Education and Crank Claims: Special Relativity

Discussion in 'Physics & Math' started by rpenner, Oct 5, 2011.

  1. Tach Banned Banned

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    They are born this way, nothing to do with the water

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  3. Magneto_1 Super Principia Registered Senior Member

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    Finally, you present some physics and some disquised answers!

    So the answer to the first question is that you "agree" with my modification of the Galilean transform, because yes the velocity in question is \(\mathbf{v} = (v_{x},v_{y},v_{z})\).

    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:
    \(\begin{pmatrix}\Delta x' \\ \Delta y' \\ \Delta z' \\ \Delta t' \end{pmatrix} = \begin{pmatrix}1 & 0 & 0 & -v \\ 0 & 1 & -v & 0 \\ 0 & -v & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} u_x \Delta t \\ u_y \Delta t \\ u_z \Delta t \\ \Delta t \end{pmatrix} = \begin{pmatrix} ( u_x - v_x ) \Delta t \\ (u_y - v_y) \Delta t \\ (u_z - v_z) \Delta t \\ \Delta t \end{pmatrix} = \begin{pmatrix} ( u_x - v_x ) \Delta t' \\ (u_y - v_y) \Delta t' \\ (u_z - v_z) \Delta t' \\ \Delta t' \end{pmatrix}\).

    So the answer to the second question is that you also "agree" with my modification of Lorentz transform besides adding the "equal in all three dimensions" constant (\( \frac{1}{sqrt{3}} \)). I agree that I left off this constant term.

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

    So once again you "agree" with me, in that RPenners choice of equations is the "Standard Configuration" for Special Relativity. Now this "Standard Configuration" for Special Relativity is accepted by mainstream physicists because when using the "Lorentz Transform Boost" method you can use the "Standard Configuration" presented by RPenner without any "LOSS OF GENERALITY".
     
    Last edited: Oct 7, 2011
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  5. Motor Daddy Valued Senior Member

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    It's really too bad that James R. has basically threatened me with a ban if I continue to talk about my example, because it clearly points out how light travels in the absolute frame. Too bad you just can't understand that. The world is not ready for the truth.
     
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  7. arfa brane call me arf Valued Senior Member

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    Do not underestimate the power of the dark side.
     
  8. Me-Ki-Gal Banned Banned

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    Wow like eyes on a painting that follow the observer . No matter were you stand the eyes are looking at you
     
  9. James R Just this guy, you know? Staff Member

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    Moderator note: chinglu has been banned for 7 days for trolling.
     
  10. AlexG Like nailing Jello to a tree Valued Senior Member

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    I'm now totally convinced that rpenner started this thread as a form of fly-paper for cranks. He knew they'd all flock here, and hopefully get so caught up in this that they'd leave other threads along.

    OR... that he knew they'd all flock here and give us someplace to laugh at.
     
  11. OnlyMe Valued Senior Member

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    He has made only four posts. Three included, Eductaion: followed by a question and his answer to it, at least once and usually more than once, in each post. He responded to only one post by DonQuixote, who is currently reading a PDF copy of Einstein's little book 1920. ("I supplied the link!")

    Seems he is supplying the Eductaion and you know what that means.., the rest is Crank Claims?
     
  12. AlexG Like nailing Jello to a tree Valued Senior Member

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    Bait.
     
  13. AlphaNumeric Fully ionized Registered Senior Member

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    You make it sound like I'm capitulating or the like. Tach, OnlyMe, Rpenner and I have been saying throughout the entire thread that you can use any velocity in the transforms, not just \(\mathbf{v} = (v_{x},0,0)\). The fact you didn't understand that when you were linked to the general Lorentz transform doesn't now mean you're getting me to agree with you where I didn't before. The fact is you failed to understand what was said to you originally.

    Except that the matrix operator should depend on \(v_{x},v_{y},v_{z}\), not v.

    It's little things like that, which you do constantly, which makes it seem like you don't understand what you're posting. You're using the LaTeX'd algebra Rpenner did and still managing to botch it.

