Motion of Massive Particles in Rindler Space and the Problem of Fall at the Centre

TL;DR

This paper explores the classical and quantum behavior of massive particles in Rindler space, revealing unique geodesic paths and a novel fall problem at the sphere's surface, differing from traditional gravitational models.

Contribution

It provides new insights into particle motion in Rindler space and formulates the quantum fall problem, highlighting differences from Schwarzschild and conventional scenarios.

Findings

Geodesics in Rindler space differ from Schwarzschild case.

Particles fall at the surface of a sphere, not at the center.

Quantum analysis shows unique fall behavior in Rindler background.

Abstract

The motion of a massive particle in Rindler space has been studied and obtained the geodesics of motion. The orbits in Rindler space are found to be quite different from that of Schwarzschild case. The paths are not like the Perihelion Precession type. Further we have set up the non-relativistic Schrodinger equation for the particle in the quantum mechanical scenario in presence of background constant gravitational field and investigated the problem of fall of the particle at the center. This problem is also treated classically. Unlike the conventional scenario, here the fall occurs at the surface of a sphere of unit radius.

Equations

$$t=\left(\frac{1}{\alpha }+x^{{}^{\prime }}\right)\sinh (\alpha t^{{}^{\prime }})$$

$$x=\left(\frac{1}{\alpha }+x^{{}^{\prime }}\right)\cosh (\alpha t^{{}^{\prime }})$$

$$d{s}^2=-d{\tau }^2=-(1+\alpha x{)}^{2}d{t}^2+d{x}^2$$

$${\int }_A^{B}d\tau ={\int }_A^{B}Ld\lambda$$

$$L={\left[(1+\alpha x{)}^2{\left(\frac{dt}{d\lambda }\right)}^2-{\left(\frac{dx}{d\lambda }\right)}^2\right]}^{\frac{1}{2}}$$

$$\frac{\partial }{\partial \lambda }\left(\frac{\partial L}{\partial (\frac{\partial q_i}{\partial \lambda })}\right)-\frac{\partial L}{\partial q_i}=0$$

$$\frac{\partial t}{\partial \tau }=\frac{C}{(1+\alpha x{)}^2}$$

$$(1+\alpha x)\alpha {\left(\frac{\partial t}{\partial \tau }\right)}^2+\frac{d^{2}x}{d{\tau }^2}=0$$

$$\frac{\alpha C}{(1+\alpha x{)}^3}+\frac{d^{2}x}{d{\tau }^2}=0$$

$$\frac{d^{2}u}{d{\tau }^2}+\frac{1}{u^3}=0$$

$$\frac{1}{2}{\left(\frac{du}{d\tau }\right)}^2-\frac{1}{2u^2}=\mathrm{constant}$$

$$\frac{p^2}{2}-\frac{C{\alpha }^2}{2u^2}=\mathrm{constant}=E$$

$$\frac{1}{2}{p}^2+V(u)=E$$

$$(1+\alpha x{)}^2{\left(\frac{dt}{d\tau }\right)}^2-{\left(\frac{dx}{d\tau }\right)}^2=1$$

$$\tau ={\int }_0^x\frac{dx}{[\frac{C}{(1+\alpha x{)}^2}-1{]}^{\frac{1}{2}}}$$

$$\alpha \tau =-\frac{C^{1/2}}{\alpha }\left[{\left\{1-{\left(\frac{1+\alpha x}{C^{1/2}}\right)}^2\right\}}^{\frac{1}{2}}-{\left(1-\frac{1}{C}\right)}^{\frac{1}{2}}\right]$$

$$\alpha x=C^{1/2}\left[1-{\left\{{\left(1-\frac{1}{C}\right)}^{1/2}-\frac{\alpha \tau }{C^{1/2}}\right\}}^2\right]-1$$

$$H\psi (u)=(\frac{p^2}{2}-\frac{\beta }{u^2})\psi (u)=-E\psi (u)$$

$$\frac{d^2\psi }{d{u}^2}+\frac{\gamma }{u^2}\psi (u)-K^2\psi (u)=0$$

$$u^2\frac{d^{2}R}{d{u}^2}-2\rho \frac{dR}{du}+(\gamma -2-K^2{u}^2)R=0$$

$$\frac{d^{2}R}{d{\rho }^2}-\frac{2}{\rho }\frac{dR}{d\rho }+\frac{\gamma -2}{{\rho }^2}R-R=0$$

