- #1

- 173

- 29

My issue here is that if you set m_1=m_2=m, i.e. set both atoms equal to each other, it doesn't automatically reduce to the old acoustic dispersion relation as the ± term doesn't disappear.

What's up with that?

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter Wminus
- Start date

- #1

- 173

- 29

My issue here is that if you set m_1=m_2=m, i.e. set both atoms equal to each other, it doesn't automatically reduce to the old acoustic dispersion relation as the ± term doesn't disappear.

What's up with that?

- #2

- 173

- 29

BTW, I got the dispersion relation from here http://en.wikipedia.org/wiki/Phonon#Dispersion_relation

- #3

DrDu

Science Advisor

- 6,148

- 818

I can't see any problem. Maybe you should fresh up your knowledge of the trigonometric equations?

- #4

- 173

- 29

OK, sorry I should've written everything out in details in the OP. My problem is that the plus-minus sign is being problematic. It gives two solutions that are phase-shifted with regards to each other. Setting the masses equal, the equation from OP reduces to:I can't see any problem. Maybe you should fresh up your knowledge of the trigonometric equations?

##\omega^2 = K(2/m) \pm K \sqrt{(2/m)^2 - (2/m)^2 sin^2{ka/2}} = \frac{2K}{m}(1 \pm cos(ka/2)) \Rightarrow \omega_+ = \sqrt{\frac{4K}{m}} |cos(ka/4)|## & ## \omega_- = \sqrt{\frac{4K}{m}} |sin(ka/4)|##.

How do I get rid of the ##\omega_+## term, the optical mode? Which physical argument makes it invalid?

- #5

DrDu

Science Advisor

- 6,148

- 818

- #6

- 173

- 29

If you graph both of them in a diagram, you get a dispersion relation with 2 modes! This is supposed to be impossible, isn't it?

- #7

DrDu

Science Advisor

- 6,148

- 818

No, it isn't. You have a basis with two atoms, so you get two modes, an acoustic one and an optical one. However, when the two atoms become identical, the two curves will touch at the zone boundaries.

Do you know the difference between a reduced and an extended zone scheme? Here it is for the example of a quasi-free electron:

http://www.pha.jhu.edu/~jeffwass/2ndYrSem/pics/Slide19.JPG [Broken]

Do you know the difference between a reduced and an extended zone scheme? Here it is for the example of a quasi-free electron:

http://www.pha.jhu.edu/~jeffwass/2ndYrSem/pics/Slide19.JPG [Broken]

Last edited by a moderator:

- #8

- 173

- 29

- #9

- 173

- 29

- #10

DrDu

Science Advisor

- 6,148

- 818

- #11

- 173

- 29

I think I get all of this now. It's got to do with the reduced/extended scheme ways of viewing the dispersion relation. thanks.

Share: