Mixing
Deriving reduction formula in Geometric Algebra
$begingroup$
I am trying to learn Geometric Algebra by going through the book "New Foundations for Classical Mechanics" by David Hestenes.
I was reading the part about reduction formula (shown below) but couldn't get the result the shown in the book.
Can someone show me how iterating (1.15) gives the reduction formula?
notation in the book:
- dot (.) is inner product
- circumflex (^) is outer product
ab mean geometric product of a and b
- inner and outer product have precedence over geometric product unless indicated by parentheses
Thank you
update 1
I can easily apply (1.15) for the first iteration and get
$a cdot (a_1 wedge a_2 wedge cdots wedge a_r) = (a cdot a_1) (a_2 wedge cdots wedge a_r) - a_1 wedge (a cdot (a_2 wedge a_3 cdots wedge a_r))$
I can see that I should apply (1.15) to the term
$- a_1 wedge (a cdot (a_2 wedge a_3 cdots wedge a_r))$
but there is a $ - a_1 wedge $ in the term which will get inherited if I apply (1.15) directly, what should I do to get rid of that?
clifford-algebras geometric-algebras
asked Feb 3 at 1:01
EricEric
134
$endgroup$
add a comment |
$begingroup$
I am trying to learn Geometric Algebra by going through the book "New Foundations for Classical Mechanics" by David Hestenes.
I was reading the part about reduction formula (shown below) but couldn't get the result the shown in the book.
Can someone show me how iterating (1.15) gives the reduction formula?
notation in the book:
- dot (.) is inner product
- circumflex (^) is outer product
ab mean geometric product of a and b
- inner and outer product have precedence over geometric product unless indicated by parentheses
Thank you
update 1
I can easily apply (1.15) for the first iteration and get
$a cdot (a_1 wedge a_2 wedge cdots wedge a_r) = (a cdot a_1) (a_2 wedge cdots wedge a_r) - a_1 wedge (a cdot (a_2 wedge a_3 cdots wedge a_r))$
I can see that I should apply (1.15) to the term
$- a_1 wedge (a cdot (a_2 wedge a_3 cdots wedge a_r))$
but there is a $ - a_1 wedge $ in the term which will get inherited if I apply (1.15) directly, what should I do to get rid of that?
clifford-algebras geometric-algebras
asked Feb 3 at 1:01
EricEric
134
$endgroup$
add a comment |
$begingroup$
I am trying to learn Geometric Algebra by going through the book "New Foundations for Classical Mechanics" by David Hestenes.
I was reading the part about reduction formula (shown below) but couldn't get the result the shown in the book.
Can someone show me how iterating (1.15) gives the reduction formula?
notation in the book:
- dot (.) is inner product
- circumflex (^) is outer product
ab mean geometric product of a and b
- inner and outer product have precedence over geometric product unless indicated by parentheses
Thank you
update 1
I can easily apply (1.15) for the first iteration and get
$a cdot (a_1 wedge a_2 wedge cdots wedge a_r) = (a cdot a_1) (a_2 wedge cdots wedge a_r) - a_1 wedge (a cdot (a_2 wedge a_3 cdots wedge a_r))$
I can see that I should apply (1.15) to the term
$- a_1 wedge (a cdot (a_2 wedge a_3 cdots wedge a_r))$
but there is a $ - a_1 wedge $ in the term which will get inherited if I apply (1.15) directly, what should I do to get rid of that?
clifford-algebras geometric-algebras
asked Feb 3 at 1:01
EricEric
134
$endgroup$
I am trying to learn Geometric Algebra by going through the book "New Foundations for Classical Mechanics" by David Hestenes.
I was reading the part about reduction formula (shown below) but couldn't get the result the shown in the book.
