Newbie question on some exercise

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Frédéric de Montmollin

Feb 28, 2021, 7:50:26 PM2/28/21
to Geometric_Algebra

I am discovering GA through New Foundations for Classical Mechanics (Hestenes).
I have difficulties playing with GA. I tried to solve some exercies, but after hours I'm still blocked.

the goal is to proove the following vector identities (p.47)
(a^b).(c^d) = b.c a.d - b.d a.c

(a^b).(c^d) = abcd - acbd -abdc + adbc - bacd + bcad + badc - bdac

on the othe hand

b.c a.d - b.d a.c = bcad + bcda + cbad + cbda - bdac -bdca - dbac - dbca

I try to develop using the rules for inner and outer product, but I can't find a way to go further.
It is surely very basic, but I can't find the other rules to apply. Or I may have misunderstood the whole process...
If anybody can give me a hint, it will be welcome.
Thanks in advance

alexander arsenovic

Mar 1, 2021, 8:36:15 AM3/1/21
to Geometric_Algebra
   those types of problems are the hardest for me as well, as it  relies mostly and algebraic trickery.    you maybe could use the appendix on spherical trig to get a good geometrical intuition, but it wasnt obvious to me.

    (a^b).(c^d) =  A.B, 
where A and B are bivectors, so its computing the scalar part of the product of two bivectors in 3 dimensions.   A.B = <AB>0.  just to keep this in mind. 
to solve, first re-arrange the first bivector, 
    a^b = 2ab-a.b   
      , (a.b).(c^d) =0 , 
    (a^b).(c^d)=(ab).(c^d) = a.(b.(c^d) = a.( - b.dc)= a.d b.d - a.c b.d

the trick between 2nd and third entry is given in between 1.20 and 1.21 ( might have changed in the editions, ) 

hope this helps.  that book is awesome, but i also found those sections and exercises really hard. this still took me like 40min, and i have been using ga for like 5 years. !


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Frédéric de Montmollin

Mar 1, 2021, 5:14:48 PM3/1/21
to Geometric_Algebra
Thanks a lot. It was very helpful. I got the trick finally.
I could resolve others problem that was thricky for me

Alan Bromborsky

Mar 2, 2021, 5:59:01 PM3/2/21
to Geometric_Algebra

Attached is the output of a galgebra program that shows the sort of identities you can derive using a general metric tensor (metric tensor where all the entries are the dot products of arbitrary vectors).  The code generating the output is:

def check_generalized_BAC_CAB_formulas():
    g4d = Ga('a b c d e')
    (a,b,c,d,e) =

    gprint('g_{ij} =',g4d.g)

    gprint('\\bm{a|(b*c)} =',a|(b*c))
    gprint('\\bm{a|(b\\wedge c)} =',a|(b^c))
    gprint('\\bm{a|(b\\wedge c\\wedge d)} =',a|(b^c^d))
    gprint('\\bm{a|(b\\wedge c)+c|(a\\wedge b)+b|(c\\wedge a)} =',(a|(b^c))+(c|(a^b))+(b|(c^a)))
    gprint('\\bm{a*(b\\wedge c)-b*(a\\wedge c)+c*(a\\wedge b)} =',a*(b^c)-b*(a^c)+c*(a^b))
    gprint('\\bm{a*(b\\wedge c\\wedge d)-b*(a\\wedge c\\wedge d)+c*(a\\wedge b\\wedge d)-d*(a\\wedge b\\wedge c)} =',a*(b^c^d)-b*(a^c^d)+c*(a^b^d)-d*(a^b^c))
    gprint('\\bm{(a\\wedge b)|(c\\wedge d)} =',(a^b)|(c^d))
    gprint('\\bm{((a\\wedge b)|c)|d} =',((a^b)|c)|d)
    gprint('\\bm{(a\\wedge b)\\times (c\\wedge d)} =',\\wedge b,c\\wedge d))
    gprint('\\bm{(a\\wedge b\\wedge c)(d\\wedge e)} =',((a^b^c)*(d^e)).Fmt(2))
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