An identity for the Lorentz cross product












0














I need some help finding the error in my proof for an identity for the Lorentz cross product.



For $x=(x_1,x_2,x_3)^T, y=(y_1,y_2,y_3)^T in R^{2,1}$ the Lorentz cross product is defined as



det[x,y,z]=$langle,x times y,zrangle$



for all $z in R^{2,1}$ where $langle,cdot, cdotrangle$ is the Lorentz scalar product of $R^{2,1}$.



In coordinates we have



$x times y=begin{pmatrix}x_2y_3-y_2x_3\y_1x_3-x_1y_3\x_1y_2-y_1x_2end{pmatrix}$.



Now let



$a=(a_1,a_2,a_3)^T, b=(b_1,b_2,b_3)^T, c=(c_1,c_2,c_3)^T in R^{2,1}$. I want to prove the identity




$a times (b times c)=clangle,a, brangle-blangle,a,crangle$




I did it in coordinates. We have



$b times c=begin{pmatrix}b_2c_3-c_2b_3\c_1b_3-b_1c_3\b_1c_2-c_1b_2end{pmatrix}$



and



$a times (b times c)=begin{pmatrix}a_2(b_1c_2-c_1b_2)-(c_1b_3-b_1c_3)a_3\(b_2c_3-c_2b_3)a_3-a_1(b_1c_2-c_1b_2)\a_1(c_1b_3-b_1c_3)-(b_2c_3-c_2b_3)a_2end{pmatrix}$



$=begin{pmatrix}-c_1b_2a_2-c_1b_3a_3+b_1a_2c_2+b_1a_3c_3\-c_2a_1b_1-c_2a_3b_3+b_2a_1c_1+b_2a_3c_3\-c_3a_1b_3-c_3a_2b_2+b_3a_1c_1+b_3a_2c_2end{pmatrix}$



We also have



$clangle,a, brangle-blangle,a, crangle$=
begin{pmatrix}c_1a_2b_2-c_1a_3b_3-b_1a_2c_2+b_1a_3c_3\c_2a_1b_1-c_2a_3b_3-b_2a_1c_1-b_2a_3c_3\c_3a_1b_1+c_3a_2b_2-b_3a_1c_1-b_3a_2c_2end{pmatrix}



So far the equations do not coincide. It seems that I have made a mistake but I don't see where. Any help would be appreciated.






geometry hyperbolic-geometry







share|cite|improve this question






















  • You'll find proofs here: math.stackexchange.com/questions/1905883/… Some, including my own, don't even require you to chug through the algebra as much as your approach does.
    – J.G.
    Dec 1 '18 at 11:59
















0














I need some help finding the error in my proof for an identity for the Lorentz cross product.



For $x=(x_1,x_2,x_3)^T, y=(y_1,y_2,y_3)^T in R^{2,1}$ the Lorentz cross product is defined as



det[x,y,z]=$langle,x times y,zrangle$



for all $z in R^{2,1}$ where $langle,cdot, cdotrangle$ is the Lorentz scalar product of $R^{2,1}$.



In coordinates we have



$x times y=begin{pmatrix}x_2y_3-y_2x_3\y_1x_3-x_1y_3\x_1y_2-y_1x_2end{pmatrix}$.



Now let



$a=(a_1,a_2,a_3)^T, b=(b_1,b_2,b_3)^T, c=(c_1,c_2,c_3)^T in R^{2,1}$. I want to prove the identity




$a times (b times c)=clangle,a, brangle-blangle,a,crangle$




I did it in coordinates. We have



$b times c=begin{pmatrix}b_2c_3-c_2b_3\c_1b_3-b_1c_3\b_1c_2-c_1b_2end{pmatrix}$



and



$a times (b times c)=begin{pmatrix}a_2(b_1c_2-c_1b_2)-(c_1b_3-b_1c_3)a_3\(b_2c_3-c_2b_3)a_3-a_1(b_1c_2-c_1b_2)\a_1(c_1b_3-b_1c_3)-(b_2c_3-c_2b_3)a_2end{pmatrix}$



$=begin{pmatrix}-c_1b_2a_2-c_1b_3a_3+b_1a_2c_2+b_1a_3c_3\-c_2a_1b_1-c_2a_3b_3+b_2a_1c_1+b_2a_3c_3\-c_3a_1b_3-c_3a_2b_2+b_3a_1c_1+b_3a_2c_2end{pmatrix}$



