1. Introduction
A significant challenge in the field of physics remains unresolved, namely how to integrate the principles of quantum mechanics and gravity into a quantum theory of gravity. The fundamental equation of gravity is Einstein's field equation [1]. The fundamental equation in quantum mechanics is Schrödinger's equation [2], while the Klein–Gordon equation [3, 4] and Dirac equation [5] are viewed as relativistic field equations. A modification to Schrödinger's equation was suggested by Diósi and Penrose [6–11] to include the force of gravity. This is known as the Schrödinger–Newton equation. This new term takes into account the gravitational potential generated by the distribution of mass in the system. The effect can be generalized for the Dirac equation, Klein–Gordon equation and Proca equation [12].
In this work we use these equations to study a specific form of gravitational potential that arises from string T-duality [13, 14]. On the other hand, a new advance was proposed by Padmanabhan who related the concept of duality and zero-point length as a tool to obtain the quantum gravity effect [15]; this was applied to black hole physics to cure the black hole singularity for regular and charged solutions [16–18]. In [19], one of the present authors argued how the zero-point length, which modifies black hole spacetime, can lead to a non-singular solution during gravitational collapse. In the context of cosmology, however, it was found that a modified Friedmann's equation [20], and in particular the zero-point length corrections, can lead to a bouncing scenario (see, [20, 21]). Such a potential is regular at r = 0 and modifies the energy of the system. In this picture, the particle is no longer concentrated into a point but is diluted along with smeared energy density. The gravitational potential implied by T-duality suggests a modification of energy that agrees with the modifications implied by generalized uncertainty principle (GUP) models [22–28] (see also [29, 30]). We realize a new connection between the spin concept of particles and these energy modifications.
Phenomenological and experimental implications of the GUP have been investigated in low- and high-energy regimes. These include atomic systems [31, 32], quantum optical systems [33], gravitational bar detectors [34], gravitational decoherence [35], gravitational waves [36–38], gravitational tests [39, 40], composite particles [41], astrophysical systems [42], condensed matter systems [43], cold atoms [44], macroscopic harmonic oscillators [45], gauge theories [46], quantum noise [47], extended GUP from deformed algebra [48], modified gravity theories [49], cosmological models [50–55] and holography [56]. Reviews of the GUP, its phenomenology, and its experimental implications can be found in [57, 58]. Our work establishes a connection between T-duality in string theory and the GUP. Among other things, in this work we would like to understand if the form of the GUP depends on the nature of particles, for example the spin. It is possible to interpret that experimental and phenomenological research on GUP provides support in favor of T-duality and string theory.
The paper is organized as follows. In section 2 , we study the modified Schrödinger–Newton equation to compute the corrections to the energy of the particle. In section 3 , we compute the self-regularized energy due to the electric field. In section 4 , we study the modified Klein–Gordon equation and the relativistic energy corrections. In section 5 , we investigate the modified Dirac equation and its energy corrections. In section 6 , we study the modified Proca equation. In section 7 , we show a link between the GUP and self-regularized energy. Importantly, here we shall elaborate on how the nature of particles, say the spin of the particle, can lead to a different form of the GUP. Finally in section 8 , we comment on our results.
2. Modified Schrödinger–Newton equation
We shall review the quantum-corrected static interaction potential according to field theory with path integral duality. We recall that the momentum space propagator induced by path integral duality for the massless case is given by [16]
$\begin{eqnarray}G(k)=-\displaystyle \frac{{l}_{0}}{\sqrt{{k}^{2}}}\,{K}_{1}\left({l}_{0}\sqrt{{k}^{2}}\right)\end{eqnarray}$
with l0 being the zero-point length and ${K}_{1}\left(x\right)$ a modified Bessel function of the second kind. One has two cases: at small momenta, i.e. for ${({l}_{0}k)}^{2}\to 0$, we obtain the conventional massless propagator G(k) = − k−2; at large momenta, interestingly, the exponential suppression is responsible for curing UV divergences [16] $\begin{eqnarray}G(k)\sim -{l}_{0}^{1/2}\,{\left({k}^{2}\right)}^{-3/4}\,{{\rm{e}}}^{-{l}_{0}\sqrt{{k}^{2}}}.\end{eqnarray}$
Consider a static external source J which consists of two point-like masses, m and M, at relative distance $\vec{r}$, then the potential was found as [16] $\begin{eqnarray}V(r)=-\displaystyle \frac{1}{m}\,\displaystyle \frac{W[J]}{T}\end{eqnarray}$
