Geodesics on metrics of Eguchi–Hanson type

Abstract

Geodesic equations are solved when at least two of $\theta$, $\phi$ and $\psi$ are constant, or r is constant, on scalar flat metrics of Eguchi–Hanson type. They can also be solved also on Eguchi–Hanson metrics which are Ricci flat if only $\phi$ is constant. However, the explicit solution of the geodesic equations is not available yet if only $\psi$ is constant.

1 Introduction

Geodesics are the fundamental geometric object in differential geometry. They also reveal nature of gravitational phenomena in curved space-time. In general relativity, null geodesics are interpreted as trajectories of light, time-like geodesics are interpreted as trajectories of massive particles, and they are studied extensively in many exact spacetimes, cf. [2, 3] and references therein. Besides enormous study of geodesics in Riemannian manifolds, Battista and Esposito recently investigated geodesic motion on Euclidean Schwarzschild metrics over $R^2 \times S^2$ and obtained its explicit form in terms of incomplete elliptic integrals of the first, the second and the third kinds [1]. As Euclidean Schwarzschild metrics belong to gravitational instantons, which play important roles in Euclidean approach of quantum gravity [7], it is interesting to investigate geodesic motion on other gravitational instantons such as Eguchi–Hanson metrics [4, 5]. This is the goal of our paper.

Eguchi–Hanson metrics are complete four-dimensional Ricci flat, anti-self-dual ALE Riemannian metrics on $R_{\ge 0} \times P^3$. The more general metrics of Eguchi–Hanson type were constructed by LeBurn [11] using the method of algebraic geometry [11] and by the second author solving an ordinary differential equation [10]. They are scalar flat on $R_{\ge 0} \times S^3 /Z_d$ $(d >2)$ and provide counter-examples of Hawking and Pope’s generalized positive action conjecture [9]. It is important that gravitational instantons always have removable singularity. And it occurs at the largest positive root of potential functions for metrics of Eguchi–Hanson type. The geodesic completeness has not been proved in the literature for these types of metrics, as pointed out in [1]. This motivates us to solve geodesic equations passing through the removable singularity.

The paper is organized as follows. In Sect. 2, we provide a brief introduction to Eguchi–Hanson metrics and scalar flat metrics of Eguchi–Hanson type. We also give geodesic equations for metrics of Eguchi–Hanson with general radial potential functions. In Sect. 3, we solve geodesic equations on scalar flat metrics of Eguchi–Hanson type in the following cases: (1) Sect. 3.1, where $\theta$, $\phi$, $\psi$ are constant, geodesics pass through the removable singularity; (2) Sect. 3.2, where $\phi$, $\psi$ are constant, geodesics pass through the removable singularity; (3) Sect. 3.3, where $\theta$, $\phi$ are constant, geodesics do not pass through the removable singularity. If geodesics pass through the removable singularity, $\psi$ must be constant and the geodesic equations reduce to Sect. 3.1; (4) Sect. 3.4, where $\theta \in (0, \pi )$, $\psi$ are constant. It yields that $\theta =\frac{\pi }{2}$ if $\phi$ is not constant; (5) Sect. 3.5, where r is constant, geodesics pass through the removable singularity; (6) Sect. 3.6, where $\phi$ is constant. It yields that either $\psi$ or $\theta$ is constant, and the geodesic equations reduce to either Sect. 3.2 or Sect. 3.3. In Sect. 4, we solve geodesic equations on Eguchi–Hanson metrics, where only $\theta \in (0, \pi )$ is constant and geodesics do not pass through the removable singularity. If geodesics pass through the removable singularity, then $\theta =\frac{\pi }{2}$, $\psi$ is constant and the geodesic equations reduce to Sect. 3.4. In Appendix 5, we provide an brief introduction to incomplete elliptic integrals of the first, the second and the third kinds.

We point out that the explicit solution of geodesic equations is not available yet if only $\psi$ is constant.

2 Metrics of Eguchi–Hanson type and geodesic equations

In this section, we introduce the metrics of Eguchi–Hanson type and provide their geodesic equations. Let $\sigma _{1}$, $\sigma _{2}$, and $\sigma _{3}$ be the Cartan–Maurer one-forms for $SU(2)\cong S^3$, defined by

$$\begin{aligned} \sigma _{1}&=\frac{1}{2}\big (\sin \psi d\theta -\sin \theta \cos \psi d\phi \big ),\\ \sigma _{2}&=\frac{1}{2}\big (-\cos \psi d\theta -\sin \theta \sin \psi d\phi \big ),\\ \sigma _{3}&=\frac{1}{2}\big (d \psi +\cos \theta d\phi \big ). \end{aligned}$$

They satisfy

$$\begin{aligned} d\sigma _{1}=2\sigma _{2}\wedge \sigma _{3},\quad d\sigma _{2}=2\sigma _{3}\wedge \sigma _{1},\quad d\sigma _{3}=2\sigma _{1}\wedge \sigma _{2}. \end{aligned}$$

Metrics of Eguchi–Hanson type are given by

$$\begin{aligned} ds^2 = f^{-1} dr ^2 +r ^2 \Big ( \sigma _1 ^2 +\sigma _2 ^2 +f \sigma _3 ^2 \Big ), \end{aligned}$$

(2.1)

where $f\ge 0$ is a smooth function, which is referred as the potential function.

If

$$\begin{aligned} f=1-\frac{B}{r ^4} \end{aligned}$$

(2.2)

for some $B>0$, and

$$\begin{aligned} r \ge \root 4 \of {B},\quad 0 \leqslant \theta \leqslant \pi , \quad 0 \leqslant \phi \leqslant 2\pi , \quad 0 \le \psi \le 2\pi ,\nonumber \\ \end{aligned}$$

(2.3)

the metrics (2.1) are Eguchi–Hanson metrics [4, 5], which are Ricci flat, geodesically complete and asymptotically local Euclidean, and $\root 4 \of {B}$ is the removable singularity. The underground manifold is, topologically,

$$\begin{aligned} R_{\ge 0} \times SU(2)/Z_2 \cong R_{\ge 0} \times SO(3) \cong R_{\ge 0} \times P _3. \end{aligned}$$

Let $n \ge 2$ be a natural number. If

$$\begin{aligned} f=1-\frac{2A}{r ^2}-\frac{B}{r ^4},\quad A=-\frac{n-2}{2}\sqrt{\frac{B}{n-1}}, \end{aligned}$$

(2.4)

for some $B>0$, and

$$\begin{aligned} r\ge \root 4 \of {\frac{B}{n-1}}, \quad 0 \leqslant \theta \leqslant \pi , \quad 0 \leqslant \phi \leqslant 2\pi , \quad 0 \le \psi \le \frac{4\pi }{n}, \end{aligned}$$

(2.5)

the metrics (2.1) are geodesically complete, asymptotically local Euclidean, and $\root 4 \of {\frac{B}{n-1}}$ is the removable singularity. The underground manifold is, topologically,

$$\begin{aligned} R_{\ge 0} \times SU(2)/Z_n \cong R_{\ge 0} \times S^3/Z_n. \end{aligned}$$

With A given by (2.4), we have

$$\begin{aligned} r^4 -2A r^2 -B=\left( r^2 - \sqrt{\frac{B}{n-1}} \right) \left( r^2 +\sqrt{(n-1)B } \right) . \end{aligned}$$

Thus $\root 4 \of {\frac{B}{n-1}}$ is the largest positive simple root of f. This fact is crucial to solve radial geodesics passing through the removable singularity.

