Question on Lagrangian Mechanics

[kakarukeys]

"There is some freedom as to what we choose for the Lagrangian in a given problem: We can add a constant, multiply by a constant, change the time scale by a multiplicative constant, or add the total time derivative ... Any of these transformations will lead to a Lagrangian that is perfectly satisfactory for describing the motion."

I could not verify the 3rd one. Why is it possible to change the time scale by a constant factor?

For example:
$$L = \frac{1}{2}m\dot{q}^2 - kqt$$

The E-L equation is
$$m\ddot{q} + kt = 0$$

If we modify the Lagrangian, multiplying all time by constant c
$$L = \frac{1}{2c^2}m\dot{q}^2 - ckqt$$

The E-L equation is
$$m\ddot{q} + c^3kt = 0$$

 

[marlon]

http://www.physik.fu-berlin.de/~kleinert/b6/psfiles/conslaw.pdf

I refer to section 6.2 on page 4 of the above link.

If the Lagrangian does NOT depend on time explicitely (which is NOT the case in your example), then this L will be invariant under time-transformations t-->t + e where e is a given constant. This symmetry yields the energy conservation law.

marlon

 

[dfan]

http://www.physik.fu-berlin.de/~kleinert/b6/psfiles/conslaw.pdf

If the Lagrangian does NOT depend on time explicitly (which is NOT the case in your example), then this L will be invariant under time-transformations t-->t + e where e is a given constant. This symmetry yields the energy conservation law.

Yeah, but that's translating time, and the excerpt he quoted was about scaling time.

Something's not quite right in kakarukeys' example, but I can't put my finger on it. Certainly it shouldn't matter whether you measure time in seconds or milliseconds. And in your $L = T - V$ example, it doesn't make any sense that $V$ became bigger but $T$ became way smaller (assuming $c > 1$) when you changed the time scale. They should stay at the same ratio.

OK, I think I know what the problem is. I think $k$ needs to be in units of mass * length / time^3 for the units to come out right. So when you did your scaling, you also needed to scale $k$, by dividing it by $c^3$. Then everything has $c^2$ on the bottom and everyone's happy again. Does that sound right?

 

[dextercioby]

Okay.It doesn't really matter.Rescaling

$$t'=:ct$$

(adimensional constant,dfan!).

$$L\left(q,\frac{dq}{dt'},t'\right)=\frac{m}{2}\left(\frac{dq}{dt'}\right)^{2}-kqt'$$

It can be easily proven that a time rescaling doesn't modify the Lagrange equation(s),so it looks (for this 1D-case)

$$\frac{\partial L}{\partial q}-\frac{d}{dt'}\left(\frac{\partial L}{\partial \left(\frac{dq}{dt'}\right)}\right) =0$$

and so,in the rescaled variables,the equation of motion is the same...

We all know that time inversion is a weird (for relativistic systems,classical & quantum fields) example of a time-rescaling.

Daniel.