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Geometric product

  1. Apr 25, 2005 #1
    Does anybody knows where can I find the resolution of the geometrical product of abc?
     
  2. jcsd
  3. Apr 26, 2005 #2

    Andrew Mason

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    Can you write the geometric product of two vectors?

    AM
     
  4. Apr 26, 2005 #3
    Dear Andrew,

    The geometrical product of two vectors is:

    ab= a.b + a^b (inner and outer product).

    I don't know if triple geometrical product is

    abc=a.b.c + a^b^c

    Or

    abc=a(b.c) + a(b^c)

    but if a(b.c) is not a·(b·c), the formulae becomes a great monster, imposible to manipulate.

    and more

    abc is a trivector???

    oh my god!!!
     
  5. Apr 26, 2005 #4

    Andrew Mason

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    If the geometric product is defined only for vectors and since the geometric product of two vectors is not a vector (ie it is a vector + a scalar), you cannot take the geometric product of a vector and a geometric product unless a and b are mutually perpendicular.

    Now if abc are all orthogonal vectors such that a^b = |a||b|[itex]\hat c[/itex] (the scalar product being 0):

    abc = (ab)c = (0 + a^b)^c = (|a||b|[itex]\hat c[/itex])^c = |a||b|[itex]\hat c[/itex]^c = |a||b||c| + [itex]\hat c[/itex] ^ [itex]\hat c[/itex] = |a||b||c|

    (Outer product is the same as the cross product)

    AM
     
  6. Apr 26, 2005 #5
    Multivectors

    But we have a vector, a scalar an a bivector. We can do a geometric multiplication of a vector (scalar + bivector)? This is multiplication of multivectors, not?
     
  7. Apr 26, 2005 #6

    Andrew Mason

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    In the case of the geometric product of two orthogonal vectors, you have no scalar component ([itex]a\cdot b = 0[/itex]). The outer product results in a 'bivector', or an area vector whose direction is perpendicular to the area. If that bivector is parallel to the third vector, its geometric product with the third vector results in a scalar but no bivector.

    If the first two vectors are perpendicular, you don't have to multiply a 'multivector' with another vector.

    Unless multiplication of multivectors is defined in some way (I am unaware of a definiton but there may be one) I don't see how you could determine the geometric product of a multivector (ie. a scalar plus a bivector) and vector.

    AM
     
  8. Apr 27, 2005 #7
    I think it could be like this:

    abc=a·b·c + a^b^c

    but I'm not sure.
     
  9. Apr 27, 2005 #8

    Andrew Mason

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    You would have to do one operation at a time: (a·b)·c or a·(b·c). Since (b·c) is a scalar, what does it mean to take the dot product of a vector, a, with a scalar (b·c)? You would have to define that first, would you not?

    AM
     
  10. Apr 28, 2005 #9

    OlderDan

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    I know nothing about this topic beyond the basic vector multiplication. From what I have read here, I'm thinking this might be

    [tex]
    \overrightarrow a \overrightarrow b \overrightarrow c = \overrightarrow a \left( {\overrightarrow b \overrightarrow c } \right) = \overrightarrow a \left( {\overrightarrow b \bullet \overrightarrow c + \overrightarrow b \times \overrightarrow c } \right) = \overrightarrow a \left( {\overrightarrow b \bullet \overrightarrow c } \right) + \overrightarrow a \bullet \left( {\overrightarrow b \times \overrightarrow c } \right) + \overrightarrow a \times \left( {\overrightarrow b \times \overrightarrow c } \right) [/tex]

    = a scaler times [itex]\overrightarrow a[/itex] + a scaler (triple product) + a vector perpendicular to [itex]\overrightarrow a [/itex]

    That would probably make it non-associative and non-commuative, but I have not checked. Is this a possibility? It seems like a natural extension of the two vector case.
     
  11. Apr 29, 2005 #10
    Trivectors

    It seems to be a vector + a trivector

    Doesn't it???
     
  12. Apr 29, 2005 #11

    OlderDan

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    I guess I'm not sure what you mean by a trivector. If you mean the term

    [tex]\overrightarrow a \times \left( {\overrightarrow b \times \overrightarrow c } \right) [/tex]

    then yes, it includes a trivector

    But when terms are combined, it looks to me like a scalar plus a vector, just like the two vector case, except that the vector is the sum of two orthogonal vectors. All that I have done is assume that the operation can be defined to include a vector "times" a scalar plus vector combination by treating the vector times scaler product in the traditional way.
     