The nonvanishing metric components of (2.1) are

$$\begin{aligned}&g_{rr}=f^{-1} , g^{rr}=f,\\&g_{\theta \theta }=\frac{r^2}{4} , g^{\theta \theta }=\frac{4}{r ^2},\\&g_{\phi \phi }=\frac{r^2}{4}\left( f\cos ^2\theta +\sin ^2\theta \right) ,\\&g^{\phi \phi }=\frac{4}{r ^2\sin ^2\theta },\\&g_{\phi \psi }=g_{\psi \phi }=\frac{r^2}{4}f\cos \theta ,g^{\phi \psi }=g^{\psi \phi }\\&\quad =-\frac{4\cos \theta }{r ^2\sin ^2\theta },\\&g_{\psi \psi }=\frac{r^2}{4}f,\\&g^{\psi \psi }=\frac{4}{r ^2}\left( \cot ^2\theta +\frac{1}{f}\right) . \end{aligned}$$

The nonvanishing Christoffel symbols are

$$\begin{aligned}&\Gamma ^r_{rr}=-\frac{f'}{2f}, \Gamma ^r_{\theta \theta }=-\frac{r}{4}f,\\&\Gamma ^r_{\phi \phi }=-\frac{f}{4}\left( \frac{\left( r^2f\right) '}{2}\cos ^2\theta +r\sin ^2\theta \right) ,\\&\Gamma ^r_{\phi \psi }=-\frac{\left( r^2f\right) '}{8f^{-1}}\cos \theta ,\\&\Gamma ^r_{\psi \psi }=-\frac{f \left( r^2f\right) '}{8}, \Gamma ^\theta _{r\theta }=\Gamma ^\phi _{r\phi }=\frac{1}{r},\\&\Gamma ^\theta _{\phi \phi }=\left( f-1\right) \sin \theta \cos \theta ,\\&\Gamma ^\theta _{\phi \psi }=\frac{f}{2}\sin \theta ,\\&\Gamma ^\phi _{\theta \phi }=y\left( 1-\frac{f}{2}\right) \cot \theta ,\\&\Gamma ^\phi _{\theta \psi }=-\frac{f}{2\sin \theta },\\&\Gamma ^\psi _{r \phi }=\frac{f'}{2f}\cos \theta ,\\&\Gamma ^\psi _{r\psi }=\frac{1}{r}+\frac{f'}{2f},\\&\Gamma ^\psi _{\theta \phi }=\frac{\left( f-1\right) \cos ^2\theta -1}{2\sin \theta },\\&\Gamma ^\psi _{\theta \psi }=\frac{f\cot \theta }{2}, \end{aligned}$$

Thus geodesic equations of (2.1) for parameter t are

$$\begin{aligned}&\frac{d^2r}{dt^2}-\frac{f'}{2f}\Big (\frac{dr}{dt}\Big )^{2}-\frac{rf}{4}\Big (\frac{d\theta }{dt}\Big )^{2}\nonumber \\&\quad -\frac{f}{4}\Big (r\sin ^2\theta +\frac{(r^2f )' }{2}\cos ^2\theta \Big )\Big (\frac{d\phi }{dt}\Big )^{2} \nonumber \\&\quad -\frac{f(r^2f) '}{4}\cos \theta \frac{d\phi }{dt}\frac{d\psi }{dt}\nonumber \\&\quad -\frac{f(r^2f) '}{8}\Big (\frac{d\psi }{dt}\Big )^{2}=0, \end{aligned}$$

(2.6)

$$\begin{aligned}&\frac{d^2\theta }{dt^2}+\frac{2}{r}\frac{dr}{dt}\frac{d\theta }{dt} +(f-1)\sin \theta \cos \theta \Big (\frac{d\phi }{dt}\Big )^{2}\nonumber \\&\quad +f\sin \theta \frac{d\phi }{dt}\frac{d\psi }{dt}=0, \end{aligned}$$

(2.7)

$$\begin{aligned}&\frac{d^2\phi }{dt^2}+\frac{2}{r}\frac{dr}{dt}\frac{d\phi }{dt}-(f-2)\cot \theta \frac{d\phi }{dt}\frac{d\theta }{dt}\nonumber \\&\quad -\frac{f}{\sin \theta }\frac{d\theta }{dt}\frac{d\psi }{dt}=0, \end{aligned}$$

(2.8)

$$\begin{aligned}&\frac{d^2\psi }{dt^2}+\frac{f'}{f}\cos \theta \frac{dr}{dt}\frac{d\phi }{dt}+\frac{(r^2f)'}{r^2f}\frac{dr}{dt}\frac{d\psi }{dt}\nonumber \\&\quad +\frac{(f-1)\cos ^2\theta -1}{\sin \theta }\frac{d\theta }{dt}\frac{d\phi }{dt} \nonumber \\&\quad +f\cot \theta \frac{d\theta }{dt}\frac{d\psi }{dt}=0. \end{aligned}$$

(2.9)

Throughout the paper, we denote $F(\alpha , k^2)$, $E(\alpha , k^2)$, and $\Pi (h,\alpha , k^2)$ the incomplete elliptic integrals of the first, second, and third kind, where $0<k<1$, $h\in \mathbb {C}$ (cf. Appendix), and

$$\begin{aligned} \varepsilon =\pm 1, \quad r_0= \root 4 \of {\frac{B}{n-1}} \end{aligned}$$

for $B>0$, $n\ge 2.$

3 Geodesics on scalar flat metrics of Eguchi–Hanson type

In this section, we shall solve the geodesic equations for the following cases when f is given by (2.4).

3.1 Geodesics for constant $\theta$, $\phi$, $\psi$

The geodesic equations reduce to

$$\begin{aligned} \frac{d^2r}{dt^2}-\frac{f'}{2f}\left( \frac{dr}{dt}\right) ^{2}=0. \end{aligned}$$

(3.1)

Theorem 3.1

The geodesics for scalar flat metrics of Eguchi–Hanson type with constant $\theta$, $\phi$ and $\psi$ and passing through $r_0$ with conditions

$$\begin{aligned}&\lim \limits _{r\rightarrow r_0} t =t_1,\\&\lim \limits _{r\rightarrow r_0} \frac{1}{\sqrt{f}}\frac{dr}{dt} = r_1 \end{aligned}$$

satisfy

$$\begin{aligned} t(r)&= t_1+\frac{\varepsilon }{r_1} \Big [\frac{\sqrt{r^4+(n-2)r_0^2r^2-B }}{r} \\&\quad -\sqrt{n}r_0 E\left( \alpha ,k^2\right) +\frac{r_0}{\sqrt{n}}F\left( \alpha , k^2\right) \Big ], \end{aligned}$$

where

$$\begin{aligned} \alpha =\arcsin \sqrt{1-\frac{r_0^2}{r^2}}, \quad k^2=1-\frac{1}{n}. \end{aligned}$$

Proof

The geodesic equation (3.1) implies that

$$\begin{aligned} \frac{d}{dt}\Big (\frac{1}{\sqrt{f}}\Big |\frac{dr}{dt}\Big |\Big )=0. \end{aligned}$$

Thus

$$\begin{aligned} \frac{dt}{dr}=\frac{ \varepsilon }{r_1\sqrt{f}}. \end{aligned}$$

