- #1

- 135

- 4

_{mn}= ∂V

_{m}/∂x

^{n}where V is a covariant vector. I am having a mental block with regard to the indeces. How is it that the derivative of a covariant vector (1st order tensor) be consider to be a 2nd order tensor?

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter nigelscott
- Start date

- #1

- 135

- 4

- #2

- 9,979

- 1,155

Does this help?

[tex]

T_{mn} = \left[ \begin{array}{cc} \frac{\partial V_0}{\partial x^0} & \frac{\partial V_0}{\partial x^1} & \frac{\partial V_0}{\partial x^2} & \frac{\partial V_0}{\partial x^3} \\

\frac{\partial V_1}{\partial x^0} & \frac{\partial V_1}{\partial x^1} & \frac{\partial V_1}{\partial x^2} & \frac{\partial V_1}{\partial x^3} \\

\frac{\partial V_2}{\partial x^0} & \frac{\partial V_2}{\partial x^1} & \frac{\partial V_2}{\partial x^2} & \frac{\partial V_2}{\partial x^3} \\

\frac{\partial V_3}{\partial x^0} & \frac{\partial V_3}{\partial x^1} & \frac{\partial V_3}{\partial x^2} & \frac{\partial V_3}{\partial x^3} \end{array} \right]

[/tex]

[tex]

T_{mn} = \left[ \begin{array}{cc} \frac{\partial V_0}{\partial x^0} & \frac{\partial V_0}{\partial x^1} & \frac{\partial V_0}{\partial x^2} & \frac{\partial V_0}{\partial x^3} \\

\frac{\partial V_1}{\partial x^0} & \frac{\partial V_1}{\partial x^1} & \frac{\partial V_1}{\partial x^2} & \frac{\partial V_1}{\partial x^3} \\

\frac{\partial V_2}{\partial x^0} & \frac{\partial V_2}{\partial x^1} & \frac{\partial V_2}{\partial x^2} & \frac{\partial V_2}{\partial x^3} \\

\frac{\partial V_3}{\partial x^0} & \frac{\partial V_3}{\partial x^1} & \frac{\partial V_3}{\partial x^2} & \frac{\partial V_3}{\partial x^3} \end{array} \right]

[/tex]

Last edited:

- #3

Nugatory

Mentor

- 13,316

- 6,242

_{mn}= ∂V_{m}/∂x^{n}where V is a covariant vector. I am having a mental block with regard to the indeces. How is it that the derivative of a covariant vector (1st order tensor) be consider to be a 2nd order tensor?

It's not. The definition is saying that each of the 16 components of this rank-2 tensor can be calculated in a given coordinate system by applying this rule for computing them. That works fine because in that coordinate system each of the V

- #4

Chestermiller

Mentor

- 20,977

- 4,605

_{mn}= ∂V_{m}/∂x^{n}where V is a covariant vector. I am having a mental block with regard to the indeces. How is it that the derivative of a covariant vector (1st order tensor) be consider to be a 2nd order tensor?

The partial derivative of a vector wrt spatial position is not a tensor, it's a vector. In curvilinear coordinates, the coordinate directions change with spatial position. Unit vectors (or coordinate basis vectors) therefore change with spatial position. Since a vector can be expressed as a linear sum of components times unit vectors (or coordinate basis vectors), if you are taking the partial derivative of a vector wrt to spatial position, you need to account for the changes in direction of the unit vectors (or coordinate basis vectors) as well as the changes in magnitude of the coordinate basis vectors. This is where the christoffel symbols come in. While the partial derivative of a vector wrt spatial position is not a tensor, the gradient vector operator applied to a vector yields a second order tensor.

- #5

PAllen

Science Advisor

- 8,280

- 1,541

The partial derivative of a vector wrt spatial position is not a tensor, it's a vector. In curvilinear coordinates, the coordinate directions change with spatial position. Unit vectors (or coordinate basis vectors) therefore change with spatial position. Since a vector can be expressed as a linear sum of components times unit vectors (or coordinate basis vectors), if you are taking the partial derivative of a vector wrt to spatial position, you need to account for the changes in direction of the unit vectors (or coordinate basis vectors) as well as the changes in magnitude of the coordinate basis vectors. This is where the christoffel symbols come in. While the partial derivative of a vector wrt spatial position is not a tensor, the gradient vector operator applied to a vector yields a second order tensor.

In general manifolds or general coordinates, the partial of a vector field with respect coordinates is a 'nothing'. The covariant derivative with respect to coordinates is rank two tensor. The covariant derivative of one vector field with respect to another vector field is yet another vector field.

In coordinates on a flat manifold chosen such that the connection vanishes globally, partial derivative of vector by coordinates

- #6

- 9,979

- 1,155

The partial derivative of a vector wrt spatial position is not a tensor, it's a vector.

True, but irrelevant. [itex]\nabla_a[/itex], the covariant derivative operator, is a different notation for the gradient operator [itex]\nabla[/itex]

The major difference is the added subscript - this makes it clearer that the covariant derivative operator [itex]nabla_a[/itex] adds one rank to any tensor it operates on.

The gradient of a vector field is a rank 2 tensor aka a matrix, sometimes called the Jacobian matrix, see for instance http://en.wikipedia.org/w/index.php?title=Gradient&oldid=508417646#Gradient_of_a_vector

For a specific gravity related example, consider the "tidal tensor". The tidal tensor is useful for getting the tidal stress on a rigid body, for instance.

http://en.wikipedia.org/w/index.php?title=Tidal_tensor&oldid=332450104

Thus if U is a scalar valued potential function , the Newtonian potential one can write

[tex] F_b = \nabla_b U[/tex]

[tex] T_{ab} = \nabla_a F_b = \nabla_a \nabla_b U [/tex]

where T is the tidal tensor.

Note that

[itex]\nabla_a = \partial_a[/itex] only in Cartesian coordinates.

- #7

- 834

- 2

Indeed, in this case, there is a well-defined (though not covariant) linear operator mapping a vector to a vector [itex]\underline T(a) = \partial [V(x) \cdot a][/itex] where [itex]\partial[/itex] is the vector partial derivative operator. This operator has components [itex]\underline T(e_m) \cdot e_n \equiv T_{mn}[/itex]. It should be clear that to fully describe this linear operator, there are two free vectors (or, just using the basis, there are two indices).

Edit: for exactly the reasons described, though, I wouldn't call this object a tensor because it does not obey the tensor transformation laws. But that's not a difficult fix, either. Just replace [itex]\partial[/itex] with [itex]\nabla[/itex], the covariant derivative.

Edit: for exactly the reasons described, though, I wouldn't call this object a tensor because it does not obey the tensor transformation laws. But that's not a difficult fix, either. Just replace [itex]\partial[/itex] with [itex]\nabla[/itex], the covariant derivative.

Last edited:

- #8

- 135

- 4

Share: