A fresh look at the influence of gravity on the quantum Hall effect

Abstract

We study the quantum Hall effect inside a gravitational field. First, we review the influence of the gravitational field of the Earth on the quantum Hall effect. Taking the gravitational field of the Earth to be uniform along the vertical direction, we compute the affected quantized Hall resistivity. Then, we investigate how the gravitational field modifies the Landau levels of a particle moving between two massive hemispheres in the presence of a constant and uniform magnetic field perpendicular to the plane of motion. We find that the familiar degeneracy of the Landau levels is removed and the spacing between the latter becomes dependent on the mass density of the hemispheres and on the gravitational constant G. We use this result to show that the quantum Hall effect in a thin conductor, sandwiched between two massive hemispheres, should yield a slightly different variation of the Hall resistivity with the applied magnetic field. We then argue that the well-known problem of the gravitationally induced electric field, that might a priori be thought to hinder the effect of gravity, has actually a beneficial role as it amplifies the latter. We finally discuss whether there is a possibility of using the quantum Hall effect to probe the inverse-square law of gravity.

The gravitational potential inside a full disk of radius R and thickness a

1.1 Newtonian potential $V(r)=-GM/r$

Let us denote the Newtonian gravitational potential between the two hemispheres of Sect. 3 at any point x away from the center by $V_{H}^N(x)$. Then, we have $V_H^N(x)=V_S^N(x)-V_D^N(x)$. Here, $V_S^N(x)$ is the Newtonian gravitational potential at any distance x from the center of a full sphere of radius R and $V_D^N(x)$ is the Newtonian gravitational potential at the distance x from the center of a full disk of thickness 2a and radius R. We are going to give here the detailed calculations that yield $V_D^N(x)$. The calculation of $V_S^N(x)$ has been done in detail in Ref.  [40], so we are only displaying here the final expression of that potential. It is given by,

$$\begin{aligned} V_S^N(x)=-2\pi G\rho \left( R^2-\frac{x^2}{3}\right) . \end{aligned}$$

(29)

The Newtonian gravitational potential at a distance x from the center of a full disk of thickness 2a, of radius R and of mass density $\rho$, can be found by integrating first the infinitesimal contributions of the mass elements $r\,\mathrm{d}\phi \,\mathrm{d}r\,\mathrm{d}z$ over the thickness 2a. These contributions should, in turn, be integrated over the whole disk by following the same strategy as the one adopted in Ref.  [42]. We find,

$$\begin{aligned} V_D^N(x)&=-4G\rho \int _0^\pi \int _0^{r(\phi )}\int _0^{a}\frac{r}{\sqrt{r^2+z^2}}\,\mathrm{d}\phi \,\mathrm{d}r\,\mathrm{d}z\nonumber \\&=-2G\rho \int _0^\pi \Bigg [a\sqrt{r^2(\phi )+a^2}+r^2(\phi ) \ln \left( \frac{\sqrt{r^2(\phi )+a^2}+a}{r(\phi )}\right) -a^2\Bigg ]\mathrm{d}\phi . \end{aligned}$$

(30)

In the second step we have integrated over the rest of the disk from $r=0$ to $r(\phi )=x\cos \phi +\sqrt{R^2-x^2\sin ^2\phi }$ [42]. For small distances x away from the center of the disk and for a small thickness 2a of the disk compared to the radius R of the latter, we may, in turn, expand the integrand in powers of the ratios x/R and a/R. This allows us then to easily integrate over the variable $\phi$. Keeping only the lower orders in the expansion, we find the following result for the Newtonian gravitational potential inside the full disk at a distance x from its center:

$$\begin{aligned} V_D^N(x)=\frac{\pi G\rho a}{R}x^2+{\mathcal {O}}\left( \frac{x^4}{R^4},\frac{x^2a^2}{R^4}\right) +\mathrm{const}. \end{aligned}$$

(31)

We have discarded here orders four and higher in x/R and a/R. Also, we have gathered all the constant terms inside the last term. Using this result, we can now find the Newtonian gravitational potential between the two hemispheres at a distance x from the center by using Eq. (29) and computing $V_H^N(x)=V_S^N(x)-V_D^N(x)$. We thus find the following result:

$$\begin{aligned} V_H^N(x)=\frac{2\pi G\rho }{3}\left( 1-\frac{3 a}{2R}\right) x^2+{\mathcal {O}}\left( \frac{x^4}{R^4},\frac{x^2a^2}{R^4}\right) +\mathrm{const}. \end{aligned}$$

(32)

1.2 Yukawa-like correction $V^{\mathrm{Y}}(r)=-\frac{1}{r}GM\alpha e^{-r/\lambda }$.

The gravitational potential inside a full sphere of radius R and mass density $\rho$ due to the Yukawa-like correction term has already been found in Ref.  [40]. The result is

$$\begin{aligned} V_S^{Y}(x)= & {} -2\pi G\rho \Bigg [\alpha \lambda ^2\left( 2-e^{-\frac{R-x}{\lambda }}-e^{-\frac{R+x}{\lambda }} \right) \nonumber \\&+\frac{\alpha \lambda ^2}{x}\bigg ((R+x+\lambda )e^{-\frac{R+x}{\lambda }} -(R-x+\lambda )e^{-\frac{R-x}{\lambda }}\Bigg )\Bigg ]. \end{aligned}$$

(33)

This potential cannot be of any use in the form given here, so we need to expand it in powers of $x/\lambda$. This expansion would be valid provided, of course, that $x<\lambda$. The result is

$$\begin{aligned} V_S^{Y}(x)&=e^{-R/\lambda }\left[ \frac{2\pi GR\rho \alpha }{3\lambda }x^2+{\mathcal {O}}\left( \frac{x^4}{R^4}\right) \right] +\mathrm{const}. \end{aligned}$$

(34)

We have kept here also only the terms up to the second order in $x/\lambda$ and gathered all the constant terms inside the last constant term.

