Understanding the Lagrangian of a Free Particle?

Click For Summary
SUMMARY

The discussion centers on deriving the Lagrangian for a free particle, specifically L = 1/2 mv^2, as presented in Landau's Mechanics. Key points include the assertion that the Lagrangian must be of the form L(v^2) due to the homogeneity and isotropy of space. The conversation highlights the relationship between L and L' under Galilean transformations, emphasizing that L' can differ from L by at most a total time derivative of a function f(q,t). The conclusion drawn is that the term ∂L/∂v^2 must be independent of velocity, confirming that the Lagrangian is indeed proportional to the square of the velocity.

PREREQUISITES
  • Understanding of Lagrangian mechanics
  • Familiarity with Galilean transformations
  • Knowledge of total time derivatives in calculus
  • Basic concepts of classical mechanics and motion
NEXT STEPS
  • Study the implications of Galilean invariance in classical mechanics
  • Explore the derivation of Lagrangians for various physical systems
  • Investigate the role of total time derivatives in Lagrangian formulations
  • Learn about the relationship between velocity and Lagrangian dynamics
USEFUL FOR

Students and professionals in physics, particularly those focusing on classical mechanics, Lagrangian dynamics, and anyone seeking to deepen their understanding of motion and forces in inertial frames.

evoluciona2
Messages
7
Reaction score
0
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 L(v'^2) = L(v^2) + (\partial L/\partial v^2) 2 v\cdot e

(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 \partial L/\partial v^2 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
 
Physics news on Phys.org
Hi.
I am not good at English, so I cannot distinguish the meanings of "must differ by at most" and "allowed to differ".

If <br /> (\partial L/\partial v^2) <br />

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

<br /> (\partial L/\partial v^2) 2 v\cdot e<br />

= 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.
 
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
 

Similar threads

  • · Replies 0 ·
Replies
0
Views
1K
  • · Replies 7 ·
Replies
7
Views
4K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 48 ·
2
Replies
48
Views
3K
  • · Replies 49 ·
2
Replies
49
Views
7K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
6
Views
2K