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Derivative of a tensor

  • Thread starter JohanL
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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?
 

dextercioby

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That's not a tensor,but a scalar.

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

Daniel.
 
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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

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[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.
 

HallsofIvy

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

robphy

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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.
 

dextercioby

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

Daniel.
 

robphy

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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].
 
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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 doesnt matter.
 

robphy

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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]
 

dextercioby

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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

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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.
 

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