I am solving a problem on geodesics with ideas from General Relativity and got stuck with one step. The simplified version is the following:

With notations $$\dot{x}\equiv \frac{dx}{dt}, \quad \mathring{x} \equiv \frac{dx}{d\lambda}, \quad x' \equiv \frac{dx}{dr}$$ $$A=A(r), B=B(r), t=t(\lambda), r=r(\lambda), \varphi=\varphi(\lambda), v= \text{ constant}$$

Given the constraints $$\mathring{t}=\frac{1}{A},\quad \mathring{\varphi}=\frac{1}{B\cdot r^2},\quad \dot{r}^2 +r^2\dot{\varphi}^2 = v^2$$

Can we find a relation between $A,B,A',B',r$?

I can convert $\dot{\varphi}$ in terms of functions of $r$. But for $\dot{r}$, I can't see any way to do it so that I can use the last constraint.

$$\mathring{\varphi}=\frac{dt}{d\lambda}\frac{d\varphi}{dt}=\frac{1}{A}\dot{\varphi}$$ $$\dot{\varphi}=A \mathring{\varphi}=\frac{A}{B}\cdot r^{-2} $$

Any hints are appreciated! Thanks.


I do not think it is possible. You have 3 equations and 3 unknowns. The first two equations are used to eliminate $t$ and $\varphi$ from the 3rd equation. There is only one way to do this - by solving them and plugging in the result, so there is no flexibility there. You get a first order ODE for $r$ and $\lambda$. This ODE can be solved by integration, and produce and integration constant. This means that $r$ that satisfies above equation has an extra degree of freedom, and thus there can't exist an equation you request that does not include this extra degree of freedom.

  • $\begingroup$ Thank you. You are right. $\endgroup$ – Shengkai Li Feb 24 '19 at 18:16

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