Associativity of Relativistic Oblique Velocity Addition












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I've encountered some information in the Wikipedia page on Lorentz transformation (https://en.m.wikipedia.org/wiki/Lorentz_transformation) that I am having difficulty reconciling with other information that I found on the page (https://en.m.wikipedia.org/wiki/Wigner_rotation) on Wigner rotation. In the Lorentz Transformation page, it says that the oblique (non-collinear) Lorentz transformation is given by a matrix $$begin{pmatrix}gamma&-gammabeta n_x&-gammabeta n_y&-gammabeta n_z\-gammabeta n_x&1+(gamma-1)n_x^2&(gamma-1)n_x n_y&(gamma-1)n_z n_x\-gammabeta n_y&(gamma-1)n_x n_y&1+(gamma-1)n_y^2&(gamma-1)n_y n_z\-gammabeta n_z&(gamma-1)n_z n_x&(gamma-1)n_y n_z2&1+(gamma-1)n_z^2end{pmatrix}$$ with the $n_q$ ($qin{x,y,z}$) being the components of the unit vector along the direction-of-motion of the moving frame. And in the Wigner Rotation page, there is given the formula for oblique velocity addition that is derived from successive application of two such transformations $$overrightarrow{beta}_uoplusoverrightarrow{beta}_v=frac{1}{1+overrightarrow{beta}_u.overrightarrow{beta}_v}left[left(1+frac{gamma_uoverrightarrow{beta}_u.overrightarrow{beta}_v}{1+gamma_u}right)overrightarrow{beta}_u+frac{overline{beta}_v}{gamma_u}right]$$$$gamma_{overrightarrow{beta}_uoplusoverrightarrow{beta}_v}=gamma_ugamma_v(1+overrightarrow{beta}_u.overrightarrow{beta}_v) .$$ But then it goes on to say that in addition to being non-commutative (which I can well-accomodate), this recipe is also non-associative!:



"Although velocity addition is nonlinear, non-associative, and non-commutative, the result of the operation correctly obtains a velocity with a magnitude less than $c$.".



Now since the addition of velocities is derived from the application of successive transformations taking the mathematical form of matrices, which are certainly associative, how then does the non -associativity of the addition of velocities come-about? For three consecutive Lorentz transformations would have result not contingent upon the association of the matrices - and the resultant velocity is derived from this product of three matrices ... so how can it come-about that the velocity addition be non-associative?










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    I've encountered some information in the Wikipedia page on Lorentz transformation (https://en.m.wikipedia.org/wiki/Lorentz_transformation) that I am having difficulty reconciling with other information that I found on the page (https://en.m.wikipedia.org/wiki/Wigner_rotation) on Wigner rotation. In the Lorentz Transformation page, it says that the oblique (non-collinear) Lorentz transformation is given by a matrix $$begin{pmatrix}gamma&-gammabeta n_x&-gammabeta n_y&-gammabeta n_z\-gammabeta n_x&1+(gamma-1)n_x^2&(gamma-1)n_x n_y&(gamma-1)n_z n_x\-gammabeta n_y&(gamma-1)n_x n_y&1+(gamma-1)n_y^2&(gamma-1)n_y n_z\-gammabeta n_z&(gamma-1)n_z n_x&(gamma-1)n_y n_z2&1+(gamma-1)n_z^2end{pmatrix}$$ with the $n_q$ ($qin{x,y,z}$) being the components of the unit vector along the direction-of-motion of the moving frame. And in the Wigner Rotation page, there is given the formula for oblique velocity addition that is derived from successive application of two such transformations $$overrightarrow{beta}_uoplusoverrightarrow{beta}_v=frac{1}{1+overrightarrow{beta}_u.overrightarrow{beta}_v}left[left(1+frac{gamma_uoverrightarrow{beta}_u.overrightarrow{beta}_v}{1+gamma_u}right)overrightarrow{beta}_u+frac{overline{beta}_v}{gamma_u}right]$$$$gamma_{overrightarrow{beta}_uoplusoverrightarrow{beta}_v}=gamma_ugamma_v(1+overrightarrow{beta}_u.overrightarrow{beta}_v) .$$ But then it goes on to say that in addition to being non-commutative (which I can well-accomodate), this recipe is also non-associative!:



    "Although velocity addition is nonlinear, non-associative, and non-commutative, the result of the operation correctly obtains a velocity with a magnitude less than $c$.".



