Lecture, Tuesday April 11
Physics 105C
We now move to how to describe deformations of a material, or strain. For the time being, we
will ignore the forces that must be applied to create those deformations though. (That will come
next week.) Strain comes in two types: stretching, where a material is elongated or compressed
in some direction, and shear, where angles within the material change. One example of a shear
deformation is flattening a cereal box before putting it in recycling. You change its originally
rectangular cross-section to a parallelogram and eventually to a line.
Assume the original position of a piece of material is X and its position after the strain is x
(in class Y was used instead of x). If you assume that the strain happens at some time t > 0,
then these are really the same variables described last week as material and spatial coordinates,
respectively. A matrix (i.e., rank two 
tensor) connects
the original and final positions: x = FX.

λ1 0 0

For example, a diagonal matrix F =  0 λ2 0 
 represents length changes along the three
0 0 λ3
basis directions. Unless λ1 = λ2 = λ3 , this is an anisotropic strain. A second example is if
F is in fact a rotation matrix R. As you’ve seen in previous classes, rotations are represented
by orthogonal matrices, RT R = I. Rotations do not in fact apply any internal strain to a
material, since they move the entire material in a uniform way. Similarly, overall translations
apply no strain. As a third example, consider F = BDB−1 , where D is a diagonal matrix and the
columns of B are orthogonal. (Actually, if the columns are orthogonal, we may as well redefine
B as having each column normalized to 1. So we can assume B is a rotation matrix.) In this
case F is not so different from the diagonal matrix; it also describes elongation or compression
along three perpendicular axes, but the matrix F doesn’t happen to be written in terms of those
three directions. The rotation matrix B alters the coordinates appropriately. Every symmetric
matrix (F = FT ) can be written as BDB−1 , so symmetric matrices represent only stretching.
Even when F is not symmetric, it has a polar decomposition. This is a generalization to
matrices of writing any complex number in polar coordinates, reiθ . In the complex case, r is
a non-negative real number and θ indicates how far the point is rotated from the positive real
axis. In our matrix case, θ will be replaced by a rotation matrix and r will be a positive-definite
matrix. A positive-definite matrix M is one where v · Mv > 0 for any non-zero vector v. If we
take v1 = 1, v2 = 0, and v3 = 0, we find that M11 > 0. Similarly, we can show that all diagonal
entries of M must be positive. (It’s harder to say anything specific about the off-diagonal entries,
which can be negative.) Another useful choice of v is to have it be one of the eigenvectors of M.
In this case v · Mv = v · λv = λv 2 , so λ > 0. Thus the eigenvalues of a positive-definite matrix
are all positive. (For the special case of a diagonal matrix, having positive eigenvalues is exactly
the same having positive diagonal entries.)
Now for the decomposition statement: given a real matrix F with non-zero determinant (i.e.,
F is invertible), there is a unique way to write F = RU, where R is a rotation matrix and U is
positive-definite and symmetric. (In class H was used instead of U.) There is also a unique way
to write F = VR, where R is a rotation matrix and V is positive-definite and symmetric, and
the same matrix R appears in both cases. I will show how to calculate U, R, and V. I’ve put
some of the “proof” parts of the construction, showing U and R have the desired properies, as
parentheticals. Skip them if you want.
Before starting, note that if there are matrices with F = RU, then FT F = (RU)T RU =
UT RT RU = UT U = U2 . The third equality is because the inverse of a rotation matrix is its
transpose, and the last is because U is symmetric. So to find the decomposition, perform the
following steps:
1) Calculate FT F.
This matrix must be symmetric, since (FT F)T = (FT )T FT = FFT . That means its eigenvalues are real, and it has a set of orthonormal eigenvectors.
2) Diagonalize, so that A−1 FT FA is a diagonal matrix. Its entries are the eigenvalues of FT F,
so they are real. Also, using the orthonormal eigenvectors as the columns of A means that
A is an orthogonal matrix, A−1 = AT .
3) Now we want the square root of AT FT FA. For that we really want to know not only that
its eigenvalues are real, but that they are positive. That holds since FT F must be positivedefinite: for any non-zero v, v · FT Fv = Fv · Fv > 0. We already showed that the
eigenvalues of any positive-definite matrix are positive.
T T
To get a positive-definite
√ U, take the positive square root of each of the entries of A F FA.
Then define U = A A√T FT FAAT .
√
T T
T
T
(As a check: U2 = A AT FT FAAT A AT FT FAAT = A(A
√ F FA)A = F F. Also,
since the square root matrix is diagonal, note that UT = (A AT FT FAAT )T = U, so U
is indeed symmetric.)
4) Now calculate R = FU−1 . Inverting U could be an ugly calculation, so you can instead do the calculation when you’re still in the diagonal basis: R = FAAT U−1 AAT =
FA(AT UA)−1 AT = FA( √AT 1FT FA )AT .
(R must be a rotation matrix because RT = (FAAT U−1 AAT )T = A(AT U−1T A)AT FT =
A(AT U−1 A)T AT FT = A(AT U−1 A)AT FT = U−1 FT , where the next-to-last step holds
because the transform of a diagonal matrix is itself. This means RT R = U−1 FT FU−1 =
U−1 U2 U−1 = I.)
5) For the decomposition in the other order, calculate V = FR−1 = FRT .
6) The above shows that the polar decomposition on each side exists, but not that it is unique.
The uniqueness argument relies on reversing part of the above calculation, in particular
step 3), so that you get a formula for FT F in terms of U. You can then show that if you
have some other positive-definite matrix U1 that works in the polar decomposition, it does
in fact have to equal U.
Physically, U and V represent pure stretch matrices. The decomposition theorem says that
any deformation can be written as a combination of a rotation and a pure stretch. Although
these operations can be done in either order, F = RU has a nicer interpretation when acting
on X. First U acts, which means you can think of the stretching in the material coordinates.
Then R rotates the sample to its new orientation. The eigenvectors of U are the principal strain
directions, and the eigenvalues are the stretch factors. Eigenvalues larger than one correspond to
elongation, and eigenvalues smaller than one indicate contraction. Negative eigenvalues would
mean the length of a vector changes sign, which doesn’t correspond to a physically realistic
deformation, so it’s good that they are mathematically unnecessary!
In a limit of very small deformation, we can write F = I + A, where √
the entries of A are much
less than 1. Then FT F = (I + AT )(I + A) ≈ I + A + AT , and U = FT F ≈ I + 12 (A + AT ).
We define the second term as E, which is known as the infinitesimal strain tensor. It is exactly
the symmetric part of the deviation A.