Non-Equilibrium Continuum Physics TA session #1 TA: Michael Aldam 16.04.2019
Mathematical preliminaries: Tensor Analysis
1 Introduction
The fundamental assumption of continuum physics is that under a wide range of condi- tions we can treat materials as continuous in space and time, disregarding their discrete structure and time-evolution at microscopic length and time scales, respectively. There- fore, we can ascribe to each point in space-time physical properties that can be described by continuous functions, i.e. fields. This implies that derivatives are well defined and hence that we can use the powerful tools of differential calculus. In order to understand what kind of continuous functions, hereafter termed fields, should be used, we should bear in mind that physical laws must be independent of the position and orientation of an observer, and the time of observation (note that we restrict ourselves to classical physics, excluding the theory of relativity). We are concerned here, however, with the mathematical objects that allow us to formulate this and related principle. Most gener- ally, we are interested in the language that naturally allows a mathematical formulation of continuum physical laws. The basic ingredients in this language are tensor fields, which are the major focus of this part of the course. Tensor fields are characterized, among other things, by their order (sometimes also termed rank). Zero-order tensors are scalars, for example the temperature field T (x, t) within a body, where x is a 3-dimensional Euclidean space and t is time. First-order tensors are vectors, for example the velocity field v(x, t) of a fluid. Why do we need to consider objects that are higher-order than vectors? The best way to answer this question is through an example. Consider a material areal element and the force acting on it (if the material areal element is a surface element, then the force is applied externally and if the material areal element is inside the bulk material, then the force is exerted by neighboring material). The point is that both the areal element and the force acting on it are basically vectors, i.e. they both have an orientation (the orientation of the areal element is usually quantified by the direction of the normal to it). Therefore, in order to characterize this physical situation one should say that a force in the ith direction is acting on a material areal element whose normal points in the jth direction. The resulting object is defined using two vectors, but it is not a vector itself. We need a higher-order tensor to describe it. Our main interest here is second-order tensors, which play a major role in continuum mechanics. A second-order tensor A can be viewed as a linear operator or a linear function that maps a vector, say u, to a vector, say v,
v = Au . (1)
Linearity implies that A(αu + v) = αAu + Av , (2) for every scalar α and vectors u and v. For brevity, second-order tensors will be usually referred to simply as tensors (zero-order tensors will be termed scalars, first-order tensors
1 will be termed vectors and higher than second-order tensors will be explicitly referred to according to their order). The most natural way to define (or express) tensors in terms of vectors is through the dyadic (or tensor) product of orthonormal base vectors {ei}
A = Aij ei ⊗ ej , (3) where Einstein summation convention is adopted, {Aij} is a set of numbers and {i, j} run over space dimensions. For those who feel more comfortable with Dirac’s Bra-Ket notation, the dyadic product above can be also written as A = Aij |ei >
0 0 0 0 T e2 ⊗ e3 = e2e3 = 1 0 0 1 = 0 0 1 . (5) 0 0 0 0
Therefore, second-order tensors can be directly represented by matrices. Thus, tensor algebra essentially reduces to matrix algebra. It is useful to note that
ei ⊗ ej ek = (ej · ek) ei . (6)