Changing variable $u=r^2$, we obtain

$$\begin{aligned} \frac{dt}{du} =\frac{ \varepsilon }{2r_1}\frac{ u}{\sqrt{u\big ( u-r_0^2\big ) \big ( u+\sqrt{(n-1)B}\big ) }}. \end{aligned}$$

The theorem follows by integrating it from $r_0^2$ to $r^2$. $\square$

3.2 Geodesics for constant $\phi$, $\psi$

The geodesic equations reduce to

$$\begin{aligned}&\frac{d^2r}{dt^2} -\frac{f'}{2f}\Big (\frac{dr}{dt}\Big )^{2}\nonumber \\&-\frac{rf}{4}\Big (\frac{d\theta }{dt}\Big )^{2}=0, \end{aligned}$$

(3.2)

$$\begin{aligned}&\frac{d^2\theta }{dt^2}+\frac{2}{r}\frac{dr}{dt}\frac{d\theta }{dt}=0. \end{aligned}$$

(3.3)

Theorem 3.2

Let constant $r_1$, $\theta _0$ satisfy

$$\begin{aligned} r_1>\frac{r_0\theta _0 }{2}. \end{aligned}$$

The geodesics for scalar flat metrics of Eguchi–Hanson type with constant $\phi$, $\psi$ and passing through $r_0$ with conditions

$$\begin{aligned}&\lim \limits _{r\rightarrow r_0} t =t_1, \quad \lim \limits _{r\rightarrow r_0} \theta =\theta _1, \quad \lim \limits _{r\rightarrow r_0}\frac{d\theta }{dt}=\theta _0, \\&\lim \limits _{r\rightarrow r_0} \frac{1}{\sqrt{f}} \left| \frac{dr}{dt} \right| =\sqrt{r_1^2-\frac{r_0^2\theta _0^2}{4}} \end{aligned}$$

satisfy

$$\begin{aligned} t(r)&=t_1+\frac{\varepsilon }{r_1} \Bigg [\frac{2r_1\sqrt{r^4+(n-2)r_0^2r^2-B }}{\sqrt{4r_1^2r^2-r_0^4\theta _0^2}}\\&\quad -\sqrt{n}r_0 E\left( \alpha ,k^2\right) +\frac{r_0}{\sqrt{n}}F\left( \alpha , k^2\right) \Bigg ],\\ \theta (r)&=\theta _1 +\frac{\varepsilon r_0\,\theta _0 }{ \sqrt{n}r_1}F\left( \alpha ,k^2\right) , \end{aligned}$$

where

$$\begin{aligned} \alpha =\arcsin \sqrt{\frac{4r_1^2 r^2-4r_0^2r_1^2}{4r_1^2 r^2-r_0^4\theta _0^2 }},\quad k^2= 1-\frac{1}{n}+\frac{r_0^2\theta _0^2}{4nr_1^2}. \end{aligned}$$

Proof

The geodesic equation (3.3) implies that

$$\begin{aligned} \frac{d}{dt}\ln \Big ( r^2 \Big | \frac{d\theta }{dt}\Big | \Big )=0. \end{aligned}$$

Thus

$$\begin{aligned} \frac{d\theta }{dt}=\frac{\varepsilon r_0^2 \theta _0}{r^{2}}. \end{aligned}$$

Substituting it into (3.2), we obtain

$$\begin{aligned} \frac{d}{dt}\Big (\frac{1}{f} \Big (\frac{dr}{dt}\Big )^{2}+\frac{r_0^4\theta _0^{2}}{4r^{2}}\Big )=0. \end{aligned}$$

Thus

$$\begin{aligned} \frac{1}{f} \Big (\frac{dr}{dt}\Big )^{2}+\frac{r_0^4\theta _0^{2}}{4r^{2}}=r_1^2. \end{aligned}$$

Therefore

$$\begin{aligned} \frac{d\theta }{dr}&=\frac{d\theta }{dt} \Big / \frac{dr}{dt}=\frac{\varepsilon r_0^2\theta _0}{r_1r^2\sqrt{\Big (1-\frac{r_0^4\theta _0^2}{4r_1^2r^2}\Big )f}},\\ \frac{dt}{dr}&=\frac{\varepsilon }{r_1\sqrt{\Big (1-\frac{r_0^4\theta _0^2}{4r_1^2r^2}\Big )f}}. \end{aligned}$$

Changing variable $u=r^2$, we obtain

$$\begin{aligned} \frac{d\theta }{du}&=\frac{\varepsilon r_0^2\theta _0}{2r_1\sqrt{P(u)}},\\ \frac{dt}{du}&=\frac{\varepsilon u}{2r_1\sqrt{P(u)}}, \end{aligned}$$

with

$$\begin{aligned} P(u)=\left( u-\frac{r_0^4\theta _0^2}{4r_1^2}\right) \left( u-r_0^2\right) \left( u+\sqrt{(n-1)B} \right) . \end{aligned}$$

The theorem follows by integrating them from $r_0^2$ to $r^2$. $\square$

3.3 Geodesics for constant $\theta$, $\phi$

The geodesic equations reduce to

$$\begin{aligned}&\frac{d^2r}{dt^2} -\frac{f'}{2f}\left( \frac{dr}{dt}\right) ^{2}- \frac{f\left( r^2f\right) '}{8}\left( \frac{d\psi }{dt}\right) ^{2}=0, \end{aligned}$$

(3.4)

$$\begin{aligned}&\frac{d^2\psi }{dt^2}+\frac{\left( r^{2}f\right) '}{r^{2}f}\frac{dr}{dt}\frac{d\psi }{dt} =0. \end{aligned}$$

(3.5)

Let $r_1$, $\psi _0$ be constant and $r_1\ne 0$. Denote

$$\begin{aligned} r_{+}{} & {} =\frac{1}{2\sqrt{2}}\\{} & {} \sqrt{4(2-n)r_0^2+\frac{\psi _0^2}{r_1^2}+\sqrt{\left( 4(2-n)r_0^2+\frac{\psi _0^2}{r_1^2}\right) ^2+64B}}. \end{aligned}$$

Lemma 3.1

With the above notations, $r_+\ge r_0$, and $r_+=r_0$ if and only if $\psi _0=0$.

Proof

A straightforward computation. $\square$

Theorem 3.3

Let $r \ge r_+> r_0$. The geodesics for scalar flat metrics of Eguchi–Hanson type with constant $\theta$, $\phi$ and conditions

$$\begin{aligned}&\lim \limits _{r\rightarrow r_+} t =t_1, \quad \lim \limits _{r\rightarrow r_+} \psi =\psi _1, \quad \lim \limits _{r\rightarrow r_+}&r^2 f \frac{d\psi }{dt}=\psi _0, \\&\lim \limits _{r\rightarrow r_+} \left( \frac{1}{f}\left( \frac{dr}{dt}\right) ^{2}+\frac{\psi _0^{2}}{4r^{2}f}\right) =r_1^2 \end{aligned}$$

