# On Landaus&Lifshitz's derivation of the lagrangian of a free particle [duplicate]

I'm reading the first pages of Landaus&Lifshitz's Mechanics tome. I'm looking for some clarification on the derivation of the Lagrange function for the mechanical system composed of a single free particle (in an inertial frame). $$\newcommand{\vec}[1]{\mathbf{#1}}$$

The book proceeds as follows. I'll write my doubts below each step.

1. Homogeneity of time and space and isotropy of space imply that the Lagrangian $$L$$ of our mechanical systems depends only on the square of the velocity $$\vec v$$ of the particle.

On reflection, I have faith that it must be so. By the way, a few more words on this would certainly be welcome. For example, how would you explain such a statement in a more mathematical way? (I know that some people asked this before on this site, but I have read their questions and looked at the answers they received, and I found them to be unclear.)

1. The authors consider now two inertial reference frames $$K$$ and $$K^\prime$$, and discuss how the Lagrangian could be affected by the corresponding change of coordinates. They denote with $$L^\prime$$ the Lagrange function that depends on the primed coordinates. Thanks to point 1) above, $$L^\prime$$ is a function only of the square of the velocity $$\vec v^\prime$$ of the particle with respect to $$K^\prime$$. In the following, $$\vec\epsilon$$ is the relative (infinitesimal[*]) the velocity of $$K^\prime$$ with respect to $$K$$, so that the velocity $$\vec v^\prime$$ of the particle with respect to $$K^\prime$$ is $$\vec v^\prime = \vec v + \epsilon$$, and $$v^{\prime2} = v^2 + 2(\vec v\cdot \vec\epsilon) + \epsilon^2$$ where $$a = \lVert \vec a\rVert$$ for every vector $$\vec a$$. They write

We have $$L^\prime = {\color{red}L}(v^{\prime 2}) = {\color{red}L}(v^2 + (\vec v\cdot \vec\epsilon) + \vec\epsilon^2)$$ and expanding this expression in powers of $$\epsilon$$ and neglecting terms above the first order, we obtain $$L^\prime(v^{\prime 2}) = L(v^2) + \frac{\partial L}{\partial v^2}2(\vec v\cdot \vec\epsilon)\text{.}$$

First of all, what is the string $$L^\prime = {\color{red}L}(v^{\prime 2}) = {\color{red}L}(v^2 + 2(\vec v\cdot \vec\epsilon) + \vec\epsilon^2)$$ supposed to mean? As I said above here $$L^\prime$$ should be the function that eats the primes coordinates and spits out the value of the Lagrange function of the system at that coordinates. If I'm not mistaken, there's a typo in the book and the formula above should read $$L^\prime(v^{\prime 2}) = L(\lVert \vec v^\prime - \epsilon \rVert)\text{.}$$

Another question: what does it mean to "expand this expression in powers of $$\vec \epsilon$$"? Is there any chance they mean like some sort of multivariable Taylor formula?

But let's go on.

1. A this point they write

The second term on the right of this equation is a total time derivative only if it is a linear function of the velocity $$\vec v$$. Hence $$\partial L/\partial v^2$$ is independent of the velocity, i.e, the Lagrangian is in this case proportional to the square of the velocity, and we write it as $$L = \frac12 mv^2\text{.}$$

I understand this sentence in the following way. We want the (coordinate representative) of the Lagrangian with respect to the primed coordinates to be equal to the (coordinate representative) of the Lagrangian with respect to the unprimed coordinates plus the total derivative with respect to time of a function that depends on position and time. In this way (one could check) the equation of motion for $$L$$ and for $$L^\prime$$ will be the same. But then such and so should happen.

Admittedly, it's not clear to me why

[The term] $$\partial L/\partial v^2$$ is independent of the velocity [and] the Lagrangian is in this case proportional to the square of the velocity.

[*] I'll just ignore the "infinitesimal" part because I don't know what it means. Can I do that every time I see things like "infinitesimal", "elementary", etc.? Forgive my naiveté: I got raised by mathematicians.

• Related: physics.stackexchange.com/q/23098/2451 , physics.stackexchange.com/q/128139/2451 and links therein. Commented May 19 at 18:12
• I read that file before posting, but I wouldn't say it answered my question. I'm still digesting the answers I've received here, but I would say that they do a better job of clarifying my doubts regarding pt. 2) in my question. Commented May 20 at 16:34
• Expecting Landau to have 'a few more words' on, well, anything, is misplaced. Commented May 20 at 16:54

In this answer I go specifically into the assertion that the Lagrangian for the free particle must be an expression that is quadratic in velocity.

