Quantum interference in the Kerr spacetime

All the formulas in this section are taken from the book [9], which gives a full introduction to TG. We will not show all the details of this theory, but only introduce the core concepts relevant to our study. Let us start with the tetrad, namely

$$h_{a}={h_{a}}^{\mu}\partial_{\mu},\qquad h^{a}={h^{a}}_{\mu}\mbox{d}x^{\mu},$$

(5)

a basis which connects the spacetime metric $g_{\mu\nu}$ to the Minkowski’s metric in the tangent space,

$$\eta_{ab}=\text{diag}(1,-1,-1,-1).$$

(6)

At each point:

$$g_{\mu\nu}=\eta_{ab}{h^{a}}_{\mu}{h^{b}}_{\nu},\qquad\eta_{ab}=g_{\mu\nu}{h_{a}}^{\mu}{h_{b}}^{\nu},$$

(7)

where the Greek letters are used to denote the coordinates in spacetime, while the Latin letters denote the coordinates in the tangent-space. The components of the tetrad satisfy the equations:

$${h^{a}}_{\mu}{h_{a}}^{\nu}=\delta^{\nu}_{\mu},\qquad{h^{a}}_{\mu}{h_{b}}^{\mu}=\delta^{a}_{b}.$$

(8)

Finally, the tetrad relates the spacetime tensors with the tangent-space tensors:

$$V^{\mu}={h_{a}}^{\mu}V^{a},\qquad V_{a}={h^{a}}_{\mu}V^{\mu}.$$

(9)

The components of the tetrad in the presence of gravity are given by:

$${h^{a}}_{\mu}=\partial_{\mu}x^{a}+\dot{A}{{}^{a}}_{b\mu}x^{b}+{B^{a}}_{\mu},$$

(10)

where $\dot{A}{{}^{a}}_{b\mu}={\Lambda^{a}}_{d}(x)\partial_{\mu}{\Lambda_{b}}^{d}(x)$ is the Lorentz connection with ${\Lambda^{a}}_{d}(x)$ a local Lorentz transformation from an inertial reference frame to a general frame, and ${B^{a}}_{\mu}$ is a gauge potential corresponding to a translational transformation $\delta x^{a}(x)=\varepsilon^{a}(x)$ on the tangent space. In TG, gravity is generated from the group of the latter transformations under which the tetrad ${h^{a}}_{\mu}$ is invariant, while the potential ${B^{a}}_{\mu}$ transforms according to

$$\delta B{{}^{a}}_{\mu}=-\partial_{\mu}\varepsilon^{a}-\dot{A}{{}^{a}}_{b\mu}\varepsilon^{b}.$$

(11)

In Eq. (10) we see that the expression of the tetrad contains three terms. The first one corresponds to a coordinates’ transformation from the spacetime to its tangent-space. As shown in [9], the second one corresponds to the inertia. And the last one corresponds to the gravitational interaction. The expression of the tetrad is obtained by combining (10) with (5), namely

$${h^{a}}=\mbox{d}x^{a}+\dot{A}{{}^{a}}_{b\mu}x^{b}\mbox{d}x^{\mu}+B{{}^{a}}_{\mu}\mbox{d}x^{\mu}.$$

(12)

Opposite to general relativity, in TG, the curvature vanishes while the torsion is non-vanishing, namely

	$\displaystyle\dot{R}^{a}{{}_{b\mu\nu}}$	$\displaystyle=$	$\displaystyle\partial_{\mu}\dot{A}^{a}{{}_{b\nu}}-\partial_{\nu}\dot{A}^{a}{{}_{b\mu}}+\dot{A}^{a}{{}_{c\mu}}\dot{A}^{c}{{}_{b\nu}}-\dot{A}^{a}{{}_{c\nu}}\dot{A}^{c}{{}_{b\mu}}=0,$		(13)
	$\displaystyle\dot{T}^{a}{{}_{\mu\nu}}$	$\displaystyle=$	$\displaystyle\partial_{\mu}h^{a}{{}_{\nu}}-\partial_{\nu}h^{a}{{}_{\mu}}+\dot{A}^{a}{{}_{c\mu}}h^{c}{{}_{\nu}}-\dot{A}^{a}{{}_{c\nu}}h^{c}{{}_{\mu}}=\dot{\mathscr{D}}_{\mu}{B^{a}}_{\nu}-\dot{\mathscr{D}}_{\nu}{B^{a}}_{\mu}\neq 0,$		(14)

where the derivative operator $\dot{\mathscr{D}}_{\mu}$ only acts on the indices in the tangent space and it is defined by:

$$\dot{\mathscr{D}}_{\mu}\phi^{a}=\partial_{\mu}\phi^{a}+\dot{A}^{a}{{}_{b\mu}}\phi^{b}.$$

(15)

The curvature and the torsion can also be expressed in terms of spacetime indices, i.e.

