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When I attempted to calculate the following by hand $$\bar{r}\times(\bar{\nabla}\times\bar{F}) - \bar{\nabla}\times(\bar{r}\times\bar{F}),$$ I noticed some of the terms I extracted looked similar to the terms that appear in the oribtal anglar momentum operator $$\bar{L} = -i\hbar(\bar{r}\times\bar{\nabla}).$$ Is there a condensed expression that uses $\bar{L}$? How does $$\bar{r}\times(\bar{\nabla}\times) - \bar{\nabla}\times(\bar{r}\times)$$ relate to the orbital angular momentum operator?

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    $\begingroup$ I added more parentheses. Did that improve clarity? $\endgroup$ – linuxfreebird Mar 17 '14 at 17:13
  • $\begingroup$ No: $\vec a \times ( \vec b \times \vec c) \neq (\vec a \times \vec b) \times \vec c$... $\endgroup$ – pressure Mar 17 '14 at 17:31
  • $\begingroup$ @pressure Did that fix the parentheses issue? $\endgroup$ – linuxfreebird Mar 17 '14 at 17:34
  • $\begingroup$ Yes, it did fix it. $\endgroup$ – pressure Mar 17 '14 at 17:36
  • $\begingroup$ @pressure I finally was able to answer the following question: physics.stackexchange.com/questions/103664/… Please investigate this question, because there is a fourth component in the field. $\endgroup$ – linuxfreebird Mar 18 '14 at 16:23
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In Einstein summation notation, we'd write $$\vec U = \vec r \times(\nabla\times\vec F) - \nabla\times(\vec r\times\vec F),$$ using a Levi-Civita symbol $\epsilon_{ijk}$ as: $$ U_a = \epsilon_{abc} ~r_b ~\epsilon_{cde} ~\partial_d ~F_e - \epsilon_{abc} ~\partial_b ~\epsilon_{cde} ~r_d ~F_e.$$Since $\epsilon$ is not varying with space we can commute it with $\partial_b$ on the right, leading to $$ U_a = \epsilon_{abc}~\epsilon_{cde}\left( r_b ~\partial_d ~F_e - \delta_{bd} F_e - r_d ~\partial_b ~ F_e\right),$$ and furthermore we have the "BAC-CAB" rule that $\epsilon_{abc}\epsilon_{cde}$ can only be nonzero when either $a = d$ and $b = e$ (with coefficient +1) or when $a = e$ and $b = d$ (with coefficient -1), so it can be rewritten as $\delta_{ad}\delta_{be} - \delta_{ae}\delta_{bd}.$ So then this expression is $$ U_a = (\delta_{ad}\delta_{be} - \delta_{ae}\delta_{bd}) \left(r_b ~\partial_d ~F_e - \delta_{bd} F_e - r_d ~\partial_b ~ F_e\right),$$whence the term $\delta_{bd} F_e$ will always vanish, as we get $\delta_{ae} F_e - \delta_{ae} F_e = 0.$ Similarly the other terms suffer from $\delta_{bd}$ forming $r_b \partial_b F_a - r_b \partial_b F_a = 0.$ So the only term left is the $\delta_{ad}\delta_{be}$ one, leading to $$U_a = r_b ~\partial_a ~F_b - r_a ~\partial_b ~ F_b.$$The first term can be rewritten as $\partial_a(r_b F_b) - F_a,$ so this can be written as,$$\vec U = \nabla(\vec r\cdot \vec F) - \vec r (\nabla \cdot \vec F) - \vec F.$$

This looks a little like a BAC-CAB rule too but it cannot be of the form $A \times (B \times C)$ because the $\nabla$ would probably not be acting on $F$, so let's instead target $(A \times B) \times C = B (A \cdot C) - A (B\cdot C)$ with $A = r$, $B = \nabla$.

In other words, let's calculate $$\vec V = (\vec r \times \nabla) \times F,$$ which we do the same way but with $$V_a = \epsilon_{abc} \epsilon_{bde} r_d \partial_e F_c.$$The relevant $\epsilon \propto \delta\delta$ analysis yields:$$V_a = (\delta_{ae} \delta_{cd} - \delta_{ad} \delta_{ce}) r_d \partial_e F_c = r_c \partial_a F_c - r_a \partial_c F_c = U_a.$$

In other words, $$\Big[\big(\vec r \times\big),~ \big(\nabla \times\big)\Big] = \frac{i}{\hbar} \big(\hat L \times\big).$$

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