The presentation in L&L isn't so much a derivation, but a consistency argument.

The demands are:
A quantity that is:
-independent of orientation
-consistent with galilean relativity

Momentum $$m\vec{v}$$ has the property that for any inertial coordinate system momentum is a conserved quantity (hence consistency with galilean relativity). But momentum is vectorial, of course, so it doesn't satisfy the independent-of-orientation demand.

For the purpose of addressing the implications of a quadratic expression for quantity of motion:

If we have two masses, $$m_1$$ and $$m_2$$, then the straightforward thing to do is to express their respective velocities with respect to some inertial coordinate system. However, it is helpful to use the Common Center of Mass (CCM) of the two masses.

Definitions:
$$V_r$$ $$\Rightarrow$$ the relative velocity between $$m_1$$ and $$m_2$$
$$V_c$$ $$\Rightarrow$$ the velocity of the CCM relative to some chosen origin

$$v_1$$ velocity of mass 1 relative to the CCM
$$v_2$$ velocity of mass 2 relative to the CCM

Then we have:

$$v_1 = V_r\frac{m_2}{m_1 + m_2} \tag{1}$$ $$v_2 = -V_r\frac{m_1}{m_1 + m_2} \tag{2}$$

Reminder: here $$v_1$$ and $$v_2$$ are defined as velocity relative to the Common Center of Mass.
Relative to the CCM: $$m_1v_1 = m_2v_2$$
Which rearranges to: $$\frac{m_1}{v_2} = \frac{m_2}{v_1}$$

This notation in term of velocity relative to the CCM embodies galilean relativity.

The notation (1) and (2) allows a way of evaluating what an quadratic expression for quantity of motion will do.

As quadratic expression for quantity of motion we take: $$mv^2$$

Then the combined quantity of motion that we attribute to $$m_1$$ and $$m_2$$ is as follows:

$$m_1\left(V_c + V_r\frac{m_2}{m_1 + m_2} \right)^2 + m_2\left(V_c - V_r\frac{m_1}{m_1 + m_2} \right)^2 \tag{3}$$

When you develop that expression quite few terms drop away against each other. The end result is as follows:

$$(m_1 + m_2) V_c^2 + \frac{m_1m_2}{m_1 + m_2} V_r^2 \tag{4}$$

Next we consider what (4) implies for the outcome of a perfectly elastic collision.

After the collision the minus sign is with the other velocity. The squaring eliminates that minus sign; other terms with the minus sign drop away, and the expression for the motion after the elastic collision comes out the same as (4)

This informs us that the quantity $$mv^2$$ has the following property: the total amount of quantity of motion $$mv^2$$ that we attribute is the same before and after an elastic collision, for any inertial coordinate system; consistency with galilean relativity.

Another way of looking at (4):
With a perfectly inelastic collision: for any inertial coordinate system the expresssion for the amount of quantity of motion $$mv^2$$ that is lost in an inelastic collision is the same: consistency with galilean relativity.

Discussion:

You wouldn't necessarily expect that an expression for quantity of motion that is quadratic is consistent with galilean relativity, but indeed we get that consistency.

It's quite a marvelous property: the squaring eliminates directional information, so the quantity is independent of orientation.

Now the question: the modern notion of quadratic quantity of motion, kinetic energy, is defined as $$\tfrac{1}{2}mv^2$$
Where does that factor 1/2 come from?

To motivate the introduction of the concept of kinetic energy we derive the work-energy theorem. If $$F=ma$$ is granted as axiom then the work-energy relation follows as theorem.

The starting point:

$$F = ma \tag{5}$$

The next step is to evaluate an integral: integrate both sides with respect to the position coordinate, integrating from starting point $$s_0$$ to final point $$s$$

$$\int_{s_0}^s F \ ds = \int_{s_0}^s ma \ ds \tag{6}$$

The work-energy theorem hinges on the fact that the two factors in the right hand side of (2), acceleration $$a$$ and position coordinate $$s$$, are not independent of each other; acceleration is the second derivative of position.

We proceed to develop the right hand side of (6).

