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?


1 Answer 1


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

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

  • $\begingroup$ 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. $\endgroup$ Commented Jul 26, 2014 at 2:39
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    $\begingroup$ @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. $\endgroup$
    – alemi
    Commented Jul 26, 2014 at 2:45
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    $\begingroup$ @PhysicsLlama also the argument must work for vanishing $\epsilon$ as you point out. $\endgroup$
    – alemi
    Commented Jul 26, 2014 at 2:48
  • $\begingroup$ Neat way to remember the coefficients of a Taylor expansion there, by the way :) +1. $\endgroup$
    – eqb
    Commented Jul 26, 2014 at 3:19
  • $\begingroup$ 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? $\endgroup$ Commented Dec 10, 2015 at 12:42

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