# Hamilton–Jacobi Equation for a Charged Test Particle in the Stäckel Space of Type (2.0)

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## Abstract

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## 1. Introduction

- (1)
- Non-null Stäckel spaces of type (3.0). The complete set includes three Killing vector fields. The coordinate hypersurface related to the non-ignored variable is non-null. (Ignored are the privileged variables (see the definition below) that occur linearly in the complete integral. Other variables are called non-ignored.)
- (2)
- Null Stäckel spaces of type (3.1). The complete set includes three Killing vector fields. The coordinate hypersurface related to the non-ignored variable is null one.
- (3)
- Non-null Stäckel spaces of type (2.0). The complete set includes two Killing vector fields. Coordinate hypersurfaces related to non-ignored variables are non-null.
- (4)
- Null Stäckel spaces of type (2.1). The complete set includes two Killing vector fields. The coordinate hypersurface belonging to one of the non-ignored variables is null.
- (5)
- Non-isotropic Stäckel spaces of type (1.0). The complete set includes one non-isotropic Killing vector field. Coordinate hypersurfaces related to non-ignored variables are non-null.
- (6)
- Null Stäckel spaces of type (1.1). The set includes one isotropic Killing vector field. The coordinate hypersurface belonging to one of the non-ignored variables is null.
- (7)
- Non-null Stäckel spaces of type (0.0). The complete set does not contain Killing vector fields. Coordinate hypersurfaces related to non-ignored variables are non-null.

