Particle shot in a flat FRW universe

[Nabeshin]

Homework Statement

The problem is Hartle 18.3 and reads: "Consider a flat FRW model whose metric is given by (18.1). Show that, if a particle is shot from the origin at time t*, with a speed v* as measured by a comoving observer (constant x,y,z), then asymptotically it comes to rest with respect to a comoving frame. Express the comoving coordinate radius at which it comes to rest as an integral over a(t)."

Homework Equations

$$ds^2=-dt^2+a^2(t)\left[dx^2+dy^2+dz^2\right]=ds^2=-dt^2+a^2(t)\left[dr^2+r^2 d\theta^2 + r^2 sin^2(\theta) d\phi^2\right]$$

The Attempt at a Solution

Consider the initial 4-velocity. Using polar coordinates,
$$u^r=V_*$$
By the normalization condition, I can find $$u^t$$, and so I get the full initial 4-velocity:

$$u^{\alpha}_0=\left(\sqrt{1+a^2(t_*)V_*},V_*,0,0\right)$$

My thought is to evolve this 4-velocity in time to derive an expression for $$u^r(t)$$ which I can integrate from t* to infinity to find the resulting final position.

So, from the geodesic equations I have:
$$\frac{d u^t}{d\tau}=-\Gamma^t_{\alpha \beta} u^\alpha u^\beta = -\frac{1}{2}g^{tt}g_{rr,t}(u^r)^2=-a(t)\dot{a}(t)(u^r)^2$$

$$\frac{d u^r}{d\tau}=-\Gamma^r_{\alpha \beta} u^\alpha u^\beta = -\frac{1}{2} g^{rr}g_{rr,t} u^t u^r = -\frac{\dot{a}(t)}{a(t)} u^t u^r = -H_0(t) u^t u^r$$
Note: I'm using the comma notation for differentiation.

I can't think of any way to solve these equations. This is further muddled by the fact that the scale factor a is a function of t, which is mixed with tau, and it's a whole big mess.

Any help?

 

[Jhero]

First, let's rewrite the initial 4-velocity in terms of the comoving coordinates:

u^{\alpha}_0=\left(\sqrt{1+a^2(t_*)V_*},V_*,0,0\right)=\left(\sqrt{1+a^2(t_*)V_*},0,0,0\right)

This is because in comoving coordinates, the particle is not moving in the x, y, or z directions.

Next, we can use the geodesic equations to find the evolution of the 4-velocity in time:

\frac{d u^t}{d\tau}=-\Gamma^t_{\alpha \beta} u^\alpha u^\beta = -\frac{1}{2}g^{tt}g_{rr,t}(u^r)^2=-a(t)\dot{a}(t)(u^r)^2

\frac{d u^r}{d\tau}=-\Gamma^r_{\alpha \beta} u^\alpha u^\beta = -\frac{1}{2} g^{rr}g_{rr,t} u^t u^r = -\frac{\dot{a}(t)}{a(t)} u^t u^r = -H(t) u^t u^r

Note that we have replaced the subscript "0" with "(t)" to indicate that these quantities are evolving in time. We have also replaced "H_0(t)" with "H(t)" to avoid confusion with the Hubble constant.

Now, we can solve these equations by using the fact that u^t is conserved along the particle's trajectory:

u^t=\frac{1}{\sqrt{1+a^2(t_*)V_*}}=\frac{1}{\sqrt{1+a^2(t)u^r}}

We can then substitute this into the equation for du^r/dtau to get:

\frac{d u^r}{d\tau}=-H(t) \frac{1}{\sqrt{1+a^2(t)u^r}} u^r

This is a separable differential equation, which we can solve to get:

\int_{u^r(t_*)}^{u^r(t)} \frac{du^r}{u^r}=\int_{t_*}^{t} -H(t') \frac{dt'}