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Motion in a rapid oscillating field

  1. Mar 25, 2014 #1
    Hello,

    I've been making my way through Landau-Lifshitz's "Mechanics" book, and I've come across a bit of math I'm not too sure about.

    What I'm confused about is here:

    https://archive.org/details/Mechanics_541

    On Page 93 (of the book, not the PDF) under Motion in a rapid oscillating field. The derivation is simple up until you get to the part "Substituting (30.3) in (30.2) and expanding in powers [itex]\xi[/itex] as far as the first order terms...".

    The equation they end up with is:

    [itex]m\ddot{X}+m\ddot{\xi}=-\frac{dU}{dx}-\xi\frac{d^2U}{dx^2}+f_{(X,t)}+\xi\frac{\partial f}{\partial X}[/itex]

    So I'm wondering how they get this using the substitution:

    [itex]x_{(t)}=X_{(t)}+\xi_{(t)}[/itex]

    into the equation:

    [itex]m\ddot{x}=-\frac{dU}{dx}+f_{(t)}[/itex]

    where [itex]f_{(t)}=f_1 \cos{\omega t} + f_2 \sin{\omega t}[/itex]

    It seems to me there is some strange derivative such as [itex]\frac{d}{d(X_{(t)}+\xi_{(t)})}[/itex] which so far I have had no luck figuring out. Also it seems as though [itex]f_{(t)} → f_{(X,t)}[/itex] which probably has something to do with the transformation. I also believe some sort of Taylor expansion is happening. If anyone has any ideas, please let me know.
     
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  3. Mar 25, 2014 #2

    Simon Bridge

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    Numbering your equations:
    $$\begin{align}\text{(1) :}& m\ddot{X}+m\ddot{\xi}=-\frac{dU}{dx}-\xi\frac{d^2U}{dx^2}+f_{(X,t)}+\xi\frac{\partial f}{\partial X}\\

    \text{(2) :}& x_{(t)}=X_{(t)}+\xi_{(t)}\\

    \text{(3) :}& m\ddot{x}=-\frac{dU}{dx}+f_{(t)}:\; f_{(t)}=f_1 \cos{\omega t} + f_2 \sin{\omega t}
    \end{align}$$ ... did you intend the (X,t) and (t) parts as subscripts there?
    (If so, what do they indicate? Partial derivatives?)

    You want to get (1) from (2) and (3).
     
  4. Mar 25, 2014 #3
    The subscripts are notation for "function-of" in this case, so [itex]f_{(t)}[/itex] would be read "[itex]f[/itex] is a function of [itex]t[/itex]".

    Definitely you cannot retrieve this equation by direct substitution, as mentioned in the book which I linked to, they expand "something" (which they don't specifically mention) in terms of [itex]\xi[/itex]. If you read the page I mentioned it will maybe explain a bit more what they are trying to accomplish (page 93 of the book).

    Thanks for your fast reply.
     
  5. Mar 25, 2014 #4

    Simon Bridge

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    OK - "function of" does not need to be in a subscript.
    "f is a function of t" would be written ##f(t)##

    I see what you mean about the passage: it says "substituting 30.3 into 30.2 and expanding in powers of ##\small \xi## we get ...

    That would certainly imply a power series expansion.

    Notice that ##f_1## and ##f_2## are functions of the coordinates only - thus the while function ##f## is a function of both the coordinates and time. The position of the mean is given by ##X## so for very small oscillations we may be able to say: ##f(x,t) = f_1(X)\cos\omega t + f_2(X)\sin\omega t = f(X,t)## ... the small oscillations in the motion are given by ##\xi(t)##

    But at this point my eyes hurt too much from the tiny text - I'll have to leave you to someone who knows it better.
     
  6. Mar 26, 2014 #5

    TSny

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    The force from ##U## on the particle at time ##t## when the particle is at ##x(t) = X(t)+\xi(t)## is given by ##-\frac{dU}{dx}## evaluated at the location of the particle: [tex]-\frac{dU}{dx} \bigg|_{x = X(t) + \xi(t)}[/tex] To first order in ##\xi##, this can be approximated as [tex]-\frac{dU}{dx}\bigg|_{x = X(t)} -\left[ \frac{d}{dx} \left(\frac{dU}{dx}\right)\right]_{x = X(t)} \cdot \xi(t) = -\frac{dU}{dx}\bigg|_{x = X(t)} -\xi(t)\frac{d^2U}{dx^2} \bigg|_{x = X(t)} [/tex]
     
    Last edited: Mar 26, 2014
  7. Mar 26, 2014 #6
    So that clears those two terms, would the other two then follow from:

    [itex]f(X,t)+\frac{\partial f}{\partial X}\xi(t)[/itex]

    in first order? this looks like some kind of expansion.

    EDIT: In other words, the evaluation of the function at the sum can be interpreted as expanding about [itex]\xi[/itex] but evaluating at [itex]X[/itex]?
     
  8. Mar 26, 2014 #7

    TSny

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    Yes, it's a typical Taylor expansion in x for a fixed time t. The usual language is to say that the expansion is "about the point ##X##", rather than about ##\xi##.

    See http://mathworld.wolfram.com/TaylorSeries.html
     
  9. Mar 26, 2014 #8
    Sorry, I do know a bit about Taylor series, and I did mean "about the point [itex]x=X[/itex]" I just didn't read over my post.

    What I don't understand about this specific example is how the derivative is with respect to [itex]X[/itex] rather than [itex]x[/itex]. Usually a Taylor series is:

    [itex]f(x)=f(x_0)+(x-x_0)\frac{df}{dx}\bigg|_{x=x_0}[/itex]

    which would translate to:

    [itex]f(x,t)=f(X,t)+(x(t)-X(t))\frac{\partial f}{\partial x}\bigg|_{x=X}[/itex]

    with [itex]x(t)-X(t)=\xi(t)[/itex], therefore:

    [itex]f(x,t)=f(X,t)+\xi(t)\frac{\partial f}{\partial x}\bigg|_{x=X}[/itex]

    So how does that derivative change from [itex]\frac{\partial}{\partial x}[/itex] to [itex]\frac{\partial}{\partial X}[/itex].

    Thanks though, this has been extremely helpful.
     
  10. Mar 26, 2014 #9

    TSny

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    I think they are just using the notation [itex]\frac{\partial f}{\partial X}[/itex] to stand for ##\frac{\partial f}{\partial x}\bigg|_{x=X}##.
     
  11. Mar 26, 2014 #10
    Sounds like something Landau and Lifshitz would do, I'll take this as the case for now. Thanks again.
     
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