## 2. Conditions for the Existence of a Complete Set of Motion Integrals

## 3. Solutions for the Case When Both Functions ${\mathit{\omega}}^{\mathit{p}}$ Are Free

## 4. The Linear Dependence of ${\mathit{\omega}}^{\mathit{p}}$ on a Free Function

#### 4.1. Finding the Metric Tensor

**I**- $\tilde{\gamma}=1.$We show that in this case all the solutions of Equation (23) for all classes ${\widehat{G}}_{\alpha}$ from the expressions (26) except ${\widehat{G}}_{5}$ can be represented as:$$\begin{array}{c}\hfill \widehat{G}=\left(\begin{array}{cc}{({a}^{0})}^{2}+\epsilon {({\alpha}^{0})}^{2}& {a}^{0}{a}^{1}+\epsilon {\alpha}^{0}{\alpha}^{1}\\ {a}^{0}{a}^{1}+\epsilon {\alpha}^{0}{\alpha}^{1}& {({a}^{1})}^{2}+\epsilon {({\alpha}^{1})}^{2}\end{array}\right).\end{array}$$By marking: ${\beta}^{pq}={\alpha}^{p}{\alpha}^{q}-{\alpha}^{pq},$ we bring the equation (23) to the form:$$\begin{array}{c}\hfill det({a}^{pq})+det({\beta}^{pq})+{a}_{1}{\beta}^{00}+{a}_{0}{\beta}^{11}-2a{\beta}^{01}=0.\end{array}$$To prove the statement, let us consider (29) for all ${\widehat{G}}_{\alpha}$ from the expressions (26) with numbers $\alpha =1,\cdots ,4$.
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**a**) - $\widehat{G}={\widehat{G}}_{1}$.Let us fix the variable ${u}^{2}$ to the point ${u}^{2}={\tilde{u}}^{2}$ in the functional Equation (24). As result, we get the expression:$$det({a}^{pq})={\tilde{\gamma}}_{pq}{a}^{pq}+\tilde{a},$$$${\beta}^{pq}=0\to {\alpha}^{pq}={\alpha}^{p}{\alpha}^{q},\phantom{\rule{1.em}{0ex}}\tilde{a}=det({a}^{pq})=0.$$From $det{a}^{(}pq)0$ we get: ${a}^{pq}={a}^{p}{a}^{q}.$ Therefore, the matrix $\widehat{G}$ can be represented as (28). The statement is proved.
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**b**) - $\widehat{G}={\widehat{G}}_{2}$.The functional equation follows from (29):$$\begin{array}{c}\hfill det({a}^{pq})+det({\beta}^{pq})+{a}_{0}({\beta}^{11}-{\beta}^{00})-2a{\beta}^{01}=0.\end{array}$$Since in ${a}_{0}$ and a are linearly independent, from (29) it follows:$${a}^{2}+{a}_{0}^{2}={\tilde{a}}^{2},\phantom{\rule{1.em}{0ex}}{\beta}^{pp}=\tilde{a},\phantom{\rule{1.em}{0ex}}{\beta}^{01}=0.$$The elements of the matrix $\widehat{G}$ can be represented as:$${a}^{00}=\tilde{a}+{a}_{0},{a}^{11}=\tilde{a}-{a}_{0},{a}^{01}=a,{\alpha}^{pq}={\alpha}^{p}{\alpha}^{q}.$$Thus, $det({a}^{pq})=det({\alpha}^{pq})=0$, The equality to zero of the coefficients before ${(\sigma )}^{b}(b=0,1,2)$ gives the system of equations: and the matrix ${\widehat{G}}_{2}$ are reduced to the form (27).
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**c**) - $\widehat{G}={\widehat{G}}_{3}$.The functional equation follows from (29):$${a}_{0}{a}_{1}+det{\beta}^{pq}+{a}_{0}{\beta}^{11}+a{\beta}^{00}=0.$$Since $a{a}_{0}\ne 0$, we get: ${\beta}^{pq}={\tilde{\beta}}^{pq},({a}_{0}+{\tilde{\beta}}^{00})({a}_{1}+{\tilde{\beta}}^{11})={\tilde{a}}^{2},$ Therefore, the matrix $\widehat{G}$ can be represented as:$$\begin{array}{c}\hfill \widehat{G}=\left(\begin{array}{cc}a+\epsilon {{\alpha}^{0}}^{2}& \tilde{a}+{\alpha}^{0}{\alpha}^{1}\\ \tilde{a}+{\alpha}^{0}{\alpha}^{1}& (\frac{{\tilde{a}}^{2}}{a}+\epsilon {{\alpha}^{0}}^{2})\end{array}\right).\end{array}$$
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**d**) - $\widehat{G}={\widehat{G}}_{4}$.The functional equation follows from (29):$${a}_{0}{\beta}^{11}-2a{\beta}^{01}+det{\beta}^{pq}=\epsilon {a}^{2}.$$Since $a,{a}_{0}\ne const$ we get: ${a}_{0}=\epsilon {a}^{2},{\alpha}^{00}={({\alpha}^{0})}^{2},{\alpha}^{01}={\alpha}^{0}{\alpha}^{1},{\alpha}^{11}=1+{({\alpha}^{1})}^{2},$ and the matrix $\widehat{G}$ can be represented as:$$\begin{array}{c}\hfill \widehat{G}=\left(\begin{array}{cc}\epsilon {a}^{2}+{({\alpha}^{0})}^{2}& a+{\alpha}^{0}{\alpha}^{1}\\ a+{\alpha}^{0}{\alpha}^{1}& \epsilon +{({\alpha}^{1})}^{2}\end{array}\right).\end{array}$$Thus, it has the form (27).
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**e**) - Now consider the case when $\widehat{G}={\widehat{G}}_{5}$. The function $a\ne const,\to \epsilon =0$, and the functional equation follows from (29):$$a({\alpha}^{11}-{{\alpha}^{1}}^{2})=({\alpha}^{00}-{{\alpha}^{0}}^{2})({\alpha}^{11}-{{\alpha}^{1}}^{2})-{({\alpha}^{01}-{\alpha}^{0}{\alpha}^{1})}^{2}.$$From here: ${\alpha}^{11}={{\alpha}^{1}}^{2},{\alpha}^{01}={\alpha}^{0}{\alpha}^{1},$ and the matrix $\widehat{G}$ takes the form:$$\begin{array}{c}\hfill \widehat{G}=\left(\begin{array}{cc}a+\alpha +{{\alpha}^{0}}^{2}& {\alpha}^{0}{\alpha}^{1}\\ {\alpha}^{0}{\alpha}^{1}& {{\alpha}^{1}}^{2}\end{array}\right).\end{array}$$