$$\frac{d^{2}R}{d{\rho }^2}-2\frac{dR}{d\rho }+(\gamma -3)R=0$$

$$u(\rho )=\exp (\rho )[A_1\exp ((4-\gamma {)}^{1/2}\rho )+A_2\exp (-(4-\gamma {)}^{1/2}\rho )]$$

$$u(\rho )=\exp (\rho )[A_1\exp \{i(\gamma -4{)}^{1/2}\rho \}+A_2\exp \{-i(\gamma -4{)}^{1/2}\rho \}]$$

$$\frac{1}{L}\left[(1+\alpha x)\alpha {\left(\frac{dt}{d\lambda }\right)}^2+\frac{d^{2}x}{d{\lambda }^2}\right]$$

$$\frac{dt}{d\lambda }=\frac{dt}{d\tau }\frac{d\tau }{d\lambda }=L\frac{dt}{d\tau }$$

$${\left(\frac{dt}{d\tau }\right)}^2=L^2{\left(\frac{dt}{d\tau }\right)}^2$$

$$\frac{d^{2}x}{d{\lambda }^2}=\frac{d}{d\lambda }\left(\frac{dx}{d\lambda }\right)=\frac{d}{d\tau }\left(\frac{dx}{d\lambda }\right)\left(\frac{d\tau }{d\lambda }\right)$$

$$\frac{d^{2}x}{d{\lambda }^2}=\frac{d}{d\lambda }\left(\frac{dx}{d\tau }\right)\left(\frac{d\tau }{d\lambda }\right)$$

$$\frac{d^{2}x}{d{\lambda }^2}=\frac{d^{2}x}{d{\tau }^2}{\left(\frac{d\tau }{d\lambda }\right)}^2=L^2\frac{d^{2}x}{d{\lambda }^2}$$

Full text

Motion of Massive Particles in Rindler Space and the Problem of Fall at the Centre

Sanchita Dasa, Soma Mitrab and Somenath Chakrabartyc

Department of Physics, Visva-Bharati, Santiniketan, India 731235

aEmail:sanchita.vbphys@gmail.com

bEmail:somaphysics@gmail.com

cEmail: somenath.chakraborty@visva-bharati.ac.in

Abstract

The motion of a massive particle in Rindler space has been studied and obtained the geodesics of motion. The orbits in Rindler space are found to be quite different from that of Schwarzschild case. The paths are not like the Perihelion Precession type. Further we have set up the non-relativistic Schrodinger equation for the particle in the quantum mechanical scenario in presence of background constant gravitational field and investigated the problem of fall of the particle at the center. This problem is also treated classically. Unlike the conventional scenario, here the fall occurs at the surface of a sphere of unit radius.

It is well known that the Lorentz transformations of space time coordinates are between the inertial frame of references [1], whereas the Rindler transformations are between an inertial frame and the frame having uniform accelerated motion [2, 3, 4, 5, 6, 7, 8, 9, 10]. The space here is called the Rindler space, exactly like the Minkowski space in the Lorentz transformation scenario. Further, the Rindler space is locally flat, whereas the flatness of the Minkowski space is global in nature. Now according to the principle of equivalence a frame undergoing accelerated motion in absence of gravity is equivalent to a frame at rest in presence of a gravitational field. The strength of the gravitational field is exactly equal to the magnitude of the acceleration of the moving frame. Therefore the Rindler space may be considered to be a space associated with a frame at rest in presence of a uniform gravitational field. The Rindler space-time transformations in natural unit ($c=\hbar=1$) and in $1+1$ dimension are given by [7, 8, 9, 10]

[TABLE]

[TABLE]

Where primed coordinates are in the non-inertial frame, whereas the unprimed are in the inertial one. The Rindler space-time transformations are therefore exactly like the Lorentz transformations but in uniformly accelerated frame. Here $\alpha$ is the local acceleration or local uniform gravitational field. Hence the line element in Rindler space in $1+1$-dimension is given by

