Vector function of vectors expansion

I am reading Landau's Mechanics. In the solution to the problem 4 on page 138, section 42, it is stated that an arbitrary vector function $$\vec f(\vec r,\vec p)$$ may be written as $$\vec f=\vec r\phi_1+\vec p\phi_2+\vec r\times\vec p\phi_3$$, where $$\phi_1$$, $$\phi_2$$, $$\phi_3$$ are scalar functions. Why so?

• It says that an arbitrary vector can be written like that. Probably an arbitrary 3-dimensional vector, from context. I think it is meant to be a vector field. Maybe r and p are in different directions? Otherwise a bit strange thing to state.
– Emil
Commented Jan 29, 2022 at 10:03
• If $r$ and $p$ are linearly independent, then they form a basis together with $r \times p$. In that case you can write every vector as a linear combination of these three. I don't see how the expansion works if they are not independent. Commented Jan 29, 2022 at 10:07
• I think the important part that should be added to this questions is the statement of problem 4: "Show that [f,Mz]=fxn, where f is a vector function of the coordinates and momentum of a particle, an n is a unit vector parallel to the z-axis". Based on the hints is want you to work thought the Poisson bracket math using the suggested relationships in the rest of the solution. Commented Jan 29, 2022 at 12:12

$$\newcommand{\bl}[1]{\boldsymbol{#1}} \newcommand{\e}{\bl=} \newcommand{\p}{\bl+} \newcommand{\m}{\bl-} \newcommand{\gr}{\bl>} \newcommand{\les}{\bl<} \newcommand{\greq}{\bl\ge} \newcommand{\leseq}{\bl\le} \newcommand{\plr}[1]{\left(#1\right)} \newcommand{\blr}[1]{\left[#1\right]} \newcommand{\vlr}[1]{\vert#1\vert} \newcommand{\Vlr}[1]{\Vert#1\Vert} \newcommand{\lara}[1]{\langle#1\rangle} \newcommand{\lav}[1]{\langle#1|} \newcommand{\vra}[1]{|#1\rangle} \newcommand{\lavra}[2]{\langle#1|#2\rangle} \newcommand{\lavvra}[3]{\langle#1|\,#2\,|#3\rangle} \newcommand{\vp}{\vphantom{\dfrac{a}{b}}} \newcommand{\hp}[1]{\hphantom{#1}} \newcommand{\x}{\bl\times} \newcommand{\qqlraqq}{\qquad\bl{-\!\!\!-\!\!\!-\!\!\!\longrightarrow}\qquad} \newcommand{\qqLraqq}{\qquad\boldsymbol{\e\!\e\!\e\!\e\!\Longrightarrow}\qquad} \newcommand{\tl}[1]{\tag{#1}\label{#1}} \newcommand{\hebl}{\bl{=\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!=}}$$

I think this Figure answers your question. It's up to you to find expressions for $$\phi_1,\phi_2,\phi_3$$ for given $$\mathbf r,\mathbf p,\mathbf f$$ with $$\mathbf r\x\mathbf p \bl\ne \bl0$$.

If $$$$\mathbf f \e \phi_1\mathbf r\p\phi_2\mathbf p\p \phi_3\plr{\mathbf r\x\mathbf p} \tl{01}$$$$ try to prove that \begin{align} \phi_1 & \e \dfrac{\plr{\mathbf f\bl\cdot\mathbf r}\Vlr{\mathbf p}^2\m\plr{\mathbf f\bl\cdot\mathbf p}\plr{\mathbf r\bl\cdot\mathbf p}}{\Vlr{\mathbf r\x\mathbf p}^2} \tl{02.1}\\ \phi_2 & \e \dfrac{\plr{\mathbf f\bl\cdot\mathbf p}\Vlr{\mathbf r}^2\m\plr{\mathbf f\bl\cdot\mathbf r}\plr{\mathbf r\bl\cdot\mathbf p}}{\Vlr{\mathbf r\x\mathbf p}^2} \tl{02.2}\\ \phi_3 & \e \dfrac{\mathbf f\bl\cdot\plr{\mathbf r\x\mathbf p}}{\Vlr{\mathbf r\x\mathbf p}^2} \tl{02.3} \end{align}

Note that $$$$\Vlr{\mathbf r\x\mathbf p}^2\e \Vlr{\mathbf r}^2\Vlr{\mathbf p}^2\m\plr{\mathbf r\bl\cdot\mathbf p}^2\e \Vlr{\mathbf r}^2\Vlr{\mathbf p}^2\sin^2\theta \tl{03}$$$$

• What if I take $\vec f(\vec r,0)=(r_x^2,r_y^2,r_z^2)=\vec r\phi_1$ , it seems to fail. Commented Jan 29, 2022 at 15:00
• What is this equation ??? I has no sense. Look at the vector $\mathbf f$ as any arbitrary 3-vector. Commented Jan 29, 2022 at 15:05
• But $\vec f$ is an arbitrary vector function. Commented Jan 29, 2022 at 15:18
• I think you spent more time writing all those newcommands than you saved using them. Commented Jan 29, 2022 at 16:42
• @Kyle Kanos : Thanks for your attention. I have the set of these ''newcommands'' in a text file and I copy-paste it in my posts in nanoseconds, so reducing the time I spend in the main body of the posts especially in cases of very long length. Commented Jan 29, 2022 at 19:50

take arbitrary vector in 3D space

$$\vec R=c_1\hat g_1+c_2\hat g_2+c_3\hat g_3\tag 1$$

where $$~\hat g_i~$$ are the independent basis vectors, and $$~c_i~$$ are the scalar components of the vector R.

$$\vec f=\phi_1\,r\,\hat r+\phi_2\,p\,\hat p+\phi_3\,|r\times p|\,\hat{r}_p\\ r_p=\vec r\times \vec p$$

comparing with equation (1)

$$c_1=\phi_1\,r\,,\hat g_1=\hat r\\ c_2=\phi_2\,p\,,\hat g_2=\hat p\\ c_3=\phi_3\,|r\times p|\,,\hat g_3=\hat r_p$$

thus there is noting unusual in this vector description

remarks:

the basis vectors must not be perpendicular to each other $$~\vec r\cdot\vec p\ne 0$$

• do you mean the basis vectors must not be parallel to each other? Commented Jan 29, 2022 at 17:43
• I mean they can be skew ? to each other $~\vec r\cdot\vec p\ne 0$
– Eli
Commented Jan 29, 2022 at 17:55