Can someone show me how iterating (1.15) gives the reduction formula?
notation in the book:
- dot (.) is inner product
- circumflex (^) is outer product
ab mean geometric product of a and b
- inner and outer product have precedence over geometric product unless indicated by parentheses
Thank you
update 1
I can easily apply (1.15) for the first iteration and get
$a cdot (a_1 wedge a_2 wedge cdots wedge a_r) = (a cdot a_1) (a_2 wedge cdots wedge a_r) - a_1 wedge (a cdot (a_2 wedge a_3 cdots wedge a_r))$
I can see that I should apply (1.15) to the term
$- a_1 wedge (a cdot (a_2 wedge a_3 cdots wedge a_r))$
but there is a $ - a_1 wedge $ in the term which will get inherited if I apply (1.15) directly, what should I do to get rid of that?
clifford-algebras geometric-algebras
clifford-algebras geometric-algebras
asked Feb 3 at 1:01
EricEric
134
asked Feb 3 at 1:01
EricEric
134
edited Feb 3 at 7:33
Eric
asked Feb 3 at 1:01
EricEric
134
asked Feb 3 at 1:01
EricEric
134
asked Feb 3 at 1:01
EricEric
134
134
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Let $b=a_1$ and $C_{r-1}=a_2wedge a_3wedgedotswedge a_r$. Then, from (1.15) it follows that
$$
acdot(a_1wedge a_2wedgedotswedge a_r)=acdot a_1 a_2wedgedotswedge a_r
$$
$$
-a_1wedge(acdot a_2wedgedotswedge a_r).
$$
Next you apply the same (1.15) to the second factor in the second piece above with $b=a_2$ and $C_{r-2}=a_3wedge a_4wedgedotswedge a_r$ to get
$$
=acdot a_1 a_2wedgedotswedge a_r-acdot a_2 a_1wedge a_3dotswedge a_r+
$$
$$
+acdot a_3 a_1wedge a_2wedge a_4wedgedotswedge a_r+textrm{next term}.
$$
One of the factors in the next term above has a structure similar to the left hand side in (1.15) and you can iterate the above procedure. In the last step, you have $C_1$ which is a vector, and the inner product with $C_1$ is the usual dot product of vectors.
Here is how you proceed:
$$
-a_1wedge(acdot (a_2wedgedotswedge a_r))=
$$
$$
=-a_1wedge[(acdot a_2a_3wedgedotswedge a_r)-a_2wedge(acdot(a_3wedge a_4dotswedge a_r))]=
$$
$$
=-(acdot a_2)a_1wedge a_3wedgedotswedge a_r+a_1wedge a_2wedge(acdot(a_3wedge a_4dotswedge a_r))
$$
Observe that the "$cdot$" operation appears in two forms. On the first of the three lines above it stands for the inner product of vector with multivector, whereas on the first term in the line right below it is just a dot product of two vectors hence a scalar. Scalars can be pulled out of any type of product in geometric algebra and that is how you end up with the first term in the third line. You can deal with the second term in the last line exactly in the same way: you get a term containing a scalar $acdot a_3$ which can be pulled out and another term of the form $a_1wedge a_2wedge (a_3cdot(a_4wedge a_5dotswedge a_r))$ where $cdot$ stands again for vector by multivector as in the left hand side of (1.15) and you repeat the procedure to pull subsequent scalars.
answered Feb 3 at 4:07
GReyesGReyes
2,50815
$endgroup$
$begingroup$
I have not problem applying (1.15) for the first time, but for the second time, there is an outer product with $a_1$, how can I get rid of that? I have edited edited my question to indicate my problem.
$endgroup$
– Eric
Feb 3 at 7:16
$begingroup$
@Eric I just improved my answer to include the details of the computations. Please let me know.
$endgroup$
– GReyes
Feb 4 at 0:44
$begingroup$
Thanks a lot for the added details, I didn't realized that I can pull out the scalar :D
$endgroup$
– Eric
Feb 4 at 6:51
add a comment |
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$begingroup$
Let $b=a_1$ and $C_{r-1}=a_2wedge a_3wedgedotswedge a_r$. Then, from (1.15) it follows that
$$
acdot(a_1wedge a_2wedgedotswedge a_r)=acdot a_1 a_2wedgedotswedge a_r
$$
$$
-a_1wedge(acdot a_2wedgedotswedge a_r).
