I'm currently studying Goldstein's Classical Mechanics book and I can't get my head around his reasoning in section 2.4. (Extending Hamilton's principle to systems with constraints). I'd like to understand the example he gives. Here it comes:

Consider a smooth solid hemisphere of radius $a$ placed with its flat side down and fastened to the Earth whose gravitational acceleration is $g$. Place a small mass $M$ at the top of the hemisphere with an infinitesimal displacement off center so the mass slides down without friction. Choose coordinate $x, y, z$ centered on the base of the hemisphere with the $z$ axis giving the vertical through the hemisphere and the $x$-$y$ plane giving the horizontal. The initial motion of the mass is in the $x$-$y$ plane.

Let $\theta$ be the angle from the top of the sphere to the mass. The Lagrangian is $L = \frac{1}{2}\cdot M \cdot (\dot x^2 + \dot y^2 + \dot z^2) - m\cdot g\cdot z$. The initial conditions allow us to ignore the $y$ coordinate, so the constraint equation is $a - \sqrt{x^2 + y^2} = 0$. Expressing the problem in terms of $r^2 = x^2+z^2$ and $x/z = \cos(\theta)$, Lagrange's equations are $$ Ma\dot\theta^2 - M g \cos(\theta) + \lambda = 0$$ and $$ Ma^2\ddot \theta + M g\, a \sin (\theta) = 0$$

Solve the second equation and then the first to obtain $$ \dot\theta^2 = -\frac{2g}{a}\cos(\theta) + \frac{2g}{a}$$ and $$ \lambda = M g (3\cos(\theta)-2)$$ So $\lambda$ is the magnitude of the force keeping the particle on the sphere and since $\lambda = 0$ when $\theta = \cos^{-1}(\tfrac{2}{3})$, the mass leaves the sphere at that angle.

I have the following questions:

  1. Shouldn't it be $x/z = \tan \theta$?

  2. Could it be that he's mixing up $r$ and $a$? My guess is that from "Lagrange's equations are" it should say $r$ instead of $a$. I get confused whether $a$ is a system parameter or a Lagrangian multiplier.

  3. Could you give me a) an explanation or b) a good read on why setting $L' = L + \lambda\cdot f$ gives us an analogue of Hamilton's principle on constraint systems? I don't understand Goldstein's derivation. ($L$ is the original Lagrangian, $f$ is the constraint and $\lambda$ is the Lagrangian multiplier.)

  4. Why can $\lambda$ be thought of as the constraint force?

When I understand 3., I understand the example -- I reverse engineered the supposedly Lagrangian equations to see that $L'$ needs to be of form $$\frac{1}{2}M r^2 \dot\theta^2 - Mrg\cos(\theta) + \lambda \cdot f$$ with generalized coordinates $\theta$ and $r$. Then everything works out just fine.

  • $\begingroup$ More Phys.SE posts on blocks sliding down hemispheres. $\endgroup$
    – Qmechanic
    Commented Sep 10, 2014 at 15:34
  • $\begingroup$ I am pretty sure that there is yet one more typographical error in this example. I believe that the correct Lagrange equation is $M\,r^2\,\ddot{\theta} - M\,g\,r\,\sin(\theta)=0$. Or, when $r$ is evaluated at $a$, the resulting Lagrange equation ought read $M\,a^2\,\ddot{\theta} - M\,g\,a\,\sin(\theta)=0$. $\endgroup$ Commented Nov 8, 2022 at 4:15

1 Answer 1

  1. Yes, it should be $x/z = tan \theta$, this is probably a typo.
  2. The constraint should be $a - \sqrt{x^2 + z^2}=0$ for the argument to make sense. $r$ is a coordinate which is variable but due to the constraint it will always be equal to $a$, so we can use $a$ in the equations instead. ($\dot{a}=0$).
  3. You know that a gradient of $f$ is always perpendicular to the surface of constant $f$, so you can understand the extra term coming from $\partial L'/\partial x_i $ as a force acting perpendicular to the surface of $f=0$ holding the particle on it. However, $\lambda$ has to be solved so that the motion of the system is only along the constant surface. But imagine now a force equal to the solved $\lambda \partial f/\partial x_i$ - it would have the same effect as the constraint, so this is the force by which the constraint actually has to be acting to hold the particle/system. (This also answers 4.)
  • $\begingroup$ OK, thanks a lot for the clarifying answer, Void! I just realized that there is one thing I don't quite understand yet: In "regular" Lagrangian equations, using new generalized coordinates incorporates all constraints. Why isn't that enough in the Hamilton approach? (i.e. why do we need Lagrangian $\lambda$s where a coordinate transformation had sufficed before?) $\endgroup$ Commented Sep 10, 2014 at 16:27
  • $\begingroup$ You could then just use only one of the equations - the one gained by variation of $\theta$ - the derivation of this fact can be done in various ways, one of them is via the d'Alembert principle. $\endgroup$
    – Void
    Commented Sep 10, 2014 at 16:52

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