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Separation of variables and the separation constant

  1. Mar 21, 2008 #1
    1. The problem statement, all variables and given/known data
    Please take a look at:


    Look at step 2.53. Can you explain to me how c^2*T'/T = k becomes T' = k*T*c^2?

    3. The attempt at a solution
    I don't get it. What am I missing here?
  2. jcsd
  3. Mar 21, 2008 #2


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    Yes, you're right!

    Hi Niles! :smile:

    Yes, you're right! :smile:

    2.5.3 is correct, but the c^2 in the following line should be 1/c^2. :frown:

    But it doesn't matter, because the c^2 in 2.5.5 is correct, since it's taken directly from 2.5.3 :smile:

    (The line after 2.5.3 was only used to get the sign of k, so it didn't matter whether k was multiplied or divided by c^2.)
  4. Mar 21, 2008 #3
    Hi Tim, thanks for replying. I hope it's OK if I ask another question.

    Please take a look at http://cow.physics.wisc.edu/~craigm/toroid/toroid/node4.html

    Here, they choose the separation constant to be -omega^2/c^2. What is the deal when finding the separation constant (SC from now on)? In my book they equal the exact same term to -omega^2. Should I include the constant in front of T''/T every time when choosing SC?

    Thanks in advance.

    Sincerely Niles.
  5. Mar 21, 2008 #4


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    Hi Niles! :smile:

    The c^2 has to be in equation (13) because (13) is taken directly from (11).

    The c^2 in equations (15) and (18) doesn't have to be there - omega is any constant, so it doesn't matter whether you choose omega or omega/c.

    I think he's done it that way for dimensional reasons. I'm inclined to agree with him.

    Don't you agree that (16) to (18) look much neater than they would if (18) included c^2? :smile:

    (Does your book use c, or does it put c = 1?)
  6. Mar 21, 2008 #5
    I agree - it does look better. So it is always a good thing to include the constant so we end up with a term T''/T = -omega^2?

    My book uses c. It is the exact same eq. as in the link.
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