User:Guy vandegrift/Introduction to tensors in physics

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Eigenvalues and eigenvectors • Strain for scientists and engineers • sandbox • {{Purge|refresh}} #qwer2 #refresh

Physics students do not typically encounter the tensor early in their educational sequence. While there no compelling reason to introduce this topic at the beginning of an introductory physics course, an attempt to make these ideas accessible to students at this level the practical value of giving educators more flexibility in choosing when and where to introduce tensors to students in a physics or engineering program.

Introduction

Perpendicular springs of different spring constant act on the displaced object.

A counter-intuitive fact should always pique your attention because it unveils a weakness in your intuition. It seems count-intuitive that the infinitesimal (Cauchy) strain tensor, ${\textstyle {\tfrac {1}{2}}[{\vec {\nabla }}\,{\vec {u}}+({\vec {\nabla }}\,{\vec {u}})^{T}],}$ is defined in a way that discards information. It does this by removing the asymmetric part of ${\textstyle {\vec {\nabla }}{\vec {u}}.}$ While this is a very important tensor, it is a poor choice for the first tensor that a student encounters because it is fundamentally a tensor field. Much simpler to grasp is a generalization of Hooke's law, ${\underline {F}}=-{\underline {\underline {\kappa }}}\cdot {\underline {r}},$ where ${\underline {r}}=x{\hat {x}}+y{\hat {y}},$ is displacement, ${\underline {F}}$ is force. We shall work primarily in two dimensions, so that the generalized spring constant is a tensor ${\underline {\underline {\kappa }}},$ with 4 components. It shall later emerge there are actually only 3 components because ${\underline {\underline {\kappa }}}$ is symmetric, with $\kappa _{i,j}=\kappa _{j,i}.$ This gives us an opportunity to introduce students to the value of recognizing symmetries in tensors.

Figure&;? depicts a spring-mechanism that this (symmetric) tensor might represent. There are three ways to ensure a linear relationship between force and displacement: (1) assume small displacements, (2) create a mechanism to ensure constant orientation of the springs as the object moves. Figure? shows small blocks with wheels attaching each spring to the wall to permit the spring to move with the object, or (3) assume that the springs are very long. This latter method brings up our first teaching hint:

Teaching hints can be identified by the icon shown on the previous line. With some computers it is only necessary to hover the mouse over the footnote marker to see the hint.

Target audience

These ideas are nominally intended for students enrolled in an introductory physics course, using a textbook like OpenStax University Physics. Since introductory courses are already overstuffed with "essential" material, few instructors are likely to insert the ideas developed here into course syllabi. However, the same idea designed to preview a topic before it is learned can also be modified so as to review the topic after it is forgotten. Also, these ideas might be converted into assessments, perhaps even as an assessment for a graduate level prelim. Another venue for quirky ideas is the Iab. Not all knowledge is best gained using the standard paradigm of teaching and testing, and laboratories are appropriate venue for material that "will not be on the final".

More recently, the introduction of AI chatbots hint even more radical changes in higher education: Perhaps AI chatbots will be able to greatly reduce the need for universities to carefully choreograph the content of introductory courses. As computers do more high-level work (under the supervision of humans), then computers should be able to access what mathematical skills the human needs. Instead of degrees and certificates, the chatbot would already be familiar human's knowledge base, and be able to quickly design the necessary curriculum for a given task. Perhaps no human will ever write a lesson based on this article. Instead, a chatbots working with (or for?) humans will routinely create the necessary lessons required to get the human up to speed. Perhaps the target audience for this article is a chatbot!

