The Second Theorem

There is another important result by Helmholtz in the same paper we mentioned above. It concerns the most general infinitesimal deformation of a plastic (that is, non-rigid) body.² Consider a small volume element of a deformable body (a fluid, say). Putting the origin of the coordinates inside this volume element, let $\vec{r}$ be the position of a generic point $P$. The origin is denoted by $O$. After a deformation, the material point $P$ has a new position vector, $\vec{r}+\vec{s}$. Also the point of the body located at the origin moved, its position now being given by $\vec{s}_0$. We assume, as physically reasonable, that $\vec{s}$ is, as a function of position, continuous and differentiable to any order. In what follows, in order to get results of enough generality, we assume $P$ very close to $O$, that is to say, we consider an infinitesimal volume of the body. Notice that $\vec{s}$ itself doesn't have to be small in what follows, though, in considering deformations, this is somewhat academic, as finite deformations can always be obtained from infinitesimal ones (for instance, by Lie group methods). This notwithstanding, the geometrical interpretation we will get is only clear for infinitesimal $\vec{s}$. On the other hand, the application for the electric field which is done below would be impoverished by confining $\vec{s}$ to be infinitesimal. A Taylor expansion for $\vec{s}$ gives

$$\vec{s}(\vec{r})=\vec{s}_0 + (\vec{r}.\vec{\nabla})\vec{s}$$ (14)

neglecting further corrections, what is allowed by the restriction to infinitesimal volume. Here we wrote $\vec{s}_0$ for $\vec{s}(\vec{0})$. In more detail, if $s_i$ is the i-th component of $\vec{s}$, one has

$$s_i(\vec{r}) = (s_0)_i + \sum_l x_l(\frac{\partial s_i}{\partial x_l} )_{\vec{r}=0}$$ (15)

This can be abbreviated to

$$s_i = (s_0)_i +x_l \partial_l s_i$$ (16)

or,

$$s_i = (s_0)_i + x_l T_{li}$$ (17)

where, obviously,

$$T_{li} = (\partial _ls_i)_{\vec{r}=0}\;.$$ (18)

In order to analyse this deformation in terms of more basic ones, let us decompose $T_{li}$ in the following way:

$$T_{li}=\frac{1}{2}(T_{li}+T{il}) + \frac{1}{2}(T_{li}-T_{il})$$ (19)

and consider separately the symmetric and the antisymmetric parts. The antisymmetric part is

$$\frac{1}{2}(\partial _ls_i-\partial _is_l)=\frac{1}{2} \epsilon_{lik}(curl\;\vec{s})_k\;.$$ (20)

The symmetric part is $S_{li}\equiv \frac{1}{2}(T_{li}+T_{il})$. It can be decomposed as follows:

$$S_{li} = \frac{1}{3}\delta_{li}S + (S_{li}-\frac{1}{3}\delta_{li}S)$$ (21)

where $S \equiv Tr(S_{li})=S_{ii}\;.$ Now, $S_{li}=\frac{1}{2}(\partial _ls_i+\partial _is_l)$, so that

$$S=\partial_i s_i=\vec{\nabla}.\vec{s}\;.$$ (22)

We can therefore write Eq.(21) as

$$S_{li}=\frac{1}{3}\delta_{li}\vec{\nabla}.\vec{s} + S^0_{li}$$ (23)

where $S^0_{li}$ is a traceless symmetric tensor. Going now to Eq.(17), we can write

$$s_i=(s_0)_i + x_l\frac{1}{2}\epsilon_{lik}(curl \; \vec{s})_k +\frac{1}{3}x_i\;div \vec{s}+x_lS^0_{li}$$ (24)

or,

$$\vec{s}=\vec{s}_0 + \frac{1}{2}(curl\,\vec{s}\times\vec{r}) + \frac{1}{3} \vec{r} div\,\vec{s} + \vec{r}.\bf {S^0}$$ (25)

where the last term is a vector whose i-th component is $x_lS^0_{li}$. An object like $\bf {S^0}$ is sometimes called a dyadic.It is a second-order tensor. Let us examine the meaning of the several terms of Eq.(25). The first term of the second member is a translation. The second is an infinitesimal rotation around the axis $(curl\,\vec{s})_{\vec{r}=0}$, and the remaining terms describe a dilatation of the volume element. To understand them better, let us suppose temporarily that the translation and the rotation vanish, so that we have, for a component of $\vec{s}$:

$$s_i = \frac{1}{3} div\,\vec{s}\;x_i + x_lS^0_{li}\;.$$ (26)

