Cobalt101 said:
why wavelength varies (inversely) with momentum for all energy and matter
Although there's always an infinite number of levels to why (As a parent knows only too well), here are some thoughts on the matter.
I think your question may be closely connected to action.
During the motion of a point particle in 1d say, the change in the action along the solution trajectory ##q(t)## is
$$dS=pdq-Hdt$$
where p is the momentum of the particle, q its position, H its energy, t is time. Notice here the pairing between momentum/space-displacement and energy/time-displacement in comparison to the de-Broglie relations. For a system of particles this generalises to a sum over degrees of freedom ##n##, the important thing being that each positional coordinate ##q_i## and momentum ##p_i## is dynamically paired in a special way. Also if you place ##t## on the same footing as the ##q_i## it turns out that the momentum paired with ##t## is in fact ##-H##.
##dq_i (i=1,2,..,n+1)## picks out a direction in extended configuration space and ##p_i## gives the flow along this direction. So this is what action differentials measure, the momentum which is carried over displacements.
Now back to 1 degree of freedom, so point particle on a line, and we imagine that time freezes but we can still investigate how ##S## would change if we virtually translated the particle by ##dq##, which is the displacement that was just about to take place during the motion i.e. ##dq=\dot{q}dt##. ##dS=pdq## is a geometric invariant so that if I deform my ##dq## in some way, ##p## will change also in a way to preserve ##dS##. So if I contract space such that the original coordinate differential ##dq## has now doubled with the new q-tics, we only need half the previous amount of flow for ##S## to tick off the same way as before and motion to be the same.
Now a wave is spread over space and not generally localised at a point. However, even without really knowing what property of the wave determines its momentum, if the wave travels with the same flow at every point along the line so that the wave just gets translated along homogeneously without saying anything about what shape the wave needs to be for this to happen. Now imagine we discretised the waveform and viewed it as being made up of infinitesimally evenly spaced points. For a given flow ##p##, there is an invariant ##dS## associated with ##dq##. Now if we contract the space in the same manner as before, the in-between spaces ##\textit{all}## grow evenly, and the flow ##p## thus decreases in order to preserve ##dS##.
A bit long winded but hope this helps..