	$\displaystyle\dot{R}^{\rho}{{}_{\lambda\nu\mu}}$	$\displaystyle=$	$\displaystyle\partial_{\nu}\dot{\Gamma}^{\rho}{{}_{\lambda\mu}}-\partial_{\mu}\dot{\Gamma}^{\rho}{{}_{\lambda\nu}}+\dot{\Gamma}^{\rho}{{}_{\eta\nu}}\dot{\Gamma}^{\eta}{{}_{\lambda\mu}}-\dot{\Gamma}^{\rho}{{}_{\eta\mu}}\dot{\Gamma}^{\eta}{{}_{\lambda\nu}},$		(16)
	$\displaystyle\dot{T}^{\rho}{{}_{\nu\mu}}$	$\displaystyle=$	$\displaystyle\dot{\Gamma}^{\rho}{{}_{\mu\nu}}-\dot{\Gamma}^{\rho}{{}_{\nu\mu}},$		(17)

where $\dot{\Gamma}^{\mu}{{}_{\rho\nu}}$ is the Weitzenböck connection defined by:

$$\dot{\Gamma}^{\mu}{{}_{\rho\nu}}={h_{a}}^{\mu}\dot{\mathscr{D}}_{\nu}{h^{a}}_{\rho}.$$

(18)

In TG, the torsion is regarded as a field strength, and from (14) we see that $B^{a}{{}_{\mu}}$ plays a role analogous to the gauge potential in electromagnetism. The torsion is gauge invariant [9] because it can be written in the following form,

$$\dot{T}^{a}{{}_{\mu\nu}}=\dot{\mathscr{D}}_{\mu}{h^{a}}_{\nu}-\dot{\mathscr{D}}_{\nu}{h^{a}}_{\mu},$$

(19)

while the tetrad is invariant under the gauge transformation (11). The action for gravity is constructed by means of the torsion tensor, which coincides with the Einstein-Hilbert action, and the field equation in TG is equivalent to the Einstein equation (all the details can be found in the book [9]).

As mentioned above, the gravitational phase is given by (4) where the gauge potential $B^{a}{{}_{\mu}}$ appears in the integrand. This is reasonable because gravity is represented by the gauge potential, as stated in Ref. [9]. This potential not only appears in the field equation, but also plays an important role in the equation of motion, which is equivalent to the geodesic equation, of a particle in the gravitational field. We now prove the latter claim and finally show that $B^{a}{{}_{\mu}}$ appears in the gravitational phase by an analogy with electromagnetism.

Let us remind the geodesic equation in general relativity, namely

$$\frac{\mbox{d}u^{\mu}}{\mbox{d}s}+\Gamma^{\mu}{{}_{\rho\nu}}u^{\rho}u^{\nu}=0.$$

(20)

In TG, the Levi-Civita connection can be written as [9]

$$\Gamma^{\mu}{{}_{\rho\nu}}=\dot{\Gamma}^{\mu}{{}_{\rho\nu}}-\dot{K}^{\mu}{{}_{\rho\nu}},$$

(21)

where $\dot{\Gamma}^{\mu}{{}_{\rho\nu}}$ is defined in (18), and $\dot{K}^{\mu}{{}_{\rho\nu}}$ is the contortion

$$\dot{K}^{\mu}{{}_{\rho\nu}}=\frac{1}{2}(\dot{T}_{\nu}{{}^{\mu}}_{\rho}+\dot{T}_{\rho}{{}^{\mu}}_{\nu}-\dot{T}^{\mu}{{}_{\rho\nu}}),$$

(22)

of the Weitzenböck torsion

$$\dot{T}^{\mu}{{}_{\rho\nu}}={h_{a}}^{\mu}\dot{T}^{a}{{}_{\rho\nu}}={h_{a}}^{\mu}(\dot{\mathscr{D}}_{\rho}{B^{a}}_{\nu}-\dot{\mathscr{D}}_{\nu}{B^{a}}_{\rho}),$$

(23)

where (14) has been used. Recalling (10), the tetrad depends on ${B^{a}}_{\mu}$. Therefore, both $\dot{\Gamma}^{\mu}{{}_{\rho\nu}}$ in (18) and $\dot{K}^{\mu}{{}_{\rho\nu}}$ in (22) depend on the gauge potential.

According to the above expressions, we can prove that the geodesic equation (20) depends on the potential ${B^{a}}_{\mu}$. Indeed, we can rewrite the geodesic equation in TG using (21),

$$\frac{\mbox{d}u^{\mu}}{\mbox{d}s}+(\dot{\Gamma}^{\mu}{{}_{\rho\nu}}-\dot{K}^{\mu}{{}_{\rho\nu}})u^{\rho}u^{\nu}=0.$$

(24)

For the last term, according to (22), we get

$$\dot{K}^{\mu}{{}_{\rho\nu}}u^{\rho}u^{\nu}=\frac{1}{2}(\dot{T}_{\nu}{{}^{\mu}}_{\rho}+\dot{T}_{\rho}{{}^{\mu}}_{\nu}-\dot{T}^{\mu}{{}_{\rho\nu}})u^{\rho}u^{\nu}=\dot{T}_{\rho}{{}^{\mu}}_{\nu}u^{\rho}u^{\nu},$$

(25)

where the last step follows from the anti-symmetry of $\dot{T}^{\mu}{{}_{\rho\nu}}$ in the last two indices (see (23)), and by re-labeling the indices of the first term. Furthermore, we rewrite (25) as:

