Motion in a rapid oscillating field

[electricspit]

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 $\xi$ as far as the first order terms...".

The equation they end up with is:

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

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

$x_{(t)}=X_{(t)}+\xi_{(t)}$

into the equation:

$m\ddot{x}=-\frac{dU}{dx}+f_{(t)}$

where $f_{(t)}=f_1 \cos{\omega t} + f_2 \sin{\omega t}$

It seems to me there is some strange derivative such as $\frac{d}{d(X_{(t)}+\xi_{(t)})}$ which so far I have had no luck figuring out. Also it seems as though $f_{(t)} → f_{(X,t)}$ 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.

 

[Simon Bridge]

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).

 

[electricspit]

The subscripts are notation for "function-of" in this case, so $f_{(t)}$ would be read "$f$ is a function of $t$".

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 $\xi$. 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.

 

[Simon Bridge]

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.

 

[TSny]

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: $$-\frac{dU}{dx} \bigg|_{x = X(t) + \xi(t)}$$ To first order in $\xi$, this can be approximated as $$-\frac{dU}{dx}\bigg|_{x = X(t)} -\left[ \frac{d}{dx} \left(\frac{dU}{dx}\right)\right]_{x = X(t)} \cdot \xi(t) = -\frac{dU}{dx}\bigg|_{x = X(t)} -\xi(t)\frac{d^2U}{dx^2} \bigg|_{x = X(t)}$$

 

[electricspit]

So that clears those two terms, would the other two then follow from:

$f(X,t)+\frac{\partial f}{\partial X}\xi(t)$

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 $\xi$ but evaluating at $X$?

 

[TSny]

$f(X,t)+\frac{\partial f}{\partial X}\xi(t)$

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 $\xi$ but evaluating at $X$?

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

 

[electricspit]

Sorry, I do know a bit about Taylor series, and I did mean "about the point $x=X$" 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 $X$ rather than $x$. Usually a Taylor series is:

$f(x)=f(x_0)+(x-x_0)\frac{df}{dx}\bigg|_{x=x_0}$

which would translate to:

$f(x,t)=f(X,t)+(x(t)-X(t))\frac{\partial f}{\partial x}\bigg|_{x=X}$

with $x(t)-X(t)=\xi(t)$, therefore:

$f(x,t)=f(X,t)+\xi(t)\frac{\partial f}{\partial x}\bigg|_{x=X}$

So how does that derivative change from $\frac{\partial}{\partial x}$ to $\frac{\partial}{\partial X}$.

Thanks though, this has been extremely helpful.

 

[TSny]

$f(x,t)=f(X,t)+\xi(t)\frac{\partial f}{\partial x}\bigg|_{x=X}$

So how does that derivative change from $\frac{\partial}{\partial x}$ to $\frac{\partial}{\partial X}$.

I think they are just using the notation $\frac{\partial f}{\partial X}$ to stand for $\frac{\partial f}{\partial x}\bigg|_{x=X}$.

 

[electricspit]

Sounds like something Landau and Lifshitz would do, I'll take this as the case for now. Thanks again.