In the steps starting with (9) the integrand $$a$$ is converted to velocity and the differential $$ds$$ is converted to $$dv$$, using the relations (7) and (8)

$$v = \frac{ds}{dt} \ \Leftrightarrow \ ds = v \ dt \tag{7}$$

$$a = \frac{dv}{dt} \ \Leftrightarrow \ dv = a \ dt \tag{8}$$

(9) corresponds to the right hand side of (6)

$$\int_{s_0}^s a \ ds \tag{9}$$ $$\int_{t_0}^t a \ v \ dt \tag{10}$$ $$\int_{t_0}^t v \ a \ dt \tag{11}$$ $$\int_{v_0}^v v \ dv \tag{12}$$

First (7) was used to change the differential from $$ds$$ to $$dt$$, with corresponding change of limits. Next (8) was used - with change of limits - to arrive at (12).

So we have the following mathematical relation:

$$\int_{s_0}^s a \ ds = \tfrac{1}{2}v^2 - \tfrac{1}{2}v_0^2 \tag{13}$$

Notice especially:
The definitions (7) and (8), in combined form, are sufficient to imply (13).
Stated differently: (13) expresses: whenever you have a quantity $$s$$, its first derivative, and its second derivative, an expression with the form of (13) is valid. That is: (13) is a mathematical property, not tied to any particular physics context.

We use (13) to go from (6) to the work-energy theorem:

$$\int_{s_0}^s F \ ds = \tfrac{1}{2}mv^2 - \tfrac{1}{2}mv_0^2 \tag{14}$$

When a quadratic expression has a factor of 1/2 in front of it: that factor 1/2 is a signature of integration.

To motivate the concept of kinetic energy I prefer the derivation of the work-energy theorem. The work-energy theorem acts as a hub, tying everything together.

Discussion:

It is not common for physiscs textbook authors to mention that the concept of kinetic energy slots in with galilean relativity. Some authors may not even be aware of the significance of that.

My guess is that Landau was prompted by an intuition that it is important and worthwhile to demonstrate that a quadratic expression for quantity of motion, $$mv^2$$, slots in with galilean relativity.

1. Homogeneity of time and space and isotropy of space imply that the Lagrangian $$L$$ of our mechanical systems depends only on the square of the velocity $$\vec v$$ of the particle.

On reflection, I have faith that it must be so. By the way, a few more words on this would certainly be welcome. For example, how would you explain such a statement in a more mathematical way? (I know that some people asked this before on this site, but I have read their questions and looked at the answers they received, and I found them to be unclear.)

The mathematical way is that there is a function $$L(s)$$ such that the Lagrangian function

$$l(\mathbf r,\mathbf v,t) = L(\mathbf v\cdot \mathbf v).$$ Thus time $$t$$ and coordinates $$\mathbf r$$ do not affect value of $$L$$.

This assumption is somewhat more than physics laws require, because, as L&L themselves point out, $$L$$ is arbitrary, as it can have any additive terms that are total time derivatives. Thus for $$l$$ obeying the above relation, we can also have $$l_2 = l + 2k\mathbf r\cdot \mathbf v$$, and that is perfectly good Lagrangian which does not obey the above condition. Thus their wording is a little inconsistent with the rest of the exposition. They assume $$l$$ is restricted to functions of $$v^2$$, when this is not required by any physics laws.

Another question: what does it mean to "expand this expression in powers of $$\vec \epsilon$$"? Is there any chance they mean like some sort of multivariable Taylor formula?

Correct, that is exactly what they mean, and write the multi-variable Taylor expansion using the Gibbs vector notation.

1. First of all, what is the string $$L^\prime = {\color{red}L}(v^{\prime 2}) = {\color{red}L}(v^2 + 2(\vec v\cdot \vec\epsilon) + \vec\epsilon^2)$$ supposed to mean? As I said above here $$L^\prime$$ should be the function that eats the primes coordinates and spits out the value of the Lagrange function of the system at that coordinates. If I'm not mistaken, there's a typo in the book and the formula above should read $$L^\prime(v^{\prime 2}) = L(\lVert \vec v^\prime - \epsilon \rVert)\text{.}$$

