# Tag Info

6

I don't think that there would be any more diagrams. Having a total derivative term in the Lagrangian leads to derivative interaction vertex, which after symmetrising gives you something like $$ig \sum_i p_i \ ,$$ where $g$ is some coupling and $p_i$ the momenta of the particles. This vertex, however, vanishes due to momentum ...

3

I have finished reading a great book called "The Theoretical Minimum" by Leonard Susskind (a famous string theorist) and George Hrabovsky. It's about classical mechanics but mainly talks about both the Lagrangian formulation and the Hamiltonian formulation of classical mechanics. It is great for beginners in physics or just about anyone. It also reviews the ...

3

If there is no external force with explicit time dependence, then the harmonic oscillator contains no explicit time dependence. Then the system has time translation symmetry, i.e. the result can only depend on the difference $T =t_b-t_a$, not on $t_a$ and $t_b$ individually.

3

Orientation. Let a lagrangian $L$ that is a local function of position and velocity be given. For a parameterized path $x$ on $M$, define a corresponding action $S$ as follows: \begin{align} S[x] = \int_a^b ds\,L(x(s), \dot x(s)). \end{align} Let $\delta x$ be a fixed-endpoint variation, then a standard computation shows that the corresponding variation ...

3

What Qmechanic said in comments is pretty solid, "Lagrangian (2) is not bounded from below because the kinetic term of $A_0$ field has the wrong sign, and hence the theory is not physical in the first place", but I think your Question needs a change of emphasis. Your Lagrangian allows us to construct four equations of motion for four non-interacting fields. ...

2

Comments to the question (v4): The question formulation seems to talk about affine parametrizations before applying the principle of stationary action. In the context of Riemannian$^1$ geometry, an affine parametrization of a (not necessarily geodesic) curve means by definition that the arc-length $s$ and the curve parameter $\lambda$ are affinely related ...

2

OP asks (v1): How one can know the gauge field emerging from the local gauge invariance is actually the EM field? Assuming that OP is pondering about gauging theoretical models (rather than concerned with our actual world and phenomenological inputs) then the answer is: One cannot know. For starters, the gauge group $G$ could be different than $U(1)$. ...

2

Yes, your equations aren't quite right. The main issue is that you're assuming a certain form for the normal force that isn't correct. What follows should illuminate why this is so in some detail. When using forces and Newton's Laws to solve this problem, it is overwhelmingly helpful to work in spherical coordinates, not just for locating the position of ...

2

The laws of physics are discovered through a mixture of heuristics, modelling and inference. In case of momentum, the story goes like this: It is possible to 'transfer motion' from one body to another. However, experiment shows that it is not velocity that is conserved during such transfers, but another 'quantity of motion'. We give that quantity the name ...

1

A simpler method would be to use spherical coordinates. To clear the notation, I define: \begin{align} z&=R \cos \theta (t)\\ y&=R \sin \theta (t) \sin \phi(t)\\ x&=R \sin \theta (t) \cos \phi(t) \end{align} The kinetic energy is then $$T=\frac{m}{2}\left(\dot x^2+\dot y^2+\dot z^2\right) =\frac{mR^2}{2}\left(\dot ... 1 Yes, I agree it looks like the second equation is missing an overall factor of \Delta t on the right. The action is a function of all the xs for the whole trajectory, but the Lagrangian is not. It is only a function of a position and a velocity. So it makes perfect sense to evaluate it using the position and velocity for a single time. Points 8 and 9 ... 1 Short answer: Already "true path" is an ugly choice of wording to say the "path that minimizes the action", taken from Hamilton's principle of least action, intuitively: Mechanical systems favor paths along which the difference between the kinetic and potential energy is as small as possible. More formally Hamilton's principle says: Given the action ... 1 An external force F_{\rm ext}(t) appears as a source term qF_{\rm ext}(t) in the Lagrangian. For example, if the equation of motion is,$$\tag{1} m\ddot{q}~=~-\frac{\partial V(q)}{\partial q} + F_{\rm ext}(t), $$then the Lagrangian reads$$\tag{2} L(q,\dot{q},t)~=~\frac{m}{2}\dot{q}^2-V(q)+ qF_{\rm ext}(t).$$1 It is there, it's just hidden by the change of coordinates. Written in Cartesian coordinates, the kinetic energy is$$ T=\frac12m_1\dot{x}_1^2+\frac12m_1\dot{y}_1^2+\frac12m_2\dot{x}_2^2+\frac12m_2\dot{y}_2^2+\frac12I_1\dot\theta_1^2+\frac12I_2\dot\theta_2^2\tag{1} where the last term is the rotational kinetic enregy. If you let \begin{align} ... 1 We start with the action S_{\Lambda,F} = \int d^4 x \left[ \frac{1}{4g^2} F_{\mu\nu} F^{\mu\nu} + a \Lambda_\mu \partial_\nu \ast F^{\mu\nu}\right] $$This action is equivalent to S_A in the sense that the equation of motion for \Lambda_\mu when plugged into S_{\Lambda,F} gives S_A. This is legit since \Lambda_\mu appears linearly and without ... 1 The Lagrangian is what is integrated over spacetime in the action, i.e. has to be a 4-form. As such, it is necessarily a (pseudo-)scalar under Lorentz transformations. When wondering about Lorentz transformations and such, the Hamiltonian is, as a non-Lorentz-covariant object, not a good starting point, by the way. It is often better to start with the ... 1 I'll get you started. It's not as bad as you might think at first glance. I think it's easiest to see how to proceed using vector notation. Let \mathbf r_1, \mathbf r_2, \mathbf r_3 denote the positions of the three masses in the plane containing the hoop. Let \ell_{ij} denote the length of the spring attaching mass i to mass j, then ... 1 The author assumed separation of variables,$$ L(u,v)=E(u)B(v) $$which leads to$$ \frac{\partial L}{\partial u}\frac{\partial L}{\partial v}=B\frac{\partial E}{\partial u}E\frac{\partial B}{\partial v}=-1 $$Rearranging,$$ E\frac{\partial E}{\partial u}=-\left(B\frac{\partial B}{\partial v}\right)^{-1}\tag{1} $$Note that$$ E\frac{\partial E}{\partial ...

1

In field theory the values of the field at every point in space are independent degrees of freedom, just like the positions of different particles in a multi-particle system. So, AFAIK to specify the initial and final configurations for an action integral you have to give the values of the field at every point in space at the initial and final times. The ...

1

In particle mechanics you integrate along a path, which is bounded by points, but in field theory you integrate over a spacetime volume, so your boundary is a hypersurface, not just points. For a typical quantum field theory process (at least the way it's formulated for calculations), there is some initial state consisting of noninteracting wavepackets, ...

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