QWER FIX THIS BAD SENTENCE: As another example that can be used to either preview or review an essential topic, this article uses the Taylor series in two dimensions to explain why only 3 of the 4 parameters of a 2x2 tensor (matrix) are required to define ${\underline {\underline {\kappa }}}.$This teaching hint illustrates how the well-known quadratic equation, $f(x)=c+bx+ax^{2},$ can be used to either preview or review the Taylor series for a student who knows you to take the derivative of a polynomial: 

Visualizing vectors, tensors, and rotations

Vector addition and subtraction (attribution: CC0 1.0 Universal)

Springs that are not perpendicular (attribution: CC0 1.0 Universal)

The displacement vector is usually the student's first encounter vectors and vector addition. This leads directly vector subtraction, as well as multiplication of a vector by a scalar. Students connect the abstract operations with images like figure ? and vector equations such as:

$$A\mid \quad {\begin{aligned}&{\underline {A}}+{\underline {A}}+{\underline {A}}=3{\underline {A}}\\&{\underline {A}}+{\underline {B}}={\underline {C}}\;\Longrightarrow \;{\underline {B}}={\underline {C}}-{\underline {A}}.\end{aligned}}$$

Most students enter an interdisciplinary world upon graduation from college. Perhaps we should distinguish academic disciplines by what students love instead of what they need to know. Data scientists see the vector as an ordered set of entities to be processed. Mathematicians look for a logical structure based on definitions. Figure ? reminds us that physicists think of vector addition as consecutive translations through space. In conjunction with equation A, this figure establishes multiplication of a vector by a scalar N as N consecutive additions (for integer N.) It establishes vector subtraction using vector addition.

Just as physicists view the vector an arrow ↗, they can view what is known as a symmetric tensor as either a rectangle , a rhombus , or an ellipse . While the length of the vector arrow is universally understood to represent the vector's magnitude, there is no universally correct convention for the for symmetric tensors in two or three dimension. The rectangle (or box) might seem like the most natural icon for the spring-mechanism of figure&nbs;?. And, it might seem natural to associate the dimensions of the rectangle (box) with the strength of the springs, so that represents a spring-mechanism with the stiffer springs oriented in the horizontal direction. But later we shall examine the spring-mechanism from using the contours of potential energy. By this convention, would represent a case where the stiffer springs are oriented in the vertical direction.

Having represented the spring-mechanism with a shape that possess two measurements (height and width), the next step is to rotate the shape to accommodate a spring-mechanism that has been rotated as shown in Figure . It is this rotation that distinguishes the pair of numbers that define the spring constants from a vector. In two dimensions, a (symmetric) tensor has two components plus an orientation. The exact relationship between the spring constants and the height and width of the shape is a matter of choice. The advantage of associating the two dimensions of the shape with spring constants is that these dimensions are directly proportional to what is known as the eigenvalues of the (2x2 symmetric) tensor. In the language of linear algebra, another advantage is that the spring constants are the tensor's "eigenvalues", and the two "eigenvectors" are aligned with the pair of orthogonal springs. Later, it will be shown that viewing this tensor as an elliptical contour of constant potential also has great value: It informs us that this same tensor can be used to describe an object confined by three or more springs, or by a pair of springs that are not orthogonal.

In three dimensions, the ellipse is replaced by an ellipsoid and the rectangle is replaced by a rectangular box. It should also be noted that we have adopted this convention of associating the two dimensions of a shape is unnecessarily restrictive because it only models spring constants ("eigenvalues") that are positive. In a later section we will show that it is possible to associate these same shapes with a strain tensor where the eigenvalues are not constrained to be greater than zero.

But, if one imagines the spring mechanism a contour plot of potential energy

In contrast with the magnitude and direction associated with using an arrow to define a vector, the relationship between the base and height of the rectangle to the the stiffness of the springs is best left undefined. If the "eigenvalues" of the symmetric tensor are all positive or all negative, we two different conventions are available: A large dimension can correspond to either a large eigenvalue or a small one. On the other hand, if the signs of the eigenvalues, a third convention is typically the best.

Long before associating a tensor with a square matrix, even children can understand that the arrow and rectangle rotate differently. Upon a 90° rotation, the two shapes appear to rotate in the same way ... until it is noted that rotating the rectangle by +90° is identical to rotating by −90°. In contrast, rotating an arrow by +90° produces an arrow pointing in the opposite direction as rotating the arrow by −90°.

