If \(U^{i}\) is a contravariant vector and \(V_{j}\) is a covariant vector, show that \(U^{i} V_{j}\) is a \(2^{\text {nd }}\) -rank mixed tensor. Hint: Write the transformation equations for \(\mathbf{U}\) and \(\mathbf{V}\) and multiply them.

Short Answer

Expert verified
By verifying the transformation properties, \(U^{i} V_{j}\) is shown to be a 2nd-rank mixed tensor.

Step by step solution

01

- Understand Contravariant and Covariant Vectors

Contravariant vectors transform with the inverse of the Jacobian matrix, while covariant vectors transform with the Jacobian matrix. So, if \(U^{i}\) transforms as \(U'^{i} = \frac{\text{d}x'^i}{\text{d}x^j} U^{j}\), and \(V_{j}\) transforms as \(V'_{j} = \frac{\text{d}x^i}{\text{d}x'^j} V_{i}\), remember these formulas for next steps.
02

- Write the Transformation Equations

Using transformation rules, express the transformed components: \(U'^{i} = \frac{\text{d}x'^i}{\text{d}x^k} U^{k}\) and \(V'_{j} = \frac{\text{d}x^l}{\text{d}x'^j} V_{l}\). We will use these transformation laws to find how \(U^{i} V_{j}\) transforms.
03

- Multiply the Transformation Equations

Now, multiply the transformation equations: \(U'^{i} V'_{j} = \left(\frac{\text{d}x'^i}{\text{d}x^k} U^{k}\right) \left(\frac{\text{d}x^l}{\text{d}x'^j} V_{l}\right)\) which simplifies to \(U'^{i} V'_{j} = \frac{\text{d}x'^i}{\text{d}x^k} \frac{\text{d}x^l}{\text{d}x'^j} U^{k} V_{l}\).
04

- Identify Transformation of the Mixed Tensor

Observe that \(U'^{i} V'_{j}\) transforms as a 2nd-rank mixed tensor, because it uses a transformation property involving both contravariant and covariant components, \(\frac{\text{d}x'^i}{\text{d}x^k}\) and \(\frac{\text{d}x^l}{\text{d}x'^j}\).
05

- Conclusion

Hence, \(U^{i} V_{j}\) indeed satisfies the condition for being a 2nd-rank mixed tensor since its transformation includes the product of transformation matrices for both contravariant and covariant indices.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Contravariant Vector
Contravariant vectors are essential objects in tensor analysis. They are vectors that transform in a specific way under a change of coordinates often represented as transformations using the inverse of Jacobian matrices.
When a coordinate system changes from \(x\) to \(x'\), a contravariant vector \(U^i\) transforms according to the following rule: \[U'^i = \frac{\text{d}x'^i}{\text{d}x^j} U^j \] Here, \(\frac{\text{d}x'^i}{\text{d}x^j}\) signifies the elements of the inverse Jacobian matrix that handles the coordinate transformation.
It's essential to note that contravariant vectors are often represented with upper indices. These upper indices indicate their nature and the way they transform.
Covariant Vector
Covariant vectors are another cornerstone in the realm of tensors, differing from contravariant vectors in their transformation behavior. Covariant vectors transform with the Jacobian matrix instead of its inverse.
For a coordinate transformation from \(x\) to \(x'\), a covariant vector \(V_j\) transforms as: \[V'_j = \frac{\text{d}x^i}{\text{d}x'^j} V_i \] In this case, \(\frac{\text{d}x^i}{\text{d}x'^j}\) represents the elements of the Jacobian matrix, directing how the vector components adjust to the new coordinates.
Covariant vectors usually have lower indices, and this index positioning signifies their transformation characteristics. Their transformation law helps in various operations and ensuring that the geometric and physical meanings are preserved across different coordinate systems.
Understanding both contravariant and covariant vectors is crucial as they combine to form more complex structures like mixed tensors.
Mixed Tensor
A mixed tensor is an advanced object in tensor calculus that combines both contravariant and covariant components. Specifically, a 2nd-rank mixed tensor has one contravariant index and one covariant index.
In this context, consider the product of a contravariant vector \(U^i\) and a covariant vector \(V_j\). This product results in a mixed tensor: \[T^i{}_j = U^i V_j \] To understand why this is a mixed tensor, we examine how it transforms under a coordinate change. Through previously discussed transformations: \[U'^i V'_j = \left( \frac{\text{d}x'^i}{\text{d}x^k} U^k \right) \left( \frac{\text{d}x^l}{\text{d}x'^j} V_l \right) \] This simplifies to: \[T'^i{}_j = \frac{\text{d}x'^i}{\text{d}x^k} \frac{\text{d}x^l}{\text{d}x'^j} T^k{}_l \] The tensor \(T'^i{}_j\) follows the combined transformation laws for both contravariant and covariant components, making it a true mixed tensor.
Understanding mixed tensors is useful for more complex mathematical and physical systems.
Transformation Equations
Transformation equations are fundamental in tensor calculus. They show how tensor quantities change under coordinate transformations, preserving the tensor's structure across different frames of reference.
To illustrate this, consider a contravariant vector \(U^i\) and a covariant vector \(V_j\). Their respective transformation formulas are:
    \[U'^i = \frac{\text{d}x'^i}{\text{d}x^k} U^k\] \[V'_j = \frac{\text{d}x^l}{\text{d}x'^j} V_l\]
When these vectors are combined to form a mixed tensor \(T^i{}_j = U^i V_j\), the transformation equation for \(T^i{}_j\) results from multiplying the individual transformation laws: \[T'^i{}_j = \left( \frac{\text{d}x'^i}{\text{d}x^k} U^k \right) \left( \frac{\text{d}x^l}{\text{d}x'^j} V_l \right) = \frac{\text{d}x'^i}{\text{d}x^k} \frac{\text{d}x^l}{\text{d}x'^j} T^k{}_l\] This shows that the transformation preserves the mixed nature of the tensor, ensuring that it properly adapts to the new coordinate system.
Mastery of transformation equations is key to working effectively with tensors, as it guarantees their correct interpretation in various coordinate frames.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Let \(\mathbf{e}_{1}, \mathbf{e}_{2}, \mathbf{e}_{3}\) be a set of orthogonal unit vectors forming a right-handed system if taken in cyclic order. Show that the triple scalar product \(\mathbf{e}_{i} \cdot\left(\mathbf{e}_{j} \times \mathbf{e}_{k}\right)=\epsilon_{i j k}\).