satisfy

$$\begin{aligned} t(r)&=t_1+ \frac{\sqrt{r^{4}-\left( \frac{\psi _0^2}{4r_1^2}+(2-n)r_0^2\right) r^2-B }}{\varepsilon r_1 r}\\&\quad -\frac{\root 4 \of {\left( 4(2-n)r_0^2r_1^2+\psi _0^2\right) ^2+64Br_1^4}}{2\varepsilon r_1^2} E\left( \alpha ,k^2\right) \\&\quad +\frac{2\varepsilon r_+^2}{\root 4 \of {\left( 4(2-n)r_0^2r_1^2+\psi _0^2\right) ^2+64Br_1^4}}F\left( \alpha , k^2\right) ,\\ \psi (r)&=\psi _1+\frac{2\varepsilon \psi _0\,r_{+}^2\,\Pi (h_1,\alpha ,k^2)}{n(r_{+}^2-r_0^2)\root 4 \of {\left( 4(2-n)r_0^2r_1^2+\psi _0^2\right) ^2+64Br_1^4}} \nonumber \\&\quad +\frac{2\varepsilon (n-1)\psi _0\,r_{+}^2\,\Pi (h_2,\alpha ,k^2)}{n(r_{+}^2+\sqrt{(n-1)B})\root 4 \of {\left( 4(2-n)r_0^2r_1^2+\psi _0^2\right) ^2+64Br_1^4}} , \end{aligned}$$

where

$$\begin{aligned} \alpha&=\arcsin \sqrt{1-\frac{4(2-n)r_0^2r_1^2+\psi _0^2+\sqrt{\left( 4(2-n)r_0^2r_1^2+\psi _0^2\right) ^2+64Br_1^4}}{8r_1^2r^2}},\\ k^2&=\frac{1}{2}\left( 1- \frac{4(2-n)r_0^2r_1^2+\psi _0^2}{\sqrt{\left( 4(2-n)r_0^2r_1^2+\psi _0^2\right) ^2+64Br_1^4}}\right) ,\\ h_1&=\frac{-8r_0^2r_1^2}{\psi _0^2-4nr_0^2r_1^2+\sqrt{\left( 4(2-n)r_0^2r_1^2+\psi _0^2\right) ^2+64Br_1^4}} ,\\ h_2&=\frac{8\sqrt{(n-1)B}r_1^2}{\psi _0^2+4nr_0^2r_1^2+\sqrt{\left( 4(2-n)r_0^2r_1^2+\psi _0^2\right) ^2+64Br_1^4}}. \end{aligned}$$

Proof

The geodesic equation (3.5) implies that

$$\begin{aligned} \frac{d}{dt}\ln \left( r^2f\left| \frac{d\psi }{dt}\right| \right) =0. \end{aligned}$$

Thus

$$\begin{aligned} \frac{d\psi }{dt}=\frac{\varepsilon \psi _0}{r^{2}f}. \end{aligned}$$

(3.6)

Substituting it into (3.4), we obtain

$$\begin{aligned} \frac{d}{dt}\left( \frac{1}{f}\left( \frac{dr}{dt}\right) ^{2}+\frac{\psi _0^{2}}{4r^{2}f}\right) =0. \end{aligned}$$

Thus

$$\begin{aligned} \frac{1}{f}\left( \frac{dr}{dt}\right) ^{2}+\frac{\psi _0^{2}}{4r^{2}f}=r_1^2. \end{aligned}$$

Therefore

$$\begin{aligned} \frac{d\psi }{dr}&=\frac{d\psi }{dt} \Big / \frac{dr}{dt}=\frac{\varepsilon \psi _0}{r_1r^2f\sqrt{f-\frac{\psi _0^2}{4r_1^2r^2}}},\\ \frac{dt}{dr}&=\frac{\varepsilon }{r_1\sqrt{f-\frac{\psi _0^2}{4r_1^2r^2}}}. \end{aligned}$$

Changing variable $u=r^2$, we obtain

$$\begin{aligned} \frac{d\psi }{du}&=\frac{\varepsilon \psi _0\, u^2}{2r_1\left( u-r_0^2\right) \left( u+\sqrt{(n-1)B}\right) \sqrt{P(u)}}, \end{aligned}$$

(3.7)

$$\begin{aligned} \frac{dt}{du}&=\frac{\varepsilon u}{2r_1 \sqrt{P(u)}}, \end{aligned}$$

(3.8)

with

$$\begin{aligned} P(u)&= u^3-\left( \frac{\psi _0^2}{4r_1^2}+(2-n)r_0^2\right) u^2-Bu \\&= u\left( u-u_0\right) \left( u-u_1\right) , \end{aligned}$$

where

$$\begin{aligned}&u_0=\frac{4(2-n)r_0^2r_1^2+\psi _0^2+\sqrt{\left( 4(2-n)r_0^2r_1^2+\psi _0^2\right) ^2+64Br_1^4}}{8r_1^2},\\&u_1=\frac{4(2-n)r_0^2r_1^2+\psi _0^2-\sqrt{\left( 4(2-n)r_0^2r_1^2+\psi _0^2\right) ^2+64Br_1^4}}{8r_1^2}. \end{aligned}$$

The theorem follows by integrating (3.7) and (3.8) from $u_0$ to $r^2$. $\square$

Theorem 3.4

The geodesics for scalar flat metrics of Eguchi–Hanson type with constant $\theta$, $\phi$ and passing through $r_0$ with conditions

$$\begin{aligned} \lim \limits _{r\rightarrow r_0} t =t_1, \quad \lim \limits _{r\rightarrow r_0} \psi =\psi _1, \quad \lim \limits _{r\rightarrow r_0} \frac{1}{\sqrt{f}}\frac{dr}{dt} = r_1 \end{aligned}$$

satisfy

$$\begin{aligned} t(r)&= t_1+\frac{\varepsilon }{r_1} \left[ \frac{\sqrt{r^4+(n-2)r_0^2r^2-B }}{r}\right. \\&\left. -\sqrt{n}r_0 E\left( \alpha ,k^2\right) +\frac{r_0}{\sqrt{n}}F\left( \alpha , k^2\right) \right] ,\\ \psi (r)&= \psi _1, \end{aligned}$$

where

$$\begin{aligned} \alpha =\arcsin \sqrt{1-\frac{r_0^2}{r^2}}, \quad k^2=1-\frac{1}{n}. \end{aligned}$$

Proof

Lemma 3.1 implies that $\psi _0=0$. Then (3.6) gives that $\frac{d \psi }{dt} =0$, and it reduces to Theorem 3.1. $\square$

3.4 Geodesics for constant $\theta \in (0, \pi )$, $\psi$

The geodesic equations (2.7), (2.9) give that

$$\begin{aligned} \left( f-1\right) \sin \theta \cos \theta \left( \frac{d\phi }{dt}\right) ^{2}=0,\quad \frac{f'}{f}\cos \theta \frac{dr}{dt}\frac{d\phi }{dt}=0. \end{aligned}$$

They imply either $\dfrac{d\phi }{dt}=0$ or $\theta =\dfrac{\pi }{2}$. It reduces to Theorem 3.1 if $\dfrac{d\phi }{dt}=0$.