We also have



$clangle,a, brangle-blangle,a, crangle$=
begin{pmatrix}c_1a_2b_2-c_1a_3b_3-b_1a_2c_2+b_1a_3c_3\c_2a_1b_1-c_2a_3b_3-b_2a_1c_1-b_2a_3c_3\c_3a_1b_1+c_3a_2b_2-b_3a_1c_1-b_3a_2c_2end{pmatrix}



So far the equations do not coincide. It seems that I have made a mistake but I don't see where. Any help would be appreciated.






geometry hyperbolic-geometry







share|cite|improve this question






















  • You'll find proofs here: math.stackexchange.com/questions/1905883/… Some, including my own, don't even require you to chug through the algebra as much as your approach does.
    – J.G.
    Dec 1 '18 at 11:59














0












0








0







I need some help finding the error in my proof for an identity for the Lorentz cross product.



For $x=(x_1,x_2,x_3)^T, y=(y_1,y_2,y_3)^T in R^{2,1}$ the Lorentz cross product is defined as



det[x,y,z]=$langle,x times y,zrangle$



for all $z in R^{2,1}$ where $langle,cdot, cdotrangle$ is the Lorentz scalar product of $R^{2,1}$.



In coordinates we have



$x times y=begin{pmatrix}x_2y_3-y_2x_3\y_1x_3-x_1y_3\x_1y_2-y_1x_2end{pmatrix}$.



Now let



$a=(a_1,a_2,a_3)^T, b=(b_1,b_2,b_3)^T, c=(c_1,c_2,c_3)^T in R^{2,1}$. I want to prove the identity




$a times (b times c)=clangle,a, brangle-blangle,a,crangle$




I did it in coordinates. We have



$b times c=begin{pmatrix}b_2c_3-c_2b_3\c_1b_3-b_1c_3\b_1c_2-c_1b_2end{pmatrix}$



and



$a times (b times c)=begin{pmatrix}a_2(b_1c_2-c_1b_2)-(c_1b_3-b_1c_3)a_3\(b_2c_3-c_2b_3)a_3-a_1(b_1c_2-c_1b_2)\a_1(c_1b_3-b_1c_3)-(b_2c_3-c_2b_3)a_2end{pmatrix}$



$=begin{pmatrix}-c_1b_2a_2-c_1b_3a_3+b_1a_2c_2+b_1a_3c_3\-c_2a_1b_1-c_2a_3b_3+b_2a_1c_1+b_2a_3c_3\-c_3a_1b_3-c_3a_2b_2+b_3a_1c_1+b_3a_2c_2end{pmatrix}$



We also have



$clangle,a, brangle-blangle,a, crangle$=
begin{pmatrix}c_1a_2b_2-c_1a_3b_3-b_1a_2c_2+b_1a_3c_3\c_2a_1b_1-c_2a_3b_3-b_2a_1c_1-b_2a_3c_3\c_3a_1b_1+c_3a_2b_2-b_3a_1c_1-b_3a_2c_2end{pmatrix}



So far the equations do not coincide. It seems that I have made a mistake but I don't see where. Any help would be appreciated.






geometry hyperbolic-geometry







share|cite|improve this question













I need some help finding the error in my proof for an identity for the Lorentz cross product.



For $x=(x_1,x_2,x_3)^T, y=(y_1,y_2,y_3)^T in R^{2,1}$ the Lorentz cross product is defined as



det[x,y,z]=$langle,x times y,zrangle$



for all $z in R^{2,1}$ where $langle,cdot, cdotrangle$ is the Lorentz scalar product of $R^{2,1}$.



In coordinates we have



$x times y=begin{pmatrix}x_2y_3-y_2x_3\y_1x_3-x_1y_3\x_1y_2-y_1x_2end{pmatrix}$.



