Instantaneous Strain Primer
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Primer on infinitesimal strain analysis in 1, 2 and 3-D
The purpose of this document is to provide some background for the infinitesimal strain analysis
we are about to undertake, making use of velocity data from GPS sites. It is best to study this
document alongside the treatment of strain in an introductory structural geology textbook.
1-dimensional analysis
Imagine a thin strip of ideal elastic material – an ideal rubber band, if you will. You measure its
initial length, and use the symbol loriginal to represent that length. Then you stretch it, measure the
stretched strip, and call its new length lfinal. Two descriptive terms are particularly useful to us in
describing your observations.
The elongation or extension of the strip (e) is defined as
extension = e =
l final - loriginal
loriginal
If the strip is longer after you change its length, the extension is a positive number, and it is
negative if you shorten the strip. If there is no change in length, the extension is zero. Extension
is a dimensionless number, because it is the quotient of a length divided by a length.
The stretch (S) of the strip is defined as
stretch = S = e + 1 = lfinal/loriginal
A positive extension results in a stretch that is greater than 1, and a negative extension or
shortening results in a stretch that is less than 1. Extension (or elongation) and stretch are terms
that an introductory student of structural geology needs to fully comprehend.
Now let’s imagine doing a similar experiment, in which we have an ideal, thin elastic strip whose
end is fixed to the “0” on one end of a ruler. We will also imagine that the ruler is along the x
axis of a Cartesian coordinate system whose origin is the “0” on the ruler. Before we change the
length of the strip, we put a black mark on the elastic strip at a distance of 5 cm from the fixed
end, and a red mark at a distance of 8 cm from the end (Figure 1).
Figure 1. Change in length of an ideal elastic strip, showing the strip’s original and final lengths. The x coordinate
axis extends parallel with the ruler. Subscripts: b = black, r = red, o = original, f = final
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The black mark has coordinates in its original position of {5, 0, 0}, but because this is a 1dimensional problem we can simplify and use just the x coordinate (xbo): {5}. In a similar
manner, the red mark has an original location (xro) of {8}. Now, we stretch the band so that the
black mark is at 6 cm along the ruler, so its final position (xbf) is {6}. The final length divided by
the initial length is the stretch (S) – the very same stretch that we just defined above.
S = lfinal/loriginal
The extension (call it eb) and stretch (call it Sb) associated with the black mark is given by
eb = (6 – 5)/5 = 0.2
Sb = eb + 1 = 1.2
or Sb = 6/5 = 1.2
The stretch will be the same for all points along the elastic strip, including the black and red
marks. The stretch is also known as the 1-dimensional Lagrangian deformation gradient in the x
coordinate direction. Without digressing into history or biography, a Lagrangian approach to
these problems (named after Joseph-Louis Lagrange) starts with an initial configuration and
projects to a final, deformed configuration, while an Eulerian approach (named after Leonhard
Euler, whose last name sounds like “oiler”) starts with the deformed and projects backward to
the original. We are only going to use the Lagrangian approach in the GPS strain problem.
If we know the initial coordinate {xoriginal} of an arbitrary point along the thin elastic strip, we
can compute the final coordinate of that point {xfinal} given the deformation gradient or stretch
(S). Elongating the strip so that the black mark moves from 5 to 6 on the ruler will result in a
deformation gradient or stretch of (6/5) = 1.2. That same elongation will cause the position of
the red mark to change from 8 to (1.2 x 8) = 9.6 if the ideal strip we are stretching is
homogeneous and has a linear-elastic rheology.
Another way of defining the deformation gradient focuses simply on the change in distance
between the two marks on our elastic strip. The initial distance between the black and red marks
(xo) is 8 – 5 = 3, while the final distance between the black and red marks (xf) is 9.6 – 6 = 3.6.
The stretch between these two points (S) is
S=
xrf - xbf Dx f
=
xro - xbo Dxo
The stretch is the slope of a line that relates the difference in initial position of two points to the
difference in position of the same two points at some other time. In the case involving our black
and red marks, S = 3.6/3 = 1.2 and so the deformation gradient is 1.2.
