# Mechanical similarity in Landau

I've read this very short paragraph from Landau & Lifshitz's Mechanics (Chap.2, Par.10) (that you can find here) about Mechanical similarity.

I was looking for some more detailed explanations of the matter, at a level like the one in the first chapters of Landau's book. I've been able to find this article but it's a little too much for me.

Can you give me some references that do not go into quantum mechanics, i.e. that refer only to classical mechanics and Lagrangian formalism, about the subject?

Content of the cited paragraph As requested in the comments, I will summarize the paragraph's content; I will give my personal understanding of Landau's explanation, so I can be corrected if I'm wrong.

Suppose a system of particles is described by a Lagrangian $$\cal L(\mathbf{r_1},\mathbf{\dot{r_1}},\mathbf{r_2},\mathbf{\dot{r_2}},...,t)$$ and suppose the potential energy is such that $$U(\alpha \mathbf{r_1},\alpha \mathbf{r_2},...)=\alpha ^k ·U(\mathbf{r_1},\mathbf{r_2},...)$$. Since multiplying the Lagrangian by a constant leaves the equations of motions unaltered, we may multiply $$\cal L$$ by $$\alpha ^k$$. In that case, the kinetical energy becomes (let's look at the kinetic energy of a generic particle): $$\alpha ^ k · T = \frac{m}{2} \left(\alpha ^{k/2}\dfrac{ \text d \mathbf r}{\text d t}\right)^2=\frac{m}{2}\left(\dfrac{\text d \,\alpha \mathbf{r}}{\text d \, \alpha^{1-k/2}t}\right)^2,$$ so letting $$\mathbf{r'}=\alpha \mathbf{r}$$ and $$t'=\alpha^{1-k/2}t$$ we get $$\alpha ^k \cal L (\mathbf{r_1},\mathbf{\dot{r_1}},\mathbf{r_2},\mathbf{\dot{r_2}},...,t)=\cal L (\mathbf{r_1'},\mathbf{\dot{r_1'}},\mathbf{r_2'},\mathbf{\dot{r_2'}},...,t').$$ In conclusion, if lenght and times are scaled respectively by a factor $$\alpha$$ and $$\alpha ^{1-k/2}$$, the resulting equations of motion are identical and the paths followed by the system of particles are similar. From this we may infer the ratios of times in similar paths from the ratios of lenghts beetween the two paths. For example: for gravitational potential $$k=-1$$, so:$$t/t'=(l/l')^{3/2},$$ which is Kepler's Third Law.

• Copy the quote here, already! Why ask every possible reader to download a fat PDF? And then hunt through it to find the paragraph you want. Why risk link-rot leaving the question incomplete in the future? Likewise, it is better to link to the abstract page for arXiv articles so that you reader can check the abstract before downloading a potentially very large file. Commented Jun 29, 2013 at 17:59
• I agree with you about link-rot, but I point out that the pdf he links to is four pages alone and contains only the relevant section (section 10 of chapter 2) of the book. Commented Jun 29, 2013 at 18:18
• Thank you @dmckee for your suggestions. I've added an explanation of the cited paragraph. I've preferred not to copy-paste Landau's derivation, but to include my own explanation, so I can be corrected if I'm wrong in interpreting Landau's. PS: first link is surely worth a reading, so I will leave it there. Commented Jun 29, 2013 at 18:37
• Sedov has a book on Similarity. Worth checking your library for a copy, if you cannot purchase one. Commented Jun 29, 2013 at 19:13
• Your interpretation of the derivation is perfectly rigth, I would say. So does there still remain anything of your question? :)
– Ilja
Commented Apr 4, 2016 at 22:11

Preface

I struggled with what Landau writes in the section on Mechanical similarity [1]. Like the OP, "I [too] was looking for some more detailed explanations of the matter, at a level like the one in the first chapters of Landau's book." I looked at a few places including [2]. I was not satisfied with what I found. So, I prepared my own detailed explanation, which I set down below. In this answer, I quote liberally and without attribution from Landau and Lifshitz; and some Wikipedia pages.

My answer is divided into three major components. In the first component, I present a short summary of an essential proposition that the reader should be familiar with by the time they reach the section on Mechanical similarity in Landau [1]. In the second component, I review what is meant of when we speak of homogeneous functions; and I present two examples. The results of both these examples are used in the third component. In the third component I first present Proposition 2 that regards what is means when we speak of mechanical similarity; and next, I prove that proposition. Equation (5) in Proposition 2 describes the exact mechanical similitude. Bear in mind that Equation (5) is not identical to Equation (10.2) given by Landau and Lifshitz in their explanation of mechanical similarity [1].