where · is the usual inner (dot) product of vectors. In the Bra-Ket notation the above simply reads |ei >
viei = v = Au = (Aijei ⊗ ej)(ukek) = Aijuk(ej · ek) ei = Aijujei , (7) which shows that the matrix representation preserves known properties of matrix algebra (vi = Aijuj). The matrix representation allows us to define additional tensorial operators. For example, we can define
tr(A) ≡ ek · (Aijei ⊗ ej) ek = Aij
AB = (Aijei ⊗ ej)(Bklek ⊗ el) = AijBklδjkei ⊗ el = AijBjlei ⊗ el . (10)
We can define the double dot product (or the contraction) of two tensors as
A : B = (Aijei ⊗ ej):(Bklek ⊗ el) ≡ AijBkl(ei · ek)(ej · el) T = AijBklδikδjl = AijBij = tr(AB ) . (11)
This is a natural way of generating a scalar out of two tensor, which is the tensorial generalization of the usual vectorial dot product (hence the name). It plays an important role in the thermodynamics of deforming bodies. Furthermore, it allows us to project a tensor on a base dyad
(ei ⊗ ej):A=(ei ⊗ ej):(Aklek ⊗ el)=Akl(ei · ek)(ej · el)=Aklδikδjl = Aij , (12)
2 i.e. to extract a component of a tensor. We can now define the identity tensor as
I = δij(ei ⊗ ej) , (13)
which immediately allows to define the inverse of a tensor (when it exists) following
AA−1 = I . (14)
The existence of the inverse is guaranteed when det A 6= 0, where the determinant of a tensor is defined using the determinant of its matrix representation. Note also that one can decompose any second order tensor to a sum of symmetric and skew-symmetric (antisymmetric) parts as 1 1 A = A + A = (A + AT ) + (A − AT ) . (15) sym skew 2 2 Occasionally, physical constraints render the tensors of interest symmetric, i.e. A = AT . In this case, we can diagonalize the tensor by formulating the eigenvalue problem
Aai = λiai , (16) where {λi} and {ai} are the eigenvalues (principal values) and the orthonormal eigenvec- tors (principal directions), respectively. This problem is analogous to finding the roots of
3 2 det(A − λI) = −λ + λ I1(A) − λI2(A) + I3(A) = 0 , (17)
where the principal invariants {Ii(A)} are given by 1 I (A) = tr(A),I (A) = tr2(A) − tr(A2) ,I (A) = det(A) . (18) 1 2 2 3 Note that the symmetry of A ensures that the eigenvalues are real and that an orthonor- mal set of eigenvectors can be constructed. Therefore, we can represent any symmetric tensor as
A = λi ai ⊗ ai , (19)
assuming no degeneracy. This is called the spectral decomposition of a symmetric tensor A. It is very useful because it represents a tensor by 3 real numbers and 3 unit vectors. It also allows us to define functions of tensors. For example, for positive definite tensors (λi > 0), we can define
ln(A) = ln(λi) ai ⊗ ai , (20) √ p A = λi ai ⊗ ai . (21)
In general, one can define functions of tensors that are themselves scalars, vectors or tensors. Consider, for example, a scalar function of a tensor f(A) (e.g. the energy
3 density of a deforming solid). Consequently, we need to consider tensor calculus. For example, the derivative of f(A) with respect to A is a tensor which takes the form ∂f ∂f = ei ⊗ ej . (22) ∂A ∂Aij The differential of f(A) is a scalar and reads ∂f ∂f df = : dA = dAij . (23) ∂A ∂Aij Consider then a tensorial function of a tensor F (A), which is encountered quite regularly in continuum mechanics. Its derivative D is defined as
∂F ∂F ∂Fkl D = = ⊗ ei ⊗ ej = ek ⊗ el ⊗ ei ⊗ ej , ∂A ∂Aij ∂Aij ∂Fkl =⇒ Dklij = , (24) ∂Aij which is a fourth-order tensor. We will now define some differential operators that either produce tensors or act on tensors. First, consider a vector field v(x) and define its gradient as
∂v ∂v ∂vi ∇v = = ⊗ ej = ei ⊗ ej , (25) ∂x ∂xj ∂xj which is a second-order tensor. Then, consider the divergence of a tensor
∂A ∂Aij ∂Aij ∇·A = ek = ei ⊗ ejek = ei , (26) ∂xk ∂xk ∂xj which is a vector. The last two objects are extensively used in continuum mechanics.
