# I Positive Mass in the Lagrangian from Landau

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1. Feb 15, 2016

### Oppie

Hello,

I'm sure most of you are already familiar with the book "Mechanics" by Landau and Lifshitz. There's a section that I do not understand.

In section 4 towards the end they mentions that "It is easy to see that the mass of the particle cannot be negative." They then give the argument that the action integral will not have a minimum if the mass is negative as can be seen if the particle leaves point 1 rapidly and approaches point 2 rapidly.

Can someone please elaborate on this?

2. Feb 21, 2016

### Greg Bernhardt

Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?

3. Feb 29, 2016

### Oppie

Thanks for the bump.

Unfortunately, I am still at a loss.

Is it as simple as seeing that v^2 is always positive and therefore if m<0 then if v at the initial and terminal points are large then you cannot have a minimum for the action integral since it can be arbitrarily negative?

Thanks.

4. Mar 1, 2016

### vanhees71

Let's calculate the 2nd variation of the action
$$A[x]=\int \mathrm{d}t \frac{m}{2} \dot{x}^2.$$
To that end we define
$$A[x,\eta]=A[x+\eta \delta x]=\int \mathrm{d} t \frac{m}{2} (\dot{x}+\eta \delta \dot{x})^2$$
and take the 1st and 2nd derivative wrt. $\eta$:
$$\frac{\mathrm{d}}{\mathrm{d} \eta} A[x,\eta]=\int \mathrm{d} t m (\dot{x}+\eta \delta \dot{x})\delta \dot{x},$$
$$\frac{\mathrm{d}^2}{\mathrm{d} \eta^2} A[x,\eta]=\int \mathrm{d} t m \eta \delta \dot{x}^2,$$
Now you want for $\eta=0$ the action to be stationary, i.e., that $x(t)$ is the trajectory of the particle according to the Hamilton principle of least action. That means
$$\frac{\mathrm{d}}{\mathrm{d} \eta} A[x,\eta]|_{\eta=0}=\int \mathrm{d} t m \dot{x} \delta \dot{x} = -\int \mathrm{d} t m \ddot{x} \delta x \stackrel{!}{=}0 \; \Rightarrow \; m \ddot{x}=0.$$
Now the second derivative is always positive if and only if $m>0$ and thus the stationary point is indeed a minimum if and only if $m>0$. Thus $m$ should be positive.

5. Mar 7, 2016

### Oppie

Thank you. I understand it now mathematically, if that makes sense.

My question now is, why require to action to be minimum? Landau and Lifshitz themselves note in a footnote that the action only needs to be stationary but now it seems that it has to be minimum for the mass to be positive.

6. Mar 9, 2016

### Jano L.

Although it is not very important, when equations of motions are derived from the Hamilton principle (often incorrectly called principle of least action), most often the action calculated for the actual trajectory beginning in $t_1,q_1$ and ending in $t_2,q_2$ is a local minimum - all nearby trajectories obeying that constraint give higher action. There are exceptions, like trajectory of a billiard ball inside elliptic pool table, when going from one focus to another in given time through encounter with the wall; due to elliptic shape of the wall, all points of the wall lead the ball to the focus in the same time, so there is no single trajectory of least action.

In Landau, I think it is a circular reasoning, they claim mass needs to be positive based on the frequency of common situation that action has a minimum, and elsewhere they will say action has to be minimum for infinitesimally long trajectories because that gives the right equations for positive mass particle.

7. Mar 9, 2016

### Oppie

Thanks for the response. I tried looking at other books, etc. to see if I can find a better explanation on this, unfortunately there isn't, at least at all the places I searched.

Also, to add, Whittaker in his Analytical Dynamics gives a proof that the action is a minimum but this requires previous knowledge of the form of the kinetic energy. Anyway, I thought it was interesting and which in turn drove me nuts trying to reason out that the mass should be positive without knowing the form of the kinetic energy beforehand.