	$\displaystyle\dot{K}^{\mu}{{}_{\rho\nu}}u^{\rho}u^{\nu}$	$\displaystyle=$	$\displaystyle g_{\rho\alpha}g^{\mu\beta}\dot{T}^{\alpha}{{}_{\beta\nu}}u^{\rho}u^{\nu}$		(26)
		$\displaystyle=$	$\displaystyle(\eta_{cd}{h^{c}}_{\rho}{h^{d}}_{\alpha})(\eta^{ef}{h_{e}}^{\mu}{h_{f}}^{\beta}){h_{a}}^{\alpha}(\dot{\mathscr{D}}_{\beta}{B^{a}}_{\nu}-\dot{\mathscr{D}}_{\nu}{B^{a}}_{\beta})u^{\rho}u^{\nu}$
		$\displaystyle=$	$\displaystyle{h_{a}}^{\mu}(\eta_{ce}{h^{c}}_{\rho})(\eta^{af}{h_{f}}^{\beta})(\dot{\mathscr{D}}_{\beta}{B^{e}}_{\nu}-\dot{\mathscr{D}}_{\nu}{B^{e}}_{\beta})u^{\rho}u^{\nu},$

where (7) and (23) have been used in the second step and (8) has been used in the last step. Then plugging (18) and (26) into (24), we get the equation of motion for a point-like particle:

$$\frac{\mbox{d}u^{\mu}}{\mbox{d}s}+{h_{a}}^{\mu}\Bigl{[}\dot{\mathscr{D}}_{\nu}{h^{a}}_{\rho}-(\eta_{ce}{h^{c}}_{\rho})(\eta^{af}{h_{f}}^{\beta})(\dot{\mathscr{D}}_{\beta}{B^{e}}_{\nu}-\dot{\mathscr{D}}_{\nu}{B^{e}}_{\beta})\Bigr{]}u^{\rho}u^{\nu}=0,$$

(27)

which is equivalent to the geodesic equation (20).

We now show by contradiction that in presence of gravity the gauge potential can not be eliminated from the equation (27). We first replace (10) in (27) and afterwards assume the gauge potential to vanish. Hence, we rewrite (27) in cartesian coordinates of an inertial frame in which the Lorentz connection $\dot{A}{{}^{a}}_{b\mu}$ vanishes and the tetrad components take the form ${h^{a}}_{\rho}={\delta^{a}}_{\rho}$ (see Ref. [9]). Therefore, the equation (27) simplifies to:

$$\frac{\mbox{d}u^{\mu}}{\mbox{d}s}+{h_{a}}^{\mu}(\partial_{\nu}{\delta^{a}}_{\rho})u^{\rho}u^{\nu}=0\qquad\Longrightarrow\quad\frac{\mbox{d}u^{\mu}}{\mbox{d}s}=0,$$

(28)

where we used the definition (15) and $\partial_{\nu}{\delta^{a}}_{\rho}=0$. Therefore, in cartesian coordinates of an inertial frame and assuming that (27) does not depend on the gauge potential, equation (27) reduces to the equation of a free particle. On the other hand, we know that in the presence of gravity (27) does not reduce to the equation of a free particle because it is equivalent to the geodesic equation (20). Therefore, in the presence of gravity we can not eliminate the gauge potential from the equation (27) and the gauge potential ${B^{a}}_{\mu}$ represents the effect of gravity on the motion of a point-like particle.

We would also emphasize the role of the Lorentz connection $\dot{A}^{a}{{}_{b\mu}}$. As stated in Ref. [9], this connection is due to the inertial effects and it appears in the tetrad when a general reference is chosen. Hence, in this case, it also appears in the equation of motion. However, if we take an inertial frame, this connection vanishes. Indeed, such connection is constructed with the local Lorentz transformation ${\Lambda^{a}}_{d}(x)$ from an inertial frame to a general frame, namely $\dot{A}{{}^{a}}_{b\mu}={\Lambda^{a}}_{d}(x)\partial_{\mu}{\Lambda_{b}}^{d}(x)$. In particular, since the Lorentz transformation from an inertial frame to another inertial frame is a global transformation, $\dot{A}{{}^{a}}_{b\mu}$ vanishes in the inertial frames.

Therefore, based on the above discussions, generally, the motion of the particle is governed by both the Lorentz connection $\dot{A}^{a}{{}_{b\mu}}$ and the gauge potential ${B^{a}}_{\mu}$. If an inertial frame is chosen, the motion is only governed by the later. These two quantities together plays a role similar to the Levi-Civita connection in general relativity. Indeed, in general relativity, the motion of the particle is governed by the Levi-Civita connection, as the equation (20) shows.

