# Physics 616

• Prof. Andrew W. Steiner
(or Andrew or "Dr. Steiner")
• Office hour: 103 South College, Thursday 11am
• Email: awsteiner@utk.edu
• Homework: Electronically as .pdf
• You may work with each other on the homework, but you must write the solution in your own words

Use down and up arrows to proceed to the next or previous slide.

## Outline

• Ch. 8 in Schutz

## Nearly Lorentz Coordinates

• Metric $$g_{\alpha \beta} = \eta_{\alpha \beta} + h_{\alpha \beta}$$ with $h_{\alpha \beta} \ll 1$
• Then Schutz shows that with $$g_{\alpha \beta} = {\Lambda^{\mu}}_{\alpha} {\Lambda^{\nu}}_{\beta} g_{\mu \nu}$$ and $${\eta}_{\alpha \beta} = {\Lambda^{\mu}}_{\alpha} {\Lambda^{\nu}}_{\beta} {\eta}_{\mu \nu}$$ we also have $${h}_{\alpha \beta} = {\Lambda^{\mu}}_{\alpha} {\Lambda^{\nu}}_{\beta} {h}_{\mu \nu}$$
• A slightly curved spacetime is a flat space time with a small correction, $h_{\mu \nu}$
• Define space in terms of $h_{\mu \nu}$, which transforms via a "background Lorentz transformation"

## Gauge Transformation

• An example of a gauge transformation is $$x^{\alpha \prime} = x^{\alpha} + \xi^{\alpha}(x^{\beta})$$
• With this definition and assuming $|{\xi^{\alpha}}_{,\beta}|\ll 1$ $${\Lambda^{\alpha \prime}}_{\beta} = \frac{\partial x^{\alpha \prime}}{\partial x^{\beta} } = \delta^{\alpha}_{\beta} + {\xi^{\alpha}}_{,\beta}$$ and (why?) $${\Lambda^{\alpha}}_{\beta \prime} \approx \frac{\partial x^{\alpha \prime}}{\partial x^{\beta} } = \delta^{\alpha}_{\beta} - {\xi^{\alpha}}_{,\beta}$$
• One can verify that (using $\xi_{\alpha} \equiv \eta_{\alpha \beta} \xi^{\beta}$) $$g_{\alpha \prime \beta \prime} = \eta_{\alpha \beta} + h_{\alpha \beta} - \xi_{\alpha,\beta} - \xi_{\beta,\alpha}$$ thus $$h_{\alpha \beta} \rightarrow h_{\alpha \beta} - \xi_{\alpha,\beta} - \xi_{\beta,\alpha}$$

## Riemann Tensor

• The form of the Riemann tensor equation is unchanged $$R_{\alpha \beta \mu \nu} = \frac{1}{2}\left( h_{\alpha \nu,\beta \mu} + h_{\beta \mu, \alpha \nu} - h_{\alpha \mu, \beta \nu} - h_{\beta \nu, \alpha \mu} \right)$$

## Weak-field Einstein Equations

• Obtain new weak-field equations, not coordinate invariant, but allow gauge transformations
• In order to get the Einstein tensor, one must define (using ${h^{\mu}}_{\beta} \equiv \eta^{\mu \alpha} h_{\alpha \beta}$ and $h \equiv {h^{\alpha}}_{\alpha}$) $$\bar{h}^{\alpha \beta} \equiv h^{\alpha \beta} - \frac{1}{2} \eta^{\alpha \beta} h$$ which implies $$h^{\alpha \beta} \equiv {\bar{h}}^{\alpha \beta} - \frac{1}{2} \eta^{\alpha \beta} \bar{h}$$

