Education and Crank Claims: Special Relativity

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

  1. Trooper Secular Sanity Valued Senior Member

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    You can apply the postulates of Special Relativity to three-dimensional motion, but the analysis gets considerably more complicated, and no essential new features emerge.
    Right, RP?

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  3. OnlyMe Valued Senior Member

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    Yup. What he said!
     
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  5. Trooper Secular Sanity Valued Senior Member

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    She...What she said!

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  7. AlphaNumeric Fully ionized Registered Senior Member

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    I can.

    After our last encounter I wrote it all up into a single pdf. Here is the overview with a link to the pdf. The explicit case I do is the (2,10,0) example but the (3,15,0) is pretty much the same. The paper you're referring to doesn't apply the Lorentz transforms to the points in question or the light sphere and thus doesn't realise his pictures are wrong.

    The refutation was easy.
     
  8. AlphaNumeric Fully ionized Registered Senior Member

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    A Lorentz boost is in a particular direction. So what direction do you pick? You can pick any but for illustration purposes people pick the direction along the positive x axis. It's 'without loss of generality' because all of the boosts are equivalent in mathematical form up to rotation. Basically it simplifies the algebra. If you picked the y axis you'd be fine, then the x and z wouldn't change. If you picked the z axis then the x and y wouldn't change. If you picked the x+z axis then x-z and y wouldn't change.

    It's not that SR says those directions aren't deformable, its that it says if you pick the x direction to boost along, as you're allowed to do without loss of generality, then they don't change.

    You really don't understand the notion of examples, do you?
     
  9. AlphaNumeric Fully ionized Registered Senior Member

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    Wow, more demonstration you have little to no experience with basic mechanics, let alone relativity.

    Given 2 objects there is a line joining their locations. That is a single direction. You are allowed to pick your coordinates such that one object is at the origin and the other is some distance along the x axis. Now when you consider a Lorentz boost along that line it is like doing a Lorentz boost along the x axis, yet it's a general case.

    That's part of the power of being able to do coordinate transforms, you can alter coordinates to vastly simplify the calculations without having to assume anything. For instance, quantum field theorists regularly work in the centre of mass frame when doing collisions, rather than moving along with an individual particle. Or in classical mechanics it is often nicer to pick a frame where the total momentum of a set of objects is zero.

    The LT have been done in full generality. It's not hard (though it is very long and boring) to work out an expression for a general Lorentz boost in any direction in N dimensional space combined with a general rotation. It's stuff covered in any SR course. Funny how you don't know it Mr Kemp.
     
  10. Trooper Secular Sanity Valued Senior Member

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    Whoa! Check it out. AN has a blog. Ought oh...cranks beware...

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    BTW, AN, your sciforums link doesn't work.
     
  11. AlphaNumeric Fully ionized Registered Senior Member

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    I only started it on Monday and mostly it's going to be a collection of slightly modified monologues I've done here before. Going round and round in circles with Chinglu reminded me of Jack_ and MD and going round and round with them so I'm saving myself future work by being able to just point to something there and say "Refuted months ago. Next!". My life isn't interesting enough to actually blog about it.
     
  12. Trooper Secular Sanity Valued Senior Member

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    It doesn’t have to be about your life just physics, math, and weirdoes. You type fast enough to pump out something once a day. That’s all it takes. You could also ask other people to contribute articles once in awhile. I think you should add a "flavor of the month" to your crank section, and if they post a link to their website...even better.

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    oke:

    “I'm saving myself future work.” From in here? I think you should reduce the time you spend in here and blog.

    I’m not judging. I’m just sayin'.

    Have fun! Nite...ZZzzz
     
  13. Magneto_1 Super Principia Registered Senior Member

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    Once again you try and disquise your lack of understanding with, oh its easy you just do a "Lorentz Transform (LT) Boost " and the problem is solved. Wrong!!!

    You are right in how you described the basic purpose of the "LT Boost." Using the "Boost Transform" allows you to switch or rotate the direction of motion axis, for an object in uniform motion.

