Gauge Transformations and the TT Gauge

1. Gauge Freedom in Linearized Gravity

The complexity of our linearized equation stems from a redundancy in our description. The metric perturbation $h_{\mu \nu }$ is not unique for a given physical spacetime. We can change our coordinates slightly, and this will change the components of $h_{\mu \nu }$ without altering the underlying geometry. This freedom to choose coordinates is what we call gauge freedom.

Let's consider an infinitesimal coordinate transformation:
$x^{\mu }\to x^{\prime \mu }=x^{\mu }+{\xi }^{\mu }(x)$
where ${\xi }^{\mu }(x)$ is a vector field of small displacements. Given your mathematical background, you will recognize this as the infinitesimal generator of a diffeomorphism. We want to see how the metric perturbation $h_{\mu \nu }$ changes under this transformation.

From this reading, you'll find the key transformation laws (to first order in $\xi$ and $h$):

• $h_{\mu \nu }^{\prime }\approx h_{\mu \nu }-{\partial }_{\mu }{\xi }_{\nu }-{\partial }_{\nu }{\xi }_{\mu }$
• ${\bar{h}}_{\mu \nu }^{\prime }\approx {\bar{h}}_{\mu \nu }-{\partial }_{\mu }{\xi }_{\nu }-{\partial }_{\nu }{\xi }_{\mu }+{\eta }_{\mu \nu }{\partial }_{\alpha }{\xi }^{\alpha }$

The crucial insight is that since we can choose ${\xi }^{\mu }$ to be any small vector field, we have the freedom to "adjust" the components of ${\bar{h}}_{\mu \nu }$. Our goal is to choose ${\xi }^{\mu }$ to make the linearized Einstein equation as simple as possible.

2. The Lorenz Gauge

The first major simplification comes from imposing the Lorenz gauge condition, named in analogy to the condition ${\partial }_{\mu }{A}^{\mu }=0$ in electromagnetism. We seek a coordinate system where:
${\partial }^{\nu }{\bar{h}}_{\mu \nu }=0$

If we can find such a system, look what happens to the linearized Einstein tensor. The first, second, and third terms all involve a divergence of ${\bar{h}}_{\mu \nu }$ and vanish, leaving only the d'Alembertian term:
$G_{\mu \nu }=-\frac{1}{2}\square {\bar{h}}_{\mu \nu }$
The linearized Einstein Field Equation in the Lorenz gauge thus becomes a set of 10 uncoupled wave equations:
$\square {\bar{h}}_{\mu \nu }=-16\pi G{T}_{\mu \nu }$
This is a profound simplification! But can we always find coordinates that satisfy the Lorenz gauge condition? Yes. We must show that we can always find a ${\xi }^{\mu }$ that takes us from an arbitrary gauge to one where ${\partial }^{\nu }{\bar{h}}_{\mu \nu }^{\prime }=0$.

The key result is that we can always move to a coordinate system where the physics of weak gravitational fields is described by a simple wave equation. In a vacuum ($T_{\mu \nu }=0$), we have $\square {\bar{h}}_{\mu \nu }=0$, whose solutions are plane waves.

3. Residual Gauge Freedom and the Path to the TT Gauge

We have simplified the dynamics, but we haven't exhausted our gauge freedom. The Lorenz gauge condition does not uniquely fix the coordinates. If we are already in a Lorenz gauge (${\partial }^{\nu }{\bar{h}}_{\mu \nu }=0$) and we perform another gauge transformation $x^{\prime \mu }=x^{\mu }+{\xi }^{\mu }$, the new field is ${\bar{h}}_{\mu \nu }^{\prime }={\bar{h}}_{\mu \nu }-({\partial }_{\mu }{\xi }_{\nu }+{\partial }_{\nu }{\xi }_{\mu }-{\eta }_{\mu \nu }{\partial }_{\alpha }{\xi }^{\alpha })$. The new field will also be in the Lorenz gauge if:
${\partial }^{\nu }{\bar{h}}_{\mu \nu }^{\prime }={\partial }^{\nu }{\bar{h}}_{\mu \nu }-{\partial }^{\nu }(\dots )=0-\square {\xi }_{\mu }=0$
This means we still have the freedom to make coordinate transformations, as long as the generating vector ${\xi }^{\mu }$ itself satisfies the homogeneous wave equation, $\square {\xi }^{\mu }=0$. This is our residual gauge freedom.