Finally, let us show that the gauge potential $B^{a}{{}_{\mu}}$ appears in the gravitational phase, though we have proved that it affects the equation of motion of the particle. As shown in the Ref. [9], the equation of motion (24) can be derived directly from the following action principle,

$$\mathscr{S}=-m\int_{p}^{q}(u_{a}\mbox{d}x^{a}+u_{a}\dot{A}{{}^{a}}_{b\mu}x^{b}\mbox{d}x^{\mu}+u_{a}B{{}^{a}}_{\mu}\mbox{d}x^{\mu}),$$

(29)

where the first term stands for the free particle, the second term relates to the inertial effects, and the last term represents the gravitational interaction. Here $u_{a}=\eta_{ab}u^{b}$ and $u^{b}$ is a four-velocity defined in the tangent space (see (33)). In presence of the electromagnetic potential $A_{\mu}$, the action (29), for a charged particle of charge $q$, should be modified by adding the term $(q/m)A_{\mu}\mbox{d}x^{\mu}$ under the integral in (29) [9]. In special relativity, the action of a particle in presence of the electromagnetic field is just the combination of a free term and the interaction term with the electromagnetic potential. Correspondingly, the electromagnetic phase factor for an Aharonov-Bohm effect [1] is given by $e^{\frac{i}{\hbar}\int qA_{\mu}dx^{\mu}}$. Thus, for a gravitational field, in strict analogy with the electromagnetism, the last two terms in (29) contribute to the gravitational phase factor [9]. Especially, if we choose an inertial frame, the second term in (29) vanishes and only the last term contributes to the gravitational phase factor. In this frame, the gauge potential $B^{a}{{}_{\mu}}$ dominates the gravitational phase. Indeed, as we see from the definition of the field strength (14), the role of the potential $B^{a}{{}_{\mu}}$ in gravity is similar to the role of the gauge potential in electromagnetism. It deserves to be mentioned that a similar discussion of the gravitational phase can be found in Ref. [15].

The gravitational phase factor for a massive particle in a generic frame is [9, 15]:

$$\Phi_{g}=\exp\Bigl{(}-\frac{i}{\hbar}\mathscr{S}_{g}\Bigr{)},$$

(30)

where

$$\mathscr{S}_{g}=-m\int_{p}^{q}u_{a}(\dot{A}{{}^{a}}_{b\mu}x^{b}\mbox{d}x^{\mu}+B{{}^{a}}_{\mu}\mbox{d}x^{\mu})$$

(31)

is the interaction part of the action

$$\mathscr{S}=-m\int_{p}^{q}\mbox{d}s,$$

(32)

and the four-velocities in spacetime and tangent-space are defined respectively as follows,

$$u^{\mu}=\frac{\mbox{d}x^{\mu}}{\mbox{d}s},\qquad u^{a}=\frac{h^{a}}{\mbox{d}s}.$$

(33)

For simplicity, we choose an inertial coordinate system $K$ in which $\dot{A}{{}^{a}}_{b\mu}=0$. Hence, according to (31), the interaction action reads:

$$\mathscr{S}_{g}=-m\int_{p}^{q}u_{a}B{{}^{a}}_{\beta}\mbox{d}x^{\beta}=-m\int_{p}^{q}g_{\mu\nu}u^{\mu}B{{}^{\nu}}_{\beta}\mbox{d}x^{\beta},$$

(34)

where the second equation in (9) is used and the function $B{{}^{\nu}}_{\beta}$ is defined as

$$B{{}^{\nu}}_{\beta}=h_{a}{{}^{\nu}}B{{}^{a}}_{\beta}=(h^{T}B){{}^{\nu}}_{\beta}.$$

(35)

Here $h^{T}$ is the transpose matrix of $h_{a}{{}^{\nu}}$, and $B$ is the matrix $B{{}^{a}}_{\beta}$. Therefore, if we have the expressions for $g_{\mu\nu}$, $u^{\mu}$ and ${B^{\nu}}_{\beta}$, we can evaluate $\mathscr{S}_{g}$. Plugging (34) into (30), we get the gravitational phase factor for a massive particle:

$$\Phi_{g}=\exp\Bigl{(}\frac{i}{\hbar}m\int_{p}^{q}g_{\mu\nu}u^{\mu}B{{}^{\nu}}_{\beta}\mbox{d}x^{\beta}\Bigr{)}.$$

(36)

For massless particles, let us consider the light firstly. For a light, its phase factor can be written as:

$$\Phi=\exp(i\psi)=\exp\Bigl{(}\frac{i}{\hbar}\int_{p}^{q}P_{\mu}\mbox{d}x^{\mu}\Bigr{)},$$

(37)

where $P_{\mu}=\hbar k_{\mu}$ is the four-momentum of the photon, and $k_{\mu}$ is the wave vector. In Ref. [8], the optical interferometry is based on (37), but for a weak gravitational field. Unlike in the Ref. [8], we extract the gravitational part from the phase factor in the framework of TG, without need of the weak field approximation. According to (37), we have:

$$\mbox{d}\psi=k_{\mu}\mbox{d}x^{\mu}=k_{a}h^{a}=k_{a}(\mbox{d}x^{a}+\dot{A}{{}^{a}}_{b\mu}x^{b}\mbox{d}x^{\mu}+B{{}^{a}}_{\mu}\mbox{d}x^{\mu})\,,$$

(38)

where Eqs. (9), (7), (8), and (12) have been used. Since we only need the interaction part, for the gravitational phase $\phi_{g}$ we have:

$$\mbox{d}\phi_{g}=k_{a}(\dot{A}{{}^{a}}_{b\mu}x^{b}\mbox{d}x^{\mu}+B{{}^{a}}_{\mu}\mbox{d}x^{\mu}).$$

