# Sum of two gauge dependents being gauge independent

I don't know much about Gauge theory but from some bits and pieces I know about it, the following problem seems to be related to it.

Consider the classical Lagrangian $$\mathcal{L}=T-U$$ Where $$T$$ is kinetic energy and $$U$$ is potential energy.

Potential energy is defined via $$\vec F=-\vec \nabla U \tag{1}$$

This comes from the fact that for a consevative force $$\oint\vec F\cdot d\vec l=0\rightarrow \vec F=\vec\nabla V \tag{2}$$

Here the minus sign is conventional in equation $$1$$. In addition, we can introduce $$U'=U+c$$ for some constant $$c$$ and nothing will change about the physics in question.

Thus the freedom to choose the sign and value $$c$$ is gauge dependence for this problem.

Now consider total energy $$E=T+U$$ Adding this to Lagrangian we get $$\mathcal{L}+E=2T$$ which is gauge-independent.

Thus when we add two gauge dependents we are getting a gauge independent.

I was wondering if there is any mathematical significance in this result?

• there's nothing that guarantees that the sum of two non-invariant things is also non-invariant. The opposite is true of course: if $A$ and $B$ are invariant, then $A+B$ must be also. It's similar to: If $v$ and $w$ are two nonzero numbers or vectors, then the sum $v+w$ need not be nonzero. Oct 19, 2023 at 23:15