Question:easy

A coordinate \(q_i\) is called cyclic (or ignorable) if:

Show Hint

Cyclic means the coordinate itself does not appear in \(L\); its velocity still can. That is \(\partial L/\partial q_i = 0\).
Updated On: Jul 2, 2026
  • \(\dfrac{\partial L}{\partial q_i} = 0\)
  • \(\dfrac{\partial L}{\partial \dot{q}_i} = 0\)
  • \(\dfrac{d}{dt}\!\left(\dfrac{\partial L}{\partial q_i}\right) = 0\)
  • \(\dfrac{d}{dt}\!\left(\dfrac{\partial L}{\partial \dot{q}_i}\right) = 0\)
Show Solution

The Correct Option is A

Solution and Explanation

Step 1: Start from the meaning of the word cyclic: the coordinate itself is absent from $L$, even though its velocity $\dot{q}_i$ may appear. Absence from $L$ is expressed mathematically as the partial derivative with respect to $q_i$ vanishing.
Step 2: Write down what each option means physically. Option (B) $\partial L/\partial \dot{q}_i = 0$ would kill the conjugate momentum, which is not the definition. The two $\dfrac{d}{dt}(\cdots)$ options are consequences or parts of the equation of motion, not the defining property.
Step 3: Only $\partial L/\partial q_i = 0$ states that $q_i$ does not enter $L$ explicitly, which is exactly the definition of an ignorable coordinate.
Step 4: As a check, this immediately gives a conservation law $p_i = \partial L/\partial\dot{q}_i = \text{const}$, a hallmark of cyclic coordinates such as an angle for a central force.
\[\boxed{\dfrac{\partial L}{\partial q_i} = 0}\]
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