$$
Next you apply the same (1.15) to the second factor in the second piece above with $b=a_2$ and $C_{r-2}=a_3wedge a_4wedgedotswedge a_r$ to get
$$
=acdot a_1 a_2wedgedotswedge a_r-acdot a_2 a_1wedge a_3dotswedge a_r+
$$
$$
+acdot a_3 a_1wedge a_2wedge a_4wedgedotswedge a_r+textrm{next term}.
$$
One of the factors in the next term above has a structure similar to the left hand side in (1.15) and you can iterate the above procedure. In the last step, you have $C_1$ which is a vector, and the inner product with $C_1$ is the usual dot product of vectors.
Here is how you proceed:
$$
-a_1wedge(acdot (a_2wedgedotswedge a_r))=
$$
$$
=-a_1wedge[(acdot a_2a_3wedgedotswedge a_r)-a_2wedge(acdot(a_3wedge a_4dotswedge a_r))]=
$$
$$
=-(acdot a_2)a_1wedge a_3wedgedotswedge a_r+a_1wedge a_2wedge(acdot(a_3wedge a_4dotswedge a_r))
$$
Observe that the "$cdot$" operation appears in two forms. On the first of the three lines above it stands for the inner product of vector with multivector, whereas on the first term in the line right below it is just a dot product of two vectors hence a scalar. Scalars can be pulled out of any type of product in geometric algebra and that is how you end up with the first term in the third line. You can deal with the second term in the last line exactly in the same way: you get a term containing a scalar $acdot a_3$ which can be pulled out and another term of the form $a_1wedge a_2wedge (a_3cdot(a_4wedge a_5dotswedge a_r))$ where $cdot$ stands again for vector by multivector as in the left hand side of (1.15) and you repeat the procedure to pull subsequent scalars.
answered Feb 3 at 4:07
GReyesGReyes
2,50815
$endgroup$
$begingroup$
I have not problem applying (1.15) for the first time, but for the second time, there is an outer product with $a_1$, how can I get rid of that? I have edited edited my question to indicate my problem.
$endgroup$
– Eric
Feb 3 at 7:16
$begingroup$
@Eric I just improved my answer to include the details of the computations. Please let me know.
$endgroup$
– GReyes
Feb 4 at 0:44
$begingroup$
Thanks a lot for the added details, I didn't realized that I can pull out the scalar :D
$endgroup$
– Eric
Feb 4 at 6:51
add a comment |
$begingroup$
Let $b=a_1$ and $C_{r-1}=a_2wedge a_3wedgedotswedge a_r$. Then, from (1.15) it follows that
$$
acdot(a_1wedge a_2wedgedotswedge a_r)=acdot a_1 a_2wedgedotswedge a_r
$$
$$
-a_1wedge(acdot a_2wedgedotswedge a_r).
$$
Next you apply the same (1.15) to the second factor in the second piece above with $b=a_2$ and $C_{r-2}=a_3wedge a_4wedgedotswedge a_r$ to get
$$
=acdot a_1 a_2wedgedotswedge a_r-acdot a_2 a_1wedge a_3dotswedge a_r+
$$
$$
+acdot a_3 a_1wedge a_2wedge a_4wedgedotswedge a_r+textrm{next term}.
$$
One of the factors in the next term above has a structure similar to the left hand side in (1.15) and you can iterate the above procedure. In the last step, you have $C_1$ which is a vector, and the inner product with $C_1$ is the usual dot product of vectors.
