# Scalar field lagrangian and potential

This question is a continuation of this Phys.SE post.

Scalar field theory does not have gauge symmetry, and in particular, $\phi\to\phi−1$ is not a gauge transformation. but why? and

I want see the mathematics that will represent the interaction potential from the Lagrangian for scalar field.

Details in the paper: see check the equation (3) please.

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## 1 Answer

Because there is no physical redundancy in the description. A gauge theory (and related gauge transformations) only occurs if there are different field configurations corresponding to a certain physical configuration. Such redundant configurations can be related by local gauge transformations (see for example the vector potential $A_\mu$ in electrodynamics). In your case (which the other question referred to), the Lagrangian is not invariant under your transformation, since the quantity $$U(\phi)= \frac{1}{8} \phi^2 (\phi -2)^2$$ changes under $\phi\rightarrow\phi-1$ to

$$U(\phi-1)= \frac{1}{8} (\phi-1)^2 (\phi -3)^2.$$

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Where from the equation $U(\phi)= \frac{1}{8} \phi^2 (\phi -2)^2$ arises or more precisely why did we write in this form? –  Unlimited Dreamer Feb 14 '13 at 11:51
That's the interaction potential you wrote down in your other question: physics.stackexchange.com/questions/52590/… –  Frederic Brünner Feb 14 '13 at 11:51
My curiosity is why did he write the equation like this? –  Unlimited Dreamer Feb 14 '13 at 11:52
Such a term (i.e. one of a higher order than 2) is necessary in order to get interactions in your theory. –  Frederic Brünner Feb 14 '13 at 12:01
arxiv.org/abs/0802.3525 in this article, check the equation(3) please. –  Unlimited Dreamer Feb 14 '13 at 12:03