Now let



$a=(a_1,a_2,a_3)^T, b=(b_1,b_2,b_3)^T, c=(c_1,c_2,c_3)^T in R^{2,1}$. I want to prove the identity




$a times (b times c)=clangle,a, brangle-blangle,a,crangle$




I did it in coordinates. We have



$b times c=begin{pmatrix}b_2c_3-c_2b_3\c_1b_3-b_1c_3\b_1c_2-c_1b_2end{pmatrix}$



and



$a times (b times c)=begin{pmatrix}a_2(b_1c_2-c_1b_2)-(c_1b_3-b_1c_3)a_3\(b_2c_3-c_2b_3)a_3-a_1(b_1c_2-c_1b_2)\a_1(c_1b_3-b_1c_3)-(b_2c_3-c_2b_3)a_2end{pmatrix}$



$=begin{pmatrix}-c_1b_2a_2-c_1b_3a_3+b_1a_2c_2+b_1a_3c_3\-c_2a_1b_1-c_2a_3b_3+b_2a_1c_1+b_2a_3c_3\-c_3a_1b_3-c_3a_2b_2+b_3a_1c_1+b_3a_2c_2end{pmatrix}$



We also have



$clangle,a, brangle-blangle,a, crangle$=
begin{pmatrix}c_1a_2b_2-c_1a_3b_3-b_1a_2c_2+b_1a_3c_3\c_2a_1b_1-c_2a_3b_3-b_2a_1c_1-b_2a_3c_3\c_3a_1b_1+c_3a_2b_2-b_3a_1c_1-b_3a_2c_2end{pmatrix}



So far the equations do not coincide. It seems that I have made a mistake but I don't see where. Any help would be appreciated.






geometry hyperbolic-geometry




geometry hyperbolic-geometry






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asked Nov 28 '18 at 21:20









Polymorph

1236




1236












  • You'll find proofs here: math.stackexchange.com/questions/1905883/… Some, including my own, don't even require you to chug through the algebra as much as your approach does.
    – J.G.
    Dec 1 '18 at 11:59


















  • You'll find proofs here: math.stackexchange.com/questions/1905883/… Some, including my own, don't even require you to chug through the algebra as much as your approach does.
    – J.G.
    Dec 1 '18 at 11:59
















You'll find proofs here: math.stackexchange.com/questions/1905883/… Some, including my own, don't even require you to chug through the algebra as much as your approach does.
– J.G.
Dec 1 '18 at 11:59




You'll find proofs here: math.stackexchange.com/questions/1905883/… Some, including my own, don't even require you to chug through the algebra as much as your approach does.
– J.G.
Dec 1 '18 at 11:59










2 Answers
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The problem is with your formula for Lorentz cross product. It should be:
$x times y=begin{pmatrix}x_2y_3-y_2x_3\y_1x_3-x_1y_3\y_1x_2- x_1y_2end{pmatrix}$.



With this formula everything fits.






share|cite|improve this answer































    0














    The correct identity is $$atimes (btimes c)=blangle,a,crangle-clangle,a, brangle.$$The right-hand-side is negative of $clangle,a, brangle-blangle,a,crangle$.



    You calculated $atimes (btimes c)$ wrongly. I calculated it as $$begin{pmatrix}-c_1b_2a_2-c_1b_3a_3+b_1a_2c_2+b_1a_3c_3\-c_2a_1b_1-c_2a_3b_3+b_2a_1c_1+b_2a_3c_3\-c_3a_1color{red}{b_1}-c_3a_2b_2+b_3a_1c_1+b_3a_2c_2end{pmatrix}$$ The highlighted symbol is where you calculated wrong. Double check that part.



    You also calculated $clangle,a, brangle-blangle,a,crangle$ wrongly. I calculated it as $$begin{pmatrix}c_1a_2b_2color{red}+c_1a_3b_3-b_1a_2c_2color{red}-b_1a_3c_3\c_2a_1b_1color{red}+c_2a_3b_3-b_2a_1c_1-b_2a_3c_3\c_3a_1b_1+c_3a_2b_2-b_3a_1c_1-b_3a_2c_2end{pmatrix}$$



    Again, the highlighted part is where you were wrong. Double check those.






    share|cite|improve this answer





















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      2 Answers
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      2 Answers
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      The problem is with your formula for Lorentz cross product. It should be:
      $x times y=begin{pmatrix}x_2y_3-y_2x_3\y_1x_3-x_1y_3\y_1x_2- x_1y_2end{pmatrix}$.



      With this formula everything fits.






      share|cite|improve this answer




























        0














        The problem is with your formula for Lorentz cross product. It should be:
        $x times y=begin{pmatrix}x_2y_3-y_2x_3\y_1x_3-x_1y_3\y_1x_2- x_1y_2end{pmatrix}$.



