# Timelike geodesics of 2-dimensional de Sitter spacetime

The de Sitter spacetime in two dimensions ($$\text{dS}_2$$) can be spanned by two coordinates : $$\tau$$ and $$\phi$$ (can be viewed as a cylinder). The metric is then defined as follows : $$ds^2=-d\tau^2+\cosh^2\tau~d\phi^2.$$ In order to compute the timelike geodesics, I tried two methods:

1. First, I use the very convenient but not very rigorous method of "dividing" both sides of the metric by $$ds^2$$. I then get $$1 = -\dot{\tau}^2+\cosh^2\tau~\dot{\phi}^2.$$ Since there a choice a parametrization and that I 'm looking for curves that are of the form $$\phi(\tau)$$, I can take $$\tau(s)=s$$. This reduce the equation to simple ODE for $$\phi(s)$$: $$\dot{\phi}(\tau)=\pm\frac{\sqrt{2}}{\cosh\tau}$$ and the solution is given by $$\phi(\tau)=\pm 2\sqrt{2}\arctan\left( \tanh\left( \frac{\tau}{2} \right) \right).$$
2. Then, I tried the more rigorous way : using Euler-Lagrange equations. I get the following system: $$\begin{cases} &\ddot{\tau}+\cosh\tau\sinh\tau~\dot{\phi}^2 = 0,\\ &\ddot{\phi}+\tanh\tau~\dot{\phi}\dot{\tau} = 0. \end{cases}$$ This time, if I take $$\tau(s)=s$$, the first equation gives $$\dot{\phi}=0$$ and the second equation is automatically satisfied.

Finally, here are my questions:

• In the first method, I only get a first-order ODE thus I don't have any choice in the "initial velocity" $$\dot{\phi}(0)$$ as I should intuitively. Same for the second method. Shouldn't I have a second order equation ?
• The second method doesn't give the same solution as the first, why is that ?
• In general, I suspect that the step where I take $$\tau(s)=s$$ is not allowed, thus how can I compute the timelike geodesics of $$\text{dS}_2$$ ?
• The first approach is wrong. The "dividing by $ds^2$" only works if $s$ is the affine parameter (proper length of the curve). The first equation in item 1 is not invariant under reparameterization, so you are not free to choose $s = \tau(s)$. Mar 6, 2021 at 2:06
• For the second approach, which action did you apply the EL equations to? Check if it is invariant under reparameterization. There is a reparameterization invariant action to get geodesics, but the geodesic equations are usually nastier than what you got. There is a simpler way with a non-invariant action, but that assumes affine parameterization again. Mar 6, 2021 at 2:10
• The action I used is the one associated to the lagrangian $\mathcal{L}=g_{\mu\nu}\dot{x}^\mu\dot{x}^\nu$, is this the one were we assume to use affine parametrization ? What should I then ?
– xpsf
Mar 7, 2021 at 8:41
• That does assume affine parameterization. You can use it. IMO it's the easier one to work with. You just can't choose a parameterization, it has already been chosen. You have to solve the coupled ODEs in the first equation in item 2 as they are. Mar 7, 2021 at 14:16

As you have already suspected: you cannot simply say $$\tau(s) = s$$.

I think you might be missing some theoretical background of how the geodesic equations are often derived.

I am going to use a bit of matrix notations, to skip some of the indexing. So $$x = \begin{bmatrix} x^i\end{bmatrix} = \begin{bmatrix} x^0\\x^1\\x^2\\x^3\end{bmatrix} \, \text{ and } \, g(x) = \Big[g_{ij}(x)\Big]_{i,j = 0}^{3} \, \text{ is the 4 by 4 metric tensor describing gravity}$$

