1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Problem getting my matrices in correct form

  1. Dec 5, 2015 #1
    1. The problem statement, all variables and given/known data
    Hello guys; I am currently dealing with a problem that I have faced before several times and I would like to know a consistent way on how to solve it. I think what I want to do is diagonalize a matrix but I'm not sure if that's exactly it. Basically I have two or three equations of motion and I want to put them in a good matrix form in order to begin finding solutions. In this particular problem I have three equations of motion and the book I'm using puts them in to a nice matrix form right away and I'm not sure how to get there. Here is a picture of what they've done:
    Classical Mechanics Matrix.jpg
    I know how to create these matrices for 'm' and 'k' when they are more symmetric if you will but I don't know how to make it nice like this. This will help me so much on other problems I also have to deal with. For some stupid reason I've ran in to this wall multiple times. Thanks in advance.

    2. Relevant equations
    3 equations of motion put in to a nice diagonalized matrix

    3. The attempt at a solution
    I just don't understand how to get it into this nice 'diagonalized(?)' matrix form. If anyone can help out, this will greatly help me in the future.
     
    Last edited by a moderator: Dec 5, 2015
  2. jcsd
  3. Dec 5, 2015 #2

    DrClaude

    User Avatar

    Staff: Mentor

    I'm not sure I understand what you want. The form you get there, which is not diagonal by the way, by tridiagonal, is due to the fact that there is only nearest neighbor coupling. If there was an additional spring between 1 and 3, the matrix would look different.

    Note also that the result is due to the Ansatz that motion is harmonic, as given by the intermediate equation. This is what transforms the problem into an eigenproblem, with ω2 the eigenvalue.
     
  4. Dec 5, 2015 #3
    Okay, thank you for your response. I guess I'm just not sure exactly how to use the Ansatz and arrive at this matrix form. It says "substitution gives"... Not sure what substitution they are making,
     
  5. Dec 5, 2015 #4

    SteamKing

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    It's just a matter of organizing and re-arranging the unknowns, x10, etc., after making the substitution of the trial solution into the equations in (1).
     
  6. Dec 5, 2015 #5
    Okay, thanks, I'll give it a shot.
     
  7. Dec 5, 2015 #6
    It's tough because after I substitute in the Ansatz, I still don't know where the [itex](k-\omega^2M)[/itex] comes from or why it's only associated with the [itex]\ddot{x}[/itex] terms. Any hints as to how that works would be excellent. Also, if someone happens to know a situation where you need to diagonalize a matrix or simultaneous diagonalize two matrices and could point me in the direction of an example of such a situation that would be really great. Thanks
     
  8. Dec 5, 2015 #7

    SteamKing

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    Well, take the first equation in (1), namely ##M\ddot{x_1} + k(x_1 - x_2) = 0## and make the substitution ##x_i = x_{i0} e^{i\omega t}##. Note that the i-subscript is not the same as i in the exponential function, which is the imaginary unit.
     
  9. Dec 5, 2015 #8
    Right, I did that and I end up with [itex]M(\ddot{x_{10}}e^{i\omega t})+k(x_{10}e^{i\omega t}-x_{20}e^{i\omega t}) = 0 [/itex]. Not sure where to go after that.
     
  10. Dec 6, 2015 #9
    Ended up figuring this out by creating a matrix for M and K and X all separately and then combining them. Still not exactly sure how to get there with the Ansatz but I appreciate the help you guys gave me. Thanks
     
  11. Dec 6, 2015 #10

    DrClaude

    User Avatar

    Staff: Mentor

    That's not correct. Remember that ##\ddot{x}## means ##\frac{d^2x}{dt^2}##, so when you do the substitution:
    $$
    \begin{align*}
    M \ddot{x_{1}} &= M \frac{d^2x}{dt^2} \\
    &= M \frac{d^2}{dt^2} ( x_{10}e^{i\omega t} ) \\
    &= -M \omega^2 x_{10} \mathbf{e^{i\omega t}}
    \end{align*}
    $$
    (##x_{10}## is independent of time; the entire time dependence of ##x_{1}## is in the exponential)

    Edit: missing exponential in bold added.
     
    Last edited: Dec 7, 2015
  12. Dec 7, 2015 #11
    That's great, thank you! why do we drop the exponential at the end though?
     
  13. Dec 7, 2015 #12

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    The general solution does not have quite as simple a form as you suggest: besides ##x_k = x_{k0} \, e^{i \omega t}## terms there are also terms of the form ##x_1 = x_2 = x_3 = a + b t##. This is due in part to the fact that when written as a ##6 \times 6## linear first-order system we get a ##6 \times 6## matrix that is not quite diagonalizable: one of its Jordan blocks is non-diagonal. I can tell you more if you are interested.
     
  14. Dec 7, 2015 #13

    DrClaude

    User Avatar

    Staff: Mentor

    It is still there, I just forgot to add it. Eventually, you multiply everything by ##e^{-i \omega t}##, so it goes away.
     
  15. Dec 7, 2015 #14
    Gotchya okay, thanks a bunch for the help.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Problem getting my matrices in correct form
Loading...