If \(P\) and \(S\) are \(2^{\text {nd }}\) -rank tensors, show that \(9^{2}=81\) coefficients are needed to write each component of \(\mathbf{P}\) as a linear combination of the components of \(\mathbf{S} .\) Show that \(81=3^{4}\) is the number of components in a \(4^{\text {th }}\) -rank tensor. If the components of the \(4^{\text {th }}\) -rank tensor are \(C_{i j k m},\) then equation (7.5) gives the components of \(P\) in terms of the components of \(S\). If \(P\) and \(S\) are both symmetric, show that we need only 36 different non-zero components in \(C_{i j k m} .\) Hint: Consider the number of different components in \(P\) and \(S\) when they are symmetric. Comment: The stress and strain tensors can both be shown to be symmetric. Further symmetry reduces the 36 components of \(\mathbf{C}\) in (7.5) to 21 or less.

Show that in 2 dimensions (say the \(x, y\) plane), an inversion through the origin (that is, \(x^{\prime}=-x, y^{\prime}=-y\) ) is equivalent to a \(180^{\circ}\) rotation of the \((x, y)\) plane about the zaxis. Hint: Compare Chapter 3, equation (7.13) with the negative unit matrix.

Let \(x=u+v, y=v .\) Find \(d \mathbf{s},\) the a vectors, and \(d s^{2}\) for the \(u, v\) coordinate system and show that it is not an orthogonal system. Hint: Show that the a vectors are not orthogonal, and that \(d s^{2}\) contains \(d u d v\) terms. Write the \(g_{i j}\) matrix and observe that it is symmetric but not diagonal. Sketch the lines \(u=\) const. and \(v=\) const. and observe that they are not perpendicular to each other.

Show that the sum of two \(3^{\mathrm{rd}}\) -rank tensors is a \(3^{\mathrm{rd}}\) -rank tensor. Hint: Write the transformation law for each tensor and then add your two equations. Divide out the \(a\) factors to leave the result \(T_{\alpha \beta \gamma}^{\prime}+S_{\alpha \beta \gamma}^{\prime}=a_{\alpha i} a_{\beta j} a_{\gamma k}\left(T_{i j k}+S_{i j k}\right)\) using summation convention.

See all solutions

Recommended explanations on Combined Science Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free