1. Jan 20, 2008

### lugita15

According to Noether's Theorem, for every symmetry of the Lagrangian there is a corresponding conservation law, and vice versa. For instance, the invariance of the Lagrangian under time translation and space translation correspond to the conservation laws of energy and momentum, respectively. Also, the invariance of the Lagrangian under rotation in space corresponds to the law of conservation of angular momentum.

In classical mechanics at least, there is also a law of conservation of (inertial) mass. What is the corresponding symmetry of the Lagrangian?

2. Jan 20, 2008

### tim_lou

I think that the "usual" Lagrangian is defined to take care of discrete point particles only, and that the mass of a particle is defined to be constant. So, it doesn't make too much sense talking about a symmetry behind such conservation. One might say that what about in certain rocket problems, where the kinetic energy of a rocket is
$$T=\frac{1}{2}M(t)\dot x^2$$

but one could equally well assert that the mass itself is constant, and take
$$T=\frac{1}{2}m\dot x^2$$
$$V\rightarrow V-\frac{1}{2}M(t)\dot x^2 + \frac{1}{2}m\dot x^2$$

However, in field theory, one has well defined differential equations for the density and current of a field, and the conservation of mass (locally) would corresponds to a verification of an equation in the form of
$$\frac{\partial \rho}{\partial t} + \nabla \cdot \mathbf{j}=0$$

which, I believe, is related to a certain gauge invariance.

Last edited: Jan 20, 2008
3. Jan 20, 2008

### jdg812

In non-relativistic mechanics, Lagrangian depends on variables (speeds) and parameters (masses). Variables are governed by equations of dynamics. Parameters are actually governed by you. You choose the parameters to be constants by definition. So no symmetry is necessary for conservation of mass.

In relativistic mechanics total mass M becomes variable. Because of the equivalence E = M*c^2, we may say that conservation of relativistic mass is linked, like energy, to time translation.

In field theories everything depends on combination of fields, which is associated with mass.

Last edited: Jan 20, 2008
4. Jan 20, 2008

### lugita15

I disagree. I base my arguments on Landau's "Mechanics," volume 1 of his "Course of Theoretical Physics." Landau bases classical mechanics on three axioms:
(1) The Newton-Laplace Principle of Determinacy
(2) Hamilton's Principle of Least Action
(3) Galileo's Principle of Relativity
Based on these three axioms, he proves that the Lagrangian of a free particle in an inertial frame is proportional to the square of the magnitude of its velocity. He then defines the mass of a particle to be 2 times the constant of proportionality. He does not in any way define the mass to be constant. That is why I would like to know how the law of conservation of mass is constant.
Any further help would be greatly appreciated.

Last edited: Jan 20, 2008
5. Jan 20, 2008

### jdg812

:rofl:

Last edited: Jan 20, 2008
6. Jan 20, 2008

### tim_lou

Unfortunately, I do not have Landau's book. It would help to pinpoint the essentials of his proof.

And, in the viewpoint you described, something needs clarification:
1. what does Landau mean by a free particle?
2. how would one define the mass of a particle that is not free?

although 1. seems trivial, it's definition truely depends on how one develops the theory. For instance, I may accept the fact that L=T-V (derived from newton's law which asserts that the mass of a fundamental particle is constant, and T by definition is Σ1/2mv^2, where m is a fundamental constant for each particle) and say a free particle is one who's Lagrangian is T, i.e. V=0. In this sense, the mass of a particle is by definition constant (which inherently comes from the fact T is 1/2 mv^2 where m is constant for a particle and L=T-V).

(Landau probably takes the following approach instead)
or one may define a free particle as one that moves in a straight line, in this sense, one does not konw that L=T-V, one merely has the Euler Lagrangian equations and one finds an L such that the motion describes a straight line (it turns out that L is proportional to v^2). In this sense then, V is not well defined, and I am not sure how one can define the mass of a non-particle (one that does not move in a straight line).

Last edited: Jan 20, 2008
7. Jan 20, 2008

### jdg812

Last edited by a moderator: Apr 23, 2017
8. Jan 20, 2008

### lugita15

Use the link jdg812 provided for a good proof.
Landau defines an inertial frame to be a frame of reference "in which space is homogeneous and isotropic and time is homogeneous." A particle is defined to be executing free motion in such an inertial frame when and only when all the properties of its motion obey these symmetries. Using this definition of a free particle, we can draw many conclusions about the Lagrangian, L(x,v,t). Due to the homogeneity of space, the Lagrangian of a free particle annot explicitly depend on position. Due to the homogeneity of time, the Lagrangian of a free particle cannot explicitly depend on time. Therefore, the Lagrangian of a free particle can only depend on its velocity. Due to the isotropy of space, the Lagrangian of a free particle cannot depend on the direction of its velocity. Therefore, the Lagrangian of a free particle can only depend on the magnitude of its velocity. Then, using Galileo's principle of relativity, Landau proves that the Lagrangian of a free particle in an inertial frame can only be proportional to the square of the magnitude of its velocity.