    Now since the addition of velocities is derived from the application of successive transformations taking the mathematical form of matrices, which are certainly associative, how then does the non -associativity of the addition of velocities come-about? For three consecutive Lorentz transformations would have result not contingent upon the association of the matrices - and the resultant velocity is derived from this product of three matrices ... so how can it come-about that the velocity addition be non-associative?










    share|cite|improve this question











    $endgroup$















      0












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      0





      $begingroup$


      I've encountered some information in the Wikipedia page on Lorentz transformation (https://en.m.wikipedia.org/wiki/Lorentz_transformation) that I am having difficulty reconciling with other information that I found on the page (https://en.m.wikipedia.org/wiki/Wigner_rotation) on Wigner rotation. In the Lorentz Transformation page, it says that the oblique (non-collinear) Lorentz transformation is given by a matrix $$begin{pmatrix}gamma&-gammabeta n_x&-gammabeta n_y&-gammabeta n_z\-gammabeta n_x&1+(gamma-1)n_x^2&(gamma-1)n_x n_y&(gamma-1)n_z n_x\-gammabeta n_y&(gamma-1)n_x n_y&1+(gamma-1)n_y^2&(gamma-1)n_y n_z\-gammabeta n_z&(gamma-1)n_z n_x&(gamma-1)n_y n_z2&1+(gamma-1)n_z^2end{pmatrix}$$ with the $n_q$ ($qin{x,y,z}$) being the components of the unit vector along the direction-of-motion of the moving frame. And in the Wigner Rotation page, there is given the formula for oblique velocity addition that is derived from successive application of two such transformations $$overrightarrow{beta}_uoplusoverrightarrow{beta}_v=frac{1}{1+overrightarrow{beta}_u.overrightarrow{beta}_v}left[left(1+frac{gamma_uoverrightarrow{beta}_u.overrightarrow{beta}_v}{1+gamma_u}right)overrightarrow{beta}_u+frac{overline{beta}_v}{gamma_u}right]$$$$gamma_{overrightarrow{beta}_uoplusoverrightarrow{beta}_v}=gamma_ugamma_v(1+overrightarrow{beta}_u.overrightarrow{beta}_v) .$$ But then it goes on to say that in addition to being non-commutative (which I can well-accomodate), this recipe is also non-associative!:



      "Although velocity addition is nonlinear, non-associative, and non-commutative, the result of the operation correctly obtains a velocity with a magnitude less than $c$.".



      Now since the addition of velocities is derived from the application of successive transformations taking the mathematical form of matrices, which are certainly associative, how then does the non -associativity of the addition of velocities come-about? For three consecutive Lorentz transformations would have result not contingent upon the association of the matrices - and the resultant velocity is derived from this product of three matrices ... so how can it come-about that the velocity addition be non-associative?










      share|cite|improve this question











      $endgroup$




      I've encountered some information in the Wikipedia page on Lorentz transformation (https://en.m.wikipedia.org/wiki/Lorentz_transformation) that I am having difficulty reconciling with other information that I found on the page (https://en.m.wikipedia.org/wiki/Wigner_rotation) on Wigner rotation. In the Lorentz Transformation page, it says that the oblique (non-collinear) Lorentz transformation is given by a matrix $$begin{pmatrix}gamma&-gammabeta n_x&-gammabeta n_y&-gammabeta n_z\-gammabeta n_x&1+(gamma-1)n_x^2&(gamma-1)n_x n_y&(gamma-1)n_z n_x\-gammabeta n_y&(gamma-1)n_x n_y&1+(gamma-1)n_y^2&(gamma-1)n_y n_z\-gammabeta n_z&(gamma-1)n_z n_x&(gamma-1)n_y n_z2&1+(gamma-1)n_z^2end{pmatrix}$$ with the $n_q$ ($qin{x,y,z}$) being the components of the unit vector along the direction-of-motion of the moving frame. And in the Wigner Rotation page, there is given the formula for oblique velocity addition that is derived from successive application of two such transformations $$overrightarrow{beta}_uoplusoverrightarrow{beta}_v=frac{1}{1+overrightarrow{beta}_u.overrightarrow{beta}_v}left[left(1+frac{gamma_uoverrightarrow{beta}_u.overrightarrow{beta}_v}{1+gamma_u}right)overrightarrow{beta}_u+frac{overline{beta}_v}{gamma_u}right]$$$$gamma_{overrightarrow{beta}_uoplusoverrightarrow{beta}_v}=gamma_ugamma_v(1+overrightarrow{beta}_u.overrightarrow{beta}_v) .$$ But then it goes on to say that in addition to being non-commutative (which I can well-accomodate), this recipe is also non-associative!:



      "Although velocity addition is nonlinear, non-associative, and non-commutative, the result of the operation correctly obtains a velocity with a magnitude less than $c$.".



