Dependence of the black-body force on spacetime geometry and topology

As shown in ref. [1], for the non-relativistic case the attractive potential associated to this force exerted on the neutral hydrogen atom in its fundamental state, placed at a distance r from the center of a spherical black body of radius R at temperature T is

$$\begin{equation} V_{bb}(r)=-\frac{3\pi^3 (\kappa T)^4}{5 (\alpha m_e)^3}\frac{\Omega_a}{4\pi}, \end{equation} \tag{ 1 }$$

where α is the fine-structure constant, m_e is the mass of the electron, and $\Omega_a$ is the solid angle subtended by the atom. About the above expression some points are worthy of comment. As pointed in ref. [1], for the hydrogen atom in an isotropic thermal bath, the dynamical Stark effect will mostly cause a negative energy shift for the ground state. If the temperature of the thermal bath is T, this shift is given by

$$\begin{equation} \Delta E_{1s}\approx -\frac{3\pi^3 (\kappa T)^4}{5 (\alpha m_e)^3}. \end{equation} \tag{ 2 }$$

Since the atom is in an isotropic thermal bath, the black-body field cannot have any directional effect on the atom motion, inducing only friction and diffusion. Those authors then argue that if a finite source is considered, this produces a net force. The point is that in this case, the atom does not see radiation coming from all directions, but only from a solid angle determined by the source. Therefore, only a fraction of the energy shift in the isotropic bath would cause the BBF. This fraction is given by the solid angle compared to $4\pi$, giving us the final potential energy of eq. (1). For the non-relativistic spherically symmetric case this solid angle is given by

$$\begin{equation} \Omega_a=2\pi\left(1-\cos\theta\right)=2\pi\left(1-\frac{\sqrt{r^2-R^2}}{r}\right). \end{equation} \tag{ 3 }$$

From the above expression we can see that the radial dependence of the potential comes only from the solid angle. However, when we consider curved static spacetimes, the effective temperature will also depend on the coordinates via Tolman temperature [14,15]. Therefore, in order to find corrections to the BBF two aspects must be considered: the first is the curvature correction to the temperature and the second is the modification in the solid angle determined by the spacetime curvature. This is the general procedure that we will follow throughout this paper.

The static and spherically symmetric exterior solution of Einstein's equation of General Relativity is initially considered, which is given by the metric

$$\begin{equation} \text{d}s^2=\left(1-\frac{r_s}{r}\right)c^2\text{d}t^2- \left(1-\frac{r_s}{r}\right)^{-1}\text{d}r^2-r^2\text{d}\Omega^2, \end{equation} \tag{ 4 }$$

where $r_s=2GM/c^2$. As said above, the spacetime modification to the temperature is given by the Tolman temperature. At the position of the atom this is given by

$$\begin{equation} T_{eff}=\frac{T}{\sqrt{g_{tt}}}=\frac{T}{\sqrt{1-\frac{r_s}{r}}}, \end{equation} \tag{ 5 }$$

where T is the temperature of the isotropic thermal bath, as measured by an observer at the infinity. The fact that the temperature depends on the radial coordinate will contribute to modify the BBF.

Now we must consider the modification in the solid angle due to the spacetime curvature which occurs in the system investigated here. This is done by considering the change $\cos\theta\to\cos\tilde{\theta}$ in eq. (3). According to ref. [16] the correct expression in Schwarzschild spacetime is given by

$$\begin{eqnarray} \cos\tilde{\theta}&= \left[1+\left(1-\frac{r_s}{r}\right)r^2 \frac{\text{d}\phi^2}{\text{d}r^2}\right]^{-\frac{1}{2}}=\nonumber\\ & \quad \left[1+\left(1-\frac{r_s}{r}\right)\tan^2\theta\right]^{-\frac{1}{2}}, \end{eqnarray} \tag{ 6 }$$

where in the last equation we used the fact that in the flat limit

$$\begin{eqnarray*} &&r^2\frac{\text{d}\phi^2}{\text{d}r^2}=\tan^2\theta=\frac{R^2}{r^2-R^2}. \end{eqnarray*}$$

Thus we obtain that the modification in the solid angle will be given by

$$\begin{equation} \Omega_a=2\pi\left[1-\frac{1}{\sqrt{1+\left(1-\frac{r_s}{r}\right) \tan^2\theta}}\right]. \end{equation} \tag{ 7 }$$

Notice that the above expression reproduces the Euclidean solid angle for large values of r and at the surface of the source is given by $2\pi$. Plugging eq. (5) and eq. (7) into eq. (1) we get the final corrected expression for the BBF potential

$$\begin{eqnarray} V_{bb}(r)&= -\frac{3\pi^3 (\kappa T)^4}{10 (\alpha m_e)^3}\left(1-\frac{r_s}{r}\right)^{-2}\nonumber\\ & \quad \times\,\left[1-\frac{1}{\sqrt{1+\left(1-\frac{r_s}{r}\right)\tan^2\theta}}\right]. \end{eqnarray} \tag{ 8 }$$

This potential corresponds to a force that also drops with the cube-inverse at large distances and is infinity when r = R, as in the flat spacetime case.

