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









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                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




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                      ComboBox Display Member on multiple fields

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                      2 I'm trying to set the .DisplayMember property of a ComboBox in C#, but I want to bind it to multiple columns in the .DataSouce . My SQL looks like this: SELECT PersNbr, PersFirstName, PersMiddleName, PersLastName FROM Pers WHERE PersNbr = :persNbr; I'm saving this query in a DataTable so each column selected has it's own column in the Datatable . I want to make the .DisplayMember a combination of PersFirstName + PersMiddleName + PersLastName so their full name appears like this: comboBox.DisplayMemeber = "PersFirstName" + "PersMiddleName" + "PersLastName" I know I can just to this on the query: SELECT PersNbr, (PersFirstName || PersMiddleName || PersLastName) PersName and then just do this: comboBox.DisplayMember = "PersName"; but I
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                      Is it possible to collect Nectar points via Trainline?

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                      2 Since I am not eligible for any Railcard, the only available discount/bonus scheme is Nectar Card for me. GWR, SWR, LNER and Virgin Trains allow to collect Nectar points and I have already linked a card to my GWR account. A friend suggested me to use Trainline App, because it can search for bus tickets either. I had a look at the application, but could not see any option for Nectar. Is there a way to collect Nectar points when I buy tickets via Trainline? uk trains loyalty-programs trainline share | improve this question asked Dec 2 '18 at 15:50 ahmedus 3,224 5 19 49
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