        With this formula everything fits.






        share|cite|improve this answer


























          0












          0








          0






          The problem is with your formula for Lorentz cross product. It should be:
          $x times y=begin{pmatrix}x_2y_3-y_2x_3\y_1x_3-x_1y_3\y_1x_2- x_1y_2end{pmatrix}$.



          With this formula everything fits.






          share|cite|improve this answer














          The problem is with your formula for Lorentz cross product. It should be:
          $x times y=begin{pmatrix}x_2y_3-y_2x_3\y_1x_3-x_1y_3\y_1x_2- x_1y_2end{pmatrix}$.



          With this formula everything fits.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 1 '18 at 11:57

























          answered Dec 1 '18 at 11:51









          S Shemyakov

          162




          162























              0














              The correct identity is $$atimes (btimes c)=blangle,a,crangle-clangle,a, brangle.$$The right-hand-side is negative of $clangle,a, brangle-blangle,a,crangle$.



              You calculated $atimes (btimes c)$ wrongly. I calculated it as $$begin{pmatrix}-c_1b_2a_2-c_1b_3a_3+b_1a_2c_2+b_1a_3c_3\-c_2a_1b_1-c_2a_3b_3+b_2a_1c_1+b_2a_3c_3\-c_3a_1color{red}{b_1}-c_3a_2b_2+b_3a_1c_1+b_3a_2c_2end{pmatrix}$$ The highlighted symbol is where you calculated wrong. Double check that part.



              You also calculated $clangle,a, brangle-blangle,a,crangle$ wrongly. I calculated it as $$begin{pmatrix}c_1a_2b_2color{red}+c_1a_3b_3-b_1a_2c_2color{red}-b_1a_3c_3\c_2a_1b_1color{red}+c_2a_3b_3-b_2a_1c_1-b_2a_3c_3\c_3a_1b_1+c_3a_2b_2-b_3a_1c_1-b_3a_2c_2end{pmatrix}$$



              Again, the highlighted part is where you were wrong. Double check those.






              share|cite|improve this answer


























                0














                The correct identity is $$atimes (btimes c)=blangle,a,crangle-clangle,a, brangle.$$The right-hand-side is negative of $clangle,a, brangle-blangle,a,crangle$.



                You calculated $atimes (btimes c)$ wrongly. I calculated it as $$begin{pmatrix}-c_1b_2a_2-c_1b_3a_3+b_1a_2c_2+b_1a_3c_3\-c_2a_1b_1-c_2a_3b_3+b_2a_1c_1+b_2a_3c_3\-c_3a_1color{red}{b_1}-c_3a_2b_2+b_3a_1c_1+b_3a_2c_2end{pmatrix}$$ The highlighted symbol is where you calculated wrong. Double check that part.



                You also calculated $clangle,a, brangle-blangle,a,crangle$ wrongly. I calculated it as $$begin{pmatrix}c_1a_2b_2color{red}+c_1a_3b_3-b_1a_2c_2color{red}-b_1a_3c_3\c_2a_1b_1color{red}+c_2a_3b_3-b_2a_1c_1-b_2a_3c_3\c_3a_1b_1+c_3a_2b_2-b_3a_1c_1-b_3a_2c_2end{pmatrix}$$



                Again, the highlighted part is where you were wrong. Double check those.






                share|cite|improve this answer
























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                  0








                  0






                  The correct identity is $$atimes (btimes c)=blangle,a,crangle-clangle,a, brangle.$$The right-hand-side is negative of $clangle,a, brangle-blangle,a,crangle$.



                  You calculated $atimes (btimes c)$ wrongly. I calculated it as $$begin{pmatrix}-c_1b_2a_2-c_1b_3a_3+b_1a_2c_2+b_1a_3c_3\-c_2a_1b_1-c_2a_3b_3+b_2a_1c_1+b_2a_3c_3\-c_3a_1color{red}{b_1}-c_3a_2b_2+b_3a_1c_1+b_3a_2c_2end{pmatrix}$$ The highlighted symbol is where you calculated wrong. Double check that part.



