# How can the derivative of this trace be constrained?

I am studying for my exam on relativity and I am going through some problems sets including ones where I was not very successful in so I want to know how to do this problem.

(Convergence of particles): Consider a freely falling reference particle in vacuum as in Exercise 1. Assume now in addition that the Hessian matrix of the gravitational potential does not vanish initially in the vicinity of the reference particle. Recall that the displacement of a nearby particle $y(t)$ is given in terms of its initial displacement $y(0)$ (we assume that the particle is intitially at rest relativie to the reference particle) by $$y(t) = A(t)y(0)$$

NOTE: Just to give some context we say that $y(t)$ is a displacement between two freely falling particles positioned $x_{0}(t)$ and $x(t)$ so $y(t) = x(t)-x_{0}(t)$. We can say that a nearby freely falling particle satisfies to first order in y the tidal equations, mainly we can see that: $$\frac{\mathrm{d}^{2}y}{\mathrm{d}t^2}=-\nabla^2\psi(x_0(t))\cdot{y(t)}.$$

Back to the problem.

We call $A$ the deformation matrix. Show that there is an initial displacement $y(0)\neq0$ and a time $t^*>0$ such that $y(t^*)=0$.

Hint. Recall the definition of the strain matrix: $$\theta=\frac{\mathrm{d}A}{\mathrm{d}t}A^{-1}.$$

Now to solve this the question breaks it down to several small parts. I was able to solve the first part so I will skip it. But here is the part I am having trouble with:

(2) Deduce from the tidal equation the following ordinary differential inequality: $$\frac{\mathrm{d}}{\mathrm{d}t}\operatorname{tr}\theta\leq-\frac{1}{3}(\operatorname{tr}\theta)^2$$ Hint: Decompose $\theta$ into its diagonal and its traceless part.

NOTE: We deduced from lecture a long time ago that $\frac{\mathrm{d}}{\mathrm{d}t}\operatorname{tr}\theta=-|\theta|^2-\operatorname{tr}M$ where $M=\nabla^2\psi(x_0(t))$ or in other words $M$ is the hessian of the gravitational potential of the particles that are freely falling. We also know that $\operatorname{tr}M=\Delta\psi(x_0(t))$ and in a vacuum this is equal to 0.

So, concerning this problem I figure if we have to break down $\theta$ to a diagonal part and a traceless part that means we can call the diagonal $\sigma$ and the traceless part $\omega$, such that $\theta=\sigma+\omega$. That means that: \begin{align} &\frac{\mathrm{d}}{\mathrm{d}t}\operatorname{tr}(\sigma+\omega) \\ &\frac{\mathrm{d}}{\mathrm{d}t}\operatorname{tr}(\sigma) \end{align} Since the trace of the omega is 0 but honestly I am not sure what more to do. I think I actually have to move that derivative and maybe I will be able to get that inequality.

• Hi InsigMath - I edited your question to make it clearer, but please review it and fix anything I messed up. – David Z Apr 13 '13 at 8:07
• @DavidZaslavsky Yeah that looks good thank you for the edit – InsigMath Apr 13 '13 at 13:52
• If you like this question you may also enjoy reading this Phys.SE post. – Qmechanic Apr 13 '13 at 15:05