Here is how you proceed:
$$
-a_1wedge(acdot (a_2wedgedotswedge a_r))=
$$
$$
=-a_1wedge[(acdot a_2a_3wedgedotswedge a_r)-a_2wedge(acdot(a_3wedge a_4dotswedge a_r))]=
$$
$$
=-(acdot a_2)a_1wedge a_3wedgedotswedge a_r+a_1wedge a_2wedge(acdot(a_3wedge a_4dotswedge a_r))
$$
Observe that the "$cdot$" operation appears in two forms. On the first of the three lines above it stands for the inner product of vector with multivector, whereas on the first term in the line right below it is just a dot product of two vectors hence a scalar. Scalars can be pulled out of any type of product in geometric algebra and that is how you end up with the first term in the third line. You can deal with the second term in the last line exactly in the same way: you get a term containing a scalar $acdot a_3$ which can be pulled out and another term of the form $a_1wedge a_2wedge (a_3cdot(a_4wedge a_5dotswedge a_r))$ where $cdot$ stands again for vector by multivector as in the left hand side of (1.15) and you repeat the procedure to pull subsequent scalars.
answered Feb 3 at 4:07
GReyesGReyes
2,50815
$endgroup$
$begingroup$
I have not problem applying (1.15) for the first time, but for the second time, there is an outer product with $a_1$, how can I get rid of that? I have edited edited my question to indicate my problem.
$endgroup$
– Eric
Feb 3 at 7:16
$begingroup$
@Eric I just improved my answer to include the details of the computations. Please let me know.
$endgroup$
– GReyes
Feb 4 at 0:44
$begingroup$
Thanks a lot for the added details, I didn't realized that I can pull out the scalar :D
$endgroup$
– Eric
Feb 4 at 6:51
add a comment |
$begingroup$
Let $b=a_1$ and $C_{r-1}=a_2wedge a_3wedgedotswedge a_r$. Then, from (1.15) it follows that
$$
acdot(a_1wedge a_2wedgedotswedge a_r)=acdot a_1 a_2wedgedotswedge a_r
$$
$$
-a_1wedge(acdot a_2wedgedotswedge a_r).
$$
Next you apply the same (1.15) to the second factor in the second piece above with $b=a_2$ and $C_{r-2}=a_3wedge a_4wedgedotswedge a_r$ to get
$$
=acdot a_1 a_2wedgedotswedge a_r-acdot a_2 a_1wedge a_3dotswedge a_r+
$$
$$
+acdot a_3 a_1wedge a_2wedge a_4wedgedotswedge a_r+textrm{next term}.
$$
One of the factors in the next term above has a structure similar to the left hand side in (1.15) and you can iterate the above procedure. In the last step, you have $C_1$ which is a vector, and the inner product with $C_1$ is the usual dot product of vectors.
Here is how you proceed:
$$
-a_1wedge(acdot (a_2wedgedotswedge a_r))=
$$
$$
=-a_1wedge[(acdot a_2a_3wedgedotswedge a_r)-a_2wedge(acdot(a_3wedge a_4dotswedge a_r))]=
$$
$$
=-(acdot a_2)a_1wedge a_3wedgedotswedge a_r+a_1wedge a_2wedge(acdot(a_3wedge a_4dotswedge a_r))
$$
Observe that the "$cdot$" operation appears in two forms. On the first of the three lines above it stands for the inner product of vector with multivector, whereas on the first term in the line right below it is just a dot product of two vectors hence a scalar. Scalars can be pulled out of any type of product in geometric algebra and that is how you end up with the first term in the third line. You can deal with the second term in the last line exactly in the same way: you get a term containing a scalar $acdot a_3$ which can be pulled out and another term of the form $a_1wedge a_2wedge (a_3cdot(a_4wedge a_5dotswedge a_r))$ where $cdot$ stands again for vector by multivector as in the left hand side of (1.15) and you repeat the procedure to pull subsequent scalars.
answered Feb 3 at 4:07
GReyesGReyes
2,50815
$endgroup$
Let $b=a_1$ and $C_{r-1}=a_2wedge a_3wedgedotswedge a_r$. Then, from (1.15) it follows that
$$
acdot(a_1wedge a_2wedgedotswedge a_r)=acdot a_1 a_2wedgedotswedge a_r
$$
$$
-a_1wedge(acdot a_2wedgedotswedge a_r).
$$
Next you apply the same (1.15) to the second factor in the second piece above with $b=a_2$ and $C_{r-2}=a_3wedge a_4wedgedotswedge a_r$ to get
$$
=acdot a_1 a_2wedgedotswedge a_r-acdot a_2 a_1wedge a_3dotswedge a_r+
$$
$$
+acdot a_3 a_1wedge a_2wedge a_4wedgedotswedge a_r+textrm{next term}.
$$
One of the factors in the next term above has a structure similar to the left hand side in (1.15) and you can iterate the above procedure. In the last step, you have $C_1$ which is a vector, and the inner product with $C_1$ is the usual dot product of vectors.