For our purposes as we build toward the GPS strain problem, it is useful to think about the 1-D
changes in the elastic strip in terms of displacements, which are commonly referenced using the
symbol u. Let us define the location vector to the initial location of the black mark as xbo, whose
1-D coordinate is {5}, and the location vector to the final location of the black mark is xbf = {6}.
The displacement vector ub is given by the vector equation
ub = xbf – xbo, so
ub = {6} – {5} = {1}
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This is called a displacement vector, because it extends from the original position to the
displaced or final position. Similarly, let’s define the location vector to the initial location of the
red mark as xro, whose 1-D coordinate is {8}, and the location vector to the final location of the
red mark is xrf = {9.6}. The displacement vector ur is given by the vector equation
ur = xrf – xro, so
ur = {9.6} – {8} = {1.6}
The displacement vector associated with the black mark that was originally located at {5} has a
length of 1, while the displacement vector associated with the red mark initially 3 cm farther
down the ruler at {8} has a length of 1.6. The magnitude of the displacement vector varies as a
function of position along the x coordinate axis.
Earlier, we defined the extension (e) as
extension = e =
l final - loriginal
loriginal
The extension is also equal to the difference between the lengths of two displacement vectors
divided by the difference between the lengths of the corresponding two initial-location vectors
e=
u r - ub
Du
=
.
xro - xbo Dx
The extension is the slope of a line that relates the difference in initial position of two points to
the difference in displacement of the same two points at some other time, or the 1-dimensional
displacement gradient. Using the 1-D example involving the black and red marks along the
elastic strip,
e=
3.6 - 3 0.6
=
= 0.2
8-5
3
The 1-D deformation gradient is equal to the stretch, and in this example has a value of 1.2.
The 1-D displacement gradient is equal to the extension, and in this example has a value of 0.2.
So far, we have thought about deformation and displacement that we can see and easily measure.
The change in length of the strip between the origin and the black mark amounted to 20% of the
original length. This kind of change is normally considered finite strain. The obvious strain that
we observe in rocks is the result of finite strain. If the change is on the order of 1% or less of the
original length, it would normally be considered to be infinitesimal strain. For example, a
change of 5 millimeters in the distance between two GPS sites that were originally 35 km apart is
an excellent example of infinitesimal strain, because the change would be ~1.4x10-5 %.
In many cases this strain is ongoing, so that it makes more sense to consider a displacement rate
or velocity rather than a single displacement. When considering velocity, which has dimensions
of length per time, the change in length becomes infinitesimal as the corresponding time interval
is reduced toward zero. Thus when time is involved, the idea of instantaneous strain or
instantaneous velocity is comparable to infinitesimal strain.
The infinitesimal 1-D deformation gradient (stretch) can be expressed as
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S=
Dx f
¶x
= lim f
Dxo
¶xo
and the infinitesimal 1-D displacement gradient (extension) is
e=
Du
¶u
= lim
Dxo
¶xo
The  symbol indicates a partial derivative, which is not really necessary in the 1-D case but
prepares us to think about the 2-D and 3-D cases in which the functions might vary in 2 or 3
dimensions.
3-dimensional analysis
The material that follows is generally confined to information about infinitesimal strain analysis
that is applicable to the analysis of GPS velocity data. More comprehensive treatments of this
material are available from several sources, including Malvern (1969), Timoshenko and Goodier
(1970), Means (1976), Oertel (1996), Pollard and Fletcher (2005), and Allmendinger and others
(2012).
If the mental image we worked with in the previous section was that of an ideal elastic string
whose length is changed in one dimension, parallel to the x coordinate axis, we are now going to
build on that experience to think of a continuous elastic sheet that can be deformed in 3 spatial
dimensions with Cartesian x, y and z axes.
Analogous to the 1-dimensional case, the 3-D Lagrangian displacement equation includes the
displacement vector {ux, uy, uz} associated with a particular reference point, a vector defining the
initial location of the reference point {xo, yo, zo}, and the 3x3 displacement gradient tensor.
