A proof about tensor invariants

In summary: Now, the only thing left to show is that ##T^\ell_\ell = \mathfrak I^T_1##. This is straightforward to see by picking out the ##\vec u = \vec v = \vec w = \hat e_i## term, which gives ##T^i_i = \mathfrak I^T_1##.
  • #1
Van Ladmon
8
0

Homework Statement


How to proof the following property of tensor invariants?
Where:
##[\mathbf{a\; b\; c}]=\mathbf{a\cdot (b\times c)} ##,
##\mathbf{T} ##is a second order tensor,
##\mathfrak{J}_{1}^{T}##is its first invariant,
##\mathbf{u, v, w}## are vectors.

Homework Equations


$$[\mathbf{T\cdot u\; v\; w}]+[\mathbf{u\; T\cdot v\; w}]+[\mathbf{u\; v\; T\cdot w}]=\mathfrak{J}_{1}^{T}[\mathbf{u\; v\; w}]$$

The Attempt at a Solution


$$T^{l}{ }_{i}u^{i}v^{j}w^{k}\epsilon_{ljk}+T^{l}{ }_{j}u^{i}v^{j}w^{k}\epsilon_{ilk}+T^{l}{ }_{k}u^{i}v^{j}w^{k}\epsilon_{ijl}
$$$$=1/6(T^{l}{ }_{i}u^{i}v^{j}w^{k}\epsilon_{ljk}\epsilon_{\alpha \beta \gamma }\epsilon^{\alpha \beta \gamma }+T^{l}{ }_{i}u^{i}v^{j}w^{k}\epsilon_{ilk}\epsilon_{\alpha \beta \gamma }\epsilon^{\alpha \beta \gamma }+T^{l}{ }_{i}u^{i}v^{j}w^{k}\epsilon_{ijl}\epsilon_{\alpha \beta \gamma }\epsilon^{\alpha \beta \gamma })=?$$[/B]
 
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  • #2
Trying to blindly chop away with tensor calculus is not the most straightforward way here. I would rather suggest that you look at the properties of the left-hand side to draw conclusions on how it has to be expressed.

Alternatively you could write out some of the sums explicitly and see several terms cancel to finally collect everything.
 
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  • #3
Orodruin said:
Trying to blindly chop away with tensor calculus is not the most straightforward way here. I would rather suggest that you look at the properties of the left-hand side to draw conclusions on how it has to be expressed.

Alternatively you could write out some of the sums explicitly and see several terms cancel to finally collect everything.
Many thanks for the hint and I've worked out this problem, but is there a more concise and elegant way to give out a proof using tensor calculus?
 
  • #4
Van Ladmon said:
Many thanks for the hint and I've worked out this problem, but is there a more concise and elegant way to give out a proof using tensor calculus?
By forum rules, full solutions cannot be provided unless the OP (in this case you) have shown that they have solved the problem (i.e., have presented their solution), so I cannot answer that question until you post your solution.
 
