Derivative of Symmetric Tensor with Respect to a Component

In summary, the derivative of a tensor with respect to a specific component can be calculated using the Leibniz rule and the summation convention. If the tensor is symmetric, then the contracted tensor product is used to select the symmetrical part, resulting in a coefficient of 2 for the final answer.
  • #1
JohanL
158
0
the derivative of a tensor

[tex]
a_{ij}x^ix^j
[/tex]

with respect to [tex]x^k[/tex], k=2 and i,j = 1,2,3.

solution:

[tex]
\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

[/tex]

is that correct?
 
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  • #2
That's not a tensor,but a scalar.

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

Daniel.
 
  • #3
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.
 
  • #4
[tex] \frac{\partial x^{i}}{\partial x^{k}} =...? [/tex]

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

Daniel.
 
  • #5
Are you given that the components of the tensor aij are constant? You seem to be assuming that.
 
  • #6
yes the components of [tex]a_{ij}[/tex] is constant.

i used that

[tex] \frac{\partial x^{i}}{\partial x^{k}} = \delta^i{}_{k}[/tex]

and Leibnitz rule and the summation convention.

i thought i had the right answer?
 
  • #7
Note that
[tex]\frac {d} {dx^k}a_{ij}x^ix^j [/tex] 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.
 
  • #8
That's actually [itex] \delta_{i}^{k} [/itex],because it is a symmetric invariant tensor.

Daniel.
 
  • #9
dextercioby said:
That's actually [itex] \delta_{i}^{k} [/itex],because it is a symmetric invariant tensor.

Daniel.

I think the indices are flipped. They should be
[tex]\displaystyle\frac{\partial x^{i}}{\partial x^{k}} = \delta_{k}^{i} [/tex]
... but it's a good idea to preserve the "slots" and write [itex] \delta_{k}{}^{i} [/itex] or [itex] \delta^{i}{}_{k} [/itex].
 
  • #10
robphy said:
Note that
[tex]\frac {d} {dx^k}a_{ij}x^ix^j [/tex] 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.
 
  • #11
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

[tex]
\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*}
[/tex]
 
  • #12
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.
 
  • #13
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:

[tex]\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*}[/tex]
So, with k=2,
[tex]\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*}[/tex]
which is equal to what is in the original post.
 

What is a tensor?

A tensor is a mathematical object that represents a physical quantity or transformation. It is a multidimensional array of numbers that follow certain transformation rules under coordinate transformations.

What is the derivative of a tensor?

The derivative of a tensor is a mathematical operation that produces a new tensor by calculating the rate of change of the original tensor with respect to one or more variables. It is used to measure how a tensor changes as its inputs change.

What is the significance of the derivative of a tensor in science?

The derivative of a tensor is essential in understanding the behavior of physical systems and predicting their future states. It is used in many fields, including physics, engineering, and machine learning, to model and analyze complex systems.

What are the different types of derivatives of a tensor?

There are several types of derivatives of a tensor, including the directional derivative, the covariant derivative, and the Lie derivative. Each type has its own specific use and is calculated differently.

How is the derivative of a tensor calculated?

The derivative of a tensor is calculated using the chain rule and the product rule, similar to the derivative of a function. However, since tensors have multiple indices, the calculation can become more complex and may require the use of tensor calculus and differential geometry.

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