é
ê
é ux ù ê
ê
ú ê
ê uy ú = ê
ê
ú ê
êë uz úû ê
ê
êë
¶u x
¶xo
¶u x
¶yo
¶u y
¶xo
¶u y
¶yo
¶uz
¶xo
¶uz
¶yo
¶u x ù
ú
¶zo ú
é xo ù
ú
¶u y ú ê
ú ê yo ú
¶zo ú ê
z ú
¶u z ú ë o û
ú
¶zo ú
û
The matrix of extensions in the displacement gradient tensor is
é
ê
ê
ê
ê
ê
ê
ê
êë
Version of 6 September 2012
¶u x
¶xo
¶u x
¶yo
¶u y
¶xo
¶u y
¶yo
¶uz
¶xo
¶uz
¶yo
¶u x ù
ú
¶zo ú
é e11 e12
¶u y ú ê
ú=ê e
e
¶zo ú ê 21 22
e
e
¶uz ú ë 31 32
ú
¶zo ú
û
e13 ù
ú
e23 ú = eij .
e33 ú
û
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The elements of the matrix along the diagonal of the displacement gradient tensor (e11, e22, e33)
are the extensions along the x, y and z coordinate axes, respectively. In other words, the diagonal
terms are related to changes in length along each of three orthogonal directions. The offdiagonal or off-axis elements are the tensor shear strains of lines originally parallel to the
coordinate axes, and reflect angular changes within the deforming material. (The tensor shear
strain is half of the engineering shear strain you might have encountered elsewhere.) Component
e21 expresses the counter-clockwise (positive) rotation of a vector that is initially parallel to the x
coordinate axis (or 1 axis) toward the y axis (or 2 axis) around the z axis (or 3 axis) during
deformation. Similarly, given a vector that is initially parallel to the y axis, component e12
expresses the rotation of that vector clockwise toward the x axis around the z axis during
deformation.
The displacement gradient tensor is an asymmetric matrix that represents both distortion (that is,
change in shape or change in volume) and rotation. A tensor that represents only rotation would
be antisymmetric, so that e12 = -e21, e13 = -e31, and e23 = -e32. The asymmetry in the displacement
gradient tensor arises because of infinitesimal rotations and distortion. An asymmetric tensor
can be decomposed into the sum of a symmetric tensor and an antisymmetric tensor. The
asymmetric displacement tensor eij shown above is the sum of the symmetric infinitesimal strain
tensor ij and the antisymmetric rotation tensor ij :
eij = ij + ij, where
é
e11
ê
ê
ê (e + e )
e ij = ê 21 12
2
ê
ê (e31 + e13 )
ê
2
ë
é
0
ê
ê
ê (e - e )
Wij = ê 21 12
2
ê
ê (e31 - e13 )
ê
2
ë
(e12 + e21 )
2
e22
(e32 + e23 )
2
(e12 - e21 )
2
0
(e32 - e23 )
2
(e13 + e31 )
2
(e23 + e32 )
2
e33
(e13 - e31 )
2
(e23 - e32 )
2
0
ù
ú
ú
ú
ú
ú
ú
ú
û
ù
ú
ú
ú
ú
ú
ú
ú
û
The antisymmetric rotation tensor ij is also known as an axial vector. The axial vector is
directed along the rotational axis, and the length of the axial vector is equal to the angle of
rotation expressed in radian measure. The Cartesian coordinates of the axial vector are given by
{rx, ry, rz}, where
æ æ (e - e )ö æ (e - e )ö ö
- ç ç 23 32 ÷ - ç 32 23 ÷ ÷
2
2
ø è
øø
èè
-(W23 - W 32 )
rx =
=
2
2
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æ æ (e - e )ö æ (e - e )ö ö
- ç ç 13 31 ÷ - ç 31 13 ÷ ÷
2
2
ø è
øø
èè
-(W13 - W 31 )
ry =
=
2
2
æ æ (e - e )ö æ (e - e )ö ö
- ç ç 12 21 ÷ - ç 21 12 ÷ ÷
2
2
ø è
øø
èè
-(W12 - W21 )
rz =
=
2
2
.
The angle of rotation in radians is the length of the axial vector,
r = rx2 + ry2 + rz2
If the angle is a positive number, the rotation is an anti-clockwise rotation as viewed looking
down the rotational axis toward the origin of the coordinate system. The computed angle only
has meaning under conditions of plane strain, and reflects the angle and direction that the
maximum principal strain axis is rotating as deformation progresses. If deformation is entirely
by pure strain (also known as irrotational strain or coaxial strain), the angle will be zero.