[TABLE]

where $\tau$ is the proper time. Then one can define the classical action integral in the form [1]

[TABLE]

where $L=\frac{d\tau}{d\lambda}$, the Lagrangian of the particle, given by [11, 12, 13]

[TABLE]

The aim of the present article is to study the particle motion of non-zero mass in Rindler space and obtain the geodesics of motion. We further investigate the problem of fall at the origin. The last one is based on both the classical motion and the quantum mechanical motion of the particle. The particlles will occupy the space near the surface of the sphere of radius unity.

Now the well known Euler-Lagrange equation is given by

[TABLE]

where $q_{i}=t$ or $x$. Let us first consider $q_{i}=t$, the time coordinate. Since $t$ is a cyclic coordinate, $\frac{\partial L}{\partial t}=0$ and therefore

[TABLE]

where $C$ is a constant and we have used the expression for $L$ as given by eqn.(5). We next consider $q_{i}=x$, then with some little algebra, we have from the Euler-Lagrange equation

[TABLE]

On substituting the expression for $\frac{\partial t}{\partial\tau}$ from eqn.(7) we have

[TABLE]

(see Appendix). Now changing the variable from $x$ to $u=1+\alpha x$ and redefining $\tau\rightarrow C^{\frac{1}{2}}\alpha\tau$, we have from the above equation,

[TABLE]

This is the equation of motion for the particle in Rindler space. Integrating both the sides with respect to $u$, this equation may be transformed to the form

[TABLE]

Rewriting $\tau$ in its original form and defining $p=du/d\tau$ as the momentum of the particle of unit mass, we have from the above equation

[TABLE]

where the first term on the left hand side is like the kinetic energy for a unit mass and the second term is an attractive inverse square potential, whereas the constant on the right hand side may be treated as the total energy of the particle. We further assume that the energy of the particle is negative in nature, i.e., we are considering bound orbit of the classical particle. Now expressing the above equation in the usual form, given by

[TABLE]

where $V(u)=-\frac{C\alpha^{2}}{2u^{2}}$, the inverse square attractive potential, then depending on the strength $0.5C\alpha^{2}$ of the attractive potential $V(u)$, the particle may fall at the centre ($x=0$), where the momentum becomes zero ($p=0$, i.e., particle is at rest) and in this situation $E=V(u=1)=-0.5C\alpha^{2}$, the maximum depth of the potential, which is also the minimum possible value of the total energy. Since it is negative it gives the maximum binding at $x=0$ or $u=1$. In fig.(1) we have shown this potential well. The coordinate $x$ is plotted along $x$-axis, whereas the magnitude of the potential $\mid V(u)\mid$ is plotted along $y$-axis. Therefore in actual scenario this is not the potential hill but is potential well. But unlike the conventional scenario of fall at the centre problem, there is no singularity at the origin ($x=0$), or at the centre of the gravitating object [14, 15, 16]. The value of $V(u)$ is $-0.5C\alpha^{2}$ at the origin and $\longrightarrow-1$ for large $u$ values. Therefore unlike the conventional case, where the potential part $\longrightarrow-\infty$ at the origin, here, since the minimum value of $u$ is positive unity, therefore $V(u)\rightarrow-0.5C\alpha^{2}$ for $x\longrightarrow 0$ or $u\longrightarrow 1$. In this case the singularity is as if covered by a sphere of unit radius. Therefore here the problem of fall at the origin is reduced to the problem of fall on the unit surface.

For further investigation of this problem of fall at the centre, to some extent elaborately, let us get the geodesics of motion for the particle in Rindler space. We start with the equation

[TABLE]

Hence we have after substituting for $\frac{dt}{d\tau}$ from eqn.(7)

[TABLE]

Using the changed variable $u(=1+\alpha x)$, we have,

[TABLE]

Since the uniform acceleration $\alpha$ of the frame is completely arbitrary, we shall study the variation of $\alpha x$ with $\alpha\tau$. The above equation can also be expressed in the following more convenient form

[TABLE]

which will give the variation of $\alpha x$ with $\alpha\tau$. Of course $\alpha$ is assumed to be constant throughout, whereas by evaluating $\alpha x$ as a function of $\alpha\tau$, we have removed the arbitrariness of $\alpha$.