$\begin{eqnarray}=\,-M\,G\,\int \displaystyle \frac{{{\rm{d}}}^{3}k}{{\left(2\pi \right)}^{3}}\,{{G}_{{\rm{F}}}(k)| }_{{k}^{0}=0}\,\exp \left(i\vec{k}\vec{r}\right)\end{eqnarray}$
$\begin{eqnarray}=\,-\displaystyle \frac{{GM}}{\sqrt{{r}^{2}+{l}_{0}^{2}}},\end{eqnarray}$
In this section, we will use the Schrödinger–Newton equation to obtain the modified energy–time uncertainty relation. The Schrödinger–Newton equation for a free particle with the self-gravitational potential V produced by a quantum source in state $\Psi$ can be written as [6–11] $\begin{eqnarray}{\rm{i}}{\hslash }\displaystyle \frac{\partial {\rm{\Psi }}(t,\vec{r})}{\partial t}=\left(-\displaystyle \frac{{{\hslash }}^{2}}{2M}{{\rm{\nabla }}}^{2}-{M}^{2}G\int \displaystyle \frac{| {\rm{\Psi }}(t,\vec{r}^{\prime} ){| }^{2}{{\rm{d}}}^{3}\vec{r}^{\prime} }{| \vec{r}-\vec{r}^{\prime} | }\right){\rm{\Psi }}(t,\vec{r}),\end{eqnarray}$
as studied by Diósi and Penrose [6–11] along with the semi-classical Poisson equation $\begin{eqnarray}{{\rm{\nabla }}}^{2}V=4\pi {GM}| {\rm{\Psi }}{| }^{2}.\end{eqnarray}$
The mass density of the source is related to the quantum probability density via $\begin{eqnarray}\rho (r,t)=M| {\rm{\Psi }}(t,\vec{r}){| }^{2}.\end{eqnarray}$
Our aim is to use a well-behaved expression without a singularity at r = 0. Such a regularized expression was found in string T-duality with the gravitational potential given by [16] $\begin{eqnarray}V(r)\to -\displaystyle \frac{{GM}}{\sqrt{{r}^{2}+{l}_{0}^{2}}},\end{eqnarray}$
where l0 is the zero-point length. The modified potential suggests a modification to the Schrödinger–Newton equation as follows: $\begin{eqnarray}\begin{array}{rcl}i{\hslash }\displaystyle \frac{\partial {\rm{\Psi }}(t,\vec{r})}{\partial t} & = & \left(-\displaystyle \frac{{{\hslash }}^{2}}{2M}{{\rm{\nabla }}}^{2}\right.\\ & & \left.-{M}^{2}G\int \displaystyle \frac{| {\rm{\Psi }}(t,\vec{r}^{\prime} ){| }^{2}{d}^{3}\vec{r}^{\prime} }{\sqrt{| \vec{r}-\vec{r}^{\prime} {| }^{2}+{l}_{0}^{2}}}\right){\rm{\Psi }}(t,\vec{r}).\end{array}\end{eqnarray}$
On the one hand, we see that when l0 = 0 the standard relation is obtained. On the other hand, we can use a coupling of gravity to matter by means of the semi-classical Einstein equation [59]
$\begin{eqnarray}{R}_{\mu \nu }+\displaystyle \frac{1}{2}{g}_{\mu \nu }R=\displaystyle \frac{8\pi G}{{c}^{4}}\,\left\langle {\rm{\Psi }}\right|{T}_{\mu \nu }\left|{\rm{\Psi }}\right\rangle .\end{eqnarray}$
In the Newtonian limit, the components of the energy–momentum tensor become the Poisson equation [59] $\begin{eqnarray}{{\rm{\nabla }}}^{2}V=\displaystyle \frac{4\pi G}{{c}^{2}}\left\langle {\rm{\Psi }}\right|{\hat{T}}_{00}\left|{\rm{\Psi }}\right\rangle .\end{eqnarray}$
In the Newtonian limit, where T00 is the dominant term of the energy–momentum tensor, the interaction Hamiltonian then becomes [59]
$\begin{eqnarray}{H}_{\mathrm{int}}=\int {{\rm{d}}}^{3}r\,V\,{T}^{00}\end{eqnarray}$
$\begin{eqnarray}=\,-{MG}{\int }_{0}^{2\pi }{\int }_{0}^{\pi }{\int }_{0}^{\infty }{{\rm{d}}}^{3}r\displaystyle \frac{\rho (\vec{r})}{\sqrt{{r}^{2}+{l}_{0}^{2}}},\end{eqnarray}$
where by using V(r) and the Poisson equation one can obtain [16] $\begin{eqnarray}\rho (r)=\displaystyle \frac{1}{4\pi G}{{\rm{\nabla }}}^{2}V(r)=\displaystyle \frac{3{l}_{0}^{2}M}{4\pi {\left({r}^{2}+{l}_{0}^{2}\right)}^{5/2}}.\end{eqnarray}$
Using spherical coordinates $\begin{eqnarray}{{\rm{d}}}^{3}r={r}^{2}\sin \theta \,{\rm{d}}\theta \,{\rm{d}}\phi \,{\rm{d}}r,\end{eqnarray}$
and integrating this equation we obtain a finite contribution $\begin{eqnarray}{H}_{\mathrm{int}}=-\displaystyle \frac{3\pi {{GM}}^{2}}{16\,{l}_{0}}.\end{eqnarray}$
Therefore the apparently nonlinear form of the Schrödinger–Newton equation reduces to the linear form
$\begin{eqnarray}{\rm{i}}{\hslash }\displaystyle \frac{\partial }{\partial t}{\rm{\Psi }}(t,\vec{r})=-\displaystyle \frac{{{\hslash }}^{2}}{2M}{{\rm{\nabla }}}^{2}{\rm{\Psi }}(t,\vec{r})+{\hat{H}}_{\mathrm{int}}{\rm{\Psi }}(t,\vec{r}),\end{eqnarray}$
or in terms of operators $\begin{eqnarray}\hat{H}{\rm{\Psi }}(t,r)=\left(-\displaystyle \frac{{{\hslash }}^{2}}{2M}{{\rm{\nabla }}}^{2}+{\hat{H}}_{{\rm{int}}}\right){\rm{\Psi }}(t,r),\end{eqnarray}$
and if we define ${\hat{H}}_{{\rm{tot}}}=\hat{H}-{\hat{H}}_{{\rm{int}}}$, we can write $\begin{eqnarray}{\hat{H}}_{{\rm{tot}}}{\rm{\Psi }}(t,r)=-\displaystyle \frac{{{\hslash }}^{2}}{2M}{{\rm{\nabla }}}^{2}{\rm{\Psi }}(t,r),\end{eqnarray}$
yielding $\begin{eqnarray}{E}_{{\rm{tot}}}{\rm{\Psi }}(t,r)=\left(E+\displaystyle \frac{3\pi {{GM}}^{2}}{16\,{l}_{0}}\right){\rm{\Psi }}(t,r).\end{eqnarray}$
Here E can be interpreted as the bare energy of the free particle, and the wave function of the particle is now $\begin{eqnarray}{\rm{\Psi }}=A\exp (-\displaystyle \frac{i}{{\hslash }}({E}_{{\rm{tot}}}t-\vec{p}\cdot \vec{r})),\end{eqnarray}$