Example. Given the following displacement gradient tensor (eij), calculate the strain tensor (ij)
and the rotation tensor (ij).
é e11 e12 e13 ù é 3 3 2 ù
ê
ú ê
ú
eij = ê e21 e22 e23 ú = ê 9 8 1 ú
ê e
e
e ú êë 6 -1 5 úû
ë 31 32 33 û
Solution. The strain tensor is
é
ê
e11
ê
ê e +e
e ij = ê ( 21 12 )
2
ê
ê (e + e )
ê 31 13
2
êë
( e12 + e21 ) ( e13 + e31 )
2
e22
( e32 + e23 )
2
2
( e23 + e32 )
2
e33
ù é
ú ê
3
ú ê
ú ê ( 9 + 3)
ú=ê
2
ú ê
ú ê (6 + 2)
ú ê
2
úû ë
( 3 + 9)
2
8
( (-1) + 1)
2
ù
ú
2
ú é
3 6 4
(1+ (-1)) ú = ê 6 8 0
ú ê
2
ú ê 4 0 5
ú ë
5
ú
û
(2 + 6)
ù
ú
ú
úû
and the rotation tensor is
é
ê
0
ê
ê e -e
Wij = ê ( 21 12 )
2
ê
ê (e - e )
ê 31 13
2
êë
( e12 - e21 ) ( e13 - e31 )
2
0
( e32 - e23 )
Version of 6 September 2012
2
2
( e23 - e32 )
2
0
ù é
ú ê
0
ú ê
ú ê ( 9 - 3)
ú=ê
2
ú ê
ú ê ( 6 - 2)
ú ê
2
úû ë
( 3 - 9)
2
0
( (-1) - 1)
2
ù
ú
2
ú é
0 -3 -2 ù
(1- (-1)) ú = ê 3 0 1 ú
ú ê
ú
2
ú ê 2 -1 0 ú
ë
û
ú
0
ú
û
( 2 - 6)
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The Cartesian coordinates of the axial vector are
- ( W23 - W32 ) -(1- (-1))
rx =
=
= -1
2
2
- ( W13 - W31 ) -((-2) - 2)
ry =
=
=2
2
2
- ( W12 - W21 ) -((-3) - 3)
rz =
=
=3
2
2
or {-1, 2, 3}. The angle of rotation is
r = rx2 + ry2 + rz2 = (-1)2 + 2 2 + 32 = 3.74 radians
Eigenvectors, eigenvalues, principle strains and shear strains
The principal strain axes are found by computing the eigenvectors of the strain tensor ij. The
eigenvectors are vectors in the directions of the principal strain axes. The eigenvalues of ij are
the principal extensions in the principal directions. The largest of the three eigenvalues is the
greatest principal extension, e1, and the length of the semi-major axis of the strain ellipsoid is
equal to the greatest principal stretch, S1 = e1 + 1. The semi-minor axis is S3 = e3 + 1, where e3 is
the smallest eigenvalue and the least principal extension, and the intermediate axis is S2 = e2 + 1,
where e2 is the intermediate principal extension.
The magnitude of the infinitesimal engineering shear strain (max) at 45° to the maximum
principal strain axis can be computed several ways.
2
æe -e ö
max = 2 ç 11 33 ÷ + ( e13 )
è
2 ø
2
max = e1 – e3
and, recalling that
(S1)2 = (e1 +1)2 and (S3)2 = (e3 +1)2,
2
2
S1 ) - ( S3 )
(
max =
.
