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When we are given a Lagrangian as $L \equiv L(\dot{x}, V)$ where $V=V(x)$, while differentiating why do we set $\frac{dV}{d\dot{x}} = 0$?

Is it just because $V$ is a function of only $x$? Since it’s a function of x, is it not possible that it is a function of $\dot{x} $ too?
For example, if $V=x^2,x=t^2$, then $\dot{x}=2t$ and $V=\frac{\dot{x}^4}{16}$.

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I came across this doubt while solving this question.

Ambica Govind
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Owaise Ahmed
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  • Welcome to PSE. In this site we don't post images in place of text and mathematical equations. Moreover, answers to homework-like questions will be deleted. – Frobenius Oct 04 '22 at 06:00

2 Answers2

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Have a look at the Lagrangian given in the problem. With the information that $V$ is a differentiable function of $x$ (and nothing else), $\partial_{\dot{x}} V \overset{!}{=} 0$ and the Langrangian's dependence on $\dot{x}$.

A hint (if you're unwilling to have faith in the authors) as to why $V \neq V(\dot{x})$ in this problem can be found in the second term of the Lagrangian as the $\dot{x}$ dependence is explicitly shown alongside the potential $V(X)$.

Song of Physics
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  1. More generally, it should be noted that Lagrangian mechanics does allow for velocity-dependent generalized potentials $U(q,\dot{q},t)$, cf. e.g. this Phys.SE post.

  2. However $V(x)$ and $V_2(x)$ in the specific problem are assumed to not depend on $\dot{x}$ as an ansatz.

  3. Concerning $\frac{\partial \dot{x}}{\partial x}=0$, see this related Phys.SE post.

Qmechanic
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