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Lagrangian of a free particle

by evoluciona2
Tags: free, lagrangian, particle
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evoluciona2
#1
Mar2-10, 01:17 AM
P: 7
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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sweet springs
#2
Mar2-10, 06:24 AM
P: 449
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.
evoluciona2
#3
Mar2-10, 08:13 PM
P: 7
Quote Quote by sweet springs View Post
= 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


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