2 Different Ways to Write Potential Energy of a Pendulum

[zachx]

I have been doing pendulum problems lately, and I have found 2 different formulations for potential energy of a pendulum.

U=mgl(1-cos(Θ)) and U=-mglcos(Θ)

The first says U=0 when Θ=0 (at the bottom). The second has U=0 when Θ=π/2 (halfway to the top).

Both give the same equation of motion, but the Lagrangians are different.

Which is better/conventional?

 

[Dale]

Either way is fine. I would tend to use the second just because it is easier to write. Nobody would object to you using the other one if you prefer.

 

[vanhees71]

Just another remark. The Lagrangian is not a physical observable but it's used to define the action for Hamilton's action principle. Thus two Lagrangians $L(q,\dot{q},t)$ and $L'(q,\dot{q},t)$ are completely equivalent if
$$L'(q,\dot{q},t)=L(q,\dot{q},t)+\frac{\mathrm{d}}{\mathrm{d} t} \Omega(q,t)=L(q,\dot{q},t)+\dot{q} \cdot \vec{\nabla}_q \Omega(q,t) + \partial_t \Omega(q,t),$$
because adding such a term leaves the variation of the action and thus the Euler-Lagrange equations invariant. This is a very important concept for the derivation of Noether's theorem.