# Question on Lagrangian Mechanics

1. Apr 7, 2005

### 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
$$m\ddot{q} + c^3kt = 0$$

2. Apr 8, 2005

### marlon

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

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

Last edited by a moderator: May 2, 2017
3. Apr 8, 2005

### dfan

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?

Last edited by a moderator: May 2, 2017
4. Apr 9, 2005

### dextercioby

Okay.It doesn't really matter.Rescaling

$$t'=:ct$$

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

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.