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3 votes
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Why can you not take $u \cdot u = c^2$ in the relativistic free massive particle Lagrangian?

In my classical electrodynamics class, we use the Lagrangian of the relativistic free massive particle as $$L = - mc\sqrt{\dot{r}\cdot\dot{r}}.$$ Where $\dot{r}^\mu = u^\mu = \frac{dr^\mu}{d \tau}$; $...
Jack's user avatar
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1 vote
1 answer
112 views

Why is integral of relativistic action $-\alpha \int_{a}^{b} \, \mathrm ds$ minimised with respect to $\mathrm ds$?

While reading some aspects concerning the conclusion that bodies follow geodesics of spacetime, I ran into relativistic action on p.24 in chapter 2 $\S8$ of the 2nd volume of Landau & Lifshitz: $$...
M. A.'s user avatar
  • 1,999
0 votes
2 answers
61 views

Lorentz force law, Action of relativistic particle in e.m. field with interaction between field and particle

This question is about the derivation of the Lorentz force law and may be answered quickly. Nonetheless, I will give some context. In lecture we defined the overall Action of a particle + field (in ...
Lenny's user avatar
  • 13
1 vote
1 answer
61 views

Particle-hole symmetry and relativistic theory

Why must a particle-hole symmetric theory always be relativistic? Let's suppose we have the following Lagrangian density: $$L = K_1 \Psi^*\frac{\partial\Psi}{\partial\tau} + K_2 \left|\frac{\partial\...
redfive's user avatar
  • 427
2 votes
1 answer
197 views

Hamiltonian for massive relativistic point particle in light-cone gauge

Given light-cone coordinates, and some worldline parameterised by some arbitrary parameter $\tau$. $$x^{+} = \frac{1}{\sqrt{2}}(x^0+x^1),$$ $$x^{-} = \frac{1}{\sqrt{2}}(x^0-x^1).\tag{2.50}$$ And the ...
jensen paull's user avatar
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1 vote
0 answers
59 views

Homogeneous Lagrangian in covariant Lagrangian formulations

I have a question in Chapter $7.10$ of Goldstein et. al.'s classical mechanics (3rd edition). Specifically, on page 320 about the discussion of covariant Lagrangian formulations. First, they introduce ...
Daren's user avatar
  • 1,505
0 votes
0 answers
36 views

For an $O(N)$ symmetric function, is the 4-derivative simply equal to the derivative w.r.t the $N$-dimensional norm?

I am working on an $O(4)$-symmetric instanton which has the Lagrangian: $$L = \frac{M^6}{4E^2T^2} \left[ \frac{1}{2}(\partial_\mu\Phi)^2 - \frac{1}{2} \Phi^2 + \frac{1}{2}\Phi^3 - \frac{\alpha}{8} \...
Adam P's user avatar
  • 357
0 votes
2 answers
98 views

Lorentz scalar Lagrangian in curved spacetime

This question might be very simple but I guess I'm missing something. We know that Lagrangian has to be a Lorentz scalar. I can see why that should be the case when dealing with inertial frames of ...
physics_2015's user avatar
0 votes
1 answer
68 views

Lorentz transformation on field theory using coordinate expressions not representations

This is probably a stupid question but, I want to show that a Lagrangian written in field theory is Lorentz invariant WITHOUT using the Lorentz transformation representation / generators. I know we ...
sars909's user avatar
0 votes
0 answers
48 views

Does the Lagrangians in different coordinate systems differ by a total derivative when converted to the same coordinate in special relativity?

According to this post, it seems like we should use the $$\mathbb{L}~=~L \mathrm{d}t~=~ L \dot{t}\mathrm{d}\lambda \tag1$$ rather than $$L(v'^2) = L(v^2)+ \frac{df}{dt}\tag2$$ when working with ...
Raffaella's user avatar
  • 413
0 votes
1 answer
122 views

Is relativistic action too restrictive?

When I was studying special relativity, I've learned that the relativistic action for a free particle is defined as $$ S = \lambda \int_{\tau_0}^{\tau_1} d\tau $$ Where $\lambda$ is a constant that is ...
Ruffolo's user avatar
  • 4,200
3 votes
2 answers
144 views

Relativistic Lagrangian for a System of Massive Particles

The standard Lagrangian $L$ written in local coordinates for a free, relativistic particle of mass $m > 0$ is given by $$L(q, \dot{q}) = -m \sqrt{-g_{\mu\nu}(q) \dot{q}^\mu \dot{q}^\nu}\tag{1}$$ ...
J_Psi's user avatar
  • 360
0 votes
0 answers
47 views

Deducing the (special) relativistic Lagrangian only using Lorentz transform [duplicate]

I feel like there must be a way to deduce the (special) relativstic Lagrangian using only the Lorentz transform (without other knowledge of special relativity). So far, all the derivations I have seen ...
Ilya Iakoub's user avatar
2 votes
2 answers
179 views

When is the Lagrangian a Lorentz scalar?

The Lagrangian $\mathcal{L}$ can be defined as the Legendre transform (when it exists) of the Hamiltonian $\mathcal{H}$, a non-Lorentz scalar quantity (as $\mathcal{H} =T^{00}$). My questions are, ...
Gabriel Ybarra Marcaida's user avatar
0 votes
0 answers
38 views

Turning a Lagrangian contains superscript and subscript indices into energy

I'm recently reading the book "Solitons and Instantons" written by R. RAJARAMAN. However, for lacking of ability, I couldn't figure out how to derivate the static solution for energy with ...
Karl's user avatar
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