2.7: Notes to Chapter 2

2.1

The term on the left of Equation \ref{2.1.2} becomes

\[\mathrm{\int_v \rho \mathbf{r} \frac{d \mathbf{u}}{d t} d V=\int_v \rho \mathbf{r} \frac{d^2 \mathbf{r}}{d t^2} d V=\int_v \rho \mathbf{r} \frac{d}{d t}\left(\frac{\mathbf{r} d \mathbf{r}}{d t}\right) d V-\int_v \rho \frac{d \mathbf{r}}{d t} \frac{d \mathbf{r}}{d t} d V}.\label{N2.1.1}\]

The third term can further be simplified So that

\[\mathrm{\int_v \rho \mathbf{r} \frac{d}{d t}\left(\frac{\mathbf{r} d \mathbf{r}}{d t}\right) d V=\frac{1}{2} \int_v \rho \frac{d^2}{d t^2}(\mathbf{r r}) d V=\frac{1}{2} \frac{d^2}{d t^2}\left(\int_v \rho \mathbf{r r}\right) d V}.\label{N2.1.2}\]

In obtaining the third term in Equation \ref{2.1.4} we have assumed that the volume V is large enough to contain all matter and the conservation of mass argument explicitly developed in Chapter I has the form

\[\frac{1}{2} \frac{\mathrm{d}^2}{\mathrm{dt}^2}\left(\int_{\mathrm{v}} \rho \mathbf{r r}\right) \mathrm{dV}=\int_{\mathrm{v}}(\rho \mathbf{u u}) \mathrm{dV}+\int_{\mathrm{v}} \rho \mathbf{r} \nabla \Phi \mathrm{dV}\label{N2.1.3}\]

2.2

If one writes Equation \ref{2.1.6} in component form, one gets

\[\frac{1}{2} \frac{\mathrm{d}^2 \mathfrak{I}_{\mathfrak{i j}}}{\mathrm{d t}^2}=2 \mathfrak{T}_{\mathfrak{i j}}+n \mathfrak{U}_{\mathfrak{i j}},\label{N2.2.1}\]

where in Cartesian Coordinates the tensor components are

\[\begin{aligned}
& \mathfrak{I}_{\mathrm{i j}}=\int_\mathrm{v} \mathrm{\rho x_i x_j d V} \\[4pt]
& \mathfrak{T}_{\mathrm{i j}}=\int_\mathrm{v} \mathrm{\rho u_i u_j d V} \\[4pt]
& \mathfrak{U}_{\mathrm{i j}}=\frac{1}{2} \int_\mathrm{v}\int_\mathrm{v’} \rho(\mathbf{r}) \rho\left(\mathbf{r}^{\prime}\right)\mathrm{\left(x_i-x_i^{\prime}\right)\left(x_j-x_j^{\prime}\right)\left(\left|\mathbf{r}-\mathbf{r}^{\prime}\right|\right)^{n-2} d V^{\prime} d V}
\end{aligned}\]

By contracting these tensors and letting the potential be that of the gravitational field, we recover the scalar form of Lagrange's identity as given in Chapter I, Equation \ref{1.4.12}.

2.3

Taking the scalar product of a space-like position vector with Equation \ref{2.3.6} and integrating over a volume sufficient to contain the system we get

\[\mathrm{c \int_v \sum_j x_j \frac{\partial\left(\rho u_j\right)}{\partial t} d V=\sum_j \sum_k \int_v x_j \frac{\partial \mathfrak{J}_{j k}}{\partial x_k} d V=0}.\label{N2.3.1}\]

From the chain rule the second integral can be written as

\[\sum_{\mathrm{j}} \sum_{\mathrm{k}} \int_{\mathrm{v}} \mathrm{x}_{\mathrm{j}} \frac{\partial \mathfrak{J}_{\mathrm{jk}}}{\partial \mathrm{x}_{\mathrm{k}}} \mathrm{dV}=\sum_{\mathrm{j}} \sum_{\mathrm{k}} \int_{\mathrm{v}} \frac{\partial}{\partial \mathrm{x}_{\mathrm{k}}}\left(\mathrm{x}_{\mathrm{j}} \mathfrak{J}_{\mathrm{jk}}\right) \mathrm{dV}-\sum_{\mathrm{j}} \sum_{\mathrm{k}} \int_{\mathrm{v}} \frac{\partial \mathrm{x}_{\mathrm{j}}}{\partial \mathrm{x}_{\mathrm{k}}} \mathfrak{J}_{\mathrm{jk}} \mathrm{dV}.\label{N2.3.2}\]

