Education and Crank Claims: Special Relativity

From post #127

So prove it already!

 

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That's the question. Are Magneto and Chinglu, Motor Daddy and Emil really so deluded they think the stuff Rpenner and I provide is just parroting and made up gibberish or are they just desperate to have the last word so they resort to "I can't heard you, laa laa, you're not answering me" with their fingers in their proverbial ears? Do they really think they are well read in relativity when they are shown wrong on basic formulae, when it is put right in front of them with a slew of links?

Magneto claims what I said wasn't physics. I can provide the lecture notes if he wants, the title of the course was "Symmetries in Particle Physics". Anyone whose done anything postgraduate in theoretical physics knows about Lie groups and Lie algebras. Continuous symmetries form one of the pillars of modern physics and they are generally described using Lie groups. I have a book called 'Lie algebras in particle physics' authored by a Physics Nobel Prize winner. Another Nobel Prize winner 't Hooft got his for studying particular Lie algebra based gauge theories. Recently several people shared a Nobel Prize for something to do with symmetries in field theory. You don't even need to know the formal maths, someone who just does layperson reading can know these things play a part in physics.

Does Magneto really think he was in the running for the head job of a department at Caltech or is he just posturing and knowingly telling himself fibs. Does Farsight really think his work is worth 4 Nobel Prizes or is he having a mid-life crisis and needs something to tell himself to reaffirm his self worth?

Sometimes I almost think it would be interesting to meet some of these people, just to see how they do when there isn't Google at their finger tips. It's easy for someone on a forum to say "Lie group? Of course I know what that is!", they can spout obscure results thanks to Wiki. But at a blackboard and in front of a dozen professors, they'd probably be less confident. Except Terry Giblin, he really believes he's done something worthwhile.

 

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First, your link did not work! And, second I did understand, you just did not answer the question adequately; however, you are satisfying your own mind claiming to have answered it! I am not sure who you think that you are fooling, your friends??

The problem is: A boost in the x direction by speed v should map a vector (0,0,0) to (v,0,0).

Initial x-axis - vx(initial) = 0 ---> Final x-axis vx(final) = v
Initial y-axis - vy(initial) = 0 ---> Final y-axis vy(final) = 0
Initial z-axis - vz(initial) = 0 ---> Final x-axis vz(final) = 0

Then you can ask "Does this alter the y or z components?". The answer is no. Go on, do the calculations yourself.

The problem is: A boost in the x direction by speed v should mapped the velocity vector (0,y,z) to anything other than (v,y,z).

Initial x-axis - vx(initial) = 0 ---> Final x-axis vx(final) = vx
Initial y-axis - vy(initial) = vy ---> Final y-axis vy(final) = vy
Initial z-axis - vz(initial) = vz ---> Final x-axis vz(final) = vz

Then you can ask "Does this alter the y or z components?". The answer is no. Pretty basic reasoning really.[/QUOTE]

I agree with ChingLu, you have not provided a correct answer to my questions like you are claiming to have done.

The question once again is, why does Special Relativity and the equations that RPenner presented assume from initial conditions that one axis dimension direction is deformable/elastic while the other two directional dimensions are rigid??


1) However, why are the y-direction and the z-direction assumed to be fixed or rigid, while only the x-direction is allowed be elastic?

2) Can you prove using postulates, and mathematics why we can't assume that the y-direction and the z-direction can also be considered elastic.

For example why not?

Further if we are talking about the world-line of a particle, we can write \(\Delta x = x(\Delta t)\) (and similarly for y and z). And if we are talking about inertial motion, we are talking about constant velocity and \(\Delta x = u_x \Delta t + \Delta x_0\) (and similarly for y and z). So we may compute:
\(\begin{pmatrix}\Delta x' \\ \Delta y' \\ \Delta z' \\ \Delta t' \end{pmatrix} = \begin{pmatrix}1 & 0 & 0 & -v \\ 0 & 1 & -v & 0 \\ 0 & -v & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} u_x \Delta t \\ u_y \Delta t \\ u_z \Delta t \\ \Delta t \end{pmatrix} = \begin{pmatrix} ( u_x - v_x ) \Delta t \\ (u_y - v_y) \Delta t \\ (u_z - v_z) \Delta t \\ \Delta t \end{pmatrix} = \begin{pmatrix} ( u_x - v_x ) \Delta t' \\ (u_y - v_y) \Delta t' \\ (u_z - v_z) \Delta t' \\ \Delta t' \end{pmatrix}\).

Why not?