(39)

Moreover, if we choose an inertial frame for which $\dot{A}{{}^{a}}_{b\mu}=0$, the gravitational phase simplifies to:

$$\phi_{g}=\int_{p}^{q}k_{a}B{{}^{a}}_{\mu}\mbox{d}x^{\mu}=\int_{p}^{q}k_{\nu}{B^{\nu}}_{\mu}\mbox{d}x^{\mu},$$

(40)

where (9) is used and $B{{}^{\nu}}_{\beta}$ is defined in (35). Finally, the gravitational phase factor for light is:

$$\Phi_{L}=\exp(i\phi_{g})=\exp\Bigl{(}\frac{i}{\hbar}\int_{p}^{q}g_{\mu\nu}P^{\mu}{B^{\nu}}_{\beta}\mbox{d}x^{\beta}\Bigr{)}.$$

(41)

Although (41) has been derived for photons, we assume it also applicable for other massless particles. Of course, this hypothesis needs a rigorous proof.

In summary, the gravitational phase for a particle (massive or massless) in an inertial frame is given by:

$$\phi_{g}=\frac{1}{\hbar}\int_{p}^{q}S_{\beta}\mbox{d}x^{\beta},$$

(42)

where the function $S_{\beta}$ is defined as

$$S_{\beta}=g_{\mu\nu}P^{\mu}{B^{\nu}}_{\beta},$$

(43)

and $P^{\mu}$ is the four-momentum. The gravitational phase factor is given by $\Phi_{g}=\exp(i\phi_{g})$.

To calculate $S_{\beta}$, we need to know ${B^{\nu}}_{\beta}$ firstly. According to (35), the expression of ${B^{\nu}}_{\beta}$ is given by ${h_{a}}^{\nu}$ and $B{{}^{a}}_{\beta}$. Thus in addition to ${h_{a}}^{\nu}$, we need to seek the expression for $B{{}^{a}}_{\beta}$. Before proceeding, let us consider the cartesian coordinate system $K^{\prime}$ in in which $\partial_{\mu^{\prime}}x^{a}=\delta{{}^{a}}_{\mu^{\prime}}$ holds [9]. Therefore, according to Eq. (10), in the coordinate $K^{\prime}$ the gauge potential can be written as:

$$B{{}^{a}}_{\mu^{\prime}}=h{{}^{a}}_{\mu^{\prime}}-\delta{{}^{a}}_{\mu^{\prime}}.$$

(44)

Moreover, the components of the tetrad in the generic coordinate $K$ can be expressed as [16]:

$$h{{}^{a}}_{\rho}=h{{}^{a}}_{\nu^{\prime}}\frac{\partial x^{\nu^{\prime}}}{\partial x^{\rho}}\,,$$

(45)

which can be derived directly by writing the second equation of (5) as:

$\displaystyle h^{a}=h{{}^{a}}_{\nu^{\prime}}\mbox{d}x^{\nu^{\prime}}=h{{}^{a}}_{\nu^{\prime}}\frac{\partial x^{\nu^{\prime}}}{\partial x^{\rho}}\mbox{d}x^{\rho}=h{{}^{a}}_{\rho}\mbox{d}x^{\rho}\,.$

(46)

Now we come back to the expression for $B{{}^{a}}_{\mu}$. We write the gravitational phase in the coordinate $K$:

$$\phi_{g}=\frac{1}{\hbar}\int_{p}^{q}g_{\mu\nu}P^{\mu}{B^{\nu}}_{\beta}\mbox{d}x^{\beta}=\frac{1}{\hbar}\int_{p}^{q}P_{a}B{{}^{a}}_{\beta}\mbox{d}x^{\beta},$$

(47)

where (9) and (8) are used. On the other hand, we write it in the cartesian coordinate $K^{\prime}$:

$$\phi_{g}=\frac{1}{\hbar}\int_{p}^{q}P_{a}B{{}^{a}}_{\mu^{\prime}}\mbox{d}x^{\mu^{\prime}}=\frac{1}{\hbar}\int_{p}^{q}P_{a}(h{{}^{a}}_{\sigma}\frac{\partial x^{\sigma}}{\partial x^{\mu^{\prime}}}-\delta{{}^{a}}_{\mu^{\prime}})\mbox{d}x^{\mu^{\prime}},$$

(48)

where (44) and (45) are used. Furthermore, we write (48) as:

$$\phi_{g}=\frac{1}{\hbar}\int_{p}^{q}P_{a}(h{{}^{a}}_{\sigma}\frac{\partial x^{\sigma}}{\partial x^{\mu^{\prime}}}-\delta{{}^{a}}_{\mu^{\prime}})\frac{\partial x^{\mu^{\prime}}}{\partial x^{\beta}}\mbox{d}x^{\beta}=\frac{1}{\hbar}\int_{p}^{q}P_{a}(h{{}^{a}}_{\beta}-\delta{{}^{a}}_{\mu^{\prime}}\frac{\partial x^{\mu^{\prime}}}{\partial x^{\beta}})\mbox{d}x^{\beta}.$$

(49)

