$l_2$, that is, at these points

            1.  $b/R \ll 1$,\[9082013e\]

            2.  $|\Phi|\ll 1$,\[9082013f\]

            3.  $|\Phi'c^2| { \put(2,0){$<$} \put(2,-5){$\sim$}\quad \ }g$, where $'\equiv d/dR$ and $g=\textrm{(Earth gravity)}$.\[9082013g\]

        3.  Trip takes less than one year from the point of view of both traveler and static observers at $-l_1$ and $l_2$. As a result, we must have $$\Delta\tau = \int_{-l_1}^{l_2}(v\gamma)^{-1}dl { \put(2,0){$<$} \put(2,-5){$\sim$}\quad \ }1\ yr, \label{9082013h}$$ $$\Delta t = \int_{-l_1}^{l_2}(ve^{\Phi})^{-1}dl { \put(2,0){$<$} \put(2,-5){$\sim$}\quad \ }1\ yr. \label{9082013i}$$

        4.  Traveler feels “less” than $g$ acceleration, $$|e^{-\Phi} d(\gamma e^{\Phi})/dl| { \put(2,0){$<$} \put(2,-5){$\sim$}\quad \ }g/c^2. \label{9082013j}$$

        5.  Tidal-gravity accelerations between different parts of traveler’s body is less than or approximately equal to $ g$: $$\left|\left(1-b/R \right)\left[ -\Phi^{\prime\prime}+\frac{1}{2}\frac{(b'-b/R)}{R-b}\Phi'-(\Phi')^2 \right] \right|{ \put(2,0){$<$} \put(2,-5){$\sim$}\quad \ }1/(10^{10}\ \textrm{cm})^2; \label{9082013l}$$ $$\left|\frac{\gamma^2}{2R^2} \left[ \frac{v^2}{c^2} (b'-b/R)+2(R-b)\Phi' \right] \right|{ \put(2,0){$<$} \put(2,-5){$\sim$}\quad \ }1/(10^{10}\ \textrm{cm})^2. \label{9082013m}$$

        6.  Traveler must not couple strongly to material that generates wormhole curvature.\[9082013n\]

2.  Properties of the material that generates wormhole curvature: \[11092013a\]

    1.  Stress-energy tensor as measured by static observers:

        1.  $$\begin{aligned}
             T_{\hat{t}\hat{t}}=\rho c^2=\textrm{(density of mass-energy)},\quad T_{\hat{R}\hat{R}}=-\tau= -\textrm{(radial tension)},
            \\ T_{\hat{\theta}\hat{\theta}}=T_{\hat{\varphi}\hat{\varphi}}=p=\textrm{(lateral pressure)}. \end{aligned}$$

            \[9082013o\]

        2.  Einstein field equations: $$\begin{aligned}
             \rho=\frac{b'}{8\pi Gc^{-2} R^2},\quad \tau=\frac{b/R-2(R-b)\Phi'}{8\pi Gc^{-4}R^2},
            \nonumber \\ p=\frac{R}{2}\left[(\rho c^2-\tau)\Phi'-\tau' \right]-\tau. \label{9082013p}\end{aligned}$$ In the throat ($R=R_m$), we have $\tau \approx 