Definition of angular frequency in nuclear structure

AI Thread Summary
The discussion focuses on the definition of angular frequency in nuclear rotation, highlighting the relationship between energy and angular frequency through two equations: E = ħω and E = (ħ²/2I) J(J+1). It clarifies that these are not conflicting definitions but rather different contexts of the same principle, with ω representing angular velocity in the rigid rotor model. The differential definition of ω as ω = (1/ħ) dE/d√(J(J+1)) is explained as specific to energy changes related to angular momentum transitions. The conversation emphasizes the importance of precision in discussing these concepts in nuclear physics.
patric44
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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
 
Last edited:
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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.
 
Hyperfine said:
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. :oops:

The equation above (post #3) looks fine to me.
 
patric44 said:
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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Transition energies indeed.

Where did I leave my coffee?
 
@patric44

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