# Clarification for Landau & Liftshitz “Mechanics” pg. 14-15 (Conservative Law)

I have two questions regarding chapter 2 of Landau & Liftshitz "Mechanics", on page 14-15.

Regarding homogeneity of time, we have:

$$L$$ is a closed system therefore $$L$$ does not depend on time explicitly

$$\frac{dL}{dt} = \sum_i \frac{\partial L}{\partial q_i}q_i + \sum_i \frac{\partial L}{\partial \dot{q_i}} \ddot{q_i}$$

$$\frac{dL}{dt} = \sum_i \dot{q_i} \frac{d}{dt}\frac{\partial L}{\partial \dot{q_i}} + \sum_i \frac{\partial L}{\partial \dot{q_i}} \ddot{q_i} = \sum_i \frac{d}{dt}(\frac{\partial L}{\partial \dot{q_i}}\dot{q_i})$$

$$\frac{d}{dt}(\sum_i \dot{q_i} \frac{\partial L}{\partial \dot{q_i}}-L) = 0$$

then $$E := \sum_i \dot{q_i} \frac{\partial L}{\partial \dot{q_i}}-L.\tag{6.1}$$

Because energy $$E$$ is a linear function of $$L$$, the additivity of $$E$$ is followed by the additivity of $$L$$

Question1: Why $$E$$ is a linear function of $$L$$?

Here is my thought: In a closed system, $$L$$ does not explicitly depend on time; therefore $$\sum_i \dot{q_i} \frac{\partial L}{\partial \dot{q_i}}$$ is a total derivative of time. But I am not sure about the linear relationship of $$E$$ with respect to $$L$$.

--------------------------------------------

Regarding the homogeneity of space:

"By virtue of this homogeneity, the mechanical properties of a closed system are unchanged by any parallel displacement of the entire system in space."

Consider the radius vector of all particles are moved by the same amount of infinitesimal distance i.e. $$\vec{r} \to \vec{r} + \vec{\epsilon}$$

$$\delta L = \sum_{\alpha}\frac{\partial L}{\partial \vec{r_{\alpha}}} \cdot \delta \vec{r_{\alpha}} = \vec{\epsilon} \cdot \sum_{\alpha}\frac{\partial L}{\partial \vec{r_{\alpha}}} = 0.\tag{7.0}$$

Question2: Why $$\delta L = \sum_{\alpha}\frac{\partial L}{\partial \vec{r_{\alpha}}} \cdot \delta \vec{r_{\alpha}}$$ ? (Sorry I am not always comfortable with the variational notation.) What algebra does the author use here?

The fact that $$E$$ as defined in (6.1) is a linear function of $$L$$ has nothing to do with physics; it's a purely mathematical fact and to understand it you just have to understand what the word "linear" means.

In general a linear transformation is (a function $$T:V\to W$$ between vector spaces over a given field) such that for all scalars $$c$$, and all $$\xi,\eta\in V$$, we have \begin{align} T(c\xi + \eta) + cT(\xi) + T(\eta). \end{align}

In your case, you have $$E[L]:= \sum \dot{q}^i\frac{\partial L}{\partial q^i} - L$$. So, saying $$E$$ is linearly related to $$L$$ means we have to check that given any real number $$c$$, and any $$L_1,L_2$$, we have $$E[cL_1 + L_2] = cE[L_1] + E[L_2]$$. But this is straight forward: \begin{align} E[cL_1 + L_2] &= \sum\dot{q}^i\frac{\partial (cL_1 + L_2)}{\partial \dot{q}^i} - (cL_1 + L_2) \\ &= \dots \\ &= c\cdot \left( \sum \dot{q}^i \dfrac{\partial L_1}{\partial q^i} - L_1\right) + \left( \sum \dot{q}^i \dfrac{\partial L_2}{\partial q^i} - L_2\right)\\ &= c\cdot E[L_1] + E[L_2] \end{align} I'm sure using standard derivative rules and algebra you can justify the $$\dots$$ above.

(If this seems abstract, just consider a simple example of $$f:\Bbb{R}\to \Bbb{R}$$, $$f(x) = mx$$; this certainly satisfies $$f(cx_1 + x_2) = cf(x_1) + f(x_2)$$... this is where the more general definition of linear transformation is motivated from).

For a differentiable function $$f:\Bbb{R}^n\to \Bbb{R}$$, are you comfortable with the idea that \begin{align} df = \sum_{i=1}^n \dfrac{\partial f}{\partial x^i}\, dx^i \end{align}

If yes, then the idea behind (7.0) is very similar. Just replace the $$f$$ with $$L$$, and $$d$$ with $$\delta$$ (but if you want a more precise answer, you would have to first properly define the variation $$\delta$$, something which isn't carefully done in L&L).

• Best regards to the user of Math.SE: – Sebastiano Aug 22 '20 at 21:26

Consider two Lagrangians $$L_1$$ and $$L_2$$ and their corresponding energies $$E_1$$ and $$E_2$$. Now consider $$L=aL_1+bL_2$$ where $$a,b$$ are real numbers. This leads to the energy following an analogous expression of: $$E=aE_1+bE_2$$

This is because energy is related to the lagrangian by a linear (differential) operator as follows: $$E := \left(\sum_i \dot{q_i} \frac{\partial }{\partial \dot{q_i}}-\text{id}\right)L$$ Where id stands for the identity operator.

The variational notation $$\delta$$ represents a change in the whole system. Essentially how has the functional form changed as a result of some small change in a parameter. As opposed to a regular derivative which informs how the function’s value has changed as a result of some small change in it’s input.

Coming to the matter at hand, we want to shift our entire system by some vector. And as a result of which we want to see how the functional form of our Lagrangian changes. Since $$r$$ and $$\dot r$$ are independent, all change must come only from $$r$$. And that’s how we get:

$$\delta L = \sum_{\alpha}\frac{\partial L}{\partial \vec{r_{\alpha}}} \cdot \delta \vec{r_{\alpha}}$$