atlas[d] - exterior differentiation 

Calling Sequence: 

    d(expr) 

Parameters: 

    expr - any expression. 

Description: 

• The d - procedure allows one to calculate the exterior derivative on an expression that is p-form.

 

• The exterior derivative is the operator  d : ->  where is p-form and is (p+1)-form. The operator has the following properties.

 

• For any 0-form   we have:   

 

• For any p-forms and constants we have:   

 

• If is p-form and is q-form then exterior derivative of their exterior product is as follows:

 

• For any p-form Poincare's lemma takes place:   

 

Examples: 

>	restart: with(atlas):

 

Declare constants:  

>	Constants(alpha,beta,p,q);

 

(2.1)

 

Declare functions:  

>	Functions(F=F(x[1],x[2],x[3]),f=f(x[k]),G=G(x,z),z=z(phi),x[k]=x[k](y[j]));

 

(2.2)

 

Declare p-forms:  

>	Forms(omega=n,sigma=m,Omega=P,omega[1]=p,omega[2]=q);

 

(2.3)

 

Using d- procedure: 

f- declared as a function with indefinite number of variables:
'd'(f)=d(f);
 

(2.4)

 

F- declared as function on three variables:
'd'(F)=d(F);
 

(2.5)

 

G- declared as function G=G(x,z) and z=z(phi) thus:
'd'(G)=d(G); 

(2.6)

 

As declared as constant thus:
'd'(alpha)=d(alpha); 

(2.7)

 

Exterior derivative is linear operation:
'd'(alpha*omega+beta*sigma)=d(alpha*omega+beta*sigma); 

(2.8)

 

There were not any declarations about x and y so they are 0-forms by defaults:
'd'(x*y)=d(x*y); 

(2.9)

 

Verify that double-d gives 0 (Poincare's lemma):
'd'(d(Omega))=d(d(Omega)); 

(2.10)

 

As y is 0-form (by defaults) and has been declared as P-form then:
'd'(y*Omega)=d(y*Omega); 

(2.11)

 

and - declared as p- and q-forms respectively:
'd'(omega[1]&^omega[2])=d(omega[1]&^omega[2]); 

(2.12)

 

>

 

See Also: & 

atlas, atlas[Constants], atlas[Functions], atlas[Forms], atlas[`&^`].