    However the problem still remains, it "Always Asumes" that one direction is deformable or elastic while the other two directions are rigid or fixed. Using the mathematical trick of the "Boost" does not resolve this problem it only rotatates the problem to another axis; and then says, "See, look I have solved the problem."

    It appears that you have solved the problem when you have not Dr. Professor Weatherhill.

    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??

    And the "Boost" is a math trick for passing the buck to something else, or kicking the can down the road!! It does not resolve why one direction of motion is prefered and deformable while the other directions remain rigid, in order to solve equations in Special Relativity.

    It's ok if you can't answer the question, there are many PhDs like you that don't know much.
     
    Last edited: Oct 6, 2011
  14. Magneto_1 Super Principia Registered Senior Member

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    Since, you repeated the same old crap pretending to answer the question, I will do the same!

    Once again you try and disquise your lack of understanding with, oh its easy you just do a "Lorentz Transform (LT) Boost " and the problem is solved. Wrong!!!

    You are right in how you described the basic purpose of the "LT Boost." Using the "Boost Transform" allows you to switch or rotate the direction of motion axis, for an object in uniform motion.

    However the problem still remains, it "Always Asumes" that one direction is deformable or elastic while the other two directions are rigid or fixed. Using the mathematical trick of the "Boost" does not resolve this problem it only rotatates the problem to another axis; and then says, "See, look I have solved the problem."

    It appears that you have solved the problem when you have not Dr. Professor Weatherhill.

    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??

    And the "Boost" is a math trick for passing the buck to something else, or kicking the can down the road!! It does not resolve why one direction of motion is prefered and deformable while the other directions remain rigid, in order to solve equations in Special Relativity.

    It's ok if you can't answer the question, there are many PhDs like you that don't know much.

    And where is RPenner, he is letting you answer his questions! He is usually on top of his posting and typically answers right away! Why so slow now!!
     
    Last edited: Oct 6, 2011
  15. AlphaNumeric Fully ionized Registered Senior Member

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    Mmm.... delicious irony. You accuse me of not understanding then you say I'm supposedly right for saying a boost allows you to rotate the direction.

    No, a rotation allows you to change direction. A boost changes velocities in a particular direction. I know it'll go over your head but I'll be more specific, since Trooper seems to like me monologing.

    Lorentz transforms in 1+3 dimensional space form the Lie group SO(1,3). This group has a Lie algebra so(1,3). This is a 6 dimensional algebra forms of two three dimensional subalgebras. The first subalgebra is the algebra of rotation generators \(J_{i}\), which correspond to the fact SO(1,3) includes the rotation group SO(3) as a subgroup. In terms of the matrix representation of a Lorentz transform this corresponds to transforms of the form

    \(\Lambda = \left( \begin{array}{cc} 1 & \mathbf{0} \\ \mathbf{0} & R \end{array}\right)\)

    where R is an element of SO(3). The generators \(J_{i}\) satisfy the su(2) algebra relations because SU(2) is a double covering of SO(3) and thus their algebras are isomorphic, \([J_{i},J_{j}] = \epsilon_{ijk}J_{k}\). Due to this block diagonal form the LT has no effect on the passage of time, as a rotation doesn't.

    Then there's the algebra of boosts, the \(K_{i}\). They too form an su(2) like structure.

    These algebras are do not commute with one another, else SO(3,1) would just be SO(3)xG for the G being the boost group. Instead SO(1,3) is a non-trivial knitting of them together.

    This is covered in courses on symmetries in theoretical physics, particularly for people studying supersymmetry and particle physics. Would you like me to provide you with some relevant lecture notes?

    A Lorentz boost affects the time coordinate and the spatial coordinate parallel to the velocity. Directions orthogonal to the velocity are not changed. This isn't an assumption, it's a derived result when you compute the general structure of SO(1,3) matrix representations.

    I'm sorry you never studied this stuff to know how it is all derived but that isn't my fault. If you spent a little less time convincing yourself you understand this stuff and actually learning this stuff we'd not be having this conversation.

    Firstly you misspelt my surname. Secondly I'm not a professor.

