ADVANCED DYNAMICS — PHY 4241/5227

HOME AND CLASS WORK – SET 2

Solution for assignment 9.

Deviation of the Euler-Lagrange equations from the least action principle in general coordinates q_{i}, q̇_{i}, i = 1, … n:

0 = δ ∫_{t_{1}}^{t_{2}} dt L(q_{i}, q̇_{i}, t) = ∫_{t_{1}}^{t_{2}} dt {L(q_{i} + δq_{i}, q̇_{i} + δq̇_{i}, t) − L(q_{i}, q̇_{i}, t)}
= ∫_{t_{1}}^{t_{2}} dt \{L(q_{i}, q̇_{i}, t) + ∑_{i} \frac{∂L}{∂q_{i}} δq_{i} + ∑_{i} \frac{∂L}{∂q̇_{i}} δq̇_{i} − L(q_{i}, q̇_{i}, t)\}
= ∫_{t_{1}}^{t_{2}} dt \{∑_{i} \frac{∂L}{∂q_{i}} δq_{i} + ∑_{i} \frac{∂L}{∂q̇_{i}} \frac{d}{dt}δq_{i}\}

= ∫_{t_{1}}^{t_{2}} dt ∑_{i} \{\frac{∂L}{∂q_{i}} − \frac{d}{dt} \frac{∂L}{∂q̇_{i}} \} δq_{i} + [\frac{∂L}{∂q̇_{i}} δq_{i}]_{t_{1}}^{t_{2}} = ∫_{t_{1}}^{t_{2}} dt ∑_{i} \{\frac{∂L}{∂q_{i}} − \frac{d}{dt} \frac{∂L}{∂q̇_{i}} \} δq_{i} .

The last equality holds because of δq_{i}(t_{1}) = δq_{i}(t_{2}) = 0. As the variations are independent, the finally obtained relation is equivalent to

\frac{∂L}{∂q_{i}} − \frac{d}{dt} \frac{∂L}{∂q̇_{i}} = 0 for i = 1, …, n .