meaning that the total energy in the wave function is modified.2.1. Significance of effects in the Schrödinger–Newton equation
The Schrödinger–Newton equation reduces to ${E}_{{tot}}{\rm{\Psi }}(t,r)\,=\tfrac{{p}^{2}}{2M}{\rm{\Psi }}(t,r)$. Consider a free particle solution described by the initial Gaussian wave function
$\begin{eqnarray}{\rm{\Psi }}(r,t)={\left(\pi \,{a}^{2}\right)}^{-3/4}{\left(1+\displaystyle \frac{{\rm{i}}\,{\hslash }\,t}{M\,{a}^{2}}\right)}^{-3/2}\,{{\rm{e}}}^{-\tfrac{{r}^{2}}{2{a}^{2}\left(1+\tfrac{{\rm{i}}\,{\hslash }\,t}{M\,{a}^{2}}\right)}},\end{eqnarray}$
and note that the peak of r2 $\Psi$*(r, t) $\Psi$(r, t) is located at $\begin{eqnarray}{r}_{p}(t)=\sqrt{{a}^{2}+{\left(\displaystyle \frac{{\hslash }\,t}{a\,M}\right)}^{2}}.\end{eqnarray}$
Now let us set the acceleration $\begin{eqnarray}{\ddot{r}}_{p}(0)=\displaystyle \frac{{{\hslash }}^{2}}{{a}^{3}\,{M}^{2}},\end{eqnarray}$
while the peak probability equal to the acceleration due to Newtonian gravity is $\begin{eqnarray}{\ddot{r}}_{p}(0)=-\displaystyle \frac{{\rm{d}}}{{\rm{d}}r}\,V({r}_{p}(0))=\displaystyle \frac{M\,a}{{\left({l}_{0}^{2}+{a}^{2}\right)}^{3/2}}.\end{eqnarray}$
From the last two equations, we get the mass $\begin{eqnarray}{M}^{3}\,a=\displaystyle \frac{{{\hslash }}^{2}}{G}{\left(1+\displaystyle \frac{{l}_{0}^{2}}{{a}^{2}}\right)}^{3/2}.\end{eqnarray}$
From the last equation, we get the standard result when l0 = 0, i.e. M3a = ℏ2/G. This equation allows us to determine a critical width for a given mass value, and vice versa. In this way, we can estimate the regime where the effects of the Schrödinger–Newton equation are important.2.2. Curved spacetime
We can consider further corrections by considering a curved spacetime background. In particular, the general solution in the case of a static, spherically symmetric source reads17 ) this only suggests an improved correction by a factor of 1/2.
$\begin{eqnarray}{\rm{d}}{s}^{2}={g}_{00}\,{\rm{d}}{t}^{2}+{g}_{{rr}}\,{\rm{d}}{r}^{2}\,+\,{r}^{2}({\rm{d}}{\theta }^{2}+{\sin }^{2}\theta \,{\rm{d}}{\phi }^{2}),\end{eqnarray}$
with g00(r) = − f(r), and $\begin{eqnarray}f(r)=1-\displaystyle \frac{2{Gm}(r)}{r}=1\,+\,2{\rm{\Phi }}(r),\end{eqnarray}$
where Φ(r) = − Gm(r)/r. Using the energy density (15) it was found that $\begin{eqnarray}m(r)=\displaystyle \frac{{{Mr}}^{3}}{{\left({r}^{2}+{l}_{0}^{2}\right)}^{3/2}},\end{eqnarray}$
along with $\begin{eqnarray}f(r)=1-\displaystyle \frac{2{{GMr}}^{2}}{{\left({r}^{2}+{l}_{0}^{2}\right)}^{3/2}}.\end{eqnarray}$
In this case, the Schrödinger–Newton equation will be $\begin{eqnarray}{\rm{i}}{\hslash }\displaystyle \frac{\partial {\rm{\Psi }}(t,\vec{r})}{\partial t}=\left(-\displaystyle \frac{{{\hslash }}^{2}}{2M}{{\rm{\nabla }}}_{{\rm{LB}}}^{2}+{H}_{{\rm{int}}}\right){\rm{\Psi }}(t,\vec{r}),\end{eqnarray}$
where ${{\rm{\nabla }}}_{{\rm{LB}}}^{2}={g}^{-1/2}{\partial }_{\mu }({g}^{\mu \nu }{g}^{1/2}{\partial }_{\nu })$ is the Laplace–Beltrami operator in curved spacetime geometry, with $g=\det {g}_{\mu \nu }$. For the self-gravitational interaction, we get $\begin{eqnarray}\begin{array}{rcl}{H}_{\mathrm{int}} & = & {\displaystyle \int }_{0}^{2\pi }{\displaystyle \int }_{0}^{\pi }{\displaystyle \int }_{0}^{\infty }{\rm{\Phi }}(r)\rho (r){{\rm{d}}}^{3}r\\ & = & -\,\displaystyle \frac{3\pi {{GM}}^{2}}{32{l}_{0}}.\end{array}\end{eqnarray}$
Compared with equation (3. Regularized self-energy due to the electric field
In T-duality, it was shown that for the electric potential of a system that consists of two point-like masses, at relative distance $\vec{r}$, one can write [60]
$\begin{eqnarray}{V}_{{\rm{em}}}(r)\to -\displaystyle \frac{Q}{4\pi {\epsilon }_{0}\sqrt{{r}^{2}+{l}_{0}^{2}}}.\end{eqnarray}$
In addition, we have Maxwell's equation, which is written as
$\begin{eqnarray}{{\rm{\nabla }}}_{\mu }{F}^{\mu \nu }={J}^{\nu },\end{eqnarray}$
where Fμν = ∇μAν − ∇νAμ and ${J}^{\mu }=(c{\rho }_{{\rm{em}}},\vec{j})$ is the four-current. One can check that the above equations lead to the charge density $\begin{eqnarray}{\rho }_{{\rm{em}}}(r)=\displaystyle \frac{3{l}_{0}^{2}Q}{{\left({r}^{2}+{l}_{0}^{2}\right)}^{5/2}}.\end{eqnarray}$
The interaction Hamiltonian then becomes $\begin{eqnarray}{H}_{\mathrm{int}}^{{\rm{em}}}=-Q{\int }_{0}^{2\pi }{\int }_{0}^{\pi }{\int }_{0}^{\infty }{{\rm{d}}}^{3}r\displaystyle \frac{{\rho }_{{\rm{em}}}(\vec{r})}{4\pi {\epsilon }_{0}\sqrt{{r}^{2}+{l}_{0}^{2}}},\end{eqnarray}$
Solving the integral we get $\begin{eqnarray}{H}_{\mathrm{int}}^{{\rm{em}}}=-\displaystyle \frac{3\pi {Q}^{2}}{16{\epsilon }_{0}{l}_{0}}.\end{eqnarray}$
3.1. Curved spacetime
If we consider again the effect of spacetime curvature, by using the mass profile with corrections due to the charge density (36 ) we obtain38 ), we obtain an improved correction by a factor of 1/2. The total energy of the particle is42 ). Such a correction was also found for a charged black hole in T-duality [17].
