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  1. What is $V_{\alpha\beta}$?

  2. And what is a symmetric, positive definite potential energy matrix?

  3. And why is there a linearized equation like this?

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I) In this answer we discuss a systematic approach to linearization and stability analysis. Imagine that the physical system under consideration is described by an autonomous Lagrangian $L=L(q,\dot{q})$ of $n$ generalized coordinates

$$\tag{1} q~=~(q^1, \ldots, q^n)~\in~ \mathbb{R}^n.$$

One of the first questions one would like to ask is, if a specific point $q_{(0)}$ is a stable equilibrium point or not? This is achieved by analyzing the effect of small displacements $u := q-q_{(0)}$ from that point, $|u|\ll 1$. Thus we expand the Lagrangian $L$ to quadratic order in $u$ and $\dot{u}$:

$$\tag{2} L(u,\dot{u}) ~=~ \frac{1}{2}M_{ij}\dot{u}^i\dot{u}^j + \frac{1}{2}B_{ij}u^i\dot{u}^j -\frac{1}{2}K_{ij}u^iu^j + f_i u^i. $$

A quadratic Lagrangian corresponds to linear equations of motion. One says that the theory has been linearized. In eq. (2) we have neglected (i) higher order terms because $|u|,|\dot{u}|\ll 1$; (ii) constant terms and total derivatives terms, which would not have affected the Euler-Lagrange equations. Similarly, we may assume without loss of generality that the real matrix

$$\tag{3} B~=~-B^{t}$$

is antisymmetric (because the symmetric part of the $B$ matrix corresponds to a total derivative in the Lagrangian). We may also assume without loss of generality that the real matrices

$$\tag{4} M~=~M^t$$

and

$$\tag{5} K~=~K^{t}$$

are symmetric. (OP denotes the $K$ matrix by $V$.) The matrix $B$ has an interpretation as a generalized dualized magnetic field;

$$\tag{6} A_j~:=~ \frac{1}{2} u^i B_{ij} $$

is a generalized magnetic potential; the $K$ matrix consists of generalized coupled spring-constants in a generalized Hooke's law; and the $M$ matrix is generalized mass. Here we will for simplicity assume that the Legendre transformation to the Hamiltonian formalism is non-singular in order to not have constraints. This means that the matrix $M$ is non-singular, $\det(M)\neq 0$. [It is always possible to rotate the displacement coordinates

$$\tag{7} u^{\prime i} ~:=~ R^i{}_j u^j$$

by an orthogonal matrix $R$ to diagonalize the mass matrix $M={\rm diag}(m_1, \ldots, m_n)$. By positive scaling transformations

$$\tag{8} u^{\prime i} ~:=~ \lambda^i u^i, \qquad \lambda^i~>~0,$$

it is possible to make all the masses $m_1, \ldots, m_n$, equal to $\pm 1$. The transformations (7) and (8) do not destroy property (3) and (5).]

II) Now let us calculate. The canonical momenta are

$$\tag{9} p_i ~=~ \frac{\partial L}{\partial \dot{u}^i} ~=~ M_{ij}\dot{u}^j+ A_i~=~ M_{ij}\dot{u}^j- \frac{1}{2} B_{ij}u^j. $$

The Euler-Lagrange equations read

$$\tag{10} \dot{p}_i = \frac{1}{2} B_{ij} \dot{u}^j- K_{ij}u^j + f_i \qquad\Leftrightarrow\qquad M_{ij}\ddot{u}^j ~=~ B_{ij} \dot{u}^j- K_{ij}u^j + f_i .$$

The energy function is defined as

$$\tag{11} h(u,\dot{u})~:=~ p_i \dot{u}^i - L ~=~ \frac{1}{2}M_{ij}\dot{u}^i\dot{u}^j +\frac{1}{2}K_{ij}u^iu^j - f_i u^i . $$

The Hamiltonian reads

$$H(u,p) ~=~\frac{1}{2}(p-A)^t M^{-1}(p-A) + \frac{1}{2}u^t K u -f^t u$$ $$\tag{12} ~=~\frac{1}{2}p^t M^{-1}p + \frac{1}{2}p^t M^{-1}Bu + \frac{1}{2}u^t (K+ \frac{1}{4}B^tM^{-1}B)u - f^t u. $$

Hamilton's equations of motion is a first-order coupled ODE

$$\tag{13} \frac{d}{dt} \begin{bmatrix} u \\ p \end{bmatrix} ~=~C \begin{bmatrix} u \\ p \end{bmatrix}, $$

where

$$\tag{14} C~:=~\begin{bmatrix}\frac{1}{2} M^{-1}B& M^{-1} \\ -K+ \frac{1}{4}BM^{-1}B & \frac{1}{2}BM^{-1}\end{bmatrix}$$

is a constant $2n\times 2n$ real matrix (which depends on the point $q_{(0)}$). Let us assume that the matrix $C$ is diagonalizable.$^1$

III) A necessary and sufficient condition for that $q_{(0)}$ is an equilibrium point $\forall t:u(t)=0$ is evidently that the source terms vanish

$$\tag{15} \forall t: f(t)~=~0,$$

cf. eq. (10). Let us assume condition (15) from now on.

IV) Let us finally discuss conditions for that the equilibrium point ${\bf q}_{(0)}$ is stable.

From a Hamiltonian perspective, a necessary (sufficient) stability condition is that the real part of $C$'s eigenvalues are non-positive (negative), respectively. (In the gap between the above necessary and sufficient conditions, it is necessary to analyze contributions from higher order terms as well.)

