# Geometric Justification of the Fundamental Interaction Fields for the Classical Long-Range Forces

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## Abstract

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## 1. Introduction

## 2. The Relativistic Particle Lagrangian

## 3. Homogeneous Lagrangians

#### 3.1. Pros and Cons of Homogeneous Lagrangians of First Order

- (1)
- First of all, the action $S=\int L(x,\frac{dx}{d\tau})d\tau $ is a reparametrization invariant. Thus, there is no fictitious acceleration due to un-proper time parametrization (see Section 4.3);
- (2)
- For any Lagrangian $L(t,{x}^{i},\frac{d{x}^{i}}{dt})$ one can construct a reparametrization-invariant Lagrangian by enlarging the space from ${x}^{i}:i=1,\cdots ,n$ to an extended space-time ${x}^{\mu}:\mu =0,1,\cdots ,n,{x}^{0}=t$ [2,18]: $L(t,{x}^{i},\frac{d{x}^{i}}{dt})\to L({x}^{\mu},\frac{d{x}^{\mu}}{d{x}^{0}})\frac{d{x}^{0}}{d\tau}$. The Euler–Lagrange equations for these two Lagrangians are equivalent as long as ${v}^{0}=dt/d\tau $ is well behaved and $\tau $ is also a reasonable “time”-parametrization choice;
- (3)
- Parameterization-independent path-integral quantization is possible since the action S is reparametrization invariant;
- (4)
- The reparametrization invariance may help in dealing with singularities [33];
- (5)
- It is easily generalized to extended objects (p-branes) that is the subject of Section 3.3.

- (1)
- There are constraints among the Euler–Lagrange equations [2] since $det\left(\frac{{\partial}^{2}L}{\partial {v}^{\alpha}\partial {v}^{\beta}}\right)=0$;
- (2)
- It follows that the Legendre transformation ($T\left(M\right)\leftrightarrow {T}^{*}\left(M\right)$), which exchanges velocity and momentum coordinates $(x,v)\leftrightarrow (x,p)$, is problematic [4];
- (3)
- There is a problem with the canonical quantization approach since the Hamiltonian function is identically ZERO ($h\equiv 0$) [5].

#### 3.2. Canonical Form of the First-Order Homogeneous Lagrangians

#### 3.3. E-dimensional Extended Objects

- The Lagrangian for a 0-brane (relativistic point particle in an electromagnetic field, $dimE=1$ and ${\omega}^{\Gamma}\to {v}^{\alpha}=\frac{d{x}^{\alpha}}{d\tau}$) is:$$\begin{array}{ccc}\hfill L\left(\overrightarrow{\varphi},\omega \right)={A}_{\Gamma}{\omega}^{\Gamma}+\sqrt{{g}_{{\Gamma}_{1}{\Gamma}_{2}}{\omega}^{{\Gamma}_{1}}{\omega}^{{\Gamma}_{2}}}\hfill & \to & \hfill L\left(\overrightarrow{x},\overrightarrow{v}\right)\hfill \\ \hfill L\left(\overrightarrow{x},\overrightarrow{v}\right)=q{A}_{\alpha}{v}^{\alpha}+m\sqrt{{g}_{\alpha \beta}{v}^{\alpha}{v}^{\beta}};\hfill \end{array}$$
- The Lagrangian for a 1-brane (strings, $dimE=2$) [5] is:$$L\left({x}^{\alpha},{\partial}_{i}{x}^{\beta}\right)=\sqrt{{Y}^{\alpha \beta}{Y}_{\alpha \beta}},$$$$\begin{array}{c}\hfill {\omega}^{\Gamma}\to {Y}^{\alpha \beta}=\frac{\partial ({x}^{\alpha},{x}^{\beta})}{\partial (\tau ,\sigma )}=det\left(\begin{array}{cc}{\partial}_{\tau}{x}^{\alpha}& {\partial}_{\sigma}{x}^{\alpha}\\ {\partial}_{\tau}{x}^{\beta}& {\partial}_{\sigma}{x}^{\beta}\end{array}\right)=\\ \hfill ={\partial}_{\tau}{x}^{\alpha}{\partial}_{\sigma}{x}^{\beta}-{\partial}_{\sigma}{x}^{\alpha}{\partial}_{\tau}{x}^{\beta}.\end{array}$$In this case, the index $\Gamma $ for labeling the components of the generalized velocity vector ${\omega}^{\Gamma}$ corresponds to the set of pairs $\{\alpha ,\beta \}$ out of m elements. For example, for $m=4$ this will be $4!/2{!}^{2}=6$ not 4 like for the standard velocity vector in
**M**; - The Lagrangian for a general p-brane has the Dirac-Nambu-Goto term (DNG) [51]:$$L\left({x}^{\alpha},{\partial}_{E}{x}^{\beta}\right)=\sqrt{{Y}^{\Gamma}{Y}_{\Gamma}}.$$