$\begin{eqnarray}f(r)=1-\displaystyle \frac{2{{Qr}}^{2}}{{\left({r}^{2}+{l}_{0}^{2}\right)}^{3/2}}=\left(1+2{{\rm{\Phi }}}_{{\rm{em}}}(r)\right),\end{eqnarray}$
which does not represent a new solution since the total mass is simply scaled to M → M + Q. The Schrödinger–Newton equation will be $\begin{eqnarray}{\rm{i}}{\hslash }\displaystyle \frac{\partial {\rm{\Psi }}(t,\vec{r})}{\partial t}=\left(-\displaystyle \frac{{{\hslash }}^{2}}{2M}{{\rm{\nabla }}}_{{\rm{LB}}}^{2}+{H}_{{\rm{int}}}+{H}_{{\rm{int}}}^{{\rm{em}}}\right){\rm{\Psi }}(t,\vec{r}).\end{eqnarray}$
The interaction Hamiltonian due to the charge density is $\begin{eqnarray}{H}_{\mathrm{int}}^{{\rm{em}}}={\int }_{0}^{2\pi }{\int }_{0}^{\pi }{\int }_{0}^{\infty }{{\rm{d}}}^{3}r\,{{\rm{\Phi }}}_{{\rm{em}}}(r){\rho }_{{\rm{em}}},\end{eqnarray}$
yielding $\begin{eqnarray}{H}_{\mathrm{int}}^{{\rm{em}}}=-\displaystyle \frac{3\pi {Q}^{2}}{32{\epsilon }_{0}{l}_{0}}.\end{eqnarray}$
Compared with ( $\begin{eqnarray}{E}_{{\rm{tot}}}=E+\displaystyle \frac{3\pi {{GM}}^{2}}{32{l}_{0}}+\displaystyle \frac{3\pi {Q}^{2}}{32{\epsilon }_{0}{l}_{0}}.\end{eqnarray}$
This equation shows that the electrostatic field also modifies the energy of the particle. Here we found the energy stored in the charge via the charged density and the curved spacetime. It is very interesting to note that the corrections to the energy due to the electromagnetic field can be found in an alternative way using the stress–energy tensor for the electromagnetic field. We will assume that the energy tensor for the electromagnetic field has the form $\begin{eqnarray}{T}_{\mu \nu }^{{\rm{em}}}={F}_{\mu \sigma }{{F}_{\nu }}^{\sigma }-\displaystyle \frac{1}{4}{g}_{\mu \nu }{F}_{\rho \sigma }{F}^{\rho \sigma },\end{eqnarray}$
where $\begin{eqnarray}{A}_{\mu }=\left(\displaystyle \frac{{V}_{{\rm{em}}}}{c},0,0,0\right).\end{eqnarray}$
It is easy to show that the ${{T}^{0}}_{0}$ component of the energy–momentum tensor for the electromagnetic field gives the energy density $\begin{eqnarray}{\rho }_{{\rm{em}}}=\displaystyle \frac{{Q}^{2}{r}^{2}}{8\pi {\left({r}^{2}+{l}_{0}^{2}\right)}^{3}}.\end{eqnarray}$
This means that the energy has to be $\begin{eqnarray}{H}_{{\rm{int}}}^{{\rm{em}}}=-\int {{\rm{d}}}^{3}r\displaystyle \frac{{Q}^{2}{r}^{2}}{8\pi {\left({r}^{2}+{l}_{0}^{2}\right)}^{3}}\end{eqnarray}$
$\begin{eqnarray}=-\,\displaystyle \frac{3\pi {Q}^{2}}{32{\epsilon }_{0}{l}_{0}},\end{eqnarray}$
which precisely matches equation (4. Modified Klein–Gordon equation
We can obtain the corrected relativistic energy for a particle with a different spin. Let us consider here the Klein–Gordon equation described by the scalar field Φ(x, t)
$\begin{eqnarray}\square {\rm{\Phi }}(x,t)-\displaystyle \frac{{M}^{2}{c}^{2}}{{{\hslash }}^{2}}{\rm{\Phi }}(x,t)-\displaystyle \frac{{H}_{{\rm{int}}}^{2}}{{{\hslash }}^{2}{c}^{2}}{\rm{\Phi }}(x,t)=0,\end{eqnarray}$
which is modified due to the self-gravitational interaction term. This equation can be further written as $\begin{eqnarray}\left({{\rm{\nabla }}}^{2}-\displaystyle \frac{1}{{c}^{2}}\displaystyle \frac{{\partial }^{2}}{\partial {t}^{2}}-\displaystyle \frac{{M}^{2}{c}^{2}}{{{\hslash }}^{2}}\right){\rm{\Phi }}(x,t)-\displaystyle \frac{{H}_{{\rm{int}}}^{2}}{{{\hslash }}^{2}{c}^{2}}{\rm{\Phi }}(x,t)=0.\end{eqnarray}$
Taking a plane wave solution for the scalar field $\begin{eqnarray}{\rm{\Phi }}(x,t)=A\exp \left(-\displaystyle \frac{{\rm{i}}}{{\hslash }}{x}^{\mu }{p}_{\mu }\right),\end{eqnarray}$
we get the energy relation given by $\begin{eqnarray}{E}_{{\rm{tot}}}^{2}={M}^{2}{c}^{4}\,+\,{p}^{2}{c}^{2}+\displaystyle \frac{9{\pi }^{2}{G}^{2}{M}^{4}}{{16}^{2}{l}_{0}^{2}}.\end{eqnarray}$
We can also rewrite this equation in a more compact form $\begin{eqnarray}{E}_{{\rm{tot}}}^{2}={M}_{{\rm{tot}}}^{2}{c}^{4}\,+\,{p}^{2}{c}^{2},\end{eqnarray}$
in terms of the new definition of the mass $\begin{eqnarray}{M}_{{\rm{tot}}}^{2}={M}^{2}\left(1+\displaystyle \frac{9{\pi }^{2}{G}^{2}{M}^{2}}{{16}^{2}{c}^{4}{l}_{0}^{2}}\right).\end{eqnarray}$
By considering a series expansion, we get $\begin{eqnarray}{M}_{{\rm{tot}}}\simeq M\left(1+\displaystyle \frac{1}{2}\displaystyle \frac{9{\pi }^{2}{G}^{2}{M}^{2}}{{16}^{2}{c}^{4}{l}_{0}^{2}}\right).\end{eqnarray}$
Using the fact that ${l}_{0}\sim {l}_{{\rm{Pl}}}=\sqrt{\tfrac{{\hslash }G}{{c}^{3}}}$, we can also write $\begin{eqnarray}{M}_{{\rm{tot}}}\simeq M\left(1+\alpha \displaystyle \frac{{M}^{2}}{{M}_{{\rm{Pl}}}^{2}}\right),\end{eqnarray}$
