Aromatic Stretches C=C in Salicylamide

I want to ask a question about the C=C streching modes of Salicylamide. 


I was presented with the following spectra and asked to comment on the bonds causing the peaks. 
[![enter image description here](https://i.stack.imgur.com/bSrZi.png)](https://i.stack.imgur.com/bSrZi.png)


I identified the molecule from this and its NMR spectra to be salicylamide. 


[![Diagram of Salicylamide](https://i.stack.imgur.com/w4Hwq.png)](https://i.stack.imgur.com/w4Hwq.png)


Here is my understanding of the IR spectra (solid state, ATR). 



> 
> * $3391.66 cm ^{-1}$ is the asymmetric stretch of the $\ce{N-H}$ in the amide.
> * $3184.80 cm^{-1}$ is the symmetric stretch of the $\ce{N-H}$ in the amide.
> 
> 
> 


From previous understanding, I was aware that the asymmetric stretch requires more energy, and hence a higher wavenumber than the symmetric stretch.



> 
> * $2736.27 cm^{-1}$ is most possibly the $\ce{O-H}$ peak of the phenol. I attributed this to a few reasons, possibly due to hydrogen bonding or due to the dissociation of the $\ce{O-H}$ bond, which is driven to the right hand side of the equilibrium by the presence of the EWG amide group.
> * $1672.98 cm^{-1}$ is due to the $\ce{C=O}$ stretch of the amide.
> * $1628.09$ and $1589.49 cm^{-1}$ are due to $\ce{C=C}$ aromatic stretches.
> 
> 
> 


I tried to find information to help me describe the $\ce{C=C}$ stretches (i.e. what orientation, any dipole arrows) but I failed to find any information to support my argument. 


I recall in ferrocene and acetalferrocene, the *ring breathing* effect is observed, whereby ferrocene has fewer IR peaks than acetalferrocene due to a lack of a net dipole moment being non-zero, and I wondered whether there was a similar vibrational mode to apply for these two $\ce{C=C}$ peaks, but I was not sure. 


What are the directions of the C=C stretches in the molecule of salicylamide in the range $1580 - 1630 cm^{-1}$?


