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In some examples I have read that if you want to find the integral of motion for some equation of motion, say on the form $\ddot{x}+ax=0$ for some constant $a$, you multiply the EOM by $$\dot{x}=q(x) \implies \ddot{x} = \frac{dq}{dt} = \frac{dq}{dx}\frac{dx}{dt}.$$ You then separate $q$ and $x$ and integrate both sides. If you then rearrange to get the integration constant (constant of motion I presume(?)) alone on either side you find some new equation on the form (in this case) $$\frac{\dot{x}^2}{2}+a\frac{x^2}{2} = C.$$

Assuming this is correct, is this a general procedure for finding integrals of motion for explicitly time independent EOMs? Is the multiplication of $\dot{x}$ just a mathematical trick or is there a physical interpretation of it?

Qmechanic
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theorem
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  • Perhaps I should have been more clear, I want to know if this method for finding the integral of motion would work for any arbitrary EOM. Maybe there are other methods of doing so? What would happen in the case of an explicitly time dependent EOM? – theorem Nov 28 '15 at 15:05
  • This wont work for every EOM. It is a special technique for this type particularly. There are different methods for solving such equations. – SchrodingersCat Nov 28 '15 at 15:14
  • Suppose we have some ODE of the form $\ddot x=f(x), x=x(t)$. Then let $U(x)=-\int f(x)\mathrm{d}x$. Then $\tfrac{1}{2}\dot x^2+U(x)=\mathrm{const.}$ One may then solve this as $$t+C=\int\frac{\mathrm{d}x}{\sqrt{2(E-U(x))}}$$ where $E$ is the first constant and $C$ is a new one. This is entirely equivalent to the situation in mechanics where $x(t)$ is the position of a particle and $f(x)$ is a force. In this case, $E$ is the conserved energy. – Ryan Unger Nov 28 '15 at 15:33

2 Answers2

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  1. A single autonomous (possibly nonlinear) 2nd-order ODE $$F(x,\dot{x},\ddot{x})~=~0. \tag{1}$$ can in principle be written as a couple of autonomous (possibly nonlinear) 1st-order ODEs of the form $$\dot{x}~=~f(x,y), \qquad \dot{y}~=~g(x,y). \tag{2}$$
  2. One may show that there always exists an integral of motion/first integral for the latter system (2), at least locally, cf. this Phys.SE post.
Qmechanic
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Separation of variables works on the following three cases:

  1. Acceleration is a function of time only $\ddot{x} = a(t)$ with solution the direct integration of time $$v(t) = \int a(t)\,{\rm d}t \\ x(t) = \int v(t)\,{\rm d}t$$
  2. Acceleration is a function of speed only $\ddot{x} = a(\dot{x})$ with solution $$ \left.\frac{{\rm d}t}{{\rm d}\dot{x}} = \frac{1}{a(\dot{x})} \right\} t(\dot{x}) = \int {\rm d}t = \int \frac{1}{a(\dot{x})} \,{\rm d}\dot{x} \\ \left. \dot{x} \frac{{\rm d}t}{{\rm d}\dot{x}} = \frac{{\rm d}x}{{\rm d}\dot{x}}= \frac{\dot{x}}{a(\dot{x})} \right\} x(\dot{x}) = \int {\rm d}x = \int \frac{\dot{x}}{a(\dot{x})} \,{\rm d}\dot{x}$$
  3. Acceleration is a function of distance only $\ddot{x} = a(x)$ with solution $$ \left. \frac{\ddot{x}}{\dot{x}} = \frac{{\rm d}\dot{x}}{{\rm d}t} \frac{{\rm d}t}{{\rm d}x} = \frac{{\rm d}\dot{x}}{{\rm d}x} = \frac{a(x)}{\dot{x}} \right\} \frac{1}{2}\dot{x}^2= \int \dot{x}\, {\rm d}\dot{x} = \int a(x) \,{\rm d}x$$ $$ t = \int \frac{1}{\dot{x}(x)}\,{\rm d}x $$

These are all mathematical "tricks". Some have physical interpretation. For Case 3 you can see equation $$\frac{1}{2} m \dot{x}^2 = {\rm Work}$$

John Alexiou
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