point, since $r_c$ is less than $r_1$. Thus, we need to compare only $f(r_1)$ with $f(r_2)$. The value of $f$ at $r_1$ is \[see Eq. (\[29012014b\])\] $$f(r_1)=\frac{4}{M\sqrt{1-v^2/c^2}}\frac{S^4}{(S+1)^4} \left(  \frac{S-1}{S+1} \right)^{S-2}, \label{08012014c}$$ where we have used $\gamma=(1-v^2/c^2)^{-1/2}$. By using $r_2=10^AM$ with $A{ \put(2,0){$>$} \put(2,-5){$\sim$}\quad \ }6$, one can easily verifies that $f(r_2)<f(r_1)$. Therefore, the condition (\[9082013j\]) becomes $$\frac{4}{M\sqrt{1-v^2/c^2}}\frac{S^4}{(S+1)^4} \left(  \frac{S-1}{S+1} \right)^{S-2} { \put(2,0){$<$} \put(2,-5){$\sim$}\quad \ }g/c^2. \label{08012014d}$$ A plot of $f(r_1)$ as a function of $S$ and $v$ is shown in Fig. \[09012014a\]. From this plot, one can see that $f(r_1)$ reaches its minimum as $S$ goes to infinity.

![This figure shows the behavior of $f(r_1)$ as a function of $S$ and $v$ in the intervals $S\in (1,10)$ and $v\in (0,1)$, where we have used $M=c=1$. It can be shown that the qualitative behavior of the above surface does not change for larger values of $S$.[]{data-label="09012014a"}](f333.jpg)

In this limit, we have $$\frac{4}{Me^2\sqrt{1-v^2/c^2}} { \put(2,0){$<$} \put(2,-5){$\sim$}\quad \ }g/c^2. \label{08012014e}$$ Note that this constraint is stronger than that imposed by condition \[9082013g\] \[see Eq. (\[15082013e\])\]. In addition, it requires low speed, which is not good for the inequalities (\[9082013h\]) and (\[9082013i\]). In what follows, we prove that the conditions (\[9082013h\]) and (\[9082013j\]) cannot 