Problem 1.

Transformation of Christoffel symbols:
(a) Show that, under a coordinate transformation, the components of the Christoffel symbol transform as follows:
$$\Gamma_{\beta^{\prime} \gamma^{\prime}}^{\alpha^{\prime}}=\frac{\partial x^{\alpha^{\prime}}}{\partial x^{\alpha}} \frac{\partial x^{\beta}}{\partial x^{\beta^{\prime}}} \frac{\partial x^{\gamma}}{\partial x^{\gamma^{\prime}}} \Gamma_{\beta \gamma}^{\alpha}-\frac{\partial^{2} x^{\alpha^{\prime}}}{\partial x^{\beta} \partial x^{\gamma}} \frac{\partial x^{\beta}}{\partial x^{\beta^{\prime}}} \frac{\partial x^{\gamma}}{\partial x^{\gamma^{\prime}}}$$
Do this by considering the form of the Christoffel symbol in terms of derivatives of the metric.
(b) Show that, using this rule, the components of the covariant derivative of a vector transform as tensors should:
$$\nabla_{\alpha^{\prime}} A^{\beta^{\prime}}=\frac{\partial x^{\alpha}}{\partial x^{\alpha^{\prime}}} \frac{\partial x^{\beta^{\prime}}}{\partial x^{\beta}} \nabla_{\alpha} A^{\beta}$$

Proof .

We ask you to show that
\begin{aligned} \Gamma_{\beta^{\prime} \gamma^{\prime}}^{\alpha^{\prime}} &=\frac{\partial x^{\alpha^{\prime}}}{\partial x^{\alpha}} \frac{\partial x^{\beta}}{\partial x^{\beta^{\prime}}} \frac{\partial x^{\gamma}}{\partial x^{\gamma^{\prime}}} \Gamma_{\beta \gamma}^{\alpha}-\frac{\partial^{2} x^{\alpha^{\prime}}}{\partial x^{\beta} \partial x^{\gamma}} \frac{\partial x^{\beta}}{\partial x^{\beta^{\prime}}} \frac{\partial x^{\gamma}}{\partial x^{\gamma^{\prime}}} \\ &=L^{\alpha^{\prime}}{ }_{\alpha} L^{\beta}{ }_{\beta^{\prime}} L^{\gamma}{ }_{\gamma^{\prime}}^{\gamma} \Gamma_{\beta \gamma}^{\alpha}-L^{\beta}{ }_{\beta^{\prime}} L^{\gamma}{ }_{\gamma^{\prime}} \partial_{\beta} L^{\alpha^{\prime}}{ }_{\gamma} \end{aligned}
You may find at the end of your calculation that you have instead derived a rule that looks like
$$\Gamma_{\beta^{\prime} \gamma^{\prime}}^{\alpha^{\prime}}=L^{\alpha^{\prime}}{ }_{\alpha} L^{\beta}{ }_{\beta^{\prime}} L^{\gamma}{ }_{\gamma^{\prime}} \Gamma_{\beta \gamma}^{\alpha}+L_{\beta^{\prime}}^{\beta} L^{\alpha^{\prime}}{ }_{\gamma} \partial_{\beta} L^{\gamma}{ }_{\gamma^{\prime}}^{\gamma}$$
This may look totally wrong $-$ the sign on the final term is incorrect. However, by inspecting this closely, you’ll see that the matrix being differentiated in the second term is not the same in the two versions $-$ the primed and unprimed indices are in opposite locations. By noting that $L^{\gamma}{ }_{\gamma^{\prime}} L^{\alpha^{\prime}} \gamma=\delta^{\alpha^{\prime}} \gamma^{\prime},$ you should be able to show that these two formulas are equivalent.

Problem 2.