First Component

Proposition 1: Multiplication of the Lagrangian by any constant does not effect the equations of motion.

Proof. \begin{align*} \left(\frac{d}{dt} \frac{\partial [kL]}{\partial{\dot{q}} } \right) -\frac{\partial [kL]}{\partial{ {q}} } =0 \\ k\left(\left(\frac{d}{dt} \frac{\partial [L]}{\partial{\dot{q}} } \right) -\frac{\partial [L]}{\partial{ {q}} }\right) =0 \end{align*} Q.E.D

This proof makes possible, in a number of important cases (cf, scattering Problem 3 [2]) some useful inferences concerning properties of the motion, without the necessity of actually integrating the equations of motions. These inferences rely on the properties of homogeneous functions.

Second Component

Definition [Cone, linear cone] A subset $$C$$ of a vector space $$V$$ over an ordered field $$F$$ is a cone (or sometimes called a linear cone) if for each $$x$$ in $$C$$ and positive scalar $$\alpha$$ in $$F$$, the product $$\alpha x$$ is in $$C$$.

Example 1:

The set $$\left \{ x \in \mathbb{R}^2 \mid x_2 \geq 0, x_1 = 0 \right \} \cup \left \{ x \in \mathbb{R}^2 \mid x_1 \geq 0, x_2 = 0 \right \}$$ is a cone, but not a convex cone.

Example 2:

The norm cone $$\left \{ (x, r) \in \mathbb{R}^{d+1} \mid \|x\| \leq r \right \}$$ is a convex cone.

Definition [Homogeneous function] Let $$V$$ and $$W$$ be two vector spaces over a field $$F$$. A homogeneous function $$f$$ from $$V$$ to $$W$$ is a function from $$V$$ to $$W$$ that has a linear cone $$C$$ as its domain, and satisfies $$f(s\,x)=s^{k}f(x)$$ for some integer $$k$$, every $$x \in C$$, and every nonzero $$s\in F$$. The integer $$k$$ is called the degree of homogeneity, or simply the degree of $$f$$.

Example 3:

The kinetic-energy function, $$T : \mathbb{R}^{3n} \to \mathbb{R}$$ is the rule that assigns to each tuple $$\left(\dot{\mathbf{q}}_1,\ldots, \dot{\mathbf{q}}_n\right)$$ in $$\mathbb{R}^{3n}$$ the value $$T{\left(\dot{\mathbf{q}}_1,\ldots, \dot{\mathbf{q}}_n\right)} = \sum\limits_{i=1}^n k_i \,\dot{\mathbf{q}}_i^2$$ in $$\mathbb{R}$$. The kinetic-energy function $$T$$ is homogeneous of degree $$2$$: $$T{\left(\frac{\alpha}{\beta}\,\dot{\mathbf{q}}_1,\ldots,\frac{\alpha}{\beta}\,\dot{\mathbf{q}}_n\right)} = \sum\limits_{i=1}^n k_i \,\left(\frac{\alpha}{\beta}\,\dot{\mathbf{q}}_i\right)^2 = \left(\frac{\alpha}{\beta}\right)^2\sum\limits_{i=1}^n k_i \,\left( \dot{\mathbf{q}}_i\right)^2 = \left(\frac{\alpha}{\beta}\right)^2 f{\left( \dot{\mathbf{q}}_1,\ldots, \dot{\mathbf{q}}_n\right)}.$$

Example 4:

Let $$k$$ be an element in the field $$\mathbb{R}$$. Let $$j$$ be an element in the field $$\mathbb{N}$$. The family of central field potential-energy functions, $$V_k : \mathbb{R}^{3n} \to \mathbb{R}$$ is the rule that assigns to each tuple $$\left( {\mathbf{q}}_1,\ldots, {\mathbf{q}}_n\right)$$ in $$\mathbb{R}^{3n}$$ the value $$V_k{\left(\mathbf{q}_1,\ldots, \mathbf{q}_n\right)} = \sum_{i=1}^{n}\sum_{l=i+1}^{n}\left|\mathbf{q}_i- \mathbf{q}_l\right|^k$$ in $$\mathbb{R}$$. The family of central field potential-energy functions $$V_k$$ is homogeneous of degree $$k$$: $$V_k {\left(\alpha\mathbf{q}_1,\ldots, \alpha\mathbf{q}_n\right)} = \sum_{i=1}^{n}\sum_{l=i+1}^{n}\left|\alpha\mathbf{q}_i- \alpha\mathbf{q}_l\right|^k = \alpha^k\sum_{i=1}^{n}\sum_{l=i+1}^{n} \left| \mathbf{q}_i- \mathbf{q}_l\right|^k = \alpha^k\,V_k {\left( \mathbf{q}_1,\ldots, \mathbf{q}_n\right)} \,.$$