      Now since the addition of velocities is derived from the application of successive transformations taking the mathematical form of matrices, which are certainly associative, how then does the non -associativity of the addition of velocities come-about? For three consecutive Lorentz transformations would have result not contingent upon the association of the matrices - and the resultant velocity is derived from this product of three matrices ... so how can it come-about that the velocity addition be non-associative?







      associativity special-relativity






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      edited Dec 15 '18 at 12:01







      AmbretteOrrisey

















      asked Dec 14 '18 at 17:15









      AmbretteOrriseyAmbretteOrrisey

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          I think I might have solved this.



          Say you have four frames of reference labelled 0 to 3, the zeroth frame being the 'base' one, if you will; and let $Q_{ab}$ be the value of $Q$ - with Q being $gamma$ or $beta$ (or any other quantity, for that matter) - of frame $b$ relative to frame $a$, with $ain{0dots3}$ & $bin{0dots3}$. This way, $beta_{01}$, $beta_{12}$ & $beta_{23}$ - & therefore $gamma_{01}$, $gamma_{12}$ & $gamma_{23}$ are the givens. If you apply the formula first to find the velocity of frame 3 relative to frame 1, you would use the addition-of-velocity formula as given with $gamma_u=gamma_{12}$, $beta_u=beta_{12}$, & $beta_v=beta_{23}$, and the result would be $beta_{13}$. Then to find the velocity of frame 3 relative to frame 0, you would simply iterate the formula with $gamma_u=gamma_{12}$, $beta_u=beta_{12}$, & $beta_v=beta_{23}$, and the result would be $beta_{03}$ - the result sought. However, associating the calculation (the calculation that is! ... not the formula itself, as will shortly be apparent), you would first calculate $beta_{02}$ by substituting $gamma_u=gamma_{01}$, $beta_u=beta_{01}$, & $beta_v=beta_{12}$. But then to calculate $beta_{03}$ you would need to substitute $gamma_u=gamma_{02}$, $beta_u=beta_{02}$, & $beta_v=beta_{23}$. Note that in this association of the calculation $gamma_{02}$ appears - which must be calculated apart using the ancilliary formula for composition of the Lorentz factor.



          So I think essentially what is being said in the Wikipedia article is that the formula is not associative as it stands, and is only applicable as it stands which is to say without invoking the ancilliary formula for composition of Lorentz factors, if the order of association is strictly from the uttermost-observed frame to the base-frame: that all other associations give the wrong result.



          That's what I'm surmising, anyway. I haven't got symbolic algebra computation handy just at the moment; and I'm not crunching through all those substitutions using pencil & paper, & I haven't got the mental calibre necessary for doing it in my head. Oblique Lorentz transformations are notorious for the the sheer algeba-load of them being formidable! I'll report back when I've investigated the matter using a symbolic algebra package. But I think I probably do have the explanation here - I think my argument is logically robust.






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            $begingroup$

            I think I might have solved this.



            Say you have four frames of reference labelled 0 to 3, the zeroth frame being the 'base' one, if you will; and let $Q_{ab}$ be the value of $Q$ - with Q being $gamma$ or $beta$ (or any other quantity, for that matter) - of frame $b$ relative to frame $a$, with $ain{0dots3}$ & $bin{0dots3}$. This way, $beta_{01}$, $beta_{12}$ & $beta_{23}$ - & therefore $gamma_{01}$, $gamma_{12}$ & $gamma_{23}$ are the givens. If you apply the formula first to find the velocity of frame 3 relative to frame 1, you would use the addition-of-velocity formula as given with $gamma_u=gamma_{12}$, $beta_u=beta_{12}$, & $beta_v=beta_{23}$, and the result would be $beta_{13}$. Then to find the velocity of frame 3 relative to frame 0, you would simply iterate the formula with $gamma_u=gamma_{12}$, $beta_u=beta_{12}$, & $beta_v=beta_{23}$, and the result would be $beta_{03}$ - the result sought. However, associating the calculation (the calculation that is! ... not the formula itself, as will shortly be apparent), you would first calculate $beta_{02}$ by substituting $gamma_u=gamma_{01}$, $beta_u=beta_{01}$, & $beta_v=beta_{12}$. But then to calculate $beta_{03}$ you would need to substitute $gamma_u=gamma_{02}$, $beta_u=beta_{02}$, & $beta_v=beta_{23}$. Note that in this association of the calculation $gamma_{02}$ appears - which must be calculated apart using the ancilliary formula for composition of the Lorentz factor.



