Definition of angular frequency in nuclear structure

[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

 

[Hyperfine]

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

 

[Hyperfine]

Reasonable question. I read right through the differential notation.

The equation above (post #3) looks fine to me.

 

[TSny]

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 $J$.

 

[Hyperfine]

Transition energies indeed.

Where did I leave my coffee?

 

[Hyperfine]

@patric44

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

 

[patric44]

@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

 

[Hyperfine]

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.