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Odd Form Of Eigenvalue - Coupled Masses

  1. Jul 4, 2013 #1
    Odd Form Of Eigenvalue -- Coupled Masses

    This isn't strictly homework, since it's something I'm trying to self-teach, but it seems to fit best here.

    1. The problem statement, all variables and given/known data
    It's an example of applying eigenvalue methods to solve (classical) mechanical systems in an introductory text to QM; specifically, 1.8.6 in Shankar. I can follow everything fine until it actually comes to calculating the eigenvalues; when I tried to do it without looking at the result, mine was out by a sign and square root.

    The exact text is "...the eigenvalue of Ω is written as -ω^2 rather than ω in anticipation of the fact that Ω has eigenvalues of the form -ω^2, with ω real". How quickly it's glossed over makes me think I'm missing something obvious, but I can't for the life of me see it. I've reproduced my work up to that point below.

    2. Relevant equations & 3. The attempt at a solution
    KPnzbHJ.png
    "Determine x_1(t) and x_2(t) given the initial displacements x_1(0) and x_2(0) and that both masses are initially at rest."
    Applying Hooke's law and F=ma:
    [itex]
    \\
    {x}_1'' = -\frac{2k}{m}x + \frac{k}{m}x \\
    {x}_2'' = \frac{k}{m}x_1 - 2\frac{k}{m}x_2 \\ \\
    \varkappa \equiv \frac{k}{m} \\
    \begin{bmatrix}
    {x}_1''\\
    {x}_2''
    \end{bmatrix} = \begin{bmatrix}
    -2\varkappa & \varkappa \\
    \varkappa & -2\varkappa
    \end{bmatrix}\begin{bmatrix}
    x_1\\
    x_2
    \end{bmatrix} \\
    \left|x''(t)\right\rangle= \Omega \left|x(t)\right\rangle
    [/itex]

    This is in the basis with one vector equal to a unit displacement of the first mass, and the other the same for the second; the coupled differential equation is a pain to solve, so I want to swap to a basis with no contribution from x_1 to x_2 and vice versa, so diagonalising the Hermitian Ω as usual (det(Ω - ωI) = 0 and making a unitary diagonalising matrix from the eigenvector components) I end up with [tex]\omega_1 = -\varkappa[/tex] and [tex]\omega_2 = -3\varkappa[/tex]. The book itself gets [tex]\omega_1 = \sqrt{\varkappa}[/tex] and [tex]\omega_2 = \sqrt{3\varkappa}[/tex], which fits with the assumed form of ω, but I don't know where that comes from.

    Sorry if this is too much information, since a lot of it seems ancillary to the question of "Why is the eigenvalue taken to be that form?", but I wanted to err on the side of caution with my first post.
     
  2. jcsd
  3. Jul 4, 2013 #2

    SteamKing

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  4. Jul 4, 2013 #3
    You're confusing yourself. What they've done is simply instead of calling the eigenvalues of the matrix eg. omega, they call them -omega^2. This should be trivial. You've called the eigenvalues of the matrix omega - so of course the two results (yours and the one in the book) for the omega are different!

    Now, you just need to think about *why* they call it -omega^2..
     
  5. Jul 4, 2013 #4
    Yeah, it's the why that's confusing me; I was meaning to illustrate with that last part of the post that my approach was right in everything but not having the minus omega^2. It was unclear: sorry.

    I'll give it some thought. It feels like a tip-of-the-tongue thing: it seems vaguely familiar, but there's some link that's not clicking. I continued on with the problem anyway and got the same basis vectors and end result, though; it's just why Shankar knows to use the (admittedly neater) -omega^2 notation from the outset that is throwing me.
     
  6. Jul 4, 2013 #5
    It's simple, really.
    The usual equation is something like x'' = -k/m x. Solving this, we identify a characterstic frequency of the solution, namely sqrt(k/m). This is usually called omega. Only now, the equation is with matrices and all that, but the idea remains the same. The eigenvalues you find are really the k/m-eigenvalues, and then you 'translate' these to some frequency.
     
  7. Jul 4, 2013 #6
    Oh, God; thanks. I was getting too caught up in the linear algebra of it all just to realise the meaning of kappa. Shouldn't have gotten lazy and combined k/m into kappa; might have seen it then.
     
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