2 Tensor transformations
Next, we should ask ourselves how do tensors transform under a coordinate transforma- tion (from x to x0)
x0 = Qx , (27)
where Q is a proper (det Q=1) orthogonal transformation matrix QT =Q−1 (note that it is not a tensor). In order to understand the transformation properties of the orthonormal base vectors {ei} we first note that
0 T 0 T 0 0 x = Qx =⇒ x = Q x =⇒ xi = Qijxj = Qjixj . (28) A vector is an object that retains its (geometric) identity under a coordinate transfor- mation. For example, a general position vector r can be represented using two different 0 base vectors sets {ei} and {ei} as
0 0 r = xiei = xjej . (29)
4 Using Eq. (28) we obtain
0 0 0 0 xiei = (Qjixj)ei = xj(Qjiei) = xjej , (30) which implies
0 ei = Qijej . (31) In order to derive the transformation law for tensors representation we first note that tensors, like vectors, are objects that retain their (geometric) identity under a coordinate transformation and therefore we must have
0 0 0 A = Aijei ⊗ ej = Aijei ⊗ ej . (32) Using Eq. (31) we obtain
0 0 0 0 0 A = Aijei ⊗ ej = AijQikek ⊗ Qjlel = (AijQikQjl)ek ⊗ el . (33) which implies
0 Akl = AijQikQjl . (34) This is the transformation law for the components of a tensor and in many textbooks it serves as a definition of a tensor. Eq. (34) can be written in terms of matrix representation as
[A] = QT [A]0Q =⇒ [A]0 = Q[A]QT , (35)
where [·] is the matrix representation of a tensor with respect to a set of base vectors. Though we did not make the explicit distinction between a tensor and its matrix rep- resentation earlier, it is important in the present context; [A] and [A]0 are different representations of the same object, the tensor A, but not different tensors. An isotropic tensor is a tensor whose representation is independent of the coordinate system, i.e.
0 0 Aij = Aij or [A] = [A] , (36) which will be discussed further in the next TA. We note in passing that in the present context we do not distinguish between covariant and contravariant tensors, a distinction that is relevant for non-Cartesian tensors (a Cartesian tensor is a tensor in three-dimensional Euclidean space for which a coordinate 0 0 0 transformation x =Qx satisfies ∂xi/∂xj =∂xj/∂xi).
3 Example: Force, torque, yada yada yada
To move all the previous discussion from abstract notions to a real and hopefully familiar realm, lets recall stuff we learned when we were just new students. In the motion of a point mass we define the momentum as
~p = m~v ⇔ pi = mvi , (37)
5 where ~p and ~v˙ are vectors, while m is a scalar. The kinetic energy, as scalar, is m m T = ~v · ~v = v v , (38) 2 2 i i which I hope is not new to any of you. The reason these equations are so simple is that a point mass by definition looks the same from all directions. If we next consider the motion of a rotating rigid object, we find a more interesting situation, where we define the angular momentum as ~ L = I~ω ⇔ Li = Iijωj , (39)
where I is the moment of inertia (not to be confused with the unit tensor), a 2nd rank tensor which was probably the first such create you have met. We see immediately that in this case the dynamics is much more complicated. For instance, a angular velocity in some direction, sayz ˆ, may lead to angular momentum with components in different directions! For a spinning rigid body the kinetic energy is 1 1 T = ~ωT I~ω = ω ω I , (40) 2 2 i j ij as is derived in any elementary mechanics text. This of course makes choosing the reference frame we want to work in much more important, since it can make life much easier. As an example take the situation described in Fig.1, taken from The Feynman Lectures on Physics which you should all know by heart1. In the figure we see a crooked wheel on an axis turning with angular frequency
1
2
Figure 1: A crooked wheel on an axis. Taken from The Feynman Lectures on Physics, Volume I, Chapter 20-4.
ω. Looking at this system, we see that there are two sensible choices for our axis. The first is to align an axis with the angular velocity ~ω, which is marked in Fig.1 by {x, y}2. In this system ω~0 =(ω, 0, 0) and the moment of inertial I0 at the time frame shown in the Fig looks quite horrible
(η2 + 3) sin2(θ) + 6 cos2(θ)(η2 − 3) sin(θ) cos(θ) 0 MR2 I0 = (η2 − 3) sin(θ) cos(θ)(η2 + 3) cos2(θ) + 6 sin2(θ) 0 , (41) 12 0 0 η2 + 3