where α = (9/2)π2/162 and ${M}_{{\rm{Pl}}}=\sqrt{\tfrac{{\hslash }c}{G}}$ is the Planck mass. This expression matches that derived in the extended uncertainty principle derived in [61]. This could be related to gravitational self-completeness with quantum mechanical mass limits. In particular, we may have two cases:| 1. | (1)When M ≪ MPl, i.e. in the particle sector, then equation ( |
| 2. | (2)When M > MPl, we have a connection between elementary particles and black holes. As was shown in [29, 30] the Compton–Schwarzschild correspondence posits a smooth transition between the Compton wavelength (RC ∼ 1/M) below the Planck mass and the Schwarzschild radius (RS ∼ M) above it. The Compton and Schwarzschild lines transform into one another under the transformation $M\to {M}_{{\rm{Pl}}}^{2}/M$, which suggests the following form for the mass: $\begin{eqnarray}{M}_{{\rm{tot}}}\simeq M\left(1+\alpha \displaystyle \frac{{M}_{{\rm{Pl}}}^{2}}{{M}^{2}}\right).\end{eqnarray}$ This relation was recently obtained in [29, 30]. |
5. Modified Dirac equation
Let us also consider here the Dirac equation described by the spinor field ψ(x, t) which is modified due to the self-gravitational interaction term
$\begin{eqnarray}{\rm{i}}{\gamma }^{\mu }{{\rm{\nabla }}}_{\mu }\psi (x,t)-\displaystyle \frac{{Mc}}{{\hslash }}\psi (x,t)+\displaystyle \frac{{H}_{{int}}}{{\hslash }c}\psi (x,t)=0,\end{eqnarray}$
where for the γμ matrices in terms of 2 × 2 Pauli sub-matrices we can write $\begin{eqnarray}{\gamma }^{t}=\left(\begin{array}{cc}{I}_{2} & 0\\ 0 & -{I}_{2}\end{array}\right),\,{\gamma }^{1}=\left(\begin{array}{cc}0 & {\sigma }_{x}\\ -{\sigma }_{x} & 0\end{array}\right),\end{eqnarray}$
$\begin{eqnarray}{\gamma }^{2}=\left(\begin{array}{cc}0 & {\sigma }_{y}\\ -{\sigma }_{y} & 0\end{array}\right),\,{\gamma }^{3}=\left(\begin{array}{cc}0 & {\sigma }_{z}\\ -{\sigma }_{z} & 0\end{array}\right).\end{eqnarray}$
If we take the solution for the spinor field
$\begin{eqnarray}\psi (x,t)=\left(\begin{array}{c}{\psi }_{1}\\ {\psi }_{2}\\ {\psi }_{3}\\ {\psi }_{4}\end{array}\right),\end{eqnarray}$
or, in a different notation, $\begin{eqnarray}\psi (x,t)=\left(\begin{array}{c}{\psi }_{A}\\ {\psi }_{B}\end{array}\right),\,{\psi }_{A}=\left(\begin{array}{c}{\psi }_{1}\\ {\psi }_{2}\end{array}\right),\,{\psi }_{B}=\left(\begin{array}{c}{\psi }_{3}\\ {\psi }_{4}\end{array}\right),\end{eqnarray}$
it is well known that the solution can be written as $\begin{eqnarray}\psi ({x}^{\mu })=A\exp \left(-\displaystyle \frac{{\rm{i}}}{{\hslash }}{x}^{\mu }{p}_{\mu }\right)\left(\begin{array}{c}{u}_{A}\\ {u}_{B}\end{array}\right).\end{eqnarray}$
Then the Dirac equation reduces to $\begin{eqnarray}\left({\gamma }^{\mu }{p}_{\mu }-{Mc}+\displaystyle \frac{{H}_{{\rm{int}}}}{c}\right)\left(\begin{array}{c}{u}_{A}\\ {u}_{B}\end{array}\right)=0,\end{eqnarray}$
From this equation, we get a system of two equations $\begin{eqnarray}\left(\begin{array}{cc}\displaystyle \frac{E}{c}-{Mc}+\displaystyle \frac{{H}_{{\rm{int}}}}{c} & -\vec{p}\cdot \vec{\sigma }\\ \vec{p}\cdot \vec{\sigma } & -\,\displaystyle \frac{E}{c}-{Mc}+\displaystyle \frac{{H}_{{\rm{int}}}}{c}\end{array}\right)\left(\begin{array}{c}{u}_{A}\\ {u}_{B}\end{array}\right)=0.\end{eqnarray}$
By solving the determinant of this system we get two solutions for the energy $\begin{eqnarray}{E}_{{\rm{tot}}}=\pm \sqrt{{M}_{{\rm{tot}}}^{2}{c}^{4}\,+\,{p}^{2}{c}^{2}},\end{eqnarray}$
where it has been defined that $\begin{eqnarray}{M}_{{\rm{tot}}}^{2}{c}^{4}={\left({{Mc}}^{2}+\displaystyle \frac{3\pi {{GM}}^{2}}{16{l}_{0}}\right)}^{2}.\end{eqnarray}$
Finally, the last equation can be further written as56 ). This is a surprising result. Again we have two cases:
$\begin{eqnarray}{M}_{{\rm{tot}}}=M\left(1+\displaystyle \frac{3\pi {GM}}{16{c}^{2}{l}_{0}}\right),\end{eqnarray}$
or $\begin{eqnarray}{M}_{{\rm{tot}}}=M\left(1+\alpha \displaystyle \frac{M}{{M}_{{\rm{Pl}}}}\right),\end{eqnarray}$
where α = 3π/16. It is rather amazing that the correction to the mass for fermions (spin 1/2) is linear, as can be seen from the last equation. On the other hand, bosons (spin 0) are quadratic, as was found in equation (| 1. | (1)When M ≪ MPl, then equation ( |
| 2. | (2)When M > MPl, we have a connection between elementary particles and black holes and we have to use the transformation $M\to {M}_{{\rm{Pl}}}^{2}/M$ [29, 30], which suggests the following form for the mass: $\begin{eqnarray}{M}_{{\rm{tot}}}\simeq M\left(1+\alpha \displaystyle \frac{{M}_{{\rm{Pl}}}}{M}\right).\end{eqnarray}$ |