The description of the motion of a point-mass particle in General Relativity starts with the relativistic Lagrangian $$L = -m\,\sqrt{- \, \dot{x}^T\,g(x)\, \dot{x}}$$ where $$\dot{x} = \cfrac{dx}{d\lambda}$$. With its help, one can define the action $$S[\gamma] = -m\, \int_{\lambda_1}^{\lambda_2} \, \sqrt{- \, \dot{x}^T(\lambda)\,g\big(x(\lambda)\big)\, \dot{x}(\lambda)}\,d\lambda$$ where $$\gamma = \{x(\lambda) \, : \, \lambda \in [\lambda_1, \lambda_2]\}$$ can be any smooth time-like curve in 4D space-time, parametrized by some arbitrary parameter $$\lambda$$ and connecting two fixed space-time points, i.e. $$x(\lambda_1) = x_1$$ and $$x(\lambda_2) = x_2$$ are fixed with $$\lambda_1$$ and $$\lambda_2$$ also fixed (this is a very important conceptual requirement that is used in the derivation of the Euler-Lagrange equations!). Fundamentally, in the philosophy of General Relativity, this parameter $$\lambda$$ is not of any importance to the theory. Only the geometric shape of the curve $$\gamma$$ in 4D space-time matters and not the specific parametrization $$\lambda$$. After all, this curve $$\gamma = \{x(\lambda)\}$$ is supposed to be a space-time time-like smooth curve for which the action functional $$S[\gamma]$$ attains a critical value (i.e. this is the definition of a time-like geodesic). This is a geometric property independent of any parametrization $$\lambda$$, so we really care about the geodesic $$\gamma$$ as a geometric curve and not as a parametrized curve, so we are free to change the parametrization as we see fit in order to simplify our calculations.

The motion of a mass-particle in General Relativity is a time-like geodesic. As per the definition of a geodesic, we need to look for the critical (non-parametrized!!!) curves that connect two fixed space-time points $$x_1$$ and $$x_2$$ and that satisfy the critical value condition
$$\delta S[\gamma] = 0$$ In coordinates $$[x^i]$$ and with respect to a generic parametrization $$\lambda$$, the equation $$\delta S[\gamma] = 0$$ is equivalent to the Euler-Lagrange differential equations $$\frac{d}{d\lambda}\left(\frac{m}{\sqrt{-\, \dot{x}^T\,g(x)\, \dot{x}}} \,\, g(x)\, \dot{x}\right) \, = \, \frac{m}{2\, \sqrt{-\, \dot{x}^T\,g(x)\,\dot{x}\,}\,}\, \left(\, \dot{x}^T\,\frac{\partial g}{\partial x}(x)\,\dot{x}\, \right)$$ where $$\dot{x}^T\, \frac{\partial g}{\partial x}(x)\, \dot{x}\, = \, \begin{bmatrix} \frac{\partial g_{ij}}{\partial x^0}(x)\,\dot{x}^i\,\dot{x}^j \\ \frac{\partial g_{ij}}{\partial x^1}(x)\,\dot{x}^i\,\dot{x}^j \\ \frac{\partial g_{ij}}{\partial x^2}(x)\,\dot{x}^i\,\dot{x}^j \\ \frac{\partial g_{ij}}{\partial x^3}(x)\,\dot{x}^i\,\dot{x}^j \end{bmatrix}$$ for short. Take a solution (time-like) $$\gamma = \{ x(\lambda)\, : \, \lambda \}$$ of the Euler-Lagrange equations above. As I have already emphasized, the parametrization of $$\gamma$$ with respect to $$\lambda$$ is not important for us. Therefore, I can define the function $$s = s(\lambda) = \int_{\lambda_0}^{\lambda}\, \sqrt{-\, \dot{x}(\zeta)^T \, g\big(\, x(\zeta)\,\big)\, \dot{x}(\zeta)\,}\, d\zeta$$ with derivative $$\frac{ds}{d\lambda} = \sqrt{-\, \dot{x}(\lambda)^T \, g\big(\, x(\lambda)\,\big)\, \dot{x}(\lambda)\,} \, > \,0$$ Thus the function $$s = s(\lambda)$$ is strictly increasing and therefore invertible, i.e. there is $$\lambda = \lambda(s)$$. Consequently, we can re-parametrize our solution curve $$\gamma$$ as $$\gamma = \{ \, x(s) \, : \, s \, \} \, \text{ where } \, x(s)= x\big(\lambda(s)\big)$$ Observe that $$\gamma = \{\,x(s)\, : \, s \,\} = \{\, x(\lambda)\, : \, \lambda \, \}$$ in other words, this is the same curve in space time, but parametrized in two different ways. Denote $$x' = \frac{dx}{ds}$$. Furthermore, $$x' = \frac{dx}{ds} =\frac{d\lambda}{ds} \frac{dx}{d\lambda} = \left( \frac{ds}{d\lambda}\right)^{-1} \frac{dx}{d\lambda} = \frac{1}{\sqrt{- \, \dot{x}^T \, g(x) \, \dot{x}}\,}\, \frac{dx}{d\lambda}$$ and in particular $$\frac{d}{ds} = \frac{1}{\sqrt{- \, \dot{x}^T \, g(x) \, \dot{x}}\,}\, \frac{d}{d\lambda}$$ Recall that the curve $$\gamma$$ is a critical curve for the action $$S[\gamma]$$, i.e. $$\delta S[\gamma] = 0$$. When $$\gamma$$ is parametrized with respect to $$\lambda$$, it's coordinate parametrization $$\gamma = \{\, x(\lambda) \, : \, \lambda\}$$ solves the Euler-Lagrange equations $$\frac{d}{d\lambda}\left(\frac{m}{\sqrt{-\, \dot{x}^T\,g(x)\, \dot{x}}} \,\, g(x)\, \dot{x}\right) \, = \, \frac{m}{2\, \sqrt{-\, \dot{x}^T\,g(x)\,\dot{x}\,}\,}\, \left(\, \dot{x}^T\,\frac{\partial g}{\partial x}(x)\,\dot{x}\, \right)$$ whose both sides I can multiply by $$\frac{1}{\sqrt{-\, \dot{x}^T\, g(x) \, \dot{x}}\,}$$ and obtain the equivalent equations $$\frac{1}{\sqrt{-\, \dot{x}^T\, g(x) \, \dot{x}}\,} \, \frac{d}{d\lambda}\left(\frac{m}{\sqrt{-\, \dot{x}^T\,g(x)\, \dot{x}}} \,\, g(x)\, \dot{x}\right) \, = \, \frac{m}{-\, 2\, \dot{x}^T\,g(x)\,\dot{x}\,}\, \left(\, \dot{x}^T\,\frac{\partial g}{\partial x}(x)\,\dot{x}\, \right)$$ It is easy to check that with the new parametrization $$\gamma = \{\, x(s) \, : \, s\, \}$$ $$\sqrt{-\, \frac{dx}{ds}^T \, g(x) \, \frac{dx}{ds}} =\sqrt{-\, (x')^T \, g(x) \, x'} = 1$$ Consequently, after the reparametrization $$\lambda = \lambda(s)$$ the Euler-Lagrange equations turn into the equivalent simplified equations $$\frac{d}{ds}\left(\, m\, g(x)\, \frac{dx}{ds}\right) \, = \, \frac{m}{2}\,\left( \frac{dx}{ds}^T\,\frac{\partial g}{\partial x}(x)\, \frac{dx}{ds}\,\right)$$ which is solved by $$\gamma = \{\, x(s)\, : \, \tau\,\}$$.

