2
$\begingroup$

I'm preparing for my masters entrance exam on pure mathematics (thought some problems are devoted to classical/lagrangian mechanics). I would be grateful to clarify some basics regarding the conversations law.

Given a point of mass $m$ that behaves oneself according to the law $$x(t) = x_{0} \log(1+ \frac{t^{2}}{T^{2}})$$ how one can find the potential energy $U(x)$ of a point? Here $t$ is the time ($t \geq 0$), $T, x_{0}$ are constant.

Using the Lagrangian method one obtain that:

$$L = \frac{m}{2} \dot{x}^{2} - U(x)$$ by Euler-Lagrange equations we get $$m \ddot{x} = - \frac{\partial U}{\partial x}$$ hence the conservation law holds.

So, is it true that in order to find $U(x)$ it is enough to calculate $$U(x) = E - \frac{m \dot{x}^{2}}{2}$$ where $\dot{x}$ is the derivative of $x := x(t)$ w.r.t to $t$? As for me, the latter looks quite confusing; for example, how to calculate the complete energy of a point $E$?

$\endgroup$
  • $\begingroup$ Btw, $E$ is a constant. You can compute it in a point where there's only KE. $\endgroup$ – FGSUZ Jul 31 '18 at 20:57
3
$\begingroup$

I'll answer the question without Lagrangian formalism. To be succinct, I'll set $m = T = x_0 = 1$.

From $$x(t) = \ln(1+t^2)$$ one can get the speed $$\dot{x}(t) = \frac{2t}{1+t^2}$$ and then the acceleration $$\ddot{x}(t) = 2\frac{1-t^2}{(1+t^2)^2}$$

We want to get the potential $U(x)$, which is, using Newton's law (or equivalently the Euler-Lagrange equation) $$\frac{\partial U}{\partial x} = -\ddot{x}$$

So we want to express $\ddot{x}$ as a function of $x$. By inverting the first equation, we can express $t(x)$ : $$t^2 = e^x-1$$ so $$\frac{\partial U}{\partial x} = 2\frac{e^x-2}{e^{2x}}$$ A final integration gives us $$U(x) = 2(e^{-2x}-e^{-x}) + C$$


Edit :

Indeed, it's much quicker to use $U(x) = E - \frac{1}{2}\dot{x}^2$. Let's choose $E=0$, because adding a constant to $U$ does not change the physics.

Using the expression for $t(x)$ above, we have $$\dot{x} = \frac{2\sqrt{e^x-1}}{e^x}$$ so $$U(x) = -\frac{1}{2}\frac{4(e^x-1)}{e^{2x}} = 2(e^{-2x}-e^{-x})$$


Keep in mind $E$, the energy of the system (as well as the kinetic energy) is an intrinsic value. It depends on initial conditions, for example. It's really a function of $t$ and not $x$. At the contrary, a potential energy can be seen as both an external quantity, independent of the system (that's why we associate a potential energy $U(x)$ to each point of space), and as a part of the energy : $$E(t) = \frac{1}{2}m\dot{x}^2(t) + U(x(t))$$ If you want a mental picture, $U(x)$ is the hill, and $E(t)$ is the total energy of the ball rolling on (here a constant).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.