2
2
2 ( S1 ) ( S3 )
The engineering shear strain really only has meaning in cases of plane strain. The volume strain,
which is also the first invariant of the strain tensor, is given by
volume strain = 11 + 22 + 33 = e1 + e2 + e3 = (S1 S2 S3) - 1
Invariants of a tensor are combinations of tensor components that do not change with changing
coordinate systems. The second invariant of the strain tensor is
( 11  22) + ( 22  33) + ( 11  33) –  122 –  232 –  132 = (e1 e2) + (e2 e3) + (e1 e3)
and the third invariant is
determinant[ ij] = e1 e2 e3
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2-dimensional analysis
It would not seem to make any sense to have a discussion of 2-D analysis after discussions of 1D and 3-D analysis. None-the-less, there are two reasons for this disorderly order. First, the 2-D
analysis is a special case of 3-D analysis, so the discussion of 2-D analysis logically follows
from prior discussions. Second, we are going to use what we learn about 2-D analysis to solve
for the horizontal infinitesimal strain rate between three GPS sites. It makes sense to focus a bit
of extra time and energy to understand the elements of the 2-D analysis.
Several first-order assumptions are made in this analysis.
• We assume plane-strain deformation in the horizontal plane.
• We assume that neglecting the vertical velocities does not significantly affect the physical
interpretation of our answer, recognizing that this might be a particularly poor assumption
in areas characterized by significant uplift or subsidence.
• We assume that strain is homogeneously distributed across the triangular area between the
three GPS sites.
Our input data are locations of GPS sites and their velocities, presented in units of length per
time where “time” is usually one year. The input velocity data are (effectively) instantaneous
velocities, so the output data should properly be expressed as rates. In the limit as time shrinks
to zero, the velocity vector and the displacement vector converge and become identical as
defined in the same reference frame. Hence, we retain the usual terminology of a displacement
gradient tensor rather than a velocity gradient tensor in the discussion that follows.
Taking one step back, the GPS data acquisition and analysis process resolves very small
displacements. The processed data available from the PBO site are daily averages of
displacements relative to the initial position of the GPS receiver as resolved in a particular
reference frame, such as the Stable North American Reference Frame (SNARF), relative to a
particular datum, such as WGS 84. The mean velocities are computed from a time series
comprised of many of these daily displacements. So we could choose to use the average
displacement between the same initial time and final time for all three GPS sites (in which case
our input data would be actual displacements) or the average displacement divided by the time
interval associated with that displacement (in which case our input data are velocities). In the
first case, our output would be a measure of the deformation during that specific time interval,
and in the second case our output is the mean deformation rate. Apart from that distinction, the
analysis described here is the same.
The 2-D analysis using data from 3 GPS sites is a perfectly constrained problem, with just the
right number of known quantities to solve for the unknown quantities exactly. Of course, all of
the input data have uncertainties associated with them, as does our “exact” solution. As a firstlook at infinitesimal crustal strain, the 3-site triangle-strain problem provides a good foundation
to build on as we learn how to work with more data. The perfectly constrained triangle-strain
problem by itself is not sufficient to provide a detailed characterization of tectonic strains in an
area. A full understanding of present-day crustal strain in a given area requires information from
all locally available GPS sites as well as other geological/geophysical data. Due consideration
would have to be paid to how a variety of non-tectonic factors might affect the GPS velocity
data, including flux of groundwater and surface water, igneous activity, deformation related to
the earthquake cycle, mass movement that might affect a given GPS site, and so on.
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After we acquire the needed information from each of the GPS sites, we will know the horizontal
coordinates of the initial site location (xo, yo) as well as the east-west instantaneous velocity (vx)
and the north-south instantaneous velocity (vy) of each site (Figure 2). The site location
coordinates expressed in the Universal Transverse Mercator (UTM) system are in meter units, so
the instantaneous velocities are expressed in meter/year units. We will have to compute the
unknown quantities: the coordinates of the translation velocity vector (tx, ty), the magnitude of
the angular velocity vector () in nano-radian per year, and the elements of the strain rate tensor
(xx, xy, yx, yy). Strain is a dimensionless quantity, but it has seemed odd to express strain rate
as some number "per year." That seems to beg the question, "What, per year." So to keep us all
happy and well adjusted, we throw in the word "strain" per year, even though "strain" has no
dimension. The units used in infinitesimal strain-rate studies using GPS data are nano-strains per
year.
Figure 2. Map of three GPS sites separated from one another by tens of kilometers. The east-west and north-south
velocities of each site relative to the Stable North American Reference Frame are indicated by arrows.