    It isn't a 'constant term', it's an important factor. \(|| (v,0,0) || = v\) but \(|| (v,v,v) || = \sqrt{3}v\) and the Lorentz factor \(\gamma\) is a function of the modulus, not of any single individual component. Again, it's just another little hint you don't understand the equations.

    Still wrong. There shouldn't be any v in there, it should be in terms of the components. For similar reasons, the numerator of \(\Delta t'\) should not just be a function of \(v_{x}\).

    As I commented, we've picked \(\mathbf{v} = (v_{x},v_{y},v_{z})\), a general vector. Since the time dilation amount depends on the components in all directions the \(\Delta t'\) term should treat each one of those equally, ie if a term with \(v_{x}\) is included then a corresponding term with \(v_{y}\) should be there and likewise for \(v_{z}\). You have just a \(v_{x}\). It would appear you have just slapped an 'x' subscript on the v, thinking that's generalised it when in fact you have to add other terms to make it general.

    You make it sound like you were the first person to say it. We've all been saying that for the entire thread. By using a rotation, ie a particular choice of basis, you can convert any Lorentz boost into a boost along the x axis. Why are you making it sound like I didn't originally agree with you or that you were saying something new?

    Have you realise that you were arguing with us when we were right all along and you want to try and make it seem differently? It's all there for people to read, you were given linked, I was the one who used the phrase 'without loss of generality'. You argued with us, asking the same questions again and again and are you now realising we were right all along?

    Yes, yes, it's a massive fearful conspiracy against you. When you learn some algebra let me know and maybe we can have an actual discussion.
     
  14. Magneto_1 Super Principia Registered Senior Member

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    You are so dishonest, that it is unbelieveable.

    The correct answers above should be more like, yes Magneto you were right. And, "no" RPenner did not introduce the "Lorentz Transform Boost" method, in his summary.

    Oh, really, then show us the other terms to make it general; using the same form of cartesian special relativity mathematics like discussed above? Since you know so well, then demonstrate for us your "pin-point" accurate mathematics?

    Oh, I know you can't do it, because you hurt your toe!!:shrug:
     
  15. Magneto_1 Super Principia Registered Senior Member

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    Come on!! Is this what you want to see. This is too trivial. I intentionally left those particular (x, y, z) indices off; because that was the math form that RPenner used in his original post (op).

    \(\begin{pmatrix}\Delta x' \\ \Delta y' \\ \Delta z' \\ \Delta t' \end{pmatrix} = \begin{pmatrix}1 & 0 & 0 & -v_x \\ 0 & 1 & -v_y & 0 \\ 0 & -v_z & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} u_x \Delta t \\ u_y \Delta t \\ u_z \Delta t \\ \Delta t \end{pmatrix} = \begin{pmatrix} ( u_x - v_x ) \Delta t \\ (u_y - v_y) \Delta t \\ (u_z - v_z) \Delta t \\ \Delta t \end{pmatrix} = \begin{pmatrix} ( u_x - v_x ) \Delta t' \\ (u_y - v_y) \Delta t' \\ (u_z - v_z) \Delta t' \\ \Delta t' \end{pmatrix}\).

    Are you really that trivial, I thought that you were a PhD.
     
    Last edited: Oct 7, 2011
  16. AlphaNumeric Fully ionized Registered Senior Member

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    Rpenner commented it was possible to convert the transform to be along the x axis. Anyone who has ever read an SR textbook will be familiar with the reason for this. Now we're spent 7 pages getting you to realise it.

    You weren't saying anything new. Everyone has been saying "You can pick another vector but the maths is nicer for the x axis" for pages and pages. I'm not being dishonest, you only need to look through the thread to see people saying it again and again.

    I like how you're so cock sure but the answer is obvious by symmetry. If we should be treating the x,y,z directions equally and fairly then if \(v_{x}\) appears in the t' term then so should the others. It's as simple as that. The time dilation factor depends on [tec]|| \mathbf{v}||[/tex], which treats each of the directions similarly, so you can't have just the x component if the y and z components are non-zero.

    Your entire argument has been "What about y and z directions?!" and now I comment you forgot them you get all snarky and dismiss my comment to include them. You're contradicting yourself!

    No, it's 9am and I have to go to work, which is 45 minutes walk away and I start at 10am. I'll reply later when I have the chance.

