# I Component derivative of a tensor

1. Jan 11, 2019

### dsaun777

This is a simple and maybe stupid question. Can you take a derivative of a vector component with respect to a vector? Or even more generally,can you take the derivative of a component of a tensor with respect to the whole tensor? For instance in the cauchy tensor could you take the xx component of the tensor and compare it to the rest of the components of the tensor by taking the derivative of xx component with respect to the rest of the tensor. I know you can take deriviatives of tensors, but can you take derivatives of components treating the rest of the components like dependent variables?

2. Jan 11, 2019

### andrewkirk

Yes you can, but it might not be meaningful. Given a point P on a differentiable manifold, a vector V in the tangent space at P, and a tensor field T and coordinate system C on a neighbourhood N of P, each of the components of T in coordinate system C is a scalar field on N. Each of those scalar fields has a directional derivative at P in direction V.

However the directional derivative will not itself be a component of a proper tensor field because it is dependent on the coordinates used, and ignores Christoffel symbols.

A tensor field also has a coordinate-independent directional derivative, which is a proper tensor field. In general the components of that will not be the same as what one gets when one simply differentiates the component field in a particular coordinate system.

3. Jan 12, 2019

### dsaun777

So using the coordinate-independent directional derivative one could proceed with taking non-trivial derivatives of the directional derivative with respect to the original tensor?

4. Jan 15, 2019

### Orodruin

Staff Emeritus
You do not take derivatives with respect to tensors.

5. Jan 15, 2019

### dsaun777

Why not? Using the appropriate directional derivatives you can take a derivative of a scalar with respect to a vector why not a derivative of a scalar with respect to a 2nd rank tensor.

6. Jan 15, 2019

### Orodruin

Staff Emeritus
Because derivatives are based on changes from one point to a nearby point (defined by some connection). This is inherently described by a displacement, which is described by a vector in the tangent space, not a tensor.

7. Jan 15, 2019

### dsaun777

You can't take a derivative of a scalar field with respect to a tensor field?

8. Jan 15, 2019

### Orodruin

Staff Emeritus
No.

9. Jan 15, 2019

### dsaun777

Matrix-by-scalarEdit
The derivative of a matrix function Y by a scalar x is known as the tangent matrix and is given (in numerator layout notation) by

{\displaystyle {\frac {\partial \mathbf {Y} }{\partial x}}={\begin{bmatrix}{\frac {\partial y_{11}}{\partial x}}&{\frac {\partial y_{12}}{\partial x}}&\cdots &{\frac {\partial y_{1n}}{\partial x}}\\{\frac {\partial y_{21}}{\partial x}}&{\frac {\partial y_{22}}{\partial x}}&\cdots &{\frac {\partial y_{2n}}{\partial x}}\\\vdots &\vdots &\ddots &\vdots \\{\frac {\partial y_{m1}}{\partial x}}&{\frac {\partial y_{m2}}{\partial x}}&\cdots &{\frac {\partial y_{mn}}{\partial x}}\\\end{bmatrix}}.}
Scalar-by-matrixEdit
The derivative of a scalar y function of a p×q matrix X of independent variables, with respect to the matrix X, is given (in numerator layout notation) by

{\displaystyle {\frac {\partial y}{\partial \mathbf {X} }}={\begin{bmatrix}{\frac {\partial y}{\partial x_{11}}}&{\frac {\partial y}{\partial x_{21}}}&\cdots &{\frac {\partial y}{\partial x_{p1}}}\\{\frac {\partial y}{\partial x_{12}}}&{\frac {\partial y}{\partial x_{22}}}&\cdots &{\frac {\partial y}{\partial x_{p2}}}\\\vdots &\vdots &\ddots &\vdots \\{\frac {\partial y}{\partial x_{1q}}}&{\frac {\partial y}{\partial x_{2q}}}&\cdots &{\frac {\partial y}{\partial x_{pq}}}\\\end{bmatrix}}.}
Important examples of scalar functions of matrices include the trace of a matrix and the determinant.

In analog with vector calculus this derivative is often written as the following.

{\displaystyle \nabla _{\mathbf {X} }y(\mathbf {X} )={\frac {\partial y(\mathbf {X} )}{\partial \mathbf {X} }}}
Also in analog with vector calculus, the directional derivative of a scalar f(X) of a matrix X in the direction of matrix Y is given by

{\displaystyle \nabla _{\mathbf {Y} }f=\operatorname {tr} \left({\frac {\partial f}{\partial \mathbf {X} }}\mathbf {Y} \right).}
It is the gradient matrix, in particular, that finds many uses in minimization problems in estimation theory, particularly in the derivation of the Kalman filter algorithm, which is of great importance in the field.

10. Jan 16, 2019

### Orodruin

Staff Emeritus
That has nothing to do with ”differentiate with respect to a tensor”.

Edit: To clarify that. Of course you can differentiate a function with respect to the components of a tensor or a vector. That is just usual differentiation. However, this has nothing to do with directional derivatives in the direction of a vector. For example, the directional derivative of $\phi$ in the direction $V$ is given by
$$\nabla_V \phi = V^a \partial_a \phi$$
and is a scalar whereas the derivative of $\phi$ with respect to the components of $V$ are given by
$$\frac{\partial \phi}{\partial V^a},$$
which is not.

Also, just inserting Wikipedia text without giving any source is presenting it as your own work. This is completely unacceptable. The proper way of doing it is to give a reference. I also note that you have not written a single word yourself in your post. That's just not constructive.

Last edited: Jan 16, 2019
11. Jan 16, 2019

### dsaun777

I'm not writing a dissertation I'm just looking for truth concisely. I appreciate your response. My question should have been clarified more, I meant taking derivative of a scalar with respect to every tensor component sorry.