# Landau's derivation of a free particle's kinetic energy- expansion of a function?

I was reading a bit of Landau and Lifshitz's Mechanics the other day and ran into the following part, where the authors are about to derive the kinetic energy of a free particle. They use the fact that the Lagrangian of this particle must be the same (or at most, differ by the total time derivative of a function of co-ordinates and time) in different inertial frames.

We have $L'=L(v'^2)=L(v^2+2\mathbf{v}\cdot\boldsymbol{\epsilon}+\boldsymbol{\epsilon}^2)$. Expanding this expression in powers of $\boldsymbol{\epsilon}$ and neglecting terms above the first order, we obtain $$L(v'^2)=L(v^2)+\frac{\partial L(v^2)}{\partial (v^2)}2\mathbf{v}\cdot\boldsymbol{\epsilon}.$$

I think I'm ok with all the physics in this section. What I don't get is just the part I quoted above (so maybe this post is better suited for the math site, but since this is book is so physics-y I thought I'd post it here). My math is pretty rusty, so I'm not really sure- how do the authors expand the function to arrive at the above expression? It reminds me a bit of a Taylor expansion, but not very much. What's the process used to arrive at it?

It is a Taylor expansion. You might be a little freaked out because they are treating $L$ as a function of $v^2$, but that doesn't matter, consider

$$L(x + \delta) \sim L(x) + \frac{\partial L}{\partial x} \delta + O(\delta^2)$$

but take $x=v^2$ and $\delta = 2 \boldsymbol{v} \cdot \boldsymbol{\epsilon} + \epsilon^2$,

\begin{align*} L(v'^2) &= L( v^2 + 2 \boldsymbol{v} \cdot \boldsymbol{\epsilon} + \epsilon^2) \\ &\sim L(v^2) + \frac{\partial L}{\partial v^2} ( 2 \boldsymbol{v} \cdot \boldsymbol{\epsilon} + \epsilon^2) + O(\epsilon^2) \\ &\sim L(v^2) + \frac{\partial L}{\partial v^2} 2 \boldsymbol{v} \cdot \boldsymbol{\epsilon} + O(\epsilon^2) \end{align*}

We can drop the other $\epsilon^2 \frac{\partial L}{\partial v^2}$ term since we are only interested in terms of first order.

### Taylor Expansion Generally

If its the taylor expansion you're having trouble with, that's pretty easy to show as well. Consider $f(x + \epsilon)$ with $\epsilon$ small, we might want to express the value of $f(x + \epsilon)$ as a power series in $\epsilon$, so we look for something of the form

$$f(x + \epsilon) = a_0 + a_1 \epsilon + a_2 \epsilon^2 + \cdots$$

How do we determine the coefficients, well we can find $a_0$ by taking the limit as $\epsilon \to 0$, obtaining $$f(x) = a_0$$

Great, but how would we get the $a_1$ coefficient? Well, just take a derivative of both sides

$$f'(x + \epsilon) = a_1 + 2 a_2 \epsilon + \cdots$$

and just take the limit again, obtaining $$f'(x) = a_1$$

Doing that process once more you get $$f''(x) = 2 a_2$$

So, so far we've got

$$f(x + \epsilon) = f(x) + f'(x) \epsilon + \frac 12 f''(x) \epsilon^2 + \cdots$$

If you think generally, you should be able to convince yourself that in general, for the $n$th term, we'll have

$$f^{(n)}(x) = n! a_n$$

giving us the general Taylor series result

$$f(x + \epsilon) = \sum_{i=0}^\infty \frac{1}{n!} f^{(n)}(x) \epsilon^n$$

• Ah, perfect- thanks. One more doubt I just had- why are they only interested in first order terms? Is it because $\epsilon$ can be arbitrarily small? It seems to me like if that was the reason, there would be some loss of generality in the derivation. – Physics Llama Jul 26 '14 at 2:39
• @PhysicsLlama They may have glossed this a bit in the text, but they use this first order difference to argue that $\frac{\partial L}{\partial v^2}$ is independent of $v^2$. But if the first partial is independent (meaning a constant) any higher partials must vanish, meaning even if we tried to go to higher order in the expansion, they would all disappear. – alemi Jul 26 '14 at 2:45
• @PhysicsLlama also the argument must work for vanishing $\epsilon$ as you point out. – alemi Jul 26 '14 at 2:48
• Neat way to remember the coefficients of a Taylor expansion there, by the way :) +1. – eqb Jul 26 '14 at 3:19
• Important subtlety that I just noticed: you expanded f(x+e) in powers of e, and then took the derivative of the polynomial with respect to e in order to find the coefficients a. So you didn't take the derivative of f with respect to x, but rather with respect to e. In your earlier development, though, you took it with respect to v^2, which you had set as your x. Isn't this inconsistent? – Physics Llama Dec 10 '15 at 12:42