The velocities measured at the three GPS sites are the result of three component velocities:
translation, rotation and distortion of the crust.
deformation rate = translation rate + rotation rate + distortion rate
The translation velocity vector has the same magnitude and direction at all three sites as the
triangle of stations simply move in space relative to the reference frame. You can think of the
translation velocity vector as extending from the centroid of the triangle of initial GPS site
locations to the centroid of the deformed triangle. (Remember that the centroid of a triangle is
the intersection of three line segments, each of which connects one apex with the midpoint of the
opposite side.)
The rotational velocity vector (also known as an axial vector or angular velocity vector) in this 2D case is a vertical vector extending from the center of the triangle. A positive rotation is a
counter-clockwise rotation. The meaning of this rotation would be clear enough if we were
thinking about a rigid body that is not distorting as it is rotating. In our case, however, virtually
all of the material lines in our deforming body (including the sides of the triangle between GPS
sites) will be affected by shear strain, which changes the trend of the line as the area is changing
shape. How do we separate the rotation of the whole area from rotations resulting from shear
strain?
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Imagine that you use the initial locations of our GPS sites, find the centroid of the triangle, and
draw a (really) big circle around that center (Fig. 3A). Then the GPS sites move, and the circle is
distorted into an ellipse (Fig. 3B,D). You draw a blue line along the longer (major) axis of the
ellipse, and a red line along the shorter (minor) axis that is 90° from the trend of the major axis.
You use your field compass to measure the trend of the major axis in the deformed state. Then
you hit the REVERSE button on any convenient time machine (remember, I started this with the
word “imagine”), and reverse the deformation. You would find that the two lines you drew
along the axes of the ellipse would still be 90° apart in the undeformed circle (Fig. 3C). You
Figure 3. Defining rotation. A. Initial configuration of GPS triangle, with circle around the centroid of the triangle.
B. Deformed triangle with circle changed to an ellipse. Major axis of the ellipse trends 60° and is blue; minor axis
is red. C. Retrodeformed triangle showing orientation of red and blue material lines prior to deformation. Trend of
blue line is 60°, so deformation shown in B did not involve rotation. D. Second example of deformed triangle.
Major axis trends 45° and the same material line in the original triangle trended 60°, so this deformed triangle has
undergone a positive (counter-clockwise) rotation.
might find that the blue line in the circle has the same trend as it had as the major axis of the
ellipse, in which case there was no bulk rotation of the triangle during deformation (Fig. 3B). If
the trend of the blue line changed during deformation, that change in angle is the bulk rotation
(Fig. 3D).
Distortion includes a change in shape (strain) as well as any change in volume/area (dilation).
Like its counterpart described for 3-D deformation, the 2-D displacement gradient tensor is an
asymmetric matrix that can be decomposed into the sum of a symmetric tensor and an
antisymmetric tensor, representing rates of both distortion and rotation. The rotational-velocity
matrix is antisymmetric.
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é 0 -W ù
Wij = ê
ú
ë W 0 û
The infinitesimal strain-rate tensor ij is symmetric,
é e xx
e ij = ê
e xy ù
ú
ê e yx e yy ú
ë
û
so that xy = yx and the 2-D strain-rate tensor thus contains only three independent variables: xx,
xy = yx, and yy. Adding these two tensors together gives us the displacement-rate gradient
tensor eij
é e xx
e xy - W ù
ê
ú
e yy ú
ê e xy + W
ë
û
to which we can add the translation terms {tx, ty} so that we account for all elements of the
deformation
é e
e xy - W ù é t x ù
xx
ê
ú+ê
ú.
ê e xy + W
ú
t
ê
ú
e
yy
ë
û ë y û
Note that the translation is independent of position. The complete matrix equation for the
deformation at an individual site, given its initial location {xo, yo} and the corresponding east
velocity (vx) and north velocity (vy) is
é v ù é e
e xy - W
xx
ê x ú=ê
ê vy ú ê e xy + W
e yy
ë
û ë
ùé x ù é t ù
úê o ú + ê x ú
úê yo ú ê t y ú
û ë
û
ûë
which can be unpacked to yield the following two equations with six unknowns: tx, ty, , xx, xy
and yy :
vx = (xo xx) + (yo xy) – (yo ) + (tx)
vy = (xo xy) + (xo ) + (yo yy) + (ty)
GPS data give us the horizontal velocities (vx and vy) for all three sites. Thus, we have six
equations (2 for each site) with six unknowns, which makes this a perfectly constrained problem
that will yield an exact answer. The six equations can then be rearranged into a single matrix
equation of the form
d=Gm
where d (for data) is the matrix of the known velocity components for our three stations, m (for
model) is the matrix of our six unknowns, and G (named after George Green) is the matrix of
coefficients that relates d to m using the equations that we just developed.
