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1Preliminaries

(1)	Twin paradox: Two twins, Alice and Bob, are seprarated on their $20^{th}$ birthday. While Alice remains on Earth (which is an inertial frame to a very good approximation), Bob departs at $80\%$ of the speed of light towards Planet $X$, $8$ light-years away from Earth. Therefore Bob reaches his destination $10$ years later (as measured on the Earth’s frame). After a short stay, he returns to Earth, again at $80\%$ of the speed of light. Consequently Alice is $40$ years old when she sees Bob again. (a) How old is Bob When they meet again? Solution:   The amount of time as measured on the ship’s clocks and the aging of the travelers during their trip will be reduced by the factor $\alpha =\sqrt{1-v^2}=0.6$, the reciprocal of the Lorentz factor (time dilation). In this case the traveler will have aged only $0.6\times 20=12$ years when he return. (b) How can the asymmetry in the twins’ ages be explained? Notice that from Bob’s point of view he is at rest in his spaceship and it is the Earth which moves away and then back again. Solution:   From the traveler’s perspective. Bob knows that the distant Planet $X$ and the Earth are moving relative to the ship at speed $v=\frac{4}{5}$ during the trip. In their rest frame the distance between the Earth and the Planet $X$ is $\alpha d=0.6\times 8=4.8$ light-years (length contraction), for both the outward and return journeys.   Each half of the journey takes $\alpha d/v=4.8/0.8=6$ years, and the round trip takes twice as long ($12$ years). Their calculations show that they will arrive home having aged $12$ years.   The traveler’s final calculation about their aging is in complete agreement with the calculations of those on Earth, though they experience the trip quite differently from Alice who stays at home. (c) Imagine that each twin watches the other through a very powerful telescope. What do they see? In particular, how much time do they experience as they see one year elapse for their twin?
(2)	A particular simple matter model is that of a smooth massless scalar field $\phi :M\to \mathbb{R}$, whose energy-momentum tensor is $$T_{\mu \nu }={\partial }_{\mu }\phi {\partial }_{\nu }\phi -\frac{1}{2}({\partial }_{\alpha }\phi {\partial }^{\alpha }\phi )g_{\mu \nu }.$$  Show that if the Lorentzian manifold $(M,g)$ satisfies the Einstein equations with this matter model then $\phi$ satisfies the wave equation: $$\square \phi =0⟺{\nabla }^{\mu }{\partial }_{\mu }\phi =0.$$
Solution:	$$\begin{array}{rl}0={\nabla }^{\mu }{T}_{\mu \nu }& ={\nabla }^{\mu }({\partial }_{\mu }\phi {\partial }_{\nu }\phi )-\frac{1}{2}{g}_{\mu \nu }{\nabla }^{\mu }({\partial }_{\alpha }\phi {\partial }^{\alpha }\phi )\\ & =({\nabla }^{\mu }{\partial }_{\mu }\phi ){\partial }_{\nu }\phi +{\partial }_{\mu }\phi ({\nabla }^{\mu }{\partial }_{\nu }\phi )-\frac{1}{2}{\nabla }_{\nu }({\partial }_{\alpha }\phi {\partial }_{\beta }\phi )g^{\alpha \beta }.\end{array}$$Considering the last two terms, $$\begin{array}{rl}{\partial }_{\mu }\phi ({\nabla }^{\mu }{\partial }_{\nu }\phi )-\frac{1}{2}{\nabla }_{\nu }({\partial }_{\alpha }\phi {\partial }_{\beta }\phi )g^{\alpha \beta }& ={\partial }_{\mu }\phi g^{\mu \alpha }{\nabla }_{\alpha }({\partial }_{\nu }\phi )-\frac{1}{2}{\nabla }_{\nu }({\partial }_{\alpha }\phi ){\partial }_{\beta }\phi g^{\alpha \beta }-\frac{1}{2}{\partial }_{\alpha }\phi ({\nabla }_{\nu }{\partial }_{\beta }\phi )g^{\alpha \beta }\\ & =g^{\mu \alpha }({\partial }_{\alpha }{\partial }_{\nu }\phi -{\Gamma }_{\alpha \nu }^{\beta }{\partial }_{\beta }\phi ){\partial }_{\mu }\phi -\frac{1}{2}({\partial }_{\nu }{\partial }_{\alpha }\phi -{\Gamma }_{\gamma \alpha }^{\theta }{\partial }_{\theta }\phi ){\partial }_{\beta }\phi g^{\alpha \beta }-\frac{1}{2}({\partial }_{\nu }{\partial }_{\beta }\phi -{\Gamma }_{\nu \beta }^{\theta }{\partial }_{\theta }\phi ){\partial }_{\alpha }\phi g^{\alpha \beta }.\end{array}$$We separate the terms with Christoffel coefficient and the terms without, then the formula becomes $0$.   Hence we have $({\nabla }^{\mu }{\partial }_{\mu }\phi ){\partial }_{\nu }\phi =0$. By the arbitrariness of $\nu$, we have ${\nabla }^{\mu }{\partial }_{\mu }\phi =0$.
(3)	The energy-momentum tensor for perfect fluid is $$T_{\mu \nu }=(\rho +p)U_{\mu }{U}_{\nu }+p{g}_{\mu \nu },$$where $\rho$ is the fluid’s rest density, $p$ is the fluid’s rest pressure, and $U$ is a unit timelike vector field tangent to the histories of the fluid particles. Show that: (a) $(T_{\mu \nu })=\mathrm{diag}(\rho ,p,p,p)$ in any orthonormal frame including $U$; Solution:   Consider any orthonormal frame with $U$ being the first unit vector, then $U$ has the component $(1,0,0,0)$. Hence $$\{\begin{array}{ll}{T}_{11}=\rho +p-p=\rho & \\ T_{\mu \mu }=p,& \mu =2,3,4\\ T_{\mu \nu }=0,& \mu \ne \nu \end{array}.$$ (b) the motion equations for the perfect fluid are $$\{\begin{array}{l}\mathrm{div}(\rho U)+p\mathrm{div}U=0\\ (\rho +p){\nabla }_{U}U=-(\mathrm{grad}p{)}^{\perp }\end{array},$$where ${}^{\perp }$ represents the orthogonal projection on the spacelike hyperplane orthogonal to $U$. Solution:   Similar to the example of pressureless perfect fluid, we have $$0=\mathrm{div}((\rho +p)U)U+(p+\rho ){\nabla }_{U}U+(\mathrm{grad}p).$$  Consider the hyperplane orthogonal to $U$, we have the second identity. And notice that $$⟨\mathrm{grad}p,U⟩+\mathrm{div}(pU)=p\mathrm{div}U,$$we get the first identity.