James,
Contracting a wave contracts it's energy. All waves still travel at c. I mentioned the multiple 'cases', and believe not distinguishing between them is why you think it 'looks wrong'. Ergo;
If two emitters, side by side, each emit one photon per nanosec, but one emits high energy (short 'blue' wavelength) and the other low (red), then because they then ARRIVE at the SAME FREQUENCY, the blue shifted photons will impart a higher charge. This is easy to be confused about as wave peak arrival rate can ALSO be expressed in 'frequency'.
It is then all about 'WAVENUMBER' 'N' (an important conserved quantity in optics). For any given amplitude it in 'N' that defines the energy. So;
In the case I've just described; where the same number of photons are emitted but at different wavelengths, the wavenumber so ENERGY of the blue ones is higher. The same number received as charge are re-emitted.
BUT; In the case previously discussed, where the same emitter photons are compressed or dilated in the new 'discrete field', (so red OR blue), the total wavenumber, so ENERGY, is the same, so is still conserved.
I hope that's distinguished OK and removed any confusion. Does that alleviate your doubts, or was it something else?
Remember there are also other differences, such as amplitude, which affects the ratio, but best to get the above fully assimilated before adding more cases!
I don'y now if anyone else is reading and keeping up with this. If so, you're doing very well! If not just ask questions.
Peter