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

  • A
  • Thread starter Thread starter Kostik
  • Start date Start date
Kostik
Messages
274
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:
OK, so this has bugged me for a while about the equivalence principle and the black hole information paradox. If black holes "evaporate" via Hawking radiation, then they cannot exist forever. So, from my external perspective, watching the person fall in, they slow down, freeze, and redshift to "nothing," but never cross the event horizon. Does the equivalence principle say my perspective is valid? If it does, is it possible that that person really never crossed the event horizon? The...
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...
ASSUMPTIONS 1. Two identical clocks A and B in the same inertial frame are stationary relative to each other a fixed distance L apart. Time passes at the same rate for both. 2. Both clocks are able to send/receive light signals and to write/read the send/receive times into signals. 3. The speed of light is anisotropic. METHOD 1. At time t[A1] and time t[B1], clock A sends a light signal to clock B. The clock B time is unknown to A. 2. Clock B receives the signal from A at time t[B2] and...
Back
Top