Far away from the field sources, where the energy-momentum tensor $$T_{mn}=0 \tag{m,n=0,1,2,3}$$ The linearized EFE becomes $$\Box \bar h_{mn}=0 \tag{1}$$

where $\bar h_{mn}$ is the trace-reverse tensor, defined by $$\bar h_{mn}=h_{mn}-\frac{1}{2}\eta_{mn}h$$ and $h_{mn}$ is the perturbation tensor.

Eq. (1) has plane-wave solutions of the form $$\bar h_{mn}=A_{mn}e^{ik_qx^q} \tag{2}$$ where $A_{mn}$ and $k_q$ are the amplitude tensor and the wave vector respectively.

The contravariant form of the gauge transformations of the perturbation tensor $$h^{mn}\rightarrow \tilde h^{mn}=h^{mn}-\partial^m\theta^n-\partial^n\theta^m \tag{3}$$

$\theta ^n$ is a four-vector which can be written in the form $$\theta^m=iC^me^{k_qx^q} \tag{4}$$

($C^m$ is a constant transverse four-vector perpendicular to the wave vector $k^p$, that is, $k_mC^m=0$)

such that the gauge transformation preserves the harmonic gauge boundary condition $$\partial_m\tilde h^{mn}=\partial_mh^{mn}=0 \tag{5}$$

(so that (4) gives $\Box \theta^n=\partial_m\theta^m=0$) The trace-reverse tensor under the gauge transformations: $$\tilde{\bar h}_{mn}=(A_{mn}+k_mC_n+k_nC_m)e^{ik_qx^q}=\tilde A_{mn}e^{ik_qx^q} \tag{6}$$ which is obtained by substituting the covariant form of $(4)$ into $$\tilde {\bar h}_{mn}=\bar h_{mn}-\partial_m\theta_n-\partial_n\theta_m+\eta_{mn}\partial_p\theta^p $$

Thus the transformed amplitude tensor takes the form $$\tilde A_{mn}=(A_{mn}+k_mC_n+k_nC_m) \tag{7}$$ By choosing $$\tilde A_{m0}=\tilde A_{0m}=A_{m0}+k_mC_0+k_0C_m=0 \tag{8}$$

all the temporal components of the tensor $A_{mn}$ will vanish.

*But how does setting the temporal components of the transformed amplitude tensor $\tilde A_{m0}$ to zero also give $A_{m0}=0$?

In other words if certain components of the (gauge) transformed tensor are zero, are the corresponding components of the original tensor necessarily zero as well? (if yes, how?)

Or is $A_{m0}$ only zero due to the presence of other conditions listed above?


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