- #1
etotheipi
Looking at L&L's solution to problem four of section §106. Lagrangian for a system of particles:\begin{align*}
L = &\sum_a \frac{m_a' v_a^2}{2} \left( 1 + 3\sum_{b}' \frac{km_b}{c^2 r_{ab}} \right) + \sum_a \frac{m_a v_a^4}{8c^2} + \sum_a \sum_b' \frac{km_a m_b}{2r_{ab}} \\
&- \sum_a \sum_b' \frac{km_a m_b}{4c^2 r_{ab}} \left[ 7 \mathbf{v}_a \cdot \mathbf{v}_b +(\mathbf{v}_a \cdot \mathbf{n}_{ab})(\mathbf{v}_b \cdot \mathbf{n}_{ab}) \right] - \sum_a \sum_b' \sum_c' \frac{k^2 m_a m_b m_c}{2c^2 r_{ab} r_{ac}}\end{align*}I think in the first approximation L&L are writing ##L = L_0 + \delta^{(1)}L + \delta^{(2)} L## with ##L_0 = \sum_a \frac{m_a v_a^2}{2}## and the first deviation ##\delta^{(1)}L## is arising from the term ##\sum_a \frac{m_a' v_a^2}{2} \sum_{b}' \frac{3km_b}{c^2 r_{ab}}##, where I guess they were considering something like\begin{align*}
\sum_{a \in \mathrm{top}} \sum_{b \in \mathrm{star}} \frac{3km_a m_b(\mathbf{V} + \boldsymbol{\omega} \times \mathbf{r})^2}{2c^2 r_{ab} } &\overset{\mathrm{continuum}}{\longrightarrow} \frac{3km'}{2c^2} \int_{\mathrm{top}} \frac{2}{R(\mathbf{r}')}(\mathbf{V} \cdot \boldsymbol{\omega} \times \mathbf{r}) dm\end{align*}having dropped the ##V^2## and neglecting anything in ##\omega^2##, also ##m'## is the mass of the star. It simplifies writing ##\int_{\mathrm{top}} x_i x_j dm = \frac{1}{2} I \delta_{ij}##. I don't know which term they used to write ##\delta^{(2)} L##; this second deviation is supposed to account for the rotation of the star; they say you can make use of the equation$$h_{03} = \frac{2kM'}{Rc^2} \sin^2{\theta}$$How do you use this? You're supposed to get$$\delta^{(2)} L = \frac{2k}{c^2} \int_{\mathrm{top}} \frac{\mathbf{M}' \cdot (\boldsymbol{\omega} \times \mathbf{r}) \times \mathbf{R}}{R^3} dm$$here ##\mathbf{M}'## is the angular momentum of the star. How do you arrive at this given the formula for ##h_{03}##? Very lost.
L = &\sum_a \frac{m_a' v_a^2}{2} \left( 1 + 3\sum_{b}' \frac{km_b}{c^2 r_{ab}} \right) + \sum_a \frac{m_a v_a^4}{8c^2} + \sum_a \sum_b' \frac{km_a m_b}{2r_{ab}} \\
&- \sum_a \sum_b' \frac{km_a m_b}{4c^2 r_{ab}} \left[ 7 \mathbf{v}_a \cdot \mathbf{v}_b +(\mathbf{v}_a \cdot \mathbf{n}_{ab})(\mathbf{v}_b \cdot \mathbf{n}_{ab}) \right] - \sum_a \sum_b' \sum_c' \frac{k^2 m_a m_b m_c}{2c^2 r_{ab} r_{ac}}\end{align*}I think in the first approximation L&L are writing ##L = L_0 + \delta^{(1)}L + \delta^{(2)} L## with ##L_0 = \sum_a \frac{m_a v_a^2}{2}## and the first deviation ##\delta^{(1)}L## is arising from the term ##\sum_a \frac{m_a' v_a^2}{2} \sum_{b}' \frac{3km_b}{c^2 r_{ab}}##, where I guess they were considering something like\begin{align*}
\sum_{a \in \mathrm{top}} \sum_{b \in \mathrm{star}} \frac{3km_a m_b(\mathbf{V} + \boldsymbol{\omega} \times \mathbf{r})^2}{2c^2 r_{ab} } &\overset{\mathrm{continuum}}{\longrightarrow} \frac{3km'}{2c^2} \int_{\mathrm{top}} \frac{2}{R(\mathbf{r}')}(\mathbf{V} \cdot \boldsymbol{\omega} \times \mathbf{r}) dm\end{align*}having dropped the ##V^2## and neglecting anything in ##\omega^2##, also ##m'## is the mass of the star. It simplifies writing ##\int_{\mathrm{top}} x_i x_j dm = \frac{1}{2} I \delta_{ij}##. I don't know which term they used to write ##\delta^{(2)} L##; this second deviation is supposed to account for the rotation of the star; they say you can make use of the equation$$h_{03} = \frac{2kM'}{Rc^2} \sin^2{\theta}$$How do you use this? You're supposed to get$$\delta^{(2)} L = \frac{2k}{c^2} \int_{\mathrm{top}} \frac{\mathbf{M}' \cdot (\boldsymbol{\omega} \times \mathbf{r}) \times \mathbf{R}}{R^3} dm$$here ##\mathbf{M}'## is the angular momentum of the star. How do you arrive at this given the formula for ##h_{03}##? Very lost.
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