    Me being a PhD has nothing to do with the fact you got it wrong. Besides, I noticed the mistake, which shows I understand it, you're the one with the error. Yes, it is a simple one but it's become clear from previous interactions with you that I shouldn't ever give you the benefit of the doubt. I honestly don't think you could pass undergrad courses in basic vector calculus, differential equations or special relativity.
     
  17. Magneto_1 Super Principia Registered Senior Member

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    Once again, you fail to see the point of my question.

    According to RPenner's original post and his mathematics he is assuming what is currently known as the "Standard Configuration"

    Case #1 -- Standard Configuration

    Time

    \( \Delta{t'} \,\,\, = \,\,\, \frac{{\Delta{t} \,\, - \,\, \frac{v \,\, \Delta{r}}{c^2_{Light}}}}{sqrt{1 \,\, - \,\, \frac{v^2}{c^2_{Light}}} \)

    Distance

    \( \Delta{x'} \,\,\, = \,\,\, \frac{{\Delta{x} - v_x \, \Delta{t}}}{sqrt{1 \,\, - \,\, \frac{v^2_x}{c^2_{Light}}} \)

    \( \Delta{y'} \,\,\, = \,\,\, \Delta{y} \)

    \( \Delta{z'} \,\,\, = \,\,\, \Delta{z} \)


    Three Dimenisonal Range - External Observer Frame of Reference

    \( \Delta{r'^2} \,\,\, = \,\,\, \Delta{x'^2} \,\,\, + \,\,\, \Delta{y'^2} \,\,\, + \,\,\, \Delta{z'^2} \)

    \( \Delta{r'^2} \,\,\, = \,\,\, {\Delta{r^2}} \,\, {({1 \,\, - \,\, \frac{v^2}{c^2_{Light}})} \)

    Three Dimenisonal Range - Center of Mass "Proper" Frame of Reference

    \( \Delta{r^2} \,\,\, = \,\,\, \Delta{x^2} \,\,\, + \,\,\, \Delta{y^2} \,\,\, + \,\,\, \Delta{z^2} \)


    Three Dimenisonal velocity - Center of Mass "Proper" Frame of Reference

    \( {v^2} \,\,\, = \,\,\, {v^2_x} \,\,\, + \,\,\, {v^2_y} \,\,\, + \,\,\, {v^2_z} \)



    The question once again then becomes why does Special Relativity assume that when one three (3) dimensional inertial frame of reference is in motion relative to another three (3) dimensional inertial frame of reference one axis of motion is elastic/deformable, while the other two axes remain fixed/rigid.

    The question now becomes why do I have to assume that the one axis is deformable while the other axes are rigid? Why can't I start with the assumption the all three (3) axes are deformable at the same time?

    If we do not assume that one axis is deformable while the other axes are rigid then we start with the initial condition equations below.


    Case#2 -- Non-Standard - But More General Configuration

    Time

    \( \Delta{t'} \,\,\, = \,\,\, \frac{{\Delta{t} \,\, - \,\, \frac{v \,\, \Delta{r}}{c^2_{Light}}}}{sqrt{1 \,\, - \,\, \frac{v^2}{c^2_{Light}}} \)

    Distance

    \( \Delta{x'} \,\,\, = \,\,\, \frac{{\Delta{x} - v_x \, \Delta{t}}}{sqrt{1 \,\, - \,\, \frac{v^2_x}{c^2_{Light}_x}} \)

    \( \Delta{y'} \,\,\, = \,\,\, \frac{{\Delta{y} - v_y \, \Delta{t}}}{sqrt{1 \,\, - \,\, \frac{v^2_y}{c^2_{Light}_y}} \)

    \( \Delta{z'} \,\,\, = \,\,\, \frac{{\Delta{z} - v_z \, \Delta{t}}}{sqrt{1 \,\, - \,\, \frac{v^2_z}{c^2_{Light}_z}} \)

    Three Dimenisonal Range - External Observer Frame of Reference

    \( \Delta{r'^2} \,\,\, = \,\,\, \Delta{x'^2} \,\,\, + \,\,\, \Delta{y'^2} \,\,\, + \,\,\, \Delta{z'^2} \)