The radial force derived from the BBF potential can be calculated via

$$\begin{equation} F_{bb}(r)=-\text{d}V_{bb}(r)/\text{d}l=-(\text{d}V_{bb}(r)/\text{d}r) (\text{d}r/\text{d}l), \end{equation} \tag{ 9 }$$

where $\text{d}r/\text{d}l=(1-r_S/r)^{1/2}$, according to the metric (4), with l being the radial ruler distance [16].

In fig. 1, we depict the graph of the relativistically corrected potential V_bb over the temperature function $f(T)=\frac{3\pi^2 (\kappa T)^4}{10 (\alpha m_e)^3}$, against the same potential energy in the flat case, both depending on the ratio $r/r_s$, for a compact sphere with $R=3r_s$, which corresponds to the dimensions of a typical neutron star with about one solar mass. The difference between both the potentials is sensible, and the magnitude of the corrected one is considerably greater than the observed in the other case. At $r=4.5r_s$, the relative difference between the potentials is $(V_{bb}^{Corr.}-V_{bb}^{Flat})/V_{bb}^{Flat}\approx 39\%$, corresponding to a force $70\%$ stronger. At shorter distances and for more compacts objects, the difference increases even more.

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Notice that by considering the Sun, this reduction in the temperature-independent part of the BBF potential is approximately 0.0004% at $r=1.5R$. For a quantum gravity correction to this force, see [17].

The case to be analyzed here is that of a global monopole, whose exterior metric is given by [4,5]

$$\begin{equation} \text{d}s^2=c^2\text{d}t^2-\text{d}r^2-\beta^2r^2\text{d}\Omega^2, \end{equation} \tag{ 10 }$$

where $\beta^2<1$ is the parameter that gives the deficit in the central solid angle, modifying the spacetime topology. In order to calculate the BBF on the atom placed in this spacetime, we redefine the polar angle so that

$$\begin{eqnarray*} &&\text{d}\overline{\theta}=\beta \text{d}\theta, \end{eqnarray*}$$

and

$$\begin{eqnarray*} &&\sin\overline{\theta}=\beta\sin{\theta}. \end{eqnarray*}$$

A solution for $\overline{\theta}$ can be found by multiplying the above expressions and integrating the result, obtaining

$$\begin{equation} \overline{\theta}=\arccos{(\beta^2\cos\theta)}, \end{equation} \tag{ 11 }$$

where we made the integration constant equal to zero. Now we can make the substitution $\theta\rightarrow\overline{\theta}$ in eq. (8), with $r_s=0$. This procedure allows us reobtaining the BBF originally found in ref. [1], since the redefinition of the polar angle makes the metric (10) locally flat. Expressing the new polar angle according to eq. (11), we arrive at

$$\begin{eqnarray} & V_{bb}(r)=\nonumber\\ & -\frac{3\pi^3 (\kappa T)^4}{10 (\alpha m_e)^3}\left[1-\frac{1}{\sqrt{1+\tan^2{[\arccos{(\beta^2 \cos\theta)}]}}}\right].\quad \end{eqnarray} \tag{ 12 }$$

Recalling that $\tan^2{\theta}=R^2/(r^2-R^2)$, we can plot the temperature-independent part of the BBF potential as a function of the atom distance to the source center, r, as shown in fig. 2 below. As in the previous subsection, the radial force can be calculated from eq. (9), with $\text{d}r/\text{d}l=1$.

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Here we have used $\beta=0.9$ and $R=10\ \text{km}$. Notice that when $\beta=1$ we come back to the flat BBF potential given by eq. (1). Notice also that the corrected BBF potential in the infinity tends to a value different from zero, equal to

$$\begin{eqnarray} V_{bb}(\infty)&= -\frac{3\pi^3 (\kappa T)^4}{10 (\alpha m_e)^3}\left[1-\frac{1}{\sqrt{1+\tan^2{(\arccos{\beta^2})}}}\right]. \end{eqnarray} \tag{ 13 }$$

From the expression without gravity (1), the only change in the potential will be due to the solid angle. Supposing that the particle is at a distance d of the rectangle center, thus the solid angle of the whole cylinder subtended by the atom is given by [18]

$$\begin{equation} \Omega_a(d)=4\pi\arccos{\sqrt{\frac{1+\alpha^2+\beta^2}{(1+\alpha^2) (1+\beta^2)}}}, \end{equation} \tag{ 14 }$$

where $\alpha=a/2d$, $\beta=b/2d$; a and b are the length and width of the rectangle, respectively. Taking an infinitely long cylinder, a → ∞, and writing the width in terms of the cylinder radius, R, and of the distance of the particle from the cylinder symmetry axis, r, which coincides with the z axis, we have

$$\begin{equation} \Omega_a(r)=4\pi\arccos{\sqrt{\frac{1}{1+\tan^2{\arcsin{\left(\frac{R}{r}\right)}}}}}, \end{equation} \tag{ 15 }$$

with r > R. The argument $\arcsin(R/r)$ is just the half-angle measured from the atom and which embraces the cylinder in the plane z = 0. In fig. 3 we depict the neutral atom (A) close to the cylindrical black body. According to it, the geometrical elements present in our analysis are: $\overline{AD}\equiv d$, $\overline{OA}\equiv r$, $\overline{BC}\equiv b$, $\overline{OB},\overline{OC}\equiv R$, and $\widehat{BOC}\equiv \phi$ (azimuthal or central angle).

