Why do negative energy solutions cause issues in Padmanabhan's Gravitation?

ShayanJ
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I'm reading Padmanabhan's Gravitation. In section 3.4, at some point he writes the Lagrangian below:
## L=-\frac 1 2 \partial_\sigma V_\rho \partial^\sigma V^\rho-J_\rho V^\rho ##
He then says that the sign of the kinetic term causes the time component of the V field to propagate as a wave with negative energy in the absence of sources which is unphysical.(Multiplying the Lagrangian by a negative sign causes the spatial components to become unphysical.) Then, to solve this problem, he says that the theory should have a gauge symmetry under ## V_\sigma \rightarrow V_\sigma+\partial_\sigma F ## so that we can get rid of the unphysical degrees of freedom.
I understand why negative energy solutions aren't desirable but I don't understand the sentence "So in the absence of the source, the ##V^0## degree of freedom will propagate as a wave carrying negative energy."
The above Lagrangian gives the equations of motion (when ## J_\rho=0 ##), ## \partial_\sigma \partial^\sigma V_\rho=0 ## for all four components. So the equations of motion are identical for all components. How is it that they are different in the sense above?
Also, as you probably noticed, the above equation of motion is our old friend ## \nabla^2 \phi=\frac{1}{c^2} \frac{\partial^2 \phi}{\partial t^2} ## which we have no problem with. So where does the above problem come from?
I'll appreciate it if the answers contain more mathematics than words.
Thanks a lot
 
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