Relativistic Euler equation
(a) Starting from the stress-energy tensor for a perfect fluid, $\mathbf{T}=\rho \vec{U} \otimes \vec{U}+P \mathbf{h}$, where $\mathbf{h}=\mathbf{g}^{-1}+\vec{U} \otimes \vec{U},$ using local energy momentum conservation, $\nabla \cdot \mathbf{T}=0$, derive the relativistic Euler equation,
$$(\rho+P) \nabla_{\vec{U}} \vec{U}=-\mathbf{h} \cdot \nabla P$$
(Note: Because both $\mathbf{T}$ and $\mathbf{h}$ are symmetric tensors, there is no ambiguity in the dot products that appear in this problem.)
(b) For a nonrelativistic fluid $\left(\rho \gg P, v^{t} \gg v^{i}\right)$ and a cartesian basis, show that this equation reduces to the Euler equation,
$$\frac{\partial v_{i}}{\partial t}+v_{k} \partial_{k} v_{i}=-\frac{1}{\rho} \partial_{i} P$$
$(i, k$ are spatial indices running from 1 to 3.) What extra terms are present if the connection is non-zero (e.g., spherical coordinates)?
(c) Apply the relativistic Euler equation to Rindler spacetime for hydrostatic equilibrium. Hydrostatic equilibrium means that the fluid is at rest in the $\bar{x}$ coordinates, i.e. $U^{\bar{x}}=0 .$ Suppose that the equation of state (relation between pressure and density) is $P=w \rho$ where $w$ is a positive constant. Find the general solution $\rho(\bar{x})$ with $\rho(0)=\rho_{0}$.
(d) Suppose now instead that $w=w_{0} /(1+g \bar{x})$ where $w_{0}$ is a constant. Show that the solution is $\rho(\bar{x})=\rho_{0} \exp (-\bar{x} / L$. Find $L,$ the density scale height, in terms of $g$ and $w_{0}$. Convert to “normal” units by inserting appropriate factors of $c-L$ should be a length.
(e) Compare your solution to the density profile of a nonrelativistic, plane-parallel, isothermal atmosphere (for which $P=\rho k T / \mu,$ where $T$ is temperature and $\mu$ is the mean molecular weight) in a constant gravitational field. [Use the nonrelativistic Euler equation with gravity: add a term $-\partial_{i} \Phi=g_{i}$, where $\Phi$ is Newtonian gravitational potential and $g_{i}$ is Newtonian gravitational acceleration, to the right hand side of Eq.
(3).] Why does hydrostatic equilibrium in Rindler spacetime – where there is no gravity – give such similar results to hydrostatic equilibrium in a gravitational field?

Proof .

relativistic Euler equation
(a) Starting from the stress-energy tensor for a perfect fluid, $\mathbf{T}=\rho \vec{U} \otimes \vec{U}+P \mathbf{h}$, where $\mathbf{h}=\mathbf{g}^{-1}+\vec{U} \otimes \vec{U},$ using local energy momentum conservation, $\nabla \cdot \mathbf{T}=0,$ derive the relativistic Euler equation,
$$(\rho+P) \nabla_{\vec{U}} \vec{U}=-\mathbf{h} \cdot \nabla P$$
(Note: Because both $\mathbf{T}$ and $\mathbf{h}$ are symmetric tensors, there is no ambiguity in the dot products that appear in this problem.)

It turns out that if you just evaluate $\nabla \cdot \mathbf{T}=0,$ you are unlikely to get a very useful answer. The result you get is, technically, an Euler equation, but it does not naturally reduce to the form we are hoping to show. In particular, you won’t recover the nonrelativistic limit in a natural way.

A more useful form is obtain by separately equating to zero the components of $\nabla \cdot \mathbf{T}$ parallel to and orthogonal to the fluid’s 4 -velocity, $\vec{u}$. In particular, if we define $j^{\nu} \equiv$ $\nabla_{\mu} T^{\mu \nu},$ then the equations
\begin{aligned} j^{\nu} u_{\nu} &=0 & &(\text { Component parallel to } \vec{u}) \\ j^{\nu} h^{\lambda}{ }_{\nu} &=0 & &(\text { Component orthogonal to } \end{aligned}
give us useful information. The second equation in particular can be considered the relativistic analog of the usual Euler equation. (It’s worth examining the first equation and seeing what it corresponds to.)

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#### Relativity and Gravitation | Part III (MMath/MASt)

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