An identity for the Lorentz cross product












0












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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







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  • $begingroup$
    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.
    $endgroup$
    – J.G.
    Dec 1 '18 at 11:59
















0












$begingroup$


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









$endgroup$












  • $begingroup$
    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.
    $endgroup$
    – J.G.
    Dec 1 '18 at 11:59














0












0








0





$begingroup$


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









$endgroup$




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









PolymorphPolymorph

1206




1206












  • $begingroup$
    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.
    $endgroup$
    – J.G.
    Dec 1 '18 at 11:59


















  • $begingroup$
    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.
    $endgroup$
    – J.G.
    Dec 1 '18 at 11:59
















$begingroup$
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.
$endgroup$
– J.G.
Dec 1 '18 at 11:59




$begingroup$
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.
$endgroup$
– 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











$endgroup$





















    0












    $begingroup$

    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
      2






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      0












      $begingroup$

      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











      $endgroup$


















        0












        $begingroup$

        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











        $endgroup$
















          0












          0








          0





          $begingroup$

          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











          $endgroup$



          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 ShemyakovS Shemyakov

          162




          162























              0












              $begingroup$

              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









              $endgroup$


















                0












                $begingroup$

                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









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  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









                  $endgroup$



                  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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                  share|cite|improve this answer










                  answered Dec 1 '18 at 14:17









                  edmedm

                  3,6231425




                  3,6231425






























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                      Title Spacing in Bjornstrup Chapter, Removing Chapter Number From Contents

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                      up vote 5 down vote favorite 2 I would like to make it so that the vertical space below the title of each chapter is the same when using the Bjornstrup package (i.e. so that the descenders do not affect the spacing - as shown by the red arrow in the picture below). I think I'm fairly happy with the spacing around "A chapter", so I'd really like titles without descenders (e.g. "Contents") to have the same vertical spacing as "A chapter" below the title. I would also like to remove the large "0" in the contents title. Any help appreciated, as always. documentclass[11pt,a4paper]{book} % Packages[![enter image description here][1]][1] usepackage[left=3.5cm, right=3.5cm, top=4.3cm, bottom=4.25cm]{geometry} usepackage[utf8]{inputenc} usepackage{libertine} usepackage[libertine]{newtx
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                      Can I use Tabulator js library in my java Spring + Thymeleaf project?

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                      0 I want to add Tabulator js to my java project. I added tabulator.min.js, tabulator.min.css to resources folder. After that added to my html page these dependencies as recommending in readme.md. <link rel="stylesheet" th:href="@{/css/tabulator.min.css}"> <script type="text/javascript" th:src="@{/js/tabulator.min.js}"></script> Now my tables view all elements, that I have in the html page, but not correct. In the documentation of Tabulator is example how to create header and table, var table = new Tabulator("#example-table", { columns:[ {title:"Name", field:"name", sorter:"string", width:200, editor:true}, {title:"Age", field:"age", sorter:"
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