            So I think essentially what is being said in the Wikipedia article is that the formula is not associative as it stands, and is only applicable as it stands which is to say without invoking the ancilliary formula for composition of Lorentz factors, if the order of association is strictly from the uttermost-observed frame to the base-frame: that all other associations give the wrong result.



            That's what I'm surmising, anyway. I haven't got symbolic algebra computation handy just at the moment; and I'm not crunching through all those substitutions using pencil & paper, & I haven't got the mental calibre necessary for doing it in my head. Oblique Lorentz transformations are notorious for the the sheer algeba-load of them being formidable! I'll report back when I've investigated the matter using a symbolic algebra package. But I think I probably do have the explanation here - I think my argument is logically robust.






            share|cite|improve this answer











            $endgroup$


















              0












              $begingroup$

              I think I might have solved this.



              Say you have four frames of reference labelled 0 to 3, the zeroth frame being the 'base' one, if you will; and let $Q_{ab}$ be the value of $Q$ - with Q being $gamma$ or $beta$ (or any other quantity, for that matter) - of frame $b$ relative to frame $a$, with $ain{0dots3}$ & $bin{0dots3}$. This way, $beta_{01}$, $beta_{12}$ & $beta_{23}$ - & therefore $gamma_{01}$, $gamma_{12}$ & $gamma_{23}$ are the givens. If you apply the formula first to find the velocity of frame 3 relative to frame 1, you would use the addition-of-velocity formula as given with $gamma_u=gamma_{12}$, $beta_u=beta_{12}$, & $beta_v=beta_{23}$, and the result would be $beta_{13}$. Then to find the velocity of frame 3 relative to frame 0, you would simply iterate the formula with $gamma_u=gamma_{12}$, $beta_u=beta_{12}$, & $beta_v=beta_{23}$, and the result would be $beta_{03}$ - the result sought. However, associating the calculation (the calculation that is! ... not the formula itself, as will shortly be apparent), you would first calculate $beta_{02}$ by substituting $gamma_u=gamma_{01}$, $beta_u=beta_{01}$, & $beta_v=beta_{12}$. But then to calculate $beta_{03}$ you would need to substitute $gamma_u=gamma_{02}$, $beta_u=beta_{02}$, & $beta_v=beta_{23}$. Note that in this association of the calculation $gamma_{02}$ appears - which must be calculated apart using the ancilliary formula for composition of the Lorentz factor.



              So I think essentially what is being said in the Wikipedia article is that the formula is not associative as it stands, and is only applicable as it stands which is to say without invoking the ancilliary formula for composition of Lorentz factors, if the order of association is strictly from the uttermost-observed frame to the base-frame: that all other associations give the wrong result.



              That's what I'm surmising, anyway. I haven't got symbolic algebra computation handy just at the moment; and I'm not crunching through all those substitutions using pencil & paper, & I haven't got the mental calibre necessary for doing it in my head. Oblique Lorentz transformations are notorious for the the sheer algeba-load of them being formidable! I'll report back when I've investigated the matter using a symbolic algebra package. But I think I probably do have the explanation here - I think my argument is logically robust.






              share|cite|improve this answer











              $endgroup$
















                0












                0








                0





                $begingroup$

                I think I might have solved this.