6. Modified Proca equation
Here we shall consider the motion of a massive vector particle of mass M with spin 1 described by the vector field ψμ, which can be studied by the Proca equation, which reads
$\begin{eqnarray}{{\rm{\nabla }}}_{\mu }{{\rm{\nabla }}}^{[\mu }{\psi }^{\nu ]}-\displaystyle \frac{{M}^{2}{c}^{2}}{{{\hslash }}^{2}}{\psi }^{\nu }-\displaystyle \frac{{H}_{{\rm{int}}}^{2}}{{{\hslash }}^{2}{c}^{2}}{\psi }^{\nu }=0,\end{eqnarray}$
where in the last term we introduce the corrections due to the self-regularized energy. Note also that $\begin{eqnarray}{{\rm{\nabla }}}_{[\mu }{\psi }_{\nu ]}=\displaystyle \frac{1}{2}({{\rm{\nabla }}}_{\mu }{\psi }_{\nu }-{{\rm{\nabla }}}_{\nu }{\psi }_{\mu }):= {\psi }_{\mu \nu }\,.\end{eqnarray}$
The solution of the equation can be written as
$\begin{eqnarray}{\psi }_{\nu }({x}^{\mu })={C}_{\nu }\exp \left(-\displaystyle \frac{{\rm{i}}}{{\hslash }}{x}^{\mu }{p}_{\mu }\right).\end{eqnarray}$
Using the four-momentum ${p}^{\mu }=(E/c,\vec{p})$ we can construct a 4 × 4 matrix ℵ, which satisfies the following matrix equation:
$\begin{eqnarray}\aleph {\left({C}_{0},{C}_{1},{C}_{2},{C}_{3}\right)}^{{\rm{T}}}=0\,.\end{eqnarray}$
Taking for simplicity only one component of the momentum $\vec{p}=({p}_{x},0,0)$, the matrix ℵhas the components
$\begin{eqnarray}\aleph =\left(\begin{array}{cccc}{\aleph }_{00} & 0 & 0 & {\aleph }_{03}\\ {\aleph }_{10} & 0 & 0 & {\aleph }_{13}\\ 0 & {\aleph }_{21} & 0 & 0\\ 0 & 0 & {\aleph }_{32} & 0\end{array}\right)\end{eqnarray}$
where $\begin{eqnarray}{\aleph }_{00}=\displaystyle \frac{{Ep}}{{c}^{2}{{\hslash }}^{2}},\end{eqnarray}$
$\begin{eqnarray}{\aleph }_{03}=-\displaystyle \frac{{M}^{2}{c}^{4}\,+\,{c}^{2}{p}^{2}-{H}_{{\rm{int}}}^{2}}{{c}^{2}},\end{eqnarray}$
$\begin{eqnarray}{\aleph }_{10}=\displaystyle \frac{{M}^{2}{c}^{4}-{E}^{2}+{H}_{{\rm{int}}}^{2}}{{c}^{2}},\end{eqnarray}$
$\begin{eqnarray}{\aleph }_{13}=-\displaystyle \frac{{Ep}}{{c}^{2}},\end{eqnarray}$
$\begin{eqnarray}{\aleph }_{21}={\aleph }_{32}=\displaystyle \frac{{M}^{2}{c}^{4}\,+\,{c}^{2}{p}^{2}-{E}^{2}+{H}_{{\rm{int}}}^{2}}{{c}^{2}}.\end{eqnarray}$
Taking the determinant of this matrix and solving for the energy we get $\begin{eqnarray}{E}_{{\rm{tot}}}=\pm \sqrt{{M}_{{\rm{tot}}}^{2}{c}^{4}\,+\,{p}^{2}{c}^{2}},\end{eqnarray}$
where $\begin{eqnarray}{M}_{{\rm{tot}}}^{2}={M}^{2}\left(1+\displaystyle \frac{9{\pi }^{2}{G}^{2}{M}^{2}}{{16}^{2}{c}^{4}{l}_{0}^{2}}\right),\end{eqnarray}$
and therefore $\begin{eqnarray}{M}_{{\rm{tot}}}\simeq M\left(1+\alpha \displaystyle \frac{{M}^{2}}{{M}_{{\rm{Pl}}}^{2}}\right).\end{eqnarray}$
In other words, the corrections to the energy and mass of the particle with spin-1 particles described by the Proca equation are similar to scalar particles with spin 0 described by the Klein–Gordon equation.7. GUP from regularized self-energy
According to the non-commutative relation between position and momentum, we have
$\begin{eqnarray}\left[{x}_{i},{p}_{j}\right]={\rm{i}}\,{\delta }_{{ij}}\,{\hslash },\end{eqnarray}$
meaning that the position and momentum cannot have a real eigenvalue for the same eigenstate. In quantum gravity theories, such as string theory [22], the quadratic form of the GUP has been suggested and has the following form [24–26]: $\begin{eqnarray}\begin{array}{rcl}{\rm{\Delta }}x{\rm{\Delta }}p & \geqslant & \displaystyle \frac{{\hslash }}{2}(1+\beta {\rm{\Delta }}{p}^{2}),\\ \left[{x}_{i},{p}_{j}\right] & = & {\rm{i}}{\hslash }\left[{\delta }_{{ij}}+\beta {\delta }_{{ij}}{p}^{2}+2\beta {p}_{i}{p}_{j}\right],\end{array}\end{eqnarray}$
where $\beta ={\beta }_{0}{l}_{{Pl}}^{2}/{{\hslash }}^{2}$, β0 is a dimensionless constant and lPl = 1.6162 × 10−35 m is the Planck length. Besides, the form of linear the GUP is also motivated by doubly spacial relativity [62, 63]. The linear GUP has the following form [23, 31, 64]: $\begin{eqnarray}[{x}_{i},{p}_{j}]={\rm{i}}{\hslash }\hspace{-0.5ex}\left[{\delta }_{{ij}}-\alpha \left(p{\delta }_{{ij}}+\displaystyle \frac{{p}_{i}{p}_{j}}{p}\right)\right],\,\,\,\end{eqnarray}$
where α = α0lPl/ℏ, and α0 is a dimensionless constant. In what follows we shall argue how these two types of GUP appear naturally from regularized self-energy.7.1. Quadratic GUP and boson particles
Let us show a very interesting result where one can obtain the GUP principle using the modified energy. Let us rewrite equations (56 ) and (83 ) as follows:87 ) on both sides by dt we get in leading order terms90 ) further as87 ) by dt we get
