Lagrangian of a free particle

In summary: Hello,You are asking about the statement "L and L' differing by a time derivative of some f(q,t) (q is a vector of generalized coordinates) does not change the solutions of the equations of motion, but the other way around." According to the statement, if an inertial frame is moving with a small velocity relative to another inertial frame, the Lagrangian will not change even though the equations of motion will. This is because the time derivative of the function is only dependent on the first order term.
  • #1
Hello,

I'm trying to follow an argument in Landau's Mechanics. The argument concerns finding the Lagrangian of a free particle moving with velocity v relative to an inertial frame K. (of course L=1/2 mv^2, which is what we have to find). I'll state the points of the argument:

(0) It has already been argued that the Lagrangian relative to an intertial frame K must be of the form L(v^2) (space is homogeneous and iostropic).

(1) If an inertial frame K is moving with infinitesimal velocity e relative to another inertial frame K', the Lagrangian L' must be of the same form because the equations of motion are unchanged under Galilean transformations.

(2) So the Lagrangian L' wrt K' must differ by L by at most a time derivative of some f(q,t).

(3) L' = L(v'^2) = L(v^2 + 2v*e + e^2) which is to first-order [tex]L(v'^2) = L(v^2) + (\partial L/\partial v^2) 2 v\cdot e[/tex]

(4) The second term in the last equation is a total time derivative only if it is a linear function of the velocity v. Hence [tex]\partial L/\partial v^2[/tex] is independent of the velocity. I.e. the Lagrangian is proportional to the square of the velocity.

I'm having trouble with (2) and (4).

Specifically, my question for (2) is that it has been proven L and L' differing by a time derivative of some f(q,t) (q is a vector of generalized coordinates) does not change the solutions of the equations of motion, but the other way around. Thus 'must differ' in (2) isn't true. I guess 'allowed to differ' is more correct.

My question for (4) is that I don't get it. :)

Thanks
-evoluciona
 
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  • #2
Hi.
I am not good at English, so I cannot distinguish the meanings of "must differ by at most" and "allowed to differ".

If [tex]
(\partial L/\partial v^2)
[/tex]

= a0 + a1v + a2v^2 + ...

[tex]
(\partial L/\partial v^2) 2 v\cdot e
[/tex]

= 2a0ve + 2a1v^2e + 2a2v^3 e+ ...

= 2a0 dr/dt e + 2a1 (dr/dt)^2 e + 2a2 (dr/dt)^3 e + ...

Only the first term is the time derivative of the function, say f=2a0re.
Regards.
 
  • #3
sweet springs said:
= 2a0 dr/dt e + 2a1 (dr/dt)^2 e + 2a2 (dr/dt)^3 e + ...

Only the first term is the time derivative of the function, say f=2a0re.
Regards.

I see. df(x,t)/dt = df/dx v + df/dt is a linear function in v so the higher order terms are eliminated.

Thank you!

-evoluciona
 

What is the Lagrangian of a free particle?

The Lagrangian of a free particle is a mathematical function that describes the kinetic and potential energy of a particle in a given system. It is used in classical mechanics to derive the equations of motion for a particle.

How is the Lagrangian of a free particle different from the Hamiltonian?

The Lagrangian and Hamiltonian are two different mathematical formulations used to describe the dynamics of a system. The Lagrangian is a function of the particle's position and velocity, while the Hamiltonian is a function of the position and momentum. They both describe the same physical system, but the equations derived from them have different forms.

What are the advantages of using the Lagrangian over the Newtonian approach?

The Lagrangian approach has several advantages over the Newtonian approach. It is more general and can be applied to systems with more complex geometries. It also simplifies the equations of motion, making it easier to solve for the particle's trajectory. Additionally, the Lagrangian is independent of the choice of coordinate system, making it more versatile.

What is the principle of least action and how is it related to the Lagrangian?

The principle of least action states that the path a particle takes between two points is the one that minimizes the action, which is the integral of the Lagrangian along the path. This means that the particle follows the path of least resistance, making it the most natural path for the particle to take. The Lagrangian is used to calculate the action and derive the equations of motion for the system.

Can the Lagrangian be used in quantum mechanics?

Yes, the Lagrangian can also be used in quantum mechanics to describe the dynamics of a system. However, it is modified to include terms for the particle's wave function and the potential energy. This allows for a more accurate description of the particle's behavior at the quantum level.

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