Do massive particles redshift the same way as photons in a gravitational field? Let's assume two observers $A$ and $B$ hovering in a gravitation field. $A$ sends a radio transmission of frequency $f_1$ to $B$. $B$ receives this transmission and finds it has frequency $f_2$.
As as second experiment $A$ sends an electron beam to $B$. They measure the energies of these electrons on emission and reception.
Particles have de Broglie frequency which is proportional to energy.
Will this frequency gets redshifted the same way as photon frequencies in a gravitational field so the rate of the original and redshift frequency/energy will be the same as in the case of photons?
In other words, for example, if a 900 keV photon, fired from $A$, gets red shifted to 850 keV when it arrives at $B$, will any massive 900 keV particle get slowed down to 850 keV after it free falls to $B$? - Given the rest mass is smaller than 850 keV, otherwise it would just fall back and never reach the other observer I guess.
I worked it out in flat spacetime between two accelerating observers that keep fixed distance between them, and $A$ just drops particle and $B$ just catches it. And in that case it seems the rate of the total energy of the received and dropped particle is exactly the rate of acceleration induced time dilation between the two observers. I'm unsure if this only works this way in this specific case or if it works in general in any gravitational or fictitious force fields.
 A: Case 1:  Convert 1 kg of matter-antimatter to energy 
Case 2: Lower 1 kg of matter-antimatter to a gravity well, convert it to energy there, beam the energy up to the original position.
From conservation of energy: Energy produced in case 1 = Energy produced in case 2
Energy in case2 = Energy beamed up + energy generated during lowering 
Energy beamed = Energy of 1 kg of matter - energy lost in redshift
It must be so that:
Energy lost in redshift = energy generated during lowering
Also: Energy generated during lowering = energy lost, or used, when the lowered thing is lifted back to the original position
So energy lost, or used, when the matter-antimatter is lifted back to the original position = energy lost in redshift 
A: They simply lose their energy. All of them. If the gravitational field is g, then raising by h in it, a particle of mass m loses
E=mgh

Of course, we are talking about the mass at the moment and changes are differential.
But the difference for photons is that their mass is their energy and vice versa. So, it can lower down to approximately 0. And the mass of an electron, for example, cannot go under 0.5Mev. So, the electron will enlarge its length of wave wore slowly.
After losing all kinetic energy, electron will start to fall back - down. And photon will continue to escape, only there is a height, at which it will lose all its energy. Their behaviour at point of turn won't be even a little bit covered by theory of relativity, but will be described by quantum physics. A standing electron so enlarge its indefinity of position, that looks really large. Almost standing atom can be large as an apple. Or more, of course. Standing still one will be big as the Universe.