$\begin{eqnarray}{E}_{{\rm{tot}}}\simeq E\left(1+\alpha \displaystyle \frac{{E}^{2}}{{c}^{4}{M}_{{\rm{Pl}}}^{2}}\right),\end{eqnarray}$
where Etot = Mtotc2 and E = Mc2. By differentiating with respect to the energy on both sides we get $\begin{eqnarray}{\rm{d}}{E}_{{\rm{tot}}}\simeq {\rm{d}}E\left(1+3\alpha \displaystyle \frac{{E}^{2}}{{c}^{4}{M}_{{\rm{Pl}}}^{2}}\right).\end{eqnarray}$
Only in the limit α → 0, for energies we have Etot = E; therefore, in general, we expect two possibilities. The first possibility is to assume $\begin{eqnarray}{\rm{d}}{E}_{{\rm{tot}}}{\rm{d}}t\sim {\hslash }.\end{eqnarray}$
If we multiply equation ( $\begin{eqnarray}{\rm{d}}E{\rm{d}}t\sim {\hslash }\left(1-3\alpha \displaystyle \frac{{E}^{2}}{{c}^{4}{M}_{{\rm{Pl}}}^{2}}\right),\end{eqnarray}$
but since $\begin{eqnarray}{l}_{0}\sim {l}_{{\rm{Pl}}}=\sqrt{\displaystyle \frac{{\hslash }G}{{c}^{3}}},\,{M}_{{\rm{Pl}}}=\sqrt{\displaystyle \frac{{\hslash }c}{G}},\end{eqnarray}$
we can write equation ( $\begin{eqnarray}{\rm{d}}E{\rm{d}}t\sim {\hslash }\left(1-\displaystyle \frac{\beta \,{l}_{{\rm{Pl}}}^{2}{E}^{2}}{{c}^{2}{{\hslash }}^{2}}\right),\end{eqnarray}$
where β = 3α. The second possibility is to assume $\begin{eqnarray}{\rm{d}}E{\rm{d}}t\sim {\hslash },\end{eqnarray}$
and, again, multiplying equation ( $\begin{eqnarray}{\rm{d}}{E}_{{\rm{tot}}}{\rm{d}}t\sim {\hslash }\left(1+\displaystyle \frac{\beta \,{l}_{{\rm{Pl}}}^{2}{E}^{2}}{{c}^{2}{{\hslash }}^{2}}\right).\end{eqnarray}$
In other words, there are two ways of expressing the GUP corrected time–energy relation with the difference in the sign. The difference in the sign before the second term is of course related to whether we choose to work with the total or the bare energy.This shows that the GUP for the time–energy can be viewed as a consequence of the modified energy of the particle due to the regularized self-energy. To obtain the momentum–time GUP relation, we simply need to use equations (56 ) and (83 ) and make use of Δptot → cMtot, along with Δp → cM (similar relations have been used for example in [29]), and multiplying both sides by Δx we get85 ), provided α = − β0/2. The second case is to assume Δp Δx ∼ ℏ, then from equation (95 ) we obtain
$\begin{eqnarray}{\rm{\Delta }}{p}_{{\rm{tot}}}\,{\rm{\Delta }}x\sim {\rm{\Delta }}p\,{\rm{\Delta }}x\left(1+\displaystyle \frac{\alpha \,{l}_{{\rm{Pl}}}^{2}\,{\rm{\Delta }}{p}^{2}}{{{\hslash }}^{2}}\right).\end{eqnarray}$
We have a similar situation where Δptot = Δp, provided α → 0. To this end, we can first assume Δptot Δx ∼ ℏ. In that case we obtain $\begin{eqnarray}{\rm{\Delta }}p\,{\rm{\Delta }}x\sim {\hslash }\left(1-\displaystyle \frac{\alpha \,{l}_{{\rm{Pl}}}^{2}\,{\rm{\Delta }}{p}^{2}}{{{\hslash }}^{2}}\right).\end{eqnarray}$
This GUP expression is consistent with the quadratic form of the GUP given by equation ( $\begin{eqnarray}{\rm{\Delta }}{p}_{{\rm{tot}}}\,{\rm{\Delta }}x\sim {\hslash }\left(1+\displaystyle \frac{\alpha \,{l}_{{\rm{Pl}}}^{2}\,{\rm{\Delta }}{p}^{2}}{{{\hslash }}^{2}}\right).\end{eqnarray}$
We therefore find that in general there are two equivalent representations of the GUP corrected momentum–position relation. In general β0 is a free parameter that can be positive or negative. This sign in the second term is also related to whether we work with Δptot or Δp.7.2. Linear GUP and Dirac particles
Consider the corrections to the mass given by equation (69 ), which can be written as98 ). If follows then that69 ), we obtain105 ), we have the modified GUP relation
$\begin{eqnarray}{E}_{{\rm{tot}}}\simeq E\left(1+\alpha \displaystyle \frac{E}{{c}^{2}{M}_{{\rm{Pl}}}}\right).\end{eqnarray}$
By differentiating with respect to the energy on both sides we get $\begin{eqnarray}{\rm{d}}{E}_{{\rm{tot}}}={\rm{d}}E\left(1+2\alpha \displaystyle \frac{E}{{c}^{2}{M}_{{\rm{Pl}}}}\right).\end{eqnarray}$
As in the above discussion, let us first consider the case $\begin{eqnarray}{\rm{d}}{E}_{{\rm{tot}}}{\rm{d}}t\sim {\hslash }.\end{eqnarray}$
We get $\begin{eqnarray}{\rm{d}}E{\rm{d}}t\sim {\hslash }\left(1-2\alpha \displaystyle \frac{E}{{c}^{2}{M}_{{\rm{Pl}}}}\right),\end{eqnarray}$
and we can write further this as $\begin{eqnarray}{\rm{d}}E{\rm{d}}t\sim {\hslash }\left(1-\displaystyle \frac{\beta \,{l}_{{\rm{Pl}}}E}{c{\hslash }}\right),\end{eqnarray}$
where β = 2α. To get the second representation we can assume $\begin{eqnarray}{\rm{d}}E{\rm{d}}t\sim {\hslash },\end{eqnarray}$
in equation ( $\begin{eqnarray}{\rm{d}}{E}_{{\rm{tot}}}{\rm{d}}t\sim {\hslash }\left(1+\displaystyle \frac{\beta \,{l}_{{\rm{Pl}}}E}{c{\hslash }}\right).\end{eqnarray}$
