# Question on Lagrangian Mechanics

• kakarukeys
In summary, the Lagrangian can be transformed in a number of ways, each of which will lead to a Lagrangian that is perfectly satisfactory for describing the motion. Any of these transformations will result in an E-L equation that is invariant under time-transformations.
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
$$m\ddot{q} + c^3kt = 0$$

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

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marlon said:
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?

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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.

## 1. What is Lagrangian Mechanics?

Lagrangian Mechanics is a mathematical framework used to describe the motion of a mechanical system. It is based on the principle of least action, which states that a system will follow the path that minimizes the action, or the difference between a system's kinetic and potential energy. It is an alternative to Newtonian mechanics and is often used in advanced physics and engineering courses.

## 2. Who developed Lagrangian Mechanics?

Lagrangian Mechanics was developed by Italian mathematician and astronomer Joseph-Louis Lagrange in the late 1700s. He published his findings in the book "Mécanique Analytique" in 1811.

## 3. What is the Lagrangian function?

The Lagrangian function is a mathematical function that describes the dynamics of a system in terms of its position, velocity, and time. It is defined as the difference between the kinetic and potential energy of a system, and it is used to derive the equations of motion in Lagrangian Mechanics.

## 4. What is the difference between Lagrangian Mechanics and Hamiltonian Mechanics?

While both Lagrangian and Hamiltonian Mechanics are used to describe the motion of a mechanical system, they differ in their approach. Lagrangian Mechanics is based on the principle of least action, while Hamiltonian Mechanics is based on the principle of least potential energy. Additionally, Hamiltonian Mechanics uses a different set of equations to describe the motion of a system.

## 5. What are some real-life applications of Lagrangian Mechanics?

Lagrangian Mechanics has various applications in physics and engineering, including celestial mechanics, fluid dynamics, and the study of rigid bodies. It is also used in the development of control systems for robots and spacecrafts, as well as in the design of bridges and other structures.

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