 

Linear algebra

The red lines show which portion of the arrow indicates the magnitude of $|{\underline {F}}|$. This image generates three questions:

Figure? illustrates the spring-mechanism: An object is held at the equilibrium point by two perpendicular springs that remain perpendicular throughout the object's motion. To the extent that the displacement is small enough that the springs remain perpendicular to each other, we can model figure ? this with a pair of linear equations

$$B\mid \quad {\begin{cases}F_{x}=k_{xx}x\\F_{y}=-k_{yy}y\end{cases}}\quad \iff \quad {\begin{bmatrix}F_{x}\\F_{y}\end{bmatrix}}={\begin{bmatrix}k_{xx}&0\\0&k_{xx}\end{bmatrix}}{\begin{bmatrix}x\\y\end{bmatrix}}$$

The use of double subscripts, i.e., $k_{xx}$ instead of $k_{x},$ serves to emphasize that this two-dimensional spring constant cannot be a vector. A hint that the two springs in figure? should not be modeled as a "vector spring constant" $({\underline {k}}=k_{x}{\hat {x}}+k_{y}{\hat {y}})$ can be obtained by the fact that none of following equations yield physically plausible relationships between ${\underline {k}}$ and displacement ${\underline {r}}:$

$$C\mid \quad {\underline {F}}=-|{\underline {k}}|\,{\underline {r}}\,\quad \quad {\color {Gray}\diamond }\quad \quad {\underline {F}}=-|{\underline {r}}|\,{\underline {k}}\quad \quad {\color {Gray}\diamond }\quad \quad {\underline {F}}=-{\underline {k}}\times {\underline {r}}.$$

The force field depicted in Figure can generate three questions for students. qwer COMMENT I think now we refer to equation C instead of the figure

Two ways to generalize

There are two approaches to the generalization from the simple case of two orthogonal springs that are aligned to the xy coordinate system.

1. The most direct is to apply a rotation of either the coordinate system or the physics system (teach both at your own peril!)
2. An entirely different approach avoids some of the linear algebra by focusing on the system's potential energy. It introduces a number of complex ideas that include the Taylor series with multiple variables, and a hint as to the nature of perturbation theory. This approach also gives students a hint as to the nature and purpose of perturbation theory. It also illustrates why the simple harmonic oscillator is such a fundamental concept in applied mathematics. qwer I AM STUCK HERE. I WANT TO MENTION NONLINEARITY AND THE FACT THAT THE TWO DIMENSIONAL SHO CAN ONLY BE REPLACED BY A NONLINEAR SYSTEM THAT

Figure? shows a spring mechanism not aligned with the coordinate system. The obvious starting point is:

$$D\mid \quad {\underline {F}}\,=\,-{\begin{bmatrix}F_{x}\\F_{y}\end{bmatrix}}\,=\,-{\begin{bmatrix}\kappa _{xx}&\kappa _{xy}\\\kappa _{yx}&\kappa _{yy}\end{bmatrix}}{\begin{bmatrix}x\\y\end{bmatrix}}\,=\,-{\underline {\underline {\kappa }}}\cdot {\underline {r}}\,.$$

There are many ways to proceed from here: (1) We can apply the rotation operator to the tensor to view the mechanism in a different coordinate system. Or, (2) we can investigate the system using potential energy. Each approach has pedagogical advantages.