Before we obtain numerically the variation of $\alpha x$ with $\alpha\tau$, we shall impose some constraints on the quantities appearing in the above equation. By inspection one can put the following restrictions: $C>1$, and if $\alpha x>0$, then $\alpha\tau>0$. Since we are measuring $x$ along positive direction and the frame is also moving with uniform acceleration $\alpha$ along positive $x$-direction, the second constraint is is also quite obvious. In fig.(2) we have shown the variation of $\alpha x$ with $\alpha\tau$ for three different $C$ values. From the curves one may infer that the geodesics are closed in nature; returned to the initial point, i.e., at $x\approx 0$ (Boomerang type geodesic). The geodesics are not like the Perihelion Precession type as has been obtained in the Schwarzschild geometry [17, 18]. For the proper justification of our arguments as given above, we next consider the equation of motion of the particle under inverse square potential, given by eqn.(10). This equation has been solved numerically for the initial conditions $\frac{du}{d\tau}=0$, i.e., the motion is assumed to be started from rest and for a number of initial positional coordinates in the scaled form given by $u_{0}$. The boundary value of $u$ is unity, i.e., on the surface of a unit sphere. In fig.(3) we have shown the geodesics of motion for three different $u_{0}$ values.

Now to complete the study of free fall at the centre in our modified formalism, let us next investigate the quantum mechanical motion of the particle in inverse square attractive potential [14, 15, 16]. We consider the bound state problem, i.e., $E<0$. Re-substituting the exact form of $\tau$, we have the eigen value equation for a unit mass particle

[TABLE]

where $\beta=\frac{\alpha^{2}C}{2}$. Hence we have

[TABLE]

where $\gamma=\frac{2\beta}{\hbar^{2}}$ and $K^{2}=\frac{2E}{\hbar^{2}}$. Therefore in the asymptotic region, i.e., $u\rightarrow\infty$, $\psi\sim e^{-Ku}$ or it goes to zero. The wave function is therefore well behaved at infinity. Next to study the nature of wave function near the unit surface, we substitute $\psi(u)=\frac{R(u)}{u}$. Then it can very easily be been shown that $R(u)$ satisfies the equation

[TABLE]

Putting $Ku=\rho$ as a new variable, we have,

[TABLE]

To investigate the nature of $R$ near unit surface we write $\rho=\rho_{0}$ in the denominator of the second and the third term and then put the limit $\rho_{0}\rightarrow 1$. Then we have

[TABLE]

Then for $\gamma<4$, we have the solution

[TABLE]

whereas for $\gamma>4$, the solution is given by

[TABLE]

The solutions are of standard stationary type and exist for $\rho=\rho_{0}\rightarrow 1$. Further, for bound state solution, i.e., for $E<0$, $E_{k}<|V|$, where $E_{k}$ and $V$ are the kinetic energy and attractive potential respectively. The quantity $|V|$ is maximum for $\rho=\rho_{0}=K$, i.e., at $x=0$ and approaches unity as $\rho$ increases to infinity. Therefore the negative value of $E$ is maximum on the surface of the unit shell. This indirectly indicates the ”fall” on the surface. The basic difference with the usual solution is that the fall is not at the centre $(\rho=0)$ which is covered by the unit shell. Therefore we may conclude that in the Rindler space the energy of the particle will minimum on the surface of the sphere of unit radius. This indirectly indicate that all the particles will fall on this surface because of the minimum values of the energy of the particles. This is true for classical as well as quantum mechanical scenarios. The other interesting finding of this work is that unlike the perihelion precession type geodesics these are closed type. We call them as boomerang geodesics.

Appendix

In this appendix we shall give a short outline to show that

[TABLE]

we have

[TABLE]

Hence

[TABLE]

[TABLE]

[TABLE]

[TABLE]

Hence we have the above equality after putting, $\frac{dt}{d\lambda}=\frac{C}{(1+\alpha x)^{2}}$