In other words we have proven that any solution $$\gamma$$ to the original Euler-Lagrange equations, after the appropriate reparametrization, solves the simplified Euler-Lagrange equations. This means that a curve $$\gamma$$ is a critical curve of the action $$S[\gamma]$$, i.e. $$\delta S[\gamma] = 0$$ if and only if it solves the simplified Euler-Lagrange differential equations $$\frac{d}{ds}\left(\,m\, g(x)\, \frac{dx}{ds}\right) \, = \, \frac{m}{2}\, \left(\, \frac{dx}{ds}^T\,\frac{\partial g}{\partial x}(x)\, \frac{dx}{ds}\,\right)$$ where the resulting parametrized solution $$\gamma = \{\, x(s)\, : \, s\,\}$$ is paremtrized with respect to proper time, i.e. $$\sqrt{ - \, x'(s)^T\, g\big(x(s)\, x'(s)\big)} = 1$$ for any $$s$$.

It is very important to note that the simplified Lagrange differential equations $$\frac{d}{ds}\left(\,m\, g(x)\, \frac{dx}{ds}\right) \, = \, \frac{m}{2}\, \left(\, \frac{dx}{ds}^T\,\frac{\partial g}{\partial x}(x)\, \frac{dx}{ds}\,\right)$$ happen to match the Euler-Lagrange equations coming from the Lagrangian $$L = \frac{m}{2}\,\frac{dx}{ds}^T g(x) \frac{dx}{ds}$$ but this latter Lagrangian is conceptually not well defined in the framework of General relativity, as it is not independent on reparametrization. So, it is probably safer to avoid it in the framework of General Relativity, as it can lead to wrong results. This Lagrangian is not a problem in the theory of classical Lagrangian mechanics, but this is because there time is an absolute variable, separate from space and there is no need for treating time and space on equal footing.