Now we focus on the case $\theta =\dfrac{\pi }{2}$. The geodesic equations reduce to

$$\begin{aligned}&\frac{d^2r}{dt^2}-\frac{f'}{2f}\left( \frac{dr}{dt}\right) ^{2}\nonumber \\&\quad -\frac{rf}{4}\left( \frac{d\phi }{dt}\right) ^{2}=0, \end{aligned}$$

(3.9)

$$\begin{aligned} \frac{d^2\phi }{dt^2}&+\frac{2}{r}\frac{dr}{dt}\frac{d\phi }{dt}=0, \end{aligned}$$

(3.10)

Theorem 3.5

Let constant $r_1$, $\phi _0$ satisfy

$$\begin{aligned} r_1>\frac{r_0\phi _0 }{2}. \end{aligned}$$

The geodesics for scalar flat metrics of Eguchi–Hanson type with constant $\theta \in (0,\pi )$, $\psi$, nonconstant $\phi$ and passing through $r_0$ with conditions

$$\begin{aligned}&\lim \limits _{r\rightarrow r_0} t =t_1, \quad \lim \limits _{r\rightarrow r_0} \phi =\phi _1, \quad \lim \limits _{r\rightarrow r_0}\frac{d\phi }{dt}=\phi _0, \\&\lim \limits _{r\rightarrow r_0} \frac{1}{\sqrt{f}} \left| \frac{dr}{dt} \right| =\sqrt{r_1^2-\frac{r_0^2\phi _0^2}{4}} \end{aligned}$$

satisfy

$$\begin{aligned} t(r)&= t_1+\frac{\varepsilon }{r_1} \Big [\frac{2r_1\sqrt{r^4+(n-2)r_0^2r^2-B }}{\sqrt{4r_1^2r^2-r_0^4\theta _0^2}}\\&\quad -\sqrt{n}r_0 E\left( \alpha ,k^2\right) +\frac{r_0}{\sqrt{n}}F\left( \alpha , k^2\right) \Big ],\\ \theta (r)&=\frac{\pi }{2}, \\ \phi (r)&= \phi _1 +\frac{\varepsilon r_0\,\phi _0 }{ \sqrt{n}r_1}F\left( \alpha ,k^2\right) , \end{aligned}$$

where

$$\begin{aligned} \alpha =\arcsin \sqrt{\frac{4r_1^2 r^2-4r_0^2r_1^2}{4r_1^2 r^2-r_0^4\phi _0^2 }},\quad k^2= 1-\frac{1}{n}+\frac{r_0^2\phi _0^2}{4nr_1^2} \end{aligned}$$

Proof

Same as the proof of Theorem 3.2. $\square$

3.5 Geodesics for constant r

The geodesic equations reduce to

$$\begin{aligned}&\frac{d^2\theta }{dt^2}+\left( f-1\right) \sin \theta \cos \theta \left( \frac{d\phi }{dt}\right) ^{2}\nonumber \\&\quad +f\sin \theta \frac{d\phi }{dt}\frac{d\psi }{dt}=0 , \end{aligned}$$

(3.11)

$$\begin{aligned}&\frac{d^2\phi }{dt^2}-\left( f-2\right) \cot \theta \frac{d\phi }{dt}\frac{d\theta }{dt}\nonumber \\&\quad -\frac{f}{\sin \theta }\frac{d\theta }{dt}\frac{d\psi }{dt}=0, \end{aligned}$$

(3.12)

$$\begin{aligned}&\frac{d^2\psi }{dt^2}+\frac{\left( f-1\right) \cos ^2\theta -1}{\sin \theta }\frac{d\theta }{dt}\frac{d\phi }{dt}\nonumber \\&\quad +f\cot \theta \frac{d\theta }{dt}\frac{d\psi }{dt}=0, \end{aligned}$$

(3.13)

$$\begin{aligned}&\frac{r}{4}f\left( \frac{d\theta }{dt}\right) ^{2}+\frac{r}{4}f\sin ^2\theta \left( \frac{d\phi }{dt}\right) ^{2}=0, \end{aligned}$$

(3.14)

As $r=r_0$ is the coordinate origin, we assume $r>r_0$.

Theorem 3.6

The geodesics for scalar flat metrics of Eguchi–Hanson type with constant $r>r_0$ and conditions

$$\begin{aligned}&\lim \limits _{r\rightarrow r_0} \theta =\theta _1, \quad \lim \limits _{r\rightarrow r_0} \phi =\phi _1, \quad \lim \limits _{r\rightarrow r_0} \psi =\psi _1, \quad \\&\lim \limits _{r\rightarrow r_0} \frac{d\psi }{dt} =\psi _0 \end{aligned}$$

satisfy

$$\begin{aligned} \theta (t)=\theta _1, \quad \phi (t)=\phi _1, \quad \psi (t)=\psi _1 +\psi _0 t. \end{aligned}$$

Proof

The geodesic equation (3.14) implies that $\dfrac{d\theta }{dt}=\dfrac{d\phi }{dt}=0$. Then (3.13) gives $\dfrac{d^2\psi }{dt^2}=0$ and the theorem follows. $\square$

3.6 Geodesics for constant $\phi$

The geodesic equation (2.8) gives

$$\begin{aligned} \frac{f}{\sin \theta }\frac{d\theta }{dt}\frac{d\psi }{dt}=0. \end{aligned}$$

Therefore $\dfrac{d\psi }{dt}=0$ or $\dfrac{d\theta }{dt}=0$, and it reduces to the theorems in Sect. 3.2 or Sect. 3.3 respectively.

4 Geodesics for constant $\theta$ on Eguchi–Hanson metrics

In this section, we solve the geodesic equations for constant $\theta \in \left( 0, \pi \right)$ on Eguchi–Hanson metrics with f given by (2.2). The geodesic equations reduce to

$$\begin{aligned}&\frac{d^2r}{dt^2}-\frac{f'}{2f}\left( \frac{dr}{dt}\right) ^{2}-\frac{f}{4}\left( r\sin ^2\theta +\frac{\left( r^2f \right) ' }{2}\cos ^2\theta \right) \left( \frac{d\phi }{dt}\right) ^{2} \nonumber \\&\quad -\frac{f\left( r^2f \right) '}{4}\cos \theta \frac{d\phi }{dt}\frac{d\psi }{dt} -\frac{f\left( r^2f \right) '}{8}\left( \frac{d\psi }{dt}\right) ^{2}=0, \end{aligned}$$

(4.1)

$$\begin{aligned}&f\frac{d\phi }{dt}\frac{d\psi }{dt}+(f-1) \cos \theta \left( \frac{d\phi }{dt}\right) ^{2}=0, \end{aligned}$$

(4.2)

$$\begin{aligned}&\frac{d^2\phi }{dt^2}+\frac{2}{r}\frac{dr}{dt}\frac{d\phi }{dt}=0, \end{aligned}$$

(4.3)

$$\begin{aligned}&\frac{d^2\psi }{dt^2} +\frac{f'}{f}\cos \theta \frac{dr}{dt}\frac{d\phi }{dt}+\frac{\left( r^2f\right) '}{r^2f}\frac{dr}{dt}\frac{d\psi }{dt}=0. \end{aligned}$$

(4.4)

Equation (4.2) implies that either $\dfrac{d\phi }{dt}=0$ or

$$\begin{aligned} \left( f-1\right) \cos \theta \frac{d\phi }{dt}+f\frac{d\psi }{dt}=0. \end{aligned}$$

(4.5)

It reduces to the theorems in Sect. 3.3 if $\dfrac{d\phi }{dt}=0$.

Now we focus on the case $\dfrac{d\phi }{dt}\ne 0$. Then (4.5) holds.

