Vibrations of Triatomic Molecules, Landau-Lifshitz

[electricspit]

Homework Statement

https://ia701205.us.archive.org/11/items/Mechanics_541/LandauLifshitz-Mechanics.pdf

Page 72 of the book itself, but Page 81 of the PDF. Problem 1.

Homework Equations

See Section 24 of the book.

The Attempt at a Solution

So far I have completed and understand the kinetic energy part of both the transverse and longitudinal modes of oscillation of the molecules for Problem 1, as well as the potential energy for the longitudinal mode. For some reason I cannot figure out, after many hours of tedious thinking why they have the potential energy they do for the transverse case.

If anyone could explain this, it would be very helpful.

 

[TSny]

Hello, electricspit.

Consider arbitrary vertical displacements of the atoms as shown in the attached figure. Please show us your attempt at expressing the angle δ in terms of y₁, y₂, y₃, S₁₂, and S₂₃.

EDIT: I was assuming your question has to do with the expression given by LL for δ. If not, are you asking why they take the potential energy to be of the form (1/2)k₂l²δ²?

 

[electricspit]

Hello TSny,

Thanks for the reply, I was asking the latter rather than the former, but if that is the correct diagram for this case I am completely wrong in my initial assumptions.

They express delta in terms of y1, y2, y3, and l alone, so I'm wondering where the S12, S13 come into play.

My attempts at finding delta were obviously wrong because I wasn't using a diagram like this because they mentioned delta being the angle ABA with y1=y3, so I assumed it was the open angle at y2.

Does this make sense?

EDIT: Well I reread the LL question definition of delta and it seems that I missed the word "deviation". How unfortunate. Now I have:

\delta=\frac{y_1-y_2}{\ell}+\frac{y_3-y_2}{\ell}

Which is correct.

Now for the potential energies I had started with:

V=\frac{1}{2}k(y_1-y_2)^2+\frac{1}{2}k(y_3-y_2)^2

Which, resting upon the fact that:

y_1-y_2=\frac{\ell\delta}{2}
y_1=y_3

Would give me:

V=\frac{1}{4}k\ell^2\delta^2

Which is half of the answer they're getting.

 

[TSny]

Now I have:

\delta=\frac{y_1-y_2}{\ell}+\frac{y_3-y_2}{\ell}

Which is correct.

OK. For arbitrary displacements, you should be able to show that $\delta = \theta_1- \theta_2$ where $\theta_1 = \sin^{-1}\frac{y_2-y_1}{S_{12}}$, etc. See attached figure. For small displacements and keeping terms only to first order, you get the expression for $\delta$.

Now for the potential energies I had started with:

V=\frac{1}{2}k(y_1-y_2)^2+\frac{1}{2}k(y_3-y_2)^2

Which, resting upon the fact that:

y_1-y_2=\frac{\ell\delta}{2}
y_1=y_3

Would give me:

V=\frac{1}{4}k\ell^2\delta^2

Which is half of the answer they're getting.

Note that they use a different force constant k₂ for the transverse displacement compared to the linear displacement constant k₁. The force constant k₂ for the bending mode will depend on the peculiar stiffness properties of the molecular bonds and does not necessarily have a simple relationship to k₁.

It seems to me that the form they take for the potential energy just comes from the initial assumption that the potential energy depends only on $\delta$: $V(\delta)$. Then, expanding: V(\delta) \approx V(0)+V’(0)\delta +\frac{1}{2} V’’(0) \delta^2 + ….
You can take $V(0) = 0$ and argue that $V’(0) = 0$.

So, $V(\delta) \approx \frac{1}{2} V’’(0) \delta^2$ = (const.) $\delta ^2$

They choose to write the constant in the last expression as $\frac{1}{2}k_2l^2$, which essentially defines $k_2$.

 

[electricspit]

Ah yes I was just lazy and wrote k when I really meant k_2. I was thinking that k_2 must have had some other definition, that they probably omitted.

I definitely understand that V(0)=0 and V'(0)=0 because it is an energy minimum. Taking the constant to be this value seems too "hand-wavy" for me at this moment in time, as I could have taken it to be anything to define some arbitrary force constant k_2. The way I'm taking it right now is that at \delta=0 the curvature of the potential function will depend on the distance between the two molecules which is given by \ell therefore V''(0)\propto\ell^2.

Would this be a correct assumption?

