You're mistaken.

The proof that you're wrong is found at

http://home.comcast.net/~peter.m.brown/gr/grav_red_shift.htmSee Eq. (12) in that page. Other sources of proof, such as the

**American Journal of Physics**, are quoted below.

Light dropping into a gravity well increases in energy... wavelength shortens/frequency increases.

This too is a common mistake. The frequency of light

*does not change* as it moves through a gravitational field as reckoned by any single observer. It’s only when you later compare measurements made by observers located at different gravitational potentials does the spectral shift manifest itself.

You’re confusing the (locally measured) kinetic energy of a light (a photon) with its

*total* energy which includes potential energy. The increase of the kinetic energy (locally measured blue shift) is compensated for a decrease in the photons potential energy. Some authors don’t like this description but its rigorous and accurate and agrees with observation.

Your mistake is not that uncommon though. Many people confuse spectral shift (aka the phenomena of gravitational redshift) with the “changing” of the “photon’s energy.” If you’re not that familiar with general relativity (GR) then you may only know the energy of a photon as its used in special relativity (SR) where the energy of the photon is all kinetic energy.

In GR any one observer will reckon the frequency of a beam of light to remain constant as if falls through a gravitational potential. See

*On the interpretation of the redshift in a static gravitational field* by L.B. Okun, K.G. Selivanov, and V.L. Telegdi, Am. J. Phys., 68(2), Feb. (2000)

The change in a photon’s kinetic energy as it moves through a gravitational field means that when the energy is measured

**locally**, i.e.

*at the same height*, the photon is located then it has different values due to spectral shift caused by the gravitational field.

Recall the analogy from Newtonian gravity: If I drop a rock from the roof of my house then the energy of the rock will remain constant as it falls. Does that mean that the kinetic energy as measured by an observer at the same height as the rock will all measure the same kinetic energy? No, it doesn't.

The same thing holds for the energy of the photon as it moves through a gravitational field. This is proven in textbooks on general relativity. E.g.

**A Short Course in General Relativity – Second Edition** by Foster and Nightingale,

*Springer*, (1994). This is proved in section 4.3

**Spectral Shift**, pages 132 to 135.

If you don’t have that text then see the proof I wrote out under my website at

http://home.comcast.net/~peter.m.brown/gr/grav_red_shift.htm