volume: The volume element

In stokes: The Exterior Calculus

The volume element

Description

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The volume element in \mjseqnn dimensions

Usage

volume(n)
is.volume(K,n=dovs(K))

Arguments

n	Dimension of the space
K	Object of class kform

Details

Spivak phrases it well (theorem 4.6, page 82):

If \mjseqnV has dimension \mjseqnn, it follows that \mjeqn\Lambda^n(V). has dimension 1. Thus all alternating \mjseqnn-tensors on \mjseqnV are multiples of any non-zero one. Since the determinant is an example of such a member of \mjeqn\Lambda^n(V). it is not surprising to find it in the following theorem:

Let \mjeqnv_1,...,v_nv_1,...,v_n be a basis for \mjseqnV and let \mjeqn\omega\in\Lambda^n(V).. If \mjeqnw_i=\sum_j=1^n a_ijv_j. then

\mjdeqn \omega\left

(w_1,...,w_n\right)=\det\left(a_ij\right)\cdot\omega\left(v_1,... v_n\right)omitted; see PDF

(see the examples for numerical verification of this).

Neither the zero \mjseqnk-form, nor scalars, are considered to be a volume element.

Value

Function volume() returns an object of class kform; function is.volume() returns a Boolean.

Author(s)

Robin K. S. Hankin

References

• M. Spivak 1971. Calculus on manifolds, Addison-Wesley

See Also

zeroform,as.1form,dovs

Examples

dx^dy^dz == volume(3) 

p <- 1
for(i in 1:7){p <- p ^ as.kform(i)}
p
p == volume(7)  # should be TRUE

o <- volume(5)
M <- matrix(runif(25),5,5)
det(M) - as.function(o)(M)   # should be zero


is.volume(d(1) ^ d(2) ^ d(3) ^ d(4))
is.volume(d(1:9))

stokes documentation built on Aug. 19, 2023, 1:07 a.m.