Third Component

Proposition 2 [Exact mechanical similitude] Let $$k$$ be a fixed element in the field $$\mathbb{R}$$. Let $$n$$ be a fixed element in the field $$\mathbb{N}$$. Consider the Lagrangian of a conservative system, $$L_k = L_k{\left( \mathbf{q}_1,\ldots, \mathbf{q}_n , \dot{\mathbf{q}}_1,\ldots, \dot{\mathbf{q}}_n \right)} \,,$$ where the potential energy is a central field given by the homogeneous power law $$V_k{\left(\mathbf{q}_1,\ldots, \mathbf{q}_n\right)} = \sum_{i=1}^{n}\sum_{j=i+1}^{n}c_{i,j}\left|\mathbf{q}_i- \mathbf{q}_j\right|^k \,.$$ The trajectories (i.e., $$\mathbf{q}_1(t),\ldots,\mathbf{q}_n(t)$$) that satisfy the equations of motion generated by the Lagrangian have the property of mechanical similarity. Namely, for any $$t \in \mathbb{R}$$, any $$i= 1,\ldots,n$$, any $$j = 1,\ldots, n$$, every $$\mathbf{q}_i \in \mathbb{R}^3$$, every $$\mathbf{q}_j \in \mathbb{R}^3$$, any integer $$k$$, and every nonzero $$\alpha \in \mathbb{R}$$, the exact mechanical similitude is given by the equation $$\left| \alpha\,\mathbf{q}_i {\left( \alpha^{\left[1-\frac{k}{2}\right]}\,t\right)} - \alpha\,\mathbf{q}_j { \left( \alpha^{\left[1-\frac{k}{2}\right]}\,t\right) } \right| = \alpha \left| \mathbf{q}_i{\left( t\right)} - \mathbf{q}_j{\left(t\right)} \right|\,\tag{5}.$$

Proof. Let $$\alpha$$ and $$\beta$$ be scalars such that $$\alpha \in \mathbb{R}$$ and $$\beta \in \mathbb{R}$$. Let $$f: \mathbb{R}^3 \to \mathbb{R}^3$$ be the rule assigns to each $$\mathbf{q} \in \mathbb{R}^3$$ the value $$\mathbf{q}' = f(\mathbf{q}) = \alpha\,\mathbf{q}\, \tag{10}$$ where $$\mathbf{q}'\in \mathbb{R}^3$$. Let $$h: \mathbb{R} \to \mathbb{R}$$ be the rule assigns to each $$t \in \mathbb{R}$$ the value $$t' = h(t) = \beta\,t\,,\tag{15}$$
where $$t'\in \mathbb{R}$$. Let $$h^{-1}: \mathbb{R} \to \mathbb{R}$$ be the rule assigns to each $$t' \in \mathbb{R}$$ the value $$t = h^{-1}(t') = \frac{1}{\beta}\,t'\,,\tag{20}$$
where $$t\in \mathbb{R}$$.