The first integral on the right is just the integral of the divergence of \(\mathrm{x}_{\mathrm{j}} \mathfrak{J}_{\mathrm{jk}}\) over V and if the volume is chosen to enclose the entire system the integral must vanish as \(\mathfrak{J}_{\mathrm{jk}}=0\) on the surface enclosing V (i.e., Gauss's Law applies). Since \(\partial \mathrm{x}_{\mathrm{i}} / \partial \mathrm{x}_{\mathrm{j}}=\delta_{\mathrm{ij}}\), the second integral becomes \(\sum_{\mathrm{j}} \int_{\mathrm{v}} \mathfrak{J}_{\mathrm{jj}} \mathrm{dV}\). Equation (N2.3.1) is then

\[\mathrm{\sum_j c \int_v x_j \frac{\partial}{\partial t}\left(\rho u_j\right) d V-\sum_j \int_v \mathfrak{J}_{j j} d V=0}\label{N2.3.3}\]

Now, \(\left(\partial \mathrm{x}_{\mathrm{j}} / \partial \mathrm{t}\right)=0\) from the orthogonality of the Lorentz frame and \(\sum_{\mathrm{j}=1}^3 \mathfrak{J}_{\mathrm{jj}}=\sum_{\alpha=0}^3 \mathfrak{J}_{\alpha \alpha}-\mathfrak{J}_{00}\), so we can write

\[\mathrm{c \int_v \frac{\partial}{\partial t}\left(\sum_j \rho x_j u_j\right) d V+\int_v \rho c^2 \gamma d V+\int_v \mathfrak{J}_{00} d V=0}.\label{N2.3.4}\]

With the sign convection \(\mathfrak{J}_{00}\) is the negative of the total engergy density, Bergmann¹⁷, among others, shows us that the "relativistic" kinetic energy density τ is given by

\[\begin{aligned}
\mathrm{\tau} & =\mathrm{\rho c^2\left(\gamma^{-1}-1\right)} \\[4pt]
\text{or}\quad\mathrm{\gamma \tau} & =\mathrm{\rho c^2-\rho c^2 \gamma}.
\end{aligned}\label{N2.3.5}\]

Thus, using Equation \ref{2.3.3} to re-writ the first term of Equation (N2.3.4), we get

\[\sum_{\mathrm{j}=1}^3 \int_{\mathrm{v}} \frac{\partial}{\partial \mathrm{t}}\left[\frac{1}{2} \rho \frac{\mathrm{d}\left(\mathrm{x}_{\mathrm{j}} \mathrm{x}_{\mathrm{j}} / \gamma\right)}{\mathrm{dt}}\right] \mathrm{dV}+\int_{\mathrm{v}}\left(\rho \mathrm{c}^2-\gamma \tau-\varepsilon-\rho \mathrm{c}^2\right) \mathrm{dV}=0,\label{N2.3.6}\]

where ε is the potential energy density. Applying Leibniz’s law for the differentiation of definite integrals¹⁸ to the first term in Equation (N2.3.6) and re-writing the second one, we get

\[\sum_{\mathrm{j}=1}^3 \int_{\mathrm{v}} \frac{\partial}{\partial \mathrm{t}}\left[\frac{1}{2} \rho \frac{\mathrm{d}\left(\mathrm{x}_{\mathrm{j}} \mathrm{x}_{\mathrm{j}} / \gamma\right)}{\mathrm{dt}}\right] \mathrm{dV}=\Omega+\mathrm{T}+\int_{\mathrm{v}} \gamma \tau \mathrm{dV}.\label{N2.3.7}\]

2.4

\[\mathfrak{J}=\frac{1}{4 \pi}\left[\mathbf{D D}+\mathbf{c}^2 \mathbf{H H}-\frac{1}{2} \mathbb{1}\left(\mathrm{D}^2+\mathrm{c}^2 \mathrm{H}^2\right)\right].\label{N2.4.1}\]

If we take the divergence of \(\mathfrak{J}\) we get

\[\nabla \cdot \mathfrak{J}=\frac{1}{4 \pi}\left\{(\mathbf{D} \cdot \nabla) \mathbf{D}+\mathbf{D}(\nabla \cdot \mathbf{D})+\left(\mathrm{c}^2 \mathbf{H} \cdot \nabla\right) \mathbf{H}+\mathrm{c}^2 \mathbf{H}(\nabla \cdot \mathbf{H})-\frac{1}{2} \nabla\left(\mathbf{D} \cdot \mathbf{D}+\mathrm{c}^2 \mathbf{H} \cdot \mathbf{H}\right)\right\},\label{N2.4.2}\]

and invoking Maxwell's laws that \(\nabla \cdot \mathbf{D}=\rho_{\mathrm{e}}\) and \(\nabla \cdot \mathbf{H}=0\), this becomes

\[\nabla \cdot \mathfrak{J}=\frac{1}{4 \pi}\left\{(\mathbf{D} \cdot \nabla) \mathbf{D}+\mathbf{D} \rho_{\mathrm{e}}+\left(\mathrm{c}^2 \mathbf{H} \cdot \nabla\right) \mathbf{H}-\frac{1}{2} \nabla\left(\mathbf{D} \cdot \mathbf{D}+\mathrm{c}^2 \mathbf{H} \cdot \mathbf{H}\right)\right\}.\label{N2.4.3}\]