A single particle, by itself, can only execute free motion, because there are no other particles in the universe which can possibly exert a force upon it. So now, consider a system of particles. Then the kinetic energy of a particle in that system is defined as the Lagrangian it would have if there were no other particles in the universe. The kinetic energy of the system is defined as the sum of the kinetic energies of the individual system. The potential function of a system is then defined as the kinetic energy of the system minus the Lagrangian of the system. As long as we're not talking about dissipative systems with friction and heat, we may make the additional assumption that the potential function of the system at a given time depends only on the positions of the particles in that system.

9. Jan 20, 2008

### lugita15

What I meant to say is that the mass is defined to be 2 times the ratio of the Lagrangian to the square of the magnitude of the velocity. The term "constant of proportionality" is just a commonly used phrase, and in that context the word "constant" does not refer to a quantity the time derivative of which is identically zero. However, when you say that the mass is defined to be a constant, what you mean is that the time derivative of the mass is defined to be zero, I don't believe is necessarily true.

Last edited: Jan 20, 2008
10. Jan 20, 2008

### jostpuur

If I consider the mass to be a variable in the usual Lagrange's function, like this

$$L(t,x,\dot{x},m,\dot{m})=\frac{1}{2}m\dot{x}^2 - V(x)$$

I immediately get something crazy out. The EOM for x is

$$m\ddot{x} = -D_xV(x) - \dot{m}\dot{x}$$

and for m it is

$$\dot{x} = 0$$

I would try to figure out a way to modify the Lagrange's function so, that we would still be getting something reasonable out, even when having the mass as a dynamic variable. Anyone having ideas about how that could be possible?

The reason why I'm asking this, that I've lately seen all kinds of tricks with different constraints and auxiliary variables, and I don't know them very well yet.

Last edited: Jan 20, 2008
11. Jan 20, 2008

### gel

Well, as the action should be minimised, you should get $m=-\infty$.

12. Jan 20, 2008

### gel

...and, if m is allowed to vary, then the EOM for x is
$$\frac{d}{dt}(m\dot{x})=-D_xV$$
But I can't think of how you would adjust the Lagrangian to treat m as a dynamical variable and get $\dot{m}=0$ for its EOM.

13. Jan 20, 2008

### jostpuur

Good point! Since my edit button has not expired, I think I'm correcting my mistake now

14. Jan 20, 2008

### jostpuur

In fact the Euler-Lagrange equations will extremize the action, so it might as well be $m=\pm\infty$, or the path between some fixed points would become

$$m(t)=\left\{\begin{array}{l} m(t_0) = m_0\\ m(t) = \pm\infty\quad\quad\quad t_0<t<t_1 \\ m(t_1) = m_0\\ \end{array}\right.$$

hmhmhmh.... this is very hypothetical anyway.... Doesn't really matter what the m(t) should be. The truth is, that the action cannot be extremized by any well defined path, unless $\dot{x}=0$.

Last edited: Jan 20, 2008
15. Jan 20, 2008

### jostpuur

$\dot{m}=0$ would follow, if we defined a third variable $\lambda$, and then added a term $\lambda\dot{m}$ into the Lagrange's function, and no $\dot{\lambda}$ at all. The EOM of lambda would result in conservation of mass. The EOM of the m itself would still be a problem, however.

16. Jan 20, 2008

### lightarrow

Sorry if I say a naive thing, but since for a free particle E = p^2/2m so m = p^2/2E, mass conservation shouldn't come from simultaneous energy and momentum conservation?

17. Jan 20, 2008

### jdg812

Men, what are you doing? :rofl: The extremum in Lagrange method must be LOCAL There is no local extremum at $m=\pm\infty$

18. Jan 20, 2008

### gel

Seems like a reasonable thing to do to me, it just shows that the Langrangian is bad.

19. Jan 20, 2008

### jdg812

20. Jan 20, 2008

### tim_lou

thank you for the clarification. since m cannot depend on x, v or t, one conclude that m is indeed a constant in the case of a free particle. So, for an object, we may define the mass as the mass that one would obtain via the above fashion (separating that object far away from other objects). So, that mass by definition is a fundamental constant. In classical mechanics, one assumes from the existence of a trajectory that particles do not vanish. Since particles in a system do not vanish, the system's mass is always constant. Unless you have another definition of mass in mind that could be varied from situation to situation, it seems clear to me that this is a fundamental postulates of classical theory (ie each particle has a constant mass, and that particles do not vanish.)

indeed, as jdg812 says, unless you can associate a "dynamical" meaning to mass (putting m as a function of the coordinates, the velocity and time), mass is a parameter that we control. Also, noether theorem only deals with dynamical variables (or constants). In field theory, the density ρ(φ,∂t φ,∇φ,t) and current j(φ,∂t φ,∇φ,t) are well defined functions of the field, φ and it's derivatives. So it makes sense to talk about conservation of these quantities, or quantities of the form
ρ(φ,∂t φ,∇φ,t)+∫∇·j(φ,∂t φ,∇φ,t) dt (which is a well defined function of (φ,∂t φ,∇φ,t)), whose conservation would imply local conservation of mass. However, in the discrete case, the mass density is in the form of ∑mδ(r-r'), where r' is the position of the particle, you could say that ∫mδ(r-r') dV is conserved but this is trivially true since m itself is defined to be a constant.

(fixed some typo...edit: i mean φ,∂t φ,∇φ,t, not x,y,z,t, since the explicit form of ρ and j is unknown)

Last edited: Jan 20, 2008