Let us now denote by $V_D^{\mathrm{Y}}(x)$ the Yukawa-like correction to the gravitational potential at a distance x from the center of a full uniform disk of radius R and of thickness 2a, having the uniform mass density $\rho$. Integrating over the disk as done for the Newtonian potential, we get,

$$\begin{aligned} V_D^Y(x)&=-4G\rho \alpha \int _0^\pi \int _0^{r(\phi )}\int _0^{a} \frac{re^{-\sqrt{r^2+z^2}/\lambda }}{\sqrt{r^2+z^2}}\,\mathrm{d}\phi \,\mathrm{d}r\,\mathrm{d}z\nonumber \\&\approx e^{-\frac{\sqrt{R^2+a^2}}{\lambda }}\left[ \frac{\pi G\rho a\alpha }{R}x^2\!+\!{\mathcal {O}}\!\left( \frac{x^4}{R^4}, \frac{x^2a^2}{R^4}\right) \!+\!\mathrm{const}.\right] \end{aligned}$$

(35)

Note that this is the lowest limit of the integral. Indeed, given that the integral in the first line of Eq. (35) does not admit any analytical expression, we have to approximate the exponential inside the integrand in order to get a rough estimate of such an integral. Then we integrated over the whole disk as we did above for the Newtonian potential by assuming that $x\ll R$. Owing to the smallness of a, we clearly see that the contribution of this lower limit of $V_D^{Y}(x)$ is very small. Finally, then, using Eqs. (32) and (34), the gravitational potential $V_H^{N+Y}(x)$ that takes into account the Yukawa-like correction at any distance x from the center between two hemispheres is found, up to the first two lowest orders, as follows:

$$\begin{aligned} V_H^{N+Y}(x)\approx V_H^N(x)+V_S^{Y}(x)\approx \frac{2\pi G\rho }{3}\left( 1-\frac{3 a}{2R}+\frac{\alpha R}{\lambda }e^{-R/\lambda }\right) x^2+\mathrm{const}. \end{aligned}$$

(36)

Note that this approximation is valid provided that $x^4/R^4\ll \alpha R/\lambda e^{R/\lambda }$. For this to be the case, one needs of course to choose the allowed range for x according to one’s estimate of $\lambda$ and $\alpha$.

1.3 Power-law correction $V^{n}(r)=-GM\,r_0^n/r^{n+1}$.

Let us denote by $V_D^{n}(x)$ the power-law correction to the gravitational potential at a distance x from the center of a full uniform disk of radius R and thickness 2a, having the uniform mass density $\rho$.

1.3.1 Case $n=1$

Only for the case $n=1$ do we have a converging potential inside the sphere [40]. Therefore, we only focus here on that case. The gravitational potential correction at a distance x from the center of a sphere of radius R and mass density $\rho$ for such a case has already been computed in detail in Ref.  [40]. Therefore, we just pick up here the result found there and expand in powers of x/R. We find,

$$\begin{aligned} V_S^{n=1}(x)&=-2\pi G\rho r_0\left[ R+\frac{R^2-x^2}{2x} \ln \frac{R+x}{R-x}\right] \nonumber \\&=-4\pi G\rho r_0R\left( 1-\frac{x^2}{3R^2}\right) +{\mathcal {O}}\left( \frac{x^4}{R^4}\right) +\mathrm{const}. \end{aligned}$$

(37)

On the other hand, we should integrate over the disk of radius R and thickness 2a as done for the Newtonian potential. We get,

$$\begin{aligned} V_D^{n=1}(x)&=-4G\rho r_0\int _0^\pi \int _0^{r(\phi )}\int _0^{a}\,\frac{r}{r^2+z^2}\,\mathrm{d}\phi \,\mathrm{d}r\,\mathrm{d}z\nonumber \\&=-2G\rho r_0 \int _0^\pi \Bigg [a\ln \left( \frac{a^2+r^2(\phi )}{a^2}\right) +r(\phi )\tan ^{-1}\left( \frac{a}{r(\phi )}\right) \Bigg ]\mathrm{d}\phi \nonumber \\&=\frac{2\pi G\rho r_0a}{R^2}\,x^2+{\mathcal {O}}\left( \frac{x^4}{R^4},\frac{x^2a^2}{R^4}\right) +\mathrm{const}. \end{aligned}$$

(38)

As the integral in the second line does not again admit any analytical solution, the best we could do is expand the integrand in powers of x/R and a/R and then integrate for the single variable $\phi$. The above result is thus valid only for $x\ll R$ as well as $a\ll R$. Therefore, the total gravitational potential between the two hemispheres reads,

$$\begin{aligned} V_H^{N+(n=1)}(x)=V_H^{N}+V_S^{n=1}-V_D^{n=1}\approx \frac{2\pi G\rho }{3}\left( 1-\frac{3 a}{2R}\right) \left( 1+\frac{2r_0}{R}\right) x^2+\mathrm{const}. \end{aligned}$$

(39)