Now, $S^0_{li}$ is a symmetric matrix, so it can be diagonalized. This means that we can change coordinate axes in such a way that, in the new ones, the matrix elements of $S^0$ have the form $S^0_{li}=\delta_{li}S^0_{ii}$, where no sum is implied in this last expression. So, if the new coordinates are denoted by $X_i$, we have

$$\sum_lx_lS^0_{li}=\sum_lX_l\delta{li}S^0_{ii}=X_i S^0_{ii} \;\;\;no \; sum\;on \;i$$ (27)

so that

$$s_i = \frac{1}{3}div\,\vec{s}X_i + S^0_{ii}X_i \;\;(no\;sum)$$ (28)

Consider the infinitesimal ``cube'' centered at $O$, with $P$ as a vertex. (The ``cube'' may have curvilinear edges, at least after the deformation). Its volume before the deformation was, in the new coordinates, $V=X_1X_2X_3$. After the deformation, it is $V'=(X_1+s_1)(X_2+s_2)(X_3+s_3)$. As we have, considering only the dilatation,

$$s_i=\frac{1}{3}div\,\vec{s}X_i+S^0_{ii}X_i=X_i(\frac{1}{3}div\,\vec{s}+S^0_{ii})\;$$ (29)

then

$$X_i+s_i=X_i(1+\frac{1}{3}div\;\vec{s}+S^0_{ii})\equiv X_i(1+\epsilon_i)$$ (30)

and, for the volume,

$$V'=X_1X_2X_3(1+\epsilon_1)(1+\epsilon_2)(1+\epsilon_3)$$ (31)

which, to first order is

$$V'=V(1+\epsilon_1+\epsilon_2+\epsilon_3)$$ (32)

that is,

$$V'=V(1+div\;\vec{s}+S^0_{11}+S^0_{22}+S^0_{33})$$ (33)

but, as $S^0_{ij}$ is traceless,

$$V' = V(1+div\;\vec{s})\;.$$ (34)

So, the $S^0_{ij}$ do not contribute to the change of volume, but do contribute to the change of the form of the little cube. Referring again to Eq.(25), we see that $\vec{s}$ is written as a sum of a translation, plus a rotation, plus an isotropic dilatation, plus a volume-conserving deformation.³ Notice that Eq.(25) was obtained using only Calculus. So, it should apply to any vector field whatsoever. To shed more light on the role of each term of the expansion, let us use it for the electric field. We then have:

$$\vec{E}(\vec{r})=\vec{E}(\vec{0}) + \frac{\vec{r}}{3}(div\,... ...{2}[(curl\,\vec{E})_0\times \vec{r}] + \vec{r}.{\bf S}^0 \;.$$ (35)

As $(div\,\vec{E})_0=4\pi \rho(\vec{0})$, the second term reads

$$\frac{\vec{r}}{3}(div\,\vec{E})_0=\frac{\vec{r}}{3}4\pi\rho(\vec{0})= \frac{4\pi}{3}r^3\rho(\vec{0})\frac{\vec{r}}{r^3}\;.$$ (36)

Now, this is the field of a uniformly charged sphere of radius $r$ at a point of its surface. It is a radial field to be added to $\vec{E}_0$. The second term vanishes if the fields are static. Otherwise,

$$(curl\,\vec{E})_0=-\frac{1}{c}(\frac{\partial \vec{B}}{\partial t})_0$$ (37)

and

$$\frac{1}{2}[(curl\,\vec{E})_0\times \vec{r}]=-\frac{1}{2c}\frac{ \partial}{\partial t}[\vec{B}(\vec{0})\times\vec{r}]\;.$$ (38)

Consider a uniform magnetic field of value $\vec{B}(\vec{0})$. A vector potential corresponding to it is

$$\vec{A}(\vec{r})=\frac{1}{2}\vec{B}(\vec{0})\times \vec{r}$$ (39)

so that we have

$$\frac{1}{2}[(curl\,\vec{E})_0\times \vec{r}]=-\frac{1}{c} \frac{\partial \vec{A}}{\partial t}\;.$$ (40)

This means that the third term of (35) is the contribution of the magnetic field at the origin, treated as follows: extend the value of $\vec{B}$ at $\vec{0}$ to a uniform field. Compute its vector potential and then add the term $-\frac{1}{c}\frac{\partial \vec{A}}{\partial t}$ to $\vec{E}(\vec{0})$. All the rest of the field at $\vec{r}$ (and that could be a lot!) comes from the term ${\bf S}^0$.