Reading 2/4/2008: 2.4-2.6

Section 2.4: In deriving the continuity equation, we're relying heavily on the continuity and differentiability of the density and velocity. Does this mean we'll need entirely new techniques when analyzing systems containing the interface between two different materials (i.e. discontinuous density)? Are there situations when continuous velocity breaks down?

I thought I'd mention with respect to equation (2.19) that one easy way to understand why a total derivative becomes a partial derivative is to write it as



Written this way, the dV = dx dy dz variables can be thought of as dummy variables, much like dummy indices, since they are being integrated over. Consequently, t is the only free variable, and so passage from total differential to partial differential is purely notational, since we do not use total differentials for functions of more than one variable. (Unless, of course, they are "secretly," as in classical/quantum mechanics, a function of just one variable - time. But we can ignore that here.)

As a side note, I'm a little surprised none of this has been adapted to be used in Math 13/61 courses. It's a fantastic showing of the power of the divergence theorem.

Section 2.5: I kind of wish the explanation had been some strictly in 2D, because it took a little while to figure out where the extra components were going. That said, Figures 2.6/2.7 are very cool - it's quite clear that with symmetry we have no torque and without symmetry there must be one.

I wish a little more could be said about the stress ellipsoid. If my understanding is correct, once we've transformed to principal axes, directions x_i in which Π_ii is large have small extent. Thus the greater the stress is one direction, the smaller the extent of the ellipsoid. Similarly, in the case of negative stress, how exactly are compressive and extensional forces differentiated? And where do the new arrows come from in FIgure 2.11?

Section 2.6: This is the physics-math barrier coming up again, but for the life of me I can't remember how physicists get away with saying that if D is an "infinitesimal rotation" about P, then δ = D x dx. I know infinitesimal rotations are elements of the lie algebra of SO(3), but there's got to be a better intuitive way of understanding where the cross product comes from.

WARNING: TANGENT

So, I'm just thinking out loud here, sorry if this is useless. But I want to see if I can explore the answer to the question I just posed. Here I'm going to assume we've rigorously defined an infinitesimal rotation as an element of the lie algebra of SO(3), the group of rotations in three dimensions, but that we don't know anything about so(3). The reason is that I can't remember offhand what the lie algebra looks like.

Okay, so lets suppose we have the identity I of SO(3) and have a path γ with γ(0) = I. We would like to calculate &gamma'(0) for an arbitrary path. First, we express &gamma as a matrix by multiplying the matrices for an xy rotation of angle &theta, a yz rotation of angle &phi, and an xz rotation of angle &psi$. Notice that if we then write each of the angles as a function of time with all angles zero at time zero, then we have an arbitrary path in SO(3) beginning at the origin. We finally differentiate the matrix and plug in t = 0. This series of calculations is extremely involved, but the result is the matrix







which represents an arbitrary element of so(3). Note that each derivative above is evaluated at t = 0. Okay, cool, so it looks like so(3) is just the set of antisymmetric matrices, since those derivatives at zero can be anything we want them to be. We've established that any infinitesimal rotation is just antisymmetric matrix. Set a = dφ/dt, b = dψ/dt. and c = dθ/dt. Then we can construct a vector D = (-a,b,-c), where the negatives will be clear momentarily. We can finally recover the antisymmetric matrix B = &gamma'(0) using a trick from the book and writing

B_ij = ε_ijk D_k

So in sum, it looks like we can legitimately make a "nice" bijection between the set of infinitesimal rotations, so(3), and individual vectors. In retrospect, I suppose the cross product comes in since if we multiple an element B of so(3) by a vector x = x_j, we get




which is essentially (2.40). So if we express an infinitesimal rotation as a vector D, we can represent the result of applying the rotation to a vector x (up to a factor of -1) by taking th cross product of D and x. That is VERY cool.

Okay, sorry if that made sense/helped absolutely no one. It was really cool to me, so I left it in. When I wrote all that I hadn't even read the next line, but now I realize that what we're doing in (2.42) is recovering the ACTUAL infinitesimal rotation, which is of course the antisymmetric matrix A_ij in the book. Even the negative sign resurfaces!

END TANGENT

I just have to say the way in which we derive &delta^S is gorgeous. Eliminate translations by defining &delta and brilliantly get rid of infinitesimal rotations by removing the antisymmetric part. Not intuitive, but both practically simple and mathematically incredible.