## 4. Classical Forces Beyond Electromagnetism and Gravity

#### 4.1. Simplest Pure ${S}_{n}\left(v\right)$ Lagrangian Systems

- Curvilinear coordinate system such that: ${S}_{n}\left(v\right)=f(t,r,w,u)$ where $w=d{x}^{0}/d\tau $ and $u=dr/d\tau $;
- Static fields, that is: ${S}_{n}\left(v\right)=f(r,w,u)$;
- Inertial coordinate system in the sense of Newtonian like space and time separation, that is: ${S}_{t\cdots tr\cdots r}=0$ except for ${S}_{t\cdots t}$ and ${S}_{r\cdots r}$ components.

#### 4.2. Choice of Proper Time Parametrization

#### 4.3. Fictitious Accelerations in Un-Proper Time Parametrization

## 5. The Background Fields and Their Lagrangians

#### 5.1. Justifying the Electromagnetic Action

#### 5.2. Justifying the Einstein–Hilbert–Cartan Action

## 6. Conclusions and Discussion

## 7. Examples and Exercises

- Show that ${g}_{\mu \nu}{v}^{\mu}{v}^{\nu}=constant$ along the trajectory of a particle is a necessary and sufficient condition for Euler–Lagrange equations corresponding to ${S}_{1}$ (1) and ${S}_{2}$ (2) to be equivalent to each other and to the geodesic equation (3). In the traditional case of velocity independent metric see [22];
- Show that for any Lagrangian $L(x,v)$ that is a homogeneous function in the velocity $\overrightarrow{v}$ of order $n\ne 1$ the corresponding Hamiltonian function $h={v}^{\alpha}\left(\frac{\partial L}{\partial {v}^{\alpha}}\right)-L$ is proportional to the Lagrangian, that is, $h=(n-1)L$;
- Show that any time independent Lagrangian $L(\overrightarrow{x},\overrightarrow{v})$, which is a homogeneous function in velocity $\overrightarrow{v}$ of order $n\ne 1$, is an integral of the motion with respect to the corresponding Euler–Lagrange equations for L;
- Consider a Lagrangian that is a constant of the motion; that is, $dL/d\tau =0$. Show that any solution of the Euler–Lagrange equations for L is also a solution for $\tilde{L}=f\left(L\right)$ under certain minor and reasonable requirements on f, such as $\tilde{L}=f\left(L\right)\ne 0$ and ${\tilde{L}}^{\prime}={f}^{\prime}\ne 0$;
- Show that if ${v}^{0}=dt/d\tau $ is well behaved (${v}^{0}\ne 0$ over the duration of the process studied) then the Euler–Lagrange equations for the reparametrization-invariant Lagrangian $L({x}^{\mu},{v}^{\mu})=L({x}^{\mu},{v}^{i}/{v}^{0}){v}^{0}$, where $i=1,\dots ,n$, $\mu =0,1,\dots ,n$ and ${x}^{0}=t,{v}^{i}=d{x}^{i}/d\tau ,{v}^{0}=dt/d\tau $, are equivalent to the Euler–Lagrange equations for coordinate-time parametrization ($\tau =t$) choice for $L(t,{x}^{i},d{x}^{i}/dt)$. Hint: Use that $L({x}^{\mu},{v}^{i}/{v}^{0})$ is a zero-order homogeneous function with respect to ${v}^{\mu}$ and notice the relationship between the Hamiltonian function h for the initial Lagrangian $L(t,{x}^{i},d{x}^{i}/dt)$ and the generalized momentum ${p}_{0}=\partial L/\partial {v}^{0}$ for the reparametrization-invariant Lagrangian $L({x}^{\mu},{v}^{\mu})=L({x}^{\mu},{v}^{i}/{v}^{0}){v}^{0}$;
- Show that ${\sum}_{\beta}{v}^{\beta}\frac{{\partial}^{2}L}{\partial {v}^{\alpha}\partial {v}^{\beta}}=0$ if L is first-order homogeneous Lagrangian. Thus, $det\left(\frac{{\partial}^{2}L}{\partial {v}^{\alpha}\partial {v}^{\beta}}\right)=0$, since in an extended space-time one usually expects ${v}^{0}\ne 0$’
