The Fall of Physical Momentum in an FRW Spacetime

[latentcorpse]

General particle motion in an FRW spacetime is governed by the geodesic equation, $\frac{du^\mu}{ds} + \Gamma^\mu{}_{\nu \alpha} u^\nu u^\alpha=0$, with $u^\mu = \frac{dx^\mu}{ds}$ and where the Christoffel symbols are given in the notes. Show that the physical momentum of any particle always falls as $|\tilde{u}| \propto a^{-1}$, where the physical velocity is related to the comoving velocity by $|\tilde{u}| = \sqrt{-g_{ij}u^iu^j}=a | \vec{u} |$. [Hint: Consider
$u^\mu u_\mu$ in the FRW metric and show that $u^0 \frac{du^0}{ds} = |\tilde{u}| \frac{d|\tilde{u}|}{ds}$ for both massive and massless particles.]

i can't get anywhere productive with this. any ideas?

 

[NateD]

Answer:Using the geodesic equation and the FRW metric, we can calculate u^μu_μ as follows:u^μu_μ = g_{00}u^0u^0 + g_{ij}u^iu^j = a(t)^2(-u^0)^2 + a(t)^2(u^i)^2 = a(t)^2(-u^0)^2(1 + (u^i)^2).Now, consider the time-derivative of the above equation:\frac{d}{ds}[u^μu_μ] = \frac{d}{ds}[a(t)^2(-u^0)^2(1 + (u^i)^2)].Using the chain rule, we get:\frac{d}{ds}[u^μu_μ] = a(t)^2(-u^0)\frac{du^0}{ds}(1 + (u^i)^2) + a(t)^2(-u^0)^2\frac{d(u^i)^2}{ds}.Now, note that for both massive and massless particles, u^0\frac{du^0}{ds} = |\tilde{u}|\frac{d|\tilde{u}|}{ds},where \tilde{u} is the physical velocity, related to the comoving velocity by |\tilde{u}| = \sqrt{-g_{ij}u^iu^j}=a|\vec{u}|.Therefore, we can rewrite the time-derivative of u^μu_μ as follows:\frac{d}{ds}[u^μu_μ] = a(t)^2|\tilde{u}|\frac{d|\tilde{u}|}{ds}(1 + (u^i)^2) + a(t)^2(-u^0)^2\frac{d(u^i)^2}{ds}.Since the second term on the RHS is always 0 (since