  13. Apr 30, 2005 #12

    Andrew Mason

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    Dan, how are you defining:

    [tex]\vec a (\vec b \cdot \vec c)[/tex] ?

    AM
     
  14. Apr 30, 2005 #13
    abc-cba

    It coudl be like this:

    abc = a(bc)=(ab)c=a(b·c+b^c)=a(b·c)+a(b^c)=
    a·(b·c)+a^(b·c)+a·(b^c)+a^(b^c).

    ???
     
  15. Apr 30, 2005 #14

    OlderDan

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    As the scalar result of the dot product times the vector a. I'm not sure this is the right way to do it. I am just speculating. If the geometric product applied to a vector yields a scalar, and you are going to take another geometric product, you have to do something with the scalar. This just seems like a reaonable something.
     
  16. Apr 30, 2005 #15

    OlderDan

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    a·(b·c)

    This term cannot work, as Andrew noted earlier. (b·c) is not a vector.
     
  17. Apr 30, 2005 #16
    ... and like scalar x vector???
     
  18. May 1, 2005 #17

    OlderDan

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    I'm not sure what you are asking or saying here.

    a·(b·c) has no meaning if the "·" means the dot product or inner product of two vectors. b·c is a scalar quantity. You can multiply a scalar times a vector, which is what I am speculating could be what is going on in the geometric product abc, but you cannot do a dot product of a vector and a scalar. The dot product is only defined between two vectors.
     
  19. May 1, 2005 #18

    OlderDan

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    Perhaps this sheds light on the subject

    http://en.wikipedia.org/wiki/Geometric_algebra#Inner_and_outer_product

    But I really doin't know what this means. To the unenlightened, this just seems to be reducable to an identity. I guess this is what you mean by a monster.

    [tex]}
    a \bullet b = \frac{1}{2}\left( {ab + ba} \right) [/tex]

    [tex] a \wedge b = \frac{1}{2}\left( {ab - ba} \right) [/tex]

    [tex] ab = a \bullet b + a \wedge b = \frac{1}{2}\left( {ab + ba} \right) + \frac{1}{2}\left( {ab - ba} \right) [/tex]

    [tex] bc = b \bullet c + b \wedge c = \frac{1}{2}\left( {bc + cb} \right) + \frac{1}{2}\left( {bc - cb} \right) [/tex]

    [tex] abc = a \bullet \left[ {\frac{1}{2}\left( {bc + cb} \right) + \frac{1}{2}\left( {bc - cb} \right)} \right] + a \wedge \left[ {\frac{1}{2}\left( {bc + cb} \right) + \frac{1}{2}\left( {bc - cb} \right)} \right] [/tex]

    [tex] abc = \frac{1}{2}\left\{ {a\left[ {\frac{1}{2}\left( {bc + cb} \right) + \frac{1}{2}\left( {bc - cb} \right)} \right] + \left[ {\frac{1}{2}\left( {bc + cb} \right) + \frac{1}{2}\left( {bc - cb} \right)} \right]a} \right\} [/tex]

    [tex] + \frac{1}{2}\left\{ {a\left[ {\frac{1}{2}\left( {bc + cb} \right) + \frac{1}{2}\left( {bc - cb} \right)} \right] - \left[ {\frac{1}{2}\left( {bc + cb} \right) + \frac{1}{2}\left( {bc - cb} \right)} \right]a} \right\} [/tex]

    [tex] abc = \frac{1}{4}\left\{ {\left[ {a\left( {bc + cb} \right) + a\left( {bc - cb} \right)} \right] + \left[ {\left( {bc + cb} \right)a + \left( {bc - cb} \right)a} \right]} \right\} [/tex]

    [tex] + \frac{1}{4}\left\{ {\left[ {a\left( {bc + cb} \right) + a\left( {bc - cb} \right)} \right] - \left[ {\left( {bc + cb} \right)a + \left( {bc - cb} \right)a} \right]} \right\} [/tex]

    I guess the bottom line is I have no idea about this stuff, so I should just be quiet.
     
  20. May 4, 2005 #19
    Thanks

    Thanks for all.
     
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