IGNOU MPH 6 2026 - The Lagrangian of a relativistic particle with charge e is given as:

Question

The Lagrangian of a relativistic particle with charge e is given as:

$$L = -mc^2 \sqrt{1 - \left( \frac{v}{c} \right)^2} - e\phi + e\vec{A} \cdot \vec{v}$

19 Jan 2026

Answer :

Word Count : 332

The given Lagrangian is: [ L = -mc^2 \sqrt{1 - \frac{v^2}{c^2}} - e\phi + e \vec{A} \cdot \vec{v} ] We are to find the equations of motion using the Euler–Lagrange equation: [ \frac{d}{dt} \left( \frac{\partial L}{\partial \vec{v}} \right) = \frac{\partial L}{\partial \vec{r}} ] Step 1: Compute (\frac{\partial L}{\partial \vec{v}}) [ \frac{\partial}{\partial \vec{v}} \Big[ -mc^2 \sqrt{1 - \frac{v^2}{c^2}} \Big] = -mc^2 \cdot \frac{1}{2} \cdot (1 - \frac{v^2}{c^2})^{-1/2} \cdot \left( -\frac{2\vec{v}}{c^2} \right) ] Simplify: [ \frac{\partial}{\partial \vec{v}} \Big[ -mc^2 \sqrt{1 - \frac{v^2}{c^2}}

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