Linked Questions

8
votes
3answers
995 views

What is the actual form of Noether current in field theory?

Let us consider $N$ independent scalar fields which satisfy the Euler-Lagrange equations of motion and are denoted by $\phi^{(i)}(x) \ ( i = 1,...,N)$, and are extended in a region $\Omega$ in a $D$-...
12
votes
2answers
987 views

Why do the Lagrangian and Hamiltonian formulations give the same conserved quantities for the same symmetries?

The connection between symmetries and conservation laws can be viewed through the lens of both Lagrangian and Hamiltonian mechanics. In the Lagrangian picture we have Noether's theorem. In the ...
7
votes
2answers
944 views

Invariance of Action vs. Lagrangian in Noether's theorem?

I have recently started studying classical field theory. Noether's theorem states that every differentiable symmetry of the action of a physical system has a corresponding conservation law. But I find ...
8
votes
3answers
2k views

What symmetry gives you charge conservation?

This is a popular question on this site but I haven't found the answer I'm looking for in other questions. It is often stated that charge conservation in electromagnetism is a consequence of local ...
8
votes
2answers
3k views

Noether current for a local gauge transformation for the Klein-Gordon Lagrangian

The Noether current corresponding to the transformation $\phi \to e^{i\alpha} \phi$ for the Klein-Gordon Lagrangian density $$\mathcal{L}~=~|\partial_{\mu}\phi|^2 -m^2 |\phi|^2$$ by finding $\...
10
votes
2answers
636 views

What does it mean for an action to be invariant under $x \to x'$, $\phi \to \phi'$?

I'm suddenly getting very confused about a basic question. Suppose somebody tells you that the action is invariant under the transformation $$x \to x', \quad \phi(x) \to \phi'(x').$$ I realize this ...
8
votes
2answers
309 views

Why are these two definitions for symmetries in the Lagrangian equivalent?

I have heard the following two definitions for a symmetry of the Lagrangian: If under a coordinate transformation the form of the Lagrangian remains unchanged then there is a symmetry. If $\delta \...
6
votes
1answer
1k views

Conservation of BRST current in QED

I am trying to understand the conservation of the BRST current in QED but am having some trouble. This is what I have so far, QED lagrangian density in Lorenz gauge is, $$L = \frac{1}{4}F_{\mu\nu}F^{\...
5
votes
3answers
229 views

What does a symmetry that changes the Lagrangian by a total derivative do to the Hamiltonian $H$?

A tiny symmetry transformation may change the Lagrangian $L$ by a total time derivative of some function $f$. This is a basic fact used in the proof of Noether's theorem. How can we see the effect of ...
8
votes
2answers
373 views

Why are symmetries in phase space generated by functions that leave the Hamiltonian invariant?

Hamilton's equation reads $$ \frac{d}{dt} F = \{ F,H\} \, .$$ In words this means that $H$ acts on $T$ via the natural phase space product (the Poisson bracket) and the result is the correct time ...
5
votes
3answers
216 views

In what sense are the equations of motion conserved by symmetries?

I am studying variational principles and I have been reading this set of notes by Townsend. In the first paragraph of Section 9, Townsend defines what it means for a transformation to be a symmetry of ...
5
votes
1answer
842 views

Big puzzle about Noether's theorem of coordinate transformation (spacetime symmetry)

For Noether theorem with only internal symmetry, I've found there has been a very clear proof. But I still struggle with the proof of coordinate transformation. Because there are so many different ...
2
votes
2answers
1k views

Detailed conditions for symmetries of Lagrangian

Edit: To clarify the question, I am asking why we are justified in calling a continuous symmetry a symmetry of a system when it changes the Lagrangian by a total derivative of a function of $t, q(t)$ ...
3
votes
1answer
627 views

Noether's Current in QFT with position dependent variations?

Setup Consider a mapping $F$ that takes every point $x$ on the manifold $M$ to the point $x'$ on the same manifold. Under this mapping the field $\phi(x)$ evaluated at the point $x$ changes to $\phi'(...
4
votes
1answer
229 views

Constrained Hamiltonian systems: spin 1/2 particle

I am trying to apply the Constrained Hamiltonian Systems theory on relativistic particles. For what concerns the scalar particle there is no issue. Indeed, I have the action \begin{equation} S=-m\int ...

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