# Sign of potential term in Lagrangian mechanics

I have heard many times that it does not matter where you put the zero to calculate the potential energy and then ##L=T-V##. But mostly what we are doing is taking potential energy negative like in an atom for electron or a mass in gravitational field and then effectively adding it to kinetic energy. What if I take zero at the centre of atom and on the surface of earth, write ##V## positive and subtract if in lagrangian. I try that and EOM is different something like $$F=ma=-mg$$ for a free fall object. If I change it for a charge an electron in atom something similar happens.

Please someone explain, does $$F=-mg \quad \textrm{vs} \quad F=+mg$$ not matter? This is just a single term but if the EOM have multiple terms, this sign would surely make a difference. And in essence to me, it feels like it voids the negative sign in ##L=T-V##, if we can choose any sign for V. That must definitely not be the case, so where am I wrong in my reasoning?

• Delta2

Staff Emeritus
Calculate the equations of motion for T - V and T - V + a constant. Any difference?

Mentor
You seem to have the mistaken idea that the arbitrariness of the zero means you can flip the sign of the potential. That is a mistake.

You can add a constant to the potential, you cannot multiply it by a constant. That includes multiplying it by -1

• vanhees71, Delta2, etotheipi and 1 other person
Homework Helper
Gold Member
2022 Award
That must definitely not be the case, so where am I wrong in my reasoning?
For GPE you have two conventions. If the object remains close to the surface, then you can write: $$V = mgh$$ where ##h## is the height above the surface.

And, in the general case we write $$V = -\frac{GMm}{r}$$.
The critical point, however, is that in both cases ##V## increases as ##r## or ##h## increases.

As a useful exercise, you could show that for ##r = R + h##, where ##h << R## we have:
$$V = -\frac{GMm}{r} = -\frac{GMm}{R + h} \approx V_0 + mgh$$ where ##V_0## is constant. This shows that the two conventions are approximately equivalent up to a constant term (in the case where the object remains close to the surface).

• vanhees71, Dale and etotheipi