Proof of Lorentz Gauge Existence: Help Understanding Schutz 8.3

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SUMMARY

The discussion centers on the proof of the existence of a Lorentz gauge as presented in Schutz 8.3. The key equation is given as $$\bar h^{(new)}_{\mu\nu} = \bar h^{(old)}_{\mu\nu} - \xi_{\mu,\nu} - \xi_{\nu,\mu} + \eta_{\mu\nu}\xi^\alpha_{,\alpha}$$, where $$\bar h$$ represents the trace reverse and $$\xi^\alpha$$ are the gauge functions. The divergence of the new gauge is expressed as $$\bar h^{(new)\mu\nu}_{\,\,\,\,\,\,\,,\nu} = \bar h^{(old)\mu\nu}_{\,\,\,\,\,\,\,\,,\nu} - \xi^{\mu,\nu}_{\,\,\,,\nu}$$. The divergence simplifies to zero due to the properties of dummy summation indices, confirming the correctness of the derivation.

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epovo
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TL;DR
In the derivation of the proof there is a step that I cannot make sense of
In Schutz 8.3, while proving that a Lorentz gauge exists, it is stated that
$$\bar h^{(new)}_{\mu\nu} = \bar h^{(old)}_{\mu\nu} - \xi_{\mu,\nu} - \xi_{\nu,\mu} + \eta_{\mu\nu}\xi^\alpha_{,\alpha}$$

where ##\bar h## is the trace reverse and ##\xi^\alpha## are the gauge functions. Then it follows with:
"Then the divergence is"
$$\bar h^{(new)\mu\nu}_{\,\,\,\,\,\,\,,\nu} = \bar h^{(old)\mu\nu}_{\,\,\,\,\,\,\,\,,\nu} - \xi^{\mu,\nu}_{\,\,\,,\nu}$$
I can't see why the divergence is that! I've tried and tried but I can't see it. Any help?
 
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epovo said:
I can't see why the divergence is that! I've tried and tried but I can't see it. Any help?
$$-\xi^{\nu,\mu}_{\,\,\,,\nu} + \eta^{\mu\nu} \xi^{\alpha}_{\,\,\,,\alpha \nu} = -\xi^{\nu,\mu}_{\,\,\,,\nu} + \xi^{\alpha,\mu}_{\,\,\,,\alpha} = 0,$$
since ##\alpha## and ##\nu## are both dummy summation indices.
 
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True! Thank you
 
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