Comparing (47) and (49), we finally obtain:

$$B{{}^{a}}_{\beta}=h{{}^{a}}_{\beta}-\delta{{}^{a}}_{\mu^{\prime}}\frac{\partial x^{\mu^{\prime}}}{\partial x^{\beta}}.$$

(50)

Summarizing. We choose an inertial coordinate system $K$. Then we find the expression for the components of the tetrad ${h^{a}}_{\beta}$, and the transformation between the coordinate $K$ and the cartesian coordinate $K^{\prime}$. Plugging the tetrad into (50), we get the expression of ${B^{a}}_{\beta}$. Hence, inserting the latter into (35), we get ${B^{\nu}}_{\beta}$. Pugging the expressions of ${B^{\nu}}_{\beta}$ and $P^{\mu}$ into (43), we get $S_{\beta}$. According to it, we calculate the integral in (42) to finally get the gravitational phase.

Using Boyer-Lindquist coordinates $K(t,r,\theta,\varphi)$ in the Kerr spacetime, the matrix form of the metric is [17]:

$$(g_{\mu\nu})=\begin{pmatrix}g_{00}&{}&{}&g_{03}\\ {}&g_{11}&{}&{}\\ {}&{}&g_{22}&{}\\ g_{30}&{}&{}&g_{33}\end{pmatrix}=\begin{pmatrix}1-r_{g}r/\rho^{2}&{0}&{0}&ar_{g}r(\mbox{s}\theta)^{2}/\rho^{2}\\ {0}&-\rho^{2}/\Delta&{0}&{0}\\ {0}&{0}&-\rho^{2}&{0}\\ ar_{g}r(\mbox{s}\theta)^{2}/\rho^{2}&{0}&{0}&-[r^{2}+a^{2}+a^{2}r_{g}r(\mbox{s}\theta)^{2}/\rho^{2}](\mbox{s}\theta)^{2}\end{pmatrix},$$

(51)

and for the inverse:

$$(g^{\mu\nu})=\begin{pmatrix}g^{00}&{}&{}&g^{03}\\ {}&g^{11}&{}&{}\\ {}&{}&g^{22}&{}\\ g^{30}&{}&{}&g^{33}\end{pmatrix}=\begin{pmatrix}\Sigma^{2}/(\rho^{2}\Delta)&{0}&{0}&ar_{g}r/(\rho^{2}\Delta)\\ {0}&-\Delta/\rho^{2}&{0}&{0}\\ {0}&{0}&-1/\rho^{2}&{0}\\ ar_{g}r/(\rho^{2}\Delta)&0&0&-[\Delta-a^{2}(\mbox{s}\theta)^{2}]/[\rho^{2}(\mbox{s}\theta)^{2}\Delta]\end{pmatrix},$$

(52)

where

$$r_{g}=2M,\quad\rho^{2}=r^{2}+a^{2}(\mbox{c}\theta)^{2},\quad\Delta=r^{2}-r_{g}r+a^{2},\quad\Sigma^{2}=(r^{2}+a^{2})^{2}-a^{2}(\mbox{s}\theta)^{2}\Delta.$$

(53)

The parameter $M$ is the mass of the black hole, and $a$ is its angular momentum per unit of mass in the units $c=1$. (In the SI units it is $a=J/(Mc)$, where $J$ is the angular momentum of the black hole [18].) Here the symbols $\mbox{s}\theta$, $\mbox{c}\theta$, $\mbox{s}\varphi$ and $\mbox{c}\varphi$ denote $\sin(\theta)$, $\cos(\theta)$, $\sin(\varphi)$ and $\cos(\varphi)$ respectively.

We will calculate the gravitational phase in the Kerr spacetime by using the last expression in (34), but before that, we derive the expression of $B{{}^{a}}_{\beta}$ according to Eq. (50). The coordinate transformation from $K^{\prime}$ to $K$ is [18]:

$$\begin{cases}t^{\prime}=t,\\ x^{\prime}=\sqrt{r^{2}+a^{2}}\,\mbox{s}\theta\mbox{c}\varphi,\\ y^{\prime}=\sqrt{r^{2}+a^{2}}\,\mbox{s}\theta\mbox{s}\varphi,\\ z^{\prime}=r\,\mbox{c}\theta.\end{cases}$$

(54)

From Eq. (54) we can get the Jacobi matrix:

$$\Bigl{(}\frac{\partial x^{\mu^{\prime}}}{\partial x^{\beta}}\Bigr{)}=\begin{pmatrix}1&0&0&0\\ 0&\frac{r}{\rho_{0}}\mbox{s}\theta\mbox{c}\varphi&\rho_{0}\mbox{c}\theta\mbox{c}\varphi&-\rho_{0}\mbox{s}\theta\mbox{s}\varphi\\ 0&\frac{r}{\rho_{0}}\mbox{s}\theta\mbox{s}\varphi&\rho_{0}\mbox{c}\theta\mbox{s}\varphi&\rho_{0}\mbox{s}\theta\mbox{c}\varphi\\ 0&\mbox{c}\theta&-r\mbox{s}\theta&0\end{pmatrix},$$