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**II**- $\gamma =0.$In case when the matrices ${\widehat{G}}_{\rho}$ belong to the first three variants ($\rho =1,\cdots ,3$), Equation (23) has no nonzero solutions and ${\omega}^{p}$ does not contain any free function. Consider the remaining options.
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**a**) - $\widehat{G}={\widehat{G}}_{4}$.From Equation (23) it follows:$$\begin{array}{c}\hfill {a}_{0}{{\alpha}^{1}}^{2}-2a{\alpha}^{0}{\alpha}^{1}+{\alpha}^{11}{{\alpha}^{0}}^{2}+{\alpha}^{00}{{\alpha}^{1}}^{2}-2{\alpha}^{01}{\alpha}^{0}{\alpha}^{1}=0.\end{array}$$From here: ${\alpha}^{1}={\alpha}^{11}=0$, and we get the matrix $\widehat{G}$ in the form:$$\widehat{G}=\left(\begin{array}{cc}{a}_{0}+{\alpha}_{0}& a+\alpha \\ a+\alpha & 0\end{array}\right).$$
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**b**) - $\widehat{G}={\widehat{G}}_{5}$.From Equation (23) it follows:$$\begin{array}{c}\hfill {{\alpha}^{1}}^{2}+\epsilon {{\alpha}^{0}}^{2}=0,\phantom{\rule{1.em}{0ex}}{\alpha}^{11}{{\alpha}^{0}}^{2}+{\alpha}^{00}{{\alpha}^{1}}^{2}-2{\alpha}^{01}{\alpha}^{0}{\alpha}^{1}=0.\end{array}$$First, let $\epsilon =-1\to {\alpha}^{0}={\alpha}^{1},{\alpha}^{11}=-{\alpha}^{00}+2{\alpha}^{01},$ and the matrix $\widehat{G}$ takes the form:$$\begin{array}{c}\hfill \widehat{G}=\left(\begin{array}{cc}({a}_{0}+{\alpha}_{0})& \alpha \\ \alpha & -({a}_{0}+{\alpha}_{0})+\alpha \end{array}\right).\end{array}$$By replacing the variables: $\widehat{{u}^{0}}\to \frac{1}{\sqrt{2}}({u}^{0}+{u}^{1}),\widehat{{u}^{1}}\to \frac{1}{\sqrt{2}}({u}^{0}-{u}^{1})$, we bring the solution to the form:$$\begin{array}{c}\hfill \widehat{G}=\left(\begin{array}{cc}\alpha & ({a}_{0}+{\alpha}_{0})\\ ({a}_{0}+{\alpha}_{0})& 0\end{array}\right).\end{array}$$Now let $\epsilon =0$. It is easy to show that in this case ${\alpha}^{1}=0$ and $\widehat{G}$ has the form:$$\widehat{G}=\left(\begin{array}{cc}{a}_{0}+{\alpha}_{0}& \alpha \\ \alpha & 0\end{array}\right),\phantom{\rule{1.em}{0ex}}{\alpha}^{1}=0.$$Thus, both solutions are special cases of the solution (35).

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#### 4.2. Building an Electromagnetic Potential

