- #1
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There must be something I'm totally missing here.
The situation is the following.
I am asked to show that given the lagrangian for the Kepler problem,
L=\frac{1}{2}\mathbf{\dot{q}}^2+q^{-1}
the k-th component of the Runge-Lenz vector,
A_k=\mathbf{\dot{q}}^2q_k-\mathbf{q}\cdot\mathbf{\dot{q}}<br /> \dot{q}_k-q_k/q
is the conserved quantity associated (in the sense of Noether's thm) with the infinitesimal coordinate transformation \mathbf{q}\rightarrow\mathbf{q}+\delta \mathbf{q}, where \delta q_i = \epsilon(\dot{q}_iq_k-\frac{1}{2}q_i\dot{q}_k-\frac{1}{2}\mathbf{q}\cdot \mathbf{\dot{q}}\delta_{ik}), epsilon being the infinitesimal parameter.
Following Noether's theorem, I know that if \delta L=L(\mathbf{q}+\delta \mathbf{q}, \mathbf{\dot{q}}+\delta \mathbf{\dot{q}},t)-L(\mathbf{q},\mathbf{\dot{q}},t)[/itex] can be written as <br /> <br /> \delta L=\epsilon \frac{d}{dt}\Lambda(\mathbf{q},\mathbf{\dot{q}},t)+\mathcal{O}(\epsilon^2)<br /> <br /> then the quantity<br /> <br /> F_k:=\sum_{i=1}^3\frac{\partial L}{\partial \dot{q}_i}(\dot{q}_iq_k-\frac{1}{2}q_i\dot{q}_k-\frac{1}{2}\mathbf{q}\cdot \mathbf{\dot{q}}\delta_{ik}) - \Lambda<br /> <br /> is conserved. By direct comparison of F_k with A_k I find that Lambda must be<br /> <br /> \Lambda = \frac{q_k}{q}<br /> <br /> (also, this is confirmed by the wiki article on the Runge-Lenz vector: <a href="http://en.wikipedia.org/wiki/Runge-Lenz#Noether.27s_theorem" target="_blank" class="link link--external" rel="nofollow ugc noopener">http://en.wikipedia.org/wiki/Runge-Lenz#Noether.27s_theorem</a> )<br /> <br /> So what remains to be done is to show by direct calculation that indeed, <br /> <br /> \delta L=\epsilon \frac{d}{dt}(\frac{q_k}{q})+\mathcal{O}(\epsilon^2)<br /> <br /> So I expand \delta L:<br /> <br /> \delta L= \frac{1}{2}(\mathbf{\dot{q}}^2+2\mathbf{\dot{q}}\cdot \delta\mathbf{\dot{q}}+(\delta\mathbf{\dot{q}})^2)+(\mathbf{q}^2+2\mathbf{q}\cdot \delta\mathbf{q}+(\delta\mathbf{q})^2)^{-\frac{1}{2}}-\frac{1}{2}\mathbf{\dot{q}}^2-(\mathbf{q}^2)^{-\frac{1}{2}}<br /> <br /> And here I find it impossible to put this in a form \delta L=\epsilon A+\mathcal{O}(\epsilon^2) because of all these guys in the numerator <i>and</i> shielded by a square root. I have also tried "cheating", i.e. say "since epsilon is arbitrarily small, I can neglect this and this term" but nothing even comes close to the form I want.<br /> <br /> So I concluded that there must be something fundamentally flawed about the reasoning laid above. Anyone sees?<br /> <br /> Thanks for reading!<u>P.S.</u> I would appreciate feedbacks, so that if I get a few feedbacks that the above is right, I will post more of my work so we can find where I'm going wrong.
The situation is the following.
I am asked to show that given the lagrangian for the Kepler problem,
L=\frac{1}{2}\mathbf{\dot{q}}^2+q^{-1}
the k-th component of the Runge-Lenz vector,
A_k=\mathbf{\dot{q}}^2q_k-\mathbf{q}\cdot\mathbf{\dot{q}}<br /> \dot{q}_k-q_k/q
is the conserved quantity associated (in the sense of Noether's thm) with the infinitesimal coordinate transformation \mathbf{q}\rightarrow\mathbf{q}+\delta \mathbf{q}, where \delta q_i = \epsilon(\dot{q}_iq_k-\frac{1}{2}q_i\dot{q}_k-\frac{1}{2}\mathbf{q}\cdot \mathbf{\dot{q}}\delta_{ik}), epsilon being the infinitesimal parameter.
Following Noether's theorem, I know that if \delta L=L(\mathbf{q}+\delta \mathbf{q}, \mathbf{\dot{q}}+\delta \mathbf{\dot{q}},t)-L(\mathbf{q},\mathbf{\dot{q}},t)[/itex] can be written as <br /> <br /> \delta L=\epsilon \frac{d}{dt}\Lambda(\mathbf{q},\mathbf{\dot{q}},t)+\mathcal{O}(\epsilon^2)<br /> <br /> then the quantity<br /> <br /> F_k:=\sum_{i=1}^3\frac{\partial L}{\partial \dot{q}_i}(\dot{q}_iq_k-\frac{1}{2}q_i\dot{q}_k-\frac{1}{2}\mathbf{q}\cdot \mathbf{\dot{q}}\delta_{ik}) - \Lambda<br /> <br /> is conserved. By direct comparison of F_k with A_k I find that Lambda must be<br /> <br /> \Lambda = \frac{q_k}{q}<br /> <br /> (also, this is confirmed by the wiki article on the Runge-Lenz vector: <a href="http://en.wikipedia.org/wiki/Runge-Lenz#Noether.27s_theorem" target="_blank" class="link link--external" rel="nofollow ugc noopener">http://en.wikipedia.org/wiki/Runge-Lenz#Noether.27s_theorem</a> )<br /> <br /> So what remains to be done is to show by direct calculation that indeed, <br /> <br /> \delta L=\epsilon \frac{d}{dt}(\frac{q_k}{q})+\mathcal{O}(\epsilon^2)<br /> <br /> So I expand \delta L:<br /> <br /> \delta L= \frac{1}{2}(\mathbf{\dot{q}}^2+2\mathbf{\dot{q}}\cdot \delta\mathbf{\dot{q}}+(\delta\mathbf{\dot{q}})^2)+(\mathbf{q}^2+2\mathbf{q}\cdot \delta\mathbf{q}+(\delta\mathbf{q})^2)^{-\frac{1}{2}}-\frac{1}{2}\mathbf{\dot{q}}^2-(\mathbf{q}^2)^{-\frac{1}{2}}<br /> <br /> And here I find it impossible to put this in a form \delta L=\epsilon A+\mathcal{O}(\epsilon^2) because of all these guys in the numerator <i>and</i> shielded by a square root. I have also tried "cheating", i.e. say "since epsilon is arbitrarily small, I can neglect this and this term" but nothing even comes close to the form I want.<br /> <br /> So I concluded that there must be something fundamentally flawed about the reasoning laid above. Anyone sees?<br /> <br /> Thanks for reading!<u>P.S.</u> I would appreciate feedbacks, so that if I get a few feedbacks that the above is right, I will post more of my work so we can find where I'm going wrong.
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