1 Not really, but it should be familiar. 2 The z axis is defined using the regular right handed convention.
6 where M is the mass of the disc, R is its radius and η is the ratio of the height to the radius. Keep in mind that this is the moment of inertia at the time frame shown in Fig1 . In general you would also need to rotate everything by ωt around the 1 axis, but this is not relevant for the moment. On the other hand, in a reference frame more befitting a disc, which is marked in Fig.1 by {1, 2} 6 0 0 cos(θ) MR2 I = 0 η2 + 3 0 , ~ω = ω sin(θ) (42) 12 0 0 η2 + 3 0 which has a much simpler form for I, albeit the bit more complicated ~ω, which is time dependent in general (as discussed earlier, these are the expression at a certain snapshot). Using I and ~ω in any of the frames, as discussed above (you can check that the results are the same if you want) in Eq. (40), we get MR2ω2 T = η2 + 3 sin2(θ) + 6 cos2(θ) , (43) 24 which has a form you will be seeing a lot: a dimensional combination, to get the right 2 2 units, MR ω in our case, followed by a function of dimensionless√ parameters, η and θ in this case. Also notice an interesting property of the result: H = 3R leads to the kinetic energy not being dependent on θ! This is actually not very surprising, since if we look back at Eq. (42) and inspect I we see immediately that for this value of η I is proportional to the unit tensor3 This means that a cylinder of the right proportions rotates just like s sphere, demonstrating the power of the spectral decomposition - let’s see you figure this out by inspecting Eq. (41)!4
4 Gauss’ integral theorem for tensors (if we make it)
Finally, you know from your undergrad studies that if ~u is a vector field in a volume Ω ⊂ 3, then R Z Z div ~udV = ~u · dS,~ (44) Ω S where S is the surface of Ω (in mathematical notation, S = ∂Ω). dS~ is a differential vector, perpendicular to a local surface. This is called Gauss’ theorem, and it also works for tensors: Z Z div A dV = A dS,~ (45) Ω ∂Ω where the right-hand-side should be understood as A operating as a tensor on dS~, exactly like the right-hand-side of (44) represented ~u operating as a tensor on dS~ , i.e. the usual dot product. Both Eqs. (44) and (45) are given here without proof. We will now see that you already know a particular case of Eq. (45). Take Ω to be a 2-dimensional sheet in a 3D space, and a vector field ~u on it. The boundary of Ω is now a curve, whose tangent vector will by denoted by ~`, Cf. Fig.2. For simplicity, we’ll assume that Ω is confined to the x − y plane, although this is not necessary. We define
3 Usually also denoted by I, but what can you do. 4 It’s actually not that bad, if you know what you are looking for.
7 Figure 2: Illustration of a surface and boundary integration. An exemplary boundary element is marked in red, together with its normal vector d~n (red) and a tangent vector d~` (blue). a new tensor A ≡ E~u, (46) where E is the Levi-Civita tensor. Index-wise, this means Aij = Eijkuk. We’ll also take ~u to be z-independent. We begin by calculating the left-hand-side of (45): Z Z Z Z ~ div AdS = ∂jEijkukdS = Eijk∂jukdS = ∇ × ~u dS . (47) Ω Ω Ω Ω The right-hand-side gives Z Z Z Ad~n = Eijkukdnj = (~u × d~n) . (48) ∂Ω ∂Ω ∂Ω Now, ~u × d~n is a vector that is perpendicular to both ~u and d~n, that is, it is directed in thez ˆ direction. Its magnitude is |~u||d~n| sin θ where θ is the angle between ~u and ~n. But the angle between ~u and ~` is α ≡ 90◦ − θ, so we can write
~ ~ |~u × d~n| = |~u| |d~n| sin θ = |~u| d` cos α = ~u · d` . (49) We conclude that ~u × d~n = ~u · d~` zˆ , (50)
The theorem (45) says that (47) and (48) are equal, so we conclude that Z I ∇~ × ~u dS = ~u · d~l , (51) Ω ∂Ω which you know well from your happy undergrad days, under the name of Stokes’ Theorem (or Green’s Theorem, sometimes).
8