What would the Second Law of Motion be in this universe?

Question

I have a universe described by the equation:

$$ds^2=c^2d\tau^2=-c^2dt^2+[dr+a_0 td\tau]^2+r^2\Omega^2$$ What would be the second law of motion in this universe?

Answer 1

First you need to do elementary algebra to get rid of $\tau$ on the right side by moving it to the left in the form of

$$ \left[A(t,r)d\tau + B(t,r)\right]^2=\,... $$

Where $A$ and $B$ are the results of your algebraic manipulations. Obviously $A=\sqrt{c^2-a_o^2t^2}$ and so on. After that you need to do a coordinate transformation by defining

$$ d\tau^{'}\equiv Ad\tau + B $$

To get the metric in the form of

$$\tau^{'2}=\,... $$

Then you could have the Lagrangian defined explicitly and write the Euler-Lagrange equations for geodesics. Depending of how complex the equations end up, you can try solving them analytically or numerically to derive the equations of motion. For the Lagrangian and geodesic equations, see chapter 7.4.3 p. 282 of Nakahara.

In the end you need to do the reverse coordinate transformation

$$ d\tau=\dfrac{d\tau^{'}-B}{A} $$

It may be necessary also to do some coordinate transformation on the right side. The calculations will quickly become complicated. If possible, you may want to consider a simpler metric to fulfill the requirements for the target behavior of your universe.

Answer 2

\begin{align*} &\text{The Line Element is (c=1) :}\\\\ &\left( 1-{{\it a_0}}^{2}{t}^{2} \right) {d\tau }^{2}-2\,{\it dr}\,{ \it a_0}\,td\tau +{{\it dt}}^{2}-{{\it dr}}^{2}-{r}^{2}{d\Omega }^{2}=0& (1)\\\\ &\text{Equation (1) is equivalent to:}\\\\ &\begin{bmatrix} d\tau & dt & dr & d\Omega \\ \end{bmatrix} \underbrace{ \left[ \begin {array}{cccc} 1-{{\it a_0}}^{2}{t}^{2}&0&-{\it a_0}\,t&0 \\ 0&1&0&0\\ -{\it a_0}\,t&0&-1&0 \\ 0&0&0&-{r}^{2}\end {array} \right]}_{=A} \underbrace{\begin{bmatrix} d\tau \\ dt \\ dr \\ d\Omega \\ \end{bmatrix}}_{\vec{x}}=0&(2)\\ &\text{To transform the matrix $A$ to a diagonal shape }\\&\text{we have to calculate the eigenvalues and the eigenvectors (Matrix $T$) of $A$ :}\\\\ &\vec{x}^T\,T^T\,A\,T\,\vec{x}=0=\vec{x'}^T\,A_d\,\vec{x'}=0&(3)&\\\text{with:}\\ &\vec{x'}=\begin{bmatrix} d\tau' & dt' & dr' & d\Omega' \\ \end{bmatrix}^T =T\,\vec{x}\quad \text{and}\\ &A_d=\begin{bmatrix} \lambda_1 & 0 & 0 & 0 \\ 0 & \lambda_2 & 0 & 0 \\ 0 & 0 & \lambda_3 & 0 \\ 0 & 0 & 0 & \lambda_4 \\ \end{bmatrix}\quad \lambda_i \quad \text{are the eigenvalues of the Matrix $A$}\\ &\lambda_1=1\quad\,,\lambda_2=-r^2\quad\,,\lambda_3= -\frac{1}{2}\,a_0^2\,t^2+\frac{1}{2}\sqrt{a_0^4\,t^4+4} \quad\,, \lambda_4=-\frac{1}{2}\,a_0^2\,t^2-\frac{1}{2}\sqrt{a_0^4\,t^4+4} \\ &\text{We can solve now equation (3) for $d\tau'^2$} \\\\ &d\tau'^2=-\lambda_2\,dt'^2-\lambda_3\,dr'^2-\lambda_4\,d\Omega'^2\\\\ &\text{To calculate the equations of motion with Lagrange we need the kinetic energy}\\\\ &T=\frac{1}{2}\,m\left[-\lambda_2\,\left(\frac{dt'}{d\eta}\right)^2 -\lambda_3\,\left(\frac{dr'}{d\eta}\right)^2-\lambda_4\,\left(\frac{d\Omega'}{d\eta}\right)^2\right] \end{align*} Remark: I use MAPLE program to do the calculations