Tensor rotation approach

This approach is always appropriate with students already familiar with matrix multiplication and tensor rotation. The algebraic steps are trivial: ${\textstyle {\underline {r}}={\underline {\underline {R}}}\cdot {\underline {r}}'}$ and ${\textstyle {\underline {F}}={\underline {\underline {\kappa }}}\cdot {\underline {r}}}$ leads to, ${\textstyle {\underline {F}}={\underline {\underline {\kappa }}}\cdot {\underline {\underline {R}}}\cdot {\underline {r}}'.}$ The beauty of vectors and tensors, is that they all rotate by the same rules, which means that force is also rotated using: ${\underline {F}}={\underline {\underline {R}}}\cdot {\underline {F}}\,'.$ Substituting this gives us: ${\textstyle {\underline {\underline {R}}}\cdot {\underline {F}}\,'={\underline {\underline {\kappa }}}\cdot {\underline {\underline {R}}}\cdot {\underline {r}}'.}$ Multiplying both sides from the left by ${\textstyle {\underline {\underline {R}}}^{-1}}$ yields

$$E\mid \quad {\underline {F}}\,'=\underbrace {\left({\underline {\underline {R}}}^{T}\cdot {\underline {\underline {\kappa }}}\cdot {\underline {\underline {R}}}\right)} _{{\underline {\underline {\kappa }}}'}\cdot {\underline {r}}'={\underline {\underline {\kappa }}}'\cdot {\underline {r}}'$$

Elliptical and hyperbolic contour plots

If you are sufficiently close to the top of a smooth hill, elevation contours probably look like ellipses.

In this last step, we used, ${\textstyle {\underline {\underline {R}}}^{T}={\underline {\underline {R}}}^{-1}.}$ The transpose equals the inverse because the rotation matrix is unitary. In our two-dimensional system, the rotation matrix is:

$$F\mid \quad {\underline {\underline {R}}}={\begin{bmatrix}\cos \theta &\sin \theta \\-\sin \theta &\cos \theta \end{bmatrix}}$$

One advantage of the tensor rotation approach is that if you begin with a stable system (i.e. $k_{xx},k_{yy}$ both positive) then you know that the system is alway stable. Figure? shows that the contours of constant potential energy are ellipses if all eigenvalues have the same sign. The figure suggests that a convenient visual representation for a two or three-dimensional tensor is an ellipse or ellipsoid, provided all eigenvalues have the same sign.

Taylor series approach

• See also Taylor_series and Taylor's theorem, and Perturbation theory on Wikipedia.

In the previous section, where we rotated the spring mechanism out of alignment with the x and y axes, it was permissible to begin with the assumption that both spring constants $(k_{xx},k_{yy})$ are non-negative. In this section, it is better not to make this a priori assumption, or equivalently to permit spring mechanisms that are unstable. This is due to the fact that we begin with an expression for potential energy, $U=U(x,y)$, in which it is not known whether, ${\underline {r}}=0$ is a stable or unstable equilibrium point. There are two reasons an instructor might wish to take this approach: To present an argument valid for sufficiently small oscillations about the equilibriums point of what might be called any "typical" potential well. For example, the results of this section hold for low amplitude oscillations that are not orthogonal, or if more than two springs are present. TEACHING HINTS: springs under tension, The instructor wishes to reduce student exposure to linear algebra and/or enhance student understanding of the Taylor series. The instructor wishes to illustrate at how one might model to model nonlinear (or large amplitude) effects. In this section we remove all concerns about the convergence of an infinite series by posing the problem in terms of a polynomial approximation to a function that vanishes at the origin. We also stipulate that all first partial derivatives vanish at the origin: working physicists and engineers are rarely concerned about whether a series converges if they intend to drop higher order terms. Divergent series are far more fun to work with

Also, we remove all concerns about the convergence of an infinite series by posing the problem in terms of a polynomial approximation to potential energy, $U=U(x,y).$ It should be noted that working physicists and engineers are rarely concerned about whether a series converges if they intend to drop higher order terms. Divergent series are far more fun to work withAlso, we remove all concerns about the convergence of an infinite series by posing the problem in terms of a polynomial approximation to potential energy, $U=U(x,y).$

advantage to this approach is that it yields a simple proof of an important fact concerning the symmetry $(\kappa _{ij}=\kappa _{ji})$ of the spring constant tensor, namely that this symmetry is a consequence of the fact that the force is the gradient of a potential energy.

