# The Lagrangian for a free particle

1. Jul 28, 2009

### qoqosz

According to Landau textbook:

Having two inertial frames K and K' moving with velocities $$\vec{v}$$ and $$\vec{v'}=\vec{v} + \vec{\epsilon}$$ where $$\vec{\epsilon}$$ is an infinitesimal. We have $$L' = L(v'^2) = L (v^2 + 2 \vec{v} \cdot \vec{\epsilon} + \epsilon^2)$$. Expanding this expression in powers of $$\vec{\epsilon}$$ and neglecting terms above first order, we obtain: $$L(v'^2) = L(v^2) + \frac{\partial L}{\partial v^2} 2 \vec{v} \cdot \vec{\epsilon}$$.

I simply don't understand how the second term on the right of the equation is of that form, i.e. why we differentiate with respect to v^2?
According to my calculations it should look like:

$$L(v'^2) \approx L(v^2) + \frac{\partial L (v^2)}{\partial \vec{\epsilon}} \vec{\epsilon} \cdot (2 \vec{v} + \vec{\epsilon} ) \approx L(v^2) + \frac{\partial L (v^2)}{\partial \vec{\epsilon}}2 \vec{\epsilon} \cdot \vec{v}$$

2. Jul 28, 2009

### Rebel

The simplest way I can put it is that in the unprimed frame the Lagrangian depends just on the square of the velocity, and if $$L(v^{2})$$ is a polynomial function or put it in terms of the Taylor expansion, the only terms that survive after taking into account those at first order in $$\overrightarrow{\varepsilon }$$ are precisely those corresponding to $$L(v^{2})$$ plus $$\frac{\partial L}{\partial v^{2}}$$ times $$2\overrightarrow{\varepsilon }\bullet \overrightarrow{v}$$..., since is the factor corresponding in this polynomial expansion. Remember Pascal's triangle?
1
1 1
1 2 1
1 3 3 1 .... the surviving terms are the first two and the factors different to 1 corresponds to $$\frac{\partial L}{\partial v^{2}}$$ in the mentioned expansion.

3. Jul 29, 2009

### qoqosz

I don't get it. Why there's $$\frac{\partial L}{\partial v^2}$$? If we expand function in Taylor series (taking $$\vec{\epsilon}$$ as an independent variable) there should be only derivatives in respect to that var (epsilon).

-edit-
I think I got it - from mean value theorem we have: $$\frac{f(x+h) - f(x)}{h} \approx f'(x), h \to 0$$, so: $$L( (v+\epsilon)^2 ) \approx L (v^2) + \frac{\partial L (v^2)}{\partial v^2} (v^2)' \epsilon$$ :)

Last edited: Jul 29, 2009
4. Jul 29, 2009

### Rebel

Right, but what I was referring with the partial derivative not was the derivatives in the taylor expansion, but to the terms that are left after this polynomial expansion. However, is much simpler the last by you.