Hot answers tagged

5

There is also the routhian formalism of mechanics which is described as being a hybrid of lagrangian and hamiltonian mechanics. The routhian is defined as $$R = \sum_{i=1}^n p_i\dot{q}_i - L$$ You can learn more about it by clicking this link for wikipedia's description of it.


3

It's worth pointing out that the Hamiltonian and Lagrangian formalisms are independent, even though they're usually taught as if the former were a filtering of the latter (here enter Legendre transforms). Both formalisms are as independent as the notions of tangent and cotangent bundles in differential geometry: independent, but intrinsically connected. ...


3

Yes, OP is right. In the field-theoretic case, the partial derivatives in OP's first formula (1) should be replaced with functional derivatives $$ \delta S~=~\int_{t_1}^{t_2}\!\mathrm{d}t\left(\frac{\delta L}{\delta q}~\delta q+\left. \frac{\delta L}{\delta v}\right|_{v=\dot{q}}~\delta \dot{q}\right),\tag{1'}$$ where the Lagrangian $$L[q(\cdot,t),v(\...


2

This is just supplementing Qmechanic's answer. I think the notations here need to be addressed. OP might be confusing Lagrangian (normal $L$) with Lagrangian density ($\mathcal{L}$). Formally, we have three fundamental relations: $$L = \displaystyle\int \mathcal{L}(\phi(x,t),\dot \phi(x,t),x,t) \mathrm d^3x$$ $$S = \displaystyle\int dt \space L = \...


2

Okay, let's give it a try. $SU(2)$ sector of Standard Model Lagrangian is rather involved, so we will take a look at something simpler. Neutron-proton interaction comes to my mind. In low energy limit it is mediated by a massive scalar particle — a pion. We will be very qualitative about this, in reality there are a lot of details. Lagrangian will look ...


1

I) The first part of OP's construction is directly related to the covariant Hamiltonian formalism for a real scalar field with Lagrangian density $$ {\cal L} ~=~ \frac{1}{2}\partial_{\alpha} \phi ~\partial^{\alpha} \phi -{\cal V}(\phi), \tag{CW4} $$ see e.g. Ref. [CW] and this Phys.SE post. [In this answer we use the $(+,-,-,-)$ Minkowski signature ...


1

No, many other couplings are possible. For example, in the very simple Lagrangian $$L = \frac{1}{2} mv^2 - mgh$$ we have coupled the particle to the gravitational field $\phi = gh$. This is already in relativistically invariant form, since both $m$ and $\phi$ are scalars. (Of course the real story for coupling to gravity is more complicated, but this works ...


1

This might help you out a bit: In what sense is a quantum field an infinite set of harmonic oscillators? From my understanding, most people think it provides a useful way to conceptualize uncoupled quantum fields physically. It doesn't, however, work for coupled quantum fields. The main problem with this seems to be that infinite harmonic oscillators give ...



Only top voted, non community-wiki answers of a minimum length are eligible