    I know you have a chip on your shoulder about people who are more successful at physics than you but do you really think highlighting the fact I have an advanced degree is going to be embarassing to me?

    Not as many comparing to internet hacks who did poorly at science in school and now roam forums proclaiming they've explained physics.

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    You keep making the mistake of thinking that because you don't know the answer then neither I nor anyone else does. You presumed your question couldn't be answered, when really it's just a matter of being familiar with how the general form of the Lorentz transform is derived and a little bit about the well known mathematical phrase "without loss of generality". But in this case it isn't surprising, you really struggle to understand the nature of 'coordinates' in differential geometry.

    Firstly, well done on the double post. Secondly Rpenner isn't here 24 hours a day, nor is he obliged to be at your beck and call. He might be doing other things, he might not think you're worth the effort, he might be happy to watch me point out your ignorance. Or perhaps he's waiting for you to start acting all tough, demanding to know where he is, before he drops a huge post of detailed mathematics proving his post correct on your head. Take a hint from Motor Daddy's experience, when he whined where my mathematics was I then posted a complete illustration of his mistake in fully general algebra and he wasn't even able to understand it. You've shown you don't understand 1st year undergraduate relativity, you should wind your neck in a bit, Mr Kemp.

    By the way, how's the job hunting going? Managed to bag yourself a head position of a major university relativity research group yet? Caltech? Harvard? Penn State? Any luck? Or are you back at that degree mill researchless University of Phoenix?
     
  16. Magneto_1 Super Principia Registered Senior Member

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    BetaNumeric

    Sorry about mispelling your name: Dr. Professor George James Weatherill

    And, no I don't think that anything embarasses you, I think that you are immune to its correcting effects!!

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    Dribble, Dribble, Dribble....:bawl: ...... still no answer!!

    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?
     
    Last edited: Oct 6, 2011
  17. Tach Banned Banned

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    They aren't. Read HERE and stop posting nonsense.
     
    Last edited: Oct 6, 2011
  18. OnlyMe Valued Senior Member

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    What she said!

    Sometimes simplicity is the best part of smart.

    I had been trying to say that same thing and continued to trip on coordinates.
     
  19. AlphaNumeric Fully ionized Registered Senior Member

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    I like how you quote me saying I'm not a professor and then you still call me it. Are you struggling to read English now?

    I know you don't understand what I said nor have you read the relevant material but that doesn't mean I didn't respond to what you said.

    I responded to this. All directions can be altered by some Lorentz transform but for a specific Lorentz transform only the direction parallel to the boost velocity will be altered, the rest are left unaffected.

    You can combine Lorentz boosts. For instance, you could do a boost in the x direction and then you could do a boost in the y direction and then you could do a boost in the direction. These would alter different directions each time, showing how all of the directions can be warped. However, each transform individually will only alter the direction the velocity is in.

    You can show it by considering the associated Lie algebra.

    SO(1,N) matrix elements are defined by the property \(\Lambda^{\top} \cdot \eta \cdot \Lambda = \eta\). If we take the derivative of this (in the usual one parameter group method to construct Lie algebras) we obtain \(\lambda^{\top} \cdot \eta + \eta \cdot \lambda = 0\) where \(\lambda\) is now an element of the Lie algebra SO(1,N). In the case of so(1,3) we find that \(\lambda = a_{i}J_{i} + b_{j}K_{j}\) where \(J_{i}\) and \(K_{j}\) are the subalgebra generators I mentioned in my previous post. We are free to choose any linear combination of generators as our basis but when considering the action of SO(1,3) on the canonical Cartesian coordinates f \(\mathbb{R}^{4}\) we select the generators to correspond to infinitesimal rotations about the x,y,z axes and the infinitesimal translations in the x,y,z directions. These are simple enough to work out in terms of their components by using the defining equation of the Lie algebra. To then compute the form of the Lie group element \(\Lambda\) corresponding to the boost in the x direction we simply exponentiate the Lie algebra generator corresponding to the infinitesimal translation in the x direction, ie \(\Lambda = \exp(K_{x})\). Matrix exponentiation is simple enough to do and when you do that you end up with a matrix whose actions on the y and z directions are trivial, ie it does nothing. This generalised to arbitrary SO(1,N), only the direction parallel to the boost velocity vector is altered, all other spatial directions are left untouched.

