1. In my example, does the laser emitter and mirror remain vertical wrt each other in every frame in relative motion, that shares the laser frame's x axis?

Pmb responded: I clearly stated - "Emitter and mirror - Yes. neither is moving so the connecting line remains perpendicular to the motion"
Didn't you understand that? The emitter-mirror forms a line which is
parallel to the y' axis. That axis is parallel to the y axis. The y and y' axes are in relative motion. The direction of motion is the
x-direction. So YES the emitter-mirror is perpendicular IN EVERY FRAME IN RELATIVE MOTION WHICH SHARES THE FRAME'S X-AXIS

(poster) Wrote: 2. In my example, does the laser beam remain vertical between the laser emitter and mirror in all frames in relative motion, that share the laser frame's x axis?

Pmb responded: I've changed my mind on this. The energy of the beam
remains parallel to the y-axis. The direction the light is moving is not parallel to the y-axis.
***********************************************
(poster) : You are wrong and refuse to admit it.
How do you split the energy of the wave from the wave?

Also, since you agree all frames in relative motion with coinciding x-axis wrt the laser frame will see the laser and emitter vertically
in line, how can the laser beam not also be vertically in line between emitter and mirror?

(poster)...
It is reasoning and faith that bind truth.
I had, of course, explained this to him before and he ignored it.

I explained that in each frame the photons coming out of the laser are aligned parallel to the y'-axis in S' and are also aligned parallel to the y-axis in S. While the photons move in the y' direction they do not move in the y direction.

I've posted this diagram to explain

http://www.geocities.com/physics_world/sr/sr23-im-04.gif

However I think his moton says it all. Seems that he doesn't like to reason since he thinks he'll miss the truth! LOL

Pete

quartodeciman
Watch out for this kind of trap:

you agree all frames...will see...vertically
in line
.

Terrell rotation

Originally posted by quartodeciman
Watch out for this kind of trap:

you agree all frames...will see...vertically
in line
.

Terrell rotation

Yep. That one caught me for a moment. But so long as particles are "simultaneously" measured to have the same x-component on one frame then they will "simultaneously" have the same x-component in all other frames in standard configuration with the first.

Pete

S = k log w
Originally posted by pmb
I don't recall seeing this derivation anywhere but its probably somewhere in the relativity literature since most things are.

A poster at sci.physics.relativity posted what he thought was proof that the Lorentz contraction doesn't exist. See the Fig. 1 in Lorentz Contraction - Version 2 -- www.geocities.com/physics_world/sr/lorentz_contraction_2.htm

Here is the poster's claim

(note: He got the angle of the mirror wrong)

I requested a proof of his claim. In fact I repeated the request many times but he refused to post a proof.

It was kind of fun to write this up so I posted it on the web. It turned out to be another way to derive the Lorentz contraction relation.

In my derivation all that is assumed is the Lorentz transformation.

So the person thought he found proof that there is no Lorentz contraction when in fact all he did was to find another derivation and he didn't even know it! :-D

Ya got to love that! :-)

Pete

You forgot to account for error of parallax.