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How to get this equation - Mechanics by Landau and Lifshitz

  1. Jul 28, 2012 #1
    In page 28 of Mechanics by Landau and Lifgarbagez, there is the following equation.

    [tex]\int_0^\alpha \frac{T(E) dE}{\sqrt{\alpha-E}}=\sqrt{2m}\int_0^\alpha \int_0^E \left[ \frac{dx_2}{dU}-\frac{dx_1}{dU}\right] \frac{dU dE}{\sqrt{[(\alpha-E)(E-U)]}}[/tex]

    Then, by changing the order of integration, it is converted to,

    [tex]\int_0^\alpha \frac{T(E) dE}{\sqrt{\alpha-E}}=\sqrt{2m}\int_0^\alpha \left[ \frac{dx_2}{dU}-\frac{dx_1}{dU}\right] dU \int_U^\alpha \frac{dE}{\sqrt{[(\alpha-E)(E-U)]}}[/tex]

    I don't really understand how this happens. Could someone please break it down for me? Thanks.

    An image of the page is attached.
     

    Attached Files:

  2. jcsd
  3. Jul 28, 2012 #2

    Simon Bridge

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    Last edited by a moderator: Sep 25, 2014
  4. Jul 28, 2012 #3

    haruspex

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    I don't like the way they've written the last line. It makes it look as though the integrals have been decoupled. Clearer would be:
    [tex]\sqrt{2m}\int_{U=0}^\alpha \left(\left[ \frac{dx_2}{dU}-\frac{dx_1}{dU}\right] dU \int_U^\alpha \frac{dE}{\sqrt{[(\alpha-E)(E-U)]}}\right)[/tex]
    For the change of order, think about the total set of values of U and E being integrated over. (Perhaps think of it as a region in the U-E plane.) Both range from 0 to α, but with the condition that U < E (a triangle in the plane). If you let E range second, that becomes "U = 0 to E for each E, while E = 0 to α". If you swap the order then it's "E = U to α for each U, while U = 0 to α"
     
  5. Jul 29, 2012 #4
    The square bracketed term doesn't contain any dependence on E, so it can be brought outside the inner integral.

    I don't think the order has actually been changed, it's just a simplification, in the same way that you might simplify d(2x^2)/dx to 2(d(x^2)/dx), because 2 is not a function of x.
     
  6. Jul 29, 2012 #5

    haruspex

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    You're overlooking the presence of E in the range for the (original) inner integral. This makes the inner integral a function of E.
     
  7. Jul 29, 2012 #6
    Ah- that makes sense! Note to self- don't second guess Landau and Lifgarbagez.
     
  8. Jul 30, 2012 #7
    Thanks for the links and the explanations. But I still don't get one small part.

    So what they mean by saying "[itex]\alpha[/itex] is a parameter" is that it is a constant, right? So if I draw the [itex]E = \alpha[/itex] line in the [itex]U - E[/itex] plane it will be a straight line parallel to the [itex]U[/itex] axis? Similarly, the [itex]U = \alpha[/itex] line will be straight line parallel to the [itex]E[/itex] axis?

    Assuming that, I get this from the original equation,
    [tex]\int_0^\alpha \frac{T(E) dE}{\sqrt{\alpha-E}}=\sqrt{2m}\int_0^\alpha \left[ \frac{dx_2}{dU}-\frac{dx_1}{dU}\right] \int_U^\alpha \frac{dE dU}{\sqrt{[(\alpha-E)(E-U)]}}[/tex]

    How did they move the [itex]dU[/itex] to the left of the second integration sign to get the final answer? Doesn't that decouple the integrals? The integration with respect to [itex]E[/itex] is still a function of [itex]U[/itex], so it cannot be taken out of the integration with respect to [itex]U[/itex].
     
    Last edited: Jul 30, 2012
  9. Jul 30, 2012 #8

    haruspex

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    That's what I was complaining about in my earlier post. The way it's written in the book, you could misread it as having decoupled the integrals. What they mean is:
    [tex]\sqrt{2m}\int_{U=0}^\alpha \left(\left[ \frac{dx_2}{dU}-\frac{dx_1}{dU}\right] dU \int_U^\alpha \frac{dE}{\sqrt{[(\alpha-E)(E-U)]}}\right)[/tex]
    Here I've deliberately made only the minimal change of putting in parentheses, but it says the same as this, which is even clearer:
    [tex]\sqrt{2m}\int_{U=0}^\alpha \left(\left[ \frac{dx_2}{dU}-\frac{dx_1}{dU}\right] \int_U^\alpha \frac{dE}{\sqrt{[(\alpha-E)(E-U)]}}dU\right)[/tex]
     
  10. Jul 31, 2012 #9
    Oh OK. I was always under the impression that when you use integration as an operation the operand always has to be between the [itex]\int[/itex] and the [itex]dU[/itex].

    But when it's written down like this, [tex]\int_0^\alpha \frac{T(E) dE}{\sqrt{\alpha-E}}=\sqrt{2m}\int_0^\alpha \left[ \frac{dx_2}{dU}-\frac{dx_1}{dU}\right] dU \int_U^\alpha \frac{dE}{\sqrt{[(\alpha-E)(E-U)]}}[/tex] how do you know that the term [itex]\frac{1}{\sqrt{[(\alpha-E)(E-U)]}}[/itex] is to be integrated with respect to [itex]U[/itex]? Because there is a [itex]U[/itex] in it?
     
  11. Jul 31, 2012 #10

    haruspex

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    That is the usual style, but in principle the dU is just a factor in the expression.
    Yes and no.
    1. If there had been no U in the second integral then it would have been independent of the first, so it wouldn't matter whether you thought it was inside or decoupled.
    2. Since there is a U in the second integral, and since the variable of integration has no meaning outside the integral, it is implicitly inside the first integral.
    3. But, it is very common to use the same symbol inside and outside an integral, e.g. [tex]y = x + \int_0^xx^2.dx[/tex]This really means [tex]y = x + \int_{t=0}^xt^2.dt[/tex]The use of x both inside and outside is a 'pun'. Since this means you cannot rely on (2), it's better to make the expression clear with suitable use of parentheses.
     
  12. Jul 31, 2012 #11
    Thanks!
     
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