[Physics] Non-uniqueness of the Lagrangian

classical-mechanicslagrangian-formalism

Goldstein, 3rd ed

$$\frac{d}{d t}\left(\frac{\partial L}{\partial \dot{q}_{j}}\right)-\frac{\partial L}{\partial q_{j}}=0\tag{1.57}$$
expressions referred to as "Lagrange's equations."

Note that for a particular set of equations of motion there is no unique choice of Lagrangian such that above equation lead to the equations of motion in the given generalized coordinates.

I'm not able to understand what does the highlighted statement mean. How can we have different Lagrangians. While deriving the above equation we went through the derivation and ended up defining $$L=T-V$$, so the Lagrangian is fixed and always $$L=T-V$$, so why talk about a different Lagrangian?

Here are the preceding steps of derivation

$$\frac{d}{d t}\left(\frac{\partial(T-V)}{\partial \dot{q}_{j}}\right)-\frac{\partial(T-V)}{\partial q_{j}}=0$$
Or, defining a new function, the Lagrangian $$L$$, as
$$L=T-V\tag{1.56}$$
the Eqs. (1.53) become
$$\frac{d}{d t}\left(\frac{\partial L}{\partial \dot{q}_{j}}\right)-\frac{\partial L}{\partial q_{j}}=0.\tag{1.57}$$

So we see that we have defined that $$L=T-V$$ so why talk of a different Lagrangian?

It may not be obvious that different Lagrangians can lead to the same equations of motion. Here is a simple example. Take theses two Lagrangians

$$L = \frac{m}{2}\dot{x}^2-\frac{k}{2}x ^2 \tag{1}$$ $$L' = \frac{m}{2}\dot{x}^2-\frac{k}{2}x ^2+ax\dot{x} \tag{2}$$ These two Lagrangians differ just by the extra term $$ax\dot{x}$$.

From the Lagrangian (1) you get the equation of motion $$\frac{d}{d t}\left(\frac{\partial L}{\partial \dot{x}}\right) = \frac{\partial L}{\partial x}$$ $$m\ddot{x}=-kx \tag{3}$$

From the Lagrangian (2) you get the equation of motion $$\frac{d}{d t}\left(\frac{\partial L'}{\partial \dot{x}}\right) = \frac{\partial L'}{\partial x}$$ $$m\ddot{x}+a\dot{x}=-kx+a\dot{x} \tag{4}$$ which is the same as (3).

The above was just an example. Actually you can add any function $$F$$ of the form $$F(x,\dot{x},t)=\frac{\partial G(x,t)}{\partial x}\dot{x}+\frac{\partial G(x,t)}{\partial t}$$ with an arbitrary function $$G(x,t)$$ to the Lagrangian, and the equation of motion will remain the same. This function $$F$$ can equivalently be written as a total time derivative $$F(x,\dot{x},t)=\frac{dG(x,t)}{dt}$$

By the way: The example above was constructed using $$G(x,t)=\frac{a}{2}x^2$$, thus giving $$F(x,\dot{x},t)=ax\dot{x}$$.