This shows again that in both cases the time–energy GUP relation can be a consequence of the modified energy of the particle due to the regularized self-energy. If we further take Δptot → cMtot and Δp → cM in equation ( $\begin{eqnarray}{\rm{\Delta }}{p}_{{\rm{tot}}}\,{\rm{\Delta }}x\sim {\rm{\Delta }}p\,{\rm{\Delta }}x\left(1+\displaystyle \frac{\alpha \,{l}_{{\rm{Pl}}}\,{\rm{\Delta }}p}{{\hslash }}\right).\end{eqnarray}$
Again, by means of Δptot Δx ∼ ℏ, we obtain $\begin{eqnarray}{\rm{\Delta }}p\,{\rm{\Delta }}x\sim {\hslash }\left(1-\displaystyle \frac{\alpha \,{l}_{{\rm{Pl}}}\,{\rm{\Delta }}p}{{\hslash }}\right).\end{eqnarray}$
Finally, if we assume Δp Δx ∼ ℏ in equation ( $\begin{eqnarray}{\rm{\Delta }}{p}_{{\rm{tot}}}\,{\rm{\Delta }}x\sim {\hslash }\left(1+\displaystyle \frac{\alpha \,{l}_{{\rm{Pl}}}\,{\rm{\Delta }}p}{{\hslash }}\right).\end{eqnarray}$
This is the linear GUP in agreement with equation (86 ). We should point out that by comparing it with equation (86 ), the correction can be positive or negative. This has to do with the fact that the free parameter α0 can be positive or negative. We therefore find that the form of the GUP depends on the nature of the particles being considered. In particular, for scalar particles with spin 0 we obtained a quadratic form of the GUP, while for fermion particles with spin 1/2 we got a linear GUP. This is consistent with the connection between spin and linear GUP that was found in [65] that has implications for explaining quantum entanglement [66] through finding a relation between uncertainty and the Bekenstein bound [67].
8. Conclusions
In the present paper we have computed the regularized self-energy of a particle using the Schrödinger–Newton equation. We used regular and well-defined gravitational and electric potentials obtained in T-duality. In this picture, a particle with mass M is no longer concentrated into a point but is diluted and can be described by a quantum-corrected stress–energy tensor with a smeared energy density, resulting in corrections to the energy of the particle that can be interpreted as the regularized self-energy of the particle. To this end, we have found the corrections to the energy due to the electrostatic field or the energy stored in the charge density of the particle.
In the second part of this work we extended the corrections to energy by incorporating relativistic effects using relativistic field equations such as the Klein–Gordon, Proca and Dirac equations. We found that the corrections to the energy can be linked to the GUP. Quite surprisingly, the form of the GUP is shown to depend on the spin of the particles: namely, for bosons having spin 0 and 1 we obtain a quadratic form of the GUP; on the other hand for fermions having spin 1/2 we obtain a linear form of the GUP. In the near future we would like to study further the phenomenological aspects of the GUP which are linked to regularized self-energy, as argued in this work. We would like to see how the regularized self-energy affects other types of particles such as spin-2 particles. Like spin-0 and spin-1 particles, one can of course also assume for gravitons a similar expression for the energy correction or modified dispersion relation
$\begin{eqnarray}{E}_{{\rm{tot}}}^{2}={\left({p}^{g}\right)}^{2}{c}^{2}+{\left({M}^{g}\right)}_{{\rm{tot}}}^{2}{c}^{4},\end{eqnarray}$
where ${M}_{{\rm{tot}}}^{g}\simeq {M}^{g}\left(1+\alpha {\left({M}^{g}/{M}_{{\rm{Pl}}}\right)}^{2}\right)$, with Mg and pg being the bare graviton mass and the graviton momentum, respectively. Using the expression for the graviton's three-velocity vg = pg c2/Etot, we can obtain $\begin{eqnarray}\displaystyle \frac{{v}^{g}}{c}=\sqrt{1-\displaystyle \frac{{\left({M}_{{\rm{tot}}}^{g}\right)}^{2}{c}^{4}}{{E}_{{\rm{tot}}}^{2}}}.\end{eqnarray}$
For the energy we may take Etot = hf, with f being the graviton's frequency, then using gravitational waves it might be interesting to see whether one can constrain the graviton mass along with the zero-point length. There is also the possibility that we cannot constrain the zero-point length from observations. This has to do with the fact that Mtotg might be the true mass being measured by observations and it simply shifts by some constant compared to the bare mass Mg, meaning that we cannot distinguish these quantities by observation. In the near future we hope to further study the implications of the GUP for graviton mass.