Given the Runge-Lenz vector


and the angular momentum


We can get


I think it is useful to write both as

$A^j=\epsilon^{abc}p_bL_c-mk\frac{r_n}{\sqrt{r^{\alpha}r_{\alpha}}}$ (maybe I'm wrong with the notation)



So the Poisson Bracket is expressed by {$L^{i},A^{j}$}$=${$\epsilon^{ijk}r_{j}p_{k},\epsilon^{abc}p_bL_c-mk\frac{r_n}{\sqrt{r^{\alpha}r_{\alpha}}}$}

and using the properties, I can distribute the components such that


Doing so, I'm getting for the first bracket

$\epsilon^{ijk}\epsilon^{abc}${$r_jp_k,p_bL_c$}=$\epsilon^{ijk}\epsilon^{abc}\epsilon^{c\mu\nu}(-r_jp_bp_{\nu}\delta^{k\mu}+p_kr_{\mu}p_{\nu}\delta^{jb}+p_kp_br_{\mu}\delta^{j\nu})$ (by defining $L_c=\epsilon^{c\mu\nu}r_{\mu}p_{\nu}$)

but for the second one, I have


I'd appreciate if you can tell me what I'm doing wrong and help me to do the math!


closed as off-topic by Emilio Pisanty, ZeroTheHero, Kyle Kanos, AccidentalFourierTransform, Cosmas Zachos Apr 29 '18 at 0:24

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  • $\begingroup$ What happens to the second term of $A^j$ when $\alpha\neq\beta$? $\endgroup$ – probably_someone Apr 27 '18 at 19:29
  • 1
    $\begingroup$ I'm using Einstein notation, which implies a summation over the indices. So, if $\alpha\neq\beta$ the term wouldn't be added. So we have, for example in 3 dimensions, $mk\frac{r_n}{\sqrt{{r_{1}}^2+{r_2}^2+{r_3}^2}}$ $\endgroup$ – Kali Apr 27 '18 at 23:32
  • $\begingroup$ When $\alpha\neq\beta$, we have that $\frac{r_n}{\sqrt{\delta^{\alpha\beta}r_\alpha r_\beta}}=\frac{r_n}{0}$. You're adding a bunch of infinities to each other. $\endgroup$ – probably_someone Apr 27 '18 at 23:35
  • $\begingroup$ And if that's not, in fact, what you're doing, then you need to fix your notation, as the summation over indices occurs at the term level, not within the square root. $\endgroup$ – probably_someone Apr 27 '18 at 23:37
  • $\begingroup$ I need to express $\frac{1}{\sqrt{{r_1}^2+{r_2}^2+{r_3}^2}}$ in terms of indices, how should I write it then? $\endgroup$ – Kali Apr 27 '18 at 23:37