We now specialize to a single plane wave solution in a vacuum, propagating in the $+z$ direction:
${\bar{h}}_{\mu \nu }(x)=A_{\mu \nu }{e}^{i{k}_{\sigma }{x}^{\sigma }}\text{with}{k}^{\mu }=(\omega ,0,0,\omega )$
The condition $\square {\bar{h}}_{\mu \nu }=0$ implies $k_{\sigma }{k}^{\sigma }=0$, which is satisfied by our choice of $k^{\mu }$. The Lorenz condition ${\partial }^{\nu }{\bar{h}}_{\mu \nu }=0$ becomes an algebraic constraint on the amplitude matrix: $k^{\nu }{A}_{\mu \nu }=0$.

We will now use our four degrees of residual gauge freedom (the four components of ${\xi }^{\mu }$) to impose four additional constraints on $A_{\mu \nu }$, with the goal of eliminating as many non-physical components as possible. This final choice of gauge is the transverse-traceless (TT) gauge.

4. The TT Gauge Matrix and Physical Degrees of Freedom

Let's see what the metric perturbation $h_{\mu \nu }$ looks like once we've imposed all these conditions for our z-propagating wave. From now on, since we are in the TT gauge, $h_{\mu \nu }={\bar{h}}_{\mu \nu }$.

1. Symmetry: $h_{\mu \nu }=h_{\nu \mu }$ (10 independent components initially).
2. Transverse Condition: We choose an observer at rest, $u^{\mu }=(1,0,0,0)$. The condition $h_{\mu \nu }{u}^{\nu }=0$ implies $h_{\mu 0}=0$ for all $\mu$.
3. Lorenz Gauge Condition: $k^{\nu }{h}_{\mu \nu }=0⟹k^0{h}_{\mu 0}+k^3{h}_{\mu 3}=0$. Since $\omega \ne 0$ and $h_{\mu 0}=0$, this forces $h_{\mu 3}=0$ for all $\mu$.
4. Traceless Condition: $h={\eta }^{\mu \nu }{h}_{\mu \nu }=h_{00}+h_{11}+h_{22}+h_{33}=0$. With our other conditions, this becomes $h_{11}+h_{22}=0$, or $h_{22}=-h_{11}$.

Let's assemble the matrix $h_{\mu \nu }^{TT}$. The transverse and Lorenz conditions tell us that any component with a 0 or a 3 index must be zero. The only potentially non-zero components are $h_{11}$, $h_{12}$, $h_{21}$, and $h_{22}$.

• Symmetry requires $h_{12}=h_{21}$.
• The traceless condition requires $h_{22}=-h_{11}$.

Remarkably, out of the original 10 components, only two are independent! We can write the general form of the TT gauge perturbation matrix as a superposition of two basis matrices:

where we have defined $h_{+}\equiv h_{11}$ and $h_{\times }\equiv h_{12}$. These two components, $h_{+}$ and $h_{\times }$, represent the two independent polarizations of a gravitational wave. They are the only physically real, propagating degrees offreedom. All other components were gauge artifacts.

Conclusion

In this lesson, we navigated the crucial concept of gauge freedom in linearized gravity. We saw that the apparent complexity of the linearized field equations was an illusion created by redundant, non-physical information in our description of the metric perturbation.

Key Takeaways:

• A small coordinate change $x^{\prime \mu }=x^{\mu }+{\xi }^{\mu }$ is a gauge transformation that alters the components of the metric perturbation $h_{\mu \nu }$.
• We can use this freedom to impose the Lorenz gauge condition (${\partial }^{\nu }{\bar{h}}_{\mu \nu }=0$), which simplifies the linearized EFE to a standard wave equation: $\square {\bar{h}}_{\mu \nu }=-16\pi G{T}_{\mu \nu }$.
• Even within the Lorenz gauge, there is residual gauge freedom for any transformation ${\xi }^{\mu }$ that satisfies $\square {\xi }^{\mu }=0$.
• For a plane wave in a vacuum, we use this residual freedom to fix the transverse-traceless (TT) gauge, which imposes $h_{0\mu }=0$ and $h=0$.
• This procedure reveals that a gravitational wave has only two physical, propagating degrees of freedom, called the plus (+) and cross (×) polarizations.

Preview of Next Lesson:

We have found the mathematical form of a gravitational wave. The next logical step is to solve the vacuum wave equation, $\square h_{\mu \nu }^{TT}=0$, to find explicit plane wave solutions. We will then identify the two polarization states, $h_{+}$ and $h_{\times }$, and understand what they physically represent.