Linked Questions

42
votes
3answers
5k views

What's the interpretation of Feynman's picture proof of Noether's Theorem?

On pp 103 - 105 of The Character of Physical Law, Feynman draws this diagram to demonstrate that invariance under spatial translation leads to conservation of momentum: To paraphrase Feynman's ...
31
votes
1answer
3k views

Do an action and its Euler-Lagrange equations have the same symmetries?

Assume a certain action $S$ with certain symmetries, from which according to the Lagrangian formalism, the equations of motion (EOM) of the system are the corresponding Euler-Lagrange equations. Can ...
20
votes
3answers
2k views

Is there a mathematical reason for the Lagrangian to be Lorentz invariant?

The Hamiltonian is the energy, which is just one component of a four-vector and therefore not Lorentz invariant. The Lagrangian is the Legendre transform of the Hamiltonian and I was wondering if ...
9
votes
1answer
770 views

Why is a theory Lorentz invariant if the Lagrangian is Lorentz invariant?

For if I started by trying to make the Hamiltonian Lorentz invariant, I would have failed. Indeed, the Hamiltonian is part of a covariant tensor. But how do I know that the Lagrangian is not a part of ...
5
votes
1answer
642 views

Explicitly show covariance of Euler Lagrange equations

I know that the Euler Lagrange equation (here only in 1D) $$ \left(\frac{d}{dt}\frac{\partial}{\partial\dot{x}}-\frac{\partial}{\partial x}\right)L\left(x,\dot{x},t\right)=0 $$ is invariant under (...
2
votes
2answers
868 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 $t, q(t)$ ...
2
votes
1answer
416 views

Does an on-shell symmetry necessarily change the Lagrangian by a total derivative?

This is a follow-up question to: Does a symmetry necessarily leave the action invariant? Qmechanic writes here: Here the word off-shell means that the Lagrangian eqs. of motion are not assumed to ...
2
votes
1answer
152 views

On-shell and off-shell transformations in Noether's theorem

For any transformation of the fields, $$\varphi\to\varphi'=\varphi+\delta\varphi$$ the change in the Lagrangian can be written as $$\delta\mathcal L = \text{EoM} + \partial_\mu j^\mu\tag{1}$$where "...
0
votes
1answer
97 views

Isotropic of Inertial frame?

My understanding of isotropic is the a particular physics law remain same no matter at what direction I look at it? Now suppose in case of inertial frame, we know that its is homogeneous and ...
1
vote
0answers
61 views

Legendre transformation and correspondance between Noether charges and quasi-symmetries

I have been trying to understand the Legendre transformation (in mechanics, in the hyperregular case: when the Legendre transformation is one-to-one) and the correspondence between symmetry $\to$ ...