13
votes
4answers
285 views

The Lagrangian as a metric

My question is, can the (classical) Lagrangian be thought of as a metric? That is, is there a meaningful sense in which we can think of the least-action path from the initial to the final ...
1
vote
1answer
78 views

What is the neatest way to describe a “non-autonomous” (lagrangian) system?

The configuration space of a system of particles $(m_i,x_i)$, $i=1,\dots,n$, subject to constraints $$\Phi (x)=0,\qquad \Phi\colon \mathbb R^{3n}\to \mathbb R ^{3n-k},\qquad x=(x_1,...,x_n),$$ if the ...
8
votes
0answers
189 views

What's the physical intuition for symplectic structures?

I always thought about symplectic forms as elements of areas in little subspaces because of the Darboux theorem, however I cannot get the physical intuition for it and for the hamiltonian vector ...
9
votes
3answers
341 views

Equation of motion of a photon in a given metric

I have this metric: $$ds^2=-dt^2+e^tdx^2$$ and I want to find the equation of motion (of x). for that i thought I have two options: using E.L. with the Lagrangian: $L=-\dot t ^2+e^t\dot x ^2 $. ...
1
vote
0answers
206 views

Trouble with calculating Christoffel symbols of FLRW metric using Lagrangian method

The FLRW metric which I am using is $$ds^2 = dt^2 - \frac{a(t)^2}{c^2} \left( dx^2 + dy^2 + dz^2 \right)$$ where $a(t)$ is the so-called 'scale factor'. I did not want to calculate the Christoffel ...
3
votes
2answers
763 views

Null geodesic given metric

I (desperately) need help with the following: What is the null geodesic for the space time $$ds^2=-x^2 dt^2 +dx^2?$$ I don't know how to transform a metric into a geodesic...! There is no need to ...
1
vote
1answer
172 views

A particular case when Lagrange equation is equivalent to equation of motion on a Riemannian manifold

Suppose a particle is moving on a surface of a sphere,then it contains a holonomic constraint and so the three Cartesian co-ordinates are available with a constraint equation(equation of surface in ...
1
vote
1answer
146 views

Symmetries of spacetime and objects over it

I guess according to mathematical didactic, we first think of spacetime as a set and we reason about elements of its topology and then it's furthermore equipped with a metric. Appearently it is this ...
1
vote
1answer
79 views

Smooth trajectory on a smooth manifold

Physicists talk about a smooth trajectory of a particle on a smooth manifold and they label it as q(t) where q_1(t)....q_n(t) are component functions coming from the homeomorphism. I don't see how we ...