Since $$\mathbf{q}' = f(\mathbf{q})$$, $$\mathbf{q} = g(t)$$, and $$t = h^{-1}(t')$$, then the chain rule is written in Leibniz notation as: \begin{align*} \frac {d\mathbf{q}'}{dt'} &= \frac {d\mathbf{q}'}{d\mathbf{q}}\,\,\frac {d\mathbf{q} }{dt} \,\,\frac {dt}{dt'} && \text{chain rule} \\ &= \alpha\,\frac {d\mathbf{q} }{dt}\,\frac {dt}{dt'} && (10) \\ &= \alpha\,\frac {d\mathbf{q} }{dt}\,\frac {1}{\beta} && (20)\\ &= \frac {\alpha}{\beta}\,\frac {d\mathbf{q} }{dt} && \text{commutation}\,. \end{align*} Thus, \begin{align*} &L_k{\left(\mathbf{q}'_1,\ldots, \mathbf{q}'_n, \frac{d\mathbf{q}'_1}{dt'},\ldots, \frac{d\mathbf{q}'_n}{dt'} ,t'\right)} \\ &\quad= L_k{\left(\alpha\,\mathbf{q}_1,\ldots, \alpha\,\mathbf{q}_n, \frac{\alpha}{\beta}\,\frac{d\mathbf{q}_1}{dt} ,\ldots, \frac{\alpha}{\beta}\,\frac{d\mathbf{q}_n}{dt} ,\beta\,t \right)} \end{align*} Yet, \begin{align*} &L_k{\left(\alpha\,\mathbf{q}_1,\ldots, \alpha\,\mathbf{q}_n, \frac{\alpha}{\beta}\,\frac{d\mathbf{q}_1}{dt} ,\ldots, \frac{\alpha}{\beta}\,\frac{d\mathbf{q}_n}{dt} ,\beta\,t \right)} \\ &\quad = T{\left( \frac{\alpha}{\beta}\,\frac{d\mathbf{q}_1}{dt} ,\ldots, \frac{\alpha}{\beta}\,\frac{d\mathbf{q}_n}{dt} \right)} - V_k{\left(\alpha\,\mathbf{q}_1,\ldots, \alpha\,\mathbf{q}_n \right)} &&\text{conservative system} \\ &\quad = \left(\frac{\alpha}{\beta}\right)^2T{\left( \frac{d\mathbf{q}_1}{dt} ,\ldots, \frac{d\mathbf{q}_n}{dt} \right)} - V_k{\left(\alpha\,\mathbf{q}_1,\ldots, \alpha\,\mathbf{q}_n \right)} &&\text{Example 3} \\ &\quad = \left(\frac{\alpha}{\beta}\right)^2T{\left( \frac{d\mathbf{q}_1}{dt} ,\ldots, \frac{d\mathbf{q}_n}{dt} \right)} - \alpha^k \, V_k{\left(\mathbf{q}_1,\ldots, \mathbf{q}_n \right)} &&\text{Example 4} \end{align*} Landau's premise is that if, and only if, $$\beta = \alpha^{1-\frac{k}{2}}$$, then we obtain a multiple of the Lagrangian by a constant. To see this, not that \begin{align*} &L_k{\left(\alpha\,\mathbf{q}_1,\ldots, \alpha\,\mathbf{q}_n, \frac{\alpha}{\beta}\,\frac{d\mathbf{q}_1}{dt} ,\ldots, \frac{\alpha}{\beta}\,\frac{d\mathbf{q}_n}{dt} ,\beta\,t \right)} \\ &\quad = \left(\frac{\alpha}{\alpha^{1-\frac{k}{2}}}\right)^2T{\left( \frac{d\mathbf{q}_1}{dt} ,\ldots, \frac{d\mathbf{q}_n}{dt} \right)} - \alpha^k \, V_k{\left(\mathbf{q}_1,\ldots, \mathbf{q}_n \right)} \\ &\quad = \alpha^k\left[ T{\left( \frac{d\mathbf{q}_1}{dt} ,\ldots, \frac{d\mathbf{q}_n}{dt} \right)} - V_k{\left(\mathbf{q}_1,\ldots, \mathbf{q}_n \right)}\right] \\ &\quad = \alpha^k\,L_k{\left(\mathbf{q}_1,\ldots, \mathbf{q}_n , \frac{d\mathbf{q}_1}{dt} ,\ldots, \frac{d\mathbf{q}_n}{dt},t \right)}\,. \end{align*}

From Proposition 1, since the transformations given in Eqs. (10) and (15) (in conjunction with the premise that necessitates that $$\beta = \alpha^{1-\frac{k}{2}}$$) produce a constant multiple of the Lagrangian, therefore these two transformations do not effect the equations of motion.

From this we can understand that the trajectories that satisfy the equations of motion generated by the Lagrangian will have the property of exact mechanical similarity [3]. Namely, the exact similitude is given by the rule in the equation $$\left| \overbrace{\mathbf{q}'_i}^{\alpha\,\mathbf{q}_i}\underbrace{\left( t'\right)}_{t\,\alpha^{\left[1-\frac{k}{2}\right]}} - \overbrace{ \mathbf{q}'_j}^{\alpha\,\mathbf{q}_j}{\underbrace{\left(t'\right)}_{t\,\alpha^{\left[1-\frac{k}{2}\right]} }} \right| = \alpha \left| \mathbf{q}_i{\left( t\right)} - \mathbf{q}_j{\left(t\right)} \right|\,.$$ This rule is valid for any $$i,j=1,\ldots,n$$ and for any time $$t$$.

Q.E.D.

Bibliography

[1] Landau, Volume 1, 3rd Edition, pp 22--23, 51.

[2] Wikipedia contributors. Mechanical similarity [Internet]. Wikipedia, The Free Encyclopedia; 2021 May 24, 09:26 UTC [cited 2022 Dec 12]. Available from: https://en.wikipedia.org/w/index.php?title=Mechanical_similarity&oldid=1024830575.