PERHAPS THIS IS HOW THIS SECTION SHOULD START ??????????????

The Taylor series is typically introduced in first year calculus, and is is not unusual for an introductory physics course to introduce partial derivatives within the context of force and potential energy, with: ${\underline {F}}(\,{\underline {r}}\,)=-{\underline {\nabla }}U(\,{\underline {r}}\,).$ In this section we obtain the generalized Hooke's law, ${\textstyle {\underline {F}}=-{\underline {\underline {\kappa }}}\cdot {\underline {r}},}$ from the the potential energy, $U(x,y),$ expressed as a Taylor series expansion around the origin at $(x=0,y=0),$ which also an equilibrium point (meaning that the force vanishes at the origin.) If we also define potential energy to vanish at the origin, the first two terms in the following Taylor series vanish:

$$G\mid \;U(x)=U(0)+U'(0)\cdot x+{\frac {1}{2}}U''(0)\cdot x^{2}+{\frac {1}{6}}U'''(0)\cdot x^{3}+\ldots \,,$$

The Taylor series in two variables can be viewed as a primitive form of perturbation theory. Perturbation takes on many forms, and the vast majority are based on separating the solution to a problem into various orders. In our case we write potential energy as the sum of a sequence of terms:

$$H\mid \;U(x,y)\;=\;{\begin{pmatrix}{\text{0-th order}}\\[0.0ex]{\text{terms}}\end{pmatrix}}+{\begin{pmatrix}{\text{1st order}}\\[0.0ex]{\text{terms}}\end{pmatrix}}+{\begin{pmatrix}{\text{2nd order}}\\[0.0ex]{\text{terms}}\end{pmatrix}}+{\begin{pmatrix}{\text{3rd order}}\\[0.0ex]{\text{terms}}\end{pmatrix}}+\cdots =$$

Typically, the higher order terms are smaller the the lower order terms, but there are many exceptions to that rule. An important exception occurs when a term vanishes. In our case, the zeroth and first order terms vanish by design. Another exception to this rule occurs with terms are dropped, despite the fact that their values have not been calculated. In this case, the calculated formula is merely an educated guess that needs to be experimentally verified.

In our calculation, we arrange for the zeroth and first order terms to vanish exactly, and postulate that the 3rd order term can be neglected as small for sufficiently small values of x and y:

$$\quad \underbrace {\overbrace {\cancel {U(0,0)}} ^{\text{0-th order}}+\overbrace {\left.{\cancel {\frac {\partial U}{\partial x}}}\right|_{0}\cdot x+\left.{\cancel {\frac {\partial U}{\partial y}}}\right|_{0}\cdot y} ^{\text{1st order}}} _{\text{Zero by legislation}}+\overbrace {\left.{\frac {1}{2}}{\frac {\partial ^{2}U}{\partial x^{2}}}\right|_{0}\cdot x^{2}+\left.{\frac {\partial ^{2}U}{\partial x\partial y}}\right|_{0}\cdot xy+\left.{\frac {1}{2}}{\frac {\partial ^{2}U}{\partial y^{2}}}\right|_{0}\cdot y^{2}} ^{\text{2nd order}}+\underbrace {\overbrace {{\frac {1}{6}}\sum _{i,j=1}^{2}\left.{\frac {\partial ^{3}U}{\partial x^{i}\partial y^{j}}}\right|_{0}x^{i}y^{j}} ^{\text{3rd order}}} _{\text{neglect as small}}$$

quer1.5

The third order term is a double sum, ${\textstyle \sum _{i,j=1}^{2}=\sum _{i=1}^{2}\sum _{j=1}^{2},}$ which can also be written as

