# Definition of angular frequency in nuclear structure

• patric44

#### patric44

Homework Statement
confusion about the definition of angular frequency in nuclear structure
Relevant Equations
E=hbar omega
Hi all
I am a little bit confused about the definition of angular frequency in the context of nuclear rotation, some times its defined in the regular way as
$$E=\hbar \omega$$
and other time from the rigid rotor formula
$$E=\frac{\hbar^{2}}{2I} J(J+1)$$
where ##I## is the moment of inertia and ##J## is the angular momentum quantum number, then I saw omega defined as:
$$\omega =\frac{1}{\hbar} \frac{dE}{d\sqrt{J(J+1)}}$$
why the two definitions? any help on that

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They are not two definitions. ##E=\hbar \omega## is a general condition.

The equation $$\omega =\frac{1}{\hbar} \frac{dE}{d\sqrt{J(J+1)}}$$
refers to a specific case of that general condition in which one is considering the rotational energy levels of a rigid rotor as you described. ##J## is the quantum number defining the total angular momentum exclusive of nuclear spin.

• patric44
They are not two definitions. ##E=\hbar \omega## is a general condition.

The equation $$\omega =\frac{1}{\hbar} \frac{dE}{d\sqrt{J(J+1)}}$$
refers to a specific case of that general condition in which one is considering the rotational energy levels of a rigid rotor as you described. ##J## is the quantum number defining the total angular momentum exclusive of nuclear spin.
but why its not simply ##\omega=E/\hbar## then
$$\omega = (\hbar/2I) J(J+1)$$
why the differential definition

Reasonable question. I read right through the differential notation. The equation above (post #3) looks fine to me.

the rigid rotor formula
$$E=\frac{\hbar^{2}}{2I} J(J+1)$$
where ##I## is the moment of inertia and ##J## is the angular momentum quantum number, then I saw omega defined as:
$$\omega =\frac{1}{\hbar} \frac{dE}{d\sqrt{J(J+1)}}$$

In the equation, ##\omega =\frac{1}{\hbar} \frac{dE}{d\sqrt{J(J+1)}}##, ##\omega## represents the angular velocity of rotation of the rigid rotor or nucleus.

This ##\omega## would not be used in the formula ##E = \hbar \omega##. I'm not sure of the context in which ##E = \hbar \omega## is being used.

If the rotor or nucleus absorbs or emits a photon so that the rotation rate of the rotor changes (i.e.., the quantum number ##J## changes), then the angular frequency ##\omega_{\rm photon}## of the photon would be given by $$E_{\rm photon} = \hbar \omega_{\rm photon}.$$The corresponding change in energy of the rotor would be given by $$\Delta E_{\rm rotor} = \frac{\hbar^2}{2I} [J_f(J_f+1) - J_i(J_i+1)]$$ where ##J_i## and ##J_f## are the initial and final values of ##J##.

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• patric44
Transition energies indeed.

Where did I leave my coffee?

• TSny
@patric44

My apologies for having confused you. I should know better than to post without thinking more carefully.

@patric44

My apologies for having confused you. I should know better than to post without thinking more carefully.
no problem bro no need to apologies, thanks for taking time to consider my question, I appreciate any help I appreciate your tolerance, however mistakes are not good; thoughtless mistakes no matter how well intended are worse; and failure to acknowledge a known mistake is inexcusable.

We are all grateful to @TSny for a timely correction.