# Component-free computation of Poisson bracket of Laplace-Runge-Lenz vector

How can the Poisson bracket $$\{A,H\}$$ be computed directly without components, where $$H$$ is the Hamiltonian for the inverse square force, $$H=\frac{p^2}{2m} - \frac{k}{|r|}\ ,$$ and $$A$$ is the integration constant called the Laplace-Runge-Lenz vector, $$\vec A = \vec p \times (\vec r \times \vec p) - \frac{mk \vec r}{|r|}\ ?$$

At first glance the issue is that traditionally the brackets are defined for scalar-valued functions of scalar values, but I don't see why they can't be interpreted as being defined on functions of vector variables and then using vector calculus and vector gradients instead of partial derivatives. I'm also fine with translating this to differential forms or using geometric algebra, or both.

The point is of course to show that the Poisson bracket $$\{A,H\}$$ is zero since $$A$$ is a constant of motion, but the usual way of using the Levi-Civita tensors is quite tedious, and I think it could be shortcuted vastly if computed directly, I just can't get it quite right.

• Could you be more specific on how exactly you think this could be done without Levi Civita symbols? I don't understand where you want to get the gradient from (or do you mean $\{f, H \} = \nabla_{\vec r} f \cdot \nabla_{\vec p} H - \nabla_{\vec p} f \cdot \nabla_{\vec r} H$) ? Commented May 23, 2023 at 17:41
• Yes this is exactly what I meant. Commented May 26, 2023 at 10:16
• Okay, that would work, but you wouldn't get rid of the Levi Civita symbols. You already need those to get the ith component of $\vec A$. Maybe it would be possible to somehow compute this defining $\vec \nabla \vec A$ as a matrix, but that definitely wouldn't make things easier. If I'll find time, maybe I'll try to calculate the poisson brackets like that and show you how I did it. But that seems like an awful lot of trouble for something normally pretty easy. Commented May 26, 2023 at 14:05
• No it's not easy using levi Civita , and it sure ist easy with matrices. I was thinking of something along the lines of using purely vector calculus identifies , expanding the double cross , or using the Leibnitz rule Commented May 26, 2023 at 20:53
• All these are well defined geometrically, you I don't think components are a must here. Commented May 26, 2023 at 20:54

This answer uses the terminology and notation of geometric algebra and geometric calculus.

## A vectorial view of the Poisson bracket

The Poisson bracket of arbitrary scalar functions $$G$$ and $$H$$ of the canonical coordinates $$(q_i,p_i)$$ is normally written as $$\{G,H\}=\sum_{i=1}^n \left(\frac{\partial G}{\partial q_i}\frac{\partial H}{\partial p_i}-\frac{\partial G}{\partial q_i}\frac{\partial H}{\partial p_i}\right)\ .$$ Instead treating $$G$$ and $$H$$ as functions of the vectors $$r=\sum q_i e_i$$ and $$p=\sum p_i e_i$$, we can alternatively write the Poisson bracket as

\begin{aligned} \{G,H\}&=(\partial_r G)\cdot(\partial_p H)-(\partial_p G)\cdot(\partial_r H)\\ &=[(\partial_p H)\cdot \partial_r-(\partial_r H)\cdot \partial_p] G\ , \end{aligned}\tag{1} where $$\partial_r$$ and $$\partial_p$$ are the vector derivatives with respect to $$r$$ and $$p$$.

Because $$H$$ is a scalar-valued function, the expression inside the $$[\ ]$$ on the RHS of (1) is just a scalar linear operator, which makes it straightforward to extend the definition of the Poisson bracket such that the left argument can be a multivector of any grade. In particular, with a vector $$A(r,p)=\sum G_i(r,p) e_i$$: \begin{aligned} \{A,H\} &=[(\partial_p H)\cdot \partial_r-(\partial_r H)\cdot \partial_p] A\\ &=\sum_{i=1}^n [(\partial_p H)\cdot \partial_r-(\partial_r H)\cdot \partial_p] G_i e_i\\ &=\sum_{i=1}^n \{G_i,H\} e_i \end{aligned}\tag{2}

Having thus recast the Poisson bracket in terms of vectors, we will no longer have any need to refer to components.

There are more powerful ways of expressing the Poisson bracket from a geometric algebra perspective than what we've done above1, but the above approach is adequate for the problem at hand.

## Poisson bracket of the Laplace-Runge-Lenz vector and the Kepler problem Hamiltonian

The question involves the Hamiltonian $$H=\frac{p^2}{2 m}- \frac{k}{|r|}$$ and the Laplace-Runge-Lenz vector, which is given in the question as $$A=p\times (r\times p)-mk\hat{r}\ ,$$ where $$\hat{r}=r/|r|$$. However, we're addressing the problem in terms of geometric algebra, which doesn't commonly use the cross product because it only works in three dimensions. So we twice use the identity $$a\times b=-I a\wedge b\ ,$$ where $$I$$ is the unit pseudoscalar, to arrive at the more geometric algebra-friendly expression $$A=-p\cdot (r\wedge p)-\frac{mk}{|r|}r\ .$$

Using the identity2 $$\partial_r \frac{1}{|r|}=-\frac{1}{|r|r^2}r\ ,\tag{3}$$ the vector derivatives of $$H$$ are seen to be the vectors $$\partial_p H=\frac{1}{m}p\tag{4}$$ and $$\partial_r H=\frac{k}{|r|r^2}r\ .\tag{5}$$

Note that in (2), it's actually the directional derivatives of $$A$$ that are needed, not the vector derivatives. Using (3) and multiple applications of the Leibniz rule, for an arbitrary vector $$v$$ the directional derivatives of $$A$$ are $$(v \cdot\partial_r) A=-p\cdot(v\wedge p)-\frac{mk}{|r|}v+\frac{mk v\cdot r}{|r|r^2}r\tag{6}$$ and $$(v \cdot\partial_p) A=-v\cdot(r\wedge p)-p\cdot(r\wedge v)\ .\tag{7}$$

Plugging (4)-(7) into (2) gives \begin{aligned} \{A,H\}&=\frac{1}{m}\left(-p\cdot(p\wedge p)-\frac{mk}{|r|}p+\frac{mkp\cdot r}{|r|r^2}r\right)\\ &\ \ \ \ -\frac{k}{|r|r^2}\left(-r\cdot(r\wedge p)-p\cdot(r\wedge r)\right)\\ &=\frac{k}{|r|r^2}\left(-r^2 p+(p\cdot r) r+r\cdot (r\wedge p)\right)\ . \end{aligned}\tag{8}

But from the identity3 $$a\cdot(b\wedge c)=(a\cdot b) c-b(a\cdot c)$$ for vectors $$a$$, $$b$$ and $$c$$ we have that $$r\cdot(r\wedge p)=r^2 p-r(r\cdot p)\ ,$$ which we can plug into (8) to give \begin{aligned} \{A,H\}&=\frac{k}{|r|r^2}\left(-r^2 p+(p\cdot r) r+r^2 p-r(r\cdot p)\right)\\ &=0 \end{aligned}

## References

1: Geometric Algebra for Physicists, section 12.3

2: Clifford Algebra to Geometric Calculus, page 57, eq. 2.32

3: Ibid., page 12, eq. 1.42

• Thank you so much ,this is perfect Commented Nov 10, 2023 at 21:55