  • #5
$$T_i^1 u^i v^2 w^3 + T_i^1 u^1 v^i w^2 + T_i^1 u^1 v^3 w^i=$$
\begin{equation*}
\left.\begin{aligned}
T_1^1 u^1 v^2 w^3 + T_1^1 u^3 v^1 w^2 + T_1^1 u^2 v^3 w^1 \\
T_2^1 u^2 v^2 w^3 + T_2^1 u^3 v^2 w^2 + T_2^1 u^2 v^3 w^2 \\
T_3^1 u^3 v^2 w^3 + T_3^1 u^3 v^3 w^2 + T_3^1 u^2 v^3 w^3 \\
\end{aligned}\right.
\end{equation*}
There're terms such as ##-T_1^1 u^1 v^3 w^2## which can be obtained by those terms by exchanging two indices once.
So terms like ##-T_2^1 u^2 v^3 w^2##, which is obtained by changing the indices of v and w in ##T_2^1 u^2 v^3 w^2## will cancel out with ##T_2^1 u^2 v^3 w^2##
and terms such as ##T_2^1 u^3 v^2 w^2## with repeated indices of u, v and w will cancel out by their own "negative" terms.
Thus the terms
\begin{equation*}
\left.\begin{aligned}
T_2^1 u^2 v^2 w^3 + T_2^1 u^3 v^2 w^2 + T_2^1 u^2 v^3 w^2 \\
T_3^1 u^3 v^2 w^3 + T_3^1 u^3 v^3 w^2 + T_3^1 u^2 v^3 w^3 \\
\end{aligned}\right.
\end{equation*}
will cancel out and the terms left are
T_1^1 u^1 v^2 w^3 + T_1^1 u^3 v^1 w^2 + T_1^1 u^2 v^3 w^1 - T_1^1 u^1 v^3 w^2 - T_1^1 u^3 v^2 w^1 - T_1^1 u^2 v^1 w^3 \\
That's exactly $$[T_1^1\left(u^i v^j w^k \varepsilon_{ijk}\right)$$
Thus $$\mathbf{T\cdot u\; v\; w}]+[\mathbf{u\; T\cdot v\; w}]+[\mathbf{u\; v\; T\cdot w}]=$$
\begin{equation*}
\left.\begin{aligned}
T_i^1 u^i v^2 w^3 + T_i^1 u^1 v^i w^2 + T_i^1 u^1 v^3 w^i= \\
T_i^2 u^i v^2 w^3 + T_i^2 u^1 v^i w^2 + T_i^2 u^1 v^3 w^i= \\
T_i^3 u^i v^2 w^3 + T_i^3 u^1 v^i w^2 + T_i^3 u^1 v^3 w^i=
\end{aligned}\right.
\end{equation*}
$$=\left(T_1^1+T_2^2+T_3^3\right)\left(u^i v^j w^k \varepsilon_{ijk}\right)=
\mathfrak{J}_{1}^{T}[\mathbf{u\; v\; w}]
$$
 
  • #6
Ok, so the easier argument is to note that your entire expression is invariant and completely anti-symmetric under exchange of the vectors ##\vec u##, ##\vec v##, and ##\vec w##. Due to this, it must be on the form ##S_{ijk} u^i v^j w^k## where ##S_{ijk}## is a completely anti-symmetric tensor. Due to this, ##S_{ijk} = \alpha \epsilon_{ijk}## for some scalar ##\alpha##. Writing down the expression for ##S## leads to
$$
\alpha \epsilon_{ijk} = T_i^\ell \epsilon_{\ell jk} + T_j^\ell \epsilon_{i\ell k} + T_k^\ell \epsilon_{ij\ell}
$$
and contraction with ##\epsilon^{ijk}## now directly leads to
$$
6\alpha = 6T^\ell_\ell,
$$
i.e., ##S_{ijk} = T^\ell_\ell \epsilon_{ijk}##, and therefore
$$
S_{ijk}u^i v^j w^k = T^\ell_\ell \epsilon_{ijk} u^i v^j w^k = \mathfrak I^T_1 [\vec u \vec v \vec w].
$$
 

Related to A proof about tensor invariants

1. What are tensor invariants?

Tensor invariants are mathematical quantities that remain unchanged under specific transformations of a tensor. They are used to characterize the properties of a tensor and are important in various fields of science and engineering, such as physics, mechanics, and computer vision.

2. How are tensor invariants used?

Tensor invariants are used to simplify equations and expressions involving tensors. They help in reducing the complexity of tensor operations and make it easier to analyze and understand the properties of a tensor.

3. What is the significance of proving tensor invariants?

Proving tensor invariants is important because it provides a better understanding of the underlying mathematical principles and properties of tensors. It also helps in developing more efficient algorithms and techniques for working with tensors.

4. Are tensor invariants unique to a particular tensor?

No, tensor invariants are not unique to a particular tensor. They are intrinsic properties of a tensor and remain the same regardless of the coordinate system or basis used to represent the tensor.

5. Can tensor invariants be used in real-world applications?

Yes, tensor invariants have various real-world applications, such as in image and signal processing, machine learning, and structural engineering. They are also used in the study of fluid mechanics, electromagnetism, and general relativity.

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