- Consider the constraint $\sqrt{{g}_{\alpha \beta}{v}^{\alpha}{v}^{\beta}}=1$ implemented via a Lagrangian multiplier $\chi $ in the Lagrangian $L=q{A}_{\alpha}{v}^{\alpha}+(m+\chi ){g}_{\alpha \beta}{v}^{\alpha}{v}^{\beta}-\chi $. Show that the value of $\chi $ is required to be $\chi =-m/2$ if $L=q{A}_{\alpha}{v}^{\alpha}+m\sqrt{{g}_{\alpha \beta}\left(x\right){v}^{\alpha}{v}^{\beta}}$ and $L=q{A}_{\alpha}{v}^{\alpha}+(m+\chi ){g}_{\alpha \beta}{v}^{\alpha}{v}^{\beta}-\chi $ are to result in the same Euler–Lagrange equations;
- Consider the Lagrangian $L=m\sqrt{{\eta}_{\alpha \beta}{v}^{\alpha}{v}^{\beta}}+\kappa \sqrt[n]{{S}_{n}\left(\overrightarrow{v},\cdots ,\overrightarrow{v}\right)}$, where ${\eta}_{\alpha \beta}=(1,-1,\dots ,-1)$ is the Lorentz invariant metric tensor. Show that in the non-relativistic limit ($v\to 0$), the Euler–Lagrange equations for the acceleration $\frac{dv}{d\tau}$ are the same up to $O\left({v}^{2}\right)$ terms whether $L=const$ or $\sqrt{{\eta}_{\alpha \beta}{v}^{\alpha}{v}^{\beta}}=const$ parametrization is imposed. Thus the non-relativistic limit cannot distinguish these two choices of trajectory parametrization;
- Show that solutions of the Euler–Lagrange equations for $L={v}^{\mu}{A}_{\mu}(x,v),$ where x is a space-time coordinate and ${v}^{\mu}$ is a world-line velocity vector (4-vector for 3+1 space-time), satisfy $\frac{d}{d\lambda}\left({v}^{\alpha}{g}_{\alpha \beta}\left(x,v\right){v}^{\beta}\right)=0$ for the velocity dependent metric ${g}_{\alpha \beta}\left(x,v\right)=\frac{1}{2}({A}_{\alpha /\beta}\left(x,v\right)+{A}_{\beta /\alpha}\left(x,v\right))$ with ${A}_{\beta /\alpha}\left(x,v\right)$ being a partial derivative with respect to ${v}^{\alpha}$ of ${A}_{\beta}\left(x,v\right)$;
- Choose a specific Lagrangian $\tilde{L}(x,v)$ that is a homogeneous function of first order in v, then consider the Lagrangian $L={v}^{\mu}{A}_{\mu}\left(x\right)+\sqrt{{g}_{\alpha \beta}{v}^{\alpha}{v}^{\beta}}$ where the fields ${g}_{\alpha \beta}\left(x\right)$ and ${A}_{\mu}\left(x\right)$ are defined via the following expressions: ${A}_{\mu}\left(x\right)=\frac{1}{2}{(\tilde{L}(x,v)-\tilde{L}(x,-v))}_{/\mu}\left|{}_{v=(1,\overrightarrow{0})}\right.$ and ${g}_{\alpha \beta}\left(x\right)=\frac{1}{4}{(\tilde{L}(x,v)+\tilde{L}(x,-v))}_{/\alpha /\beta}^{2}\left|{}_{v=(1,\overrightarrow{0})}\right.$. Compare the corresponding Euler–Lagrange equations of motions for $\tilde{L}$ and L. At what order k of $O\left({v}^{k}\right)$ are the differences?

## Author Contributions

## Funding

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## Informed Consent Statement

## Data Availability Statement

## Acknowledgments

## Conflicts of Interest

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Gueorguiev, V.G.; Maeder, A.
Geometric Justification of the Fundamental Interaction Fields for the Classical Long-Range Forces. *Symmetry* **2021**, *13*, 379.
https://doi.org/10.3390/sym13030379

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Gueorguiev VG, Maeder A.
Geometric Justification of the Fundamental Interaction Fields for the Classical Long-Range Forces. *Symmetry*. 2021; 13(3):379.
https://doi.org/10.3390/sym13030379

**Chicago/Turabian Style**

Gueorguiev, Vesselin G., and Andre Maeder.
2021. "Geometric Justification of the Fundamental Interaction Fields for the Classical Long-Range Forces" *Symmetry* 13, no. 3: 379.
https://doi.org/10.3390/sym13030379