Coming back to the problem at hand, there are various ways one can calculate the geodesics of 1,1 de Sitter space. One way is to think of the 1,1 de Sitter space as a one-sheeted hyperboloid embedded in 2, 1 Minkowski space, invariant under the action of the Lorentz group, and by using a symmetry argument, to show that the geodesics are the curves obtained by the intersection of 2D planes through the origin with the de Sitter hyperboloid. Another approach could be to observe that the Lagrangian $$L = -m\sqrt{\left(\frac{d\tau}{d\lambda}\right)^2 - \,\cosh^2(\tau)\left(\frac{d\phi}{d\lambda}\right)^2}$$ does not depend explicitly on the variable $$\phi$$, which means the Lagrangian is invariant under the action of the one-parametr group of symmetries $$\phi \mapsto \phi + \sigma$$ so one can apply Noether's theorem and obtain $$\frac{\partial }{\partial \dot{\phi}} \sqrt{\left(\frac{d\tau}{d\lambda}\right)^2 - \,\cosh^2(\tau)\left(\frac{d\phi}{d\lambda}\right)^2 } \, = \, c_0 \,(= \text{const})$$ then perform the differentiation in the latter identity and after that express $$\phi = \phi(\tau)$$ as a function of $$\tau$$. However, I think all of these methods are not first principle methods, so I would just use the direct definitions of geodesics.

