Noethers theorem, transformations of the Lagrange density

[PreposterousUniverse]

TL;DR Summary

Noethers theorem, confusion about transformation in the lagrange density

I'm so confused here. If we make the transformation of the coordinates x -> x', are we not suppose to consider the transformation of the coordinates only
$$ \phi(x) \rightarrow \phi(x') $$ ? Then why are they writing $$ \phi(x) \rightarrow \phi'(x') $$ ? If $$ \phi(x) $$ is a scalar function then by definition $$ \phi'(x') = \phi(x) $$ for any transformation. Then we have by definition no change in the fields and therefore no change in the lagrangian. So I don't understand why they put a prime on the fields here.

 

[Orodruin]

While $\phi$ is a scalar field, the relevant transformations of the Lagrangian may involve additional field transformations apart from the coordinate transformations. These are not covered by the coordinate transformation properties of the field itself.

One of the more down-to-Earth examples: Take an infinite string vibrating in two dimensions. The relevant fields $u_1$ and $u_2$ being the orthogonal displacements of the string away from the equilibrium and the base space is one-dimensional described by a single coordinate $x$. While you have translation symmetry $x \to x’ = x+s$, you will also have some field transformation symmetries such as
$$
u_1 \to u_1’ = u_1\cos(s) + u_2 \sin(s),\qquad
u_2 \to u_2’ = -u_1\sin(s) + u_2 \cos(s)
$$

 

[PreposterousUniverse]

While $\phi$ is a scalar field, the relevant transformations of the Lagrangian may involve additional field transformations apart from the coordinate transformations. These are not covered by the coordinate transformation properties of the field itself.

One of the more down-to-Earth examples: Take an infinite string vibrating in two dimensions. The relevant fields $u_1$ and $u_2$ being the orthogonal displacements of the string away from the equilibrium and the base space is one-dimensional described by a single coordinate $x$. While you have translation symmetry $x \to x’ = x+s$, you will also have some field transformation symmetries such as
$$
u_1 \to u_1’ = u_1\cos(s) + u_2 \sin(s),\qquad
u_2 \to u_2’ = -u_1\sin(s) + u_2 \cos(s)
$$

Not sure I understood really. But what are you actually doing when you make a coordinate transformation of the fields $$ \phi(x) $$ in general in the context of Noethers theorem? What I imagine is you just take your field evaluated at x on the manifold and you evaluate the same field it at some other point x'. Where x and x' are related by x' = f(x). If the scalar field take the same value at x and x', then this transformation is a symmetry transformation.

 

[Orodruin]

Not sure I understood really. But what are you actually doing when you make a coordinate transformation of the fields $$ \phi(x) $$ in general in the context of Noethers theorem? What I imagine is you just take your field evaluated at x on the manifold and you evaluate the same field it at some other point x'. Where x and x' are related by x' = f(x). If the scalar field take the same value at x and x', then this transformation is a symmetry transformation.

What you describe is only a coordinate transformation without a field transformation. Noether’s theorem allow both or even a combination of the two.

 

[Orodruin]

If the scalar field take the same value at x and x', then this transformation is a symmetry transformation.

This is also incorrect. You are looking for symmetries of the action, not the field. In other words, transformations that leave the action invariant. It is not necessary to have the same field values at different points (or even well defined as at this point we are not considering a particular solution for the fields).

 

[PreposterousUniverse]

This is also incorrect. You are looking for symmetries of the action, not the field. In other words, transformations that leave the action invariant. It is not necessary to have the same field values at different points (or even well defined as at this point we are not considering a particular solution for the fields).

You are correct. What I imagined was to evaluate the action at the different points on the manifold. If the action remains invariant under translation along the direction described by x -> x' so that is takes the same value at those different points. Then this is a symmetry transformation. Is this correct?