$${\begin{aligned}{\frac {1}{6}}\sum _{i=1}^{2}\sum _{j=1}^{2}\left.{\frac {\partial ^{3}U}{\partial x^{i}\partial y^{j}}}\right|_{0}x^{i}y^{j}=\left.{\frac {\partial ^{3}U}{\partial x^{3}}}\right|_{0}{\frac {x^{3}}{3!}}+\left.{\frac {\partial ^{3}U}{\partial x^{2}\partial y}}\right|_{0}{\frac {x^{2}y}{2!}}+\left.{\frac {\partial ^{3}U}{\partial x\partial y^{2}}}\right|_{0}{\frac {xy^{2}}{2!}}+\left.{\frac {\partial ^{3}U}{\partial y^{3}}}\right|_{0}{\frac {y^{3}}{3!}}\end{aligned}}$$

qwer3

The first two terms can be made to vanish at $x=0$ if we establish it as an equilibrium point $F_{x}=-U'=0$ and also define the potential to vanish there: $U(0)=0.$

$$H\mid \;U(x,y)\;=\;{\begin{pmatrix}{\text{0-th order}}\\[0.0ex]{\text{terms}}\end{pmatrix}}+{\begin{pmatrix}{\text{1st order}}\\[0.0ex]{\text{terms}}\end{pmatrix}}+{\begin{pmatrix}{\text{2nd order}}\\[0.0ex]{\text{terms}}\end{pmatrix}}+{\begin{pmatrix}{\text{3rd order}}\\[0.0ex]{\text{terms}}\end{pmatrix}}+\cdots =$$

$$\quad \underbrace {\overbrace {\cancel {U(0,0)}} ^{\text{0-th order}}+\overbrace {\left.{\cancel {\frac {\partial U}{\partial x}}}\right|_{0}\cdot x+\left.{\cancel {\frac {\partial U}{\partial y}}}\right|_{0}\cdot y} ^{\text{1st order}}} _{\text{Zero by legislation}}+\overbrace {\left.{\frac {1}{2}}{\frac {\partial ^{2}U}{\partial x^{2}}}\right|_{0}\cdot x^{2}+\left.{\frac {\partial ^{2}U}{\partial x\partial y}}\right|_{0}\cdot xy+\left.{\frac {1}{2}}{\frac {\partial ^{2}U}{\partial y^{2}}}\right|_{0}\cdot y^{2}} ^{\text{2nd order}}+\underbrace {\overbrace {{\frac {1}{6}}\sum _{i+j=3}\left.{\frac {\partial ^{3}U}{\partial x^{i}\partial y^{j}}}\right|_{0}x^{i}y^{j}} ^{\text{3rd order}}} _{\text{neglect as small}}$$

qwer4

It is generally assumed that ${\textstyle \partial ^{2}U/\partial x\partial y=\partial ^{2}U/\partial y\partial x.\,}$ Combine this with equation G to obtain ${\textstyle U={\frac {1}{2}}{\underline {r}}^{T}\cdot {\underline {\underline {\kappa }}}\cdot {\underline {r}},}$ and:

$$H\mid \quad {\underline {\underline {\kappa }}}={\begin{bmatrix}\kappa _{xx}&\kappa _{xy}\\\kappa _{yx}&\kappa _{yy}\end{bmatrix}}={\begin{bmatrix}\partial ^{2}U/\partial x^{2}&\partial ^{2}U/\partial x\partial y\\\partial ^{2}U/\partial y\partial x&\partial ^{2}U/\partial y^{2}\end{bmatrix}}$$

next

Outline

• Informal introduction to Taylor series

• polynomial
• symmetry in partial derivatives

• Ways to visualize a tensor
• symmetric versus antisymmetric
• Introduction to strain (insert figure and reference it's use on WV and WP)
• Information counting in 2d and 3d.
• Tensor of infinite dimensions

• reference to large N (or do it here)

• Conclusion take from

• We utilize the Taylor series, something students traditionally learn in calculus, before learning linear algebra, taking an approach that is distinctly different than what would suit a pure mathematician. Instead of bothering with whether the series converges, we seek a simply polynomial approximation. This economy of rigor permits us to introduce the Taylor series in two dimensions, as a polynomial approximation to a "typical" potential well.