EDIT: Also thank you for your help!

of course I was just think and maybe he expanded at \ell\delta:

V(\ell\delta)=\frac{1}{2}V''(0)\ell^2\delta^2

and assuming:

V(\ell\delta)\propto \delta^2

we have:

V(\ell\delta)=\frac{1}{2}k_2\ell^2\delta^2

 

[TSny]

I suppose your argument is OK for the presence of $l^2$ in $V$. I tend to think of $l^2$ as being there so that $k_2$ will have the same units as $k_1$. $l$ is the only natural length in the problem.

But your argument might be fine.

 

[electricspit]

Ah I see where you're coming from that actually makes more sense than my expansion which might be more arbitrary of a choice than LL.

I appreciate your insight. If you have the patience and time, I was wondering if you knew how they use 24.2 to obtain the expression in Problem 2:

(y_1-y_3)\sin{\alpha}-(x_1+x_3)\cos{\alpha}=0

I'm a bit confused about the displacement vectors u_1 and u_2 they are using.

Sorry for all the questions but this is my first real mechanics course (my university doesn't have a second or first year course) and my professor is literally copying his notes from LL without going into any detail of what anything means, so I'm basically following this book by myself in my spare time.

 

[TSny]

In applying 24.2, note that $\vec{r}_{10} = l\sin\alpha\hat{i} + l\cos\alpha\hat{j}$ and $\vec{u}_{10} = x_1\hat{i} + y_1\hat{j}$, etc.

 

[electricspit]

Thank you so much for your help!

 

[ldzambra]

Hello, TSny. I realize it has been a long time since this publication, but perhaps you can assist me in understanding something related to the same 24.2 exercise (Page 73, triangle molecule, Landau-Lifshitz). Specifically:

I'm not sure how these three conditions came about.

​

 

[TSny]

Hello, TSny. I realize it has been a long time since this publication, but perhaps you can assist me in understanding something related to the same 24.2 exercise (Page 73, triangle molecule, Landau-Lifshitz). Specifically:

I'm not sure how these three conditions came about.

​

I'm fairly sure that a good number of my brain cells that were working way back then are now kaput.

But, let's start with how to get $$\delta l_1 = (x_1 - x_2) \sin \alpha - (y_1 - y_2)\cos \alpha.$$ Consider the case where only particle 1 is displaced in the horizontal direction from its equilibrium position by the amount $x_1$ as shown:

Can you derive an expression for the change in length, $\delta l_1$, of $l_1$ in terms of $x_1$ and $\alpha$.

Then consider how $l_1$ would change for a vertical displacement $y_1$ of particle 1.

How would horizontal and vertical displacements $x_2$ and $y_2$ of particle 2 change $l_1$?

 

[ldzambra]

Thank you for answering to my question. Using your illustration, I found the following, but I don't see how a displacement of point two can influence it. Since the angle alpha would changing if the point 2 is displaced

 

[TSny]

If you keep particle 1 fixed and displace particle 2 horizontally by $x_2$, then you can see that $\delta l_1 = -x_2 \sin \alpha$, etc.

But, you are right to worry about whether we can treat the individual displacements $x_1$, $y_1$, $x_2$, and $y_2$ as acting independently in changing $l_1$. We can as long as the displacements are "small" compared to $l_1$ and we only want expressions accurate to first order in the displacements.

If you want to make this more rigorous, let $(X_1, Y_1)$ be the coordinates of particle 1 and let $(X_2, Y_2)$ be the coordinates for particle 2. Then $l_1 = \sqrt{(X_1-X_2)^2 + (Y_1 - Y_2)^2}$.

For small displacements, $$\delta l_1 \approx \frac{\partial l_1}{\partial X_1} \delta X_1 + \frac{\partial l_1}{\partial X_2} \delta X_2 + \frac{\partial l_1}{\partial Y_1} \delta Y_1 + \frac{\partial l_1}{\partial Y_2} \delta Y_2 = \frac{\partial l_1}{\partial X_1} x_1 + \frac{\partial l_1}{\partial X_2} x_2 + \frac{\partial l_1}{\partial Y_1} y_1 + \frac{\partial l_1}{\partial Y_2} y_2$$

The partial derivatives are evaluated for the equilibrium position and can be expressed in terms of trig functions of $\alpha$. The result is accurate to first order in the small displacements $x_1$, $y_1$, $x_2$, and $y_2$.