(55)

where $\rho_{0}=\sqrt{r^{2}+a^{2}}$. The tetrad in the Kerr spacetime is [9, 16]:

$$(h{{}^{a}}_{\beta})=\begin{pmatrix}\gamma_{00}&0&0&\eta\\ 0&\gamma_{11}\mbox{s}\theta\mbox{c}\varphi&\gamma_{22}\mbox{c}\theta\mbox{c}\varphi&-\zeta\mbox{s}\varphi\\ 0&\gamma_{11}\mbox{s}\theta\mbox{s}\varphi&\gamma_{22}\mbox{c}\theta\mbox{s}\varphi&\zeta\mbox{c}\varphi\\ 0&\gamma_{11}\mbox{c}\theta&-\gamma_{22}\mbox{s}\theta&0\end{pmatrix},$$

(56)

where²²2In Ref. [16] the authors only give the expression $\zeta^{2}=\eta^{2}-g_{33}$. We believe $\zeta=\sqrt{\eta^{2}-g_{33}}$ also holds, which is in accordance with the tetrad in Schwardschild space time (see (29) in Ref. [16]).

$$\eta=g_{03}/\gamma_{00},\quad\zeta=\sqrt{\eta^{2}-g_{33}},\quad\gamma_{00}=\sqrt{g_{00}},\quad\gamma_{jj}=\sqrt{-g_{jj}}.$$

(57)

Inserting Eqs. (56) and (55) into (50), we get the gauge potential in the Kerr spacetime:

$$(B{{}^{a}}_{\beta})=\begin{pmatrix}\gamma_{00}-1&0&0&\eta\\ 0&(\gamma_{11}-\frac{r}{\rho_{0}})\mbox{s}\theta\mbox{c}\varphi&(\gamma_{22}-\rho_{0})\mbox{c}\theta\mbox{c}\varphi&(\rho_{0}\mbox{s}\theta-\zeta)\mbox{s}\varphi\\ 0&(\gamma_{11}-\frac{r}{\rho_{0}})\mbox{s}\theta\mbox{s}\varphi&(\gamma_{22}-\rho_{0})\mbox{c}\theta\mbox{s}\varphi&(\zeta-\rho_{0}\mbox{s}\theta)\mbox{c}\varphi\\ 0&(\gamma_{11}-1)\mbox{c}\theta&(r-\gamma_{22})\mbox{s}\theta&0\end{pmatrix}.$$

(58)

It is easy to check that ${B^{a}}_{\beta}=0$ in flat spacetime, namely for $a=0$ and $M=0$. The matrix form for the inverse of the tetrad is³³3We think that these are some typos in equation (14.37) in [9]. This equation should be modified as (59).:

$$(h_{a}{{}^{\nu}})=\begin{pmatrix}\gamma_{00}^{-1}&0&0&0\\ \zeta g^{03}\mbox{s}\varphi&\gamma_{11}^{-1}\mbox{s}\theta\mbox{c}\varphi&\gamma_{22}^{-1}\mbox{c}\theta\mbox{c}\varphi&-\zeta^{-1}\mbox{s}\varphi\\ -\zeta g^{03}\mbox{c}\varphi&\gamma_{11}^{-1}\mbox{s}\theta\mbox{s}\varphi&\gamma_{22}^{-1}\mbox{c}\theta\mbox{s}\varphi&\zeta^{-1}\mbox{c}\varphi\\ 0&\gamma_{11}^{-1}\mbox{c}\theta&-\gamma_{22}^{-1}\mbox{s}\theta&0\end{pmatrix}.$$

(59)

One can check that (56) and (59) indeed satisfy $g_{\mu\nu}=\eta_{ab}{h^{a}}_{\mu}{h^{b}}_{\nu}$, $\eta_{ab}=g_{\mu\nu}{h_{a}}^{\mu}{h_{b}}^{\nu}$ and ${h^{a}}_{\mu}{h_{b}}^{\mu}=\delta^{a}_{b}$. Inserting Eqs. (58) and (59) into (35), we obtain

$$(B{{}^{\nu}}_{\beta})=\begin{pmatrix}1-\gamma_{00}^{-1}&0&0&\eta\gamma_{00}^{-1}+(\rho_{0}\mbox{s}\theta-\zeta)\zeta g^{03}\\ 0&1-\gamma_{11}^{-1}[r\rho_{0}^{-1}(\mbox{s}\theta)^{2}+(\mbox{c}\theta)^{2}]&\gamma_{11}^{-1}\mbox{s}\theta\mbox{c}\theta(r-\rho_{0})&0\\ 0&\gamma_{22}^{-1}\mbox{s}\theta\mbox{c}\theta(1-r/\rho_{0})&1-\gamma_{22}^{-1}[r(\mbox{s}\theta)^{2}+\rho_{0}(\mbox{c}\theta)^{2}]&0\\ 0&0&0&1-\zeta^{-1}\rho_{0}\mbox{s}\theta\end{pmatrix}.$$

(60)