    \( \Delta{r'^2} \,\,\, = \,\,\, {\Delta{r^2}} \,\, {({1 \,\, - \,\, \frac{v^2}{c^2_{Light}})} \)

    Three Dimenisonal Range - Center of Mass "Proper" Frame of Reference

    \( \Delta{r^2} \,\,\, = \,\,\, \Delta{x^2} \,\,\, + \,\,\, \Delta{y^2} \,\,\, + \,\,\, \Delta{z^2} \)

    Three Dimenisonal velocity - Center of Mass "Proper" Frame of Reference

    \( {v^2} \,\,\, = \,\,\, {v^2_x} \,\,\, + \,\,\, {v^2_y} \,\,\, + \,\,\, {v^2_z} \)


    So which case is the more general case?

    I argue that "case #2" is more general!
     
    Last edited: Oct 7, 2011
  18. DonQuixote Registered Senior Member

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    I will try to state my question.

    It is commonly said in SR discussions that, according to SR, a moving clock (B) runs slow "as vieved from" or "as seen from" or "as judged from" a stationary system (A). It pops up in the twin paradox threads a lot.

    I am a bit confused by the phrase "as vieved from".
    If it was to be taken litterally as viewing/seeing the moving clock, then this is exactly what I would expect without invoking any relativity. It would simply be the result of the increasing distance light have to travel from B to A, and would probably be reversed if the movement was decreasing the distance between A and B.

    Can I safely just assume that this completely "apparent" phenomenon has nothing to do with relativity, and that the slowing down of clocks in SR is something else entirely?

    My motive here is just getting rid of misconceptions.

    Thank you
     
  19. funkstar ratsknuf Valued Senior Member

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    Nnnnnnngh!

    Speed. Velocity. Get it right.
     
  20. Tach Banned Banned

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    Absolutely. Shame on him.

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    You mean "alone".

    Who knows, he's a mean person

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  21. OnlyMe Valued Senior Member

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    The simple answer is yes and no.

    There are two kinds of time and length contraction involved in SR.

    The first is exactly as you stated above and involves the time of light delay on information and observations made from inertial frames of reference in motion relative to one another.

    The second is the effect that an objects velocity has on both its length in line with its linear velocity (and the rate of clocks in motion). This is almost a direct adoption of the original Lorentz-Fitzgerald contraction. Einstein also recognized the impact this would have on a moving clock.., that it runs slow compared to a similar clock would at rest.

    The first example, involving the time of light delay and apparent time and length contraction lies at the heart of Special Relativity. At the heart of the realization that there is no at rest frame of reference and that.., essentially what we "see" from any inertial frame of reference, is no more valid than what may be "seen" from any other inertial frame of reference. Everything is relative to the frame of reference from which it is observed and the laws of physics apply equally, to all inertial frames of reference.

    In the second case, time dilation resulting from a clock's velocity has been established, observed to occur in experiment and practice, in both airplanes and satellites. The GPS system being a common example.

    The length contraction aspect has never been observed at common lengths and velocities, mostly because our ability to measure, at velocities currently practical at everyday scales is not sufficient to result in a measurable change. There have been experiments involving subatomic particles, in particle accelerators that does support length contraction at relativistic velocities.

    So, what you have referred to as the apparent "phenomena" is actually the best demonstration of SR. SR holds that there is no observable stationary at rest frame and that all inertial frames of reference are equally valid. Recognizing the time of light delay and its impact on how we perceive events from different inertial frames of reference are central to the recognition that the laws of physics are the same in all inertial frames of reference.

    I may have over talked that it is something I am prone to.
     
  22. funkstar ratsknuf Valued Senior Member

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    Yes. You're essentially describing the Doppler effect. That's not what's happening in str.
     
  23. OnlyMe Valued Senior Member

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    I may have misunderstood DQ's intent here. I took it to be referring to the difference in the time of light delay between A and B, rather than any Doppler effect. The distance between A and B as well as whether they are moving toward one another or away from one another, will have an effect on the order and timing of events, an observer in one frame "sees" occurring in the other frame.
     

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