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The Lagrangian of fractional classical mechanics $L_\alpha(\dot{q},q)$ is defined as $$ L_\alpha(\dot{q}, q)=p\dot{q}-H_\alpha(p,q) $$ where $$ \begin{split} p & =\frac{\partial L_\alpha(\dot{q}, q)}{\partial \dot{q}}, \\ H_\alpha(p, q)&=D_\alpha |p|^\alpha + V(q), \qquad 1 < \alpha \le 2 \end{split} $$ How to obtain the following using above definitions? $$ L_\alpha(\dot{q}, q)=\left(\frac{1}{\alpha D_\alpha}\right)^{\frac{1}{\alpha-1}} \frac{\alpha-1}{\alpha}|\dot{q}|^{\frac{\alpha}{\alpha-1}}, \qquad 1 < \alpha \le 2 $$

Source: Fractional Quantum Mechanics, Nick Laskin, World Scientific, 2018, page 260, MR3821542, Zbl 1425.81007.

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  • $\begingroup$ Apart from a missing $-V$, you can just verify this works by inserting the final expression for $L$ into the first three defining relations, no? $\endgroup$ Commented Jul 20, 2023 at 14:15
  • $\begingroup$ Yes, by placing the last relation in the previous three relations, I can confirm the authenticity of the relationship. But I want to compute the last relation from previous three relations. $\endgroup$
    – Dante
    Commented Jul 20, 2023 at 14:33
  • $\begingroup$ Well, you're essentially solving the differential equation $L=xL' - (L')^{\alpha } $ for the function $L(x)$. So the first thing I would do is try a power ansatz $L=cx^b $ ... I don't think there's more to it than that. $\endgroup$ Commented Jul 20, 2023 at 14:47
  • $\begingroup$ I am looking for a mathematical proof. $\endgroup$
    – Dante
    Commented Jul 20, 2023 at 17:36
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    $\begingroup$ Well, now you're changing what you're looking for. Then just go back to the start, verify the relationship holds, that's your proof. $\endgroup$ Commented Jul 20, 2023 at 18:17

1 Answer 1

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How to obtain the following using above definitions?

Referring to this note by Nick Laskin:

  1. Get $|p|$ from the first of Eqations (15);

  2. Substitute this $|p|$ into Eq. (4) assuming $V(q)=0$ in the definition of $H(p,q)$ given by Eq.(2);

  3. After trivial algebra we obtain Eq. (7), which is the equation in question.

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