# Linearization of Einstein’s equation

The general mathematical form of Einstein’s equation is

$\mathbf{G}_{\alpha\beta}=\mathbf{T}_{\alpha\beta}$

where $\mathbf{G}$ is the Einstein tensor and $\mathbf{T}$ is the energy-momentum tensor.

One problem in solving this equation lies in the high non-linearity of this differential equation. This means that, considering the Riemann tensor, $\mathbf{R}$ involves the products of the metric tensor $g$ and its first order derivatives, second order derivatives. So, some times it is highly desired that we can linearize this differential equation. But with respect to which starting point? Perhaps the most natural setting is the Minkowski space, $\mathbb{R}^{1+n}$ with metric $m(m_{00}=-1)$. By linearization, it means that we have a metric $g$ that differs from $m$ slightly, and we see what will happen. In other words, we assume

$g_{\alpha\beta0}=m_{\alpha\beta}+h_{\alpha\beta}$

with $|h_{\alpha\beta}|<<1$.

With this condition, we see that the Christofell symbol writes as

$\Gamma_{\alpha\beta}^{\gamma}=1/2m^{\gamma\rho}(\frac{\partial h_{\alpha\rho}}{\partial x^{\beta}}+\frac{\partial h_{\beta\rho}}{\partial x^{\alpha}}-\frac{\partial h_{\alpha\beta}}{\partial x^{\rho}})+O(h^2)$

So, after some complicated yet routine calculations, we get that

$\mathbf{G}_{\alpha\beta}=1/2(-\Box \bar{h}_{\alpha\beta}+\partial^{\rho}(\partial_{\alpha}\bar{h}_{\beta\rho}+\partial_{\beta}\bar{h}_{\alpha\rho})-m_{\alpha\beta}\partial^{\rho}\partial^{\gamma}\bar{h}_{\rho\gamma})$

where $\bar{h}_{\alpha\beta}=h_{\alpha\beta}-1/2m_{\alpha\beta}m^{\gamma\rho}h_{\gamma\rho}$.

One important observation is that, the Einstein tensor is invariant under a transformation like

$h_{\alpha\beta}\rightarrow h_{\alpha\beta}+\partial_{\alpha}X_{\beta}+\partial_{\beta}X_{\alpha}$ where $X$ is any vector field on this manifold. In other words, Einstein tensor is invariant under the transformation in terms of $\bar{h}$, $\bar{h}_{\alpha\beta}\rightarrow\bar{h}_{\alpha\beta}+\overline{\partial_{\alpha}X_{\beta}+\partial_{\beta}X_{\alpha}}=\bar{h}_{\alpha\beta}+\partial_{\alpha}X_{\beta}+\partial_{\beta}X_{\alpha}-\partial^{\rho}\partial_{\rho}m_{\alpha\beta}$

Notice that, if we take a derivative, we get that

$\partial^{\beta}(\bar{h}_{\beta\alpha}+\partial_{\alpha}X_{\beta}+\partial_{\beta}X_{\alpha}-\partial^{\rho}\partial_{\rho}m_{\alpha\beta})=\partial^{\beta}\bar{h}_{\alpha\beta}+\Box X_{\alpha}$. This means that we can choose some vector field $X$ such that the last term is zero. Thus, a priori, we can pose the condition on $h$, $\partial^{\beta}\bar{h}_{\alpha\beta}=0$.

This having been done, we see readily that

$\mathbf{G}_{\alpha\beta}=-1/2\Box \bar{h}_{\alpha\beta}+)(h^2)$

So, if we ignore the higher order terms, we get a differential equation like

$\Box\bar{h}_{\alpha\beta}+2\mathbf{T}_{\alpha\beta}=0$

So, at last we get an equation totally linear in the metric $h$. In fact, it is a wave equation, and we all know how to solve such equations.

One problem remains: what does the invariance of the Einstein tensor under this transformation mean? In fact, this comes from the invariance of the Riemann tensor under this gauge transformation. For details, see here.