Since the eigenspaces of t are all one dimensional

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Unformatted text preview: · · 1 0 0 · · · −2 xn We can write this as where x= 1 V (x1 , . . . , xn ) = − k xT Ax, 2 x1 x2 x3 · xn , A= −2 1 0 · 1 1 −2 1 · 0 0 1 −2 · 0 ··· ··· ··· ··· ··· or equivalently as n n 1 V (x1 , . . . , xn ) = − k aij xi xj , 2 i=1 j =1 54 1 0 0 · −2 , where aij denotes the (i, j )-component of the matrix A. Just as in the preceding section, the force acting on the i-th cart can be calculated as minus the derivative of the potential energy with respect to the position of the i-th cart, the other carts being held fixed. Thus for example, n F1 = − n n ∂V 1 1 =k a1j xj + k ai1 xi = k a1j xj , ∂x1 2 j =1 2 i=1 j =1 the last step obtained by using the symmetry of A. In general, we obtain the result: n ∂V Fi = − =k aij xj , ∂xi j =1 which could be rewritten in matrix form as F = kAx. (2.16) On the other hand, by Newton’s second law of motion, m d2 x = F. dt2 Substitution into (2.16) yields m d2 x = kAx dt2 or d2 x k = Ax, dt2 m (2.17) where A is...
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