Version of 6 September 2012
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Instantaneous Strain Primer
é
ê
ê
ê
ê
ê
ê
ê
ê
ê
êë
Comments/corrections to Vince_Cronin@baylor.edu or presor@wesleyan.edu
vx ù é
ú ê
site1
vy ú ê
ú ê
site2
vx ú ê
ú=ê
site2
vy ú ê
ú
site3
vx ú ê
ê
site3
vy úú êë
û
site1
1 0
0 1
- site1 yo
site1
1 0 0 1
1 0 0 1
yo
0
site1
xo
site2
site2
yo
yo
0
site2
xo
site3
site3
yo
site3
xo
xo
site3
site3
site1
xo
site2
site2
site1
yo
xo
xo
xo
xo
0
ùé
úê
site1
yo ú ê
ú
0 ú êê
ú
site2
yo ú ê
ê
0 úê
úê
site3
yo ú ë
û
0
tx ù
ú
ty ú
ú
W ú
e xx ú
e xy úú
e yy ú
û
Because we already know all of the elements of the data matrix d, but do not know the model
parameters in matrix m, we will need to solve the inverse problem so that all of the known
quantities are collected on one side and the unknowns are on the other side of the equation.
m = G-1d
Once we have computed matrix m, its six components include the coordinates of the
translational velocity vector (in m/yr), the rotational velocity (in radian/yr), and the three
independent elements of the strain matrix.
The principal strain axes are found by computing the eigenvectors of the 2-D strain tensor ij.
The eigenvectors are unit vectors in the directions of the principal strain axes. The eigenvalues
of ij are the principal extensions in the principal directions. The larger of the two eigenvalues is
the greater principal extension, e1, and the length of the semi-major axis of the horizontal strain
ellipse is equal to the greater principal strain, S1 = e1 + 1. The semi-minor axis is S2 = e2 + 1,
where e2 is the lesser principal extension.
In the specific case in which we have a 2-D strain tensor
é e11
e12 ù
ú
e 22 ú
e ij = ê
êë e 21
û
the eigenvalues are given by
l=
(e11 + e 22 ) ±
( 4e12e 21 ) + ( e11 - e 22 )2 .
2
The larger eigenvalue is 1 (also known as e1) and the smaller eigenvalue is 2 (also known as
e2). An eigenvector corresponding to 1 is
ì l1 - e11 ü
í1,
ý,
e12 þ
î
and an eigenvector associated with 2 is
ì l2 - e11 ü
í1,
ý.
e12 þ
î
Version of 6 September 2012
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Instantaneous Strain Primer
Comments/corrections to Vince_Cronin@baylor.edu or presor@wesleyan.edu
The unit eigenvectors can then be determined by dividing each of the components of these
vectors by their length or norm. For example, the unit eigenvector associated with eigenvalue 1
is
ì
( l1 - e11 )
ï
ï
1
e12
,
í
2
2
ï
æ ( l1 - e11 ) ö
æ ( l1 - e11 ) ö
1+ ç
ï 1+ ç e
÷ø
è
è e12 ÷ø
12
î
ü
ï
ï
ý
ï
ï
þ
The magnitude of the infinitesimal shear strain (max) at 45° to the maximum principal strain axis
can be computed a couple of different ways, given the data we now have in hand.
æe -e ö
max = 2 ç 11 22 ÷ + (e12 )2
è
2 ø
2
max = e1 - e2
The area strain, which is also the first invariant of the 2-D strain tensor, is given by
area strain = 11 + 22 = e1 + e2 = (S1 S2) - 1
The second invariant of the 2-D strain tensor is
( 11  22) –  122 = e1 e2
and the third invariant is
determinant[ ij] = e1 e2
Remember that the invariants of a tensor are combinations of tensor components that do not
change with changing coordinate systems.