This is indeed confusing. They mean probably this: when we change the description using the variables $$\mathbf r,\mathbf v$$ and Lagrangian $$l(\mathbf r,\mathbf v)=L(v^2)$$ into a description using the variables $$\mathbf r',\mathbf v'$$ and Lagrangian $$m(\mathbf r',\mathbf v') = M(v'^2)$$ (appropriate to be used in the primed frame), the value $$M(v'^2)$$ can be expressed as

$$M(v'^2) = M(v^2 + 2\mathbf v \cdot \boldsymbol \epsilon + \epsilon^2).$$

Then the physics argument is this: since all inertial frames are equivalent, the functions $$L,M$$ could be chosen to be the same function, as there is nothing in the frame $$K'$$ pointing to a need to have a different function than the one used in frame $$K$$. When this is assumed, we have

$$L(v'^2) = L(v^2 + 2\mathbf v \cdot \boldsymbol \epsilon + \epsilon^2).$$

So this is why $$v'^2$$ appears as argument of $$L$$, although $$L$$ for its main use is not usually thought of as function of $$v'$$. It is just the result of the realization that we are allowed to use single function in all frames, and thus refer to it by the same letter $$L$$, even though in $$K'$$ its value for the primed coordinates is used, and in $$K$$ its value for the unprimed coordinates is used.

Now to find some restrictions on the function $$L$$, using the Taylor expansion, we get

$$L(v'^2) = L(v^2) + \frac{\partial L}{\partial v^2}[v^2]~ 2\mathbf v\cdot \boldsymbol\epsilon ~+~...$$ where dots mean further terms proportional to higher powers of $$\epsilon$$.

Then the argument is: if function $$L(s)$$ is such that in $$K$$ for Lagrangian $$l=L(v^2)$$ the action principle is obeyed: $$\delta \int l dt = 0,$$ then it will be also obeyed in $$K'$$ for $$l'=L(v'^2)$$ if the term $$G=\frac{\partial L}{\partial v^2}[v^2]~ 2\mathbf v\cdot \boldsymbol\epsilon$$ is a total time derivative $$dg/dt$$, because then variation of its time integral is zero: $$\delta \int G dt = \delta \int \frac{dg}{dt}dt = 0$$ and thus

$$\delta \int l' dt = 0$$ holds.

So the L&L (totally made up) condition that $$G$$ is a total time derivative is an interesting condition to analyse, and it can only be true if $$L(v^2) = Cv^2$$ for any $$C$$, which turns out to be the well-known standard Lagrangian for a free particle.

However, I don't think it has been shown that $$G$$ has to be a total time derivative in order to get a valid description of the actual motion. For $$L=Cv^2$$, $$G$$ is a total time derivative, hence L&L were able to reverse the argument. But on the other hand, let's check what would happen if we assumed $$L(v^2) = Dv^4$$ in all frames. The Euler-Lagrange equations of motion would admit two kinds of solution: either $$\mathbf v= const.$$, or $$\mathbf v=0$$. The first solution is correct in all frames, the second is correct only in one frame, so we have a defect of this Lagrangian - it allows, in all frames, for the particle to be at rest. That isn't consistent with the Galilei transformation, because particle can be at rest only in single reference frame, in others it has to move. But if we do not demand that all solutions of the Euler-Lagrange equations have to be realized, and will be content with $$L(v^2)$$ if the actual physical motion obeys the Euler-Lagrange equations derived from $$L$$, then the Lagrangian $$l= L(v^2) = Dv^4$$ is fine. The Euler-Lagrange equations in all frames actually give the same equations of motion, thus they have the same form, and they all admit the correct solution. So I suspect the L&L condition that $$G$$ be a total time derivative is not well motivated in their text, and it is not really necessarily obeyed by Lagrangians that admit correct physical solutions.

1. [...] Admittedly, it's not clear to me why [The term] $$\partial L/\partial v^2$$ is independent of the velocity [and] the Lagrangian is in this case proportional to the square of the velocity.

If it was a function of $$\mathbf v$$, then the whole term could not be a total time derivative of a function of position and time.

If you find the L&L argumentation/derivation suspect and weird, you're not alone. Essentially trivial result is being derived from hard-to-motivate mathematical requirements. But their argument does reach the correct result, so maybe there is something interesting to those hard-to-motivate requirements.

My impression is that this is all interesting viewpoint and methods, but not really a good way to teach basic facts from analytic mechanics, such as what Lagrangian function to use in Newtonian mechanics of particles (we know that from much clearer and well-motivated traditional arguments) as much as it is a simple application of the kinds of arguments one may use ( equivalence of reference frames, preserving the action principle) to invent new theories within the doctrine of the action principle.

I'll just ignore the "infinitesimal" part because I don't know what it means. Can I do that every time I see things like "infinitesimal", "elementary", etc.? Forgive my naiveté: I got raised by mathematicians.

It usually means $$\epsilon$$ is so small that only the first non-zero Taylor term is important for the argument and all the rest can be neglected and dropped.