Anyone care to explain the TA's note?

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Homework Help Overview

The discussion revolves around the relationship between frequency, wavelength, and angular frequency in the context of wave mechanics. Participants are examining the formulas related to these concepts and questioning the validity of certain expressions.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants are exploring the derivation of the wavelength formula and its connection to angular frequency. There are questions about the validity of using lambda = c/f in this context and the implications of using angular frequency instead.

Discussion Status

Some participants have provided insights into the relationship between angular frequency and energy, while others are reflecting on their misunderstandings. There is an ongoing exploration of the definitions and relationships between the variables involved.

Contextual Notes

There are indications of prior confusion regarding the definitions of frequency and angular frequency, as well as the impact of previous homework performance on participants' willingness to seek further clarification from the TA.

Shackleford
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Omega here is frequency.

I used lambda = c/f. Why is it (2*pi*c)/omega and not c/f or c/omega?

http://i111.photobucket.com/albums/n149/camarolt4z28/2010-10-04203213.jpg?t=1286245112

http://i111.photobucket.com/albums/n149/camarolt4z28/2010-10-04203118.jpg?t=1286245065
 
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It looks like the formula you used was derived as

\omega_{ij} = \frac{E_j - E_i}{\hbar},

which is an angular frequency as can be seen from the presence of \hbar rather than h.
 
fzero said:
It looks like the formula you used was derived as

\omega_{ij} = \frac{E_j - E_i}{\hbar},

which is an angular frequency as can be seen from the presence of \hbar rather than h.

Yeah. But h-bar is already in the denominator in the formula I used to calculate omega_{ij}.
 
The wavelength is still 2\pi c/\omega.
 
Why not ask your TA for clarification?
 
fzero said:
The wavelength is still 2\pi c/\omega.

Why is lambda = c/f not valid?
 
cristo said:
Why not ask your TA for clarification?

I forgot. I already bugged him about two other problems. I did poorly on this damn homework.
 
Shackleford said:
Why is lambda = c/f not valid?

It is, but \omega = 2\pi f.
 
fzero said:
It is, but \omega = 2\pi f.

Well, that's my mistake. I wasn't taking omega to be an angular frequency. Why? I don't know. Son of a *****.
 

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