NET/JRF Physics : Relativistic Lagrangian and equation of motion



Relativistic Lagrangian and equation of motion

Here I have tried to solve problem in 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

L=T+V

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

Thus,

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)

where,

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

and

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

v^{2}=\dot{x}^{2}+\dot{y}^{2}+\dot{z}^{2}

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

From equation (1)

\partial L=\gama 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

where,

\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 relativistic equation of motion.



Lagrangian equation of motion for x co-ordinate is

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

\Rightarrow

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

and

{{\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)

or,

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

Thus

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

or,

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

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}

and

{{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.

 

 

 

 





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