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

Question

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 $-\infty<t<\infty$.

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

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

Answer 1

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.

Usually the real problem is noticing if the paper uses "natural" units (seconds and light-seconds) rather than metric units.