                Say you have four frames of reference labelled 0 to 3, the zeroth frame being the 'base' one, if you will; and let $Q_{ab}$ be the value of $Q$ - with Q being $gamma$ or $beta$ (or any other quantity, for that matter) - of frame $b$ relative to frame $a$, with $ain{0dots3}$ & $bin{0dots3}$. This way, $beta_{01}$, $beta_{12}$ & $beta_{23}$ - & therefore $gamma_{01}$, $gamma_{12}$ & $gamma_{23}$ are the givens. If you apply the formula first to find the velocity of frame 3 relative to frame 1, you would use the addition-of-velocity formula as given with $gamma_u=gamma_{12}$, $beta_u=beta_{12}$, & $beta_v=beta_{23}$, and the result would be $beta_{13}$. Then to find the velocity of frame 3 relative to frame 0, you would simply iterate the formula with $gamma_u=gamma_{12}$, $beta_u=beta_{12}$, & $beta_v=beta_{23}$, and the result would be $beta_{03}$ - the result sought. However, associating the calculation (the calculation that is! ... not the formula itself, as will shortly be apparent), you would first calculate $beta_{02}$ by substituting $gamma_u=gamma_{01}$, $beta_u=beta_{01}$, & $beta_v=beta_{12}$. But then to calculate $beta_{03}$ you would need to substitute $gamma_u=gamma_{02}$, $beta_u=beta_{02}$, & $beta_v=beta_{23}$. Note that in this association of the calculation $gamma_{02}$ appears - which must be calculated apart using the ancilliary formula for composition of the Lorentz factor.



                So I think essentially what is being said in the Wikipedia article is that the formula is not associative as it stands, and is only applicable as it stands which is to say without invoking the ancilliary formula for composition of Lorentz factors, if the order of association is strictly from the uttermost-observed frame to the base-frame: that all other associations give the wrong result.



                That's what I'm surmising, anyway. I haven't got symbolic algebra computation handy just at the moment; and I'm not crunching through all those substitutions using pencil & paper, & I haven't got the mental calibre necessary for doing it in my head. Oblique Lorentz transformations are notorious for the the sheer algeba-load of them being formidable! I'll report back when I've investigated the matter using a symbolic algebra package. But I think I probably do have the explanation here - I think my argument is logically robust.






                share|cite|improve this answer











                $endgroup$



                I think I might have solved this.



                Say you have four frames of reference labelled 0 to 3, the zeroth frame being the 'base' one, if you will; and let $Q_{ab}$ be the value of $Q$ - with Q being $gamma$ or $beta$ (or any other quantity, for that matter) - of frame $b$ relative to frame $a$, with $ain{0dots3}$ & $bin{0dots3}$. This way, $beta_{01}$, $beta_{12}$ & $beta_{23}$ - & therefore $gamma_{01}$, $gamma_{12}$ & $gamma_{23}$ are the givens. If you apply the formula first to find the velocity of frame 3 relative to frame 1, you would use the addition-of-velocity formula as given with $gamma_u=gamma_{12}$, $beta_u=beta_{12}$, & $beta_v=beta_{23}$, and the result would be $beta_{13}$. Then to find the velocity of frame 3 relative to frame 0, you would simply iterate the formula with $gamma_u=gamma_{12}$, $beta_u=beta_{12}$, & $beta_v=beta_{23}$, and the result would be $beta_{03}$ - the result sought. However, associating the calculation (the calculation that is! ... not the formula itself, as will shortly be apparent), you would first calculate $beta_{02}$ by substituting $gamma_u=gamma_{01}$, $beta_u=beta_{01}$, & $beta_v=beta_{12}$. But then to calculate $beta_{03}$ you would need to substitute $gamma_u=gamma_{02}$, $beta_u=beta_{02}$, & $beta_v=beta_{23}$. Note that in this association of the calculation $gamma_{02}$ appears - which must be calculated apart using the ancilliary formula for composition of the Lorentz factor.



                So I think essentially what is being said in the Wikipedia article is that the formula is not associative as it stands, and is only applicable as it stands which is to say without invoking the ancilliary formula for composition of Lorentz factors, if the order of association is strictly from the uttermost-observed frame to the base-frame: that all other associations give the wrong result.



                That's what I'm surmising, anyway. I haven't got symbolic algebra computation handy just at the moment; and I'm not crunching through all those substitutions using pencil & paper, & I haven't got the mental calibre necessary for doing it in my head. Oblique Lorentz transformations are notorious for the the sheer algeba-load of them being formidable! I'll report back when I've investigated the matter using a symbolic algebra package. But I think I probably do have the explanation here - I think my argument is logically robust.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Dec 15 '18 at 12:05

























                answered Dec 15 '18 at 11:19









                AmbretteOrriseyAmbretteOrrisey

                54210




                54210






























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