Lemma 4.1

Let function T(x)

$$\begin{aligned} T(x)=\frac{x \left( 27\sin ^2\theta - 2x^2 -9\right) }{2(x^2+3)^\frac{3}{2}}, \quad x \ge 1 \end{aligned}$$

(4.6)

for fixed $\theta \in (0, \pi )$, then it is monotonically decaying and

$$\begin{aligned} -1<T(x)\le 1, \quad x=1 \Longleftrightarrow \theta =\frac{\pi }{2}. \end{aligned}$$

(4.7)

Proof

It is straightforward that

$$\begin{aligned} T'(x)=-\frac{27\left[ (2x^2-3)\sin ^2 \theta +1\right] }{2 (x^2+3)^\frac{5}{2}}\le 0, \end{aligned}$$

and

$$\begin{aligned} T(1)=\frac{27 \sin ^2 \theta -11}{16} \le 1,\quad T(\infty )=-1. \end{aligned}$$

Therefore the lemma follows. $\square$

Let $r_1$, $\phi _0$ be constant such that

$$\begin{aligned} r_1\ne 0, \quad \left| \phi _0\right| \ge \frac{2|r_1|}{\root 4 \of {B}}. \end{aligned}$$

(4.8)

By Lemma 4.1, we can define

$$\begin{aligned} T=T\left( \frac{\sqrt{B}\phi _0 ^2}{4r_1 ^2} \right) =\frac{\sqrt{B}\phi _0^2\left( 216r_1^4\sin ^2\theta -B\phi _0^4-72r_1^4\right) }{\sqrt{(B\phi _0^4+48r_1^4)^3}}. \end{aligned}$$

and

$$\begin{aligned}&\bar{r}_{0}=\frac{1}{2\sqrt{3}} \sqrt{B\frac{\phi _0^2}{r_1^2}-2\sqrt{B^2\frac{\phi _0^4}{r_1^4}+48B}\cos \left( \frac{\eta }{3}+ \frac{2\pi }{3}\right) }, \\&\quad \eta =\arccos T. \end{aligned}$$

Lemma 4.2

Suppose that (4.8) holds, then

$$\begin{aligned} \bar{r} _0\ge \root 4 \of {B}. \end{aligned}$$

Equality holds if and only if

$$\begin{aligned} \left| \phi _0\right| =\frac{2|r_1|}{\root 4 \of {B}},\quad \theta =\frac{\pi }{2}. \end{aligned}$$

Proof

It is obvious that $\eta \in [0, \pi )$. A straightforward calculation shows that

$$\begin{aligned} -1< \cos \left( \frac{\eta }{3}+ \frac{2\pi }{3}\right) \le -\frac{1}{2}. \end{aligned}$$

Thus

$$\begin{aligned} \bar{r}_{0}&=\frac{1}{2\sqrt{3}} \sqrt{B\frac{\phi _0^2}{r_1^2}-2\sqrt{B^2\frac{\phi _0^4}{r_1^4}+48B}\cos \left( \frac{\eta }{3}+ \frac{2\pi }{3}\right) }\\&\ge \frac{1}{2\sqrt{3}} \sqrt{B\frac{\phi _0^2}{r_1^2}+\sqrt{B^2\frac{\phi _0^4}{r_1^4}+48B}}\\&\ge \root 4 \of {B}. \end{aligned}$$

Equality holds if and only if

$$\begin{aligned} \frac{\phi _0^2}{r_1^2}=\frac{4}{\sqrt{B}}, \quad \eta =0 \Longleftrightarrow \left| \phi _0\right| =\frac{2|r_1|}{\root 4 \of {B}},\quad \theta =\frac{\pi }{2}. \end{aligned}$$

Therefore, the lemma follows. $\square$

Theorem 4.1

Let $r \ge \bar{r}_0 >\root 4 \of {B}$. The geodesics for Eguchi–Hanson metrics with only constant $\theta \in \left( 0, \pi \right)$ and conditions

$$\begin{aligned}&\lim \limits _{r\rightarrow \bar{r}_0} t =t_1, \quad \lim \limits _{r\rightarrow \bar{r} _0}\phi =\phi _1, \quad \lim \limits _{r\rightarrow \bar{r} _0}\psi =\psi _1, \\&\lim \limits _{r\rightarrow \bar{r} _0} r^2\frac{d\phi }{dt} =\sqrt{B}\phi _0 ,\\&\lim \limits _{r\rightarrow \bar{r} _0} \left( \frac{1}{f}\left( \frac{dr}{dt}\right) ^{2} +\frac{B\phi _0^2\cos ^2\theta }{4r^2f}\right) = r_1^2-\frac{B\phi _0^2\sin ^2\theta }{4\bar{r} _0^2} \end{aligned}$$

satisfy

$$\begin{aligned} t(r)&= t_1+ \frac{\sqrt{r^4-\left( \frac{B\phi _0^2}{4r_1^2}-r_-\right) r^2-\frac{B^2\phi _0^2\sin ^2\theta }{4r_1^2r_-}}}{\varepsilon r_1 \sqrt{r^2-r_-}}\\&\quad -\frac{\root 4 \of {16Br_1^4+\frac{B^2\phi _0^4}{3}}\sqrt{\sin \left( \frac{\eta }{3}+\frac{\pi }{3}\right) }}{\varepsilon \sqrt{2} r_1^2} E\left( \alpha ,k^2\right) \\&\quad +\frac{\varepsilon \sqrt{2} \bar{r}_0^2}{\root 4 \of {16Br_1^4+\frac{B^2\phi _0^4}{3}}\sqrt{\sin \left( \frac{\eta }{3}+\frac{\pi }{3}\right) }}F\left( \alpha , k^2\right) ,\\ \phi (r)&= \phi _1 +\frac{ \varepsilon \sqrt{2B}\phi _0 }{\root 4 \of {16Br_1^4+\frac{B^2\phi _0^4}{3}}\sqrt{\sin \left( \frac{\eta }{3}+\frac{\pi }{3}\right) }}F\left( \alpha ,k^2\right) ,\\ \psi (r)&= \psi _1 +\frac{ \varepsilon \sqrt{2}B^{\frac{3}{2}}\phi _0\cos \theta \,F(\alpha ,k^2)}{(r_-^2-B)\root 4 \of {16Br_1^4+\frac{B^2\phi _0^4}{3}}\sqrt{\sin \left( \frac{\eta }{3}+\frac{\pi }{3}\right) }} \nonumber \\&\quad -\frac{ \varepsilon \sqrt{2} B\,\phi _0\,\root 4 \of {16Br_1^4+\frac{B^2\phi _0^4}{3}}\,\cos \theta \,\sin \frac{\eta }{3}\,\Pi (h_1,\alpha ,k^2)}{(r_--\sqrt{B})(\bar{r} _0^2-\sqrt{B})\sqrt{\sin \left( \frac{\eta }{3}+\frac{\pi }{3}\right) }} \nonumber \\&\quad +\frac{ \varepsilon \sqrt{2}B\,\phi _0\root 4 \of {16Br_1^4+\frac{B^2\phi _0^4}{3}}\,\cos \theta \,\sin \frac{\eta }{3}\,\Pi (h_2,\alpha ,k^2)}{(r_-+\sqrt{B})(\bar{r} _0^2+\sqrt{B})\sqrt{\sin \left( \frac{\eta }{3}+\frac{\pi }{3}\right) }} , \end{aligned}$$

