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I need to derive the Hamiltonian from the Lagrangian defined in the following way:

$$L[x, \dot{x}] = \int_{t_0}^{t_1} f(x(t), t) \sqrt{1 + \dot{x}^2} \mathrm{d}t.\tag{1}$$

The usual method is to compute the momentum $p$:

$$p = \frac{\partial L}{\partial \dot{x}} = \int_{t_0}^{t_1} f(x(t), t) \frac{\dot{x}}{\sqrt{1 + \dot{x}^2}} \mathrm{d}t,\tag{2}$$

invert the relation and write $\dot{x} = \dot{x}(x, p)$ and then compute the Hamiltonian:

$$H(x, p) = \dot{x}(x, p) p - L[x, \dot{x}(x, p)].\tag{3}$$

But in this case it is not possible to simply get the formula for $\dot{x}$. It is still possible to get the Hamiltonian?

Qmechanic
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lemurman
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2 Answers2

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The $L(x,\dot{x},t)$ that you've written is actually the action $S[x(t)]$, the Lagrangian is the integrand. In other words $$L=f(x,t)\sqrt{1+\dot{x}^2}$$ and so the conjugate momentum to $x$ is given by $$p=f(x,t)\frac{\dot{x}}{\sqrt{1+\dot{x}^2}}$$ thus $$\dot{x}=\pm\frac{p}{\sqrt{f(x,t)^2-p^2}}$$ and finally $$H=\frac{\pm~ p^2 -f(x,t)^2}{\sqrt{f(x,t)^2-p^2}}$$ I'm not really sure which sign we're supposed to pick but if we pick the plus sign this simplifies nicely to $$H=-\sqrt{f(x,t)^2-p^2}$$

AfterShave
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In this answer, we take OP's integral (1) literal. There are 2 possibilities:

  1. If $\dot{x}$ is supposed to be the time derivative of $x$, then it does not make sense to list $\dot{x}$ as an independent argument of the action functional (1) [which we will call $S$], cf. e.g. this & this Phys.SE posts. It also renders OP's further manipulations meaningless.

  2. If $\dot{x}$ is supposed to be an independent variable [which we will call $v$], then it does make sense to define $$p(t)~:=~\frac{\delta S[x,v]}{\delta v(t)}\tag{A}$$ as a functional derivative, and define a functional Legendre transformation/Hamiltonian functional via $$\mathbb{H}[x,p]~:=~\sup_{v}\left( \int_{t_i}^{t_f}\! dt~ p(t) v(t) - S[x,v] \right),\tag{B} $$ cf. e.g. this Phys.SE post.

    It is straightforward to calculate eqs. (A) & (B) for OP's local functional (1). Eqs. (A) & (B) even make sense if $S[x,v]$ is a non-local functional.

Qmechanic
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