Making use of the vector identity

\[\nabla(\mathbf{A} \cdot \mathbf{G})=\mathbf{A} \times(\nabla \times \mathbf{G})+(\mathbf{A} \cdot \nabla) \mathbf{G}+\mathbf{G} \times(\nabla \times \mathbf{A})+(\mathbf{G} \cdot \nabla) \mathbf{A},\label{N2.4.4}\]

Equation (N2.4.3) takes on the more familiar form

\[\nabla \cdot \mathfrak{J}=\frac{1}{4 \pi}\left\{\left(\mathbf{D} \rho_{\mathrm{e}}-\mathbf{D} \times(\nabla \times \mathbf{D})+\mathrm{c}^2[\mathbf{H} \times(\nabla \times \mathbf{H})]\right\} .\right.\label{N2.4.5}\]

2.5

By use of the vector identity Equation (N2.4.4), it is clear that

\[\frac{1}{2} \nabla(\boldsymbol{w} \times \mathbf{r})^2=(\boldsymbol{w} \times \mathbf{r}) \times[\nabla \times(\boldsymbol{w} \times \mathbf{r})]+[(\boldsymbol{w} \times \mathbf{r}) \cdot \nabla](\boldsymbol{w} \times \mathbf{r}),\label{N2.5.1}\]

but

\[\nabla \times(\boldsymbol{w} \times \mathbf{r})=(\boldsymbol{w} \cdot \nabla) \mathbf{r}-(\mathbf{r} \cdot \nabla) \boldsymbol{w}-\boldsymbol{w}(\nabla \cdot \mathbf{r})+\mathbf{r}(\nabla \cdot \boldsymbol{w}).\label{N2.5.2}\]

The constancy of \(\boldsymbol{w}\) causes the second and fourth term to vanish, while the first third terms are equal but of opposite sign. Again, since \(\boldsymbol{w}\) is constant.

\[[(\boldsymbol{w} \times \mathbf{r}) \cdot \nabla](\boldsymbol{w} \times \mathbf{r})=\boldsymbol{w} \times[(\boldsymbol{w} \times \mathbf{r}) \cdot \nabla](\mathbf{r}).\label{N2.5.3}\]

In component form

\[[(\boldsymbol{w} \times \mathbf{r}) \cdot \nabla](\mathbf{r})=\sum_{\mathrm{i}} \sum_{\mathrm{j}} \sum_{\mathrm{k}} \mathfrak{e}_{\mathrm{ijk}} \omega_{\mathrm{j}} \mathrm{r}_{\mathrm{k}} \frac{\partial \mathrm{x}_{\ell}}{\partial \mathrm{x}_{\mathrm{i}}}=\sum_{\mathrm{j}} \sum_{\mathrm{k}} \mathfrak{e}_{\ell \mathrm{jk}} \omega_{\mathrm{j}} \mathrm{r}_{\mathrm{k}}=(\boldsymbol{w} \times \mathbf{r}),\label{N2.5.4}\]

where \(\mathfrak{e}_{\mathrm{ijk}}\) in a completely antisymmetric tensor of rank 3 sometimes called the Levi Civita tensor density. Thus

\[\frac{1}{2} \nabla(\boldsymbol{w} \times \mathbf{r})^2=\boldsymbol{w} \times(\boldsymbol{w} \times \mathbf{r}).\label{N2.5.5}\]

2.6

\[\int_{\mathrm{v}} \rho \mathbf{r} \frac{\mathrm{d} \mathbf{w}}{\mathrm{dt}} \mathrm{dV}+2 \int_{\mathrm{v}} \rho \mathbf{r}(\boldsymbol{w} \times \mathbf{w}) \mathrm{dV}=-\int_{\mathrm{v}} \rho \nabla\left[\Phi+(\boldsymbol{w} \times \mathbf{r})^2+\mathrm{P} / \rho\right] \mathrm{dV}-\frac{1}{4 \pi} \int_{\mathrm{v}} \mathbf{r}[\mathbf{H} \times(\nabla \times \mathbf{H})] \mathrm{dV}.\label{N2.6.1}\]

