# Phonon dispersion

1. May 21, 2013

### Lindsayyyy

Hi everyone

1. The problem statement, all variables and given/known data

The frequence for phonons between two atoms with mass M1 and M2 is given by:

$$\Omega^2 = C (\frac{1}{M_1} + \frac {1} {M_2}) \pm C*[(\frac{1}{M_1} + \frac {1} {M_2})^2 - \frac {4} {M_1 M_2} sin^2(\frac {Ka}{2})]^{\frac 1 2}$$

Show that for Ka <<1 the solutions are:

$$\Omega^2= 2C(\frac{1}{M_1} + \frac {1} {M_2})$$

and
$$\Omega^2= \frac {C}{2(M_1 +M_2)} (Ka)^2$$

2. Relevant equations

-

3. The attempt at a solution

I tried to approximate sin^2(ka/2) as (ka/2)^2 but that didn't work. I have troubles finding the second solution. If I do it this way it works for the first solution but I don't know how I can get the factor (Ka)^2 in the second solution.

2. May 21, 2013

### Staff: Mentor

Where did it stop working?
For the first solution, you used "+" at the ± sign. How do you get the second solution?

You could try a Taylor approximation for the square root.

3. May 21, 2013

### Lindsayyyy

When I approximate it I get to:

$$\Omega^2 = C (\frac{1}{M_1} + \frac {1} {M_2}) \pm C*[(\frac{1}{M_1} + \frac {1} {M_2})^2 - \frac {4} {M_1 M_2} (\frac {Ka}{2})^2]^{\frac 1 2}$$

I thought I can say if Ka <<1 the last term disappears but as you can see I only get the first solution when I add the remaining terms.

I tried doing Taylor approximatin writing sin^2 as 0,5*(1-cos(2x)) and stumbled upon the same problem :(

4. May 21, 2013

### Staff: Mentor

Well if you neglect Ka/2 completely, the second solution is zero.
The approximation sin(x)=x is okay, but the square root needs a taylor approximation (or similar methods) to account for the Ka-expression.

5. May 21, 2013

### Lindsayyyy

Ok I'll try tomorrow again. It's late over here and if I have trouble doing so. I will post here again.
Just one more thing: When I do Taylor approxmation here my variable is K (because task said if K*a <<1)and the point where I approximate is zero. Is that correct?

6. May 21, 2013

### Staff: Mentor

I would treat Ka as single variable (they are not used individually) and use 0 as point of the approximation.

7. May 22, 2013

### Lindsayyyy

When I do it I have the following:

$$sin^2(\frac {Ka}{2}) \approx sin^2(0) + sin(\frac {Ka}{2}) cos (\frac {Ka}{2}) (Ka-0) + [\frac {cos(\frac {Ka}{2})}{2} cos(\frac {Ka}{2}) -\frac {sin^2(\frac {Ka}{2})}{2} (Ka-0)^2+...$$

first two terms become zero, aswell as the last sin^2,so finally:

$$sin^2 \approx \frac {1}{2} (Ka)^2$$

Is that correct? If so, the next problem came up. I have a squareroot which I don't know how to ease up a bit. Looks to me like I could do something with binomial formula but the (ka)^2 doesn't let me do it that easy.

8. May 22, 2013

### tia89

Actually as far as I see, it was good to approximate
$$\sin x\sim x\quad\to\quad \sin^2 \left( \frac{Ka}{2} \right)\sim \left( \frac{Ka}{2} \right)^2$$
Once you do so, you have to cast the square root (collecting the first term) to be in the form
$$\sqrt{1-x}$$
Where $x$ will be something like $\frac{(Ka)^2}{M_1 M_2}\left( \frac{1}{M_1}+\frac{1}{M_2} \right)^{-2}$.
Then you can expand also the square root ($\sqrt{1-x}\sim1-\frac{1}{2}x$) and the trick is done. Taking the upper sign, you can disregard $(Ka)^2$ as it is higher order, while in the second sign, it is the only term remaining and you are then obliged to keep it. According to my calculations, they come out the same as they should in your post

9. May 22, 2013

### Lindsayyyy

Thank, I just flew over your reply and I'll try it later, have to do some chemistry stuff first.
Also while you wrote your approximation down I realised that I did a mistake in my previus post. I forgot the 2! in my Taylor polynom which leads me to the same approximation you said.

10. May 22, 2013

### tia89

Right... indeed it sounded strange to me that there was such difference, in principle both approximations should be the same (at first order at least). Anyway it is nothing difficult, just boring calculations
Have fun anyway

11. May 22, 2013

### Lindsayyyy

Ok I tried it now, but I have problems unterstanding how to get to the squareroot(1-x). How do I know I have to get there? And I don't understand how to get there finally.

edit: It's not clear for me how you get your x in your post. I can exclude (ka)^2/M1M2 but that doesn't seem to help

edit2: ah nvm I got this now, but I still don't understand how I know that I have to get to the square root. Is that just about experience or is that plain logical :P? (because for me it isn't :( )

Last edited: May 22, 2013
12. May 22, 2013

### tia89

Mainly the point is that, as you can't take out the $(Ka)^2$ from the root, you need to expand also that one... and the expansion is easy when you have $\sqrt{1\pm x}$ with $x\ll 1$. The expansion of course is just another Taylor expansion. Anyway this one you should find tabulated somewhere, perhaps as
$$(1\pm x)^\alpha\qquad\text{for }x\ll 1$$
Simply use the form they give you with $\alpha=\frac{1}{2}$.

As for the logical step, well, being the argument small you will want to expand to simplify the expression (polynomials are ALWAYS easier than strange functions), of course taking in account that you are doing an approximation, so you have to keep in mind that it is not precise. Anyway if you decide that expanding is good enough for you, because (as in this case) the error you do in expanding is compensated by the fact that calculations are much simpler, then do it. As for trying to get that very form, well this is just for the sake of simplicity. Experience teaches in this case, we always do so...

p.s. you don't have to GET to the square root, you already have it, and as you can't take it out trivially, you try to find a form for the argument that is suitable to be expanded in a kind way

13. May 22, 2013

### Lindsayyyy

Ok thank you. I think I understand it a bit better now.