atlas[L] - Lie derivative

Calling Sequence:

     L[V1, V2, V3..., Vn](expr)
     L(V1, V2, V3..., Vn, expr)
     atlas[L](V1, V2, V3..., Vn, expr)

Parameters:

         expr - any expression (on which Lie derivative is defined) .
         V1, V2, V3..., Vn  - vector fields.

Description:

• The L  - procedure calculates the Lie derivative of an expression along given vector field. The derivative has the following properties.
• For any vector fields  and expression  we have:
• For any vector field X and 0-form  we have:   
• For vector fields X and Y we have:   
• For any vector field X and tensor fields  and  the Leibniz rule for Lie derivative takes place:
• For any vector field X and p-form  we have:

Examples:
restart:
with(atlas):

Declare constants:
Constants(Lambda);

Declare functions:
Functions(F=F(x,y));

Declare p-forms:
Forms(omega=p);

Declare vectors:
Vectors(X,Y,Z);

Declare tensors:
Tensors(T=[n,k],Omega=[l,m]);

Using L- procedure:

Just definition for "long" Lie operator:
'L[X,Y,Z](T)'=L[X,Y,Z](T);
'L(X,Y,Z,T)'=L(X,Y,Z,T);
'L[X](Y,Z,T)'=L[X](Y,Z,T);

As h is 0-form by defaults then:
'L[X](h)'=L[X](h);

F- declared as function F=F(x,y) thus:
'L[X]'(F)=L[X](F);

As  declared as constant thus:
'L[X]'(Lambda)=L[X](Lambda);

Lie derivative is linear with respect to any argument:
'L[X]'(Y+Z)=L[X](Y+Z);

And:
'L[X+Y]'(Z)=L[X+Y](Z);

More complex examples:
'L[f*X+h*Y]'(Z)=L[f*X+h*Y](Z);

And:
'L[Z]'(f*X+h*Y)=L[Z](f*X+h*Y);

 - declared as p-form thus:
'L[X]'(omega)=L[X](omega);

Verify that Leibniz rule takes place for tensor product:
'L[X]'(T&.Omega)=L[X](T&.Omega);

See Also:

atlas , atlas[d] , atlas[cov] , atlas[`&.`] , atlas[`&^`] , atlas[iota] .