From a Lagrangian perspective, consider the characteristic equation

$$\tag{16} \det(K-\lambda B + \lambda^2M)~=~0. $$

A necessary (sufficient) stability condition is that the real part of the roots $\lambda$ are non-positive (negative), respectively.

The stability condition simplifies considerably if we assume that there are no magnetic fields $B=0$. In that case a necessary stability condition is that $M^{-1}K$'s eigenvalues are real and non-negative. (Again, to obtain sufficient stability conditions, it is necessary to analyze contributions from higher order terms as well.)

V) In physics we usually also impose that the Hamiltonian (12) is bounded from below, cf. unitarity. In the general case (with $B$ included), unitarity (at the linearized level) is equivalent to that the matrix $M$ is positive definite and the matrix $K$ is semipositive. (Recall that we have previously assumed that $M$ has no zero-directions, while $K$ in principle could have zero-directions/zero-modes. E.g. free particles would have $K=0$.)

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$^1$ I have a vague feeling that diagonalizability of $C$ is an unnecessary assumption, i.e. that the matrix $C$ is automatically diagonalizable.

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  • $\begingroup$ Who can tell me how the equation in my question is derived? $\endgroup$ – park ning Sep 27 '13 at 1:10
  • $\begingroup$ Eq. (10) is essentially your eq. $\endgroup$ – Qmechanic Sep 27 '13 at 3:22
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What's $V_{ab}$? Well, $V_{ab}$ is the "symmetric, positive definite potential energy matrix".

Ok lol I'm trolling here, but as the name suggests, $V_{ab}$ describes the strength of the (linearized) interaction between particles $a$ and $b$. To be precise, it is the second derivative of the potential energy function of the system with respect to $u_a$ and $u_b$, evaluated at the equilibrium point. So, $V_{ab}$ is a collection of $N \times N$ real numbers which are constants.

What does it mean to linearize an equation? It means to keep only terms that are proportional to the distance $u$ and below, in the Taylor expansion of the actual function.

For example, how do we linearize the function $y(x) = e^x$ about the point $x_0=0$? This means, we wish to describe $y(x)$ up to terms in $(x-x_0) = x-0= x$ only. Now the Taylor expansion of $e^x$ is $e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots$ so the linearized equation is $y(x) = 1+x$. So, $y(x)$ can be described by $1 + x$, and the agreement between $e^x$ and $1 + x$ becomes arbitrarily good as $x \to 0$.

So in your case of $N$ particles, the forces between any two particles is given by some potentially (no pun intended) complicated function of the coordinates, but we wish to only look at very small deviations from the equilibrium position, hence we 'linearize' it. So the total forces of all the other particles on particle $a$ must look like $\sum_b V_{ab} u_b$. Now, why we care about only small deviations from the equilibrium is a totally different question. It is a physics question. The reason is because we wish to characterize what kind of equilibrium it is, whether it's stable or unstable, and what frequency of oscillation it admits, and it suffices to just look at $V_{ab}$ which dominates the interaction assuming your deviation from equilibrium is small.

As for your question on what is a symmetric, positive definite matrix? These terms only apply for a square matrix. A matrix $M$ (with real coefficients) is symmetric if $M = M^T$, i.e. its transpose. In components, it means $M_{ij} = M_{ji}$. A positive definite matrix (usually applied only to symmetric/hermitian matrices) means that for any vector $w$, $w^T M w > 0$. This is also equivalent to the fact that the eigenvalues of $M$ are all $> 0$. Physically this guarantees that the equilibrium we have a is a stable one (the frequency of oscillation is related to the square root of the eigenvalues).

Now from what I've written you should be able to figure out the form of $V_{ab}$ and hence show why it is a positive definite, symmetric, matrix.

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  • $\begingroup$ Positivity seems to need more constraints. For instance, with $V_{11}=V_{22}=0$, and $V_{12}=V_{21}$, the determinant of $V$ is negative. $\endgroup$ – Trimok Sep 25 '13 at 10:00
  • $\begingroup$ Sorry,I also can't understand your explaination.Could you tell me what "To be precise, it is the second derivative of the potential energy function of the system with respect to ua and ub, evaluated at the equilibrium point. So, Vab is a collection of N×N real numbers which are constants." and"So the total forces of all the other particles on particle a must look like ∑bVabub" is ? $\endgroup$ – park ning Sep 25 '13 at 10:17
  • $\begingroup$ @Trimok Indeed, positivity needs the extra physical constraint that we are describing a stable equilibrium. But that's given in the paragraph park ning quoted. $\endgroup$ – nervxxx Sep 25 '13 at 10:24
  • $\begingroup$ @parkning The stuff in the first quotation is something for you to figure out by yourself. Once you do it everything will be clear to you. Start with this: we know that there is a potential energy of the entire system, which is a scalar function of all the particles' displacements, i.e. $V = V(u_1, \cdots, u_N)$. What is the force on particle $a$? Linearize this force. The stuff in the second quotation is saying that, because we linearize the force, the force on particle $a$ looks like $v_1 u_1 + v_2 u_2 + \cdots + v_N u_N$ (this is LINEAR in $u$), for $v_i$ constants. $\endgroup$ – nervxxx Sep 25 '13 at 10:30
  • $\begingroup$ But each $v_i$ should actually carry the index $a$, since this coefficients change depending on which particle you are talking about. So really it should be $v_{a1} u_1 + v_{a2} u_2 + \cdots + v_{aN} u_N$, but hey, this is just $\sum_b v_{ab} u_b$. $\endgroup$ – nervxxx Sep 25 '13 at 10:32

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