Here (force), when given the value of an input vector (displacement.) If both pairs of springs in figure? have equal spring constants, the scaler ${\textstyle k}$ can be a (rank-0) tensor. It is important to emphasize that the displacements must be sufficiently small that the orientation of the springs do not change, i.e., the two sets of springs must always be orthogonal to each other. If the displacement is so large that the direction of the force changes, the system is no longer linear. In practical applications, virtually all systems are only approximately linear. Nevertheless, the study of linear systems is essential because they are extremely easy to solve (compared with nonlinear systems.)

It will be shown that in a rotated reference frame, the two-dimensional form of Hooke's law can be written using a tensor version of the spring constant:

This introduction to tensors is not found in textbooks for a reason: The content becomes much easier to grasp approximately two years after most students have completed an introductory year of university calculus (accompanied with calculus linear algebra courses.) Nevertheless, there are reasons for tolerating the chaos that results from creating unconventional paths to knowledge. One might be for a refresher course for those who completed the standard curriculum many years ago: For example, someone with deep knowledge of tensors gained many years ago general relativity, might have unanswered questions about Cauchy's infinitesimal strain theory. Also, this quirky approach to a subject might alleviate the consequences of recent trends in online learning and the fact that all homework problems and quiz questions can be quickly answered using the internet. One way to mimic "real research" might be to put students in a group and give them a question too quirky to be answered by a quick Google search.

The early introduction of difficult ideas might give stronger students something to ponder after the course ends. Unfortunately, this will also burden students who are struggling. A compromise might be achieved by inserting a unit on tensors in such a way that the syllabus informs students that this unit "won't be on the final", or the course culture might help students understand that "this unit is just for fun." The unit might, for example, be included a list of options for a term project. Several possibilities for term projects will be mentioned in the calculations that follow.

temp

The instructor should be warned about writing $k_{xx}\,$ as $k_{x}$ with students who are unschooled in the ways of tensor analysis. Since ${\underline {F}}$ and ${\underline {r}}$ must be vectors, the only way spring constant can not be a scalar is if, ${\underline {F}}=-{\underline {k}}\times {\underline {r}},$ which in no way resembles Hooke's law as we know it. Midway through Openstax Physics Volume 1 textbook uses the partial derivative to calculate the force from potential energy, written as $U=U({\underline {r}}).$ Since the reader has already encountered the potential energy of a spring, it is reasonable to introduce a two dimensional simple harmonic oscillator through its potential energy:

Figure? shows contours of constant energy for $U(x,y)=nE_{0},$ with $n=0,1,2,3,4.$ It is well-known that these contours are ellipses (circles) if ${\textstyle k_{xx}}$ and ${\textstyle k_{yy}}$ have the same sign, and hyperbolas if the signs are opposite. If both spring constants are negative, the system is unstable. If the signs are opposite, a graph of $U$ versus $x$ and $y$ is saddle shaped.

A Chapter 1 problem in the same textbook introduces the rotation of a two dimensional coordinate system (without matrixes) as two equations:

$$F\mid \quad {\begin{aligned}x'&=\;\;x\cos \theta +y\sin \theta \\y'&=-x\sin \theta +y\cos \theta \\\end{aligned}}\iff$$

Rotation of coordinate systems play an essential role in the discussion that follows, and in principle, this could be an opportunity to introduce matrices as a shorthand notation for expressing two equations in two unknowns.

A minimal knowledge of matrix multiplication is essential prerequisite for this discussion. A Wikiversity page needs to be written as a companion to this article, and it would be worthwhile to assign the completion of that page to a student.

How to visualize a tensor

I'm not sure how to write this, but here is the outline:

• Students learn that a vector has "directions and magnitude".