Let's see how the theoretical procedure outlined above applies to your example. In your case, we can assume that the time-like geodesics $$\big(\tau = \tau(s), \, \phi = \phi(s)\big)$$ are parametrized by the proper time $$s$$ and thus, satisfy the simplified Euler-Lagrange equations: \begin{align} &\frac{d}{ds} \Big( -\, \frac{d\tau}{ds}\Big) \, = \, \frac{1}{2} \frac{\partial}{\partial \tau} \left( - \, \Big(\, \frac{d\tau}{ds}\,\Big)^2 \, + \, \cosh^2(\tau)\Big(\, \frac{d\phi}{ds}\,\Big)^2\right)\\ &\frac{d}{ds} \Big( \,\cosh^2(\tau)\, \frac{d\phi}{ds}\Big) \, = \, \frac{1}{2} \frac{\partial}{\partial \phi} \left( - \, \Big(\, \frac{d\tau}{ds}\,\Big)^2 \, + \, \cosh^2(\tau)\Big(\, \frac{d\phi}{ds}\,\Big)^2\right) \end{align} After performing most of the differentiations, we obtain the system \begin{align} & -\, \frac{d^2\tau}{ds^2} \, = \, \cosh(\tau)\sinh(\tau)\Big(\, \frac{d\phi}{ds}\,\Big)^2\\ &\frac{d}{ds} \Big( \,\cosh^2(\tau)\, \frac{d\phi}{ds}\Big) \, = \, 0 \end{align} The second equation can be immediately integrated once with respect to $$s$$, yielding the system \begin{align} & -\, \frac{d^2\tau}{ds^2} \, = \, \cosh(\tau)\sinh(\tau)\, \Big(\, \frac{d\phi}{ds}\,\Big)^2\\ &\cosh^2(\tau)\, \frac{d\phi}{ds} \, = \, c_0 \end{align} where $$c_0$$ is a constant. Thus \begin{align} & \frac{d^2\tau}{ds^2} \, = \, -\, \cosh(\tau)\sinh(\tau)\Big(\, \frac{d\phi}{ds}\,\Big)^2\\ &\frac{d\phi}{ds} \, = \, \frac{c_0}{\cosh^2(\tau)} \end{align} One can plug the second equation in the first \begin{align} & \frac{d^2\tau}{ds^2} \, = \, -\, \cosh(\tau)\sinh(\tau)\Big(\, \frac{c_0}{\cosh^2(\tau)}\,\Big)^2 \\ &\frac{d\phi}{ds} \, = \, \frac{c_0}{\cosh^2(\tau)} \end{align} and simplify \begin{align} & \frac{d^2\tau}{ds^2} \, = \, -\, c_0^2\,\frac{\sinh(\tau)}{\cosh^3(\tau)}\\ &\frac{d\phi}{ds} \, = \, \frac{c_0}{\cosh^2(\tau)} \end{align} The first equation decouples from the second because it is an equation only for the variable $$\tau$$. If you multiply both sides of the first equation by $$\frac{d\tau}{ds}$$ then you can integrate it once and obtain \begin{align} & \frac{1}{2}\left(\frac{d\tau}{ds} \right)^2 \, = \, \frac{c_1}{2} \, -\, c_0^2\, \int \frac{\sinh(\tau)}{\cosh^3(\tau)} d\tau\\ &\frac{d\phi}{ds} \, = \, \frac{c_0}{\cosh^2(\tau)} \end{align} After integrating the right-hand side, the equations become \begin{align} & \frac{1}{2}\left(\frac{d\tau}{ds} \right)^2 \, = \, \frac{c_1}{2} \, + \, \frac{c_0^2}{2\,\cosh^2(\tau)}\\ &\frac{d\phi}{ds} \, = \, \frac{c_0}{\cosh^2(\tau)} \end{align} divide by two and take a square root \begin{align} &\frac{d\tau}{ds} \, = \, \pm \sqrt{c_1 \, + \, \frac{c_0^2}{\cosh^2(\tau)} \, }\\ &\frac{d\phi}{ds} \, = \, \frac{c_0}{\cosh^2(\tau)} \end{align} From here, if you decide to express the proper time $$s = s(\tau)$$ as a function of $$\tau$$, then the variable $$\phi = \phi(s) = \phi\big(s(\tau)\big)$$ becomes also a function of $$\tau$$ and by the chain rule $$\frac{d\phi}{d\tau} = \frac{d\phi}{ds} \, \frac{ds}{d\tau} = \frac{\,\,\left( \frac{d\phi}{ds}\right)\,\,}{\left(\frac{d\tau}{ds}\right)} = \, \pm\,\frac{\,\,\left( \frac{c_0}{\cosh^2(\tau)}\right)\,\,}{\left(\sqrt{c_1 \, + \, \frac{c_0^2}{\cosh^2(\tau)} \, }\right)}$$ so we end up with a differential equation that is readily integrable $$\frac{d\phi}{d\tau} = \,\frac{\left(\frac{c_0}{\sqrt{c_1}\,\cosh^2(\tau)}\right)}{\sqrt{1 \, + \, \frac{c_0^2}{c_1\, \cosh^2(\tau)} \, }}$$ where I have taken only the plus version of the equation. Consequently, $$\phi = \phi_0 \, + \, \int \, \frac{\left(\frac{c_0}{\sqrt{c_1}\,\cosh^2(\tau)}\right)}{\sqrt{1 \, + \, \frac{c_0^2}{c_1\, \cosh^2(\tau)} \, }}\, d\tau$$ to solve the integral, you can for example set $$a_0 = \frac{c_0}{\sqrt{c_1}}$$ and perform the substitution $$z = a_0\,\tanh(\tau) \,\, , \,\,\,\,\, dz = \frac{a_0}{\cosh^2(\tau)} \, d\tau$$ $$\frac{a_0^2}{\cosh^2(\tau)} = a_0^2\,\frac{\cosh^2(\tau) - \sinh^2(\tau)}{\cosh^2(\tau)} = a_0^2 - a_0^2\,\tanh^2(\tau) = a_0^2 - z^2$$ so the integral becomes $$\phi = \phi_0 \, + \, \int \, \frac{1}{\sqrt{(1 \, + \, {a_0^2}) \,-\, z^2}}\, dz = \arcsin\left(\frac{z}{\sqrt{1+a_0^2}}\right)$$ Thus, finally we find the formula $$\phi = \phi \pm \arcsin\big(\,k_0\tanh(\tau)\,\big)$$ where $$k_0 = \frac{a_0}{\sqrt{1 + a_0^2}}$$ is a constant.

• I don't get the same result for the final integration, could your change of variable contain an error (in the expression of $dz$ in particular) ? But thank you very much for your more than complete et very understandable answer.
– xpsf
Mar 8, 2021 at 10:44
• @xpsf Yes, the substitution is not the right one. It's good idea to redo the calculations carefully by yourself, just following the ideas outline in the answer not the actual calculations, because I was not very careful and was skipping steps and directly writing the answer, instead of doing it carefully in advance. So, there could be errors, but the ideas are fine. Mar 8, 2021 at 13:54
• @xpsf I think I fixed the integral substitution, if you still need to double check. Mar 8, 2021 at 18:03

