Trace-reverse field in linearized GR

[michael879]

Ok I feel like there is a really simple answer to this but I've been trying to get this to work for days and I just can't. Here is the basic problem:
The "free-field" lagrangian for weak perturbations in the metric on a flat space is:
$$L = \frac{1}{4}(\partial^\sigma h_{\mu\nu}\partial_\sigma h^{\mu\nu} - \partial^\mu h \partial_\mu h)+\frac{1}{2}\partial_\mu h^{\mu\nu}(\partial_\nu h - \partial^\sigma h_{\nu\sigma})$$
Solving for the E-L equations, and using the gauge $$\partial_\nu h^{\mu\nu}=\frac{1}{2}\partial^\mu h$$, I get the following equation:
$$\partial^\sigma\partial_\sigma h_{\mu\nu} - \frac{1}{2} \eta_{\mu\nu}\partial^\mu\partial_\mu h = 0$$
and plugging in the trace-reverse of h, it is easy to get the final result:
$$\partial^\sigma\partial_\sigma \overline{h}_{\mu\nu} = 0$$

So far this all agrees with my GR book (which doesn't take the lagrangian approach so its a good check). The problem I'm having is when I try to use the trace reverse from step 1. The lagrangian I get (I pre-set the same gauge as above to save myself some typing) is:
$$L = \frac{1}{4}\partial^\sigma \overline{h}_{\mu\nu}\partial_\sigma \overline{h}^{\mu\nu} - \frac{1}{8}\partial^\mu \overline{h} \partial_\mu \overline{h}$$
However, this is EXACTLY the same lagrangian as the normal field (again, only after the gauge I mentioned has been set), and will therefore give the same EOM. So in the end I get:
$$\partial^\sigma\partial_\sigma \overline{h}_{\mu\nu} - \frac{1}{2} \eta_{\mu\nu}\partial^\sigma\partial_\sigma \overline{h} = 0$$
Am I going crazy?? My logic seems sound but this is an OBVIOUS inconsistancy. I'd be tempted to just ignore the trace-reversed field but it makes a lot of things simpler and I'd rather use it...

 

[michael879]

^bump, anyone??

 

[Physics Monkey]

Perhaps I misunderstand your notation, but if $$\bar{h} = \eta^{ab} \bar{h}_{ab}$$ then doesn't your last equation imply (by contracting with $$\eta$$) that $$\partial^a \partial_a \bar{h} = 0$$ and hence that $$\partial^a \partial_a \bar{h}_{bc} = 0$$.

I haven't checked anything else you said, but I'll be back later if this doesn't help.

 

[michael879]

wow, your right, that completely resolved the "inconsistency". Thanks, I've been staring at these equations for waaaay too long. Now I'm just wondering why my GR book even bothered with the trace-reversed field since you can get the same equation for the regular one.