Continuum mechanics/Curl of a gradient of a vector

From Wikiversity
Jump to navigation Jump to search

Curl of the gradient of a vector - 1[edit | edit source]

Let be a vector field. Show that

Proof:

For a second order tensor field , we can define the curl as

where is an arbitrary constant vector. Substituting into the definition, we have

Since is constant, we may write

where is a scalar. Hence,

Since the curl of the gradient of a scalar field is zero (recall potential theory), we have

Hence,

The arbitrary nature of gives us


Curl of the transpose of the gradient of a vector[edit | edit source]

Let be a vector field. Show that

Proof:

The curl of a second order tensor field is defined as

where is an arbitrary constant vector. If we write the right hand side in index notation with respect to a Cartesian basis, we have

and

In the above a quantity represents the -th component of a vector, and the quantity represents the -th components of a second-order tensor.

Therefore, in index notation, the curl of a second-order tensor can be expressed as

Using the above definition, we get

If , we have

Therefore,