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In normal modes, we often refer the total potential energy of the system to be:

$$V = q^T B q$$

where $V$ is the total potential energy, $q$ is the coordinates of the system and $B$ is just some matrix.

In this problem I am working on, the total potential energy I have come out to is:

$$V =k/4[(-x_1+x_4+y_1-y_4)^2 + (x_2-x_3+y_2-y_3)^2+2(x_1-x_2)^2 + 2(x_3-x_4)^2 + 2(x_3-x_4)^2 + 2(x_3-x_4)^2 + 2(y_1-y_3)^2 + 2(y_2-y_4)^2] \tag{1}$$

Now, I am trying to find the matrix B, and the method I am familiar with is to compute:

$$V = {\begin{pmatrix} x_1 \\ y_1 \\\vdots \\x_4 \\y_4 \end{pmatrix}}^T \begin{pmatrix} B_{11} & B_{22} & \cdots & B_{18} \\\vdots & \vdots & \ &\vdots \\B_{81} & B_{82} & \cdots & B_{88} \end{pmatrix} \begin{pmatrix} x_1 \\ y_1 \\\vdots \\x_4 \\y_4 \end{pmatrix} \tag{2}$$

in algebraic form and then compare the coefficients of $(2)$ and $(1)$ in order to determine the individual elements $B_{ij}$ noting of course, that for $i \neq j$, $B_{ij} = B_{ji}$ (B is symmetric) as the potential energy is a quadratic form. However, this is a very tedious process and i was wondering if there were any alternative methods to find the matrix B? Thank you very much!

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  • $\begingroup$ Note that for example, if you multiply out your expression for $V$ then the quadratic term in $x_1$ is $B_{11}x_1^2$. So $B_{11}$ is the co-efficient of $x_1^2$ which should be reasonable simple to eyeball from your expression for $V$. $\endgroup$ – jacob1729 Jun 8 at 10:45
  • $\begingroup$ Just to be clear: would you be able to do this quickly if your other form for $V$ looked like $V=3x^2+2xy+y^2+...$ (ie if you expanded all the brackets and grouped like terms, which does look a bit tedious in this case)? $\endgroup$ – jacob1729 Jun 8 at 10:46
  • $\begingroup$ @jacob1729 I see. But what about the other elements? are there alternatives? $\endgroup$ – Akira D.Soul Jun 8 at 23:46
  • $\begingroup$ @jacob1729 I actually already expanded out everything. which is sort of the "normal" but I was wondering if there are better alternatives $\endgroup$ – Akira D.Soul Jun 8 at 23:48
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Given an expression in the form of your equation (1) you should be able to go directly to equation (2) without any extra work. Note: the following assumes in equation (1) everything is expanded fully.

Let's use a smaller example so I can type it more easily:

$$V(x,y) = (x-y)^2+2(x+y)^2 \tag{1}$$

this is easier to work with in the expanded form:

$$V(x,y) = 3x^2 + 2xy + 3y^2 \tag{1*}$$

for every mixed entry (in this case just $2xy$) we split it into two reversed terms: $2xy = xy + yx$ to get:

$$ V(x,y) = 3x^2 + xy + yx + 3y^2 $$

(you can do this step mentally, or just recognise where it's going to lead to factors of two in a moment).

But this is in the form:

$$V(x,y) = \begin{pmatrix} x & y \end{pmatrix}\begin{pmatrix} 3 & 1 \\ 1 & 3 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} \tag{2}$$

where the entry in the $i-j$'th slot in the matrix is simply the co-efficient of $x_ix_j$ in our previous expression for $V$.

To sum up: diagonal entries are coefficients of squared terms, off diagonal entries are half the coefficients of mixed quadratic terms. The factor of two comes from when we split the $2xy=xy+yx$ up, you should be able to convince yourself that it's correct.

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  • $\begingroup$ sorry forgot to upvote this $\endgroup$ – Akira D.Soul Jun 23 at 7:53

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