So \(u'_x = u_x - v_x\), and \(u'_y = u_y - v_y\), and \(u'_z = u_z - v_z\) is how inertial world lines transform under the Galilean transform.

Or why not?

\(\begin{pmatrix}\Delta x' \\ \Delta y' \\ \Delta z' \\ \Delta t' \end{pmatrix} = \begin{pmatrix} \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} & 0 & 0 & \frac{-v}{\sqrt{1 - \frac{v^2}{c^2}}} \\ 0 & 1 & \frac{-v}{\sqrt{1 - \frac{v^2}{c^2}}} & 0 \\ 0 & \frac{-v}{\sqrt{1 - \frac{v^2}{c^2}}} & 1 & 0 \\ \frac{-v}{c^2 \sqrt{1 - \frac{v^2}{c^2}}} & 0 & 0 & \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} \end{pmatrix} \begin{pmatrix} u_x \Delta t \\ u_y \Delta t \\ u_z \Delta t \\ \Delta t \end{pmatrix} = \begin{pmatrix} \frac{u_x - v_x}{\sqrt{1 - \frac{v^2_x}{c^2}}} \Delta t \\ \frac{u_y - v_y}{\sqrt{1 - \frac{v^2_y}{c^2}}} \Delta t \\ \frac{u_z - v_z}{\sqrt{1 - \frac{v^2_z}{c^2}}} \Delta t \\ \frac{1 - \frac{u_x v}{c^2}}{\sqrt{1 - \frac{v^2}{c^2}}} \Delta t \end{pmatrix} = \begin{pmatrix} \frac{u_x - v_x}{1 - \frac{u_x v_x}{c^2}} \Delta t' \\ \frac{u_y - v_y}{1 - \frac{u_y v_y}{c^2}} \Delta t' \\ \frac{u_z - v_z}{1 - \frac{u_z v_z}{c^2}} \Delta t' \\ \Delta t' \end{pmatrix}\)

3) Why is the y-axis and the z-axis treated differently than the x-axis? Can you prove it using mathematics and postulates or is this just parroting?

 

Huh...Someone misses me?

 

The dishonour is yours. I provided mathematics, you made a comment about it. I'm asking you to back it up. Why is it you expect everyone to prove their understanding but when I ask it of you you ignore me?

I repeat my request/questions from this post :

If you won't answer questions you're trolling. I've provided plenty of maths and I'm asking you a question. You haven't provided any maths and you won't answer questions.

I'll happily address your new issues after you answer my questions about your old issues. The ball is in your court. They are simple enough questions, I want you to justify your assertion I'm wrong. I claim you are wrong. I have provided the algebra and the numbers.

If you can't find a mistake in my work at least admit it. Either admit it or answer my questions. I refuse to allow you to do a gish gallop. We stick to this issue until you can either answer the question or admit you cannot find a mistake.

 

I agree ChingLu, what is taking so long??:shrug:

 

This is a typical AlphaNumeric "mathematical" function; when being "retortically" defeated, retreat, and deflect derogatory attention to other members on the forum.

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I have said: the only objection is that your math is naive because it does not take into account that speed is a vector and that objects have size.

 

http://panda.unm.edu/Courses/Finley/P495/handouts/PoincareLieAlgebra.pdf

If you understood this stuff you would understand what I said and see I did. I get it, you're ignorant. Can we move on?

Project much?

All of this can be answered by considering what you said afterwards.

You make a number of mistakes or misconceptions.

1. A Galilean transform wouldn't have the effect as you claim, namely "tex]u'_x = u_x - v_x[/tex], and \(u'_y = u_y - v_y\), and \(u'_z = u_z - v_z\)" unless the velocity in question was \(\mathbf{v} = (v_{x},v_{y},v_{z})\). Of course if you used that velocity in a Lorentz transform then you would indeed get a similar thing, but with the relevant relativistic shifts.

2. Your transform isn't a Lorentz transform. You can do a Lorentz boost in the direction \(\mathbf{v} = (v_{x},v_{y},v_{z}) = (v,v,v)\), that is fine. But then the velocity value in the \(\gamma\) factor is not \(v^{2}\) but \(\Vert \mathbf{v} \Vert^{2} = 3v^{2}\). So you've done your basic vector calculus wrong, the squared norm \((v,v,v)\) is not \(v^{2}\).