    If you were really the whiz at relativity you claim to be you'd know this. This isn't particularly advanced, it's covered in a number of 4th year courses at Cambridge. After all, you thought you were in the running to get a head of department at Caltech, so you should know material which is 4th year Cambridge standard inside out, right? Right?

    Funny how you're asking questions you should know the answer to. It's almost like you don't know anything about this stuff....
     
  20. Magneto_1 Super Principia Registered Senior Member

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    AlphaGeneric

    I agree that the method that you describe above can be used to resolve that the x-axis, y-axis, and z-axis is not any more special than any other axis/direction dimension and should not be treated differently. Although your initial conditions treat one axis as special or elastic/deformable and the other two (2) of three (3) dimensions of space as fixed/rigid.

    Using your method, what you do is essentially:

    1) first iteration - compute along the x-axis ignoring y-axis and z-axis
    2) second iteration - "LT Boost" then compute along the y-axis ignoring x-axis and z-axis
    3) third iteration - "LT Boost" then compute along the z-axis ignoring y-axis and x-axis

    What this really means is that your initial condtions were "conditional" and not "general."


    Once again, the same goes for using your method above, what you do is essentially:

    1) first iteration - compute along the x-axis ignoring y-axis and z-axis
    2) second iteration - "LT Boost" then compute along the y-axis ignoring x-axis and z-axis
    3) third iteration - "LT Boost" then compute along the z-axis ignoring y-axis and x-axis

    What this really means is that the initial condtions presented by RPenner in the OP are "conditional" and not "general."

    The method of solution that you present works because it resolves the fact that the initial conditions of your Special Relativity equations are not symmetrical.


    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?

    AlphaGeneric

    You should have named your new blog/website - "Leader of the Cranks"

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  21. Tach Banned Banned

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    This is not how is done. Take an introductory class in relativity.
     
  22. billvon Valued Senior Member

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    Yes, the Lorentz transform takes into account three dimensional motion. The math is a lot simpler with motion in one axis which is why it is so often presented that way.

    Exactly! But someone might see a derivation based on a purely X axis perspective which included the scaling factors Yp=Y/X and Zp=Z/X. (Indeed, this is a fundamental operation performed during a 3D to 2D transform.) And they might ask "why does this just consider the X axis and not the other two?"

    The answer is of course that perspective generation CAN consider all axes, but the above example is a lot easier to explain.
     
  23. AlphaNumeric Fully ionized Registered Senior Member

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    It really shows your level of maturity that you resort to such things. Besides, you call me 'generic' but I can't help but notice you are not actually replying to what I've said and you just repeat yourself. Struggling to come up with a response to what I've said?

    I was illustrating how you can combine transforms to warp all directions. But a single Lorentz boost will only alter one spatial direction. There is no loss in generality in deciding for that direction to be the x axis.

    What about this don't you understand? Oh, I suppose it was my entire reply. My reply addressed your questions. Do you understand what a Lie group and a Lie algebra are? Do you understand what a generator is? Do you understand the subalgebras of SO(1,N)? Do you understand how to convert elements of the algebra into elements of the group? Do you understand how to compute matrix representations of these? Do you know how to do matrix exponentiation?

    If you understand these things you'll see I gave answers to your question. I'm sorry you don't have the same understanding as myself or anyone whose competent at special relativity or group theory but that isn't my fault.

    You're asking questions you should know the answer to. If you're so good at relativity you deserve a position at Caltech then you should be able to work this stuff out yourself. You should be able to compute the subalgebras of so(1,3) yourself. Actually, if you want to find the commutation relations for the generators of so(1,3) I suggest you look in chapter 6 of my thesis, I give the 1-form dual of them, as well as for all other isotropic 6 dimensional simple Lie algebras! Would you like me to walk you through the relevant section or can you do it yourself?
     

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