Here is how you proceed:
$$
-a_1wedge(acdot (a_2wedgedotswedge a_r))=
$$
$$
=-a_1wedge[(acdot a_2a_3wedgedotswedge a_r)-a_2wedge(acdot(a_3wedge a_4dotswedge a_r))]=
$$
$$
=-(acdot a_2)a_1wedge a_3wedgedotswedge a_r+a_1wedge a_2wedge(acdot(a_3wedge a_4dotswedge a_r))
$$
Observe that the "$cdot$" operation appears in two forms. On the first of the three lines above it stands for the inner product of vector with multivector, whereas on the first term in the line right below it is just a dot product of two vectors hence a scalar. Scalars can be pulled out of any type of product in geometric algebra and that is how you end up with the first term in the third line. You can deal with the second term in the last line exactly in the same way: you get a term containing a scalar $acdot a_3$ which can be pulled out and another term of the form $a_1wedge a_2wedge (a_3cdot(a_4wedge a_5dotswedge a_r))$ where $cdot$ stands again for vector by multivector as in the left hand side of (1.15) and you repeat the procedure to pull subsequent scalars.
answered Feb 3 at 4:07
GReyesGReyes
2,50815
edited Feb 4 at 0:44
answered Feb 3 at 4:07
GReyesGReyes
2,50815
answered Feb 3 at 4:07
GReyesGReyes
2,50815
answered Feb 3 at 4:07
GReyesGReyes
2,50815
2,50815
$begingroup$
I have not problem applying (1.15) for the first time, but for the second time, there is an outer product with $a_1$, how can I get rid of that? I have edited edited my question to indicate my problem.
$endgroup$
– Eric
Feb 3 at 7:16
$begingroup$
@Eric I just improved my answer to include the details of the computations. Please let me know.
$endgroup$
– GReyes
Feb 4 at 0:44
$begingroup$
Thanks a lot for the added details, I didn't realized that I can pull out the scalar :D
$endgroup$
– Eric
Feb 4 at 6:51
add a comment |
$begingroup$
I have not problem applying (1.15) for the first time, but for the second time, there is an outer product with $a_1$, how can I get rid of that? I have edited edited my question to indicate my problem.
$endgroup$
– Eric
Feb 3 at 7:16
$begingroup$
@Eric I just improved my answer to include the details of the computations. Please let me know.
$endgroup$
– GReyes
Feb 4 at 0:44
$begingroup$
Thanks a lot for the added details, I didn't realized that I can pull out the scalar :D
$endgroup$
– Eric
Feb 4 at 6:51
$begingroup$
I have not problem applying (1.15) for the first time, but for the second time, there is an outer product with $a_1$, how can I get rid of that? I have edited edited my question to indicate my problem.
$endgroup$
– Eric
Feb 3 at 7:16
$begingroup$
I have not problem applying (1.15) for the first time, but for the second time, there is an outer product with $a_1$, how can I get rid of that? I have edited edited my question to indicate my problem.
$endgroup$
– Eric
Feb 3 at 7:16
$begingroup$
@Eric I just improved my answer to include the details of the computations. Please let me know.
$endgroup$
– GReyes
Feb 4 at 0:44
$begingroup$
@Eric I just improved my answer to include the details of the computations. Please let me know.
$endgroup$
– GReyes
Feb 4 at 0:44
$begingroup$
Thanks a lot for the added details, I didn't realized that I can pull out the scalar :D
$endgroup$
– Eric
Feb 4 at 6:51
$begingroup$
Thanks a lot for the added details, I didn't realized that I can pull out the scalar :D
$endgroup$
– Eric
Feb 4 at 6:51
add a comment |
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