where

$$\begin{aligned} r_-&= \frac{1}{12r_1^2}\left( B\phi _0^2-2\sqrt{B^2\phi _0^4+48Br_1^4}\cos \left( \frac{\eta }{3}-\frac{2\pi }{3}\right) \right) ,\\ \alpha&= \arcsin \sqrt{\frac{ 12r_1^2r^2-B\phi _0^2+2\sqrt{B^2\phi _0^4+48Br_1^4}\cos \left( \frac{\eta }{3}+\frac{2\pi }{3}\right) }{12r_1^2r^2-B\phi _0^2+2\sqrt{B^2\phi _0^4+48Br_1^4}\cos \left( \frac{\eta }{3}-\frac{2\pi }{3}\right) }},\\ k^2&= \frac{\sin \left( \frac{\eta }{3}+\frac{\pi }{3}\right) }{\sin \left( \frac{\pi }{3}-\frac{\eta }{3}\right) },\\ h_1&= \frac{B\phi _0^2-2\sqrt{B^2\phi _0^4+48Br_1^4}\cos \left( \frac{\eta }{3}-\frac{2\pi }{3}\right) -12\sqrt{B}r_1^2}{B\phi _0^2-2\sqrt{B^2\phi _0^4+48Br_1^4}\cos \left( \frac{\eta }{3}+\frac{2\pi }{3}\right) -12\sqrt{B}r_1^2} , \\ h_2&= \frac{B\phi _0^2-2\sqrt{B^2\phi _0^4+48Br_1^4}\cos \left( \frac{\eta }{3}-\frac{2\pi }{3}\right) +12\sqrt{B}r_1^2}{B\phi _0^2-2\sqrt{B^2\phi _0^4+48Br_1^4}\cos \left( \frac{\eta }{3}+\frac{2\pi }{3}\right) +12\sqrt{B}r_1^2}. \end{aligned}$$

Proof

The geodesic equation (4.3) implies that

$$\begin{aligned} \frac{d}{dt}\ln \Big ( r^2 \Big | \frac{d\phi }{dt}\Big | \Big )=0. \end{aligned}$$

Thus

$$\begin{aligned} \frac{d\phi }{dt}=\frac{ \varepsilon \sqrt{B}\phi _0}{r^{2}}. \end{aligned}$$

(4.9)

By (4.5) and (4.9), we derive

$$\begin{aligned} \frac{d\psi }{dt}=\frac{ \varepsilon \sqrt{B}\phi _0\cos \theta (1-f)}{r^{2}f}. \end{aligned}$$

(4.10)

Substituting (4.9) and (4.10) into (4.1), we obtain

$$\begin{aligned} \frac{d}{dt}\left( \frac{1}{f} \left( \frac{dr}{dt}\right) ^{2}+ \frac{B\phi _0^2\sin ^2\theta }{4r^2}+\frac{B\phi _0^2\cos ^2\theta }{4r^2f} \right) =0, \end{aligned}$$

Thus

$$\begin{aligned} \frac{1}{f}\left( \frac{dr}{dt}\right) ^{2}+\frac{B\phi _0^2\sin ^2\theta }{4r^2}+\frac{B\phi _0^2\cos ^2\theta }{4r^2f}=r_1^2. \end{aligned}$$

Therefore

$$\begin{aligned} \frac{d\phi }{dr}&=\frac{d\phi }{dt} \Big / \frac{dr}{dt}=\frac{\varepsilon \sqrt{B}\phi _0}{r_1r^2}\frac{1}{\sqrt{f-\frac{B\phi _0^2\sin ^2\theta }{4r_1^2r^2}f-\frac{B\phi _0^2\cos ^2\theta }{4r_1^2r^2}}},\\ \frac{d\psi }{dr}&=\frac{d\psi }{dt} \Big / \frac{dr}{dt}\\&=\frac{\varepsilon \sqrt{B}\phi _0 (1-f)\cos \theta }{r_1r^2f}\frac{1}{\sqrt{f-\frac{B\phi _0^2\sin ^2\theta }{4r_1^2r^2}f-\frac{B\phi _0^2\cos ^2\theta }{4r_1^2r^2}}},\\ \frac{dt}{dr}&=\frac{\varepsilon }{r_1}\frac{1}{\sqrt{f-\frac{B\phi _0^2\sin ^2\theta }{4r_1^2r^2}f-\frac{B\phi _0^2\cos ^2\theta }{4r_1^2r^2}}}. \end{aligned}$$

Changing variable $u=r^2$, we obtain

$$\begin{aligned} \frac{d\phi }{du}&=\frac{\varepsilon \sqrt{B}\phi _0}{2r_1\sqrt{P(u)}},\end{aligned}$$

(4.11)

$$\begin{aligned} \frac{d\psi }{du}&=\frac{\varepsilon B^{\frac{3}{2}}\phi _0\cos \theta }{2r_1(u^2-B)}\frac{1}{\sqrt{P(u)}},\end{aligned}$$

(4.12)

$$\begin{aligned} \frac{dt}{du}&=\frac{\varepsilon u}{2r_1\sqrt{P(u)}}, \end{aligned}$$

(4.13)

with

$$\begin{aligned} P(u)&=u^3-\frac{B\phi _0^2}{4r_1^2}u^2-Bu+\frac{B^2\phi _0^2}{4r_1^2}\sin ^2\theta \nonumber \\&=(u-u_2)(u-u_3)(u-u_4). \end{aligned}$$

The cubic equation $P(u)=0$ can be denoted in canonical form [6]

$$\begin{aligned} w^3+pw+q=0, \end{aligned}$$

with

$$\begin{aligned}&p=-B\left( 1+\frac{B\phi _0^4}{48r_1^4}\right) ,\\&q=\frac{B^2\phi _0^2}{4r_1^2}\left( \sin ^2\theta -\frac{1}{3}-\frac{B\phi _0^4}{216r_1^4}\right) . \end{aligned}$$

Thus the discriminant is

$$\begin{aligned} \Delta&=-(4p^3+27q^2)\\&=\frac{B^5\phi _0^8}{256r_1^8}\left( 4\left( \frac{16r_1^4}{B\phi _0^4} \right) ^2\right. \\&\quad \left. -(27\sin ^4\theta -18\sin ^2\theta -1)\frac{16r_1^4}{B\phi _0^4}+4 \sin ^2\theta \right) . \end{aligned}$$

Denote

$$\begin{aligned}{} & {} Y=4m^2-(27\sin ^4\theta -18\sin ^2\theta -1)m+4 \sin ^2\theta ,\\{} & {} m=\frac{16r_1^4}{B\phi _0^4}>0.\end{aligned}$$

It is obvious that

$$\begin{aligned} \theta \ne \frac{\pi }{2} \Longrightarrow Y>0 \Longrightarrow \Delta >0. \end{aligned}$$

Therefore, the algebraic equation $P(u)=0$ has three real roots as follows

$$\begin{aligned}&u_2=\frac{1}{12r_1^2}\left( B\phi _0^2-2\sqrt{B^2\phi _0^4+48Br_1^4}\cos \left( \frac{\eta }{3}+\frac{2\pi }{3}\right) \right) ,\\&u_3=\frac{1}{12r_1^2}\left( B\phi _0^2-2\sqrt{B^2\phi _0^4+48Br_1^4}\cos \left( \frac{\eta }{3}-\frac{2\pi }{3}\right) \right) ,\\&u_4=\frac{1}{12r_1^2}\left( B\phi _0^2-2\sqrt{B^2\phi _0^4+48Br_1^4}\cos \frac{\eta }{3} \right) . \end{aligned}$$

The theorem follows by integrating (4.11)–(4.13) from $\bar{r}_0^2$ to $r^2$ respectively.