Estimating the uncertainty of our strain solution
The uncertainty of our strain estimate is a function of the individual uncertainties in the velocity
estimates (which are available from the same source as the GPS data) and how these
uncertainties map into the model space.
The data uncertainty can be expressed as a covariance matrix (covd) with diagonal values equal
to the individual variances (2), assuming that the error in each station’s velocity estimate is
independent of the other stations.
For a linear equation of the form m = G-1d (our inverse solution). The covariance of the model
(cov m) can then be expressed as (Menke, 1989)
[ covm] = G-1 [ cov d ]G-1
T
Version of 6 September 2012
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Instantaneous Strain Primer
Comments/corrections to Vince_Cronin@baylor.edu or presor@wesleyan.edu
or more simply as
[ cov m] = éëGT éëcov d-1 ùû G ùû
-1
where the inverse of the data covariance matrix (cov d)-1 is simply a diagonal matrix with values
equal to 1/2 for each of the velocity estimates.
æ
ç
ç
ç
ç
ç
ç
ç
covd -1 = ç
ç
ç
ç
ç
ç
ç
ç
è
1
site1vx
s2
0
1
0
0
0
0
0
0
0
0
0
0
0
0
0
0
site1vy
0
0
0
0
0
site2vy
0
0
0
0
0
0
0
0
s2
1
site2vx
s2
1
s2
1
site3vx
0
s2
0
1
site3vy
s2
ö
÷
÷
÷
÷
÷
÷
÷
÷
÷
÷
÷
÷
÷
÷
÷
ø
The diagonal elements of the 6x6 model covariance matrix (cov m) are the variances (2) for
each of the model parameters we have estimated. Their standard deviation () can be
approximated by taking the square root of variance. (Note that the off-diagonal elements of the
model covariance matrix (cov m) might not be zero, indicating some covariance between the
various model parameters.) The diagonal values in rows 1 and 2 of matrix (cov m) are the
variances for the x and y coordinates, respectively, of the translation vector. The diagonal value
in row 3 is the variance of the angular velocity vector, . The diagonal values in rows 4-6 are
the variances for xx, xy and yy, respectively.
Now we are ready to calculate the translation, rotation, and strain rate for a set of GPS stations as
well as estimate the uncertainty of our results.
Version of 6 September 2012
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Instantaneous Strain Primer
Comments/corrections to Vince_Cronin@baylor.edu or presor@wesleyan.edu
References
Allmendinger, R.W., Cardozo, N, and Fisher, D.M., 2012, Structural geology algorithms,
vectors and tensors: Cambridge University Press, 289 p., ISBN 978-1-107-40138-9.
Cardozo, N., and Allmendinger, R.W., 2009, SSPX -- A program to compute strain from
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10.1016/j.cageo.2008.05.008.
Jaeger, J.C., 1964, Elasticity, fracture and flow: Methuen, New York, 212 p.
Malvern, L.E., 1969, Introduction to the mechanics of a continuous medium: Englewood
Cliffs, New Jersey, Prentice-Hall, Inc., 713 p.
Means. W.D., 1976, Stress and strain – Basic concepts of continuum mechanics for
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Menke, W., 1984, Geophysical data analysis – Discrete inverse theory: Orlando, Florida,
Academic Press, 260 p.
Menke, W., 2012, Geophysical data analysis – Discrete inverse theory MATLAB edition:
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Oertel, G., 1996, Stress and deformation – A handbook on tensors in geology: New York,
Oxford University Press, 292 p., ISBN 0-19-509503-0.
Pollard, D.D., and Fletcher, R.C., 2005, Fundamentals of structural geology: Cambridge
University Press, 500 p.
Savage, J.C., Gan, W., and Svarc, J.L., 2001, Strain accumulation and rotation in the Eastern
California Shear Zone: Journal of Geophysical Research, v. 106, B10, P. 21,995-22,007.
Timoshenko, S.P., and Goodier, J.N., 1970, Theory of elasticity [3rd edition]: New York,
McGraw-Hill Book Company, 567 p., ISBN 07-064720-8.
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