## Einstein Tensor

• The Einstein tensor is $$G_{\alpha \beta} = - \frac{1}{2} \left[ {\bar{h}_{\alpha \beta,\mu}}^{\mu} + \eta_{\alpha \beta}{\bar{h}_{\mu \mu}}^{\mu \nu} - {\bar{h}_{\alpha \mu,\beta}}^{\mu} - {\bar{h}_{\beta \mu,\alpha}}^{\mu} + {\cal O}\left(h^2_{\alpha \beta}\right) \right]$$ where $f^{,\mu} \equiv \eta^{\mu \nu} f_{, \nu}$
• More definitions $$\Box f \equiv {f^{,\mu}}_{,\mu} = \eta^{\mu \nu} f_{,\mu \nu} = \left( - \frac{\partial^2}{\partial t^2} + \nabla^2 \right)f$$
• If we choose (not the only possible choice) $${\bar{h}^{\mu \nu}}_{,\nu} = 0$$ (the Lorentz gauge, aka harmonic gauge, etc.) then $$G^{\alpha \beta} = -\frac{1}{2} \Box \bar{h}^{\alpha \beta} = 8 \pi T^{\alpha \beta}$$
• That is, Einstein's field equations reduce down to wave equations

## Lorentz Gauge in E&M

• From Jackson, the Lorentz gauge is $$\begin{eqnarray} \mathbf{A} &\rightarrow& \mathbf{A} + \nabla \Lambda \\ \Phi &\rightarrow& \Phi - \frac{1}{c} \frac{\partial \Lambda}{\partial t} \end{eqnarray}$$ where $$\nabla^2 \Lambda - \frac{1}{c^2} \frac{\partial^2 \Lambda} {\partial t^2} =0$$ in which case the wave equations are $$\Box A^{\alpha} = \frac{4 \pi}{c} J^{\alpha}$$ and the Lorentz condition is $$\partial_{\alpha} A^{\alpha} = 0$$
• There is a connection with covariant derivatives: often one defines a covariant derivative in a way as to ensure that it is invariant under a gauge transformation $${\cal L}_{\mathrm{E\&M}} = - \frac{1}{4} F_{\mu \nu} F^{\mu \nu} + \bar{\psi} \left[ \left(i \partial_{\mu}- e A_{\mu}\right)\gamma^{\mu} - m\right] \psi$$

## Solution of the wave equations

• In nearly Newtonian systems, $|\mathbf{v}|\ll 1$ and $P \ll \rho$ so $|T^{00}| \gg |T^{0i}| \gg |T^{ij}|$
• If we also assume that $|\bar{h}^{00}|$ dominates, then $$\Box \bar{h}^{0 0} = - 16 \pi \rho$$
• And for small velocities we have $$\frac{\partial}{\partial t} \sim v \frac{\partial}{\partial x} \ll \frac{\partial}{\partial x}$$ so $\Box \approx \nabla^2$
• Thus $$\nabla^2 \bar{h}^{00} = -16 \pi \rho$$ but Newtonian result is $\nabla^2 \phi = 4 \pi \rho$, so $\bar{h}^{00} = - 4 \phi$.
• This implies $h^{00} = h^{ij} = -2 \phi$ thus $$ds^2 = -\left( 1+2 \phi\right)dt^2 + \left(1-2\phi\right) \left( dx^2 + dy^2 + dz^2 \right)$$

## Far Field

• Far away from a static source, $$\phi = -1/4 \bar{h}^{00}$$ and the Newtonian result (in natural units) is $$-M/r$$ so our metric is approximately $$ds^2 = -\left( 1-2 M/r\right)dt^2 + \left(1+2 M/r\right) \left( dx^2 + dy^2 + dz^2 \right)$$
• A critical point is that we use this to define what we mean by mass, i.e. "gravitational mass"
• This is not the same as adding up all the component masses, and an important correction for strong-field systems

## Lagrangian Formulation

• One may also define a GR action $$S_H = \int d\vec{x} {\cal L}_{H}$$ as an integral over the Lagrangian $${\cal L}_H = \sqrt{-g} R$$ where $R$ is the Ricci scalar defined earlier
• One can then derive Einstein's field equations from the Euler-Lagrange equations
• Scalar tensor theories: $$S = \int d\vec{x}~\sqrt{-g}\left[ f(\lambda) R + \frac{1}{2} g^{\mu \nu} \left( \partial_{\nu} \lambda \right) \left( \partial_{\nu} \lambda \right) - V(\lambda) \right]$$

## Group Work

• Complete 7.10 in Schutz (hint, take a look at Carroll's book for help)