A The Lagrangian of a free particle ##L=-m \, ds/dt##

  • A
  • Thread starter Thread starter Kostik
  • Start date Start date
Kostik
Messages
269
Reaction score
32
TL;DR Summary
How to show that ##L=-m \, ds/dt## for a free particle.
In Dirac's "General Theory of Relativity" (p. 52), he postulates that the action for a free particle of mass ##m## is $$I=-m \int ds$$ hence the Lagrangian is $$L=-m\frac{ds}{dt} = -m\frac{\sqrt{\eta_{\mu\nu}dx^\mu dx^\nu}}{dt}
\, .$$ To confirm that ##-m## is the correct coefficient, he assumes flat spacetime (special relativity) and calculates $$\frac{\partial L}{\partial \dot{x}^k} = -m\frac{\partial}{\partial \dot{x}^k}\left( \frac{ds}{dt}\right) = m\frac{ \dot{x}^k }{ ds/dt } = m \frac{dx^k}{ds}$$ which is the correct formula for relativistic 4-momentum ##p^k##. ("As it ought to be", says Dirac.)

##\qquad## But doesn't this assume that $$p^k = \frac{\partial L}{\partial \dot{x}^k} \quad ?$$ This is true for the non-relativistic Lagrangian for a free particle $$L=T-U = \frac{1}{2}mv^2 \qquad (*)$$ but is it true for a relativistic particle?

##\qquad## (Landau-Lifshitz give what seems to be a more convincing confirmation. They also postulate that ##L= \kappa \, ds/dt##, and assume flat spacetime (special relativity) where ##ds^2 = \eta_{\mu\nu}dx^\mu dx^\nu##. Then $$L=\kappa \frac{ds}{dt}=\kappa \sqrt{1-v^2} \, .$$ Hence, for velocities ##v \ll 1##, we have $$L = \kappa - \frac{1}{2}\kappa v^2 + O(v^4) \, .$$ The nonrelativistic Lagrangian is shown above in ##(*)##, so to get the correct kinetic energy term, we must have ##\kappa = -m##.)

##\qquad## I'd like to understand Dirac's confirmation that ##\kappa = -m##. How does he know in advance that $$p^k = \frac{\partial L}{\partial \dot{x}^k} $$ for a relativistic particle? It seems like he's using circular reasoning.

##\qquad## I think what Dirac forgot to add was "Since ##p^k = \partial L / \partial \dot{x}^k## in the case ##v \ll 1##, we see that ##\kappa = -m##."
 
Last edited:
Physics news on Phys.org
IIRC, Dirac uses Greek for 0-3, while Latin for 1-3, unlike L-L who use Greek for 1-3, and Latin for 0-3.

So if we try to switch from the Lagrangian density \mathcal{L} to the Hamiltonian formalism, we define the 4-momentum as

\begin{equation} p_\mu =: \frac{\partial \mathcal{L} \left(x,\dot{x}\right)}{\partial \dot{x}^\mu}\end{equation},

where the dot is the worldline parameter (most commonly chosen as the proper time).

Why would going Greek > Latin be circular reasoning?
 
I think it's circular reasoning because ##p^k = \partial L / \partial \dot{x}^k## is derived for a non-relativistic free particle, where ##L=mv^2/2##. But, otherwise, ##p^k \equiv \partial L / \partial \dot{x}^k## is the definition of the "conjugate momentum". We cannot simultaneously define ##p^k \equiv \partial L / \partial \dot{x}^k## and ##p^k \equiv m\, dx^k/ds = mv^k##.

I think Dirac needs to reduce to the non-relativistic case (like L-L) in order to determine ##\kappa##.

##\qquad## I think what Dirac should have done is this: Consider the case ##v \ll 1##, where ##L = T = mv^2/2##. Then we confirm by calculation that $$\frac{\partial L}{\partial \dot{x}^k} = \frac{m}{2}\frac{\partial v^2}{\partial \dot{x}^k}=m\dot{x}^k \,. $$ Comparing this with $$\frac{\partial L}{\partial \dot{x}^k} = -\kappa \frac{dx^k}{ds} $$ and using ##ds=dt## when ##v \ll 1##, we obtain ##\kappa = -m##.
 
Last edited:
Kostik said:
TL;DR Summary: How to show that ##L=-m \, ds/dt## for a free particle.

But doesn't this assume that pk=∂L∂x˙k? This is true for the non-relativistic Lagrangian for a free particle
This is also true for SR Lagrangian. For SR Lagrangian pk has the factor \frac{1}{\sqrt{1-v^2/c^2}} as we see in experiments. Analytical mechanics holds also in special relativity only by changing the Lagrangian functions.
 
Last edited:
In this video I can see a person walking around lines of curvature on a sphere with an arrow strapped to his waist. His task is to keep the arrow pointed in the same direction How does he do this ? Does he use a reference point like the stars? (that only move very slowly) If that is how he keeps the arrow pointing in the same direction, is that equivalent to saying that he orients the arrow wrt the 3d space that the sphere is embedded in? So ,although one refers to intrinsic curvature...
I started reading a National Geographic article related to the Big Bang. It starts these statements: Gazing up at the stars at night, it’s easy to imagine that space goes on forever. But cosmologists know that the universe actually has limits. First, their best models indicate that space and time had a beginning, a subatomic point called a singularity. This point of intense heat and density rapidly ballooned outward. My first reaction was that this is a layman's approximation to...
So, to calculate a proper time of a worldline in SR using an inertial frame is quite easy. But I struggled a bit using a "rotating frame metric" and now I'm not sure whether I'll do it right. Couls someone point me in the right direction? "What have you tried?" Well, trying to help truly absolute layppl with some variation of a "Circular Twin Paradox" not using an inertial frame of reference for whatevere reason. I thought it would be a bit of a challenge so I made a derivation or...
Back
Top