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

  • A
  • Thread starter Thread starter Kostik
  • Start date Start date
Kostik
Messages
250
Reaction score
28
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:
Thread 'Can this experiment break Lorentz symmetry?'
1. The Big Idea: According to Einstein’s relativity, all motion is relative. You can’t tell if you’re moving at a constant velocity without looking outside. But what if there is a universal “rest frame” (like the old idea of the “ether”)? This experiment tries to find out by looking for tiny, directional differences in how objects move inside a sealed box. 2. How It Works: The Two-Stage Process Imagine a perfectly isolated spacecraft (our lab) moving through space at some unknown speed V...
Does the speed of light change in a gravitational field depending on whether the direction of travel is parallel to the field, or perpendicular to the field? And is it the same in both directions at each orientation? This question could be answered experimentally to some degree of accuracy. Experiment design: Place two identical clocks A and B on the circumference of a wheel at opposite ends of the diameter of length L. The wheel is positioned upright, i.e., perpendicular to the ground...
According to the General Theory of Relativity, time does not pass on a black hole, which means that processes they don't work either. As the object becomes heavier, the speed of matter falling on it for an observer on Earth will first increase, and then slow down, due to the effect of time dilation. And then it will stop altogether. As a result, we will not get a black hole, since the critical mass will not be reached. Although the object will continue to attract matter, it will not be a...
Back
Top