As in section 3, the first term on the left becomes

\[\mathrm{\int_V \rho \mathbf{r} \frac{d \mathbf{w}}{d t} d V=\frac{1}{2} \frac{d^2}{d t^2}\left(\int_v \rho(\mathbf{r r}) d V\right)-\int_v \rho(\mathbf{w w}) d V}.\label{N2.6.2}\]

The second term is more difficult to simplify. Let the local velocity field \(\mathbf{w}=\boldsymbol{\varpi} \times \mathbf{r}\) by defining a local angular velocity field \(\boldsymbol{\varpi}\). Then, by expanding the resulting vector triple product in the second term we can write

\[\begin{aligned}
& 2 \int_{\mathrm{v}} \rho \mathbf{r}(\boldsymbol{w} \times \mathbf{w}) \mathrm{dV}=2 \int_{\mathrm{v}} \rho \mathbf{r}[\boldsymbol{w} \times(\boldsymbol{\varpi} \times \mathbf{r})] \mathrm{dV}=2 \int_{\mathrm{v}} \rho \mathbf{r}[\boldsymbol{\varpi}(\boldsymbol{w} \cdot \mathbf{r})-\mathbf{r}(\boldsymbol{w} \cdot \boldsymbol{\varpi})] \mathrm{dV} \\[4pt]
\text{or}\quad& 2 \int_{\mathrm{v}} \rho \mathbf{r}(\boldsymbol{w} \times \mathbf{w}) \mathrm{dV}=2 \boldsymbol{w} \cdot \int_{\mathrm{v}} \rho[\mathbf{r}(\mathbf{r} \boldsymbol{\varpi})-\boldsymbol{\varpi}(\mathbf{r r})] \mathrm{dV}=2 \boldsymbol{w} \boldsymbol{\mathfrak{L}},
\end{aligned}\label{N2.6.3}\]

where the tensor \(\boldsymbol{\mathfrak{L}}\) is an angular moment-like three index tensor representing the various components of volume net angular momentum within the body. Thus, \(\boldsymbol{w}\cdot \boldsymbol{\mathfrak{L}}\) is a kinetic energy like tensor resulting from the net motions induced by the coriolis forces. However, since there can be no net motions about any axis other than that defined by the total angular momentum of the body all components of \(\boldsymbol{w}\cdot \boldsymbol{\mathfrak{L}}\) must be zero except those associated with the axis of rotation. In addition we can choose our rotating frame \(\boldsymbol{w}\) so that in that frame the net angular momentum is zero and all contributions from the term \(\boldsymbol{w}\cdot \boldsymbol{\mathfrak{L}}\) will vanish. For the sake of generality we shall keep the term for the present. The first term on the right of Equation (N2.6.1) is given by Equation \ref{2.1.4}. Since we just went to some length to show that, \(\frac{1}{2} \nabla(\boldsymbol{w} \times \mathbf{r})^2=\boldsymbol{w} \times(\boldsymbol{w} \times \mathbf{r})\), we shall use the earlier version and an expansion of the vector triple product to evaluate the second term on the right.

Thus

\[\begin{aligned}
& \frac{1}{2} \int_{\mathrm{v}} \rho \mathbf{r} \nabla(\boldsymbol{w} \times \mathbf{r})^2 \mathrm{dV}=\int_{\mathrm{v}} \rho \mathbf{r}[\boldsymbol{w} \times(\boldsymbol{w} \times \mathbf{r})] \mathrm{dV}=\int_{\mathrm{v}} \rho \mathbf{r} \boldsymbol{w}(\boldsymbol{w} \cdot \mathbf{r}) \mathrm{dV}-\int_{\mathrm{v}} \rho \mathbf{r r} \boldsymbol{\omega}^2 \mathrm{dV}, \\[4pt]
\text{or}\quad& \frac{1}{2} \int_{\mathrm{v}} \rho \mathbf{r} \nabla(\boldsymbol{w} \times \mathbf{r})^2 \mathrm{dV}=\boldsymbol{w} \cdot \left(\int_{\mathrm{v}} \rho \mathbf{rr} \boldsymbol{w} \mathrm{dV}\right)-\omega^2 \int_{\mathrm{v}} \rho \mathbf{r r d V} .
\end{aligned}\label{N2.6.4}\]

It is worth noting at this point that all terms dealt with so far involve only the volume integrals \(\mathrm{\int_v \rho \mathbf{r r} d V \text { and } \int_v \rho \mathbf{w w} d V}\) which are the same as the tensors defined in Section 1. Thus by combining Equations (N2.6.2), (N2.6.3), (N2.6.4), and Equation \ref{2.1.4}, we may assess our progress in simplifying Equation (N2.6.1) so far [see Equation \ref{2.5.10}].