Linked Questions

22
votes
5answers
3k views

Is Fermat's principle only an approximation?

Fermat's principle says that the path taken between two points by a ray of light is the path that can be traversed in the least time. It occurred to me today that maybe the path is actually the one ...
43
votes
3answers
8k views

How are classical optics phenomena explained in QED (Snell's law)?

How is the following classical optics phenomenon explained in quantum electrodynamics? Reflection and Refraction Are they simply due to photons being absorbed and re-emitted? How do we get to Snell'...
10
votes
2answers
4k views

Geodesics equations via variational principle

I would like to recover the (timelike) geodesics equations via the variational principle of the following action: $$ \mathcal{S}[x] = -m \int d\tau = -m \int \sqrt{-g_{\mu\nu}\,dx^{\mu}\,dx^{\nu}} $$ ...
11
votes
2answers
2k views

Constraints of relativistic point particle in Hamiltonian mechanics

I try to understand constructing of Hamiltonian mechanics with constraints. I decided to start with the simple case: free relativistic particle. I've constructed hamiltonian with constraint: $$S=-m\...
10
votes
4answers
4k views

Lagrangian for relativistic massless point particle

For relativistic massive particle, the action is $$S ~=~ -m_0 \int ds ~=~ -m_0 \int d\lambda ~\sqrt{ g_{\mu\nu} \dot{x}^{\mu}\dot{x}^{\nu}} ~=~ \int d\lambda \ L,$$ where $ds$ is the proper time of ...
11
votes
1answer
1k views

Infinitesimal transformations for a relativistic particle

The action of a free relativistic particles can be given by $$S=\frac{1}{2}\int d\tau \left(e^{-1}(\tau)g_{\mu\nu}(X)X^\mu(\tau)X^\nu(\tau)-e(\tau)m^2\right).$$ If we then make an infinitesimal ...
6
votes
4answers
1k views

How to find a particle's dynamics in general relativity?

About a year ago, I took a course on general relativity. It isn't until now that I realize that, given a metric, I am unsure how to find a particle's dynamics. What I mean by that is, normally I ...
4
votes
2answers
1k views

What is the relativistic action of a massive particle?

all Lorentz observers watching a particle move will compute the same value for the quantity $$ds^2 = -(c \, dt)^2 + dx^2 + dy^2 + dz^2,$$ $$ds^2 = g_{\mu\nu}dx^{\mu}dx^{\nu},$$ and ''ds/c'' is then ...
8
votes
3answers
824 views

General relativity: For massless particles, is the momentum and velocity 4-vectors equal?

I am following Carroll's GR book. He explain that it is convention to parameterize geodesics of photons by a parameter $\lambda$ such that $$p^\mu ~=~ \frac{d x^{\mu}}{d \lambda}.\tag{3.62}$$ But ...
3
votes
3answers
452 views

Is Hamilton's principle compatible with the relativity principle?

The principle of Hamilton in classical mechanics is a fundamental one. It states that the real trajectory of a particle extremize the action $$ \int_{t_1}^{t_2} d \tau L (q , \dot{q}, \tau ) . $$ ...
1
vote
2answers
431 views

Why can we choose affine parameterization?

In general relativity when deriving the geodesic equation $$\ddot{x}^\mu + \Gamma^\mu_{\alpha\beta}\dot{x}^\alpha\dot{x}^\beta = 0\tag{1}$$ from the action $$S = \int d\tau \sqrt{|g_{\mu\nu} \dot{x}...
3
votes
2answers
121 views

Geodesic equations from action with auxiliary field

A textbook says that the geodesic equations (for both massive and massless) can be derived from the following action: $$ S = -\frac{1}{2} \int d\tau \:\eta \: (\eta^{-2} \dot{x}^\mu \dot{x}^\nu g_{\...
2
votes
2answers
174 views

Invariance under local diffeomorphisms

In the context of the Polyakov action, the action for a relativistic point particle $$ S_P = \frac{1}{2} \int \mathrm{d}\tau \, e(\tau) \left(\frac{1}{e^2(\tau)}\left(\frac{\mathrm{d} X^\mu(\tau)}{\...
1
vote
2answers
138 views

How to get the fourth component of EOM in a relativistic formulation of a charged particle in an electromagnetic field?

We consider in Lorentz spacetime, $(x^0,x^1,x^2,x^3)=(t,x,y,z)$, choose the unit of time such that $c=1$. Given a four vector $A_\mu$, and let the Lagrangian $$L(x^i,\dot x^i,t)=-m\sqrt{1-\dot x_i\...
3
votes
1answer
91 views

Principled approach to arrive at geodesic Hamiltonian $H = g^{\mu \nu} p_\mu p_\nu$?

The background: If we have a spacetime path $x^\mu(t)$ parameterized by arbitrary parameter $t$, the proper time along the path between $t_1$ and $t_2$ is $$ \int_{t_1}^{t_2} (g_{\mu \nu} \dot x^\mu \...

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