Theorem 4.2

Let constant $r_1$, $\phi _0$ satisfy

$$\begin{aligned} r_1>\frac{\root 4 \of {B}\phi _0 }{2}. \end{aligned}$$

The geodesics for Eguchi–Hanson metrics constant $\theta \in (0, \pi )$, nonconstant $\phi$ and passing through $\root 4 \of {B}$ with conditions

$$\begin{aligned}&\lim \limits _{r\rightarrow \root 4 \of {B}} t =t_1, \quad \lim \limits _{r\rightarrow \root 4 \of {B}} \phi =\phi _1,\\&\lim \limits _{r\rightarrow \root 4 \of {B}} \psi =\psi _1,\quad \lim \limits _{r\rightarrow \root 4 \of {B}} \frac{d\phi }{dt}=\phi _0, \\&\lim \limits _{r\rightarrow \root 4 \of {B}} \frac{1}{\sqrt{f}}\left| \frac{dr}{dt} \right| =\sqrt{r_1^2-\frac{\sqrt{B}\phi _0^2}{4}} \end{aligned}$$

satisfy

$$\begin{aligned} t(r)&= t_1+\frac{\varepsilon }{r_1} \Big [\frac{2r_1\sqrt{r^4-B }}{\sqrt{4r_1^2r^2-r_0^4\theta _0^2}}\\&\quad -\sqrt{2}r_0 E\left( \alpha ,k^2\right) +\frac{r_0}{\sqrt{2}}F\left( \alpha , k^2\right) \Big ],\\ \theta (r)&= \frac{\pi }{2},\\ \phi (r)&= \phi _1 +\frac{\varepsilon \root 4 \of {B}\,\phi _0 }{ \sqrt{2}r_1}F\left( \alpha ,k^2\right) ,\\ \psi (r)&= \psi _1,\\ \end{aligned}$$

where

$$\begin{aligned} \alpha =\arcsin \sqrt{\frac{4r_1^2 r^2-4\sqrt{B}r_1^2}{4r_1^2 r^2-B\phi _0^2 }},\quad k^2= \frac{1}{2}+\frac{\sqrt{B}\phi _0^2}{8r_1^2}. \end{aligned}$$

Proof

Lemma 4.2 implies that $\theta =\frac{\pi }{2}$. Then (4.10) gives that $\frac{d \psi }{dt}=0$, and it reduces to Theorem 3.5. $\square$

Appendix: Elliptic integrals

In [8], the incomplete elliptic integrals of the first, the second, and the third kinds are given by

$$\begin{aligned} F(\alpha , k^2)&=\int _{0}^{\alpha }\frac{d\varrho }{\sqrt{1-k^2\sin ^2\varrho }},\\ E(\alpha , k^2)&=\int _{0}^{\alpha }\sqrt{1-k^2\sin ^2\varrho }\,d\varrho ,\\ \Pi (h,\alpha , k^2)&=\int _{0}^{\alpha }\frac{d\varrho }{(1-h\sin ^2\varrho )\sqrt{1-k^2\sin ^2\varrho }}, \end{aligned}$$

respectively, where $0<k<1$, $h\in \mathbb {C}$.

The following propositions can be derived straightforwardly.

Proposition 5.1

Let $a,b,c,x\in \textrm{R}$, $x>a>b>c$, and denote

$$\begin{aligned} f_1(z)=\frac{1}{\sqrt{(z-a)(z-b)(z-c)}}, \end{aligned}$$

then

$$\begin{aligned} \int _{a}^{x}f_1(z)dz=\frac{2}{\sqrt{a-c}}F\left( \alpha ,k^2\right) , \end{aligned}$$

where

$$\begin{aligned} \alpha = \arcsin \sqrt{\frac{x-a}{x-b}}, \quad k^2= \frac{b-c}{a-c}. \end{aligned}$$

Proposition 5.2

Let $a,b,c,x\in \textrm{R}$, $x>a>b>c$, and denote

$$\begin{aligned} f_2(z)=\frac{z}{\sqrt{(z-a)(z-b)(z-c)}}, \end{aligned}$$

then

$$\begin{aligned} \int _{a}^{x}f_2(z)dz&=\frac{2\sqrt{(x-a)(x-c)}}{\sqrt{x-b}}-2\sqrt{a-c}E\left( \alpha ,k^2\right) \\&\quad +\frac{2a}{\sqrt{a-c}}F\left( \alpha ,k^2\right) , \end{aligned}$$

where

$$\begin{aligned} \alpha = \arcsin \sqrt{\frac{x-a}{x-b}}, \quad k^2= \frac{b-c}{a-c}. \end{aligned}$$

Proposition 5.3

Let $a,b,c,d, x\in \textrm{R}$, $x>a>b>c$, $a>d$, and denote

$$\begin{aligned} f_3(z)=\frac{1}{(z^2-d^2)\sqrt{(z-a)(z-b)(z-c)}}, \end{aligned}$$

then

$$\begin{aligned} \int _{a}^{x}f_3(z)dz&=\frac{2F(\alpha ,k^2)}{(b^2-d^2)\sqrt{a-c}}-\frac{(a-b)\Pi (h_1,\alpha ,k^2)}{d(b-d)(a-d)\sqrt{a-c}}\\&\quad +\frac{(a-b)\Pi (h_2,\alpha ,k^2)}{d(b+d)(a+d)\sqrt{a-c}}, \end{aligned}$$

where

$$\begin{aligned}&\alpha = \arcsin \sqrt{\frac{x-a}{x-b}}, \quad k^2= \frac{b-c}{a-c}, \quad h_1=\frac{b-d}{a-d},\\&h_2=\frac{b+d}{a+d}. \end{aligned}$$

Proposition 5.4

Let $a_1, a_2, a_3, a_4, a_5, x\in \textrm{R}$, $x>a_1>a_2>a_3$, $a_1>a_4 \ge a_5$, and denote

$$\begin{aligned} f_4(z)=\frac{z^2}{(z-a_4)(z-a_5)\sqrt{(z-a_1)(z-a_2)(z-a_3)}}, \end{aligned}$$

then

$$\begin{aligned} \int _{a}^{x}f_4(z)dz&=\frac{2a_2^2}{(a_2-a_4)(a_2-a_5)\sqrt{a_1-a_3}}F(\alpha ,k^2)\nonumber \\&\quad -\frac{2a_4^2(a_1-a_2)}{(a_4-a_5)(a_1-a_4)(a_2-a_4)\sqrt{a_1-a_3}}\Pi (h_1,\alpha ,k^2)\nonumber \\&\quad +\frac{2a_5^2(a_1-a_2)}{(a_4-a_5)(a_1-a_5)(a_2-a_5)\sqrt{a_1-a_3}}\Pi (h_2,\alpha ,k^2), \end{aligned}$$

where

$$\begin{aligned}&\alpha = \arcsin \sqrt{\frac{x-a_1}{x-a_2}}, \quad k^2= \frac{a_2-a_3}{a_1-a_3}, \\&h_1=\frac{a_2-a_4}{a_1-a_4}, \quad h_2=\frac{a_2-a_5}{a_1-a_5}. \end{aligned}$$