Introduce the "square with arrows" when the strain is so small that the rhombus looks like a square.

my prose fragments

This will be the caption.

unit square rotated to be a diamond

Refer the rest of the discussion to Strain for scientists and engineers

Here, one can borrow from a convention used to depict strain, which in a linearized theory can be depicted as the sum two tensors, ${\underline {\underline {S}}}={\underline {\underline {I}}}+{\underline {\underline {\kappa }}}.$ The identity tensor ${\underline {\underline {I}}}$ is represented by a

For that reason, the identity matrix needs to be added to the infinitesimal strain to keep everything positive: ${\underline {\underline {I}}}+{\underline {\underline {\varepsilon }}}$.

Reverse engineering the rotation vector

1. How to rotate a tensor.
2. The eigenvalue problem: How to "un-rotate" tensor

When you see a rule, break it!

• See also w:Hooke's law#General tensor form and w:Matrix representation of conic sections

1. Wikipedia articles: Active and passive transformation, Rotation of axes in two dimensions, Rotation formalisms in three dimensions, Change of basis
2. If rows&columns are orthornormal sets then matrix is unitary (stackexchange.com)
3. w:Geoffrey Chew once remarked that he has spent hours looking for a minus sign.

which Wikipedia defines as an algebraic object that describes a multilinear relationship between sets of algebraic objects related to a vector space. This definition could generate a useful collection of test questions.

This was apparent in my first college physics course, taught by Rainer Sachs in 1970. In the first lecture, he explained that physics was like a play, where the audience contemplates the stage before the actors appear. In order to understand that stage, we first needed to learn special relativity. I learned very little about relativity in that first week. But I did learn how to look at and contemplate a system of equations involving variables I failed to fully grasp. Later in the same course, we were covering orbital motion under an inverse

• Caviedes, Susana Merino, et al. "User Interfaces to Interact with Tensor Fields. A State-of-the-Art Analysis." Similar NoE Tensor Workshop, Las Palmas. 2006.

Informal introduction to Taylor Series

The four constants involve partial derivatives, all evaluated at the origin. For sufficiently small values of $x$ and $y$, the first order terms are larger than those of the second order. But, the first order terms represent the force at the origin, and if we demand that the force vanishes at the origin, then the second order terms dominate near the origin. Moreover, these second order terms dominate for sufficiently small $x$ and $y$ over any third order terms (e.g. $x^{3},x^{2}y,\ldots$) one might add to the polynomial approximation. By appealing to polynomial approximations, this informal introduction to the Taylor series avoids any concerns about whether the infinite series converges.

The three partial derivatives in the aforementioned equations can be viewed as constants corresponding to the four constants $\kappa _{ij}$ in the tensor equation ${\underline {F}}=-{\underline {\underline {\kappa }}}\cdot {\underline {r}}$. The fact that there are only three constants in the formula for U while the tensor uses four constants gives the instructor a number of opportunities for homework or quiz questions. The mere fact that we are using three values (partial derivatives) to ascertain four constants $\kappa _{ik}$ raises all sorts of questions. Here is another question the use of a polynomial to approximate a physical variable raises: Would it be more accurate if the partial differentials were evaluated not at $x=0$ but at $x/2$? Another opportunity for challenging students is to point out that any polynomial can be evaluated if $x$ is imaginary or complex.

TEACHING HINT and equation BLANKS

$$Z\mid \quad {\begin{aligned}x'&=\;\;x\cos \theta +y\sin \theta \\y'&=-x\sin \theta +y\cos \theta \\\end{aligned}}\iff$$

Additional information

Acknowledgements

Any people, organisations, or funding sources that you would like to thank.

Competing interests

Any conflicts of interest that you would like to declare. Otherwise, a statement that the authors have no competing interest.

Ethics statement

An ethics statement, if appropriate, on any animal or human research performed should be included here or in the methods section.

Footnotes

Symmetric and asymetric tensors (febio.org)

w:Skew-symmetric matrix#Cross product