Relativistic Lagrangian and equation of motion : Classical Mechanics

Relativistic Lagrangian and equation of motion

Here I have tried to solve a problem in Classical mechanics which is about relativistic Lagrangian and equation of motion.
Question Our problem is to show that relativistic Lagrangian give the equation of motion $F_{i}=-\frac{\partial V}{\partial \dot{x_{i}}}$

Solution From non-relativistic Lagrangian


where $T$ is the kinetic energy and $V$ is the potential energy and $\frac{\partial L}{\partial \dot{q}}$ is the momentum of the particle.


$p_{x}=\frac{\partial L}{\partial \dot{x}}$ , $p_{y}=\frac{\partial L}{\partial \dot{y}}$ , $p_{z}=\frac{\partial L}{\partial \dot{z}}$

If we assume similar equations of motion in relativistic mechanics then,

$p_{x}=\gamma m_{0} \dot{x}=\frac{\partial L}{\delta \dot{x}}$                               (1)

$p_{y}=\gamma m_{0} \dot{y}=\frac{\partial L}{\delta \dot{y}}$                               (2)

$p_{z}=\gamma m_{0} \dot{z}=\frac{\partial L}{\delta \dot{z}}$                               (3)


$\gamma =\frac{1}{\sqrt{1-\beta^{2}}}$


$\beta=\frac{v}{c}$ ,


and $v$ is the speed of the particle in the inertial frame under consideration.

From equation (1)

$\partial L=\gamma m_{0}\dot{x}\partial \dot{x}$

On integrating it we get

$L=\int \frac{m_{0}\dot{x}d\dot{x}}{\sqrt{1-\beta^{2}}}=-m_{0}c^{2}\sqrt{1-\beta^{2}}-V$


$\beta=\frac{v}{c}$ , $v=\left ( \dot{x}^{2}+\dot{y}^{2}+\dot{z}^{2} \right )^{1/2}$ and $V$ is a constant of integration and may be taken as the potential energy of the particle and $V=V(x,y,z)$ .

Now we use above definition of Lagrangian for obtaining the relativistic equation of motion.

The Lagrangian equation of motion for x coordinate is

$\frac{d}{dt}\left (\frac{\partial L}{\partial x}  \right )=0$

$\Rightarrow $

$\frac{\partial L}{\partial x}=-\frac{\partial V}{\partial x}$


${{\partial L} \over {\partial \dot x}} = {\partial  \over {\partial \dot x}}\left( { – {m_0}{c^2}\sqrt {1 – {{{{\dot x}^2} + {{\dot y}^2} + {{\dot z}^2}} \over {{c^2}}}}  – V} \right)$


${{\partial L} \over {\partial \dot x}} = {{{m_0}\dot x} \over {\sqrt {1 – {\beta ^2}} }} = {p_x}$


$\frac{dp_{x}}{dt}=\frac{\partial V}{\partial x}=0$


${{d{p_x}} \over {dt}} =  – {{\partial V} \over {\partial x}} = {F_x}$

The same way we can find the equation of motion in $y$ and $z$ direction as

${{d{p_y}} \over {dt}} =  – {{\partial V} \over {\partial y}} = {F_y}$


${{d{p_z}} \over {dt}} =  – {{\partial V} \over {\partial z}} = {F_z}$

Thus in generalized form we can write,

$F_{i}=-\frac{\partial V}{\partial \dot{x_{i}}}$

where $q_{i}$ is the generalized co-ordinates.

This is the required relativistic equation of motion.