Actually, maybe I forgot to mention an important point and a somewhat more direct (obvious) approach to the problem. When one thinks of the Lagrangian and the action, one needs to think of the Lagrangian as a one-form rather than a function, i.e. $$L = \sqrt{ \left(\frac{d\tau}{d\lambda}\right)^2 \, - \, \cosh^2(\tau)\left(\frac{d\phi}{d\lambda}\right)^2} \, d\lambda$$ If you want to change parametrization, say you want to write $$\lambda = \lambda(\tau)$$ as a funciton of $$\tau$$, then $$\frac{d\phi}{d\lambda} = \frac{d\phi}{d\tau} \, \frac{d\tau}{d\lambda}$$ and $$d\lambda = \frac{d\lambda}{d\tau} \, d\tau$$ so when you substitute in the Lagrangian, you get $$L = \sqrt{ \left(\frac{d\tau}{d\lambda}\right)^2 \, - \, \cosh^2(\tau)\left(\frac{d\phi}{d\tau} \frac{d\tau}{d\lambda}\right)^2} \, \, \frac{d\lambda}{d\tau}\,d\tau$$ $$L = \sqrt{ \left(1 \, - \, \cosh^2(\tau)\left(\frac{d\phi}{d\tau} \right)^2 \, \right) \left(\frac{d\tau}{d\lambda}\right)^2\,} \,\, \frac{d\lambda}{d\tau}\,d\tau$$ $$L = \sqrt{ \left(1 \, - \, \cosh^2(\tau)\left(\frac{d\phi}{d\tau} \right)^2 \, \right) }\,\left(\frac{d\tau}{d\lambda}\right) \,\, \frac{d\lambda}{d\tau}\,d\tau$$ $$L = \sqrt{ 1 \, - \, \cosh^2(\tau)\left(\frac{d\phi}{d\tau} \right)^2 \, }\,d\tau$$ If you try to do the same thing with the classical Lagrangian $$L = \frac{1}{2} \left( \, \left(\frac{d\tau}{d\lambda}\right)^2 \, - \, \cosh^2(\tau)\left(\frac{d\phi}{d\lambda}\right)^2 \,\right) d\lambda$$ the reparametrization described above is not going to work!

Now, to find the geodesics, you can directly derive the Euler-Lagrange equations for $$L = \sqrt{ 1 \, - \, \cosh^2(\tau)\left(\frac{d\phi}{d\tau} \right)^2 \, }$$ which yields $$\frac{d}{d\tau} \left(\frac{\partial}{\partial {\phi'}} \sqrt{ 1 \, - \, \cosh^2(\tau)\left(\frac{d\phi}{d\tau} \right)^2 \, } \right) \, = \, \frac{\partial}{\partial {\phi}} \sqrt{ 1 \, - \, \cosh^2(\tau)\left(\frac{d\phi}{d\tau} \right)^2 \, }$$ However, the Lagrangian does not depend explicitly on $$\phi$$, so $$\frac{d}{d\tau} \left(\frac{\partial}{\partial {\phi'}} \sqrt{ 1 \, - \, \cosh^2(\tau)\left(\frac{d\phi}{d\tau} \right)^2 \, } \right) \, = \, 0$$ so you can integrate it once with respect to $$\tau$$ and obtain the equation $$\frac{\partial}{\partial {\phi'}} \sqrt{ 1 \, - \, \cosh^2(\tau)\left(\frac{d\phi}{d\tau} \right)^2 \,} \, = \, a_0$$ which becomes $$-\, \left(\frac{\cosh^2(\tau)}{\sqrt{ 1 \, - \, \cosh^2(\tau)\left(\frac{d\phi}{d\tau} \right)^2 \,}}\,\right) \frac{d\phi}{d\tau} \, = \, a_0$$ and now you can raise both side to power $$2$$ and get

$$\left(\,\frac{\cosh^4(\tau)}{{ 1 \, - \, \cosh^2(\tau)\left( \frac{d\phi}{d\tau} \right)^2 \,}}\right) \left( \frac{d\phi}{d\tau} \right)^2 \, = \, a_0^2$$

$${\cosh^4(\tau)} \left( \frac{d\phi}{d\tau} \right)^2 \, = \, a_0^2 \, - \, a_0^2\,\cosh^2(\tau)\left( \frac{d\phi}{d\tau} \right)^2$$ and again you get the equation $$\left( \frac{d\phi}{d\tau} \right)^2 \, = \, \frac{a_0^2}{ \cosh^4(\tau) + a_0^2 \, \cosh^2(\tau)}$$