# Q & A: mass and frequency

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Q:
Using e= (mc2) and e= hf, then mc2 = hf, then f= (mc2)/h. c2= (299,792,458 meteres/second) squared h= 6.626075 x 10^-34 Js (joule seconds) If these are correct, then Frequency= Mass x [8.987551787 x 1016 (m/s)squared]/6.626075 x 10^-34 Joule seconds. How do I get these units to cancel and leave me cycles per second. What am I missing? How do I derive a formula to give me the theoretical frequency of a known mass? If I have Carbon 12.011, what is it’s relative frequency in cycles per second? Albert :-)
- Albert
Music and Science, Sacramento CA
A:
You're just about there. A Joule is a kg-m^2/s^2, so what your formulas says is that the frequency in Hz or inverse seconds is 1.36x10^50 x mass, where mass is given in kg.

Now you can figure out the frequency of a carbon 12 atom, remembering that atomic mass of 12 means that one mole (6x10^23 atoms) has mass of 12 gm, so one atom has mass of 2x 10^-23 gm, or 2x 10^-26 kg. I get a frequency of 2.72 x10^24 Hz.

Of course you may wonder what in the world we are talking about here- what's the 'frequency' of an atom? It's really a quantum mechanical property. Quantum mechanics allows systems to exist in a mixture of different states, all at the same time. If a system is prepared so that it is in a mixture of two different states with different frequencies, then some of the properties of the system will oscillate with a frequency that's the difference of the frequencies of the two ingredient states. The common name for this is "beat frequencies".

An example of this from particle physics is neutrino oscillation. Neutrinos come in three types, or "flavors" -- electron, muon, and tau. These we now know to have small masses which are different from each other. These small masses correspond to different frequencies according to your formula above. Neutrinos are made as an electron neutrino, a muon neutrino, or a tau neutrino, but these three states are mixtures of the states with definite mass, and so the neutrinos slowly oscillate from one flavor to another and back again as they travel through space. Measurements of this oscillation rate gives us indications about the differences of the masses of the neutrinos.

Mike W. (and Tom)

(republished on 07/21/06)

## Follow-Up #1: photon inertia

Q:
Hi, I'm having difficulty reverse engineering this formula. I'd like to solve for m I reset it to m = (c)2 / hf I'm actually looking for the kinetic inertia of a photon. More specifically what mass inertia would a specific frequency be equal to?
- Jon (age 37)
Columbus, Ohio
A:
The formula you typed looks wrong, probably because some symbols didn't reproduce properly. Anyway, even in classical electromagnetism it was possible to show that the momentum of a wave traveling in some direction was just its energy divided by the speed of light c (in standard units). So a photon with energy hf has momentum p=hf/c. Since the frequency, f, is related to the wavelength lambda by lambda= c/f, you can also write the momentum as p= h/lambda. It turns out this expression for momentum works for all sorts of particles, not just massless ones that travel at speed c.

Mike W.

(published on 01/26/09)

## Follow-Up #2: mass-frequency relations

Q:
i was pondering Einstein's and Planck's equations, m=E/c squared and E=hf, respectively. when i rewrote them together i got h (Planck's constant)/c squared= mass/frequency= 7.372503276491 e -51. if this is true, then mass= h/c squared/frequency and frequency=m/h/c squared. i was curious to see if this is true.
- judas (age 14)
andover, mn, us
A:
Yes, so long as by "mass" you mean the inertial mass, the basic relation you describe is true. However, the number you gave for the m/f ratio is meaningless unless you specify some units. I think you're using kg/Hz.

In practice, we usually write equations (e.g. for atoms) in which we ignore the contribution of the rest mass of the particles to the frequency. That's convenient because otherwise the frequency difference between the ground state and the first excited state of a hydrogen atom would just be a tiny fraction of the total frequency. We can get away with doing that so long as those rest masses don't change. In quantum mechanics adding fixed frequency doesn't really change anything physical.

Mike W.

(published on 06/30/11)