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**a**) - Matrix $\widehat{{G}_{1}}$.We substitute the matrix $\widehat{{G}_{1}}$ into the equation (24). After some transformation we get the equation:$${a}^{0}{h}^{1}-{a}^{1}{h}^{0}={a}^{0}(\gamma {\alpha}^{1}-{\sigma}^{1})-{a}^{1}(\gamma {\alpha}^{0}-{\sigma}^{0}).$$This implies:$${a}^{0}({h}^{1}-{\tilde{c}}^{1})={a}^{0}({h}^{0}-{\tilde{c}}^{0}),\phantom{\rule{1.em}{0ex}}{\sigma}^{p}=\gamma {\alpha}^{p}.$$By the admissible gradient transformations of the potential, the values ${\tilde{c}}^{p}$ and $\gamma $ can be set to zero. The solution has the form:$${A}^{p}=\frac{{a}^{p}h+{\alpha}^{p}\omega}{\Delta},\phantom{\rule{1.em}{0ex}}{A}^{\nu}=0,$$$$\begin{array}{cc}\hfill \phantom{\rule{1.em}{0ex}}& ({g}^{ij})=\left(\begin{array}{cccc}\frac{{({a}^{0})}^{2}+\epsilon {({\alpha}^{0})}^{2}}{\Delta}& \frac{{a}^{0}{a}^{1}+\epsilon {\alpha}^{0}{\alpha}^{1}}{\Delta}& 0& 0\\ \frac{{a}^{0}{a}^{1}+\epsilon {\alpha}^{0}{\alpha}^{1}}{\Delta}\phantom{\rule{1.em}{0ex}}& \frac{{({a}^{1})}^{2}+\epsilon {({\alpha}^{1})}^{2}}{\Delta}& 0& 0\\ 0& 0& \frac{{\epsilon}_{2}}{\Delta}& 0\\ 0& 0& 0& \frac{{\epsilon}_{3}}{\Delta}\end{array}\right).\hfill \end{array}$$This result was first obtained by Carter [38].
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**b**) - Matrix ${\widehat{G}}_{3}$.We substitute the matrix ${\widehat{G}}_{3}$ into Equation (24). After some transformations in Equation (24) we get: ${h}^{1}+{\omega}^{1}=\gamma ({\alpha}^{01}+{a}^{01}).$ Hence, $\gamma =\tilde{\gamma}$. By the admissible gradient transformation of the potential $\tilde{\gamma}$ can be set to zero. The final solution is:$${A}^{0}=\frac{h+\omega}{\Delta}\phantom{\rule{1.em}{0ex}}{A}^{1}={A}^{\nu}=0,\phantom{\rule{1.em}{0ex}}{g}^{ij}=\left(\begin{array}{cccc}\frac{{a}_{0}+{\alpha}_{0}}{\Delta}& \frac{a+\alpha}{\Delta}& 0& 0\\ \frac{a+\alpha}{\Delta}& 0& 0& 0\\ 0& 0& \frac{{\epsilon}_{2}}{\Delta}& 0\\ 0& 0& 0& \frac{{\epsilon}_{3}}{\Delta}\end{array}\right).$$
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**c**) - Matrix ${\widehat{G}}_{5}$.Substitute it in Equation (24). Denote: ${H}_{p}={h}^{p}+{\sigma}^{p}$. After the reduction, we get:$${H}_{1}={\sigma}^{1}\to {h}^{1}=0.$$The Equation (25) can be reduced to the form:$$\begin{array}{c}\hfill {({H}_{0}-{\alpha}^{0}\sigma )}^{2}=(\rho +p)A,\phantom{\rule{1.em}{0ex}}(A=a+\alpha ).\end{array}$$Equation (42) has a unique solution: ${H}_{0}={\alpha}^{0}\sigma +\widehat{e}A.$ By the admissible gradient transformation of the potential we vanish $\tilde{e}$. The solution has the form$${A}^{0}=\frac{{a}^{0}\omega}{\Delta},\phantom{\rule{1.em}{0ex}}{A}^{1}={A}^{\nu}=0.$$

## 5. Quadratic Dependence between Free Functions

## 6. Discussion

**I**- $$\left(\right)$$
**II**- $$\left(\right)$$
**III**- $$\left(\right)$$
**IV**- $$\left(\right)$$
**V**- $$\left(\right)$$

## 7. Conclusions

## Funding

## Acknowledgments

## Conflicts of Interest

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Obukhov, V.
Hamilton–Jacobi Equation for a Charged Test Particle in the Stäckel Space of Type (2.0). *Symmetry* **2020**, *12*, 1289.
https://doi.org/10.3390/sym12081289

**AMA Style**

Obukhov V.
Hamilton–Jacobi Equation for a Charged Test Particle in the Stäckel Space of Type (2.0). *Symmetry*. 2020; 12(8):1289.
https://doi.org/10.3390/sym12081289

**Chicago/Turabian Style**

Obukhov, Valeriy.
2020. "Hamilton–Jacobi Equation for a Charged Test Particle in the Stäckel Space of Type (2.0)" *Symmetry* 12, no. 8: 1289.
https://doi.org/10.3390/sym12081289