How does one calculate the tidal force in the throat of a wormhole?

If you were to attempt to open a wormhole from Earth to Alpha Centauri (4.37 LY), and its throat were to be a radius of 1m, while its mouth were to be 1.5m radius, how many newtons of force would be acting inwards from the throat of the wormhole? (also assume that $$t$$ is constant and $$\theta=\pi/2$$)I'm aware that one would need to use the Morris-Thorne metric to solve for this. The metric and parameters can be found here:

$$ds^2=-c^2dt^2+dl^2+(b_0^2+l^2)(d\theta^2+\sin^2\theta\; d\phi^2)\tag{18}$$

where

• $$t$$ is the time measured by a static observer, and $$-\infty.

• $$\theta$$, $$\phi$$ are spherical polar coordinates, and $$0\le\theta\le\pi$$, $$0\le\phi\le2\pi$$

• $$l$$ is the radial coordinate, and $$-\infty.

The tidal tensor in general relativity (see also this for more details) is $$\Phi_{ij}=R_{i0j0}$$ where $$R_{abcd}$$ is the Riemann tensor. The components tell you what accelerations $$A^j$$ in direction $$j$$ are experienced in direction $$i$$: $$A^j = -g^{jk}\Phi_{ki}x^k$$. Section D of the linked paper discuss this for an object moving at velocity $$v$$, finding the formula in equation 35: $$\Phi_{22}=\Phi_{33}=-\left(\frac{\gamma v}{c}\right)^2\frac{b_0^2}{(b_0^2+l^2)^2}.$$ So your question seems to mostly be about how to plug the numbers in?
Note that the throat at $$l=0$$ and $$r=b_0$$. The circumference of the throat is just s=$$\int_0^{2\pi} b_0 d\theta =2\pi b_0$$ (from equation 18) so $$b_0$$ really does act as a radius of the throat. So if I put in $$b_0=1.5$$ m for $$l=0$$ I get an acceleration of $$-(\gamma v/c b_0)^2$$. For $$v=1$$ m/s this is $$\sim -10^{-18}$$ m/s$$^2$$, for $$v=1000$$ km/s it is $$\sim 10^{-6}$$ m/s$$^2$$ - it only starts to become significant once $$v$$ is close to lightspeed. For this particular wormhole metric and its assumptions tidal effects are negligible.