3. Tach has provided Chinglu with a link to a general Lorentz transform in a general direction. The vector \(\mathbf{v} = (v,v,v)\) is in the \(\frac{1}{\sqrt{3}}(1,1,1)\) unit vector direction. If \(R(\mathbf{v},\mathbf{w})\) is the rotation in SO(3) which rotates \(\mathbf{v}\) to be parallel to \(\mathbf{w}\) then the proper (ie not your incorrect botched version) Lorentz boost in that direction can be written as \(R(\mathbf{e}_{x},\mathbf{v})\cdot \Lambda \cdot R(\mathbf{v},\mathbf{e}_{x})\) where \(\Lambda\) is the Lorentz transform in the x axis direction Rpenner gave. By using the fact all Lorentz boosts involving the same \(\Vert \mathbf{v}\Vert\) are similar via said rotations we obtain the result that Rpenner's consideration of that particular Lorentz boost is without loss of generality.

I've provided lengthy high level explanations and answers. I've provided detailed correct mathematics. I've provided my own responses. You're the one failing to understand what I'm telling you. You're the one who has copy/edited Rpenner's algebra and botched it. You're the one showing ignorance.

I know you want to think I'm parroting, that it's easier to dismiss what I say, the errors of yours I point out, if you tell yourself I don't understand this stuff and I'm parroting but that simply isn't reality. Remember, I get paid to do this stuff so clearly I've convinced others I'm able to produce original work. How's the job hunting going? Bagged any interviews for high level research groups yet? Or you still convinced people are stealing your work? Your work which can't even get the SC metric right. Your work which doesn't know what a tensor is. Your work which isn't reviewed by physicists.

My offer to provide some lecture notes and engage in discussion with you on them stands. I've got nothing to hide about my understanding because I actually understand this stuff. Hell, this stuff is fun. I enjoy talking about Lie algebras, Lie groups, representations, tangent spaces, symmetries etc. How about it? You have nothing to lose but your ignorance.

 

Yes.

This is true if you are describing one dimensional motion.

However when describing motion in three dimensions, and then you decide that one axis is deformable and the other two axes are rigid; you might want to explain why this is so for initial conditions?

Why not make all three axes deformable from initial conditions?

Deformable means that lengths contract as the velocity of the relative inertial frame of reference increases.

 

What is taking Chinglu so long to answer my questions? He says my pdf is wrong. I asked him which line. Is that too hard? I even numbers the important equations.

Once he answers my simple question the discussion can move on. If I allow him to make claims he won't justify then it's a waste of time to discuss things with him.

Perhaps you could help him out?

No, I was replying to someone else. I still replied to you and Chinglu. There are other people here besides you two. I know it's hard for you to accept but you aren't the centre of my universe.

 

You believe in some conspiracy?

 

If I understand your intent, it would seem the answer is simple.

If you address length contraction in all three axis'/directions you must have more than two frames of reference involved.

The Lorentz Transformations can only be applied to the relative motion of two objects at a time. And yes if you have three objects all inertial and in motion relative to one another, it is possible that all three the x, y and z axis'/directions could be contracted, but not all from a single frame of reference.

Call each object A, B and C. You can perform LT for AB, AC and BC and the axis or combinations of axis' contracted could be different in each case. But each case represents a unique frame of reference and any contraction observed or present within AB, will not be observable or present in BC or AC etc.

Both SR and GR are essentially limited to two body comparisons. So if three bodies are involved three sets of transformations will also be involved. In each length contraction will be observed or real in only one direction.

And when expressed in the conventional mathematical form, only one axis will be contracted, usually the x axis.

 

In a conspiracy of ignorance? No, unfortunately people can be ignorant all by themselves.

 

Pfft, I told you that you are unable to read. The transformation that you, Magneto_1, Emil and the other trolls can't read and understand, is the GENERAL Lorentz transform for ARBITRARY motion in ALL 3 directions. You can find its derivation in any textbook. Now, why you idiots persist in questioning this issue is beyond any reason. But, then again, it is expected from the ones like you.

 

I provided Magneto_1 with the same link much earlier on. Did not stop him from asking the same idiocy over and over again. Now, as expected, chinglu is doing the same. The "fault" lies with rpenner since he named the thread in such a way it GUARANTEED it will attract the cranks

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Is there something in the drinking water at the moment? Chinglu, Emil, Motor Daddy and Magneto have all recently engaged in this activity, asking a question and then repeating it again and again, ignoring all responses. Chinglu's last thread was locked as he wouldn't answer any question. Emil and Motor Daddy just yammered on about relativity and now Magneto doesn't get Lorentz transforms.

Why is entering into rational, coherent discussion so difficult for them?