                  You also calculated $clangle,a, brangle-blangle,a,crangle$ wrongly. I calculated it as $$begin{pmatrix}c_1a_2b_2color{red}+c_1a_3b_3-b_1a_2c_2color{red}-b_1a_3c_3\c_2a_1b_1color{red}+c_2a_3b_3-b_2a_1c_1-b_2a_3c_3\c_3a_1b_1+c_3a_2b_2-b_3a_1c_1-b_3a_2c_2end{pmatrix}$$



                  Again, the highlighted part is where you were wrong. Double check those.






                  share|cite|improve this answer












                  The correct identity is $$atimes (btimes c)=blangle,a,crangle-clangle,a, brangle.$$The right-hand-side is negative of $clangle,a, brangle-blangle,a,crangle$.



                  You calculated $atimes (btimes c)$ wrongly. I calculated it as $$begin{pmatrix}-c_1b_2a_2-c_1b_3a_3+b_1a_2c_2+b_1a_3c_3\-c_2a_1b_1-c_2a_3b_3+b_2a_1c_1+b_2a_3c_3\-c_3a_1color{red}{b_1}-c_3a_2b_2+b_3a_1c_1+b_3a_2c_2end{pmatrix}$$ The highlighted symbol is where you calculated wrong. Double check that part.



                  You also calculated $clangle,a, brangle-blangle,a,crangle$ wrongly. I calculated it as $$begin{pmatrix}c_1a_2b_2color{red}+c_1a_3b_3-b_1a_2c_2color{red}-b_1a_3c_3\c_2a_1b_1color{red}+c_2a_3b_3-b_2a_1c_1-b_2a_3c_3\c_3a_1b_1+c_3a_2b_2-b_3a_1c_1-b_3a_2c_2end{pmatrix}$$



                  Again, the highlighted part is where you were wrong. Double check those.







                  share|cite|improve this answer












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                  answered Dec 1 '18 at 14:17









                  edm

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                  3,6131425






























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                      Verónica Boquete

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                      Verónica Boquete im Trikot der Nationalmannschaft (2013) Personalia Name Verónica Boquete Giadans Geburtstag 9. April 1987 Geburtsort Santiago de Compostela, Spanien Größe 159 cm Position Mittelfeld / Sturm Juniorinnen Jahre Station 2001–2005 Xuventú Aguiño Frauen Jahre Station Spiele (Tore) 1 2005–2008 Prainsa Zaragoza 83 (40) 2008–2010 Espanyol Barcelona 53 (30) 2010 Buffalo Flash 9 0 (9) 2010 Chicago Red Stars 3 0 (1) 2010–2011 Espanyol Barcelona 26 (41) 2011 Philadelphia Independence 13 0 (5) 2011 Energija Woronesch 5 0 (1) 2012–2014 Tyresö FF 39 (14) 2014 Portland Thorns FC 15 0 (4) 2014–2015 1. FFC Frankfurt 23 0 (7) 2015–2016 FC Bayern München 9 0 (0) 2016–2018 Paris Saint-Germain 31 0 (8) 2018 Peking BG Phoenix 2019– Utah Royals FC 0 0 (0) Nationalmannschaft Jahre Auswahl ...
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                      Ida-Boy-Ed-Garten

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                      Die Lage des Ida-Boy-Ed-Gartens, rot markiert Der Ida-Boy-Ed-Garten Der Ida-Boy-Ed-Garten, Blick von der Wakenitzmauer durch das Tor in der Stadtmauer Der Ida-Boy-Ed-Garten ist eine Straße der Lübecker Altstadt. Inhaltsverzeichnis 1 Lage 2 Geschichte 3 Literatur 4 Weblinks Lage | Der etwa 90 Meter lange Ida-Boy-Ed-Garten befindet sich an der nördlichen Spitze der Altstadtinsel, vor dem Burgtor. Die in einem Bogen durch eine Grünanlage hangaufwärts führende Straße verbindet die Wakenitzmauer mit der Burgtorbrücke. Geschichte | Der heutige Ida-Boy-Ed-Garten wurde ursprünglich als Verlängerung der Wakenitzmauer angelegt. 1903 wurde in den erhaltenen Teil der mittelalterlichen Stadtmauer, an dem die Wakenitzmauer bei der Einmündung der Kaiserstraße begann, ein Tor mit einem großen Bogen für die Durchführung der Straße und zwei kleineren seitlichen Öffnungen für Fußgänger gebrochen, um für die neu errichteten Wohnhäuser in diesem Bereich ...
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