Sign of potential term in Lagrangian mechanics

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The discussion centers on the implications of choosing different reference points for potential energy in Lagrangian mechanics. It highlights that while one can add a constant to potential energy, one cannot arbitrarily change its sign, as this affects the equations of motion (EOM). The example of gravitational potential energy illustrates that potential energy must be defined consistently to maintain the correct relationship between kinetic and potential energy. The conversation emphasizes that the potential energy should increase with height or distance, regardless of the reference point. Ultimately, the critical takeaway is that the choice of zero for potential energy does not allow for sign changes that would alter the fundamental equations of motion.
Admiralibr123
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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?
 
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Calculate the equations of motion for T - V and T - V + a constant. Any difference?
 
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
 
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Admiralibr123 said:
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).
 
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