In terms of Eq. (43), the first expression in Eq. (51), and Eq. (60), the matrix $S_{\beta}$ can be written as:

$$(S_{\beta})=(P^{\mu}g_{\mu\nu}B{{}^{\nu}}_{\beta})=\begin{pmatrix}(1-\gamma_{00}^{-1})(P^{0}g_{00}+P^{3}g_{30})\\ P^{1}g_{11}\{1-\gamma_{11}^{-1}[r\rho_{0}^{-1}(\mbox{s}\theta)^{2}+(\mbox{c}\theta)^{2}]\}+P^{2}g_{22}\gamma_{22}^{-1}\mbox{s}\theta\mbox{c}\theta(1-r/\rho_{0})\\ P^{1}g_{11}\gamma_{11}^{-1}\mbox{s}\theta\mbox{c}\theta(r-\rho_{0})+P^{2}g_{22}\{1-\gamma_{22}^{-1}[r(\mbox{s}\theta)^{2}+\rho_{0}(\mbox{c}\theta)^{2}]\}\\ (P^{0}g_{00}+P^{3}g_{30})[\eta\gamma_{00}^{-1}+(\rho_{0}\mbox{s}\theta-\zeta)\zeta g^{03}]+(P^{0}g_{03}+P^{3}g_{33})(1-\zeta^{-1}\rho_{0}\mbox{s}\theta)\end{pmatrix}.$$

(61)

The latter result can be further simplified by using the following conserved quantities in the Kerr spacetime [17],

	$\displaystyle E$	$\displaystyle=$	$\displaystyle\Bigl{(}1-\frac{r_{g}r}{\rho^{2}}\Bigr{)}\frac{\mbox{d}t}{\mbox{d}\xi}+\frac{ar_{g}r(s\theta)^{2}}{\rho^{2}}\frac{\mbox{d}\varphi}{\mbox{d}\xi}=u^{0}g_{00}+u^{3}g_{30},$
	$\displaystyle-L$	$\displaystyle=$	$\displaystyle\frac{ar_{g}r(s\theta)^{2}}{\rho^{2}}\frac{\mbox{d}t}{\mbox{d}\xi}-\Bigl{[}r^{2}+a^{2}+\frac{r_{g}r}{\rho^{2}}a^{2}(s\theta)^{2}\Bigr{]}(s\theta)^{2}\frac{\mbox{d}\varphi}{\mbox{d}\xi}=u^{0}g_{03}+u^{3}g_{33},$		(62)

where $\xi$ is an affine parameter (for massive particles it is the proper time), and $E$ and $L$ are defined as

$$E=\begin{cases}\mathcal{E}m^{-1},&\text{for massive particles,}\\ \mathcal{E},&\text{for massless particles,}\end{cases}\qquad L=\begin{cases}\mathcal{L}m^{-1},&\text{for massive particles,}\\ \mathcal{L},&\text{for massless particles.}\end{cases}$$

(63)

Notice that for massive particles we have $P^{\mu}=mu^{\mu}$, while for massless particles we have $P^{\mu}=u^{\mu}$. Here the quantity $\mathcal{E}$ has the meaning of energy, while the quantity $\mathcal{L}$ has the meaning of angular momentum along the spin of the black hole. Plugging (62) into (61), we get

$$(S_{\beta})=\begin{pmatrix}(1-\gamma_{00}^{-1})\mathcal{E}\\ P^{1}g_{11}\{1-\gamma_{11}^{-1}[r\rho_{0}^{-1}(\mbox{s}\theta)^{2}+(\mbox{c}\theta)^{2}]\}+P^{2}g_{22}\gamma_{22}^{-1}\mbox{s}\theta\mbox{c}\theta(1-r/\rho_{0})\\ P^{1}g_{11}\gamma_{11}^{-1}\mbox{s}\theta\mbox{c}\theta(r-\rho_{0})+P^{2}g_{22}\{1-\gamma_{22}^{-1}[r(\mbox{s}\theta)^{2}+\rho_{0}(\mbox{c}\theta)^{2}]\}\\ \mathcal{E}[\eta\gamma_{00}^{-1}+(\rho_{0}\mbox{s}\theta-\zeta)\zeta g^{03}]-\mathcal{L}(1-\zeta^{-1}\rho_{0}\mbox{s}\theta)\end{pmatrix}.$$

(64)

Finally, recalling (42), the gravitational phase in the Kerr spacetime is given by:

$$\phi=\frac{1}{\hbar}\int_{p}^{q}S_{\beta}\mbox{d}x^{\beta}.$$

(65)

We will study an interference experiment in the region $r\gg r_{g}$ with the size of the setup much smaller than its distance from the black hole. Let us start with a review of the Colella-Overhauser-Werner (COW) experiment on the earth [11]. The principle of this experiment is shown in FIG. 1, where the parallelogram is vertical and its base AB is parallel to the surface of the earth. A beam of neutrons is split into two beams along the paths ABC and ADC respectively, and, afterwards they interfere. Since of the presence of gravity, the phase accumulated along the path ABC is different from the phase accumulated along ADC. The theoretical predictions for this experiment were given in Ref. [10], in which the gravitational phase difference between these two paths was found to be:

$$\delta\phi=\frac{m^{2}gl\lambda_{d}}{2\pi\hbar^{2}}s,$$

(66)

where $m$ is the mass of the neutron, $\lambda_{d}=2\pi\hbar/(mv)$ is its de Broglie wavelength, $g$ is the gravitational acceleration, $l$ is the height of the parallelogram, and $s$ is the length of AB.