SCIENCE CHINA Physics, Mechanics & Astronomy, Volume 60 , Issue 5 : 050411(2017) https://doi.org/10.1007/s11433-017-9012-8

## Dispersion relation and surface gravity of universal horizons

ChiKun Ding 1,2,*,
• AcceptedFeb 24, 2017
• PublishedMar 17, 2017
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### Acknowledgment

This work was supported by the National Natural Science Foundation of China (Grant No. 11247013), Hunan Provincial Natural Science Foundation of China (Grant Nos. 2015JJ2085, and QSQC1203), and the Special Fund of the National Natural Science Foundation of China (Grant No. 11447168). We would like to thank Prof. Stefano Liberati for supporting this work.

### Supplement

Appendix

Review of the incorrect trajectory

In ref. [29], the authors calculate the value of surface gravity of the universal horizon which is the relevant one for the rays propagating with infinite group velocity $v_g$. However, there is a mathematical error in their deriving process which ultimately leads to an incorrect result. We firstly review their original process.

Generically any particles propagating in the Einstein aether spacetime will have a four-velocity that can be given in the orthogonal frame provided by $u^a$ and $s^a$ as: \begin{align}V^a=u^a+v_gs^a. \tag{35} \end{align} The trajectory for an instantaneously propagating ray would then be given by \begin{align} \frac{{\rm d}v}{{\rm d}r}=\frac{V^v}{V^r}=\lim_{v_g\rightarrow\infty}\frac{u^v+v_gs^v} {u^r+v_gs^r}=\frac{s^v} {s^r}. \tag{36} \end{align} The group velocity $v_g$ is dependent on the specific form of dispersion relation $\omega=f(k_s)$, and $v_g=\partial\omega/\partial k_s=f'(k_s)$. So one can see that above trajectory is indeed independent of the specific superluminal dispersion. Then the surface gravity of the universal horizon is \begin{align} \kappa_{uh}=\frac{1}{2}\frac{\rm d}{{\rm d}r}\Big(\frac{s^v}{s^r}\Big)\Big|_{uh}, \tag{37} \end{align} which is also independent of the specific superluminal dispersion relation.

But, there is a mathematical error in eq. (36). In the numerator, both $u^v$ and $s^v$ are finite and nonzero at the universal horizon, so the term $u^v$ can be ignored in the limit $v_g\rightarrow\infty$. However, in the denominator, $s^r$ is zero at the universal horizon, therefore the product of $v_gs^r$ maybe finite and the term $u^r$ cannot be ignored In their paper, $u^r,v_g$, and $s^r$ are given by, respectively \begin{align}\begin{array}{l} u^r=-(s\cdot\chi),\;s^r=(-u\cdot\chi), \\ v_g=1+3l^2k_s^2 \sim1+3\dfrac{(s\cdot\chi)|_{uh}}{(u\cdot\chi)}. \end{array} \tag{38} \end{align} Hence, the product is \begin{align}v_gs^r=(-u\cdot\chi)-3(s\cdot\chi)|_{uh}, \tag{39} \end{align} which is indeed finite at the universal horizon $(u\cdot\chi)=0$ and, the quantity $u^r$ cannot be ignored. Ultimately the result eq. (37) is incorrect.

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