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We find that the attractive potential on the hydrogen atom in the ground state due to the cylindrical black body at absolute temperature T is given by [1]

$$\begin{eqnarray} V_{bb}(r)&= \Delta E_{1s}(T)\frac{\Omega_a(r)}{4\pi}=\nonumber\\ & \quad -\,\frac{3\pi^3(\kappa T)^4}{5\alpha^3m_e^3}\arccos{\sqrt{\frac{1}{1+\tan^2{\arcsin{\left(\frac{R}{r}\right)}}}}},\qquad \end{eqnarray} \tag{ 16 }$$

where $\Delta E_{1s}(T)$ is the thermal shift in the absorption lines of the spectrum of the unexcited hydrogen atom. Thus the magnitude of the radial force exerted on this atom is given by

$$\begin{equation} F(r)=-\frac{3\pi^2(\kappa T)^4}{5\alpha^3m_e^3}\frac{R}{r^2\sqrt{1-R^2/r^2}}, \end{equation} \tag{ 17 }$$

where α is now the fine-structure constant. Notice that the BBF is attractive, falling with a square-inverse law at large distances (or $R\rightarrow0$). That is different from what happens in the spherical black bodies where the force drops with the cubic-inverse law at those distances [1].

The fact that the cosmic string spacetime has a locally flat geometry but a non-trivial topology raises interesting points. First of all, due to the local flatness, the effective temperature is not changed and therefore the only possible correction comes from the solid angle like in the previous case. The interesting point is that the solid angle now depends on the deficit angular parameter ν, incorporating then topological features in the BBF.

Let us model a static cosmic string, or its realization as a disclination in a liquid crystal, by the cylinder above considered. First, the conical geometry of the space around of the defect has to be incorporated. Such a geometry is described through the metric

$$\begin{equation} \text{d}s^2=-\text{d}t^2+\text{d}r^2+\frac{r^2}{\nu^2}\text{d}\phi^2 +\text{d}z^2. \end{equation} \tag{ 18 }$$

This expression describes the space around an ideal cosmic string of zero thickness, or the exterior region of that one endowed with an internal structure [19]. For this defect, $\nu>1$ (angular deficit parameter), and $0<\phi\leq2\pi$. For a disclination, ν can also be between zero and one, meaning that we have an angular excess parameter. Then all we have to do is rewrite the half-angle from which the atom sees the cylinder, $\arcsin{(R/r)}$, in terms of the correspondent half central (azimuthal) angle, $\phi/2=\arccos{(R/r)}$, such that

$$\begin{eqnarray} &&\arcsin{\left(\frac{R}{r}\right)}=\frac{\pi}{2}-\arccos{\left(\frac{R}{r}\right)}, \end{eqnarray} \tag{ 19 }$$

and to insert the factor $\nu^{-1}$ which multiplies the azimuthal angles, according to the metric (18). Thus the expression (16) becomes

$$\begin{eqnarray} V_{bb}(r)&= -\frac{3\pi^3(\kappa T)^4}{5\alpha^3m_e^3}\arccos\nonumber\\ & \quad \times\,{\sqrt{\frac{1}{1+\tan^2{\left[\frac{\pi}{2}-\nu^{-1}\arccos{\left(\frac{R}{r}\right)}\right]}}}}, \end{eqnarray} \tag{ 20 }$$

and calculating the force via $F(r)=-\text{d}V_{bb}/\text{d}r$, we get

$$\begin{equation} F(r)=\Delta E_{1s}(T)\frac{R}{r^2 \nu \sqrt{1-\frac{R^2}{r^2}}}. \end{equation} \tag{ 21 }$$

It is worth noticing that, for the neutral hydrogen atom in the Minkowsky space (i.e., when $\nu=1$) we have the expression (17) again, and when one takes the limit $R\rightarrow 0$ valid for an ideal cosmic string, the BBF vanishes. Then, an important conclusion is that ideal cosmic strings irradiating do not exert BBFs.

For non-zero thickness and $R\ll r$, we can approximate eq. (21) by

$$\begin{equation} F(r)\approx\Delta E_{1s}(T)\frac{R}{r^2 \nu}, \end{equation} \tag{ 22 }$$

the force is attractive, due to the factor $\Delta E_{1s}$ to be negative for neutral hydrogen atoms, and it still falls off with the inverse square of the radial distance. Such a law is the same one that manages the self-force on neutral (or charged) massive particles situated in the spacetime of a Gott-Hiscock cosmic string, when the particle is far away from the string [20].