GrassmannCalculus`
ExpandAndSimplifyScalars  |
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Details and Options
Examples
(1)
Basic Examples
(1)
In[1]:=
<<GrassmannCalculus`
Set the book default to establish scalars and vectors.
In[2]:=
★A;
ExpandScalars
SimplifyScalars
ExpandAndSimplifyScalars
In[3]:=
X=(a+b)(x+a⋀b)X//
%//
X//
 | ExpandScalars |
| SimplifyScalars |
| ExpandAndSimplifyScalars |
Out[3]=
(a+b)(x+a⋀b)
Out[3]=
(a+b)x+(a+b)a⋀b
Out[3]=
ab(a+b)+(a+b)x
Out[3]=
ab(a+b)+(a+b)x
Here is a more complex expression. distributes the scalar expression over the sum (which is not scalar).
ExpandScalars
(a+b+c)
x⋀x+(2⋀f⋀-c⋀-a⋀f⋀a)
In[4]:=
1+(a+b+c)(x⋀x+(2⋀f⋀-c⋀-a⋀f⋀a))X=
[%]
 | ExpandScalars |
Out[4]=
1+(a+b+c)(x⋀x+2⋀f⋀-c⋀-a⋀f⋀a)
Out[4]=
1+(a+b+c)x⋀x+(a+b+c)2⋀f⋀-c⋀-a⋀f⋀a
SimplifyScalars
(2⋀f⋀-c⋀-a⋀f⋀a)
In[5]:=
| SimplifyScalars |
Out[5]=
1+2c(a+b+c)+(a+b+c)x⋀x
2
a
2
f
ExpandAndSimplifyScalars
x⋀x
★
SimplifyExteriorProducts
In[6]:=
1+(a+b+c)(x⋀x+(2⋀f⋀-c⋀-a⋀f⋀a))
[%]%//
| ExpandAndSimplifyScalars |
| ★ |
Out[6]=
1+(a+b+c)(x⋀x+2⋀f⋀-c⋀-a⋀f⋀a)
Out[6]=
1+2c(a+b+c)+(a+b+c)x⋀x
2
a
2
f
Out[6]=
1+2c(a+b+c)
2
a
2
f
You can also use new symbols as long as you assert their grades, or you can override the grades of currently declared symbols. Here we assert that is of grade 0, making the exterior product with itself non-zero. Here, is a new symbol.
x
A
In[7]:=
1+(a+b+c)(x⋀x+(2⋀f⋀-c⋀-A⋀f⋀A))
%,{A,x}∈
| ExpandAndSimplifyScalars |
★Λ
0
Out[7]=
1+(a+b+c)(x⋀x+2⋀f⋀-c⋀-A⋀f⋀A)
Out[7]=
1+2c(a+b+c)+(a+b+c)
2
A
2
f
2
x
In[8]:=
Clear[X]
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""