Derivative of Symmetric Tensor with Respect to a Component

[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:

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

is that correct?

 

[dextercioby]

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]

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

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

Daniel.

 

[HallsofIvy]

Are you given that the components of the tensor a_ij 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.

i thought i had the right answer?

 

[robphy]

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]

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

Daniel.

 

[robphy]

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]

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 doesn't matter.

 

[robphy]

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

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

 

[dextercioby]

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]

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:

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