# Derivative of a tensor

#### JohanL

the derivative of a tensor

$$a_{ij}x^ix^j$$

with respect to $$x^k$$, k=2 and i,j = 1,2,3.

solution:

$$\frac {d} {dx^k}a_{ij}x^ix^j = a_{ij}\frac {dx^i} {dx^k}x^j + a_{ij}x^i \frac {dx^j} {dx^k} = a_{2j}x^j + a_{i2}x^i = a_{21}x^1 + a_{22}x^2 + a_{23}x^3 + a_{12}x^1 + a_{22}x^2 + a_{32}x^3$$

is that correct?

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#### dextercioby

Homework Helper
That's not a tensor,but a scalar.

Are u familiar with the delta-Kronecker invariant tensor ??

Daniel.

#### JohanL

My misstake...the problem was to calculate the derivative of that _expression_...not tensor.
Did i do it correct?
yes...i am familiar with the delta-Kronecker invariant tensor.

#### dextercioby

Homework Helper
$$\frac{\partial x^{i}}{\partial x^{k}} =...?$$

If you know that,u could just apply the Leibniz rule and then the summation convention.

Daniel.

#### HallsofIvy

Homework Helper
Are you given that the components of the tensor aij are constant? You seem to be assuming that.

#### JohanL

yes the components of $$a_{ij}$$ is constant.

i used that

$$\frac{\partial x^{i}}{\partial x^{k}} = \delta^i{}_{k}$$

and Leibnitz rule and the summation convention.

#### robphy

Homework Helper
Gold Member
Note that
$$\frac {d} {dx^k}a_{ij}x^ix^j$$ has two "dummy-indices [involved in summation]", namely, i and j, and one "free-index [not involved in summation]", k. So, the right-hand side must also have the free-index k.

#### dextercioby

Homework Helper
That's actually $\delta_{i}^{k}$,because it is a symmetric invariant tensor.

Daniel.

#### robphy

Homework Helper
Gold Member
dextercioby said:
That's actually $\delta_{i}^{k}$,because it is a symmetric invariant tensor.

Daniel.
I think the indices are flipped. They should be
$$\displaystyle\frac{\partial x^{i}}{\partial x^{k}} = \delta_{k}^{i}$$
... but it's a good idea to preserve the "slots" and write $\delta_{k}{}^{i}$ or $\delta^{i}{}_{k}$.

#### JohanL

robphy said:
Note that
$$\frac {d} {dx^k}a_{ij}x^ix^j$$ has two "dummy-indices [involved in summation]", namely, i and j, and one "free-index [not involved in summation]", k. So, the right-hand side must also have the free-index k.
k=2...but maybe that doesnt matter.

#### robphy

Homework Helper
Gold Member
Ok, I see now that it's a issue of clarity of presentation.
You are correct, but I would have written your initial post as

\begin{align*} \frac {d} {dx^k}a_{ij}x^ix^j &= a_{ij}\frac {dx^i} {dx^k}x^j + a_{ij}x^i \frac {dx^j} {dx^k} \\ &=a_{ij} \delta^i{}_k x^j + a_{ij}x^i \delta^j{}_k\\ &=a_{kj} x^j + a_{ik}x^i \mbox{ which is a co-vector.} \\ \intertext{Now, with k=2,} \frac {d} {dx^2}a_{ij}x^ix^j &= a_{2j}x^j + a_{i2}x^i \\&= a_{21}x^1 + a_{22}x^2 + a_{23}x^3 + a_{12}x^1 + a_{22}x^2 + a_{32}x^3 \mbox{ which is the 2-component of that co-vector} \end{align*}

#### dextercioby

Homework Helper
Incidentally that matrix "a" is symmetrical (or if it isn't,the contracted tensor product selects the symmetrical part),so in the final result there should be a 2...

Daniel.

#### robphy

Homework Helper
Gold Member
dextercioby said:
Incidentally that matrix "a" is symmetrical (or if it isn't,the contracted tensor product selects the symmetrical part),so in the final result there should be a 2...

Daniel.
I think what you mean is this:

\begin{align*}\frac {d} {dx^k}a_{ij}x^ix^j &=a_{ij}\frac {dx^i} {dx^k}x^j + a_{ij}x^i \frac {dx^j} {dx^k} \\ &=a_{ij} \delta^i{}_k x^j + a_{ij}x^i \delta^j{}_k\\ &=a_{kj} x^j + a_{ik}x^i\\ &=a_{k\color{red}{i}} x^{\color{red}{i}} + a_{ik}x^i\\ &=\left( a_{ki} + a_{ik} \right) x^i\\ &=\left( 2 a_{ik}^S \right) x^i = 2 a_{ik}^Sx^i \\ \end{align*}
So, with k=2,
\begin{align*} \frac {d} {dx^2}a_{ij}x^ix^j &= 2 a_{i2}^Sx^i = 2\left( a_{12}^Sx^1 + a_{22}^Sx^2 + a_{32}^Sx^3\right) \end{align*}
which is equal to what is in the original post.